Another Bogus Claim by the SR Experts
Ken H. Seto
source is in a car moving relative to the highway at a connstant speed. The
detector makes frequency measurements at regular intervals. The car is
approaching the detector from a distance. It passes the detector and
continues to reced from the detector.
SR experts Tom Roberts and Paul Andersen claimed the following for this
thought experiment:
1. The detector will initially detect an arbituary high frequency---call
this Fh.
2. Subsequent measurements will show a decrease in frequency--from Fh. The
frequency rate will continue to decrease until it is no longer detectable.
Their claims are obviously bogus. In real life, the initial frequency would
be lowest detectable. It will increase to a max. when the car is at the
nearest point of approach ---call this Fm.
When the car passes the detector, the detected frequency will decrease from
Fm until it is no longer detectable.
:
What do you think???
Ken Seto
Eric Prebys
"Ken H. Seto" wrote:
>
> A frequency detector is at rest on the side of a straight highway. A radar
> source is in a car moving relative to the highway at a connstant speed. The
> detector makes frequency measurements at regular intervals. The car is
> approaching the detector from a distance. It passes the detector and
> continues to reced from the detector.
>
> SR experts Tom Roberts and Paul Andersen claimed the following for this
> thought experiment:
> 1. The detector will initially detect an arbituary high frequency---call
> this Fh.
> 2. Subsequent measurements will show a decrease in frequency--from Fh. The
> frequency rate will continue to decrease until it is no longer detectable.
>
The only one who ever said that is you, Ken, and whenever you do say it,
someone corrects you and you ignore it.
You're just too stupid to understand the algebra.
> Their claims are obviously bogus. In real life, the initial frequency would
> be lowest detectable. It will increase to a max. when the car is at the
> nearest point of approach ---call this Fm.
> When the car passes the detector, the detected frequency will decrease from
> Fm until it is no longer detectable.
> :
> What do you think???
>
That you're a moron, but it's not just me, Ken, the entire world knows
you're
a moron.
> Ken Seto
--
-------------------------------------------------------------------
Eric Prebys, Fermi National Accelerator Laboratory
Office: 630-840-8369, Email: pre...@fnal.gov
WWW: http://home.fnal.gov/~prebys
-------------------------------------------------------------------
Ken H. Seto
"Eric Prebys" <pre...@fnal.gov> wrote in message
news:3C34BA4C...@fnal.gov...
>
>
> "Ken H. Seto" wrote:
> >
> > A frequency detector is at rest on the side of a straight highway. A
radar
> > source is in a car moving relative to the highway at a connstant speed.
The
> > detector makes frequency measurements at regular intervals. The car is
> > approaching the detector from a distance. It passes the detector and
> > continues to reced from the detector.
> >
> > SR experts Tom Roberts and Paul Andersen claimed the following for this
> > thought experiment:
> > 1. The detector will initially detect an arbituary high frequency---call
> > this Fh.
> > 2. Subsequent measurements will show a decrease in frequency--from Fh.
The
> > frequency rate will continue to decrease until it is no longer
detectable.
> >
>
> The only one who ever said that is you, Ken, and whenever you do say it,
> someone corrects you and you ignore it.
You are a fucking idiot. If he didn't say that where did I got it grom?? My
interpretation of the situation is below.
>
> You're just too stupid to understand the algebra.
You are an SR runt---a fucking stupid runt..You think that your use of
mutual time dilation to proof reciprocity is king shit algebra. Get a life
Eric. Go do some more circular algebra to impress some of your SR runt buddy
>
> > Their claims are obviously bogus. In real life, the initial frequency
would
> > be lowest detectable. It will increase to a max. when the car is at the
> > nearest point of approach ---call this Fm.
> > When the car passes the detector, the detected frequency will decrease
from
> > Fm until it is no longer detectable.
> > :
> > What do you think???
> >
>
> That you're a moron, but it's not just me, Ken, the entire world knows
> you're
> a moron.
So are you saying that what I said is not correct?? You are even more stupid
than I give you credit for.
Ken Seto
Eric Prebys
"Ken H. Seto" wrote:
>
> "Eric Prebys" <pre...@fnal.gov> wrote in message
> news:3C34BA4C...@fnal.gov...
> >
> >
> > "Ken H. Seto" wrote:
> > >
> > > A frequency detector is at rest on the side of a straight highway. A
> radar
> > > source is in a car moving relative to the highway at a connstant speed.
> The
> > > detector makes frequency measurements at regular intervals. The car is
> > > approaching the detector from a distance. It passes the detector and
> > > continues to reced from the detector.
> > >
> > > SR experts Tom Roberts and Paul Andersen claimed the following for this
> > > thought experiment:
> > > 1. The detector will initially detect an arbituary high frequency---call
> > > this Fh.
> > > 2. Subsequent measurements will show a decrease in frequency--from Fh.
> The
> > > frequency rate will continue to decrease until it is no longer
> detectable.
> > >
> >
> > The only one who ever said that is you, Ken, and whenever you do say it,
> > someone corrects you and you ignore it.
>
> You are a fucking idiot. If he didn't say that where did I got it grom?? My
> interpretation of the situation is below.
How should I know where you got it "grom". If you got it from him, you
would have quoted it. Since you didn't I can only assume you
misunderstood
something he said - as always.
> >
> > You're just too stupid to understand the algebra.
>
> You are an SR runt---a fucking stupid runt..You think that your use of
> mutual time dilation to proof reciprocity is king shit algebra. Get a life
This had nothing to do with "mutual time dilation"; this had to do with
your badly misquoting someone, and now apparently resorting to
obscenities
to "prove" your point.
> Eric. Go do some more circular algebra to impress some of your SR runt buddy
> >
> > > Their claims are obviously bogus. In real life, the initial frequency
> would
> > > be lowest detectable. It will increase to a max. when the car is at the
> > > nearest point of approach ---call this Fm.
> > > When the car passes the detector, the detected frequency will decrease
> from
> > > Fm until it is no longer detectable.
> > > :
> > > What do you think???
> > >
> >
> > That you're a moron, but it's not just me, Ken, the entire world knows
> > you're
> > a moron.
>
> So are you saying that what I said is not correct?? You are even more stupid
Yes, I am saying that what you said was not correct. You seldom say
anything
that is correct. You also seldom understand what anyone else says.
Then you asked what I thought. I think you are a moron. THAT is
correct.
> than I give you credit for.
>
> Ken Seto
--
Paul Cardinale
completely clueless about the Doppler effect. But this isn't a surprise;
the kenseto is completely clueless about everything. The really funny part
is that he has named his pet not-a-real-theory-that-makes-no-preditions
"Doppler Relativity Theory", AKA DiRT. It isn't a surprise that DiRT
can't make any predictions because the kenseto is completely clueless
about what a prediction is (four-syllable words are at least one order
of magnitude beyond the comprehension of the kenseto).
Paul Cardinale
Bruce Richmond
Perhaps you are thinking of the amplitude of the signal. That does
start low and builds to a max when the source is nearest to the
detector.
When approaching, the radar source gets closer to the receiver as each
crest is transmitted. So each crest takes a little less time than the
one before it to reach the receiver. That crowds the crests together,
giving a higher frequency. If the source is headed straight toward
the detector the higher frequency will remain constant. After the
source passes, each crest has to travel further to get back to the
detector, so the frequency ends up lower.
Ken H. Seto
"Paul Cardinale" <pcard...@volcanomail.com> wrote in message
news:64050551.02010...@posting.google.com...
Hey runt, I suggest that you spend more time to find out why you failed the
Mars Lander missions. If I want any shit from you I'll squeeze your head.
Ken Seto
Paul Cardinale
> "Ken H. Seto" <ken...@erinet.com> wrote in message news:<3c34a0d6$0$35612$4c5e...@news.erinet.com>...
> > A frequency detector is at rest on the side of a straight highway. A radar
> > source is in a car moving relative to the highway at a connstant speed. The
> > detector makes frequency measurements at regular intervals. The car is
> > approaching the detector from a distance. It passes the detector and
> > continues to reced from the detector.
> >
> > SR experts Tom Roberts and Paul Andersen claimed the following for this
> > thought experiment:
> > 1. The detector will initially detect an arbituary high frequency---call
> > this Fh.
> > 2. Subsequent measurements will show a decrease in frequency--from Fh. The
> > frequency rate will continue to decrease until it is no longer detectable.
> >
> > Their claims are obviously bogus. In real life, the initial frequency would
> > be lowest detectable. It will increase to a max. when the car is at the
> > nearest point of approach ---call this Fm.
> > When the car passes the detector, the detected frequency will decrease from
> > Fm until it is no longer detectable.
> > :
> > What do you think???
> >
> > Ken Seto
>
> Perhaps you are thinking of the amplitude of the signal.
>
The kenseto isn't "thinking of the amplitude of the signal".
The kenseto is not capable of thinking at all.
Paul Cardinale
Eric Prebys
Bruce Richmond wrote:
>
> "Ken H. Seto" <ken...@erinet.com> wrote in message news:<3c34a0d6$0$35612$4c5e...@news.erinet.com>...
> > A frequency detector is at rest on the side of a straight highway. A radar
> > source is in a car moving relative to the highway at a connstant speed. The
> > detector makes frequency measurements at regular intervals. The car is
> > approaching the detector from a distance. It passes the detector and
> > continues to reced from the detector.
> >
> > SR experts Tom Roberts and Paul Andersen claimed the following for this
> > thought experiment:
> > 1. The detector will initially detect an arbituary high frequency---call
> > this Fh.
> > 2. Subsequent measurements will show a decrease in frequency--from Fh. The
> > frequency rate will continue to decrease until it is no longer detectable.
> >
> > Their claims are obviously bogus. In real life, the initial frequency would
> > be lowest detectable. It will increase to a max. when the car is at the
> > nearest point of approach ---call this Fm.
> > When the car passes the detector, the detected frequency will decrease from
> > Fm until it is no longer detectable.
> > :
> > What do you think???
> >
> > Ken Seto
>
> Perhaps you are thinking of the amplitude of the signal. That does
> start low and builds to a max when the source is nearest to the
> detector.
>
You're giving Ken way too much credit; he's simply a moron.
> When approaching, the radar source gets closer to the receiver as each
> crest is transmitted. So each crest takes a little less time than the
> one before it to reach the receiver. That crowds the crests together,
> giving a higher frequency. If the source is headed straight toward
> the detector the higher frequency will remain constant. After the
> source passes, each crest has to travel further to get back to the
> detector, so the frequency ends up lower.
--
Ken H. Seto
This is some more of your bull shit. You claimed that Roberts didn't make
the claim that uniform acceleration will produce a gravity gradient and now
he didn't deny making such a claim in the thread "Tom Roberts Admitted
Indirectly That He Was Wrong".
You are such a gutless wonder that you will go all lengths to defend your SR
brothers. Now that Roberts admits that he made the claim that uniform
acceleration will produce a gravity gradient how do you defend him now??
>
> > >
> > > You're just too stupid to understand the algebra.
> >
> > You are an SR runt---a fucking stupid runt..You think that your use of
> > mutual time dilation to proof reciprocity is king shit algebra. Get a
life
>
> This had nothing to do with "mutual time dilation";
It had everthing to do with mutual time dilation. You used it to do your
algebra to get reciprocity between two relatively moving frames. This means
that your point that my not understanding algebra is full of crap.
>this had to do with
> your badly misquoting someone, and now apparently resorting to
> obscenities
> to "prove" your point.
You are such a runt that obscenities are the only thintg that you can
understand. You can check with Tom Roberts if I misquoted him.
>
> > > That you're a moron, but it's not just me, Ken, the entire world knows
> > > you're
> > > a moron.
> >
> > So are you saying that what I said is not correct?? You are even more
stupid
>
> Yes, I am saying that what you said was not correct. You seldom say
> anything
> that is correct. You also seldom understand what anyone else says.
> Then you asked what I thought. I think you are a moron. THAT is
> correct.
So what is your correct interpretation of the experiment I described?? If
you don't come up with one then shut the fuck up.
Ken Seto
Ken H. Seto
"Bruce Richmond" <bsr...@my-deja.com> wrote in message
news:747a5d11.02010...@posting.google.com...
No I am not thinking of the amplitude of the signal. I specifically said
that the frequency start low and builds to a max.when the source is nearest
to the detector.
>
> When approaching, the radar source gets closer to the receiver as each
> crest is transmitted. So each crest takes a little less time than the
> one before it to reach the receiver. That crowds the crests together,
> giving a higher frequency. If the source is headed straight toward
> the detector the higher frequency will remain constant. After the
> source passes, each crest has to travel further to get back to the
> detector, so the frequency ends up lower.
What you said here is exactly what I said in my original post. So what is
your point??
Ken Seto
Eric Prebys
He never did, and you have yet to produce a quote where he did.
> and now
> he didn't deny making such a claim in the thread "Tom Roberts Admitted
> Indirectly That He Was Wrong".
To my knowlegde, you've never denied making the claim that you
shot JFK, either. Does that mean that you did?
What he said is he wasn't going to waste any more time arguing with you.
> You are such a gutless wonder that you will go all lengths to defend your SR
> brothers. Now that Roberts admits that he made the claim that uniform
> acceleration will produce a gravity gradient how do you defend him now??
Nowhere does he admit this, as anyone else reading the thread can see.
I suggest you add the English language to the very, very long list of
things you have trouble understanding.
> >
> > > >
> > > > You're just too stupid to understand the algebra.
> > >
> > > You are an SR runt---a fucking stupid runt..You think that your use of
> > > mutual time dilation to proof reciprocity is king shit algebra. Get a
> life
> >
> > This had nothing to do with "mutual time dilation";
>
> It had everthing to do with mutual time dilation. You used it to do your
> algebra to get reciprocity between two relatively moving frames. This means
> that your point that my not understanding algebra is full of crap.
>
I meant that had nothing to do with this thread. That topic
was put to rest to the satisfaction of everyone who can
grasp simple math - which I'm afraid leaves you out.
> >this had to do with
> > your badly misquoting someone, and now apparently resorting to
> > obscenities
> > to "prove" your point.
>
> You are such a runt that obscenities are the only thintg that you can
> understand. You can check with Tom Roberts if I misquoted him.
You didn't quote him at all. You paraphrased him. Seen through
the filter of your own misonceptions, even the TV guide would
come out unrecognizable.
> >
> > > > That you're a moron, but it's not just me, Ken, the entire world knows
> > > > you're
> > > > a moron.
> > >
> > > So are you saying that what I said is not correct?? You are even more
> stupid
> >
> > Yes, I am saying that what you said was not correct. You seldom say
> > anything
> > that is correct. You also seldom understand what anyone else says.
> > Then you asked what I thought. I think you are a moron. THAT is
> > correct.
>
> So what is your correct interpretation of the experiment I described?? If
> you don't come up with one then shut the fuck up.
>
I've answered it, Tom has answered it, Daryl has answered it, and others
have answered it, but I'll do it one more time.
Light going toward the front of the ship will be red-shifted by a
fractional amount a*L/c^2, where a is the acceleration and L is the
distance of propagation. Light going toward the rear will be
blue-shifted
by the same fractional amount. This first order approximation will
be valid until a*L becomes more than a miniscule fraction of c^2.
Also, as many have pointed out, this first order approximation
is valid not only for SR, but for just about any model you can
think of.
Also, as many have pointed out, accelerations do not "cause"
gravity or "gravity gradients". I never said that; Tom Roberts
never said that; the only one who ever said that was you.
Ken H. Seto
>
>
> "Ken H. Seto" wrote:
> >
> > "Eric Prebys" <pre...@fnal.gov> wrote in message
> > news:3C34E262...@fnal.gov...
> > >
> > >
> > > "Ken H. Seto" wrote:
> > > >
> > > > "Eric Prebys" <pre...@fnal.gov> wrote in message
> > > > news:3C34BA4C...@fnal.gov...
> >
> > So what is your correct interpretation of the experiment I described??
If
> > you don't come up with one then shut the fuck up.
> >
>
> I've answered it, Tom has answered it, Daryl has answered it, and others
> have answered it, but I'll do it one more time.
It is a waste of time to argue with an idiot like you. For your information
we were not talking about the accelerating spaceship in this thread. So what
is your correct interpretation that the detector will detect at regular
intervals as the car is approaching and receding from it??
>
> Light going toward the front of the ship will be red-shifted by a
> fractional amount a*L/c^2, where a is the acceleration and L is the
> distance of propagation. Light going toward the rear will be
> blue-shifted
> by the same fractional amount. This first order approximation will
> be valid until a*L becomes more than a miniscule fraction of c^2.
>
> Also, as many have pointed out, this first order approximation
> is valid not only for SR, but for just about any model you can
> think of.
I already pointed out to you that this bull shit is wrong. Your math is
based on the wrong assumption that light takes a different transit time to
traverse the same distance L in the spaceship. This assumption violates the
isotropy of light in the spaceship. So for you to keep on repeating the same
wrong math to proof your point is laughable.
Ken Seto
Eric Prebys
"Ken H. Seto" wrote:
>
> "Eric Prebys" <pre...@fnal.gov> wrote in message
> news:3C361588...@fnal.gov...
> >
> >
> > "Ken H. Seto" wrote:
> > >
> > > "Eric Prebys" <pre...@fnal.gov> wrote in message
> > > news:3C34E262...@fnal.gov...
> > > >
> > > >
> > > > "Ken H. Seto" wrote:
> > > > >
> > > > > "Eric Prebys" <pre...@fnal.gov> wrote in message
> > > > > news:3C34BA4C...@fnal.gov...
>
> > >
> > > So what is your correct interpretation of the experiment I described??
> If
> > > you don't come up with one then shut the fuck up.
> > >
> >
> > I've answered it, Tom has answered it, Daryl has answered it, and others
> > have answered it, but I'll do it one more time.
>
> It is a waste of time to argue with an idiot like you. For your information
> we were not talking about the accelerating spaceship in this thread. So what
> is your correct interpretation that the detector will detect at regular
> intervals as the car is approaching and receding from it??
Sorry! You say so many stupid things at once that it's hard to keep
track
of them all. For the record, you actually *had* just finished making
comments about the accelerating spaceship in this thread. You
had also made comments about things I put to rest a long time ago,
so don't blame me if I lost track. But back to this thread...
The answer is that as long as the speed is constant, the frequency
will be shifted up by a CONSTANT amount while the transmitter is moving
(directly) toward you, and down by a CONSTANT amount, when it is moving
(directly) away from you.
Unless the speed or relative direction changes, the received frequency
does not change, and no one has ever said it does except you.
This has been explained to you in nauseating detail mathematically,
but of course you can't do math, and with words (small, simple words you
should have been able to understand but didn't). My advice to you:
just give up; you're just too darn stupid to understand this topic,
or any other physics topic.
If the transimitter is not moving directly toward or away from
you, then the relationship becomes a bit more complicated, but
you have to walk before you can run, and so far *you* can't even
crawl.
> >
> > Light going toward the front of the ship will be red-shifted by a
> > fractional amount a*L/c^2, where a is the acceleration and L is the
> > distance of propagation. Light going toward the rear will be
> > blue-shifted
> > by the same fractional amount. This first order approximation will
> > be valid until a*L becomes more than a miniscule fraction of c^2.
> >
> > Also, as many have pointed out, this first order approximation
> > is valid not only for SR, but for just about any model you can
> > think of.
>
> I already pointed out to you that this bull shit is wrong. Your math is
> based on the wrong assumption that light takes a different transit time to
> traverse the same distance L in the spaceship. This assumption violates the
No, it assumes the time it takes is L/c in either direction. I'm sorry
I overestimated you; I should have done that it 4 or 5 separate steps
so that *maybe* you could have followed it.
-Eric
Ken H. Seto
ROTFL. So your fuck up is my fault??
>
> The answer is that as long as the speed is constant, the frequency
> will be shifted up by a CONSTANT amount while the transmitter is moving
> (directly) toward you, and down by a CONSTANT amount, when it is moving
> (directly) away from you.
So how is this different than what I said in my original post that started
this thread?? Now do you see why I used obscenities on you??
_________________________________________________________
Their claims are obviously bogus. In real life, the initial frequency would
be lowest detectable. It will increase to a max. when the car is at the
nearest point of approach ---call this Fm.
When the car passes the detector, the detected frequency will decrease from
Fm until it is no longer detectable.
______________________________________________________
>
> Unless the speed or relative direction changes, the received frequency
> does not change, and no one has ever said it does except you.
>
> This has been explained to you in nauseating detail mathematically,
> but of course you can't do math, and with words (small, simple words you
> should have been able to understand but didn't). My advice to you:
> just give up; you're just too darn stupid to understand this topic,
> or any other physics topic.
You are not just stupid you are also a fucking liar.. No math was presented
by anybody in this thread.
