Three spaceships
Dieter Linz
second spaceship is exactly in the middle between the other spaceships. Now the
middle spaceship sends per radio the following order: The rockets shall catch
fire for one minute, when the order arrives. Therefore the first and the last
ships get the same acceleration and the same endvelocity.
This is my question: Is there a change in the distance between the first and
the last ships (centre of gravity) for the crews of the ships?
Dieter Linz
Jonathan Doolin
"Dieter Linz" <delam...@aol.com> wrote in message
news:20020104162613...@mb-fm.aol.com...
Hi Dieter.
This is one of my favorite thought experiments. You need to go one more
step to see how it works though.
You've got them all moving at the same speed, which is good. The middle
ship matches speed and travels halfway between them. Now they are all moving
along at the same rate, and the command from the middle ship is given again.
The people on the ships see the same thing happen as before. But the people
watching see the signal reach the back ship before the front ship. The back
ship begins accelerating first, and the three ships end up closer together.
But from the point of view of the ships, the front and back ship accelerated
again at the same time. From their own perspective, they did not come
closer together.
This is really well explained in Epstein's "Relativity Visualized"
http://www.appliedthought.com/InsightPress/RelativityVisualized.html
Russell Blackadar
>
> "Dieter Linz" <delam...@aol.com> wrote in message
> news:20020104162613...@mb-fm.aol.com...
> > Three identical spaceships fly one after another with the same velocity.
> The
> > second spaceship is exactly in the middle between the other spaceships.
> Now the
> > middle spaceship sends per radio the following order: The rockets shall
> catch
> > fire for one minute, when the order arrives. Therefore the first and the
> last
> > ships get the same acceleration and the same endvelocity.
> > This is my question: Is there a change in the distance between the first
> and
> > the last ships (centre of gravity) for the crews of the ships?
For the crews of *those* ships, yes there is a change. They are
further apart at the end (i.e. in their final inertial frame) than
when they started.
> >
> > Dieter Linz
> >
>
> Hi Dieter.
>
> This is one of my favorite thought experiments. You need to go one more
> step to see how it works though.
>
> You've got them all moving at the same speed, which is good. The middle
> ship matches speed and travels halfway between them.
Isn't that exactly what the OP said?
Now they are all moving
> along at the same rate, and the command from the middle ship is given again.
> The people on the ships see the same thing happen as before. But the people
> watching
Which people?? The OP asked about the crews of the ships. Why
bring in other people, especially ones you tell us nothing about.
see the signal reach the back ship before the front ship. The back
> ship begins accelerating first, and the three ships end up closer together.
True for a frame in which the rockets are already moving at a
positive velocity in the direction they are facing. But there
are other frames -- in particular it's *not* true in the final
inertial frame of the rockets.
>
> But from the point of view of the ships, the front and back ship accelerated
> again at the same time. From their own perspective, they did not come
> closer together.
Yes, they started accelerating at the same time and hence did not
get closer together in their original inertial frame. But the rockets
(and their crews) are no longer in that frame after the acceleration.
In the inertial frame they end up in, the rockets are further apart.
We've already been through this ad nauseam, Jonathan; I thought you
understood it.
>
> This is really well explained in Epstein's "Relativity Visualized"
> http://www.appliedthought.com/InsightPress/RelativityVisualized.html
I hope it was just you and not Epstein that got this wrong.
and...@attglobal.net
Take a look at the FAQ entry for Bell's "paradox".
Accelerating extended objects has some very non-intuitive
consequences.
Reply if you don't get the point.
John Anderson
Tom Roberts
Jonathan Doolin
"Russell Blackadar" <rus...@REMOVEmdli.com> wrote in message
news:3C3652BD...@REMOVEmdli.com...
> Jonathan Doolin wrote:
> >
> > "Dieter Linz" <delam...@aol.com> wrote in message
> > news:20020104162613...@mb-fm.aol.com...
> > > Three identical spaceships fly one after another with the same
velocity.
> > The
> > > second spaceship is exactly in the middle between the other
spaceships.
> > Now the
> > > middle spaceship sends per radio the following order: The rockets
shall
> > catch
> > > fire for one minute, when the order arrives. Therefore the first and
the
> > last
> > > ships get the same acceleration and the same endvelocity.
> > > This is my question: Is there a change in the distance between the
first
> > and
> > > the last ships (centre of gravity) for the crews of the ships?
>
> For the crews of *those* ships, yes there is a change. They are
> further apart at the end (i.e. in their final inertial frame) than
> when they started.
>
I have to disagree with Russel here. For the crews of the front and end
ships, there would be no change.
The middle ship would actually see the front and rear ship slightly length
contracted, until he matched speeds.
Jonathan Doolin
"Jonathan Doolin" <spence...@earthlink.net> wrote in message
news:dvrZ7.3810$zw3.5...@newsread1.prod.itd.earthlink.net...
Here is another way of looking at it. Instead of sending out one signal
that says "accelerate for one minute from 0 to 600 km per second" lets
divide the signal up into 60 signals, each saying "accelerate by 10 km per
second"
Is this equivalent to the thought experiment you put together? No. Because
the center ship is not accelerating for the first 60 seconds. In this
situation the rear ship should be getting the signals to accelerate faster
than the front ship, and it will creep up on the front ship. This effect
will be observable from inside the ships' reference frame.
Can we make a thought experiment using 60 signals which will do the trick?
Yes. I think so. Have the center ship send out an accelerating beacon
which will send out the signals which will tell the other ships to
accelerate. In this case the ships would not get closer in their own
reference frame, but they would be seen to get closer from outside. You
should be able to verify this mentally.
Jonathan Doolin
Oh, I see where you got this idea. From reading the relativity FAQ.
http://www.public.iastate.edu/~physics/sci.physics/faq/spaceship_puzzle.html
.
I think its most likely that the disagreement comes from 'ambiguities in the
formulation'
Jonathan Doolin
"Jonathan Doolin" <spence...@earthlink.net> wrote in message
news:uzEZ7.4854$zw3.6...@newsread1.prod.itd.earthlink.net...
>
I found the problem.
In the relativity FAQ:
http://www.public.iastate.edu/~physics/sci.physics/faq/spaceship_puzzle.html
Michael Weiss uses these equations to describe the positions of the two
space-ships in the lab frame.
x=sqrt(1+t^2)
and
x=k+sqrt(1+t^2)
With this formulation, the two ships stay the same distance from one another
in the lab frame (distance =k) Thus in their own frame of reference they
would get further apart.
However, in Dieter Linz's description the positions of the two ships should
look something more like this:
x=integral(1+(t/g1(t))^2)dt
and
x=k+integral(1+t/g2(t)^2)dt
The construction might not be quite accurate, but I know there would have to
be an integral involved, and a factor of time-dependent gamma.
Eli Botkin
Reply to Dieter Linz:
Based on your question, I just worked the following problem:
The 3 spaceships are initially at rest on the X-axis, each separated from
the adjacent one by 0.1 lightyears (ly). At time zero all three ships
accelerate in the +X direction at 1.03 ly/sec^2 (that's the same as 9.8
m/s^2, due to Earth's surface gravity).
Take the event on the center-ship where its clock reads 1.0 year. An
observer who remained at rest on the X-axis would say his own clock reads
1.186 years when that center-ship's clock reads 1.0 year. In his rest frame
all three ships are then moving at 0.774c and are still separated by 0.1 ly
distances.
Now let's look at things as an observer on the center-spaceship.
When that ship's clock reads 1.0 year, in that ship's instantaneous,
comoving inertial frame the leading-ship is 0.166 ly ahead of the
center-ship and the trailing-ship is 0.150 ly behind. In that comoving
frame, the leading-ship has a greater relative velocity (relative to the
center-ship) than does the trailing-ship.
Happy New Year
Eli Botkin
Russell Blackadar
So you assert. Care to give your reasoning? Mine is already in
the FAQ.
> > The middle ship would actually see the front and rear ship slightly length
> > contracted, until he matched speeds.
> >
>
> Oh, I see where you got this idea. From reading the relativity FAQ.
> http://www.public.iastate.edu/~physics/sci.physics/faq/spaceship_puzzle.html
> .
Ya know, just because two people happen to get the same correct
answer, that doesn't necessarily mean one person copied the other.
But yes, as it happens, I *have* read the FAQ, and that page in
particular. Quite some time ago. I hope that your mentioning it
here is an indication that you are now reading it too.
> I think its most likely that the disagreement comes from 'ambiguities in the
> formulation'
Sorry, no; it comes from your mistake. The ambiguities mentioned
in the FAQ are not present in the OP's statement of *this* problem.
The main trickiness in the FAQ page is that it tries to deal with
acceleration that goes on and never stops; in that case the ships
are never again in the same comoving inertial frame once they have
started accelerating. In contrast, the present scenario has an
acceleration lasting only one minute (of proper time on each ship)
and the OP's question can be interpreted as referring only to the
final state *after* that minute has passed. That final distance
is unambiguously gamma*L.
If you want to ask what the crew sees *during* the acceleration,
yes that's ambiguous because there is no natural way to specify
accelerating coordinates; it's a matter of convention. But in
the coords of the measuring crew's comoving inertial frame, the
distance is slightly *greater* than gamma*L while the acceleration
is in progress. See the abovementioned FAQ page.
Russell Blackadar
Exactly. Why do you call that a problem?
>
> However, in Dieter Linz's description the positions of the two ships should
> look something more like this:
> x=integral(1+(t/g1(t))^2)dt
> and
> x=k+integral(1+t/g2(t)^2)dt
>
> The construction might not be quite accurate, but I know there would have to
> be an integral involved, and a factor of time-dependent gamma.
What makes you think an integral, and a time-dependent gamma,
aren't involved in Weiss's treatment? Where did you think those
two hyperbolas came from? (In case you didn't notice, he never
said. They also appear, again without derivation, in the FAQ
page for the relativistic rocket.)
Jonathan Doolin
"Russell Blackadar" <rus...@REMOVEmdli.com> wrote in message
Well, in a way, it's actually rather nice. He unambiguously states the
assumptions he is using in a mathematical fashion, so it is very easy to
understand what he is saying. He is describing a situation where the two
objects remain at the same distance from one another in the *lab* frame, and
is using this to define his problem. My own treatment of the problem
assumed that the space ships would preserve their distance in their *own*
frame. When I read Linz's question, I thought this was a reasonable
assumption.
The description of this situation was: "the middle spaceship sends per
radio the following order: The rockets shall catch fire for one minute,
when the order arrives. Therefore the first and the last ships get the
same acceleration and the same endvelocity."
There important variables here are the acceleration, and the time. The crew
of each ship will rely on the clocks inside the ship to measure the time.
They will also rely on their own engines to create a constant acceleration.
As long as they do this, they should remain at the same distance in their
*own* frame, just as though they were attached to each other by a strong
tether.
>
> >
> > However, in Dieter Linz's description the positions of the two ships
should
> > look something more like this:
> > x=integral(1+(t/g1(t))^2)dt
> > and
> > x=k+integral(1+t/g2(t)^2)dt
> >
> > The construction might not be quite accurate, but I know there would
have to
> > be an integral involved, and a factor of time-dependent gamma.
>
> What makes you think an integral, and a time-dependent gamma,
> aren't involved in Weiss's treatment? Where did you think those
> two hyperbolas came from? (In case you didn't notice, he never
> said. They also appear, again without derivation, in the FAQ
> page for the relativistic rocket.)
>
Where do I think those two hyperbolas came from? I honestly have no idea.
But are they accurate? Do they describe an object under constant
acceleration? Maybe we can reverse engineer his solution. It is proving
too difficult an equation for me to reverse engineer, with my pathetic math
abilities. I would much rather have them tell me where they got it.
==Warning: Poor attempt at reverse engineering, below==
sqrt(1+t^2) is position.
Using my tiny little bit of calculus that I remember
(f(g(x)))'=f'(g(x))g'(x)
dx/dt=t/sqrt(1+t^2) is velocity.
Hmmm. I think he's doing the old unity trick. Define 1 foot as one light
nanosecond. Use units of feet for distance, nanoseconds for time, feet/ns
for velocity, and feet/ns^2 for acceleration, and assume a tremendous
acceleration of 1 foot per nanosecond per nanosecond.
So dx/dt =t/sqrt(1+t^2) is the velocity at time t in the laboratory
reference frame.
And the acceleration in the lab reference frame would be
d^2x/dt^2=(ddn-ndd)/dd=the reason they made programs like Maple.
Now if I could remember the two Lorentz equations...
x'= x-Vt/sqrt(1-(V/c)^2),
t'= (t-Vx/c^2/sqrt(1-(V/c)^2
http://zebu.uoregon.edu/~kevan/ph251/101695/101695.html
t'= t-Vx/sqrt(1-(V/c)^2) because of the unity trick.
So now we solve for the acceleration, d^2x'/dt'^2 and see if it is a
constant for the two spaceships.
I leave this as an exercise for the reader. (Just kidding.)
===
Russell Blackadar
>
> "Russell Blackadar" <rus...@REMOVEmdli.com> wrote in message
> news:3C3A2A16...@REMOVEmdli.com...
> > Jonathan Doolin wrote:
[snip]
> > > I found the problem.
> > > In the relativity FAQ:
> > >
> http://www.public.iastate.edu/~physics/sci.physics/faq/spaceship_puzzle.html
> > >
> > > Michael Weiss uses these equations to describe the positions of the two
> > > space-ships in the lab frame.
> > >
> > > x=sqrt(1+t^2)
> > > and
> > > x=k+sqrt(1+t^2)
> > >
> > > With this formulation, the two ships stay the same distance from one
> another
> > > in the lab frame (distance =k) Thus in their own frame of reference
> they
> > > would get further apart.
> >
> > Exactly. Why do you call that a problem?
>
> Well, in a way, it's actually rather nice. He unambiguously states the
> assumptions he is using in a mathematical fashion, so it is very easy to
> understand what he is saying. He is describing a situation where the two
> objects remain at the same distance from one another in the *lab* frame, and
> is using this to define his problem.
No, read it again. The statement of the problem in the FAQ
is that they "each accelerate at the same constant rate".
That's all. Then he goes on to make two apparently conflicting
conclusions from that statement of the problem, the first of
which is that they remain at a constant distance in the original
frame. That is a correct conclusion.
The other conclusion, that there is Lorentz contraction in that
frame, is *also* correct! The hidden assumption, that the ships
remain the same distance apart in their own frames, is what's
wrong.
My own treatment of the problem
> assumed that the space ships would preserve their distance in their *own*
> frame. When I read Linz's question, I thought this was a reasonable
> assumption.
Reasonable, perhaps. But you should know by now, in SR many
reasonable assumptions turn out to be false. Btw the FAQ page
even tells why there's no such thing as the rockets' (plural)
own frame (singular) while they are accelerating.
>
> The description of this situation was: "the middle spaceship sends per
> radio the following order: The rockets shall catch fire for one minute,
> when the order arrives. Therefore the first and the last ships get the
> same acceleration and the same endvelocity."
> There important variables here are the acceleration, and the time. The crew
> of each ship will rely on the clocks inside the ship to measure the time.
> They will also rely on their own engines to create a constant acceleration.
Yes, clearly you are talking about the same interpretation of
the problem that I am: the ships have the same proper acceleration
for the same proper time. IMHO that's the only reasonable
interpretation one can make of Linz's statement. In any case I
won't let you get away now with telling me that we differ because
of ambiguity in the formulation, since our interpretations are the
same.
(Actually you don't even have to assume that the acceleration
is constant -- you can just note that it must be identical on
the two ships as parameterized by their respective proper times,
since the ships are said to be identical.)
> As long as they do this, they should remain at the same distance in their
> *own* frame, just as though they were attached to each other by a strong
> tether.
But they *aren't* attached by a tether. Note, you are using
words like "should" and "would". In other words, you are
guessing, right? In reality the answer is known -- and it's
not what you think.
>
> >
> > >
> > > However, in Dieter Linz's description the positions of the two ships
> should
> > > look something more like this:
> > > x=integral(1+(t/g1(t))^2)dt
> > > and
> > > x=k+integral(1+t/g2(t)^2)dt
> > >
> > > The construction might not be quite accurate, but I know there would
> have to
> > > be an integral involved, and a factor of time-dependent gamma.
> >
> > What makes you think an integral, and a time-dependent gamma,
> > aren't involved in Weiss's treatment? Where did you think those
> > two hyperbolas came from? (In case you didn't notice, he never
> > said. They also appear, again without derivation, in the FAQ
> > page for the relativistic rocket.)
> >
>
> Where do I think those two hyperbolas came from? I honestly have no idea.
> But are they accurate? Do they describe an object under constant
> acceleration? Maybe we can reverse engineer his solution. It is proving
> too difficult an equation for me to reverse engineer, with my pathetic math
> abilities. I would much rather have them tell me where they got it.
Yes, IMHO the FAQ be changed here. If you read it quickly it's
not obvious that they just plunked down a rather tricky equation
with no explanation at all.
There's a reference in the *other* FAQ page to MTW section 6.2.
The derivation has also appeared recently in this n.g., I think.
I might even be coaxed to type it in myself but not today.
[snip]
Jonathan Doolin
"Russell Blackadar" <rus...@REMOVEmdli.com> wrote in message
Excellent. Now we can really fight. :)
>
> (Actually you don't even have to assume that the acceleration
> is constant -- you can just note that it must be identical on
> the two ships as parameterized by their respective proper times,
> since the ships are said to be identical.)
Wonderful, this leads me to a very simple explanation as to why the two
ships must move closer together.
If the accelerations are identical by *their* respective proper times, then
they will not occur simultaneously in the lab frame.
In the lab frame, once the ships are under way, each burst of acceleration
will be performed first by the rear ship, then by the front ship, causing
the two ships to move closer together.
>
> > As long as they do this, they should remain at the same distance in
their
> > *own* frame, just as though they were attached to each other by a strong
> > tether.
>
> But they *aren't* attached by a tether. Note, you are using
> words like "should" and "would". In other words, you are
> guessing, right? In reality the answer is known -- and it's
> not what you think.
Oops. Sorry. I meant "As long as they do this, they *will* remain at the
same distance in their *own* frame, just as though they were attached to
each other by a strong tether."
>
> >
> > >
> > > >
They are working with a very tricky subject and not getting paid for it,
after all.
I hope you can find the derivation and let me know where it is, it might be
easier.
(If you could provide a link to the *other* FAQ, or MTW section 6.2,)
Russell Blackadar
>
> "Russell Blackadar" <rus...@REMOVEmdli.com> wrote in message
> > Jonathan Doolin wrote:
[snip]
> > (Actually you don't even have to assume that the acceleration
> > is constant -- you can just note that it must be identical on
> > the two ships as parameterized by their respective proper times,
> > since the ships are said to be identical.)
>
> Wonderful, this leads me to a very simple explanation as to why the two
> ships must move closer together.
> If the accelerations are identical by *their* respective proper times, then
> they will not occur simultaneously in the lab frame.
If the ships were both in the same (instantly comoving inertial)
frame, that would be a correct conclusion.
But during their acceleration, there is no such frame! I.e. there
is never an inertial frame such that the ships both have v=0, except
before and after the acceleration. The FAQ explains why this is so;
all you need to do is come to see why the FAQ applies to this problem,
which I assure you it does.
> In the lab frame, once the ships are under way, each burst of acceleration
> will be performed first by the rear ship, then by the front ship, causing
> the two ships to move closer together.
I am curious how you decided which one had to fire first in the
lab frame -- what I suspect is that you "knew" what the answer "had
to be" and worked backwards. In any case you are wrong.
But one thing you said is right -- it *is* very simple.
Look at it this way. You tell me there is a translational
asymmetry here, in the lab frame -- the rear ship does it first.
Yet clearly that is not so at the very start of the acceleration;
the first "burst" occurs simultaneously in the lab frame. And of
course that burst is the same for both rockets since they are
identical. So, in all respects the result of that acceleration
burst will be translationally symmetrical in the lab frame; both
ships will have the same resulting velocity and the same dilation
of their clocks in the lab frame, after that first burst. So, the
second burst will also occur simultaneously for both rockets in
the lab frame, and of course will be equal. And so on. Your
asymmetry never develops at all, in the lab frame.
[snip]
> > There's a reference in the *other* FAQ page to MTW section 6.2.
> > The derivation has also appeared recently in this n.g., I think.
> > I might even be coaxed to type it in myself but not today.
> >
> > [snip]
>
> They are working with a very tricky subject and not getting paid for it,
> after all.
> I hope you can find the derivation and let me know where it is, it might be
> easier.
> (If you could provide a link to the *other* FAQ, or MTW section 6.2,)
The one I was remembering on Usenet is scattered about in a huge
thread with a somewhat low signal-to-noise ratio. The closest
thing I could find in a single posting was
(you may need to remove line breaks) which derives v as a function
of t, in the lab frame -- but note, it contains a serious typo:
the exponent 3 should be 3/2 everywhere it appears. (It's just
gamma*3.) The integral in that posting is a bit tricky, but you
can verify that his final result is correct. You can also fairly
easily integrate his result to get x as a function of t, and it's
indeed a hyperbola.
MTW box 6.1 is much, much cleaner, but it involves concepts of
the four-velocity which might seem strange if you just jumped
straight into it. If your local bookstore has this book, you
could take a quick peek at this section without having to spend
the big bucks.
Finally, all pages of the FAQ can be found by just going up from
the page that you posted and following the links; i.e.
http://www.public.iastate.edu/~physics/sci.physics/faq/relativity.html
They are excellent reading.
Russell Blackadar
[snip]
> MTW box 6.1 is much, much cleaner
Sorry, that should be 6.2.
Jonathan Doolin
"Russell Blackadar" <rus...@REMOVEmdli.com> wrote in message
> Jonathan Doolin wrote:
> >
> > "Russell Blackadar" <rus...@REMOVEmdli.com> wrote in message
> > news:3C3B9DAB...@REMOVEmdli.com...
> > > Jonathan Doolin wrote:
>
> [snip]
>
> > > (Actually you don't even have to assume that the acceleration
> > > is constant -- you can just note that it must be identical on
> > > the two ships as parameterized by their respective proper times,
> > > since the ships are said to be identical.)
> >
> > Wonderful, this leads me to a very simple explanation as to why the two
> > ships must move closer together.
> > If the accelerations are identical by *their* respective proper times,
then
> > they will not occur simultaneously in the lab frame.
>
> If the ships were both in the same (instantly comoving inertial)
> frame, that would be a correct conclusion.
>
> But during their acceleration, there is no such frame! I.e. there
> is never an inertial frame such that the ships both have v=0, except
> before and after the acceleration. The FAQ explains why this is so;
> all you need to do is come to see why the FAQ applies to this problem,
> which I assure you it does.
Once the first particle of jet-fuel breaks away from the fuel tank in the
rockets, the space-ships are moving. After that, they are moving relative
to the lab frame. From the perspective of the lab frame, the second
particle of jet fuel comes out of the hind rocket before the second particle
of jet fuel comes out of the front rocket. From the perspective of the
space-ships, these two events are simultaneous.
>
> > In the lab frame, once the ships are under way, each burst of
acceleration
> > will be performed first by the rear ship, then by the front ship,
causing
> > the two ships to move closer together.
>
> I am curious how you decided which one had to fire first in the
> lab frame -- what I suspect is that you "knew" what the answer "had
> to be" and worked backwards. In any case you are wrong.
>
Lorentz's equations are pretty clear on this point.
let gamma=sqrt(1-(v/c)^2)
t'=(t-vx/c^2)/gamma
thus, t=gamma*(t'+vx/c^2)
Let x0=0, x1=1, v>0 at time t'=1.
Assuming an event occurs at time t'=1.
t(x0)=gamma*(1)
t(x1)=gamma*(1+v/c^2)>1
Thus from the lab frame, the event occurs at x0 before it occurs at x1.
> But one thing you said is right -- it *is* very simple.
>
> Look at it this way. You tell me there is a translational
> asymmetry here, in the lab frame -- the rear ship does it first.
> Yet clearly that is not so at the very start of the acceleration;
> the first "burst" occurs simultaneously in the lab frame. And of
> course that burst is the same for both rockets since they are
> identical.
True, but the first burst is precisely one collision's worth. Only one
particle escapes before the space-ships are under way. (Or a plane of
particles leaving simultaneously.) For that matter, Isn't there something
in quantum mechanics that says all collisions have to take a finite amount
of time? The burst might be *less* than one collision's worth, depending on
whether the acceleration begins before the collision is over.
<snip>
> The one I was remembering on Usenet is scattered about in a huge
> thread with a somewhat low signal-to-noise ratio. The closest
> thing I could find in a single posting was
>
>
http://groups.google.com/groups?as_umsgid=%3C48658b64.0112181825.2ac1a9e1%40
posting.google.com%3E&hl=en
I'll look at this some more later. Thanks for posting the link.
I am reading the newsgroup through Outlook Express, so I don't come to a
web-page to look at it. Is this the page you were referring to?
http://www.public.iastate.edu/~physics/sci.physics/faq/relativity.html
Also, what does MTW stand for?
Russell Blackadar
Just because they are moving? Show me your math.
Here's mine. Suppose that the rockets are such that one particle
is emitted at each nanosecond. (Of the rocket's proper time, of
course.) Rocket A and Rocket B emit their first particle, let's
say at labframe t=0, so now they are both moving at some v in the
lab frame. The *same* v. So, the on-board clocks of Rocket A and
Rocket B are dilated in the lab frame. By the *same* gamma. That
means, in the lab frame they will both read 1 nanosecond at the
same time, t = (gamma * 1)ns. Rocket B emits its second particle
at that exact time in the lab frame. And I think you can see, if
Rocket A emits its particle any later than that same time in the
lab frame, it is by definition *broken*, because Rocket A's clock
will not read 1ns at any later time.
We only use unbroken rockets in this thought experiment. ;-)
From the perspective of the
> space-ships, these two events are simultaneous.
But it's easy to see that they are not! They are in a frame
now in which their initial burst did not occur simultaneously.
It occurred simultaneously in the *lab* frame. The lead rocket
has been in this frame longer than the trailing rocket, according
to this frame's clocks. It will send its second burst first in
this frame, causing it to go even further forward.
>
> >
> > > In the lab frame, once the ships are under way, each burst of
> acceleration
> > > will be performed first by the rear ship, then by the front ship,
> causing
> > > the two ships to move closer together.
> >
> > I am curious how you decided which one had to fire first in the
> > lab frame -- what I suspect is that you "knew" what the answer "had
> > to be" and worked backwards. In any case you are wrong.
> >
>
> Lorentz's equations are pretty clear on this point.
> let gamma=sqrt(1-(v/c)^2)
> t'=(t-vx/c^2)/gamma
>
> thus, t=gamma*(t'+vx/c^2)
> Let x0=0, x1=1, v>0 at time t'=1.
> Assuming an event occurs at time t'=1.
That's your problem, right there.
There are *two* events (one for each rocket). You have assumed
that these events occur simultaneously in the primed frame.
They don't. They occur when the onboard clocks of the respective
ships read the same *proper time*. And as I showed above, the
rocket's onboard clocks read tau=0 at *different* times in the
primed frame, so naturally they must read tau=1 at different t'
times as well.
Accelerations are tricky.
> t(x0)=gamma*(1)
> t(x1)=gamma*(1+v/c^2)>1
> Thus from the lab frame, the event occurs at x0 before it occurs at x1.