> > > Light going toward the front of the ship will be red-shifted by a
> > > fractional amount a*L/c^2, where a is the acceleration and L is the
> > > distance of propagation. Light going toward the rear will be
> > > blue-shifted
> > > by the same fractional amount. This first order approximation will
> > > be valid until a*L becomes more than a miniscule fraction of c^2.
> > >
> > > Also, as many have pointed out, this first order approximation
> > > is valid not only for SR, but for just about any model you can
> > > think of.
> >
> > I already pointed out to you that this bull shit is wrong. Your math is
> > based on the wrong assumption that light takes a different transit time
to
> > traverse the same distance L in the spaceship. This assumption violates
the
>
> No, it assumes the time it takes is L/c in either direction.
Hey stupid when you assumed that the light is frequency shifted the shift is
caused by the front is moving relative to the light differently than the
rear is moving relative to the same light. This means that L/c to the front
will take a longer transit time than L/c to the rear. In case you are too
stupid to understand the assumption of frequency shift within the same frame
implies the detection of absolute motion of the frame.
I'm sorry
> I overestimated you; I should have done that it 4 or 5 separate steps
> so that *maybe* you could have followed it.
I suggest that you take time from your job and study physics before you give
me any shit. <ahrug>
Ken Seto
Ken H. Seto
Ooops, I fuck up. What you said is wrong. The frequency will be shifted up
but not by a constant amount while the transimitter is moving toward you.
The frequency will shift down but not by a constant amount while the
transmitter is receding away from you.
Ken Seto
Eric Prebys
I knew it! I just knew you couldn't accidentally say something correct
without coming back and "fixing" it!! My faith is restored!
> but not by a constant amount while the transimitter is moving toward you.
> The frequency will shift down but not by a constant amount while the
> transmitter is receding away from you.
>
> Ken Seto
Gershon
Ken,
The problems depends on how far "on the side" of the highway the
detecor is. If it were in the exact line of the emitter on the
highway, the frequency would be shifted higher and be constant the
entire approach, and then be shifted lower when the car passed by and
and remain constant for the entire time the car travels away (of
course the detector and emitter would have to collide for this to be
100% true). But if the detector is offset from the emitter by any
amount (as the original problem asserts), there will be a change in
the effective rate of approach as the car approaches. The frequency
will be shifted highest when the emitter is farthest away and approach
the emitted frequency as it approaches the detector, then shift lower
as it travels away. The rate of change is fastest when the emitter is
closest to the detector.
To visualize this better draw a 7 inch highway and the detector 1 inch
off to the side, then draw lines from the highway to the detector from
equal (1/4 inch) intervals along the highway. Farther away from the
detector, the lines will be increasing at ever more "regular"
intervals, while the lines closest to the detector will be almost
equal in length. The difference in length of two adjacent lines
represents the relative velocity of the emitter to the detector.
I'd say that the claim by the "SR Experts" is accurate. I can find no
basis for your claim that the frequency would, "increase to a max.
when the car is at the nearest point of approach", that is simply
incorrect even if the detector is in line with the emitter.
If someone can show the formula for the rate of change as a function
of distance for a given offset of emitter to detector, that would be
cool.
Paul B. Andersen
"Ken H. Seto" <ken...@erinet.com> wrote in message news:3c34a0d6$0$35612$4c5e...@news.erinet.com...
> A frequency detector is at rest on the side of a straight highway. A radar
> source is in a car moving relative to the highway at a connstant speed. The
> detector makes frequency measurements at regular intervals. The car is
> approaching the detector from a distance. It passes the detector and
> continues to reced from the detector.
>
> SR experts Tom Roberts and Paul Andersen claimed the following for this
> thought experiment:
> 1. The detector will initially detect an arbituary high frequency---call
> this Fh.
> 2. Subsequent measurements will show a decrease in frequency--from Fh. The
> frequency rate will continue to decrease until it is no longer detectable.
No, we didn't claim that.
Please don't try to paraphrase me, Ken.
You always get it stupidly wrong.
> Their claims are obviously bogus.
Sure the claims above are obviously bogus.
But they are not mine. Neither are they Tom's.
> In real life, the initial frequency would
> be lowest detectable. It will increase to a max. when the car is at the
> nearest point of approach ---call this Fm.
> When the car passes the detector, the detected frequency will decrease from
> Fm until it is no longer detectable.
> :
> What do you think???
I think what anybody will think.
That you have no clue.
BTW, Ken.
Most people learn how your scenario works in the real world
by experimental evidence while they are still kids.
Have you never passed a sound source while travelling by train, Ken?
Paul
Ken H. Seto
"Paul B. Andersen" <paul.b....@hia.no> wrote in message
news:a1ed7p$ncn$1...@snipp.uninett.no...
>
> "Ken H. Seto" <ken...@erinet.com> wrote in message
news:3c34a0d6$0$35612$4c5e...@news.erinet.com...
> > A frequency detector is at rest on the side of a straight highway. A
radar
> > source is in a car moving relative to the highway at a connstant speed.
The
> > detector makes frequency measurements at regular intervals. The car is
> > approaching the detector from a distance. It passes the detector and
> > continues to reced from the detector.
> >
> > SR experts Tom Roberts and Paul Andersen claimed the following for this
> > thought experiment:
> > 1. The detector will initially detect an arbituary high frequency---call
> > this Fh.
> > 2. Subsequent measurements will show a decrease in frequency--from Fh.
The
> > frequency rate will continue to decrease until it is no longer
detectable.
>
> No, we didn't claim that.
Yes you did. Or at least Tom Roberts did and you agreed. I will spent some
time to dig this up.
>
> Please don't try to paraphrase me, Ken.
> You always get it stupidly wrong.
I was not trying to paraphase you. That was the biggest disagreement I had
with you and Tom. In the end you complimented me for my explanation why the
detected frequency is at a max when the transmitter is at the closest point
of approach. My explanation was that the detected frequency is dependent on
the transit time and the state of absolute motion of the detector. The
farther away is the transmitter the more transit time is required for the
light to reach the detector . This means that the detector will have more
time to move away (due to its own absolute motion) and thus miss detecting
some of the wave crests and thus a lower detected frequency. When the
transimitter is at the closest point of approach, the detector will have the
least transit time to move away and thus detect virtually all the wave crest
out put by the treansmitter and thus the detected frequency is at a Max. at
the closest point of approach.
>
> > Their claims are obviously bogus.
>
> Sure the claims above are obviously bogus.
> But they are not mine. Neither are they Tom's.
Yes they are Tom's and you agreed.
>
> > In real life, the initial frequency would
> > be lowest detectable. It will increase to a max. when the car is at the
> > nearest point of approach ---call this Fm.
> > When the car passes the detector, the detected frequency will decrease
from
> > Fm until it is no longer detectable.
> > :
> > What do you think???
>
> I think what anybody will think.
> That you have no clue.
So you think what I said is wrong?? Or are you just letting off steam??
>
> BTW, Ken.
> Most people learn how your scenario works in the real world
> by experimental evidence while they are still kids.
> Have you never passed a sound source while travelling by train, Ken?
Ah so you agree that my description works in the real world!!! But
unfortunately for you that would mean that you are disagreeing with your SR
brother Eric Prebys. He said:
"The answer is that as long as the speed is constant, the frequency
will be shifted up by a CONSTANT amount while the transmitter is moving
(directly) toward you, and down by a CONSTANT amount, when it is moving
(directly) away from you."
So I guess you will have to fight with your SR brother to settle this issue.
Unless of course you are claiming that he is also correct by reason of
authority.
Ken Seto
Eric Prebys
"Paul B. Andersen" wrote:
>
> "Ken H. Seto" <ken...@erinet.com> wrote in message news:3c34a0d6$0$35612$4c5e...@news.erinet.com...
> > A frequency detector is at rest on the side of a straight highway. A radar
> > source is in a car moving relative to the highway at a connstant speed. The
> > detector makes frequency measurements at regular intervals. The car is
> > approaching the detector from a distance. It passes the detector and
> > continues to reced from the detector.
> >
> > SR experts Tom Roberts and Paul Andersen claimed the following for this
> > thought experiment:
> > 1. The detector will initially detect an arbituary high frequency---call
> > this Fh.
> > 2. Subsequent measurements will show a decrease in frequency--from Fh. The
> > frequency rate will continue to decrease until it is no longer detectable.
>
> No, we didn't claim that.
>
> Please don't try to paraphrase me, Ken.
> You always get it stupidly wrong.
>
> > Their claims are obviously bogus.
>
> Sure the claims above are obviously bogus.
> But they are not mine. Neither are they Tom's.
>
This has been pointed out to him many, many times, and he is never
able to produce any actual quotes.
> > In real life, the initial frequency would
> > be lowest detectable. It will increase to a max. when the car is at the
> > nearest point of approach ---call this Fm.
> > When the car passes the detector, the detected frequency will decrease from
> > Fm until it is no longer detectable.
> > :
> > What do you think???
>
> I think what anybody will think.
> That you have no clue.
>
> BTW, Ken.
> Most people learn how your scenario works in the real world
> by experimental evidence while they are still kids.
> Have you never passed a sound source while travelling by train, Ken?
>
I doubt it. He might hear the ambulance approach, but when it
leaves, he's always in it :)
-Eric
> Paul
Daryl McCullough
>> BTW, Ken.
>> Most people learn how your scenario works in the real world
>> by experimental evidence while they are still kids.
>> Have you never passed a sound source while travelling by train, Ken?
>
>Ah so you agree that my description works in the real world!!! But
>unfortunately for you that would mean that you are disagreeing with your SR
>brother Eric Prebys. He said:
>"The answer is that as long as the speed is constant, the frequency
>will be shifted up by a CONSTANT amount while the transmitter is moving
>(directly) toward you, and down by a CONSTANT amount, when it is moving
>(directly) away from you."
>So I guess you will have to fight with your SR brother to settle this issue.
>Unless of course you are claiming that he is also correct by reason of
>authority.
I'm sure that Paul agrees with Eric's description of the Doppler
shift. What is your description of the Doppler shift?
--
Daryl McCullough
CoGenTex, Inc.
Ithaca, NY
Eric Prebys
"Ken H. Seto" wrote:
>
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message
> news:a1ed7p$ncn$1...@snipp.uninett.no...
> >
> > "Ken H. Seto" <ken...@erinet.com> wrote in message
> news:3c34a0d6$0$35612$4c5e...@news.erinet.com...
> > > A frequency detector is at rest on the side of a straight highway. A
> radar
> > > source is in a car moving relative to the highway at a connstant speed.
> The
> > > detector makes frequency measurements at regular intervals. The car is
> > > approaching the detector from a distance. It passes the detector and
> > > continues to reced from the detector.
> > >
> > > SR experts Tom Roberts and Paul Andersen claimed the following for this
> > > thought experiment:
> > > 1. The detector will initially detect an arbituary high frequency---call
> > > this Fh.
> > > 2. Subsequent measurements will show a decrease in frequency--from Fh.
> The
> > > frequency rate will continue to decrease until it is no longer
> detectable.
> >
> > No, we didn't claim that.
>
> Yes you did. Or at least Tom Roberts did and you agreed. I will spent some
> time to dig this up.
No, Tom Roberts never said it, Paul Andersen never agreed, and on
repeated requests, you've never been able to produce a quote to
back up either of these claims, so shut up until you do.
All that's happened is your usual incorrect paraphrasing of things
you don't understand. Since you understand nothing, you should
probably stick to direct quotes.
> >
> > Please don't try to paraphrase me, Ken.
> > You always get it stupidly wrong.
>
> I was not trying to paraphase you. That was the biggest disagreement I had
> with you and Tom. In the end you complimented me for my explanation why the
Now *that* one you'll have to back up with a quote. I have a hard
time imagining anyone ever complimenting you about anything.
> detected frequency is at a max when the transmitter is at the closest point
> of approach. My explanation was that the detected frequency is dependent on
> the transit time and the state of absolute motion of the detector. The
> farther away is the transmitter the more transit time is required for the
> light to reach the detector . This means that the detector will have more
> time to move away (due to its own absolute motion) and thus miss detecting
> some of the wave crests and thus a lower detected frequency. When the
> transimitter is at the closest point of approach, the detector will have the
> least transit time to move away and thus detect virtually all the wave crest
> out put by the treansmitter and thus the detected frequency is at a Max. at
> the closest point of approach.
>
Certainly no one would have complimented you on something as stupid
as that.
> >
> > > Their claims are obviously bogus.
> >
> > Sure the claims above are obviously bogus.
> > But they are not mine. Neither are they Tom's.
>
> Yes they are Tom's and you agreed.
No, that never happened. If it did, you would have been
able to quote it by now.
> >
> > > In real life, the initial frequency would
> > > be lowest detectable. It will increase to a max. when the car is at the
> > > nearest point of approach ---call this Fm.
> > > When the car passes the detector, the detected frequency will decrease
> from
> > > Fm until it is no longer detectable.
> > > :
> > > What do you think???
> >
> > I think what anybody will think.
> > That you have no clue.
>
> So you think what I said is wrong?? Or are you just letting off steam??
So far, everyone is in unanimous agreement that you are wrong. How
many times are you going to ask that before you understand the
answer?
> >
> > BTW, Ken.
> > Most people learn how your scenario works in the real world
> > by experimental evidence while they are still kids.
> > Have you never passed a sound source while travelling by train, Ken?
>
> Ah so you agree that my description works in the real world!!! But
> unfortunately for you that would mean that you are disagreeing with your SR
> brother Eric Prebys. He said:
> "The answer is that as long as the speed is constant, the frequency
> will be shifted up by a CONSTANT amount while the transmitter is moving
> (directly) toward you, and down by a CONSTANT amount, when it is moving
> (directly) away from you."
This statement is correct for relativistic Doppler shift as well as
non-relativistic Doppler shift. To be totally correct, I probably
should have written "constant factor" rather than "constant amount".
Then it would remain correct even if the transmitted frequency were
changing.
You also snipped the part when I said things got more complicated if
it wasn't moving *directly* toward or away from you (e.g. when it
is passing), but I guess that's because even the concept was over your
head.
> So I guess you will have to fight with your SR brother to settle this issue.
At this point, this is not really an SR issue. My above statement
was also true for non-relativistic Doppler shifts. You seem to place
the
label "SR" on anything you don't understand - which is just about
everything.
> Unless of course you are claiming that he is also correct by reason of
> authority.
>
There is no disagreement between what I said and anything Mr. Roberts
or Mr. Andersen said. The fact that you think there is, is just an
indication that, as usual, you don't understand what anybody said.
Ken H. Seto
news:3c34a0d6$0$35612$4c5e...@news.erinet.com...
> > A frequency detector is at rest on the side of a straight highway. A
radar
> > source is in a car moving relative to the highway at a connstant speed.
The
> > detector makes frequency measurements at regular intervals. The car is
> > approaching the detector from a distance. It passes the detector and
> > continues to reced from the detector.
> >
> > SR experts Tom Roberts and Paul Andersen claimed the following for this
> > thought experiment:
> > 1. The detector will initially detect an arbituary high frequency---call
> > this Fh.
> > 2. Subsequent measurements will show a decrease in frequency--from Fh.
The
> > frequency rate will continue to decrease until it is no longer
detectable.
>
> No, we didn't claim that.
>
> Please don't try to paraphrase me, Ken.
> You always get it stupidly wrong.
OK while I am searching do you deny that you made the following claims??
__________________________________________________________
A space ship under goes a constant acceleration of 1g and a light source is
located in the middle of the ship. They claimed that the frequency of that
light source will be red shifted at the front of the ship and blue shifted
at the rear of the ship. Also they claimed that a standard weigh will
weight less at the front of the ship and more at the rear of the ship. They
said that these claims are supported by the Equivalent Principle.
___________________________________________________
If no response on this post I will assume that you did make the claims.
Ken Seto
Ken H. Seto
"Daryl McCullough" <da...@cogentex.com> wrote in message
news:a1f5i...@drn.newsguy.com...
No they don't agree. Paul said that Doppler shift of light is like a sound
source in a moving train and you are on the embankment. The frequency of
the sound shifts to a higher pitch as the train approaches you but the
shift is not constant and it is dependent on the distance between you and
the source. The frequency of the sound shifts to a lower pitch as the train
is receding away from you but the shift is not constant and it is dependent
on the distance between you and the source.
Eric Prebys said:
The answer is that as long as the speed is constant, the frequency
will be shifted up by a CONSTANT amount while the transmitter is moving
(directly) toward you, and down by a CONSTANT amount, when it is moving
(directly) away from you.
Clearly Paul and Eric disagree with each other..
What is your description of the Doppler shift?
My description agrees with Paul's sound analogy as follows:
The initial frequency would be lowest detectable. It will increase to a max.
when the car is at the nearest point of approach ---call this Fm. When the
car passes the detector, the detected frequency will decrease from Fm until
it is no longer detectable.
Ken Seto
Ken H. Seto
ROTFLOL. Is there no limit to your stupidity?? You said the shift is
CONSTANT for both legs of the trip. I said that the measured frequency will
increase to a max at the nearest point of approach. Thiis does not mean that
the shift is by a CONSTANT amount.as you asserted.
Ken Seto
Eric Prebys
Everybody reading this knows he never made that claim, if only because
his spelling, grammar, and command of the English language are too
good.
You're stupid enough when you get the quotes right, so please stop
making stuff up.
> If no response on this post I will assume that you did make the claims.
>
Ken, do you deny that you made the following claims???
________________________________________________________
The bats in my belfry keep me awake at night. So do the
aliens living under my mattress.
________________________________________________________
If no response on this post I will assume that you did make the claims.
> Ken Seto
--
Ken H. Seto
>
>
> "Paul B. Andersen" wrote:
> >
> > "Ken H. Seto" <ken...@erinet.com> wrote in message
news:3c34a0d6$0$35612$4c5e...@news.erinet.com...
>
>
> > > In real life, the initial frequency would
> > > be lowest detectable. It will increase to a max. when the car is at
the
> > > nearest point of approach ---call this Fm.
> > > When the car passes the detector, the detected frequency will decrease
from
> > > Fm until it is no longer detectable.
> > > :
> > > What do you think???
> >
> > I think what anybody will think.
> > That you have no clue.
> >
> > BTW, Ken.
> > Most people learn how your scenario works in the real world
> > by experimental evidence while they are still kids.
> > Have you never passed a sound source while travelling by train, Ken?
> >
>
> I doubt it. He might hear the ambulance approach, but when it
> leaves, he's always in it :)
Anything is better than your assertion that the shift is a CONSTANT for both
legs of the trip. <shrug>
Ken Seto
Eric Prebys
I'm well aware of what you said. I'm also well aware that it was wrong
-
as is anyone who took freshman physics in college, or even a decent
high school. But by now I've learned you've never studied physics.
The equation for relativistic doppler shift for an approaching
source is (check any physics book)
f' = f*sqrt{(c+v)/(c-v)}
Constant v -> frequency *increased* by a constant factor.
For a receding source
f' = f*sqrt{(c-v)/(c+v)}
Constant v -> frequency *decreased* by a constant factor.
Now unless you can produce an equation to back up your ridiculous
claim, shut the hell up.
Eric Prebys
You keep overlooking the word "directly", Ken. When you're
on the embankment, the train is not moving directly toward you.
The changing frequency comes as the train passes (as I said and
you snipped, twice). When the train is far enough away in either
direction that the impact parameter is negligible, then the
frequency shift is constant, as anyone who has listened to passing
trains knows.
Actually, on second thought, try the experiment while standing
on the track and you'll see my point.
> Clearly Paul and Eric disagree with each other..
>
No, we don't. As usual, you are just too stupid to understand
what either of us said.
> What is your description of the Doppler shift?
>
> My description agrees with Paul's sound analogy as follows:
> The initial frequency would be lowest detectable. It will increase to a max.
> when the car is at the nearest point of approach ---call this Fm. When the
> car passes the detector, the detected frequency will decrease from Fm until
> it is no longer detectable.
>
You clearly hear trains differently than anybody else in world.
Probably confused by the voices in your head.
Paul B. Andersen
"Ken H. Seto" <ken...@erinet.com> wrote in message news:3c3b1276$0$37097$4c5e...@news.erinet.com...
>
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message
> news:a1ed7p$ncn$1...@snipp.uninett.no...
> >
> > "Ken H. Seto" <ken...@erinet.com> wrote in message
> news:3c34a0d6$0$35612$4c5e...@news.erinet.com...
> > > A frequency detector is at rest on the side of a straight highway. A radar
> > > source is in a car moving relative to the highway at a connstant speed. The
> > > detector makes frequency measurements at regular intervals. The car is
> > > approaching the detector from a distance. It passes the detector and
> > > continues to reced from the detector.
> > >
> > > SR experts Tom Roberts and Paul Andersen claimed the following for this
> > > thought experiment:
> > > 1. The detector will initially detect an arbituary high frequency---call
> > > this Fh.