>
> > But one thing you said is right -- it *is* very simple.
> >
> > Look at it this way. You tell me there is a translational
> > asymmetry here, in the lab frame -- the rear ship does it first.
> > Yet clearly that is not so at the very start of the acceleration;
> > the first "burst" occurs simultaneously in the lab frame. And of
> > course that burst is the same for both rockets since they are
> > identical.
>
> True, but the first burst is precisely one collision's worth. Only one
> particle escapes before the space-ships are under way. (Or a plane of
> particles leaving simultaneously.) For that matter, Isn't there something
> in quantum mechanics that says all collisions have to take a finite amount
> of time? The burst might be *less* than one collision's worth, depending on
> whether the acceleration begins before the collision is over.
None of the details matter. All that matters is that the ships
are identical, and start their acceleration programs (whatever
they happen to be) simultaneously in the lab frame.
>
> <snip>
>
> > The one I was remembering on Usenet is scattered about in a huge
> > thread with a somewhat low signal-to-noise ratio. The closest
> > thing I could find in a single posting was
> >
> >
> http://groups.google.com/groups?as_umsgid=%3C48658b64.0112181825.2ac1a9e1%40
> posting.google.com%3E&hl=en
>
> I'll look at this some more later. Thanks for posting the link.
It's ok, though the ASCII is not pretty to read and there's the
typo I mentioned. Some earlier posts in the thread (click on the
links) give a prettier derivation of the gamma^3 in the integrand,
but the cited post is the only one I saw that shows the integration
to v explicitly. Nobody showed the integration to x but that's
pretty easy.
>
> I am reading the newsgroup through Outlook Express, so I don't come to a
> web-page to look at it. Is this the page you were referring to?
>
> http://www.public.iastate.edu/~physics/sci.physics/faq/relativity.html
Yes. Happy browsing.
>
> Also, what does MTW stand for?
Misner, Thorne, and Wheeler's "Gravitation". There is an FAQ page
on relativity books, you'll find all the particulars there.
Jonathan Doolin
"Russell Blackadar" <rus...@REMOVEmdli.com> wrote in message
> Jonathan Doolin wrote:
> >
> > "Russell Blackadar" <rus...@REMOVEmdli.com> wrote in message
> > news:3C3D198F...@REMOVEmdli.com...
> > > Jonathan Doolin wrote:
> > > >
> > > > "Russell Blackadar" <rus...@REMOVEmdli.com> wrote in message
> > > > news:3C3B9DAB...@REMOVEmdli.com...
<SNIP>
> > > I am curious how you decided which one had to fire first in the
> > > lab frame -- what I suspect is that you "knew" what the answer "had
> > > to be" and worked backwards. In any case you are wrong.
> > >
> >
> > Lorentz's equations are pretty clear on this point.
> > let gamma=sqrt(1-(v/c)^2)
> > t'=(t-vx/c^2)/gamma
> >
> > thus, t=gamma*(t'+vx/c^2)
> > Let x0=0, x1=1, v>0 at time t'=1.
> > Assuming an event occurs at time t'=1.
>
> That's your problem, right there.
>
> There are *two* events (one for each rocket). You have assumed
> that these events occur simultaneously in the primed frame.
> They don't.
You are wrong.
> They occur when the onboard clocks of the respective
> ships read the same *proper time*.
You are right. This means exactly the same thing as "The events occur
simultaneously in the primed frame." The ship's *proper time* is in their
own frame. NOT the LAB FRAME.
> And as I showed above, the
> rocket's onboard clocks read tau=0 at *different* times in the
> primed frame,
The primed frame *is* the rocket's frame.
> so naturally they must read tau=1 at different t'
> times as well.
>
> Accelerations are tricky.
Let's fix that and use a simpler problem (see below.)
>
> > t(x0)=gamma*(1)
> > t(x1)=gamma*(1+v/c^2)>1
> > Thus from the lab frame, the event occurs at x0 before it occurs at x1.
> >
<SNIP>
We are saying the same thing here and somehow arriving at different
conclusions.
You are using tau when Lorentz and I use t', to represent the *proper time*
of events on the rocketships.
If acceleration is tricky, let's make it a bit simpler. Have the two ships
traveling without acceleration, and schedule to eat lunch at the same time:
t'=1. Now, using the same math as I did above, I claim that in the lab
frame, the crew of the rear ship will end up eating before the crew of the
front ship. Do you disagree?
Jonathan Doolin
"Russell Blackadar" <rus...@REMOVEmdli.com> wrote in message
> Jonathan Doolin wrote:
> >
> > "Russell Blackadar" <rus...@REMOVEmdli.com> wrote in message
> > news:3C3D198F...@REMOVEmdli.com...
> > > Jonathan Doolin wrote:
> > > >
> > > > "Russell Blackadar" <rus...@REMOVEmdli.com> wrote in message
> > > > news:3C3B9DAB...@REMOVEmdli.com...
> > > > > Jonathan Doolin wrote:
> > >
<SNIP>
> > Once the first particle of jet-fuel breaks away from the fuel tank in
the
> > rockets, the space-ships are moving. After that, they are moving
relative
> > to the lab frame. From the perspective of the lab frame, the second
> > particle of jet fuel comes out of the hind rocket before the second
particle
> > of jet fuel comes out of the front rocket.
>
> Just because they are moving? Show me your math.
>
> Here's mine. Suppose that the rockets are such that one particle
> is emitted at each nanosecond. (Of the rocket's proper time, of
> course.) Rocket A and Rocket B emit their first particle, let's
> say at labframe t=0, so now they are both moving at some v in the
> lab frame. The *same* v. So, the on-board clocks of Rocket A and
> Rocket B are dilated in the lab frame.
And desynchronized. Don't forget desynchronized. Oh, but I forget, the
desynchronization effect is *so obvious* in the Lorentz equations that it
doesn't bear mentioning.
> By the *same* gamma. That
> means, in the lab frame they will both read 1 nanosecond at the
> same time, t = (gamma * 1)ns.
Wrong answer. Didn't you get after me before about not knowing Lorentz
equations?
t=gamma*(t'+vx/c^2).
The clocks on the two ships will be separated by gamma*vx/c^2.
> Rocket B emits its second particle
> at that exact time in the lab frame. And I think you can see, if
> Rocket A emits its particle any later than that same time in the
> lab frame, it is by definition *broken*, because Rocket A's clock
> will not read 1ns at any later time.
>
> We only use unbroken rockets in this thought experiment. ;-)
They'll be broken by the time we're done if we try to keep them the same
length in the lab frame!
<SNIP>
Russell Blackadar
Not in the lab frame, surprisingly for you. In their *own* frame (or
rather I should say in their own respective frames).
Don't forget desynchronized. Oh, but I forget, the
> desynchronization effect is *so obvious* in the Lorentz equations that it
> doesn't bear mentioning.
Before making a smart answer, it would behoove you to make sure it is
really smart.
>
> > By the *same* gamma. That
> > means, in the lab frame they will both read 1 nanosecond at the
> > same time, t = (gamma * 1)ns.
>
> Wrong answer. Didn't you get after me before about not knowing Lorentz
> equations?
Yes, I did. And you still don't know them, apparently.
>
> t=gamma*(t'+vx/c^2).
No, there should be a prime on that x. That makes a big difference!
Also you don't know what x' is yet. Don't just tell me it is k; that
would be assuming what you are trying to prove. (And it ain't true,
but I'm getting ahead of myself.)
See, the proper thing to do is transform from coords you *do* know --
the lab coords -- to the primed frame. We can certainly do that at
t=0. The transform this direction uses a -v. The two initial firing
events have the following primed coordinates (origin at rear ship):
Front ship: t' = gamma(0 - vk/c^2) = -gamma * vk/c^2
x' = gamma(k - v*0) = gamma * k
Rear ship: t' = gamma(0 - 0) = 0
x' = gamma(0 - 0) = 0
The ships' clocks are of course both reading 0 at these events, because
they were synchronized in the lab frame. That means only the rear
ship's
clock is set correctly in the new (primed) frame. The two ship clocks
are
therefore desynchronized in the primed frame. And they will still be
desynchronized at the second firing, and so on.
> The clocks on the two ships will be separated by gamma*vx/c^2.
Actually yes (ignoring sign) -- but in the primed frame, not the lab
frame.
>
> > Rocket B emits its second particle
> > at that exact time in the lab frame. And I think you can see, if
> > Rocket A emits its particle any later than that same time in the
> > lab frame, it is by definition *broken*, because Rocket A's clock
> > will not read 1ns at any later time.
> >
> > We only use unbroken rockets in this thought experiment. ;-)
>
> They'll be broken by the time we're done if we try to keep them the same
> length in the lab frame!
I don't know about you, but I'm done. Get that copy of Spacetime
Physics and read it.
>
> <SNIP>
Bilge
>
>Also, what does MTW stand for?
"Mighty Thick and Wordy"
Jonathan Doolin
"Russell Blackadar" <rus...@mdli.com> wrote in message
news:3C3FC7AA...@mdli.com...
Well, yes, if there were a prime on the x, it would make a pretty big
difference.
"Modern Physics" Chapter 2, Equation 2.8d by Kenneth Krane say's your
wrong.
"Fundamentals of Physics" Chapter 42, Equation 15 by Halliday & Resnick,
says your wrong.
And I had one other book I was looking at last night that said you were
wrong too, but I can't find it now.
<SNIP>
> I don't know about you, but I'm done. Get that copy of Spacetime
> Physics and read it.
>
Well, it might be interesting to read the book that says
t'=(t-vx'/c^2)/gammabut I'm not quite jumping out of my seat just yet, since
t'=(t-vx/c^2)/gamma happens to make a lot more sense.
Stephen Speicher
Jonathan, for someone so mistaken, you have an unearned haughty
attitude. You seem so interested in defending your views that
you do not listen, nor do you understand, what Russell tries to
teach you. I'm amazed that Russell has hung in this long before
he finally said "I'm done." Instead of all your "visualizations,"
take Russell's advice and see with your mind -- read a good book
on relativity, such as _Spacetime Physics_. Maybe you will
actually learn something about the subject, before you presume to
be teaching others.
Stephen
s...@compbio.caltech.edu
Welcome to California. Bring your own batteries.
Printed using 100% recycled electrons.
--------------------------------------------------------
Jonathan Doolin
"Stephen Speicher" <s...@compbio.caltech.edu> wrote in message
news:Pine.LNX.4.10.102011...@photon.compbio.caltech.edu...
The argument at issue here is the Lorentz time-dilation equation:
t'=(t-vx/c^2)/sqrt(1-(v/c)^2)
I have solved for t and found that
t=sqrt(1-(v/c)^2)*(t'+vx/c^2).
Russell said that the x should be an x' in the equations (which is clearly a
algebraic error on his part) and then said he would no longer answer any
further posts. If I reacted with a haughty tone after seeing this behavior,
I assure you, it was warranted.
Stephen Speicher
> The argument at issue here is the Lorentz time-dilation equation:
>
> t'=(t-vx/c^2)/sqrt(1-(v/c)^2)
>
> I have solved for t and found that
>
> t=sqrt(1-(v/c)^2)*(t'+vx/c^2).
>
> Russell said that the x should be an x' in the equations (which is clearly a
> algebraic error on his part) and then said he would no longer answer any
> further posts. If I reacted with a haughty tone after seeing this behavior,
> I assure you, it was warranted.
>
No. The fact that you continue to defend utterly simple and basic
errors on your part, clearly demonstrates that your attitude is
completely unwarranted.Your solution for t (just above) is wrong
-- perhaps you do not know how to solve simple algebraic
equations. Here is the correct solution for going from t' to t.
t'=(t-vx/c^2)/sqrt(1-(v/c)^2)
t' * sqrt(1-(v/c)^2) = t - vx/c^2
t' * sqrt(1-(v/c)^2) + vx/c^2 = t (equ. a)
That is what you should get by solving for t. Note that the
result is different from the form as shown in the texts. I
suspect you do not even understand how to solve such a simple
equation, and instead just copied the standard form the text,
replacing x' with x.
To put my solution for t (equ. a) into standard form, we note
that you must also make use of the Lorentz transformation x' -> x.
That transformation is given as x' = (x -vt)/sqrt(1-(v/c)^2). So
solving for x we get x = x'*sqrt(1-(v/c)^2) +vt. So, by replacing
x with x' in (equ. a), we get
t' * sqrt(1-(v/c)^2) + v * [x'*sqrt(1-(v/c)^2) +vt] = t
----------------------------
c^2
t' * sqrt(1-(v/c)^2) + v * [x'*sqrt(1-(v/c)^2)] + t*(v/c)^2 = t
------------------------
c^2
t' * sqrt(1-(v/c)^2) + x'v *sqrt(1-(v/c)^2) = t - t*(v/c)^2
--------------------
c^2
t' * sqrt(1-(v/c)^2) + x'v *sqrt(1-(v/c)^2) = t (1-(v/c)^2)
--------------------
c^2
t' * sqrt(1-(v/c)^2) + x'v *sqrt(1-(v/c)^2) = t
------------------- --------------------
(1-(v/c)^2) (1-(v/c)^2)
t' x'v = t
--------------- + ---------------
sqrt(1-(v/c)^2) sqrt(1-(v/c)^2)
t' +x'v
--------------- = t (equ. b)
sqrt(1-(v/c)^2)
Now, (equ. b) is in the standard form for t, and, please note,
just as Russell said up above, the expression has an x', not an
x. As long as you do not try to use magic to arrive at the
result, or as long as you do not pretend by taking the result
from a text without understanding how it was arrived at, then you
get the correct expession and you do not go around presuming to
tell others who know better, that they are wrong.