> > > 2. Subsequent measurements will show a decrease in frequency--from Fh. The
> > > frequency rate will continue to decrease until it is no longer detectable.
> >
> > No, we didn't claim that.
>
> Yes you did. Or at least Tom Roberts did and you agreed. I will spent some
> time to dig this up.
We NEVER said that "the frequency will continue to decrease until
it is no longer detectable." It won't.
When the source is infinitely far away and approaching,
the frequency will be fH = fo*sqrt((c+v)/(c-v))
When the source is infinitely far away and receding,
the frequency will be fL = fo*sqrt((c-v)/(c+v))
As the source is passing, the frequency will monotonously decrease
from fH to fL. The rate of decrease will be highest when the source
is closest to the detector.
THAT's what I said.
Next time, please "dig it up" FIRST, then quote me literally.
When you paraphrase, you always get it wrong because you never
understand what you are paraphrasing.
Insisting that people have said what they never have said is lying, Ken.
Stop it, please.
> > > In real life, the initial frequency would
> > > be lowest detectable. It will increase to a max. when the car is at the
> > > nearest point of approach ---call this Fm.
> > > When the car passes the detector, the detected frequency will decrease from
> > > Fm until it is no longer detectable.
> > > :
> > > What do you think???
> >
> > I think what anybody will think.
> > That you have no clue.
>
> So you think what I said is wrong?? Or are you just letting off steam??
Of course is what you said wrong. It always is.
> > BTW, Ken.
> > Most people learn how your scenario works in the real world
> > by experimental evidence while they are still kids.
> > Have you never passed a sound source while travelling by train, Ken?
>
> Ah so you agree that my description works in the real world!!! But
> unfortunately for you that would mean that you are disagreeing with your SR
> brother Eric Prebys. He said:
> "The answer is that as long as the speed is constant, the frequency
> will be shifted up by a CONSTANT amount while the transmitter is moving
> (directly) toward you, and down by a CONSTANT amount, when it is moving
> (directly) away from you."
Of course this is right. Everybody knows that.
But tell me, Ken:
Are you deaf, or have you never travelled by train? :-)
> So I guess you will have to fight with your SR brother to settle this issue.
> Unless of course you are claiming that he is also correct by reason of
> authority.
There is no issue to settle.
Everybody but you know how Doppler shift works.
And you don't count.
Paul
Ken H. Seto
So are you saying that you can detect a receding source indefintiely?? In
other words it will never get out of range?? You got one powerful
transimitter there Bud. <shrug>
>
> When the source is infinitely far away and approaching,
> the frequency will be fH = fo*sqrt((c+v)/(c-v))
Where do you get fo from?? The experiment I described has only one moving
source in the car. and one detector on the side of the road. The detector
measures the frequency of the frequency of the source in the moving car at a
regular interval.
BTW you never gave the above formula as part of your previous argument.
> When the source is infinitely far away and receding,
> the frequency will be fL = fo*sqrt((c-v)/(c+v))
Again where do you get fo from?? Also you never gave this formula as part of
your previous argument. Is it because Prebys said so and you changed your
position?
> As the source is passing, the frequency will monotonously decrease
> from fH to fL. The rate of decrease will be highest when the source
> is closest to the detector.
>
> THAT's what I said.
You are a liar. In the previous post you said:
"BTW, Ken.
Most people learn how your scenario works in the real world
by experimental evidence while they are still kids.
Have you never passed a sound source while travelling by train, Ken?"
Here you were equating the frequency of the train whistle with the
experiment that I described with a moving radar source in the car. The
frequency of the train whistle increases as the train is approaching you and
decreases as the train is receding away from you. So are you now saying
that what you said here was a lie?
>
> Next time, please "dig it up" FIRST, then quote me literally.
> When you paraphrase, you always get it wrong because you never
> understand what you are paraphrasing.
> Insisting that people have said what they never have said is lying, Ken.
> Stop it, please.
You changed your position within the same thread so you credibility is shot.
>
> > > > In real life, the initial frequency would
> > > > be lowest detectable. It will increase to a max. when the car is at
the
> > > > nearest point of approach ---call this Fm.
> > > > When the car passes the detector, the detected frequency will
decrease from
> > > > Fm until it is no longer detectable.
> > > > :
> > > > What do you think???
> > >
> > > I think what anybody will think.
> > > That you have no clue.
> >
> > So you think what I said is wrong?? Or are you just letting off steam??
>
> Of course is what you said wrong. It always is.
Of course you are full of shit.
>
> > > BTW, Ken.
> > > Most people learn how your scenario works in the real world
> > > by experimental evidence while they are still kids.
> > > Have you never passed a sound source while travelling by train, Ken?
> >
> > Ah so you agree that my description works in the real world!!! But
> > unfortunately for you that would mean that you are disagreeing with your
SR
> > brother Eric Prebys. He said:
> > "The answer is that as long as the speed is constant, the frequency
> > will be shifted up by a CONSTANT amount while the transmitter is moving
> > (directly) toward you, and down by a CONSTANT amount, when it is moving
> > (directly) away from you."
>
> Of course this is right. Everybody knows that.
> But tell me, Ken:
> Are you deaf, or have you never travelled by train? :-)
So you changed your position from the train whistle analogy to Prebys'
stupid assertion. But unfortunately for you his assertion is stupidly wrong.
Ken Seto
Daryl McCullough
>
>
>"Daryl McCullough" <da...@cogentex.com> wrote
>> >He [Eric Prebys] said:
>> >"The answer is that as long as the speed is constant, the frequency
>> >will be shifted up by a CONSTANT amount while the transmitter is moving
>> >(directly) toward you, and down by a CONSTANT amount, when it is moving
>> >(directly) away from you."
>> >So I guess you will have to fight with your SR brother to settle
>> >this issue.
>> >Unless of course you are claiming that he is also correct by reason of
>> >authority.
>>
>> I'm sure that Paul agrees with Eric's description of the Doppler
>> shift.
>
>No they don't agree. Paul said that Doppler shift of light is like a sound
>source in a moving train and you are on the embankment. The frequency of
>the sound shifts to a higher pitch as the train approaches you but the
>shift is not constant and it is dependent on the distance between you and
>the source.
That agrees with what Eric said. The problem is this: unless
you are standing right on the railroad track, then the train's
closing rate (the rate at which the distance between you and
the train decreases) is not constant. Mathematically, if you
are standing a distance x from the railroad track, and the
train is a distance y down the track, then the distance between
you and the train is
L = square-root(x^2 + y^2)
the rate at which L is changing is
dL/dt = (x dx/dt + y dy/dt)/L
Since x is constant, and dy/dt = v (the speed of the train),
this gives:
V = dL/dt = y v/L
The velocity that is used in the Doppler shift formula is V,
and you can see that it depends on the distance y.
>The frequency of the sound shifts to a lower pitch as the train
>is receding away from you but the shift is not constant and it is dependent
>on the distance between you and the source.
That's because the separation rate V is dependent on the distance.
When people write the (nonrelativistic) Doppler shift formula as
f' = f/(1 - V/v_sound)
the meaning of V is the component of the velocity in the direction
of the observer. For a train approaching, that changes with time,
even if the velocity of the train is constant.
Eric Prebys
The only mistake Paul made was to appeal to *your* experience -
which tacitly assumes that you hear the same sounds
(and see the the same shapes and colors) as mentally stable people.
I'm sure he won't make that mistake again.
> Here you were equating the frequency of the train whistle with the
> experiment that I described with a moving radar source in the car. The
> frequency of the train whistle increases as the train is approaching you and
> decreases as the train is receding away from you. So are you now saying
No it doesn't, Ken. You're (once again) confusing *volume* with
*frequency*. Have you ever actually listened to a train? When it's
approaching from far away, the frequency is shift higher. As it passes
by, the frequency shifts lower. At the point of closest approach, the
frequency shift is excactly zero (for sound, that is. In the
relativistic
case, there's a slight transverse red-shift, which has been measured
in experiment).
Now, please restrict yourself to reality.
> that what you said here was a lie?
> >
> > Next time, please "dig it up" FIRST, then quote me literally.
> > When you paraphrase, you always get it wrong because you never
> > understand what you are paraphrasing.
> > Insisting that people have said what they never have said is lying, Ken.
> > Stop it, please.
>
> You changed your position within the same thread so you credibility is shot.
>
No one has changed their position on anything. Specifically, they've
been quite consistent in
(1) Describing relativistic Doppler shift
(2) Describing non-relativistic Doppler shift
(3) Telling you you're an idiot.
There is no disagreement between my assertion and Paul's. You see,
we apparently both took physics in college.
Doppler shifts are not considered an "advanced" topic, except by
you.
Ken H. Seto
Hey jerk face why don't you let Paul answer this?
Ken Seto
Ken H. Seto
Hey stupid, the frequency of the ambulance's siren increases as it
approaches you. This means that the frequency detected is not CONSTANT as
you asserted.
Ken Seto
Paul B. Andersen
"Ken H. Seto" <ken...@erinet.com> wrote in message news:3c3c55d6$0$35620$4c5e...@news.erinet.com...
I won't bother to participate in this stupid discussion where
everybody but you know the correct answers.
And as you so vividly have demonstrated in this posting of yours you
never understand what you are told anyway, so there is no point.
But please remember to quote me literally the next time you
feel the need to tell the NG about what I have said.
Never paraphrase me!
That's all.
Paul
Ken H. Seto
>
>
> "Ken H. Seto" wrote:
> >
> > "Eric Prebys" <pre...@fnal.gov> wrote in message
> > news:3C39DA6A...@fnal.gov...
> > >
> > >
> > > "Ken H. Seto" wrote:
> > > >
> > > > "Ken H. Seto" <ken...@erinet.com> wrote in message
> > > > news:3c3715c2$0$35612$4c5e...@news.erinet.com...
> > > > >
> > > > > "Eric Prebys" <pre...@fnal.gov> wrote in message
> > > > > news:3C363714...@fnal.gov...
> > > > > >
> > > > > >
> > > > > > "Ken H. Seto" wrote:
> > > > > > >
> > > > > > > "Eric Prebys" <pre...@fnal.gov> wrote in message
> > > > > > > news:3C361588...@fnal.gov...
> > > > > > > >
> > > > > > > >
> > > > > > > > "Ken H. Seto" wrote:
> > > > > > > > >
> > > > > > > > > "Eric Prebys" <pre...@fnal.gov> wrote in message
> > > > > > > > > news:3C34E262...@fnal.gov...
> > > > > > > > > >
> > > > > > > > > >
> > > > > > > > > > "Ken H. Seto" wrote:
> > > > > > > > > > >
> > > > > > > > > > > "Eric Prebys" <pre...@fnal.gov> wrote in message
> > > > > > > > > > > news:3C34BA4C...@fnal.gov...
> > > > > > >
This equation does not apply to the transverse Doppler situation that I
described in my original post. In the tranverse Doppler situation the
frequency detected is not a constant as asserted by you.
Ken Seto
Ken H. Seto
I didn't overlook anything. The scenario I described is a transverse Doppler
situation. In transeverse Doppler the frequency detected is dependent on the
radial length of the approaching source. In case you are too stupid to
understand this means that the frequency detected by the detector on the
side of the highway is not a constant amount as you asserted.
Ken Seto
Eric Prebys
No, it doesn't. You must have been passed out when the ambulance came
for
you.
> you asserted.
>
> Ken Seto
I'm signing off. It's worthless to argue with someone who invents
quotes,
doesn't understand basic mathematics, and has never heard an ambulance
or
train whistle.
As far as I can tell, everyone on this thread is in perfect agreement
on the facts except you, and history has shown your wrong about
everything
in the world, so I'm clearly wasting my time.
Enjoy your monumental ignorance.
-Eric
Eric Prebys
No this does not apply to the transverse Doppler shift, nor were you
talking about the transverse Doppler shift in your original
post. The transverse Doppler shift only becomes important
when the object is close.
I was quite careful to make this distinction in my response, but
you snipped that part since you didn't understand it.
> frequency detected is not a constant as asserted by you.
>
There is no transverse Doppler shift in sound. In sound, the
Doppler shift is determined only be the component of velocity
of the source toward or away from the observer, and so it
goes to precisely zero at the point of closest approach (NOT
a maximum as you claim).
The relativisity, there is an additionaly redshift component
coming from the relative slowing of the moving source's clock.
At the point of closest approach, this is the only effect, and
so the frequency at that point will be shifted *DOWN* by a
factor or 1/gamma, NOT *up* as you keep asserting. The
relativistic transverse Doppler shift has been well measured in
experiment. See
http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html
section 4 befor you say anything else stupid.
You're just wrong on all counts, Ken. You get your physics wrong,
you get your math wrong, you get your quotes wrong. A while back
you accidentally said something right - and then corrected yourself.
-Eric
Ken H. Seto
1. The situation I described in my origial post is a transverse Doppler
situation. The frequency detected in a transverse Doppler situation is not
constant. It depends on the radial length of the transimitter from the
detector.
2. The situation Eric decribe is a direct Doppler situation---the
transmitter is moving directly toward the detector. The frequency detected
by the detector is a constant according to the following formula:
f'=fo*sqrt(c+v/c-v)
3. The situatyion Paul initially described (the train example) is a
transverse Doppler situation. Therefore the frequency detected is not a
constant as Eric asserted. That means that Paul did not agree with Eric
initially.
4. Now Paul is changing his position to Eric's postion.
BTW did you see my comment regarding the accelerating space ship?
_______________________________________________________
In real life, an accelerating rocketship of 1g is equivalent to a rocketship
lying horizontally at sea level in earth's gravitational field. In other
words, no frequency shift or weight change throughout the length of the
ship. All your calculations are based on the faulty assumption that the
different heights in the ship is equivalent to the different heights in
earth's gravitational field.
______________________________________________________
Ken Seto
"Daryl McCullough" <da...@cogentex.com> wrote in message
news:a1hl6...@drn.newsguy.com...
Joe Fischer
: The problem is as follows:
: 1. The situation I described in my origial post is a transverse Doppler
: situation. The frequency detected in a transverse Doppler situation is not
: constant. It depends on the radial length of the transimitter from the
: detector.
Ken, I would like to see you agree with somebody,
or somebody, anybody, agree with you. Ir is not helpful
to anyone for continued misunderstandings and disagreements.
I am not sure what "transverse doppler" would mean.
Yes, there is a difference between situations where
an object, emitter or receiver, is not moving directly
toward or away from each other.
But it requires trig, and very few people seem
to use trig, and there is no way to tell if theu understand
or can use trig.
: 2. The situation Eric decribe is a direct Doppler situation---the
: transmitter is moving directly toward the detector. The frequency detected
: by the detector is a constant according to the following formula:
: f'=fo*sqrt(c+v/c-v)
: 3. The situatyion Paul initially described (the train example) is a
: transverse Doppler situation. Therefore the frequency detected is not a
: constant as Eric asserted. That means that Paul did not agree with Eric
: initially.
So include the correct formulas for both cases,
including the trig functions.
: 4. Now Paul is changing his position to Eric's postion.
Please try to avoid this type of rebuttal.
: BTW did you see my comment regarding the accelerating space ship?
I saw it, and I think you are not aware if which
way the crew compartment is oreinted within the type of
rockets in current use, which is a good way to reference
discussions.
: ______________________________________________________
: In real life, an accelerating rocketship of 1g is equivalent to a rocketship
: lying horizontally at sea level in earth's gravitational field. In other
: words, no frequency shift or weight change throughout the length of the
: ship. All your calculations are based on the faulty assumption that the
: different heights in the ship is equivalent to the different heights in
: earth's gravitational field.
: ______________________________________________________
: Ken Seto
Please either state whether or not you are using
a different type of crew compartment orientation than the
way the Space Shuttle is constructed.
It is made that way so that when the shuttle returns
and flies like an airplane the crew is sitting like they
would in an airplane.
I am not sure this is the best situation, but
it is essentially the only one used by the USA now, but
I don't know about the Russian crew modules.
Joe Fischer
--
3
Eric Prebys
"Ken H. Seto" wrote:
>
> The problem is as follows:
> 1. The situation I described in my origial post is a transverse Doppler
> situation. The frequency detected in a transverse Doppler situation is not
Not true, here is what you described
== Ken Seto wrote:
-> A frequency detector is at rest on the side of a straight highway. A
radar
-> source is in a car moving relative to the highway at a connstant
speed. The
-> detector makes frequency measurements at regular intervals. The car
is
-> approaching the detector from a distance. It passes the detector and
-> continues to reced from the detector.
The transverse doppler shift is only applicable when the car is
more or less directly beside you. When it is "approaching" or
"receding", it is reasonably correct to use the equation for motion
directly toward or away from the observer. This, anyway, is the
principle
which radar guns employ.
BTW, radar guns are not sensitive enough to detect the transverse
doppler shift.
You then went on to claim:
==Ken Seto wrote:
-> SR experts Tom Roberts and Paul Andersen claimed the following for
this
-> thought experiment:
-> 1. The detector will initially detect an arbituary high
frequency---call
-> this Fh.
-> 2. Subsequent measurements will show a decrease in frequency--from
Fh. The
-> frequency rate will continue to decrease until it is no longer
detectable.
This is not true. Neither Tom nor Paul claimed this, and on repeated
requests
you have been unable to produce a quote where they did.
You then completely made a fool of yourself by makeing the following
claim
==Ken Seto wrote:
-> Their claims are obviously bogus. In real life, the initial
frequency would
-> be lowest detectable. It will increase to a max. when the car is at
the
-> nearest point of approach ---call this Fm.
-> When the car passes the detector, the detected frequency will
decrease from
-> Fm until it is no longer detectable.
which cannot be supported by any physics or mathematical equation and
doesn't
apply to ANY of the situations being discussed.
> constant. It depends on the radial length of the transimitter from the
> detector.
Again, incorrect. It depends on the total velocity of the source
and the angle between the path of the source and a direct path
to the observer. It does NOT depend on the distance of the source
from the observer.
> 2. The situation Eric decribe is a direct Doppler situation---the
> transmitter is moving directly toward the detector. The frequency detected
> by the detector is a constant according to the following formula:
> f'=fo*sqrt(c+v/c-v)
Which is approximately true except when the transmitter is directly
beside the observer.
At non relativistic speeds, the approximation:
\Delta f/f = v/c
where v is the component of velocity toward from the observer
will be valid in all cases - leading to no frequency shift
when the source is directly alongside.
On no case does the radial position have anything to do with it.
> 3. The situatyion Paul initially described (the train example) is a
> transverse Doppler situation. Therefore the frequency detected is not a
> constant as Eric asserted. That means that Paul did not agree with Eric
> initially.
No, it means that Paul and I both understand physics and each understood
what the other was talking about. You do not understand physic, math,
or anything else, to my knowledge.
If we do disagree about something, we'll discuss it between ourselves.
We certainly don't need mediation from someone who doesn't understand
what either of us said.
> 4. Now Paul is changing his position to Eric's postion.
>
No one has changed their position. You merely do not understand
what anybody said. Since everybody else in the entire world was
smart enough to understand it, I don't think anyone's going to
lose sleep over that.
Ken H. Seto
>
> "Ken H. Seto" <ken...@erinet.com> wrote in message
news:3c3c55d6$0$35620$4c5e...@news.erinet.com...
> >
> > "Paul B. Andersen" <paul.b....@hia.no> wrote in message
> > news:a1g0ul$50a$1...@snipp.uninett.no...
> >
> > stupid assertion. But unfortunately for you his assertion is stupidly
wrong.
>
> I won't bother to participate in this stupid discussion where
> everybody but you know the correct answers.
> And as you so vividly have demonstrated in this posting of yours you
> never understand what you are told anyway, so there is no point.
The problem is as follows:
1. The situation I described in my origial post is a transverse Doppler
situation. The frequency detected in a transverse Doppler situation is not
constant. It depends on the radial length of the transimitter from the
detector.
2. The situation Eric decribe is a direct Doppler situation---the
transmitter is moving directly toward the detector. The frequency detected
by the detector is a constant according to the following formula:
f'=fo*sqrt(c+v/c-v)
3. The situatyion Paul initially described (the train example) is a
transverse Doppler situation. Therefore the frequency detected is not a
constant as Eric asserted. That means that Paul did not agree with Eric
initially.
4. Now Paul is changing his position to Eric's postion.
>But please remember to quote me literally the next time you
>feel the need to tell the NG about what I have said.
>Never paraphrase me!
>That's all.
I did quote you correctly. Your denial now is laughable. If you are so sure
that I mis-quoting you why didn't you provide us with the correct quotes??
BTW did you or did you not make the following claims??
________________________________________________________
A space ship under goes a constant acceleration of 1g and a light source is
located in the middle of the ship. They claimed that the frequency of that
light source will be red shifted at the front of the ship and blue shifted
at the rear of the ship. Also they claimed that a standard weigh will
weight less at the front of the ship and more at the rear of the ship. They
said that these claims are supported by the Equivalent Principle.
________________________________________________________
Ken Seto
Ken H. Seto
1. The situation I described in my origial post is a transverse Doppler
situation. The frequency detected in a transverse Doppler situation is not
constant. It depends on the radial length of the transimitter from the
detector.
2. The situation Eric decribe is a direct Doppler situation---the
transmitter is moving directly toward the detector. The frequency detected
by the detector is a constant according to the following formula:
f'=fo*sqrt(c+v/c-v)
3. The situatyion Paul initially described (the train example) is a
transverse Doppler situation. Therefore the frequency detected is not a
constant as Eric asserted. That means that Paul did not agree with Eric
initially.
4. Now Paul is changing his position to Eric's postion.
Ken Seto
"Eric Prebys" <pre...@fnal.gov> wrote in message
news:3C3B6088...@fnal.gov...
Ken H. Seto
I certainly was talking about the transverse Doppler shift. The detector is
at the side of ther highway and the transmitter is in a car moving at a
constant velocity in the straight higway. If that's not transerverse Doppler
what is it??
>
> I was quite careful to make this distinction in my response, but
> you snipped that part since you didn't understand it.
I understand everything. It was you who kept on using the wrong Doppler
situation to explain my proposed experiment. In the transverse Doppler
situation the detected frequency is never constant as you asserted.
Ken Seto
Dirk Van de moortel
"Ken H. Seto" <ken...@erinet.com> wrote in message news:3c3ca752$0$37108$4c5e...@news.erinet.com...
This is a pure Seto quote.
First appearance on 10-dec-2001
http://groups.google.com/groups?as_umsgid=3c15afc5%240%2437100%244c5...@news.erinet.com
The quote is preceeded by
|________________________________________________________
| In the past, top dog SR experts such as Tom Roberts and
| Paul Andersen made the following bogus claims:
|________________________________________________________
Dirk Vdm
Daryl McCullough
>
>The problem is as follows:
>1. The situation I described in my origial post is a transverse Doppler
>situation. The frequency detected in a transverse Doppler situation is not
>constant. It depends on the radial length of the transimitter from the
>detector.
By "transverse Doppler situation", I assume that you mean that
the velocity of the sound source is perpendicular to the line
connecting the source from the observer. For a train, that
occurs *only* when the train whistle passes you by.
Classically, there is no transverse Doppler effect, but relativistically,
the frequency is shifted lower by square-root(1-(v/c)^2).
>2. The situation Eric decribe is a direct Doppler situation---the
>transmitter is moving directly toward the detector.
For the case of a train, that only happens if you are sitting directly
in the path of the train---that is, if you are on the track. If you
are off the track, then the situation will go from being an approximately
"direct Doppler situation" when the train is approaching from far
away, to being an approximately "transverse Doppler situation" when
the train is passing you by, and then will return to a "direct
Doppler situation" long after the train has passed, when it is
receding from you.
>The frequency detected by the detector is a constant according
>to the following formula: f'=fo*sqrt(c+v/c-v)
Yes. If you are actually sitting on the track, that is the formula
that is appropriate. If you are some distance from the track, the
formula is more complicated.
>3. The situation Paul initially described (the train example) is a
>transverse Doppler situation.
No, it's not. Only right at the moment when the train is passing
you does the transverse Doppler formula apply.
>Therefore the frequency detected is not a constant as Eric
>asserted.
That's because the component of velocity that is directed
towards the observer is constantly changing. It is not
because the Doppler formula depends on distance.
>That means that Paul did not agree with Eric initially.
I don't think so.
>4. Now Paul is changing his position to Eric's postion.
>
>BTW did you see my comment regarding the accelerating space ship?
>_______________________________________________________
>In real life, an accelerating rocketship of 1g is equivalent to a rocketship
>lying horizontally at sea level in earth's gravitational field. In other
>words, no frequency shift or weight change throughout the length of the
>ship. All your calculations are based on the faulty assumption that the
>different heights in the ship is equivalent to the different heights in
>earth's gravitational field.
>______________________________________________________
I don't know what you mean by "in real life". Do you mean that
this situation has been experimentally tested?
My post gave Special Relativity prediction for the g-force
felt on a rocket undergoing constant proper acceleration.
If the front end and the back end accelerate in just such
a way that the length of the rocket remains constant (in
the frame in which the rocket is instantaneously at rest)
then the g-force felt by the front end must be less than
the g-force felt by the rear end.
I didn't assume this, I derived it from Special Relativity.
Joe Fischer
: The transverse doppler shift is only applicable when the car is
: "receding", it is reasonably correct to use the equation for motion
: directly toward or away from the observer. This, anyway, is the
: principle
: which radar guns employ.
:
: BTW, radar guns are not sensitive enough to detect the transverse
: doppler shift.
That isn't true in the sense that as you pass
the unit, your registered speed drops, and it is very
easy to see in the big display speed warning setups.
Joe Fischer
--
3
Eric Prebys
That's a different thing. What speed guns measure is the speed of
the car toward or away from speed gun. When you're reasonably
far away from the gun, that speed is approximately your
speed. As you pass, the speed drops as
the car changes from moving toward the gun to moving away
from the gun, so that velocity is instantaneously zero, which
is what a speed gun will measure if you point it directly
at a car as it passes (i.e. you're pointing it at 90 degrees
to the path of th car). That is NOT what's referred
to as "transverse Doppler shift".
The "transverse Doppler shift" is a relativistic effect
in which a fast-moving source - in your frame of reference -
will be redshifted as it passes you, because its clock
is running slowly in your frame. Radar guns do not
detect this because
(1) It's a higher order effect.
(2) It does not affect reflected signals, only
signals which are *transmitted* at a
particular frequency by the moving source.
Ken doesn't know the difference, and gets both wrong
anyway.
> Joe Fischer
>
> --
> 3
Ken H. Seto
> Ken says...
> >
> >The problem is as follows:
> >1. The situation I described in my origial post is a transverse Doppler
> >situation. The frequency detected in a transverse Doppler situation is
not
> >constant. It depends on the radial length of the transimitter from the
> >detector.
>
> By "transverse Doppler situation", I assume that you mean that
> the velocity of the sound source is perpendicular to the line
> connecting the source from the observer. For a train, that
> occurs *only* when the train whistle passes you by.
No, the set up I described is that the car is not headed directly into the
detector and the detector makes frequency measurements at a regular
intervals. My interpretation of these measurements is that they are NOT a
constant as Prebys asserted.
>
> Classically, there is no transverse Doppler effect, but relativistically,
> the frequency is shifted lower by square-root(1-(v/c)^2).
This is just one point in a series of measurements made by the detector.
>
> >2. The situation Eric decribe is a direct Doppler situation---the
> >transmitter is moving directly toward the detector.
>
> For the case of a train, that only happens if you are sitting directly
> in the path of the train---that is, if you are on the track. If you
> are off the track, then the situation will go from being an approximately
> "direct Doppler situation" when the train is approaching from far
> away, to being an approximately "transverse Doppler situation" when
> the train is passing you by, and then will return to a "direct
> Doppler situation" long after the train has passed, when it is
> receding from you.
I am not interested in this situation. I don't understand why you SR experts
keep on wanting to use it to explain the situation I described in item #1.
>
> >The frequency detected by the detector is a constant according
> >to the following formula: f'=fo*sqrt(c+v/c-v)
>
> Yes. If you are actually sitting on the track, that is the formula
> that is appropriate. If you are some distance from the track, the
> formula is more complicated.
The situation I am interested in is exactly that--the detector is at some
distance from the track and making a series of frequency measurements. My
interpretation of these measurements is nort of the same value (not a
constant) as asserrted by Prebys.
>
> >3. The situation Paul initially described (the train example) is a
> >transverse Doppler situation.
>
> No, it's not. Only right at the moment when the train is passing
> you does the transverse Doppler formula apply.
Again the detector is at a distance from the track and makes a series of
frequency measurements. I said that these measurements are not constant and
Prebys claims that they are constant.
>
> >Therefore the frequency detected is not a constant as Eric
> >asserted.
>
> That's because the component of velocity that is directed
> towards the observer is constantly changing. It is not
> because the Doppler formula depends on distance.
That's what I been telling you SR experts. But the Runt Prebys went berserk
and accused me of incompetent.
>
> >That means that Paul did not agree with Eric initially.
>
> I don't think so.
I think so.
>
> >4. Now Paul is changing his position to Eric's postion.
> >
> >BTW did you see my comment regarding the accelerating space ship?
> >_______________________________________________________
> >In real life, an accelerating rocketship of 1g is equivalent to a
rocketship
> >lying horizontally at sea level in earth's gravitational field. In other
> >words, no frequency shift or weight change throughout the length of the
> >ship. All your calculations are based on the faulty assumption that the
> >different heights in the ship is equivalent to the different heights in
> >earth's gravitational field.
> >______________________________________________________
>
> I don't know what you mean by "in real life". Do you mean that
> this situation has been experimentally tested?
The Equivalent Principle means that you can't distinguish between
gravtational mass and inertial mass. That's all. There is no gravitational
gradient or frequency shift within an accelerating spaceship.
>
> My post gave Special Relativity prediction for the g-force
> felt on a rocket undergoing constant proper acceleration.
> If the front end and the back end accelerate in just such
> a way that the length of the rocket remains constant (in
> the frame in which the rocket is instantaneously at rest)
> then the g-force felt by the front end must be less than
> the g-force felt by the rear end.
That's not true. Even Prebys will disagree with you on that. He said that
there is no gravtational gradient within the accelerating spaceship. But he
said the there is frequency shift within the spaceship.
Ken Seto
Ken H. Seto
>
>
> "Ken H. Seto" wrote:
> >
> > The problem is as follows:
> > 1. The situation I described in my origial post is a transverse Doppler
> > situation. The frequency detected in a transverse Doppler situation is
not
>
> Not true, here is what you described
> == Ken Seto wrote:
> -> A frequency detector is at rest on the side of a straight highway. A
> radar
> -> source is in a car moving relative to the highway at a connstant
> speed. The
> -> detector makes frequency measurements at regular intervals. The car
> is
> -> approaching the detector from a distance. It passes the detector and
> -> continues to reced from the detector.
>
> The transverse doppler shift is only applicable when the car is
> more or less directly beside you.
I specified that the detector makes frequency measurements at a regular
intervals. I said that these measurements are not constant and you asserted
that they are. That's the whole argument between us..
>When it is "approaching" or
> "receding", it is reasonably correct to use the equation for motion
> directly toward or away from the observer. This, anyway, is the
> principle
> which radar guns employ.
I didn't say anything about a radar gun. I said a radar source.
>
> BTW, radar guns are not sensitive enough to detect the transverse
> doppler shift.
There are detectors that are sensitive enough to detect the frequency of a
radar source.
Ken Seto
Daryl McCullough
>Again the detector is at a distance from the track and makes a series of
>frequency measurements. I said that these measurements are not constant and
>Prebys claims that they are constant.
I think that you misunderstood what he was saying.
Decompose the velocity v of the source into
a velocity v_parallel toward the detector, and a velocity
v_perpendicular at right angles towards the line
connecting the source and the detector. Classically, the
Doppler shift only depends on v_parallel. If v_parallel is
constant, then so is the Doppler shift.
But a train does *not* have a constant v_parallel
(unless you are sitting on the track).
I don't think that Eric would disagree with that.
Ask him.
>> >In real life, an accelerating rocketship of 1g is equivalent to a
>rocketship
>> >lying horizontally at sea level in earth's gravitational field. In other
>> >words, no frequency shift or weight change throughout the length of the
>> >ship. All your calculations are based on the faulty assumption that the
>> >different heights in the ship is equivalent to the different heights in
>> >earth's gravitational field.
>> >______________________________________________________
>>
>> I don't know what you mean by "in real life". Do you mean that
>> this situation has been experimentally tested?
>
>The Equivalent Principle means that you can't distinguish between
>gravitational mass and inertial mass.
That's true.
>There is no gravitational gradient or frequency shift within an
>accelerating spaceship.
The frequency shift in an accelerated spaceship is derivable
from Special Relativity alone. The fact that the proper
acceleration at the front of the spaceship is slightly
less than at the rear of the spaceship is derivable from
Special Relativity alone. The equivalence principle doesn't
affect that conclusion.
>> My post gave Special Relativity prediction for the g-force
>> felt on a rocket undergoing constant proper acceleration.
>> If the front end and the back end accelerate in just such
>> a way that the length of the rocket remains constant (in
>> the frame in which the rocket is instantaneously at rest)
>> then the g-force felt by the front end must be less than
>> the g-force felt by the rear end.
>
>That's not true.
It's *derivable* from SR.
>Even Prebys will disagree with you on that.
He may not have gone through the derivation.
>He said that there is no gravitational gradient within
>the accelerating spaceship. But he said the there is
>frequency shift within the spaceship.
There are two different sorts of "gradients" involved
in the accelerating ship, and I'm not sure which one
you are calling the "gravitational gradient".
(1) First, the "gravitational potential" changes in going
from the rear of the ship to the front of the ship.
It takes work to move something from the rear to the
front.
(2) Second, the "force of gravity" changes in going from
the rear to the front. The front has a slightly smaller
proper acceleration than the rear.
The frequency shift depends only on (1), not on (2).
For most purposes, you can neglect effect (2) (it's
a small effect) but you can't neglect effect (1).
Paul Cardinale
>BTW, radar guns are not sensitive enough to detect the transverse doppler shift.
The kenseto replied:
> There are detectors that are sensitive enough to detect the frequency of a radar source.
Does anyone doubt that this clearly shows the quality of the kenseto's mind?
Paul Cardinale
Eric Prebys
Daryl McCullough wrote:
>
> Ken says...
>
> >Again the detector is at a distance from the track and makes a series of
> >frequency measurements. I said that these measurements are not constant and
> >Prebys claims that they are constant.
>
> I think that you misunderstood what he was saying.
> Decompose the velocity v of the source into
> a velocity v_parallel toward the detector, and a velocity
> v_perpendicular at right angles towards the line
> connecting the source and the detector. Classically, the
> Doppler shift only depends on v_parallel. If v_parallel is
> constant, then so is the Doppler shift.
>
You're wasting your time. Ken does not understand math
this advanced.
> But a train does *not* have a constant v_parallel
> (unless you are sitting on the track).
>
But if you are anywhere near the track v_parallel ~ v ~
constant until the train is very close.
> I don't think that Eric would disagree with that.
> Ask him.
>
He doesn't have to. I was quite clear about it to begin
with.
Besides, Ken's assertion - that the frequency *increases*
as the source approaches - is impossible to reconcile
with any of these scenarios.
> >> >In real life, an accelerating rocketship of 1g is equivalent to a
> >rocketship
> >> >lying horizontally at sea level in earth's gravitational field. In other
> >> >words, no frequency shift or weight change throughout the length of the
> >> >ship. All your calculations are based on the faulty assumption that the
> >> >different heights in the ship is equivalent to the different heights in
> >> >earth's gravitational field.
> >> >______________________________________________________
> >>
> >> I don't know what you mean by "in real life". Do you mean that
> >> this situation has been experimentally tested?
> >
> >The Equivalent Principle means that you can't distinguish between
> >gravitational mass and inertial mass.
>
> That's true.
>
> >There is no gravitational gradient or frequency shift within an
> >accelerating spaceship.
>
> The frequency shift in an accelerated spaceship is derivable
> from Special Relativity alone. The fact that the proper
As several people pointed out, the first order
result is consistent not only with special relativity,
but just about any model you can think of. In fact,
it's *extremely* difficult to invent a model that *doesn't*
give you these shifts in an frame.
> acceleration at the front of the spaceship is slightly
> less than at the rear of the spaceship is derivable from
> Special Relativity alone. The equivalence principle doesn't
> affect that conclusion.
>
> >> My post gave Special Relativity prediction for the g-force
> >> felt on a rocket undergoing constant proper acceleration.
> >> If the front end and the back end accelerate in just such
> >> a way that the length of the rocket remains constant (in
> >> the frame in which the rocket is instantaneously at rest)
> >> then the g-force felt by the front end must be less than
> >> the g-force felt by the rear end.
> >
> >That's not true.
>
> It's *derivable* from SR.
>
> >Even Prebys will disagree with you on that.
>
> He may not have gone through the derivation.
>
I was trying to restrict myself to the first order correction
to avoid having to think about this, but on reflection,
I agree with your interpretation. My main
point was that if the acceleration was 1g, it's effectively
uniform for any conceivable experiment. When talking about
the red shift, bringing up the *difference* in acceleration
from the back to the front of the ship is a red herring.
Nevertheless, I believe I mispoke on that particular issue.
Ken is fixated on the notion that it's the *difference*
in gravitational acceleration that's responsible for
gravitational redshift - in other words, he believes
that what was responsible for the Rebka-Pound result
was that "g" was slightly different at the top and
bottom of the tower.
> >He said that there is no gravitational gradient within
> >the accelerating spaceship. But he said the there is
> >frequency shift within the spaceship.
>
> There are two different sorts of "gradients" involved
> in the accelerating ship, and I'm not sure which one
> you are calling the "gravitational gradient".
>
which was my point. Ken seems to think that acceleration
*causes* gravity. I was trying to get him to keep
the two separate.
I was also trying to make it clear that it's
NOT the difference in the gravitational acceleration
that's responsible for gravitational redshift, but
rather the difference in gravitational potential.
> (1) First, the "gravitational potential" changes in going
> from the rear of the ship to the front of the ship.
> It takes work to move something from the rear to the
> front.
>
> (2) Second, the "force of gravity" changes in going from
> the rear to the front. The front has a slightly smaller
> proper acceleration than the rear.
>
> The frequency shift depends only on (1), not on (2).
> For most purposes, you can neglect effect (2) (it's
> a small effect) but you can't neglect effect (1).
>
which was my point.
> --
> Daryl McCullough
> CoGenTex, Inc.
> Ithaca, NY
--
Paul B. Andersen
"Ken H. Seto" <ken...@erinet.com> wrote in message news:3c3ca752$0$37108$4c5e...@news.erinet.com...
>
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message
> news:a1hn7i$fkd$1...@snipp.uninett.no...
> >
> > "Ken H. Seto" <ken...@erinet.com> wrote in message
> news:3c3c55d6$0$35620$4c5e...@news.erinet.com...
> > >
> > > "Paul B. Andersen" <paul.b....@hia.no> wrote in message
> > > news:a1g0ul$50a$1...@snipp.uninett.no...
> > >
> > > So you changed your position from the train whistle analogy to Prebys'
> > > stupid assertion. But unfortunately for you his assertion is stupidly
> wrong.
> >
> > I won't bother to participate in this stupid discussion where
> > everybody but you know the correct answers.
> > And as you so vividly have demonstrated in this posting of yours you
> > never understand what you are told anyway, so there is no point.
>
> The problem is as follows:
> 1. The situation I described in my origial post is a transverse Doppler
> situation. The frequency detected in a transverse Doppler situation is not
> constant. It depends on the radial length of the transimitter from the
> detector.
Indeed.
And I described what would happen in this scenario:
When the source is infinitely far away and approaching,
the frequency will be fH = fo*sqrt((c+v)/(c-v))
When the source is infinitely far away and receding,
the frequency will be fL = fo*sqrt((c-v)/(c+v))
As the source is passing, the frequency will monotonously decrease
from fH to fL. The rate of decrease will be highest when the source
is closest to the detector.
And I might add:
f = fo*gamma when observed angle is v/c radians from transverse
f = fo when observed angle is 0.5*v/c rads from transverse
f = fo/gamma when observed angle is transverse
> 2. The situation Eric decribe is a direct Doppler situation---the
> transmitter is moving directly toward the detector. The frequency detected
> by the detector is a constant according to the following formula:
> f'=fo*sqrt(c+v/c-v)
Eric Prebys's statement wich you asked if I agee to, was:
"The answer is that as long as the speed is constant, the frequency
will be shifted up by a CONSTANT amount while the transmitter is moving
(directly) toward you, and down by a CONSTANT amount, when it is moving (directly) away from you."
This statement is perfectly correct, so of course I agree.
> 3. The situatyion Paul initially described (the train example) is a
> transverse Doppler situation. Therefore the frequency detected is not a
> constant as Eric asserted. That means that Paul did not agree with Eric
> initially.
The statement of Eric Prebey does not describe the same
situation as I described above.
> 4. Now Paul is changing his position to Eric's postion.
I haven't changed anything.
The only thing I said was that Eric Prepey's statement as
quoted above is correct. I never said it was description
of your scenario.
> >But please remember to quote me literally the next time you
> >feel the need to tell the NG about what I have said.
> >Never paraphrase me!
> >That's all.
>
> I did quote you correctly.
You didn't quote me at all, Ken.
You paraphrased me.