Jonathan, I bothered to write out all the little steps here, just
to show you that the result is derived from a mathematical
process and is not intuited by "visualization." This is utterly
simple high school mathematics, and aside from the likes of
Robert Winn (who still has trouble figuring out where the "1"
comes from in the Lorentz transformation) should be accessible to
most anyone who bothers to learn.
So, again, for the final time, get out of your "visualiztion"
loop and use your mind to actually learn these things
conceptually. As has been suggested to you by a dozen people, a
dozen times, read a book such as _Spacetime Physics_ and actually
learn this stuff before you presume to teach it to others.
Eli Botkin
>
> The argument at issue here is the Lorentz time-dilation equation:
>
> t'=(t-vx/c^2)/sqrt(1-(v/c)^2)
>
> I have solved for t and found that
>
> t=sqrt(1-(v/c)^2)*(t'+vx/c^2).
>
> Russell said that the x should be an x' in the equations (which is clearly
a
> algebraic error on his part) and then said he would no longer answer any
> further posts. If I reacted with a haughty tone after seeing this
behavior,
> I assure you, it was warranted.
>
>
Jonathan:
Firstly your algebraic manipulation is faulty. What you would get from
solving for t in the equation t'=(t-vx/c^2)/sqrt(1-(v/c)^2) is
t=sqrt(1-(v/c)^2)*t'+vx/c^2, not t=sqrt(1-(v/c)^2)*(t'+vx/c^2).
Secondly, so what? Transformation equations, by definition, should give you
(t',x') for input (t,x) and vice versa, not t for input (t',x). Your
equation, even when algebraically corrected, serves no useful purpose.
Eli Botkin
shuba
> The argument at issue here is the Lorentz time-dilation equation:
>
> t'=(t-vx/c^2)/sqrt(1-(v/c)^2)
>
> I have solved for t and found that
>
> t=sqrt(1-(v/c)^2)*(t'+vx/c^2).
That's nice. Did you check your work? Even if you redid it
correctly, there is a fundamental problem with what you are trying to
do. The transformation equations define (not surprisingly) a
transformation of coordinates between two relatively moving systems,
one written with primes and the other without primes. What use is an
equation for t as a function of t' and x?
A better way of thinking about the transformations, for me at least,
is by using matrices. Let's de-uglify the equation for t' above,
setting c=1 and introducing g = \gamma = 1/sqrt(1-v^2). Now we get:
t' = g*(t-vx) [eq 1]
This isn't exactly a time dilation equation, rather it gives the time
coordinate in the primed system for an event with coordinates (t,x)
in the unprimed system. The cooresponding equation for x' is
x' = g*(x-vt) [eq 2]
We now have the information needed to construct the matrix for a
Lorentz transformation from the unprimed to the primed systems.
|t'| |g -vg| |t|
| | = | | | | [matrix eq 1M]
|x'| |-vg g| |x|
Now, had you wanted to solve the original equation to find the t
coordinate in the unprimed frame for an event having (t',x') in the
primed frame, you should have also solved for x in eq 2. You'd get:
|t| |g vg| |t'|
| | = | | | | [matrix eq 2M]
|x| |vg g| |x'|
or t = g*(t' + vx') [eq 3]
x = g*(vt' + x') [eq 4]
or *if* you'd prefer to re-uglify it,
t = (t' + vx'/c^2)/sqrt(1-(v/c)^2)
x = (vt' + x')/sqrt(1-(v/c)^2)
But enough of that, let's see what equations 1M and 2M say.
First, the 2x2 transformation matrices in 1M and 2M are the same,
except for having opposite signs for the relative velocity. This
makes sense, because e.g. if you are moving south wrt me, I am moving
north wrt you. Second, the matrices are inverses of each other,
which also is a good sign intuitively. Try multiplying them and see;
their product is the 2x2 identity matrix. It turns out that all of
these so-called boosts have a transformation matrix that can be
written in a mathematically interesting form involving the hyperbolic
trig functions.
"Boosts? What the huh?", I hear someone say. It refers to Lorentz
transformations of the form given in 1M and 2M above. A more general
definition of LTs can be given, where a matrix L defines a LT if the
following holds (L^T is the transpose of L):
(L^T)(M)(L) = M [matrix eq 3M]
where M is the dreaded thing called the metric of SR, with components:
|-1 0|
| |
|0 1|
It's easy to see that eq 3M holds for the L matrices in 1M and 2M.
---Tim Shuba---
Russell Blackadar
>
> "Stephen Speicher" <s...@compbio.caltech.edu> wrote in message
> news:Pine.LNX.4.10.102011...@photon.compbio.caltech.edu...
> > On Sat, 12 Jan 2002, Jonathan Doolin wrote:
> >
> > >
> > > "Russell Blackadar" <rus...@mdli.com> wrote in message
> > > news:3C3FC7AA...@mdli.com...
> > > > Jonathan Doolin wrote:
> > > > >
> > > > >
> > > > > t=gamma*(t'+vx/c^2).
[snip]
> The argument at issue here is the Lorentz time-dilation equation:
>
> t'=(t-vx/c^2)/sqrt(1-(v/c)^2)
Correct equation, but it's NOT the "Lorentz time-dilation equation".
It's the transform.
The time dilation equation is
(t_2' - t_1') = (t_1 - t_2)/gamma
where t_1 and t_2 are times of two events occuring at the same x in
the unprimed frame, and t_1' and t_2' are the coords of those same
events in the primed frame. Of course it follows immediately from
the transform equation you wrote above (but not vice versa).
>
> I have solved for t and found that
>
> t=sqrt(1-(v/c)^2)*(t'+vx/c^2).
I have two comments about that:
(1) It's not what you wrote in your previous post. Hint: gamma is
not equal to sqrt(1 - (v/c)^2). Easy mistake for a newbie to make;
even I, a relative oldbie, occasionally flub it. A good way to
remember the correct formula is to remember that gamma is sometimes
a big number, and thus always >= 1.
(2) Your formula isn't even right! Take a moment and do your
algebra again. Take your original expression, multiply both
sides by sqrt(1-(v/c)^2), and then add vx/c^2 to both sides.
I get:
t = sqrt(1-(v/c)^2)*t' + vx/c^2
>
> Russell said that the x should be an x' in the equations (which is clearly a
> algebraic error on his part)
It should be primed in the equation of your *previous* post, the
one quoted here way at the top. And I made no mistake.
Before you jump on me again for being wrong, I suggest you reread
the textbooks you cited, and pay very careful attention to where
the primes are. For example, the following two equations are
*both* correct:
t' = (t - vx/c^2)/sqrt(1 - (v/c)^2)
t = (t' + vx'/c^2)/sqrt(1 - (v/c)^2)
The second eqn is the one that I said last time was true, and it
is! It is the inverse Lorentz transform, i.e. the transform from
primed coordinates to unprimed. You can prove it follows algebraically
with scarcely more work than you already did (incorrectly; ok I'll
stop harping on that) but it does require you to solve two equations
in two unknowns. IOW you need to plug in a rearrangement of the
transform for x, which I didn't write above but which you of course
know.
If you think for just a minute, you will see that I have to be right --
after all, why should the inverse Lorentz transform (from primed to
unprimed) be any different *in form* from the transform from unprimed
to primed? My formula just swaps t for t', and x for x', everywhere,
and reverses the sign on v which of course makes sense. It's right.
Relativity says the two frames must obey the same physics, and that
includes the transforms themselves.
I notice also you snipped without comment my calculations using the
Lorentz transform with *no* algebraic manipulations, for front and
back ship. Those calculations are correct, and they prove that you
are wrong.
and then said he would no longer answer any
> further posts. If I reacted with a haughty tone after seeing this behavior,
> I assure you, it was warranted.
Well, if you'd been reading this group longer, you'd know I often go
back on such threats. You're normally a nice guy -- an intelligent
newbie I'd call you -- so I think it's worth my while to post at least
this one time more. But *this* may be my last. ;-)
Also I'm sorry I succumbed to the temptation to twist the knife, last
time. I was hoping to teach you a lesson, but you didn't catch on.
Probably my tone got in the way. Let *that* be a lesson to you, when
you are tempted to be haughty yourself.
Russell Blackadar
> The time dilation equation is
> (t_2' - t_1') = (t_1 - t_2)/gamma
Oops, swap the 1 and 2 on the rhs, of course.
Russell Blackadar
Oops again, it should be *gamma unless I reverse the frames in the
description I made in my original post. As I said in that post, I
sometimes do confuse gamma with 1/gamma, hoary oldbie that I am.
Russell Blackadar
[snip]
You're normally a nice guy -- an intelligent
> newbie I'd call you -- so I think it's worth my while to post at least
> this one time more. But *this* may be my last. ;-)
I just remembered something else I wanted to say. And that is,
don't be discouraged that you were wrong this time. You took on
a problem that many people find confusing; e.g. Dirk van de moortell
recently got it wrong, too, and he's no dummy. (He also eventually
saw his error when it was pointed out.)
It's important to realize that two untethered rockets, accelerating
independently, behave in a different way from tethered rockets that
are constrained to move in a common frame. You don't have to unlearn
what you already know about SR; just realize that extra special care
is needed when you apply the Lorentz transform to cases of acceleration.
Russell Blackadar
of snipping my calculation without comment. (But you're still wrong.)
Yes. But you have assumed, without proof, that the ships
are in the *same* frame. They are, of course, when they start
out, at t=0. (What we have called the lab frame.) From that,
you can *calculate* whether they remain in the same frame
after they begin accelerating. It turns out they do not.
NOT the LAB FRAME.
>
> > And as I showed above, the
> > rocket's onboard clocks read tau=0 at *different* times in the
> > primed frame,
>
> The primed frame *is* the rocket's frame.
Yes, but tau is the time reading on the respective rockets' shipboard
clocks, and those clocks were synchronized in the LAB frame.
>
> > so naturally they must read tau=1 at different t'
> > times as well.
> >
> > Accelerations are tricky.
>
> Let's fix that and use a simpler problem (see below.)
>
> >
> > > t(x0)=gamma*(1)
> > > t(x1)=gamma*(1+v/c^2)>1
> > > Thus from the lab frame, the event occurs at x0 before it occurs at x1.
> > >
>
> <SNIP>
>
> We are saying the same thing here and somehow arriving at different
> conclusions.
> You are using tau when Lorentz and I use t', to represent the *proper time*
> of events on the rocketships.
No, Lorentz doesn't say anything about proper time. Proper time, when
acceleration is present, is determined from an integral. Not from a
Lorentz transform. (At least, not directly from one.)
For us, t' is simply the time coordinate of an inertial frame which we
have constructed to be comoving with the rockets after their first
burst of acceleration. Strictly speaking, we don't even know whether
both rockets are ever in this frame at the same time; in general they
will not be, but it doesn't matter for the calculation I did. (It does
matter for the calculation you tried to do, but that was wrong as I've
pointed out elsewhere.)
>
> If acceleration is tricky, let's make it a bit simpler. Have the two ships
> traveling without acceleration, and schedule to eat lunch at the same time:
> t'=1. Now, using the same math as I did above, I claim that in the lab
> frame, the crew of the rear ship will end up eating before the crew of the
> front ship. Do you disagree?
Yes I disagree. If there is no acceleration, then we are in the lab
frame still, and t' = t.
Russell Blackadar
...
> >
> > "Russell Blackadar" <rus...@REMOVEmdli.com> wrote in message
> > news:3C3E4821...@REMOVEmdli.com...
> > > They occur when the onboard clocks of the respective
> > > ships read the same *proper time*.
> >
> > You are right. This means exactly the same thing as "The events occur
> > simultaneously in the primed frame."
I should have said here, NO it does NOT mean the same thing. As should
be clear in what I wrote that followed.
The ship's *proper time* is in their
> > own frame.
>
> Yes. But you have assumed, without proof, that the ships
> are in the *same* frame.
In other words, that they are both simultaneously in the primed frame
and that their clocks are synchronized in that frame. That is a
matter to be determined by calculation, and it turns out to be false.
There are good reasons why people use tau, rather than t', for proper
time. One is that you are less likely to think you can blindly apply
the Lorentz transform to tau, which is a *local* coordinate in an
*accelerated* frame. The Lorentz transform only applies to *inertial*
frames.
Jonathan Doolin
"Eli Botkin" <ebo...@suffolk.lib.ny.us> wrote in message
news:_O608.6861$0g.1...@iad-read.news.verio.net...
> [SNIP]
> >
> > The argument at issue here is the Lorentz time-dilation equation:
> >
> > t'=(t-vx/c^2)/sqrt(1-(v/c)^2)
> >
> > I have solved for t and found that
> >
> > t=sqrt(1-(v/c)^2)*(t'+vx/c^2).
> >
> > Russell said that the x should be an x' in the equations (which is
clearly
> a
> > algebraic error on his part) and then said he would no longer answer
any
> > further posts. If I reacted with a haughty tone after seeing this
> behavior,
> > I assure you, it was warranted.
> >
> >
>
> Jonathan:
> Firstly your algebraic manipulation is faulty. What you would get from
> solving for t in the equation t'=(t-vx/c^2)/sqrt(1-(v/c)^2) is
> t=sqrt(1-(v/c)^2)*t'+vx/c^2, not t=sqrt(1-(v/c)^2)*(t'+vx/c^2).
Ooops. Sorry. That's correct. But that isn't the point I was making. You
make my point for me, below.
>
> Secondly, so what? Transformation equations, by definition, should give
you
> (t',x') for input (t,x) and vice versa, not t for input (t',x). Your
> equation, even when algebraically corrected, serves no useful purpose.
>
> Eli Botkin
>
>
>
I strongly disagree with this idea. And you may feel free to take that for
what its worth, since I only have a measley electrical engineering bachelors
degree, and I'm six hours short of a bachelors in mathematics. But Lorentz
equations can be used for any two inputs and give any two outputs.
Jonathan Doolin
There are many times that I am wrong, and I am usually terribly embarrassed.