Wrongly.
> Your denial now is laughable. If you are so sure
> that I mis-quoting you why didn't you provide us with the correct quotes??
When YOU quote somebody, it is YOUR bloody job to ensure
that it is right.
Paul
kenseto
> Ken says...
>
> >Again the detector is at a distance from the track and makes a series of
> >frequency measurements. I said that these measurements are not constant
and
> >Prebys claims that they are constant.
>
> I think that you misunderstood what he was saying.
No I didn't.
> Decompose the velocity v of the source into
> a velocity v_parallel toward the detector, and a velocity
> v_perpendicular at right angles towards the line
> connecting the source and the detector. Classically, the
> Doppler shift only depends on v_parallel. If v_parallel is
> constant, then so is the Doppler shift.
But v_parallel is never constant when the train is moving transerversely
relative to the detector on the embankment.
>
> But a train does *not* have a constant v_parallel
> (unless you are sitting on the track).
You are not sitting on the track. You are sitting on the embankment beside
the track.
>
> I don't think that Eric would disagree with that.
> Ask him.
You ask him. He is a fanatic and he will never give me a true answer. He
even said to me tthat an increase in the pitch of an ambulance's siren is
not an increase in frequency.
>
> >> >In real life, an accelerating rocketship of 1g is equivalent to a
> >rocketship
> >> >lying horizontally at sea level in earth's gravitational field. In
other
> >> >words, no frequency shift or weight change throughout the length of
the
> >> >ship. All your calculations are based on the faulty assumption that
the
> >> >different heights in the ship is equivalent to the different heights
in
> >> >earth's gravitational field.
> >> >______________________________________________________
> >>
> >> I don't know what you mean by "in real life". Do you mean that
> >> this situation has been experimentally tested?
> >
> >The Equivalent Principle means that you can't distinguish between
> >gravitational mass and inertial mass.
>
> That's true.
>
> >There is no gravitational gradient or frequency shift within an
> >accelerating spaceship.
>
> The frequency shift in an accelerated spaceship is derivable
> from Special Relativity alone.
Your derivation is based on the erroneous assumption that light from the
middle will take a longer transit time to the front than to the rear. If it
is true we should be able to measure this effect in the lab--the earth's
surface is in a state of continuing acceleration.
>The fact that the proper
> acceleration at the front of the spaceship is slightly
> less than at the rear of the spaceship is derivable from
> Special Relativity alone. The equivalence principle doesn't
> affect that conclusion.
Again your derivation is based on a false assumption.
>
> >> My post gave Special Relativity prediction for the g-force
> >> felt on a rocket undergoing constant proper acceleration.
> >> If the front end and the back end accelerate in just such
> >> a way that the length of the rocket remains constant (in
> >> the frame in which the rocket is instantaneously at rest)
> >> then the g-force felt by the front end must be less than
> >> the g-force felt by the rear end.
> >
> >That's not true.
>
> It's *derivable* from SR.
>
> >Even Prebys will disagree with you on that.
>
> He may not have gone through the derivation.
>
> >He said that there is no gravitational gradient within
> >the accelerating spaceship. But he said the there is
> >frequency shift within the spaceship.
>
> There are two different sorts of "gradients" involved
> in the accelerating ship, and I'm not sure which one
> you are calling the "gravitational gradient".
>
> (1) First, the "gravitational potential" changes in going
> from the rear of the ship to the front of the ship.
> It takes work to move something from the rear to the
> front.
There is no gravitational potential in the ship. If there is an extension
from the rear of the ship it will take work to move something from that
extension to the rear of the ship. In other words, the rear and the front
have the same potential. The gravitational potential on earth is cauised by
the rotation of the earth. The rotation of the earth causes different
heights to have a different state of absolute motion and thus a different
gravitational
potentials at different heights .
>
> (2) Second, the "force of gravity" changes in going from
> the rear to the front. The front has a slightly smaller
> proper acceleration than the rear.
Again thus is based on the erroneous assumption that the transit time for
light is different from the front to the rear than the transit time from the
rear to the front.
>
> The frequency shift depends only on (1),
No gravitational potential difference in any part of the ship and therefore
no frequency shift in any part of the ship.
Ken Seto
Robert Andary
> Ken says...
>
> >Again the detector is at a distance from the track and makes a series of
> >frequency measurements. I said that these measurements are not constant
and
> >Prebys claims that they are constant.
>
> I think that you misunderstood what he was saying.
No I didn't.
> Decompose the velocity v of the source into
> a velocity v_parallel toward the detector, and a velocity
> v_perpendicular at right angles towards the line
> connecting the source and the detector. Classically, the
> Doppler shift only depends on v_parallel. If v_parallel is
> constant, then so is the Doppler shift.
But v_parallel is never constant when the train is moving transerversely
relative to the detector on the embankment.
>
> But a train does *not* have a constant v_parallel
> (unless you are sitting on the track).
You are not sitting on the track. You are sitting on the embankment beside
the track.
>
> I don't think that Eric would disagree with that.
> Ask him.
You ask him. He is a fanatic and he will never give me a true answer. He
even said to me tthat an increase in the pitch of an ambulance's siren is
not an increase in frequency.
>
> >> >In real life, an accelerating rocketship of 1g is equivalent to a
> >rocketship
> >> >lying horizontally at sea level in earth's gravitational field. In
other
> >> >words, no frequency shift or weight change throughout the length of
the
> >> >ship. All your calculations are based on the faulty assumption that
the
> >> >different heights in the ship is equivalent to the different heights
in
> >> >earth's gravitational field.
> >> >______________________________________________________
> >>
> >> I don't know what you mean by "in real life". Do you mean that
> >> this situation has been experimentally tested?
> >
> >The Equivalent Principle means that you can't distinguish between
> >gravitational mass and inertial mass.
>
> That's true.
>
> >There is no gravitational gradient or frequency shift within an
> >accelerating spaceship.
>
> The frequency shift in an accelerated spaceship is derivable
> from Special Relativity alone.
Your derivation is based on the erroneous assumption that light from the
middle will take a longer transit time to the front than to the rear. If it
is true we should be able to measure this effect in the lab--the earth's
surface is in a state of continuing acceleration.
>The fact that the proper
> acceleration at the front of the spaceship is slightly
> less than at the rear of the spaceship is derivable from
> Special Relativity alone. The equivalence principle doesn't
> affect that conclusion.
Again your derivation is based on a false assumption.
>
> >> My post gave Special Relativity prediction for the g-force
> >> felt on a rocket undergoing constant proper acceleration.
> >> If the front end and the back end accelerate in just such
> >> a way that the length of the rocket remains constant (in
> >> the frame in which the rocket is instantaneously at rest)
> >> then the g-force felt by the front end must be less than
> >> the g-force felt by the rear end.
> >
> >That's not true.
>
> It's *derivable* from SR.
>
> >Even Prebys will disagree with you on that.
>
> He may not have gone through the derivation.
>
> >He said that there is no gravitational gradient within
> >the accelerating spaceship. But he said the there is
> >frequency shift within the spaceship.
>
> There are two different sorts of "gradients" involved
> in the accelerating ship, and I'm not sure which one
> you are calling the "gravitational gradient".
>
> (1) First, the "gravitational potential" changes in going
> from the rear of the ship to the front of the ship.
> It takes work to move something from the rear to the
> front.
There is no gravitational potential in the ship. If there is an extension
from the rear of the ship it will take work to move something from that
extension to the rear of the ship. In other words, the rear and the front
have the same potential. The gravitational potential on earth is cauised by
the rotation of the earth. The rotation of the earth causes different
heights to have a
different state of absolute motion and thus a different gravitational
potentials at different heights .
>
> (2) Second, the "force of gravity" changes in going from
> the rear to the front. The front has a slightly smaller
> proper acceleration than the rear.
Again thus is based on the erroneous assumption that the transit time for
light is different from the front to the rear than the transit time from the
rear to the front.
>
> The frequency shift depends only on (1),
No gravitational ptotential difference in any part of the ship and therefore
Robert Andary
>
>
> Daryl McCullough wrote:
>
> > But a train does *not* have a constant v_parallel
> > (unless you are sitting on the track).
> >
>
> But if you are anywhere near the track v_parallel ~ v ~
> constant until the train is very close.
v_parallel is never constant. BTW if you stand on the track you would still
hear the increase in the pitch (frequency) of the train's whistle as the
train approaches you. This means that the frequency detected by you is also
dependent on the distance of separation between you and the train's whistle.
>
> > I don't think that Eric would disagree with that.
> > Ask him.
> >
>
> He doesn't have to. I was quite clear about it to begin
> with.
ROTFL. you were using the direct Dopppler formula to explain the transmitter
travelling transversely relative to the detector.
>
> Besides, Ken's assertion - that the frequency *increases*
> as the source approaches - is impossible to reconcile
> with any of these scenarios.
The pitch of the whistle increase means the frequency increase. Everybody
knows that except an SR runt fanatic like you.
They are wrong. Their idea was based on the faulty assumption that
light takes a longer transit time moving from the middle to the front than
to the rear.
>
This is a lie. My idea is that there is no gradient of any kind in the ship
and therefore no frequency shift of any kind in the ship.
- in other words, he believes
> that what was responsible for the Rebka-Pound result
> was that "g" was slightly different at the top and
> bottom of the tower.
This again is a lie. I believe that the shift was due to the different
states of absolute motion between the top and bottom of the tower. The
different states of absolute motion between the top and bottom of the tower
is, in turn, caused by the rotation of the earth.
>
> > >He said that there is no gravitational gradient within
> > >the accelerating spaceship. But he said the there is
> > >frequency shift within the spaceship.
> >
> > There are two different sorts of "gradients" involved
> > in the accelerating ship, and I'm not sure which one
> > you are calling the "gravitational gradient".
> >
>
> which was my point. Ken seems to think that acceleration
> *causes* gravity. I was trying to get him to keep
> the two separate.
This is also a lie. I did not think that acceleration causes gravity. In
fact I think the opposite--acceleration will not cause frequency shift or
weight shift as gravity does.
Ken Seto
kenseto
Ah.... does this sound like: fH (the first detectable frequency measurement)
of the approaching source has the highest value?
When the transmitter is at the closest point of approach (middle between the
two legs) we know that measured frequency is fM=fo*sqrt(1-v^2/c^2). This
means that fH>fM. Do you agree?
When the transmitter passes the detector the measured frequency is fL. This
means that fH>fM>fL. Do you agree?
Now here's what I paraphrase Tom Roberts said and you agreed with him::
1. The detector will initially detect an arbituary high frequency---call
this Fh.
2. Subsequent measurements will show a decrease in frequency--from Fh. The
frequency rate will continue to decrease until it is no longer detectable.
How is this different than what you said above? So are you now admitting
that I didn't paraphrase you wrongly?? In that case I demand an apology from
you.
In real life what you said is wrong. The first detectable frequency is NOT
fH as you asserted. Why? At infinity there is no frequency detected. The
first detectable frequency is the lowest detectable. Subsequent frequency
measurements will show an increase until it reached a max at the closest
point of approach. From that point on the measured frequency is decrease
until it is far enough away that it is no longer detectable.
>
> >
> > I did quote you correctly.
>
> You didn't quote me at all, Ken.
> You paraphrased me.
> Wrongly.
It seems that you have provided proof that I paraphrased you and Tom Roberts
correctly. :-)
>
> > Your denial now is laughable. If you are so sure
> > that I mis-quoting you why didn't you provide us with the correct
quotes??
>
> When YOU quote somebody, it is YOUR bloody job to ensure
> that it is right.
BTW since you refused to answer my question regarding your claim of a
gravity gradient in an accelerating spaceship I assumed that you did made
that claim. That means that you and Prebys are at odd. So please be a man
and correct him. He accused me of paraphrasing you wrongly in this case
too..
Ken Seto
Paul B. Andersen
"Eric Prebys" <pre...@fnal.gov> wrote in message news:3C3B6C63...@fnal.gov...
>
>
> "Ken H. Seto" wrote:
> >
> >
> > My description agrees with Paul's sound analogy as follows:
> > The initial frequency would be lowest detectable. It will increase to a max.
> > when the car is at the nearest point of approach ---call this Fm. When the
> > car passes the detector, the detected frequency will decrease from Fm until
> > it is no longer detectable.
> >
>
> Probably confused by the voices in your head.
Ken is living in Kenland, Alice is his neighbour.
You must have that in mind when he says "in the real world."
Kenland is where two light fronts simultaneously meet at two
different locations, where light detectors always recede from
the light sources (they recede upwards if they receive light from
two opposite directions at the same time), where you have to
make the seconds of a clock longer to make the clock run faster,
where the time you use to pass a bus moving in the opposite
direction of you is longer than if the bus were stationary,
and where the sound from trains behave like Ken describe above.
Paul
Daryl McCullough
>BTW if you stand on the track you would still
>hear the increase in the pitch (frequency) of the train's whistle as the
>train approaches you.
I don't think that's true. But assuming you are right, what
is the formula relating distance to pitch?
kenseto
>
> "Eric Prebys" <pre...@fnal.gov> wrote in message
news:3C3B6C63...@fnal.gov...
> >
> >
> > "Ken H. Seto" wrote:
> > >
> > >
> > > My description agrees with Paul's sound analogy as follows:
> > > The initial frequency would be lowest detectable. It will increase to
a max.
> > > when the car is at the nearest point of approach ---call this Fm. When
the
> > > car passes the detector, the detected frequency will decrease from Fm
until
> > > it is no longer detectable.
> > >
> >
> > You clearly hear trains differently than anybody else in world.
> > Probably confused by the voices in your head.
>
> Ken is living in Kenland, Alice is his neighbour.
> You must have that in mind when he says "in the real world."
Kenland is real. Pauland is based on math constructs and abstractions. Why
don't you ask Disney to make a cartoon of your abstraction so that we can
watch it at our leisure?
As the train approach you the pitch of its whistle gets higher. Does this
not suggest to you that the frequency is increased? Or are you agreeing with
the SR Runt Prebys that the increase in the pitch is merely a higher
volume---IOW, no frequency shift??
>
> Kenland is where two light fronts simultaneously meet at two
> different locations, where light detectors always recede from
> the light sources (they recede upwards if they receive light from
> two opposite directions at the same time),
This interpretation is derived from the SR postulates and the Lorentz
transformations. Also this interpretation is confirmed by experiments---the
speed of light is isotropic as per the MMX. Your naive understanding how
light move from the source to the target is laughable. Furthermore this
interpretation is a tad more realistic than the following absurd Einsteinian
assertion:
1. A rod is moving left to right with a light source on the left end
2. The tranasit time is T1
3. Now we turn the rod around --the source is now at the right end of the
rod.
4. The transit time is now T2.
5. Einsein's claim and your claim: T1 is larger than T2. This claim is
obviously false. Why? The SR postulate says that the speed of light is
isotropic in all directions.
>where you have to
> make the seconds of a clock longer to make the clock run faster,
This is a lie. I claimed that a clock second in a higher state of absolute
motion will have a higher absolute time content than a clock. second that is
in a lower state of absolute motion. This means that the clock that is in a
higher state of absolute motion will run faster when the two clocks are
compared side by side. The relationship between the two clock is according
to the Lorentz transformations. IOW, the Lorentz Transformations are merely
the math used to calculate the equivalent clock time value in all frames for
a specific interval of absolute time.
> where the time you use to pass a bus moving in the opposite
> direction of you is longer than if the bus were stationary,
> and where the sound from trains behave like Ken describe above.
ROTFL. This again shows your naive understanding how light move from the
source to the target. I'll give you a hint: light will take the same transit
time to travverse the same distance in all directions and all frames of
reference.
Ken Seto
Daryl McCullough
>
>"Daryl McCullough" <da...@cogentex.com> wrote
>> Decompose the velocity v of the source into
>> a velocity v_parallel toward the detector, and a velocity
>> v_perpendicular at right angles towards the line
>> connecting the source and the detector. Classically, the
>> Doppler shift only depends on v_parallel. If v_parallel is
>> constant, then so is the Doppler shift.
>
>But v_parallel is never constant when the train is moving
>transerversely relative to the detector on the embankment.
The formula for v_parallel is this:
If Y is the distance from the observer to the train track,
and X is the distance of the train from the point of closest
approach, then v_parallel = vX/D (where D = distance to the
train = square-root(X^2 + Y^2)). The rate of change of
v_parallel is d v_parallel/dt = v^2 Y/D^3.
Let's plug in some typical values:
v = 30 meters/second, D = 0.5 kilometers, Y = 10 meters. This
gives
v_parallel = 29.994,
d v_parallel/dt = 0.000072 meters/second^2.
The rate of change of v_parallel is pretty small.
Practically speaking, v_parallel is constant until the train is
quite close.
>> The frequency shift in an accelerated spaceship is derivable
>> from Special Relativity alone.
>
>Your derivation is based on the erroneous assumption that light from the
>middle will take a longer transit time to the front than to the rear.
That was not an assumption. I derived that fact from Special Relativity.
>> There are two different sorts of "gradients" involved
>> in the accelerating ship, and I'm not sure which one
>> you are calling the "gravitational gradient".
>>
>> (1) First, the "gravitational potential" changes in going
>> from the rear of the ship to the front of the ship.
>> It takes work to move something from the rear to the
>> front.
>
>There is no gravitational potential in the ship.
That's why I put it in quotes. It's not really a
gravitational potential, but it acts like one.
It requires work to move an object from the rear
of the ship to the front of the ship, just as it
takes work to move something from the bottom of
a building to the top of the building. According
to the equivalence principle, this effect is just
like a gravitational potential.
>In other words, the rear and the front
>have the same potential. The gravitational
>potential on earth is cauised by the rotation
>of the earth.
You are seriously in error here. Gravitational
potential has nothing to do with rotation. Gravitational
potential is defined as follows:
Potential Difference between A and B =
(Work required to move an object from A to B)/(The mass of the object)
That gives a nonzero potential between the front and rear of a rocket,
and gives a nonzero potential for a nonrotating planet.
Paul B. Andersen
"kenseto" <ken...@erinet.com> wrote in message news:3c41a3a1$0$35612$4c5e...@news.erinet.com...
>
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message
> news:a1q03d$bf9$1...@snipp.uninett.no...
> >
> > "Ken H. Seto" <ken...@erinet.com> wrote in message
> news:3c3ca752$0$37108$4c5e...@news.erinet.com...
> > >
> > > "Paul B. Andersen" <paul.b....@hia.no> wrote in message
> > > news:a1hn7i$fkd$1...@snipp.uninett.no...
> > > >
> > > > "Ken H. Seto" <ken...@erinet.com> wrote in message
> > > news:3c3c55d6$0$35620$4c5e...@news.erinet.com...
> > > > >
> > > 1. The situation I described in my origial post is a transverse Doppler
> > > situation. The frequency detected in a transverse Doppler situation is not
> > > constant. It depends on the radial length of the transimitter from the
> > > detector.
> >
> > Indeed.
> > And I described what would happen in this scenario:
> > When the source is infinitely far away and approaching,
> > the frequency will be fH = fo*sqrt((c+v)/(c-v))
> > When the source is infinitely far away and receding,
> > the frequency will be fL = fo*sqrt((c-v)/(c+v))
> > As the source is passing, the frequency will monotonously decrease
> > from fH to fL. The rate of decrease will be highest when the source
> > is closest to the detector.
>
> Ah.... does this sound like: fH (the first detectable frequency measurement)
> of the approaching source has the highest value?
Of course.
I have stated this quite a number of times now, havent I?
> When the transmitter is at the closest point of approach (middle between the
> two legs) we know that measured frequency is fM=fo*sqrt(1-v^2/c^2). This
> means that fH>fM. Do you agree?
Not quite. I gave the answer to this in my previous posting:
#1: f = fo*gamma when observed angle is v/c radians from transverse
#2: f = fo when observed angle is 0.5*v/c rads from transverse
#3: f = fo/gamma when observed angle is transverse
#1 is "at the point of closest approach", but the source is not
visually observed to be transverse at that time, the light is
still sligtly blue shifted.
#3 is when the light is visually observed to be transverse,
e.g. the source is seen at the point of closest approach,
but has really (as measured in detector frame) passed it.
At this time the light is slightly red shifted.
THIS is the transverse Doppler shift. It does NOT happen
when the source is at the point of closest approach due
to aberration.
> When the transmitter passes the detector the measured frequency is fL. This
> means that fH>fM>fL. Do you agree?
Of course.
I have stated this quite a number of times now, havent I?
> Now here's what I paraphrase Tom Roberts said and you agreed with him::
> 1. The detector will initially detect an arbituary high frequency---call
> this Fh.
> 2. Subsequent measurements will show a decrease in frequency--from Fh. The
> frequency rate will continue to decrease until it is no longer detectable.
>
> How is this different than what you said above? So are you now admitting
> that I didn't paraphrase you wrongly?? In that case I demand an apology from
> you.
Yes, you paraphrased me wrongly, and I have stated exactly
what part was wrong before:
I NEVER SAID:
"The frequency rate will continue to decrease until
it is no longer detectable."
This is at best imprecise and ambiguous, at worst plain wrong.
The frequency will decrease asymptotically to fL.
Why the hell do I always have to repeat the same thing over and over?
Can't you read?
> In real life what you said is wrong. The first detectable frequency is NOT
> fH as you asserted. Why? At infinity there is no frequency detected. The
> first detectable frequency is the lowest detectable. Subsequent frequency
> measurements will show an increase until it reached a max at the closest
> point of approach. From that point on the measured frequency is decrease
> until it is far enough away that it is no longer detectable.
You do not have to repeat it, Ken.
I know you have no clue.
> > > I did quote you correctly.
> >
> > You didn't quote me at all, Ken.
> > You paraphrased me.
> > Wrongly.
>
> It seems that you have provided proof that I paraphrased you and Tom Roberts
> correctly. :-)
It seems like you have been unable to quote me literally
and show that your paraphrase is correct.
> BTW since you refused to answer my question regarding your claim of a
> gravity gradient in an accelerating spaceship I assumed that you did made
> that claim.
Your paraphrase is again too imprecise to be called correct,
so if I should answer with one word, it would have to be "no".
But OK, I will answer in more detail:
Ken Seto wrote:
| "A space ship under goes a constant acceleration of 1g and
| a light source is located in the middle of the ship. They
| claimed that the frequency of that light source will be red shifted
| at the front of the ship and blue shifted at the rear of the ship."
Yes.
The light from the source will be red shifted in the front
and blue shifted at the rear of the ship.
| "Also they claimed that a standard weigh will weight less at
| the front of the ship and more at the rear of the ship."
I am sure I never claimed that. Others might have, however, because
it is right if "weight less" mean "have a lower proper acceleration".
If the ship is accelerated in such a way that there is no strain
in the ship, e.g. it retains its length as measured with metre sticks
which are not mechanically distorted by the acceleration,
then the proper acceleration (what an accelerometer shows) will
vary slightly along the length of the ship. This is a very small
effect unless the ship is extremely long, though.
| "They said that these claims are supported by the Equivalent Principle."
No. I never said that because it is plain wrong.
The above follows from SR only in flat space time, the effects
of gravitating masses do not come into it at all.
The EP is irreleavant to the above phenomena.
> That means that you and Prebys are at odd. So please be a man
> and correct him.
I have not read all the postings in this thread, and do
not consider it my responsibility to correct errors
Prepys or others might have done. IF they have done any.
> He accused me of paraphrasing you wrongly in this case
> too..
And you did.
Paul
Paul B. Andersen
"Daryl McCullough" <da...@cogentex.com> wrote in message news:a1url...@drn.newsguy.com...
> Robert says...
> >BTW if you stand on the track you would still
> >hear the increase in the pitch (frequency) of the train's whistle as the
> >train approaches you.
>
> I don't think that's true. But assuming you are right, what
> is the formula relating distance to pitch?
I strongly suspect that Robert wouldn't agree. :-)
I wonder if this "Robert Andary" knows what he is doing when
he let Ken post from his account and use his name.
Paul
kenseto
> Robert says...
> >BTW if you stand on the track you would still
> >hear the increase in the pitch (frequency) of the train's whistle as the
> >train approaches you.
>
> I don't think that's true.
If you don't think so then go ahead and find out by standing directly in
front of an incoming ambulance. The pitch of the ambulance's siren will
increase in pitch as it approaches you.
>But assuming you are right, what
> is the formula relating distance to pitch?
Your following formula will do.
v_parallel is d v_parallel/dt = v^2 Y/D^3.
The Y in this formula is due to the absolute motion of the observer
(detector) moving in the vertical direction. Please note that this
interpretation applies to all locations of the detector.
Ken Seto
kenseto
My computer was not working for the last few days. Robert knew exactly what
I was doing and he didn't care. So I guess all your slight remarks are just
a waste of time. :-)
Ken Seto
Bilge
Re: Another Bogus Claim by the SR Experts to usenet:
>
>"Daryl McCullough" <da...@cogentex.com> wrote in message
>news:a1url...@drn.newsguy.com...
>> Robert says...
>> >BTW if you stand on the track you would still
>> >hear the increase in the pitch (frequency) of the train's whistle as the
>> >train approaches you.
>>
>> I don't think that's true.
>
>If you don't think so then go ahead and find out by standing directly in
>front of an incoming ambulance. The pitch of the ambulance's siren will
>increase in pitch as it approaches you.
David Evens
wrote:
Do not forget that Y has nothing whatsoever to do with how anything
anywhere is seen to be moving by anyone.
-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
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kenseto
> "kenseto" says...
> >
> >"Daryl McCullough" <da...@cogentex.com> wrote
>
> >> Decompose the velocity v of the source into
> >> a velocity v_parallel toward the detector, and a velocity
> >> v_perpendicular at right angles towards the line
> >> connecting the source and the detector. Classically, the
> >> Doppler shift only depends on v_parallel. If v_parallel is
> >> constant, then so is the Doppler shift.
> >
> >But v_parallel is never constant when the train is moving
> >transerversely relative to the detector on the embankment.
>
> The formula for v_parallel is this:
>
> If Y is the distance from the observer to the train track,
> and X is the distance of the train from the point of closest
> approach, then v_parallel = vX/D (where D = distance to the
> train = square-root(X^2 + Y^2)). The rate of change of
> v_parallel is d v_parallel/dt = v^2 Y/D^3.
> Let's plug in some typical values:
> v = 30 meters/second, D = 0.5 kilometers, Y = 10 meters. This
> gives
> v_parallel = 29.994,
> d v_parallel/dt = 0.000072 meters/second^2.
> The rate of change of v_parallel is pretty small.
>
> Practically speaking, v_parallel is constant until the train is
> quite close.
Your assumption that Y = 10 meters is faulty. In real life, Y=Va*X/c
Where Va=absolute motion of the detector. Since Va is thought to be approx.
370 km /sec then the rate of change of v_parallel is significant and thus
v_paraelel is not constant as you claimed.. Also this means that if you
stand directly in front of the track you would still hear the pitch change.
>
> >> The frequency shift in an accelerated spaceship is derivable
> >> from Special Relativity alone.
> >
> >Your derivation is based on the erroneous assumption that light from the
> >middle will take a longer transit time to the front than to the rear.
>
> That was not an assumption. I derived that fact from Special Relativity.
No SR's postulates says that the transit time in any frame (including the
accelerated frames) for equal distance must be the same. So I don't know how
you can derive your claim (assumption) from SR.
>
> >> There are two different sorts of "gradients" involved
> >> in the accelerating ship, and I'm not sure which one
> >> you are calling the "gravitational gradient".
> >>
> >> (1) First, the "gravitational potential" changes in going
> >> from the rear of the ship to the front of the ship.
> >> It takes work to move something from the rear to the
> >> front.
> >
> >There is no gravitational potential in the ship.
>
> That's why I put it in quotes. It's not really a
> gravitational potential, but it acts like one.
> It requires work to move an object from the rear
> of the ship to the front of the ship, just as it
> takes work to move something from the bottom of
> a building to the top of the building. According
> to the equivalence principle, this effect is just
> like a gravitational potential.
But in a gravitational field it will take more work to take something from
the basement to the ground floor than to take something from the ground
floor to the second floor. This is what a gravitational potential (gravity
gradient) means. In an accelerating spaceship it take the same amount of
work to move something vertically for an equal distance from any location of
the ship. IOW, it takes the same amount of work to take something fdrom the
rear to the middle as to take something from the middle to the front. This
means that there is no gravtational potential (gradient) in any part of the
ship.
>
> >In other words, the rear and the front
> >have the same potential. The gravitational
> >potential on earth is cauised by the rotation
> >of the earth.
>
> You are seriously in error here.
No I am not. Gravitational potential is caused by different state of
absolute motion. The rotation og the earth gives rise to different state of
absolute motion at different heights.
>Gravitational
> potential has nothing to do with rotation. Gravitational
> potential is defined as follows:
>
> Potential Difference between A and B =
> (Work required to move an object from A to B)/(The mass of the
object)
On earth, you will find that different heights will have different
gravitational potential---a standard weight will have different
gravitational mass at different heights. In the spaceship a standard weight
will have the same inertial mass in all parts of the ship.
>
Ken Seto
kenseto
Not quite?? The last time I check the text book it confirmsed that
fM=fo*sqrt(1-v^2/c^2).
But fL is not detectable because it is only observable at infinity. Are you
forgetting that we are talking about actual measurements? IOW, we are not
talking about your absurd assumptions.
>
> Why the hell do I always have to repeat the same thing over and over?
> Can't you read?
Because you were wrong??
>
> > In real life what you said is wrong. The first detectable frequency is
NOT
> > fH as you asserted. Why? At infinity there is no frequency detected. The
> > first detectable frequency is the lowest detectable. Subsequent
frequency
> > measurements will show an increase until it reached a max at the closest
> > point of approach. From that point on the measured frequency is decrease
> > until it is far enough away that it is no longer detectable.
>
> You do not have to repeat it, Ken.
> I know you have no clue.
ROTFLOL. The clueless accuse others of having no clue!!!!
>
> > > > I did quote you correctly.
> > >
> > > You didn't quote me at all, Ken.
> > > You paraphrased me.
> > > Wrongly.
> >
> > It seems that you have provided proof that I paraphrased you and Tom
Roberts
> > correctly. :-)
>
> It seems like you have been unable to quote me literally
> and show that your paraphrase is correct.
I paraphrased you and Tom correctly. The first detectable frequency is
indeed the highest detecable and then decrease from that to the lowest
frequency detectable. Remember---fH>fM>fL???
BTW you paraphrased me wrongly as follows:
"Kenland is where two light fronts simultaneously meet at two
different locations"
Kenland said: Two light fronts simultaneously meet at two different
locations at two different times.
>
> > BTW since you refused to answer my question regarding your claim of a
> > gravity gradient in an accelerating spaceship I assumed that you did
made
> > that claim.
>
> Your paraphrase is again too imprecise to be called correct,
> so if I should answer with one word, it would have to be "no".
Parsing words again eh Paul?? Is your real name Bill Clinton??
>
> | "Also they claimed that a standard weigh will weight less at
> | the front of the ship and more at the rear of the ship."
>
> I am sure I never claimed that.
Yes you did. You were the only one participated in that discussion.
>Others might have, however, because
> it is right if "weight less" mean "have a lower proper acceleration".
So are you claiming that the front of the ship is have a lower proper
acceleration??
> If the ship is accelerated in such a way that there is no strain
> in the ship, e.g. it retains its length as measured with metre sticks
> which are not mechanically distorted by the acceleration,
> then the proper acceleration (what an accelerometer shows) will
> vary slightly along the length of the ship. This is a very small
> effect unless the ship is extremely long, though.
This is just some more of your erroneous assumptions. :-)
>
> | "They said that these claims are supported by the Equivalent Principle."
>
> No. I never said that because it is plain wrong.
> The above follows from SR only in flat space time, the effects
> of gravitating masses do not come into it at all.
> The EP is irreleavant to the above phenomena.
>
> > That means that you and Prebys are at odd. So please be a man
> > and correct him.
>
> I have not read all the postings in this thread, and do
> not consider it my responsibility to correct errors
> Prepys or others might have done. IF they have done any.
>
> > He accused me of paraphrasing you wrongly in this case
> > too..
>
> And you did.
No I didn't.
Ken Seto
Daryl McCullough
>Your assumption that Y = 10 meters is faulty. In real life, Y=Va*X/c
>Where Va=absolute motion of the detector.
You are talking about a theory that I'm not familiar with. I have
no comment about that. I was only talking about the standard explanation
of the Doppler shift.
>> >> The frequency shift in an accelerated spaceship is derivable
>> >> from Special Relativity alone.
>> >
>> >Your derivation is based on the erroneous assumption that light from the
>> >middle will take a longer transit time to the front than to the rear.
>>
>> That was not an assumption. I derived that fact from Special Relativity.
>
>No SR's postulates says that the transit time in any frame (including the
>accelerated frames) for equal distance must be the same.
No, it doesn't say that. It says that as measured in any *inertial*
frame (not accelerated frame) light travels at speed c. From that
fact, you can derive the conclusion that for an accelerated rocket,
it takes longer for a light signal to travel from the rear to the
front than from the front to the rear.
>But in a gravitational field it will take more work to take something from
>the basement to the ground floor than to take something from the ground
>floor to the second floor. This is what a gravitational potential (gravity
>gradient) means.
No, that's not what it means. As I said, gravitational potential
is defined as follows:
Potential Difference between A and B =
(Work required to move an object from A to B)/(The mass of the object)
--
kenseto
> "kenseto" says...
>
> >Your assumption that Y = 10 meters is faulty. In real life, Y=Va*X/c
> >Where Va=absolute motion of the detector.
>
> You are talking about a theory that I'm not familiar with. I have
> no comment about that. I was only talking about the standard explanation
> of the Doppler shift.
Then how does SR explain the fact that you can still hear the pitch of the
train's whistle increases as it gets closer to you?? This is so even if you
are standing on the track??
>>>in an accelerated spaceship is derivable
> >> >> from Special Relativity alone.
> >> >
> >> >Your derivation is based on the erroneous assumption that light from
the
> >> >middle will take a longer transit time to the front than to the rear.
> >>
> >> That was not an assumption. I derived that fact from Special
Relativity.
> >
> >No SR's postulates says that the transit time in any frame (including the
> >accelerated frames) for equal distance must be the same.
>
> No, it doesn't say that. It says that as measured in any *inertial*
> frame (not accelerated frame) light travels at speed c. From that
> fact, you can derive the conclusion that for an accelerated rocket,
> it takes longer for a light signal to travel from the rear to the
> front than from the front to the rear.
In that case what you said is outside the realm of SR postulates So how
can you say that your derivation is based on SR?
>> floor than to take something from the ground
> >floor to the second floor. This is what a gravitational potential
(gravity
> >gradient) means.
>
> No, that's not what it means. As I said, gravitational potential
> is defined as follows:
>
> Potential Difference between A and B =
> (Work required to move an object from A to B)/(The mass of the object)
In that case the middle of the ship has the same gravity potential as the
front of the ship--the same amount of work is required to move 1 kg from
the rear to the middle than from the middle to the front. This would destroy
your assertion that the front has a higher potential than the middle and
the rear.
Ken Seto
Daryl McCullough
>
>
>"Daryl McCullough" <da...@cogentex.com> wrote in message
>news:a24m5...@drn.newsguy.com...
>> "kenseto" says...
>>
>> >Your assumption that Y = 10 meters is faulty. In real life, Y=Va*X/c
>> >Where Va=absolute motion of the detector.
>>
>> You are talking about a theory that I'm not familiar with. I have
>> no comment about that. I was only talking about the standard explanation
>> of the Doppler shift.
>
>Then how does SR explain the fact that you can still hear the pitch of the
>train's whistle increases as it gets closer to you?? This is so even if you
>are standing on the track??
I don't think that happens. When I've listened to sirens and train
whistles, the pitch starts out high, and gets *lower*.
>> No, it doesn't say that. It says that as measured in any *inertial*
>> frame (not accelerated frame) light travels at speed c. From that
>> fact, you can derive the conclusion that for an accelerated rocket,
>> it takes longer for a light signal to travel from the rear to the
>> front than from the front to the rear.
>
>In that case what you said is outside the realm of SR postulates.
No, it's not. It follows from the postulates of SR.
>So how can you say that your derivation is based on SR?
Because you can write down the postulates of SR, do some
mathematical derivations, and voila! The answer comes out.
Let me just write down the postulates from which the conclusion
follows. Let S be an inertial frame in which the rocket is initially
at rest. Suppose that (as measured in frame S) at times t=0 and
t=T, a light pulse is sent from the rear of the rocket to
the front of the rocket. Let t=T1 and t=T2 be the times at
which these light pulses are received in the front of the
rocket. Let L be the length of the rocket, and let the acceleration
be a. Then write down what you know.
1. The position of the rear of the rocket at any time t > 0
is given by x_r = 1/2 a t^2. (Note, this is only approximately
true, valid for small a and for small t.)
2. The position of the front of the rocket at any time t > 0
is approximately given by x_f = L + 1/2 a t^2.
3. The position of the first pulse of light at any time t > 0
is given by x_l1 = c t.
4. The first pulse of light will catch up to the front of
ship at a time T1 in which their locations are equal. Therefore,
x_l1 = x_f at time T1. Using 2 and 3, we know that
c T1 = L + 1/2 a T1^2.
5. Solving for T1 gives T1 = L/c + 1/2 a L^2/c^3 + ...
(keeping only the lowest powers of a).
6. The position of the second pulse of light at any time t > T
is given by x_l2 = 1/2 a T^2 + c(t - T). (The general formula
for a pulse of light travelling in the x-direction is
x = A + ct. To find the constant A, we use the fact that at
t=T, the pulse is at the rear of the ship, which at that time
has location 1/2 a T^2. Therefore, we have at time t=T:
1/2 a T^2 = A + cT. Solving for A gives A = 1/2 aT^2 - cT.)
7. The second pulse of light will catch up to the front of
ship at a time T2 in which their locations are equal. Therefore,
x_l2 = x_f at time T2. Using 2 and 6, we know that
1/2 a T^2 + c(T2 - T) = L + 1/2 a T2^2.
8. Solving for T2 gives T2 = L/c + T + 1/2 a (L^2/c^3 + 2LT/c^2) + ...
(keeping only the lowest powers of a).
9. Therefore, the time between the receiving of the first pulse
and the receiving of the second pulse is
T' = T2 - T1
= [L/c + T + 1/2 a (L^2/c^3 + 2LT/c^2)] - [L/c + 1/2 a L^2/c^3] + ...
= T + aLT/c^2 + ...
= T (1 + aL/c^2) + ...
10. Therefore, the rate at which the front receives signals is
less than the rate at which the rear sends the signals, by a factor
of (1 + aL/c^2).
Nothing is assumed in this derivation except that (1) light travels
at speed c in coordinate system S, (2) the front and rear are
undergoing constant acceleration of magnitude a, and (3) aL/c^2 is
much less than 1 (so we can ignore higher-order terms).
>> No, that's not what it means. As I said, gravitational potential
>> is defined as follows:
>>
>> Potential Difference between A and B =
>> (Work required to move an object from A to B)/(The mass of the object)
>
>In that case the middle of the ship has the same gravity potential as the
>front of the ship
No, it doesn't. The potential difference between the middle
and the front is the amount of work required to move a 1kg
mass from the middle to the front. If that is zero, then they
have the same potential. If that is nonzero, they have different
potentials.
>--the same amount of work is required to move 1 kg from
>the rear to the middle than from the middle to the front.
That means that the potential *difference* between the rear
and the middle is equal to the potential *difference* between
the middle and the front. Potential difference means the
*difference* in the potential.
In other words, letting Phi(x) be the potential at height x,
we have:
Phi(L) - Phi(L/2) = Phi(L/2) - Phi(0) = W
where W = work required to raise a 1 kg mass to
a height L/2 in the ship. So,
Phi(L/2) = Phi(0) + W
Phi(L) = Phi(L/2) + W
Phi(L) is not equal to Phi(L/2) except when W = 0.
kenseto
> "kenseto" says...
> >
> >
> >"Daryl McCullough" <da...@cogentex.com> wrote in message
> >news:a24m5...@drn.newsguy.com...
> >> "kenseto" says...
> >>
> >> >Your assumption that Y = 10 meters is faulty. In real life, Y=Va*X/c
> >> >Where Va=absolute motion of the detector.
> >>
> >> You are talking about a theory that I'm not familiar with. I have
> >> no comment about that. I was only talking about the standard
explanation
> >> of the Doppler shift.
> >
> >Then how does SR explain the fact that you can still hear the pitch of
the
> >train's whistle increases as it gets closer to you?? This is so even if
you
> >are standing on the track??
>
> I don't think that happens. When I've listened to sirens and train
> whistles, the pitch starts out high, and gets *lower*.
No the pitch starts out low--the first detectable pitch is very faint (you
can hardly hear it). The pitch and volume increase as the siren approaches
you then they decrease as the siren passes you.
>
> >> No, it doesn't say that. It says that as measured in any *inertial*
> >> frame (not accelerated frame) light travels at speed c. From that
> >> fact, you can derive the conclusion that for an accelerated rocket,
> >> it takes longer for a light signal to travel from the rear to the
> >> front than from the front to the rear.
> >
> >In that case what you said is outside the realm of SR postulates.
>
> No, it's not. It follows from the postulates of SR.