However, you do not realize just how fundamental this particular piece is.
If I am wrong about this, then everything, *EVERYTHING* I think I know about
Special Relativity is wrong. If length contraction is not caused by the
desynchronization effect, then everything else I know falls apart--becoming
separate memorized details which do not fit into a single all-encompassing
whole.
You are saying that acceleration makes it an entirely different problem,
with complexities which I have not yet fathomed. I am saying to chop the
problem up, like you would if you were doing calculus, and it is the same
problem, but slowed down.
And absolutely *do not* pull out the space-time diagrams. Solve the problem
in your head.
The FAQ on Bell's Paradox has some problems. I know it as sure as I know
that the speed of light is constant. This really isn't one of those things
that I'm *wondering* about. It's absolute certainty, as far as I'm
concerned. I can understand, with my limited education, that you would be
skeptical of me. But for this one occasion, please turn off your PhD litmus
test and consider the arguments on their own merit.
From the perspective of the space-ships, they will begin their acceleration
simultaneously. They will end their acceleration simultaneously. They will
end their acceleration and find the distance between them to be the same.
(There is admittedly, some confusion during the acceleration. If they
analyze their launch afterwards, they will realize that from their new
perspective, it appears that the front ship left *before* the back ship.)
From the perspective of the lab, they will begin their acceleration
simultaneously, but the rear ship will continue accelerating after the front
ship has stopped. At the end, the two ships are length contracted and will
be closer together.
Jonathan Doolin
If there is a velocity t' is not equal to t.
Jonathan Doolin
"Jonathan Doolin" <spence...@earthlink.net> wrote in message
news:qD718.34732$zw3.3...@newsread1.prod.itd.earthlink.net...
You probably won't believe this, but I had a really apologetic tone in my
voice when I typed that.
Russell Blackadar
>
> "Russell Blackadar" <rus...@mdli.com> wrote in message
> news:3C435E47...@mdli.com...
> > Russell Blackadar wrote:
> >
> > [snip]
> >
> > You're normally a nice guy -- an intelligent
> > > newbie I'd call you -- so I think it's worth my while to post at least
> > > this one time more. But *this* may be my last. ;-)
> >
> > I just remembered something else I wanted to say. And that is,
> > don't be discouraged that you were wrong this time. You took on
> > a problem that many people find confusing; e.g. Dirk van de moortell
> > recently got it wrong, too, and he's no dummy. (He also eventually
> > saw his error when it was pointed out.)
> >
> > It's important to realize that two untethered rockets, accelerating
> > independently, behave in a different way from tethered rockets that
> > are constrained to move in a common frame. You don't have to unlearn
> > what you already know about SR; just realize that extra special care
> > is needed when you apply the Lorentz transform to cases of acceleration.
> >
>
> There are many times that I am wrong, and I am usually terribly embarrassed.
> However, you do not realize just how fundamental this particular piece is.
> If I am wrong about this, then everything, *EVERYTHING* I think I know about
> Special Relativity is wrong.
No! And that is exactly why I wrote this clarification to my
earlier post. You need to see it otherwise. You know a lot about
relativity, except you've made a mistake (which, amazingly, you
still don't see) in applying what you know. That most definitely
does not mean throwing out everything.
If length contraction is not caused by the
> desynchronization effect, then everything else I know falls apart--becoming
> separate memorized details which do not fit into a single all-encompassing
> whole.
Length contraction *does* occur in this problem. Your mistake was
to assume that the rockets don't get physically further apart. If
they didn't, they couldn't remain the same distance apart in the
lab frame under length contraction.
>
> You are saying that acceleration makes it an entirely different problem,
> with complexities which I have not yet fathomed.
Not at all. A trickier problem, yes. But you know all of the
machinery already, to do it right. Your many algebraic errors,
and the fact that you did not recognize the inverse Lorentz
transform for what it was, suggest to me that you jumped to a
conclusion and would reach a different one if you started the
entire calculation over, and did it with great care. Start
with what you *know*, and work out what you don't know.
I am saying to chop the
> problem up, like you would if you were doing calculus, and it is the same
> problem, but slowed down.
>
> And absolutely *do not* pull out the space-time diagrams. Solve the problem
> in your head.
NO! Do it with algebra on paper. Your head is playing tricks
on you; it's not reliable. Do it carefully, and don't drop any
primes.
>
> The FAQ on Bell's Paradox has some problems. I know it as sure as I know
> that the speed of light is constant. This really isn't one of those things
> that I'm *wondering* about. It's absolute certainty, as far as I'm
> concerned.
A little humility would help you right about now. There are a
*lot* of people looking at these FAQ pages. You seriously think
there is a fundamental error in one of them, that you are the
first to find? It can happen, of course, but the odds are such
that you really ought to put all of your mathematical tools to
work, explicitly, on paper, to make sure it's not you who are
wrong.
I can understand, with my limited education, that you would be
> skeptical of me. But for this one occasion, please turn off your PhD litmus
> test and consider the arguments on their own merit.
Why do you assume that's *not* what I've been doing? Btw I don't
have a PhD myself, and I agree completely with you that it's
irrelevant to the issue.
>
> From the perspective of the space-ships, they will begin their acceleration
> simultaneously. They will end their acceleration simultaneously. They will
> end their acceleration and find the distance between them to be the same.
Why? Calculate it. You know how far they are apart in the lab
frame at t=0. You know the Lorentz transform. So, get to work!
Don't wave your hands.
> (There is admittedly, some confusion during the acceleration. If they
> analyze their launch afterwards, they will realize that from their new
> perspective, it appears that the front ship left *before* the back ship.)
>
> From the perspective of the lab, they will begin their acceleration
> simultaneously, but the rear ship will continue accelerating after the front
> ship has stopped.
You've changed your tune, I see. You began by saying it started
accelerating first, but that is obviously wrong. This latest attempt
of yours, to save your idea, won't fly either. Do the calculation.
At the end, the two ships are length contracted and will
> be closer together.
In the lab frame, they will be measured to be closer together than
their "real" distance apart (as measured in their own frame). So
you see, I'm saying that length contraction occurs exactly as
advertised. (IOW you don't have to give up anything you know!!)
But since their real distance apart is now gamma*k, it happens that
in the lab frame, with length contraction, their distance is measured
to be k.
Russell Blackadar
>
> "Eli Botkin" <ebo...@suffolk.lib.ny.us> wrote in message
> news:_O608.6861$0g.1...@iad-read.news.verio.net...
> > [SNIP]
> > >
> > > The argument at issue here is the Lorentz time-dilation equation:
> > >
> > > t'=(t-vx/c^2)/sqrt(1-(v/c)^2)
> > >
> > > I have solved for t and found that
> > >
> > > t=sqrt(1-(v/c)^2)*(t'+vx/c^2).
> > >
> > > Russell said that the x should be an x' in the equations (which is
> clearly
> > a
> > > algebraic error on his part) and then said he would no longer answer
> any
> > > further posts. If I reacted with a haughty tone after seeing this
> > behavior,
> > > I assure you, it was warranted.
> > >
> > >
> >
> > Jonathan:
> > Firstly your algebraic manipulation is faulty. What you would get from
> > solving for t in the equation t'=(t-vx/c^2)/sqrt(1-(v/c)^2) is
> > t=sqrt(1-(v/c)^2)*t'+vx/c^2, not t=sqrt(1-(v/c)^2)*(t'+vx/c^2).
>
> Ooops. Sorry. That's correct. But that isn't the point I was making.
Jonathan, the point is you were defending an incorrect formula, and
I told you one perfectly correct way to fix it. The incorrect formula
that you wrote was crucial to your whole argument; therefore your
argument has collapsed and you have to start over. IOW you made hand
waving arguments; I called you on it; you responded with some math;
and I have shown that your math is wrong. Ball is in your court.
You
> make my point for me, below.
How so?
>
> >
> > Secondly, so what? Transformation equations, by definition, should give
> you
> > (t',x') for input (t,x) and vice versa, not t for input (t',x). Your
> > equation, even when algebraically corrected, serves no useful purpose.
Right. (His point, Jonathan, is usefulness -- not mere correctness.)
> >
> > Eli Botkin
> >
> >
> >
>
> I strongly disagree with this idea. And you may feel free to take that for
> what its worth, since I only have a measley electrical engineering bachelors
> degree, and I'm six hours short of a bachelors in mathematics. But Lorentz
> equations can be used for any two inputs and give any two outputs.
Of course. But for them to be useful, the two inputs must be known,
and presumably you are using the the equations to determine two
unknowns. Normally the knowns are both coordinates in the *same*
frame. Can you think of any practical calculations you might do,
where you know only the x coordinate of an event in one frame
(but not its t in that frame) and only the time coordinate in a
*different* frame (and not its position in that frame)? I guess
with some effort you could, but they would be pretty contrived
I think.
Russell Blackadar
>
> Jonathan Doolin wrote:
> >
> > "Russell Blackadar" <rus...@mdli.com> wrote in message
[snip]
> > From the perspective of the space-ships, they will begin their acceleration
> > simultaneously. They will end their acceleration simultaneously. They will
> > end their acceleration and find the distance between them to be the same.
>
> Why? Calculate it. You know how far they are apart in the lab
> frame at t=0. You know the Lorentz transform. So, get to work!
> Don't wave your hands.
Here's another hint. The first two sentences you wrote above are
correct; your problem is that you believe, incorrectly, that the
third must follow from them. If you try to put that belief into
explicit math and show that it is really true, you will find there
is a problem.
(Oh yes, the third sentence *does* follow in Galilean relativity.)
>
> > (There is admittedly, some confusion during the acceleration. If they
> > analyze their launch afterwards, they will realize that from their new
> > perspective, it appears that the front ship left *before* the back ship.)
"Some confusion" can actually be described quite well by known
mathematics; it's done in the FAQ. You can supply the missing
piece (derivation of the relativistic rocket eqn) by reading other
sources, as I've said.
But you don't have to try to figure that out; we both agree that
the essence of the problem can be seen quite well if we approximate
the process by a sequence of burstwise accelerations, i.e. consider
the velocity to be a step function. In fact, you only need to consider
*one* burst, to see what will happen in general.
[snip]
Russell Blackadar
>
> "Russell Blackadar" <rus...@mdli.com> wrote in message
> news:3C43662F...@mdli.com...
> > Sorry, I didn't see this post of yours earlier. I falsely accused you
> > of snipping my calculation without comment. (But you're still wrong.)
> >
> > Jonathan Doolin wrote:
[snip]
> > > If acceleration is tricky, let's make it a bit simpler. Have the two
> ships
> > > traveling without acceleration, and schedule to eat lunch at the same
> time:
> > > t'=1. Now, using the same math as I did above, I claim that in the lab
> > > frame, the crew of the rear ship will end up eating before the crew of
> the
> > > front ship. Do you disagree?
> >
> > Yes I disagree. If there is no acceleration, then we are in the lab
> > frame still, and t' = t.
> >
>
> If there is a velocity t' is not equal to t.
If there is a velocity, then acceleration has occurred! Don't
forget, we are given that the rockets start out in the lab frame,
with clocks synchronized in that frame.
(Or, if you insist on changing the problem, then don't expect anything
I've said to apply to that different problem.)
Jonathan Doolin
I left out an important piece of my argument. Lorentz equations have an
intersection point. The (x',t') refrence frame meets the (x,t) reference
frame at (0,0). I don't know what to call this intersection point, so I'll
call it a pseudo-event. A point in space and time where an event may or may
not have happened.
Below is a nice contrived example of a practical calculation you might do
where one knows the x coordinate in one frame and the t coordinate in
another:
Just for arguments sake, lets say an event does occur at this point in
space-time. Let's say the front wheel of a train passes over a rail-road
tie at this intersection point.
Let's say we know the length of the train, L. And we know the time, T=0.
The length of the train should be lorentz contracted, therefore we know that
x'=L, and t=0.
From this information, we can calculate the position of the back of the
train in our reference frame, and the time one would see on a clock at the
back of the train, (which would not match the clock at the front of the
train from our reference frame)
= = = = = = =
Jon Doolin
P.S.
I created a demo to show Lorentz equations, after I realized that my
"animated space-time diagram" did not preserve the intersection-point,
(x',t')=(x,t)=(0,0) which is vital to Lorentz equations. You may find it
less confusing.
http://home.earthlink.net/~spencerdoolin/relativity/velrotation.swf
In the demo, the horizontal axis represents a freeze-frame of a physical
object. The three time axes intersect the horizontal and show the different
times along the hypothetical train. Though t0=0, t1=0, t2=0, The times on
the clocks in the other reference frame depend on the velocity of the
object.
Jonathan Doolin
> Russell Blackadar wrote:
> >
> > Jonathan Doolin wrote:
> > >
> > > "Russell Blackadar" <rus...@mdli.com> wrote in message
>
> [snip]
>
> > > From the perspective of the space-ships, they will begin their
acceleration
> > > simultaneously. They will end their acceleration simultaneously.
They will
> > > end their acceleration and find the distance between them to be the
same.
> >
> > Why? Calculate it. You know how far they are apart in the lab
> > frame at t=0. You know the Lorentz transform. So, get to work!
> > Don't wave your hands.
How should I calculate it? I would have to develop a model to approximate
the acceleration in order to figure out what is going on mathematically.
Here are three equivalent models that I am willing to use. Which one do you
prefer? Or are all of these unsatisfactory?
= = =
1. The two ships attach themselves by a fairly rigid pole. An engine at
the center of the pole accelerates the two ships for one minute, assuring
that both ships experience the same acceleration.
2. An accelerating surface, such as the side of a much larger rocketship
passes by. Both ships synchronize their clocks, grab hold of the rockets
surface as it matches speed with them, hold on, and release as soon as their
clocks read one minute.