No it does not follow from the SR postulates. The constant light speed
postulate demands that the transit time for the same length to be equal in
all directions. Your calculations assumed that the transit time to be
different in different directions.
Ken Seto
Daryl McCullough
>> >> No, it doesn't say that. It says that as measured in any *inertial*
>> >> frame (not accelerated frame) light travels at speed c. From that
>> >> fact, you can derive the conclusion that for an accelerated rocket,
>> >> it takes longer for a light signal to travel from the rear to the
>> >> front than from the front to the rear.
>> >
>> >In that case what you said is outside the realm of SR postulates.
>>
>> No, it's not. It follows from the postulates of SR.
>
>No it does not follow from the SR postulates.
I *derived* it from those postulates! That's a proof
that it follows from those postulates.
>The constant light speed postulate demands that
>the transit time for the same length to be equal in all directions.
Light speed is constant as measured in any *inertial*
reference frame.
>Your calculations assumed that the transit time to be
>different in different directions.
You don't seem to understand the difference between
an assumption and a conclusion. The assumption is that
light speed is c as measured in any inertial reference
frame. The *conclusion* is that for an accelerating
rocket, the time required for light to travel from
the rear to the front is greater than the time required
to travel from the front to the rear.
Dirk Van de moortel
"Daryl McCullough" <da...@cogentex.com> wrote in message news:a278t...@drn.newsguy.com...
Daryl, he hasn't even *looked* at your derivation.
Surely you can't expect such an effort from someone who goes
to sci.math to argue that
sqrt(-25) = -5
and that
(A-B)^2 = A^2 - B^2
Nice try though.
Dirk Vdm
kenseto
> "kenseto" says...
> >
> >
> >"Daryl McCullough" <da...@cogentex.com> wrote
>
> >> >> No, it doesn't say that. It says that as measured in any *inertial*
> >> >> frame (not accelerated frame) light travels at speed c. From that
> >> >> fact, you can derive the conclusion that for an accelerated rocket,
> >> >> it takes longer for a light signal to travel from the rear to the
> >> >> front than from the front to the rear.
> >> >
> >> >In that case what you said is outside the realm of SR postulates.
> >>
> >> No, it's not. It follows from the postulates of SR.
> >
> >No it does not follow from the SR postulates.
>
> I *derived* it from those postulates! That's a proof
> that it follows from those postulates.
>
> >The constant light speed postulate demands that
> >the transit time for the same length to be equal in all directions.
>
> Light speed is constant as measured in any *inertial*
> reference frame.
Since the ship is not an inertial frame therefore it is outside the realm of
the SR postulate.
>
> >Your calculations assumed that the transit time to be
> >different in different directions.
>
> You don't seem to understand the difference between
> an assumption and a conclusion. The assumption is that
> light speed is c as measured in any inertial reference
> frame.
So the "assumption" here is the SR postulate.
>The *conclusion* is that for an accelerating
> rocket, the time required for light to travel from
> the rear to the front is greater than the time required
> to travel from the front to the rear.
This "conclusion" is not derivable from the SR postulates. This "CONCLUSION"
is an assertion by you. It is not derivable from any current theory. To
derive anything using SR you need relative velocity. Since there is no
measureable relative velocity within the ship there is no frequency shift
according to SR.
Ken Seto
Daryl McCullough
>> You don't seem to understand the difference between
>> an assumption and a conclusion. The assumption is that
>> light speed is c as measured in any inertial reference
>> frame.
>
>So the "assumption" here is the SR postulate.
>
>>The *conclusion* is that for an accelerating
>> rocket, the time required for light to travel from
>> the rear to the front is greater than the time required
>> to travel from the front to the rear.
>
>This "conclusion" is not derivable from the SR postulates.
I *derived* it. That proves that it is derivable.
Paul B. Andersen
"kenseto" <ken...@erinet.com> wrote in message news:3c45b4ab$0$37102$4c5e...@news.erinet.com...
> BTW you paraphrased me wrongly as follows:
> "Kenland is where two light fronts simultaneously meet at two
> different locations"
> Kenland said: Two light fronts simultaneously meet at two different
> locations at two different times.
I am very sorry, Ken.
Kenland is an even weirder place than I could imagine.
Paul
kenseto
Yeah that's because you have no clue how light move from the source to the
target.
I'll give you a hint: The train guy (K') and the track guy (K) are coincide
with each other when the lightings strike simultaneously. The direction of
absolute motion for both K and K' is in the vertical direction. The state of
absolute motion of K' is higher.than K. When the light fronts arrive at K
simultaneously, K' is no longer in that location because of its highere
state of absolute motion. The light fronts will have to traverse the extra
distance to reach K'. That means that the light fronts meet at K'
simultaneously at a later time.
BTW, that also explains the null result of the MMX.
Ken Seto
Ken Seto
>
> Paul
>
>
kenseto
So it doesn't bother you that your derivation is based on your erroneous
assumption?? You should know that if you assumed frequency shift--assuming
that the transit time for light is different in different directions--in
your derivation then you will get frequency shift in your final equation.
Don't you think that's a bit circular??
Ken Seto
Paul Cardinale
<snip>
> Kenland said: Two light fronts simultaneously meet at two different
> locations at two different times.
Wow. Is this a new high for the ratio of contradictions per word? Judges?
Paul Cardinale
Daryl McCullough
>> >>The *conclusion* is that for an accelerating
>> >> rocket, the time required for light to travel from
>> >> the rear to the front is greater than the time required
>> >> to travel from the front to the rear.
>> >
>> >This "conclusion" is not derivable from the SR postulates.
>>
>> I *derived* it. That proves that it is derivable.
>
>So it doesn't bother you that your derivation is based on your erroneous
>assumption??
The only assumption was that light has speed c in all directions,
as measured in any inertial frame.
>You should know that if you assumed frequency shift
I didn't. The only assumption is that light travels at speed c
as measured in any inertial reference frame.
--
Paul B. Andersen
"kenseto" <ken...@erinet.com> wrote in message news:3c482d0f$0$37097$4c5e...@news.erinet.com...
As I said.
Paul
kenseto
> "kenseto" says...
>
> >> >>The *conclusion* is that for an accelerating
> >> >> rocket, the time required for light to travel from
> >> >> the rear to the front is greater than the time required
> >> >> to travel from the front to the rear.
> >> >
> >> >This "conclusion" is not derivable from the SR postulates.
> >>
> >> I *derived* it. That proves that it is derivable.
> >
> >So it doesn't bother you that your derivation is based on your erroneous
> >assumption??
>
> The only assumption was that light has speed c in all directions,
> as measured in any inertial frame.
But the ship is not in an inertial frame.
>
> >You should know that if you assumed frequency shift
>
> I didn't. The only assumption is that light travels at speed c
> as measured in any inertial reference frame.
Your calculations assume that the transit time from the front to the rear is
shorter than the transit time from the rear to the front. That means that
the speed of light from the front to the rear is different than that from
the rear to the front. So how is this mean that light is traveling at c in
all directions in the ship???
Ken Seto
Daryl McCullough
>> The only assumption was that light has speed c in all directions,
>> as measured in any inertial frame.
>
>But the ship is not in an inertial frame.
I didn't derive it in the frame of the ship. I derived
it in an inertial reference frame in which the ship
is initially at rest.
>> >You should know that if you assumed frequency shift
>>
>> I didn't. The only assumption is that light travels at speed c
>> as measured in any inertial reference frame.
>
>Your calculations assume that the transit time from the front to the rear is
>shorter than the transit time from the rear to the front.
No, I didn't assume that. I derived that.
The assumptions were: (S is an inertial frame
in which the rocket initially is at rest)
1. Light has speed c as measured in S.
2. The location of the rear of the ship,
as measured in S, is x_r = 1/2 a t^2.
3. The location of the front of the ship,
as measured in S, is x_f = L + 1/2 a t^2.
4. A light signal is sent from the rear of
the ship once every T seconds.
The conclusion is
5. A light signal is received by the front end
every T' seconds, with T' > T.
>That means that the speed of light from the front to the rear
>is different than that from the rear to the front.
No, it doesn't mean that. As measured in any inertial frame,
light travels a greater distance in going from the rear to
the front than in going from the front to the rear.
>So how is this mean that light is traveling at c in
>all directions in the ship???
The ship is not inertial. Light *only* travels at speed c
as measured in *inertial* reference frames.
kenseto
That you have no clue?? :-)
Ken Seto
kenseto
> "kenseto" says...
> >
> >
> >"Daryl McCullough" <da...@cogentex.com> wrote
>
>
> >> The only assumption was that light has speed c in all directions,
> >> as measured in any inertial frame.
> >
> >But the ship is not in an inertial frame.
>
> I didn't derive it in the frame of the ship. I derived
> it in an inertial reference frame in which the ship
> is initially at rest.
>
> >> >You should know that if you assumed frequency shift
> >>
> >> I didn't. The only assumption is that light travels at speed c
> >> as measured in any inertial reference frame.
> >
> >Your calculations assume that the transit time from the front to the rear
is
> >shorter than the transit time from the rear to the front.
>
> No, I didn't assume that. I derived that.
>
> The assumptions were: (S is an inertial frame
> in which the rocket initially is at rest)
>
> 1. Light has speed c as measured in S.
We are not at S.
> 2. The location of the rear of the ship,
> as measured in S, is x_r = 1/2 a t^2.
This assumes that acceleration will change the transit time for light in the
accelerating ship. You have absolutely no proof of this assumption. This
assumption is not based on any theory.
> 3. The location of the front of the ship,
> as measured in S, is x_f = L + 1/2 a t^2.
This calculation is based on the erroneous assumption at item#2
> 4. A light signal is sent from the rear of
> the ship once every T seconds.
>
> The conclusion is
>
> 5. A light signal is received by the front end
> every T' seconds, with T' > T.
This conclusion is based on the erroneous assumption of item #2.
>
> >That means that the speed of light from the front to the rear
> >is different than that from the rear to the front.
>
> No, it doesn't mean that. As measured in any inertial frame,
> light travels a greater distance in going from the rear to
> the front than in going from the front to the rear.
This is just plain wrong. In any inertial frame light travel the same
distance in going fromthe rear to the front than going from the front to the
rear.
I think what you are trying to say is that the accelerating ship as measured
in an inertial ship. But even that is wrong.
>
> >So how is this mean that light is traveling at c in
> >all directions in the ship???
>
> The ship is not inertial. Light *only* travels at speed c
> as measured in *inertial* reference frames.
Sorry you are talking in circle.
Ken Seto
David Evens
wrote:
>"Daryl McCullough" <da...@cogentex.com> wrote in message
>news:a2aa2...@drn.newsguy.com...
>> "kenseto" says...
>> >"Daryl McCullough" <da...@cogentex.com> wrote
>>
>> >> The only assumption was that light has speed c in all directions,
>> >> as measured in any inertial frame.
>> >
>> >But the ship is not in an inertial frame.
>>
>> I didn't derive it in the frame of the ship. I derived
>> it in an inertial reference frame in which the ship
>> is initially at rest.
>>
>> >> >You should know that if you assumed frequency shift
>> >>
>> >> I didn't. The only assumption is that light travels at speed c
>> >> as measured in any inertial reference frame.
>> >
>> >Your calculations assume that the transit time from the front to the rear
>is
>> >shorter than the transit time from the rear to the front.
>>
>> No, I didn't assume that. I derived that.
>>
>> The assumptions were: (S is an inertial frame
>> in which the rocket initially is at rest)
>>
>> 1. Light has speed c as measured in S.
>
>We are not at S.
You forgot to support your assumption that this matters.
>> 2. The location of the rear of the ship,
>> as measured in S, is x_r = 1/2 a t^2.
>
>This assumes that acceleration will change the transit time for light in the
>accelerating ship. You have absolutely no proof of this assumption. This
>assumption is not based on any theory.
You forgot to support your assumption that the motion of the ship
depends on the velocity of light in a non-inertial frame.
>> 3. The location of the front of the ship,
>> as measured in S, is x_f = L + 1/2 a t^2.
>
>This calculation is based on the erroneous assumption at item#2
You forgot to support your assumption that the position of the ship in
frame S is in some way dependent on the speed of light in the ship.
>> 4. A light signal is sent from the rear of
>> the ship once every T seconds.
>>
>> The conclusion is
>>
>> 5. A light signal is received by the front end
>> every T' seconds, with T' > T.
>
>This conclusion is based on the erroneous assumption of item #2.
You forgot to support your assumption that the position of the ship
depends on the speed of light in the ship.
>> >That means that the speed of light from the front to the rear
>> >is different than that from the rear to the front.
>>
>> No, it doesn't mean that. As measured in any inertial frame,
>> light travels a greater distance in going from the rear to
>> the front than in going from the front to the rear.
>
>This is just plain wrong. In any inertial frame light travel the same
>distance in going fromthe rear to the front than going from the front to the
>rear.
You forgot to support your assumption that the accelerating ship is
not accelerating.
>I think what you are trying to say is that the accelerating ship as measured
>in an inertial ship. But even that is wrong.
Yes, your assumption is wrong. You know that. That is why you made
it.
>> >So how is this mean that light is traveling at c in
>> >all directions in the ship???
>>
>> The ship is not inertial. Light *only* travels at speed c
>> as measured in *inertial* reference frames.
>
>Sorry you are talking in circle.
Why is it talking in a circle to not contradict yourself, Ken?
Paul B. Andersen
"kenseto" <ken...@erinet.com> wrote in message news:3c48f79a$0$37111$4c5e...@news.erinet.com...
Right.
Kenland physics is beyond me.
But since I live in the real world, it doesn't bother me much.
You sure are a funny guy, Ken.
But you will have to take my word for it because what's
most funny is that you don't understand what's so funny.
Paul
Daryl McCullough
>> The assumptions were: (S is an inertial frame
>> in which the rocket initially is at rest)
>>
>> 1. Light has speed c as measured in S.
>
>We are not at S.
What does that mean? S is a coordinate system
used to describe the rocket's motion. What does
it mean to be "at" a coordinate system?
>> 2. The location of the rear of the ship,
>> as measured in S, is x_r = 1/2 a t^2.
>
>This assumes that acceleration will change the transit
>time for light in the accelerating ship.
No, it assumes that the rear of the ship
is at position x_r = 1/2 a t^2 at time t.
Assumption 2 doesn't say *anything* about
light.
Item number 2 is simply putting into mathematical
form the assumption that the rocket is accelerating
at rate a.
>> No, it doesn't mean that. As measured in any inertial frame,
>> light travels a greater distance in going from the rear to
>> the front than in going from the front to the rear.
>
>This is just plain wrong. In any inertial frame light travel the same
>distance in going from the rear to the front than going from the front
>to the rear.
No, it's not. Think about it, Ken. In travelling
from the rear to the front, the light signal first
must travel a distance L (the length of the ship).
That takes a time equal to L/c seconds. But during
that time, the front of the rocket is not staying
still. It is moving. So after the light signal has
travelled a distance L, the front of the rocket has
moved a little bit. Therefore, the light must travel
a little bit more than L to reach the front of the
rocket. The light signal will reach the front in
time slightly more than L/c.
In travelling from the front to the rear, the light
travels *less* than L. The light signal is travelling
towards the rear at speed c, but in the meanwhile,
the rear is moving toward the light signal, as well.
The light signal will reach the rear in time slightly
less than L/c.
The only coordinate system in which light takes
the same time to travel rear to front as it does
to travel front to rear is one in which the rocket
is at rest. But if the rocket is accelerating, then
there is *no* inertial coordinate system in which
the rocket is at rest.
kenseto
> "kenseto" says...
> >
> >
> >"Daryl McCullough" <da...@cogentex.com> wrote
>
> >> The assumptions were: (S is an inertial frame
> >> in which the rocket initially is at rest)
> >>
> >> 1. Light has speed c as measured in S.
> >
> >We are not at S.
>
> What does that mean? S is a coordinate system
> used to describe the rocket's motion. What does
> it mean to be "at" a coordinate system?
Why do you need the S coordinate system? The guy in the rocket can measure
his own acceleration and frequency shift. The use of the S system to
describe an accelerating rocket is not within the realm of the SR
postulates.
>
> >> 2. The location of the rear of the ship,
> >> as measured in S, is x_r = 1/2 a t^2.
> >
> >This assumes that acceleration will change the transit
> >time for light in the accelerating ship.
>
> No, it assumes that the rear of the ship
> is at position x_r = 1/2 a t^2 at time t.
It"s the same thing as saying that acceleration will change the transit
time. You are imagining that the rear is moving relative to light. This is
forbidden in SR.
> Assumption 2 doesn't say *anything* about
> light.
Yes it does. You assumption leads to the erroneous notion that x_r is
changing in the rocket with time. This means that the light path length
length is changing in the rocket. This means that the transit time in the
rocket is changing. This also means that the frequency in the roocket is
changing.
>
> Item number 2 is simply putting into mathematical
> form the assumption that the rocket is accelerating
> at rate a.
No item #2 contains the assumption that acceleration will affect the
frequency rate. This means that your math already contains the information
that the frewquency rate is changing.
>
> >> No, it doesn't mean that. As measured in any inertial frame,
> >> light travels a greater distance in going from the rear to
> >> the front than in going from the front to the rear.
> >
> >This is just plain wrong. In any inertial frame light travel the same
> >distance in going from the rear to the front than going from the front
> >to the rear.
>
> No, it's not. Think about it, Ken. In travelling
> from the rear to the front, the light signal first
> must travel a distance L (the length of the ship).
> That takes a time equal to L/c seconds. But during
> that time, the front of the rocket is not staying
> still. It is moving.
This movement is not detectable as per the SR postulate. If what you said is
true then you should be able to determine different speed of light in
different directions. I know that you experts think that aceleration will
affect the speed of light but this has not been proven experimentally.
>So after the light signal has
> travelled a distance L, the front of the rocket has
> moved a little bit. Therefore, the light must travel
> a little bit more than L to reach the front of the
> rocket. The light signal will reach the front in
> time slightly more than L/c.
This assumption is forbidden by thew SR postulate.
>
> In travelling from the front to the rear, the light
> travels *less* than L. The light signal is travelling
> towards the rear at speed c, but in the meanwhile,
> the rear is moving toward the light signal, as well.
> The light signal will reach the rear in time slightly
> less than L/c.
This assumption is forbidden by the SR postulates.
>
> The only coordinate system in which light takes
> the same time to travel rear to front as it does
> to travel front to rear is one in which the rocket
> is at rest. But if the rocket is accelerating, then
> there is *no* inertial coordinate system in which
> the rocket is at rest.
What you are saying is that relative motion will not affect the transit time
but acceleration will affect the transit time. This is an assumption. There
is no experimental proof of this assumption. The earth's surface is in a
state of constant acceleration and yet we measure constant light speed in
all directions on earht's surface. So how do you explain this
contradiction???
Ken Seto
Daryl McCullough
>
>
>"Daryl McCullough" <da...@cogentex.com> wrote in message
>news:a2haj...@drn.newsguy.com...
>> "kenseto" says...
>> >
>> >
>> >"Daryl McCullough" <da...@cogentex.com> wrote
>>
>> >> The assumptions were: (S is an inertial frame
>> >> in which the rocket initially is at rest)
>> >>
>> >> 1. Light has speed c as measured in S.
>Why do you need the S coordinate system?
Because the usually stated laws of Special Relativity
only apply in an inertial coordinate system.
>The guy in the rocket can measure his own acceleration
>and frequency shift.
Yes, but the guy is not in an inertial coordinate
system, so we cannot apply the simple rules of SR
(for example, we cannot apply the rule that light
has speed c in all directions).
>The use of the S system to describe an accelerating
>rocket is not within the realm of the SR postulates.
Sure it is. Special Relativity doesn't make any assumptions
about whether rockets accelerate or not. It only assumes
that they are *described* using an inertial coordinate
system.
>> >> 2. The location of the rear of the ship,
>> >> as measured in S, is x_r = 1/2 a t^2.
>> >
>> >This assumes that acceleration will change the transit
>> >time for light in the accelerating ship.
>>
>> No, it assumes that the rear of the ship
>> is at position x_r = 1/2 a t^2 at time t.
>
>It"s the same thing as saying that acceleration will change
>the transit time.
No, it is the same thing as saying "the rear of the
ship is accelerating".
>You are imagining that the rear is moving relative to light.
That assumption doesn't mention anything about light.
It just formalizes the assumption that the rear is accelerating.
Do you want to consider an accelerating rocket, or not?
>> Assumption 2 doesn't say *anything* about
>> light.
>
>Yes it does.
Show me where in the sentence "The position of the
rear of the rocket as a function of time is given
by x_r = 1/2 a t^2" does the word "light" appear.
>> Item number 2 is simply putting into mathematical
>> form the assumption that the rocket is accelerating
>> at rate a.
>
>No item #2 contains the assumption that acceleration will affect the
>frequency rate.