3. From a central location send out an accelerating beacon
which will send out the signals which will tell the other ships to
accelerate. In this case the ships would not get closer in their own
reference frame, but they would be seen to get closer from outside. You
should be able to verify this mentally.
= = =
>
> Here's another hint. The first two sentences you wrote above are
> correct; your problem is that you believe, incorrectly, that the
> third must follow from them. If you try to put that belief into
> explicit math and show that it is really true, you will find there
> is a problem.
>
> (Oh yes, the third sentence *does* follow in Galilean relativity.)
>
> >
> > > (There is admittedly, some confusion during the acceleration. If they
> > > analyze their launch afterwards, they will realize that from their new
> > > perspective, it appears that the front ship left *before* the back
ship.)
>
> "Some confusion" can actually be described quite well by known
> mathematics; it's done in the FAQ. You can supply the missing
> piece (derivation of the relativistic rocket eqn) by reading other
> sources, as I've said.
>
> But you don't have to try to figure that out; we both agree that
> the essence of the problem can be seen quite well if we approximate
> the process by a sequence of burstwise accelerations, i.e. consider
> the velocity to be a step function. In fact, you only need to consider
> *one* burst, to see what will happen in general.
>
> [snip]
>
As long as you don't require me to use the *first* burst as the one burst,
this is trivial, and you already know what I would say: For each burst, in
the lab frame, the back rocket moves first, causing the rockets to come
closer together.
If you want to use the *first* burst, then it becomes a different problem,
and I'm in over my head. However, I can still wave my hands above the
water: If I am not mistaken, the uncertainty principle: (delta)E(delta)T>=
h/(4Pi).) indicates that the burst must take a finite amount of time. If so,
the back rocket will finish its acceleration first, causing the rockets to
come closer together.
On the other hand, if I consider short bursts of infinite acceleration, it
might be that I would arrive at your conclusion after all. As a matter of
fact, if you *only* consider the first acceleration--say a jump from 0 to
80% of the speed of light in an extremely short amount of time (on the order
of
t=h/(E*4Pi) ~ 0) the two ships *would* stay at the same distance in the lab
frame and *would* be further apart in their own reference frame. Do you
know what? This may be the key of our disagreement.
http://home.earthlink.net/~spencerdoolin/relativity/length.swf (This demo
is related, because it shows infinite acceleration, though not completely
accurately)
Russell Blackadar
>
> "Russell Blackadar" <rus...@mdli.com> wrote in message
> news:3C45F9A2...@mdli.com...
> > Russell Blackadar wrote:
> > >
> > > Jonathan Doolin wrote:
> > > >
> > > > "Russell Blackadar" <rus...@mdli.com> wrote in message
> >
> > [snip]
> >
> > > > From the perspective of the space-ships, they will begin their
> acceleration
> > > > simultaneously. They will end their acceleration simultaneously.
> They will
> > > > end their acceleration and find the distance between them to be the
> same.
> > >
> > > Why? Calculate it. You know how far they are apart in the lab
> > > frame at t=0. You know the Lorentz transform. So, get to work!
> > > Don't wave your hands.
>
> How should I calculate it? I would have to develop a model to approximate
> the acceleration in order to figure out what is going on mathematically.
> Here are three equivalent models that I am willing to use. Which one do you
> prefer? Or are all of these unsatisfactory?
> = = =
> 1. The two ships attach themselves by a fairly rigid pole. An engine at
> the center of the pole accelerates the two ships for one minute, assuring
> that both ships experience the same acceleration.
That changes the problem.
I already told you, if the rockets are tethered together and
the tether does not permanently deform, the system behaves as
you say. (But in that case, the rockets don't really end up
accelerating at exactly the same rate as measured by their
onboard accelerometers. The tether influences their motion.)
>
> 2. An accelerating surface, such as the side of a much larger rocketship
> passes by. Both ships synchronize their clocks, grab hold of the rockets
> surface as it matches speed with them, hold on, and release as soon as their
> clocks read one minute.
That's the same as your #1. The big surface serves as your
pole.
>
> 3. From a central location send out an accelerating beacon
> which will send out the signals which will tell the other ships to
> accelerate. In this case the ships would not get closer in their own
> reference frame, but they would be seen to get closer from outside. You
> should be able to verify this mentally.
Use paper, Jonathan!!! Take it from me, don't trust your
mind to do it unaided. If you have a third rocket supposedly
accelerating along with the other two, then you have only
created two more problems, namely, how does the distance
between outer rockets and central rocket change with time.
But why can't you use the first burst? Do you deny that it
occurs?
>
> If you want to use the *first* burst, then it becomes a different problem,
> and I'm in over my head.
Not at all! The only problem is that the answer you get
is not the one you believe must be true. Your belief is
the problem. Furthermore, you are definitely not in over
your head; you prove below that you can get the right
answer.
However, I can still wave my hands above the
> water: If I am not mistaken, the uncertainty principle: (delta)E(delta)T>=
> h/(4Pi).) indicates that the burst must take a finite amount of time. If so,
> the back rocket will finish its acceleration first, causing the rockets to
> come closer together.
You are mistaken. And you are clutching at straws. The
theory of relativity does not depend on the Uncertainty
Principle in any way. It's about the geometry of spacetime.
It is classical.
What you are saying, basically, is that it is not valid to
do calculus (which deals with limits of the very small) in
relativity. Well, I guess if that were true you could still
have special relativity -- though it begs the question what
rules you would use for acceleration if you cannot use
calculus. But general relativity would be out the window.
Not likely, is it?
>
> On the other hand, if I consider short bursts of infinite acceleration, it
> might be that I would arrive at your conclusion after all.
Yes! You are coming around. Good show.
As a matter of
> fact, if you *only* consider the first acceleration--say a jump from 0 to
> 80% of the speed of light in an extremely short amount of time (on the order
> of
> t=h/(E*4Pi) ~ 0) the two ships *would* stay at the same distance in the lab
> frame and *would* be further apart in their own reference frame. Do you
> know what? This may be the key of our disagreement.
Indeed. Not to put too fine a point on it, it's the key to
why you are *wrong*.
I said nothing about going 80% of the speed of light. And the
infinite acceleration was just a convenient fiction, but you
don't need it at all; the equations are the same in the limit
as you go from finite step function to an infinitesimal. As a
soon-to-be math grad, you should be able to see what's going on
with all that and phrase it rigorously in terms of epsilons and
deltas if you really want to.
But, if you still don't believe me, just try a believable real
acceleration on a pair of ships separated by a big distance,
let's say 10^16 m. Suppose they each get a modest, constant
acceleration of 100 m/s^2 (about 10G's) for exactly one second,
starting at t=0. At the end, they are moving 100 m/s in the lab
frame. The front rocket is at x = (10^16 + 50)m and the rear
rocket is at x = 50m.
(You can correct these for relativity if you like, but you will
need to use the equations from the FAQ, which you don't believe!
Hmm. Well, in any case the correction is going to be small,
because we're only talking 100 m/s. It's a second-order effect,
so let's ignore it. BTW don't come to me and say it's really
some big number compared to 10^16m or even 50m. Do you really
think the lead rocket will be observed to *jump* to a totally new
position in the lab frame, i.e. enough to make a difference in
the calculation I give below?)
Okay, now use the Lorentz transform to get the coordinates in the
frame moving at 100m/s wrt the lab frame. t = 1 in the lab frame
for both rockets; x_1 = 10^16 + 50 and x_2 = 50. Therefore:
t'_1 = (1 - 100*(10^16+50)/(3*10^8)^2)*gamma = -11.1
t'_2 = (1 - 100*50/(3*10^8)^2)*gamma = 1.0
In that frame, the front rocket *finishes* its acceleration more
than 11 seconds before the rear rocket even *begins* to accelerate!
And we didn't do anything at all funny.
Please don't try to tell me relativity works one way if the rockets
are 10^16m apart and a different way if they are closer together.
>
> http://home.earthlink.net/~spencerdoolin/relativity/length.swf (This demo
> is related, because it shows infinite acceleration, though not completely
> accurately)
It's an ok demo, but it's a different problem because the
ships are not starting their accelerations at the same time
in the lab frame. What you want is to have two "red lines",
one for each ship, moving in parallel. And of course when
you do that, you will see each ship get contracted, but
their distance apart will remain the same (in the lab
frame). A good mental picture for you to think about.
Russell Blackadar
[snip]
> Okay, now use the Lorentz transform to get the coordinates in the
> frame moving at 100m/s wrt the lab frame. t = 1 in the lab frame
> for both rockets; x_1 = 10^16 + 50 and x_2 = 50. Therefore:
> t'_1 = (1 - 100*(10^16+50)/(3*10^8)^2)*gamma = -11.1
Oops, make that -10.1.
> t'_2 = (1 - 100*50/(3*10^8)^2)*gamma = 1.0
>
> In that frame, the front rocket *finishes* its acceleration more
> than 11 seconds before the rear rocket even *begins* to accelerate!
Actually only 10 seconds before. ;-)
My point is that no matter how long your "infinitesimal" acceleration
element takes to complete -- and I agree with you that by Heisenberg
there is in reality some question about that time -- you can find
a distance of separation for the rockets that totally swamps any
significance that that uncertain time might have. The calculation
works out the same as if it were instantaneous. In QM terms this is
simply a manifestation of the correspondence principle. It says we
don't expect QM to matter at all for large rockets shooting many
particles out in their thrust. A continuous approximation will work
fine.
Larry Richardson
> From the perspective of the space-ships, they will begin their acceleration
> simultaneously. They will end their acceleration simultaneously. They will
> end their acceleration and find the distance between them to be the same.
> (There is admittedly, some confusion during the acceleration. If they
> analyze their launch afterwards, they will realize that from their new
> perspective, it appears that the front ship left *before* the back ship.)
From the perspective of an observer in the ships' intitial frame, immediately
upon their simultaneous (in that frame only) initiation of acceleration, their
individual worldlines and lines of simultaneity will commence shearing toward
one another. Their individual lines of simultaneity will remain parallel from
that perspective, but they won't remain congruent. This will result in a time
offset among their lines of simultaneity which the occupants of the ships
themselves will experience and the original-frame observer can corroborate. In
essense, each ship's line of simultaneity will intersect a ship ahead of it in
the direction of motion at a slightly later time than will the original-frame
observer's line of simultaneity, so that ship is measured to be further advanced
in the motion direction than the original-frame observer would say it is.
Similarly, each ship's line of simultaneity will intersect a ship behind it at a
slightly earlier time than will the original-frame observer's line of
simultaneity, so that ship is measured to be less advanced in the motion
direction than the original-frame observer would say it is. The separation of
the ships increases for the occupants of the ships themselves precisely because
of the relativity of simultaneity (the way that each differently-moving observer
divides spacetime into space and time).
LR
Russell Blackadar
followup to my own article. Btw I also endorse Larry Richardson's
excellent contribution to this thread, wherein he very clearly
reiterates a point made by the FAQ.
Russell Blackadar wrote:
>
> Jonathan Doolin wrote:
[snip]
> > As long as you don't require me to use the *first* burst as the one burst,
> > this is trivial, and you already know what I would say: For each burst, in
> > the lab frame, the back rocket moves first, causing the rockets to come
> > closer together.
And I replied
>
> But why can't you use the first burst? Do you deny that it
> occurs?
Also I shouldn't have let you get away with calling the other
bursts trivial. Triviality is in the eye of the beholder,
and unfortunately your eye is not seeing this problem right.
*All* of the acceleration bursts are like the first, in that
each results in the front rocket's increasing its lead over
the rear. The front rocket always enters the "next" inertial
frame before the rear rocket does (in the coords of that next
frame) and by an amount that increases cumulatively. You would
see this if you went to the trouble of calculating a few cases,
but you have to start at #1, hence my comment.
>
> >
> > If you want to use the *first* burst, then it becomes a different problem,
> > and I'm in over my head.
>
> Not at all! The only problem is that the answer you get
> is not the one you believe must be true. Your belief is
> the problem. Furthermore, you are definitely not in over
> your head; you prove below that you can get the right
> answer.
>
> However, I can still wave my hands above the
> > water: If I am not mistaken, the uncertainty principle: (delta)E(delta)T>=
> > h/(4Pi).) indicates that the burst must take a finite amount of time. If so,
> > the back rocket will finish its acceleration first, causing the rockets to
> > come closer together.
>
> You are mistaken. And you are clutching at straws. The
> theory of relativity does not depend on the Uncertainty
> Principle in any way. It's about the geometry of spacetime.
> It is classical.
[snip]
Moreover, as I neglected to say, you never show why a finite
time of acceleration favors your position in any way. I gave
you an example of an obviously classical acceleration which
occurs over a finite time, and which supports my position
unambiguously.
I also neglected the real reason why your invoking Heisenberg
is so bizarre and so wrong. What Heisenberg does is to replace
the classical prediction of a single number -- say, the time
at which a particle of some specific energy is emitted -- by
a statistical distribution: a Gaussian, centered about the
classically predicted value. I argued, essentially, that the
width of the Gaussian is too small to matter here, which is
true. But a more salient point is that, whatever the width,
the predicted mean value of a lot of measurements is the *same*
as the classical prediction. You're trying to turn uncertainty
into a bias in your favor, but that's not at all how it works.
Finally, I want to clarify my position about not doing the
calculation in your head. When you have worked through enough
problems dealing with acceleration, you will probably develop
a good intuition for them and can once again shoot from the
hip. Perhaps soon! But not yet. Keep thinking of that mod
to your demo that I suggested at the end of my previous article
(here snipped). The mod is necessary if you want to solve this
*current* problem, which is a *different* problem from one shown
correctly in your demo.
Russell Blackadar
>
> "Russell Blackadar" <rus...@mdli.com> wrote in message
> news:3C45E539...@mdli.com...
> > Jonathan Doolin wrote:
> > >
> > > "Eli Botkin" <ebo...@suffolk.lib.ny.us> wrote in message
> > > news:_O608.6861$0g.1...@iad-read.news.verio.net...
[assertion that *useful* Lorentz equations have all primed,
or all unprimed, coordinates on the rhs, and Jonathan's
disagreement with that assertion]
> I left out an important piece of my argument. Lorentz equations have an
> intersection point. The (x',t') refrence frame meets the (x,t) reference
> frame at (0,0). I don't know what to call this intersection point, so I'll
> call it a pseudo-event.
It's an event. An event (in the context of SR) is simply by
definition a point in spacetime. Nothing particular has to
happen there/then.
A point in space and time where an event may or may
> not have happened.
>
> Below is a nice contrived example of a practical calculation you might do
> where one knows the x coordinate in one frame and the t coordinate in
> another:
>
> Just for arguments sake, lets say an event does occur at this point in
> space-time. Let's say the front wheel of a train passes over a rail-road
> tie at this intersection point.
>
> Let's say we know the length of the train, L. And we know the time, T=0.
>
> The length of the train should be lorentz contracted, therefore we know that
> x'=L, and t=0.
OK, I think I follow you, but in a way you've already
proved Eli's point. When you say it "should be lorentz
contracted" you are claiming that you already know x=L/gamma
at the event you are considering (the back end of the train
at track time t=0). IOW you can't claim that you don't know
x prior to x'. You wrote down x'=L by doing what amounts to
a Lorentz transform in your head!
Another way to say it is, how did you know that x'=L and t=0
are simultaneously true for an event at which the back end
of the train is present? It does happen to be true, but
either you got lucky, or you did an implicit Lorentz transform.
But I am sure we can argue for days on this, and what a silly
argument that would be, so I won't press the point.
>
> From this information, we can calculate the position of the back of the
> train in our reference frame, and the time one would see on a clock at the
> back of the train, (which would not match the clock at the front of the
> train from our reference frame)
Right, no argument with you there.
>
> = = = = = = =
> Jon Doolin
> P.S.
>
> I created a demo to show Lorentz equations, after I realized that my
> "animated space-time diagram" did not preserve the intersection-point,
> (x',t')=(x,t)=(0,0) which is vital to Lorentz equations. You may find it
> less confusing.
> http://home.earthlink.net/~spencerdoolin/relativity/velrotation.swf
> In the demo, the horizontal axis represents a freeze-frame of a physical
> object. The three time axes intersect the horizontal and show the different
> times along the hypothetical train. Though t0=0, t1=0, t2=0, The times on
> the clocks in the other reference frame depend on the velocity of the
> object.
Sorry, I've no time to check it out. It's probably fine.
While I have your attention, though, I just stumbled onto a quote
from John Baez (a true relativity guru, in case you didn't know)
in sci.physics.research, Message-ID: <a0kv72$9fk$1...@glue.ucr.edu>
that is relevant to our ongoing discussion:
| take
| two rockets at rest, one in front of the other, and connect
| them by a taut string. Fire them up simultaneously and let
| them both maintain the same constant blast of exhaust, so that
| they both "accelerate at the same constant rate" as measured
| by little guys with accelerometers sitting at the tail pipes.
|
| Then the string connecting them will stretch and ultimately snap!
|
| I remember being aghast at this when I first learned it. Now
| it seems perfectly obvious, with the help of a spacetime diagram.
You seem to be stuck at the aghast state. Time to move on to
enlightenment! ;-)
Russell Blackadar
a simple argument occurred to me, that shows how wrong Jonathan
Doolin's position is.
Build a linear accelerator pointed west-to-east in Stanford, CA,
and an identical one in Brookhaven, NY, also pointed west-to-east.
Start an electron in the Stanford accelerator at t = 00:00:00 UTC,
and at the same time, start an electron at Brookhaven. Some
nanoseconds later, let's say the Stanford electron is moving at
a modest gamma=2. It's gone down the tube a bit, but obviously
it is still in Stanford. Similarly, the Brookhaven electron is
moving at gamma=2 sometime around then. (Obviously, *exactly*
then, neglecting Earth's rotation, but that will only be obvious
to Jonathan sometime real soon.) My question is merely, *where*
is the Brookhaven electron at that time?
By Doolin's line of reasoning, we should be looking for it
somewhere in Kansas.
Jonathan Doolin
Thanks for this example. Sorry I took so long responding. I have been kind
of busy, and this example kind of threw me for a loop.
Two electrons in linear accelerators separated by a long distance are fired
at the same time. Your argument is that these two electrons will retain
their distance in our reference frame, and be separated by a much greater
distance in their own. I agree.
Lets set up your argument with some numbers and see what is going on from
the perspective of the final electron reference frame.
You have two electrons separated by 1000 miles (Approximately 5 light
milliseconds)
Both electrons are placed in 1000 foot linear accelerators and accelerated
from 0 to .866c in about a microsecond.
From the perspective of the initial electron reference frame, (the lab
frame), the two accelerations take place simultaneously.
By symmetry, (and this is the point I was missing before), the two electrons
are undergoing the exact same acceleration.
However, from the perspective of the final electron reference frame, the
front electron gun is only 500 miles from the rear gun, and was fired 8.6
milliseconds (.866c*8.6 milliseconds~7.5 million feet=1500 miles) before
the rear gun. The front electron ends up being 2000 miles ahead of the
rear electron instead of the original 1000 miles (assuming straight line
movement).
(Why 8.6 milliseconds?
c=Approximately 1 million feet/ millisecond.
1000 miles ~5 million feet
v=.866 million feet/millisecond, x=5 million miles
desynchronization=(vx/c^2)/sqrt(1-(v/c)^2)
0.866*5 / 0.5
(This is the desynchronization part of the Lorentz equations.)
Now I could point out that if three spaceships underwent this kind of
acceleration, (From 0 to .866c in a matter of microseconds) everyone aboard
the ships would be splattered against the floor. You could, however,
increase the distance between the ships, and lower the acceleration, and
make the argument sound again.
But regardless, there is a big difference between my formulation of this
problem and your formulation of this problem:
You have been assuming that the distance between the ships is large and/or
the acceleration is large. This brings into effect relativistic tidal
forces causing the ships to drift apart. Regardless of the distance, this
tidal force will be present for any nonzero acceleration.
However, at small distances and small accelerations, (100 miles and 1
gravity, for instance) the tidal effect is not as great. (There is time
between acceleration impulses for the two ships to re-synchronize.)
I have been assuming that the distance between the ships was small (within a
mile or so), and the acceleration was small (around 10 m/s^2 or so).
Though I didn't realize I was doing this, two of the three possible
formulations I developed (ships tethered together, ships glomming onto a
larger accelerating spaceship) assured that the tidal effect was small.
However, the third formulation (accelerating central ship broadcasting
signals to execute acceleration impulses) would result in the tidal
expansion (non- Lorentz contraction) if the signals were sent close enough
together. However, if the acceleration signals were far enough apart,
Lorentz contraction would occur.
Jonathan Doolin
Russell Blackadar
>
> "Russell Blackadar" <rus...@mdli.com> wrote in message
> news:3C4C740F...@mdli.com...
> > In case anyone is still reading this thread, over the weekend
> > a simple argument occurred to me, that shows how wrong Jonathan
> > Doolin's position is.
> >
> > Build a linear accelerator pointed west-to-east in Stanford, CA,
> > and an identical one in Brookhaven, NY, also pointed west-to-east.
> > Start an electron in the Stanford accelerator at t = 00:00:00 UTC,
> > and at the same time, start an electron at Brookhaven. Some
> > nanoseconds later, let's say the Stanford electron is moving at
> > a modest gamma=2. It's gone down the tube a bit, but obviously
> > it is still in Stanford. Similarly, the Brookhaven electron is
> > moving at gamma=2 sometime around then. (Obviously, *exactly*
> > then, neglecting Earth's rotation, but that will only be obvious
> > to Jonathan sometime real soon.) My question is merely, *where*
> > is the Brookhaven electron at that time?
> >
> > By Doolin's line of reasoning, we should be looking for it
> > somewhere in Kansas.
> >
>
> Thanks for this example. Sorry I took so long responding. I have been kind
> of busy, and this example kind of threw me for a loop.
As well it should. I am sorry I didn't think of it in my
very first posting; the thread would have ended sooner and
you'd have had less opportunity to display your foolishness.
(And I'd have looked less like a chump.)
Actually I had thought you felt too chastened to respond,
so in a way I regretted being so hard on you. I don't want
to dampen your enthusiasm for the subject, or to stop you
from contributing to this group if you find that enjoyable.
As I said before, you know a good deal about SR, and you
do make valuable contributions to the group. But you don't
know *everything* about SR, and it seems we've identified a
sore spot here, where you need some work.
>
> Two electrons in linear accelerators separated by a long distance are fired
> at the same time. Your argument is that these two electrons will retain
> their distance in our reference frame, and be separated by a much greater
> distance in their own. I agree.
>
> Lets set up your argument with some numbers and see what is going on from
> the perspective of the final electron reference frame.
>
> You have two electrons separated by 1000 miles (Approximately 5 light
> milliseconds)
> Both electrons are placed in 1000 foot linear accelerators and accelerated
> from 0 to .866c in about a microsecond.
>
> From the perspective of the initial electron reference frame, (the lab
> frame), the two accelerations take place simultaneously.
>
> By symmetry, (and this is the point I was missing before), the two electrons
> are undergoing the exact same acceleration.
Exactly. And the rockets are no different. Just slower.
>
> However, from the perspective of the final electron reference frame, the
> front electron gun is only 500 miles from the rear gun, and was fired 8.6
> milliseconds (.866c*8.6 milliseconds~7.5 million feet=1500 miles) before
> the rear gun. The front electron ends up being 2000 miles ahead of the
> rear electron instead of the original 1000 miles (assuming straight line
> movement).
Yes. The distance increases in the final frame,
exactly as I've been saying all along. You have
finally arrived at the correct answer for the
rockets; you just don't see yet that you have.
(Btw, why did you say "however"? Are you saying
anything that wouldn't be true for rockets of
sufficient power, under the same circumstances?)
>
> (Why 8.6 milliseconds?
> c=Approximately 1 million feet/ millisecond.
> 1000 miles ~5 million feet
> v=.866 million feet/millisecond, x=5 million miles
> desynchronization=(vx/c^2)/sqrt(1-(v/c)^2)
> 0.866*5 / 0.5
> (This is the desynchronization part of the Lorentz equations.)
>
> Now I could point out that if three spaceships underwent this kind of
> acceleration, (From 0 to .866c in a matter of microseconds) everyone aboard
> the ships would be splattered against the floor. You could, however,
> increase the distance between the ships, and lower the acceleration, and
> make the argument sound again.
The argument is sound. The physical properties of
human beings are not relevant to the argument. Let
them splatter.
(A joke! I'm not really that heartless.)
>
> But regardless, there is a big difference between my formulation of this
> problem and your formulation of this problem:
>
> You have been assuming that the distance between the ships is large and/or
> the acceleration is large. This brings into effect relativistic tidal
> forces causing the ships to drift apart. Regardless of the distance, this
> tidal force will be present for any nonzero acceleration.
Where did these "tidal forces" come from? You invented
them. As it happens, there are no tidal forces at all
in either of the accelerated frames (the gravitational
field in both is uniform and constant), and in any case
we weren't even using those frames for our analysis. You
are trying to save your wrong belief (and perhaps your
pride?) with buzzwords.
>
> However, at small distances and small accelerations, (100 miles and 1
> gravity, for instance) the tidal effect is not as great. (There is time
> between acceleration impulses for the two ships to re-synchronize.)
Re-sychronize? Do you mean that somebody on board
one of the ships resets its clock? Well then of course
you could get the result you wanted, but that would be
changing the problem, wouldn't it. It makes no difference
at all how far apart the ships are, if you're going to
reset the clocks. The difference between your formulation
and mine is not the distance, it's the fact that you are
resetting clocks.
Btw, you are wrong if you think that your reformulation
doesn't have big numbers in it. If you make the distance
and acceleration small, all that does is make the *time*
greater for the ships to reach a velocity such that any
measurable difference will appear. But eventually, a
difference will become measurable. You can make that
difference disappear (in a calculation) if you ignore all
of the tiny differences accumulated along the way, at
each pulse, but that would simply be a mistaken calculation.
Remember, in spacetime, locality involves time as well as
space. You can't invoke locality if your experiment spans
many seconds of time.
>
> I have been assuming that the distance between the ships was small (within a
> mile or so), and the acceleration was small (around 10 m/s^2 or so).
> Though I didn't realize I was doing this, two of the three possible
> formulations I developed (ships tethered together, ships glomming onto a
> larger accelerating spaceship) assured that the tidal effect was small.
Tethering the ships together is a *different* problem.
It affects how one or the other of the ships accelerates.
Apparently you still don't understand this. Go back and
reread the passage I quoted from John Baez: the string
breaks. That's caused by a real force. If the string is
strong enough to resist that force, it affects the rockets.
>
> However, the third formulation (accelerating central ship broadcasting
> signals to execute acceleration impulses) would result in the tidal
> expansion (non- Lorentz contraction) if the signals were sent close enough
> together. However, if the acceleration signals were far enough apart,
> Lorentz contraction would occur.
There are no tides here, unless you insist upon using
the accelerated frame. (But if you do use that frame, the
result is the same unless you ignore it. You can't ignore
it completely even if the distances are short.)
Also, Lorentz contraction *does* occur (when you transform
between frames). It always occurs -- it's not an either-or.
The question here is whether the proper separation of the
ships increases or remains the same. It increases.
Russell Blackadar
[snip my earlier posting]
Jonathan, I am guessing that your lack of response means
that you have retreated to your lair, to do some further
pondering and calculation. That's a good thing. But let
me stand at the entrance and shout a few more words of
encouragement.