Show me where the word "frequency" appears in the
sentence "The position of the rear of the rocket
as a function of time is given by x_r = 1/2 a t^2".
>> >> No, it doesn't mean that. As measured in any inertial frame,
>> >> light travels a greater distance in going from the rear to
>> >> the front than in going from the front to the rear.
>> >
>> >This is just plain wrong. In any inertial frame light travel the same
>> >distance in going from the rear to the front than going from the front
>> >to the rear.
>>
>> No, it's not. Think about it, Ken. In travelling
>> from the rear to the front, the light signal first
>> must travel a distance L (the length of the ship).
>> That takes a time equal to L/c seconds. But during
>> that time, the front of the rocket is not staying
>> still. It is moving.
>
>This movement is not detectable as per the SR postulate.
On the contrary, that is the whole basis for "relativity
of simultaneity" in SR. If two signals are sent simultaneously
from the middle of a rocket towards both ends, the signals
will arrive at the same time in an inertial
coordinate system in which the rocket is at rest, but in
other coordinate systems, the signals will arrive at *different*
times.
For an accelerating rocket, there is no inertial coordinate
system in which the rocket is at rest, so the light signals
arrive at different times in *every* inertial coordinate system.
>If what you said is true then you should be able to determine
>different speed of light in different directions.
How? For a rocket moving at a constant velocity, light
takes the same length of time to go from rear to front
as from front to rear, as measured in the coordinate
system in which the rocket is at rest. For other
coordinate systems, the times are different.
>I know that you experts think that aceleration will
>affect the speed of light but this has not been proven
>experimentally.
We're not talking about experiment here, we are talking
about the predictions of SR. SR predicts that the time
to go from the rear to the front of an accelerating rocket
will be greater than the time to go from the front to the
rear, as measured in any inertial coordinate system.
>What you are saying is that relative motion will not affect
>the transit time but acceleration will affect the transit time.
No. *Motion* affects the transit time. In any coordinate system
in which the rocket is moving, a light signal will take longer
to go from rear to front than from front to rear. It doesn't
matter whether the rocket is accelerating or not.
What is different about acceleration is that for an accelerating
rocket, there *is* no coordinate system in which the rocket is
at rest.
>This is an assumption.
No, the assumption was that light travels at speed c in
any inertial coordinate system. That light takes longer to
travel from the rear to the front than from the front to
the rear is a *conclusion*, not an assumption.
>There is no experimental proof of this assumption.
Right now, I'm just talking about what are the predictions
of SR.
>The earth's surface is in a state of constant acceleration
>and yet we measure constant light speed in all directions
>on earth's surface.
No, we don't. If you send a light around the world east-to-west,
it will take a different amount of time than if you send it
west to east.
Only as measured in an inertial reference frame does light
always travel at speed c.
kenseto
> "kenseto" says...
> >
> >
> >"Daryl McCullough" <da...@cogentex.com> wrote in message
> >news:a2haj...@drn.newsguy.com...
> >> "kenseto" says...
> >> >
> >> >
> >> >"Daryl McCullough" <da...@cogentex.com> wrote
> >>
> >> >> The assumptions were: (S is an inertial frame
> >> >> in which the rocket initially is at rest)
> >> >>
> >> >> 1. Light has speed c as measured in S.
>
> >Why do you need the S coordinate system?
>
> Because the usually stated laws of Special Relativity
> only apply in an inertial coordinate system.
>
> >The guy in the rocket can measure his own acceleration
> >and frequency shift.
>
> Yes, but the guy is not in an inertial coordinate
> system,
So what? He can measure whether there is frequency shift or not.
>so we cannot apply the simple rules of SR
> (for example, we cannot apply the rule that light
> has speed c in all directions).
This is an assumption on your part. That's where you get your frequency
shift from.
>
> >The use of the S system to describe an accelerating
> >rocket is not within the realm of the SR postulates.
>
> Sure it is. Special Relativity doesn't make any assumptions
> about whether rockets accelerate or not. It only assumes
> that they are *described* using an inertial coordinate
> system.
>
> >> >> 2. The location of the rear of the ship,
> >> >> as measured in S, is x_r = 1/2 a t^2.
> >> >
> >> >This assumes that acceleration will change the transit
> >> >time for light in the accelerating ship.
> >>
> >> No, it assumes that the rear of the ship
> >> is at position x_r = 1/2 a t^2 at time t.
> >
> >It"s the same thing as saying that acceleration will change
> >the transit time.
>
> No, it is the same thing as saying "the rear of the
> ship is accelerating
It is the same. Why? When you establish position x_r you are saying that
light move from that position to the front of the ship. Since light from the
front is moving toward that position therefore the transit time from the
front to the rear is shorter.
>
> >I know that you experts think that aceleration will
> >affect the speed of light but this has not been proven
> >experimentally.
>
> We're not talking about experiment here, we are talking
> about the predictions of SR. SR predicts that the time
> to go from the rear to the front of an accelerating rocket
> will be greater than the time to go from the front to the
> rear, as measured in any inertial coordinate system.
So it doesn't bother you that this assumption doesn't fit actual
measurements in the ship??
>
> >What you are saying is that relative motion will not affect
> >the transit time but acceleration will affect the transit time.
>
> No. *Motion* affects the transit time. In any coordinate system
> in which the rocket is moving, a light signal will take longer
> to go from rear to front than from front to rear. It doesn't
> matter whether the rocket is accelerating or not.
The following previous post will show that you are wrong.
______________________________________________
Observed relative velocity has no effect on tranit timeof light. This is
supported by actual measurements in the train.
Einstein used this erroneous concept to get Relativity of Simultaneity
(RoS). In the train gedanken he said that the reason that the train observer
sees the lightning not to be simultaneous is that the train observer is
rushing toward the light front from the front and moving away from the light
from from the rear. This statement violate the constancy of the speed of
light postulate in the train..
In real life the transit time for light is affected only by the state of
absolute motion of the observer. Since the train observer is in a higher
state of absolute motion than the track observer therefore the light path
length in the train will be longer than in the track and therefore the light
fronts will meet in the train at a different later time.
Another way of looking at this is as follows: The lightnings are two light
spheres. They meet at infinite locations as they expand. The direction of
absolute motion of the train and track observers are along the line where
these light
spheres meet. Since the train observer is in a higher state of absolute
motion than the track observer then these two light spheres will meet at a
different later time in the train.
From the track observer's point of view, he also conclude that the train
observer sees the lightnings to be simultaneous at a different later time as
follows:
The light path length in the train in any direction is:
L'=L*gamma
The transit time in the train for the lightnings to meet is as
follows:
T'=L*gamma/c
The transit time for the lightnings to meet in the track frame is:
T=L/c
Clearly T'>T. Therefore the lightnings meet at a different later time in
the train.
_____________________________________________________
>
> What is different about acceleration is that for an accelerating
> rocket, there *is* no coordinate system in which the rocket is
> at rest.
>
> >This is an assumption.
>
> No, the assumption was that light travels at speed c in
> any inertial coordinate system. That light takes longer to
> travel from the rear to the front than from the front to
> the rear is a *conclusion*, not an assumption.
If relative motion has no effect on transit time. Acceleration also has no
affect on transit time as measured in the ship.
>
> >The earth's surface is in a state of constant acceleration
> >and yet we measure constant light speed in all directions
> >on earth's surface.
>
> No, we don't. If you send a light around the world east-to-west,
> it will take a different amount of time than if you send it
> west to east.
You are talking about one way light speed. SRT asserts that two way light
speed is isotropic. This is experimentally proven with the MMX.
Ken Seto
David Evens
wrote:
It doesn't explain how you claim that two light frontsthat are unable
to meet more than once because they are moving in opposite directions
not only meet at two different places at two different times but these
times are actually the SAME time.
David McAnally
>"Eric Prebys" <pre...@fnal.gov> wrote in message
>news:3C363714...@fnal.gov...
>>
>Their claims are obviously bogus. In real life, the initial frequency would
>be lowest detectable. It will increase to a max. when the car is at the
>nearest point of approach ---call this Fm.
>When the car passes the detector, the detected frequency will decrease from
>Fm until it is no longer detectable.
This has already been explained to you. You are the only person who has
experienced this strange phenomenon. When the source is approaching you,
then successive wavefronts emitted by the source have to travel shorter
distances than their predecessors to reach you. This means that each
wavefront takes a shorter time to reach you from the source than its
predecessors, and so the time difference between your receipt of
successive wavefronts is less than the time difference between their
emissions by the source. Since the frequency that you receive is the rate
at which successive wavefronts reach you, and the frequency of the source
is the rate at which it emits successive wavefronts, then the frequency
that you receive is higher than that emitted by the source (since there is
less time between receipt of successive wavefronts than between successive
emissions, then in one second, you receive more wavefronts than is
emitted). In practical terms, this means that the frequency that you hear
from an approaching source is higher than that which is emitted.
>You are not just stupid you are also a fucking liar.. No math was presented
>by anybody in this thread.
Okay. Here is a mathematical presentation. As we are discussing pitch,
then we will restrict ourselves to sound, and as the velocity relative to
c is extremely small, then Newtonian mechanics is a good enough
approximation to SR that we can safely adopt it here.
Denote the speed of sound in air by V, and suppose the air to be perfectly
calm (we don't want any complications due to wind). Take a stationary
observer, and a source emitting at frequency fo travelling directly
towards the observer at speed v, so we are describing the case of source
approaching observer, which is appropriate to the case under discussion:
the pitch of an approaching brain. Let the distance between source and
observer be given by L at time t=0, so that the distance at time t is
L-vt. Position the x-axis so that the origin is at the observer, and the
x-axis is oriented so that the direction of motion of the source is in the
positive x-direction. The position of the source at time t is therefore
-L+vt, and we are interested only in t < L/v (since the source passes the
observer at t = L/v, after which it is receding from the observer). The
phase of the sound, as a function of x and t, for sound moving in the
*positive* x-direction, is given by At-Bx+C, where A and B are positive
constants, and C is a constant (you can take both A and B negative, but
that can be renormalized to positive to reversing the phase - A and B
*must* have the same sign since we are taking sound travelling in the
direction of *increasing* x). The phase is a linear function since the
source is travelling at a constant speed and it is emitting at a constant
frequency. Since the speed of sound is V, then A/B = V. This can be seen
as follows. V is the speed of a wavefront of given phase, so that in time
interval tau (from t to t+tau), the wavefront moves from x to x+V tau, so
that A(t+tau)-B(x+V tau)+C = At-Bx+C (this is just an assertion that the
phase is the same, as already pointed out), so A tau - BV tau = 0, and so,
since this is true for arbitrary tau, A = BV, and so A/B = V (as
asserted). This means that the functional form of the phase is given by
B(Vt - x) + C.
The functional form of the phase means that the phase at the source (the
point of emission), where x = -L+vt, is given by
BVt - B(-L+vt) + C
= (BV - Bv)t + BL + C
= B (V-v) t + BL + C.
Since the frequency of emission from the source is fo, then the time
between the emission successive wavefronts is 1/fo, and so the phase
of emission increases by 2 pi when the t is increased by 1/fo. It
follows that
2 pi = B (V-v)/fo,
and so fo = B (V-v)/(2 pi). Solving for B, it follows that
B = 2 pi fo/(V-v),
and so
A = BV = 2 pi V fo/(V-v) = 2 pi fo/(1-v/V).
Now, let's go to what the observer sees. The functional form of the phase
is given by
2 pi fo/(1-v/V) (t - x/V) + C.
Since the observer remains at rest at x = 0, then the phase for the
observer is given by
2 pi fo/(1 - v/V) t + C.
Since this is linear in t, then the observer is receiving a constant
frequency (i.e. constant pitch), which I will denote by f. Since the
frequency that the observer hears is f, then the time between successive
wavefronts is 1/f, so that the phase changes by 2 pi when t increases by
1/f. It follows that
2 pi = 2 pi fo/(1 - v/V) (1/f)
= 2 pi fo/[f (1 - v/V)],
and so, solving for f,
f = fo/(1 - v/V).
Note specifically that f > fo (so that the pitch of an approaching source
is higher).
>> > > Light going toward the front of the ship will be red-shifted by a
>> > > fractional amount a*L/c^2, where a is the acceleration and L is the
>> > > distance of propagation. Light going toward the rear will be
>> > > blue-shifted
>> > > by the same fractional amount. This first order approximation will
>> > > be valid until a*L becomes more than a miniscule fraction of c^2.
>> > >
>> > > Also, as many have pointed out, this first order approximation
>> > > is valid not only for SR, but for just about any model you can
>> > > think of.
>> >
>> > I already pointed out to you that this bull shit is wrong. Your math is
>> > based on the wrong assumption that light takes a different transit time
>to
>> > traverse the same distance L in the spaceship. This assumption violates
>the
No. Let's lay this to rest. We have an *accelerating* ship. In the
inertial frame in which the ship is at rest when it emits the light, the
light travels at speed c is all directions (including front and back).
The ship itself is accelerating, so that it moves slightly during the time
that the light is in transit. Taking a*L to be much smaller than c^2, we
can again assume that the ship acts according to Newtonian mechanics as a
good approximation, and that each part of the ship has position determined
by 1/2 A*t^2 + C in the given inertial frame, where the x-axis is oriented
so that it points in the direction of acceleration of the ship, and the
wavefront in question is emitted at time t = 0 in the given inertial
frame. Let's take the middle of the ship at x = 0 at the time of
emission. This means that the location of the middle of the ship is given
by x = 1/2 A*t^2, the location of the front of the ship is given by
x = L + 1/2 A*t^2, and the location of the back is given by
x = - L + 1/2 A*t^2.
When does the wavefront reach the front of the ship? Suppose that the
wavefront reaches the front of the ship at time Tf, then, since light is
travelling at speed c in the inertial frame,
c*Tf = L + 1/2 A*Tf^2
(the light is travelling in the direction of increasing x), so that
1/2 A*Tf^2 - c*Tf + L = 0.
Using the well-known formula for the solution of a quadratic equation, we
find that
Tf = [c + sqrt(c^2 - 2*A*L)]/A,
where square root can be taken to be positive or negative (in order to
spare confusion, it will use the notation Sqrt when I want to discuss the
positive square root). Since the argument of the square root is less than
c^2, then Sqrt(c^2 - 2*A*L) < c, and so we take the negative square root
(incidentally, the positive square root has to eliminated any way since
then the time would be approximately 2c/A, a time when the Newtonian
approximation is no longer valid). It follows that the time for light to
reach the front of the ship is given by
Tf = [c - Sqrt(c^2 - 2*A*L)]/A.
Since A*L << c^2, then Sqrt(c^2 - 2*A*L) is very close to
c - A*L/c - A^2*L^2/(2*c^3),
using the Taylor series expansion for Sqrt, and so Tf is very close to
L/c + A^2*L^2/(2*c).
When does the wavefront reach the back of the ship? Suppose that the
wavefront reaches the front of the ship at time Tb, then, since light is
travelling at speed c in the inertial frame,
- c*Tb = - L + 1/2 A*Tb^2
(the light is travelling in the direction of decreasing x), so that
1/2 A*Tb^2 + c*Tb - L = 0.
Again using the well-known formula for the solution of a quadratic
equation, we find that
Tb = [- c + sqrt(c^2 + 2*A*L)]/A.
Now, the negative square root can be eliminated from consideration since
we require that Tb be positive. It follows that the time for light to
reach the back of the ship is given by
Tb = [- c + Sqrt(c^2 + 2*A*L)]/A.
Since A*L << c^2, then Sqrt(c^2 + 2*A*L) is very close to
c + A*L/c - A^2*L^2/(2*c^3),
again using the Taylor series expansion for Sqrt, and so Tb is very close
to
L/c - A^2*L^2/(2*c).
Note that the second order approximations already indicate that the time
taken to the front of the accelerating ship is more than the time taken to
the back. The fact that this hold true under the Newtonian approximation
can also be seen by the observation that as a function, Sqrt is concave
down (so that any chord between points on the graph of the function
actually lies *beneath* the curve).
Now, we came to the question of what happens to the frequency. The front
is moving away from the original location of the source by the time the
light reaches the front (with a speed of approximately A*L/c). This means
that the light observed by the front is red-shifted (Doppler shift) by an
amount of A*L/c^2 (i.e. an emitted frequency of f is observed at
frequency f - A*L*f/c^2 to first order).
Similarly, the back is moving towards the original location of the source
by the time the light reaches the back (with a speed of approximately
A*L/c). This means that the light observed by the back is blue-shifted
(Doppler shift) by an amount of A*L/c^2 (i.e. an emitted frequency of f is
observed at frequency f + A*L*f/c^2 to first order).
This gives the reason why the acceleration gives rise to the shifts: the
shifts are Doppler shifts due to the fact that ship is moving by the time
the light reaches front and back, and the reason why the ship is moving at
that time is because it is accelerating.
>> No, it assumes the time it takes is L/c in either direction.
>Hey stupid when you assumed that the light is frequency shifted the shift is
>caused by the front is moving relative to the light differently than the
>rear is moving relative to the same light. This means that L/c to the front
>will take a longer transit time than L/c to the rear. In case you are too
>stupid to understand the assumption of frequency shift within the same frame
>implies the detection of absolute motion of the frame.
Frequency is not assumed to shift. The above demonstrates that the shift
is a necessary consequence of the acceleration (red-shift for front,
blue-shift for back).
>I'm sorry
>> I overestimated you; I should have done that it 4 or 5 separate steps
>> so that *maybe* you could have followed it.
>I suggest that you take time from your job and study physics before you give
>me any shit. <ahrug>
It seems a strange thing to tell a person who is working at the Fermi Lab
(and whose job is therefore presuambly doing physics) to take time off
their job (i.e. doing physics) so that they can study physics (i.e. what
they are already doing, and what they would have studied for years at
undergraduate and postgraduate level). You *do* know who Fermi was, don't
you?
David McAnally
David McAnally
>"Paul B. Andersen" <paul.b....@hia.no> wrote in message
>news:a1ed7p$ncn$1...@snipp.uninett.no...
>>
>> "Ken H. Seto" <ken...@erinet.com> wrote in message
>news:3c34a0d6$0$35612$4c5e...@news.erinet.com...
>I was not trying to paraphase you. That was the biggest disagreement I had
>with you and Tom. In the end you complimented me for my explanation why the
>detected frequency is at a max when the transmitter is at the closest point
>of approach.
I find that difficult to believe.
>My explanation was that the detected frequency is dependent on
>the transit time and the state of absolute motion of the detector.
No. Frequency depende on the rate at which successive wavefonts hit the
detector. It is demontrable mathematically that this rate is greatest
when the source is directly approaching the detector, and certainly not
when it is passing the detector.
>The
>farther away is the transmitter the more transit time is required for the
>light to reach the detector.
This is irrelevant. What is relevant is the *difference* is transit times
between successive wavefronts. The actual time required has no bearing on
the matter.
>This means that the detector will have more
>time to move away (due to its own absolute motion) and thus miss detecting
>some of the wave crests and thus a lower detected frequency.
How would it miss detecting some of the wavecrests? The wavecrests are
spherical about their point of emission. There is no way to evade them.
>When the
>transimitter is at the closest point of approach, the detector will have the
>least transit time to move away and thus detect virtually all the wave crest
>out put by the treansmitter and thus the detected frequency is at a Max. at
>the closest point of approach.
See comments above.
>> > In real life, the initial frequency would
>> > be lowest detectable. It will increase to a max. when the car is at the
>> > nearest point of approach ---call this Fm.
>> > When the car passes the detector, the detected frequency will decrease
>from
>> > Fm until it is no longer detectable.
>> > :
>> > What do you think???
>>
>> I think what anybody will think.
>> That you have no clue.
>So you think what I said is wrong?? Or are you just letting off steam??
Have you actually observed your phenomenon in real life? Where is your
evidence?
>> BTW, Ken.
>> Most people learn how your scenario works in the real world
>> by experimental evidence while they are still kids.
>> Have you never passed a sound source while travelling by train, Ken?
>Ah so you agree that my description works in the real world!!!
No, he didn't say that. He said that most people learn what actually
happens by experimental evidence while they are still kids. What they
learn is not what you claim.
>But
>unfortunately for you that would mean that you are disagreeing with your SR
>brother Eric Prebys. He said:
>"The answer is that as long as the speed is constant, the frequency
>will be shifted up by a CONSTANT amount while the transmitter is moving
>(directly) toward you, and down by a CONSTANT amount, when it is moving
>(directly) away from you."
The above is correct. If the path misses you (i.e. not dirctly towards or
away from you, then the frequency decreases more steadily). Incidentally,
all this is First Year Physics at most.
>So I guess you will have to fight with your SR brother to settle this issue.
>Unless of course you are claiming that he is also correct by reason of
>authority.
No, he is correct by reason of the fact that the results can be derived.
I think your problem is that you misinterpret what you read.
David McAnally