First, please ignore the next-to-last paragraph of my
last posting. I wrote it earlier than the rest -- before
remembering that there are no tidal forces at all in this
problem -- and I forgot to revise it later.
Second, I think I know why you are so tenaciously holding
onto your position, against reason. You are under the
mistaken impression -- common enough, but mistaken -- that
the Lorentz transformation equations (LTE) apply directly
to the change from some state A (e.g. rockets pre-acceleration)
to a different state B (rockets post-acceleration) in the same
frame. In other words, you believe that if the rockets have
some distance of separation L in a frame, and later they are
moving at speed v in the same frame, then they *must* have
separation L/gamma in that frame or SR is wrong. Probably
my just putting it that way shows up the nature of the mistake
clearly enough to you.
You see, that's not at all what the LTE are about. They are
about how some event or set of events, known to have certain
coordinates in one frame, are measured to take place in a
different frame. That's all! IOW it's about switching
observers, not about something evolving in time. For the
latter, in general, you have to do an integral (in which
the LTE of course play a role, but other things do as well,
viz. the nature of the acceleration).
Yes, it can happen that the change from state A to state B
in some frame is described by the same equation for Lorentz
contraction that is useful in transforming between frames.
When that is true, the acceleration is called "Born-rigid".
But disconnected rockets firing simultaneous and equal burns
are definitely *not* Born-rigid.
Don't worry. As I said, you are in very good company indeed;
John Baez himself says he had the wrong view when he started
out. I know I did. It's almost a rite of passage when you
come to see this issue correctly.
Jonathan Doolin
If you want to know the ridiculous assumption that I had made which made
your arguments difficult for me to swallow it is this:
False assumption: When I hit the accelerator of my car, I expect that the
front of the car and the rear of the car will accelerate at the same rate.
I was rejecting anything that falsified this idea, though I hadn't formally
realized it was stuck in my head. This false assumption was causing me, as
accused, to manually synchronize the clocks between the ships for each
impulse. I was convinced that it happened naturally at low accelerations,
until I began to type up a letter telling you why. This re-synchronization
only happens to things that are attached together.
You misunderstood what I meant by the word "tide," as I meant it to describe
the effect that would stretch the two space ships apart *while they were
accelerating* and *break any tether between them.* The "buzzword" was
intended to save your correct belief, not my incorrect one. I thought I had
explained better what I had meant. I thought it was an appropriate
description of what happens, since it streches things (in their proper
accelerated reference frame), just like tidal forces strech the oceans.
As for the tethering experiment:
Two space ships sit 100 feet from each other and agree to accelerate at one
gravity to .866c. They decide to do this experiment twice: Once free, and
once tethered together.
In the first experiment, the space ships complete the acceleration about one
year later. They measure the distance between themselves and find that they
have drifted apart to 200 feet.
They re-establish their distance at one hundred feet and attach themselves
with the 100 foot thether.
This time, they accelerate again another year. They do not detect any
tension on the rope because it is too small to measure. However, they do
end up retaining their distance, because of that finite but undetectable
tension in the rope.
Jonathan Doolin
> Russell Blackadar wrote:
>
> [snip my earlier posting]
>
> Jonathan, I am guessing that your lack of response means
> that you have retreated to your lair, to do some further
> pondering and calculation. That's a good thing. But let
> me stand at the entrance and shout a few more words of
> encouragement.
>
> First, please ignore the next-to-last paragraph of my
> last posting. I wrote it earlier than the rest -- before
> remembering that there are no tidal forces at all in this
> problem -- and I forgot to revise it later.
Ahh, but the effect which streches and breaks the tether between the
two space ships. What is this effect called? When it came clear to me that
this actually happened, I thought it seemed very similar to a tidal effect,
but I will admit that it is not a "tide" in the common
earth-ocean-moon-centripital-motion-gravity-sense of the word, as it is a
streching due to linear acceleration instead of orbital acceleration.
>
> Second, I think I know why you are so tenaciously holding
> onto your position, against reason. You are under the
> mistaken impression -- common enough, but mistaken -- that
> the Lorentz transformation equations (LTE) apply directly
> to the change from some state A (e.g. rockets pre-acceleration)
> to a different state B (rockets post-acceleration) in the same
> frame.
Some state of the rockets? You might want to be a little more explicit
there.
> In other words, you believe that if the rockets have
> some distance of separation L in a frame, and later they are
> moving at speed v in the same frame, then they *must* have
> separation L/gamma in that frame or SR is wrong.
The nature of my mistake was believing that "uniformly applied acceleration
causes Lorentz contraction" In fact uniformly applied acceleration simply
causes things to pull apart from each other. Only objects which are
attached become Lorentz contracted.
> Probably
> my just putting it that way shows up the nature of the mistake
> clearly enough to you.
To be honest, if I didn't know you were right already, it probably would
have gone right over my head. Or I would have objected that there are some
states (i.e. length of the rockets) which the Lorentz transformation do
apply to. But, as you point out below, the rockets are "Born-rigid"
>
> You see, that's not at all what the LTE are about. They are
> about how some event or set of events, known to have certain
> coordinates in one frame, are measured to take place in a
> different frame. That's all! IOW it's about switching
> observers, not about something evolving in time. For the
> latter, in general, you have to do an integral (in which
> the LTE of course play a role, but other things do as well,
> viz. the nature of the acceleration).
>
> Yes, it can happen that the change from state A to state B
> in some frame is described by the same equation for Lorentz
> contraction that is useful in transforming between frames.
> When that is true, the acceleration is called "Born-rigid".
> But disconnected rockets firing simultaneous and equal burns
> are definitely *not* Born-rigid.
>
> Don't worry. As I said, you are in very good company indeed;
> John Baez himself says he had the wrong view when he started
> out. I know I did. It's almost a rite of passage when you
> come to see this issue correctly.
>
I can see that. It seems to be a good step in understanding General
Relativity.
Jonathan Doolin
> Jonathan Doolin wrote:
> >
> > "Russell Blackadar" <rus...@mdli.com> wrote in message
> > news:3C4C740F...@mdli.com...
Incidentally, if this interpretation of the two electron gun experiment is
correct, the two electrons are (or will be) separated by 2000 miles in their
reference frame, but we "see" them separated only by 1000 miles in our
reference frame.
If these two electrons are matching pace with a 2000 mile long space-ship
traveling by, it would see the front electron fired as its front end was
passing the front electron gun, and it would see the back electron fired as
its back end were passing by.
From earth, the space ship would appear to be only 1000 miles long--it's
front end would pass the front electron gun at the same moment that the back
end passed the rear electron gun.
However, Penrose and Terrel proved in 1959 that I am in error. The ship
would appear to be 2000 miles long.
If you set up a camera where both electron guns could be observed firing at
the same time, the image of the space ship would be slightly skewed, so you
could see a portion of the back end, but I can't understand how the ship
could appear to be 2000 miles long when its ends are 1000 miles from each
other.
Russell Blackadar
[snip]
> Incidentally, if this interpretation of the two electron gun experiment is
> correct, the two electrons are (or will be) separated by 2000 miles in their
> reference frame, but we "see" them separated only by 1000 miles in our
> reference frame.
>
> If these two electrons are matching pace with a 2000 mile long space-ship
> traveling by, it would see the front electron fired as its front end was
> passing the front electron gun, and it would see the back electron fired as
> its back end were passing by.
Yes.
>
> From earth, the space ship would appear to be only 1000 miles long--it's
> front end would pass the front electron gun at the same moment that the back
> end passed the rear electron gun.
Yes.
>
> However, Penrose and Terrel proved in 1959 that I am in error. The ship
> would appear to be 2000 miles long.
*And* rotated, so that the aspect it appears to present
to you is only 1000 miles from front to back, exclusive
of the view of the rear that you can also see. In
other words, the electron guns *will* appear to be at
exactly the forward end (side toward you) and back end
(side toward you) of the ship.
>
> If you set up a camera where both electron guns could be observed firing at
> the same time, the image of the space ship would be slightly skewed, so you
> could see a portion of the back end, but I can't understand how the ship
> could appear to be 2000 miles long when its ends are 1000 miles from each
> other.
You would see *all* of the back end (assuming the ship is
a cylinder and you are looking at it exactly side-on).
That back end would look like an ellipse. The side
"toward you" will be contracted and will fit exactly
between your gun "markers".
Penrose-Terrell rotation is really a combination of
contraction and skew. At sufficient distance (so that
perspective isn't important) it looks exactly like a
rotation. Closer up, it doesn't quite.
You seem to have included the skew but forgotten the
contraction.
Russell Blackadar
>
> "Russell Blackadar" <rus...@mdli.com> wrote in message
> news:3C58523D...@mdli.com...
> > Russell Blackadar wrote:
> >
> > [snip my earlier posting]
> >
> > Jonathan, I am guessing that your lack of response means
> > that you have retreated to your lair, to do some further
> > pondering and calculation. That's a good thing. But let
> > me stand at the entrance and shout a few more words of
> > encouragement.
> >
> > First, please ignore the next-to-last paragraph of my
> > last posting. I wrote it earlier than the rest -- before
> > remembering that there are no tidal forces at all in this
> > problem -- and I forgot to revise it later.
>
> Ahh, but the effect which streches and breaks the tether between the
> two space ships. What is this effect called?
If you are thinking about how to describe it in
(say) the rear rocket's accelerated coordinates,
I would call that effect "front rocket is moving
faster than I am". Or, "front rocket has been
accelerating for a longer time than I have, and
has done work on the tether, stretching it to its
breaking point." In GR terms, that longer time is
due to the gravitational blue shift of front ship
as measured by the rear ship.
In your other post, you talked about breaking any
tether "no matter how strong", as if you think
there is some magic going on here. The only magic
is that we have set up the problem with the specific
requirement that the acceleration be identical.
If something is impeding the front ship (tether
pulling it back) or impelling the rear ship on
(tether pulling it forward) then the problem setup
requires that each ship adjust its thrust to
compensate. So, by requirement of the problem,
there is always sufficient force provided by the
difference in rocket thrust to pull the tether
apart.
(And if it is a very very weak tether, even the
*lack* of a slight difference in thrust in the
other direction, is sufficient to break it.)
When it came clear to me that
> this actually happened, I thought it seemed very similar to a tidal effect,
> but I will admit that it is not a "tide" in the common
> earth-ocean-moon-centripital-motion-gravity-sense of the word, as it is a
> streching due to linear acceleration instead of orbital acceleration.
Tides are not due to orbital acceleration! That's
a common myth. In Newtonian physics, they are due
to nonuniformity of the gravitational field. The
lunar tides on Earth as they are -- with Earth
orbiting at constant distance from the moon -- are
pretty much the same as they would be if the Earth
were not orbiting, but rather falling directly
toward the moon at its current distance away.
In GR, tides are due to the curvature of spacetime,
which causes neighboring geodesics to deviate along
one spatial direction, and come together along the
other.
I do not think it is correct to use the term "tide"
in flat spacetime. Perhaps someone more experienced
in GR can correct me if I am wrong.
>
> >
> > Second, I think I know why you are so tenaciously holding
> > onto your position, against reason. You are under the
> > mistaken impression -- common enough, but mistaken -- that
> > the Lorentz transformation equations (LTE) apply directly
> > to the change from some state A (e.g. rockets pre-acceleration)
> > to a different state B (rockets post-acceleration) in the same
> > frame.
>
> Some state of the rockets? You might want to be a little more explicit
> there.
What I mean is this. You have the starting state:
Rocket F and Rocket R with full tanks, ready to go,
proper distance apart = L. You have the ending state:
Rocket F and Rocket R with empty tanks and people
splattered all over the wall. ;-) In other words,
something physical happened in between. You can't
just apply the LTE willy-nilly to compare starting
and ending states.
>
> > In other words, you believe that if the rockets have
> > some distance of separation L in a frame, and later they are
> > moving at speed v in the same frame, then they *must* have
> > separation L/gamma in that frame or SR is wrong.
>
> The nature of my mistake was believing that "uniformly applied acceleration
> causes Lorentz contraction" In fact uniformly applied acceleration simply
> causes things to pull apart from each other. Only objects which are
> attached become Lorentz contracted.
No, you're saying that wrong, and that's my point.
Lorentz contraction is a result of observing something
from a different frame. It's an inevitable consequence
of SR. You can't turn it off! If after a physical
process some object has a different v (in your frame)
than it had previously (in your frame) but the same
length (in your frame) as it had previously (in your
frame), then you can conclude that the object's proper
length changed. Or pair of objects, in the present
case. You can make that inference precisely because
you know it *is* Lorentz contracted in your frame, vs.
its own.
By the way, you don't need physical attachment. All
you need is the right kind of acceleration. E.g. you
could adjust the rockets' thrust to simulate the
effect of tension in the tether.
>
> > Probably
> > my just putting it that way shows up the nature of the mistake
> > clearly enough to you.
>
> To be honest, if I didn't know you were right already, it probably would
> have gone right over my head. Or I would have objected that there are some
> states (i.e. length of the rockets) which the Lorentz transformation do
> apply to. But, as you point out below, the rockets are "Born-rigid"
I said they are *not*. (Meaning, the pair take
together are not. You could certainly contrive to
make each rocket individually accelerate Born-rigidly,
or close enough not to matter.)
[snip rest]
Jonathan Doolin
Skewed, not rotated. The back end of the ship and the front end of the ship
would appear to be the same distance from the electron guns. "Rotated in
spacetime" does not mean the same thing as rotated in space.
>
> >
> > If you set up a camera where both electron guns could be observed firing
at
> > the same time, the image of the space ship would be slightly skewed, so
you
> > could see a portion of the back end, but I can't understand how the ship
> > could appear to be 2000 miles long when its ends are 1000 miles from
each
> > other.
>
> You would see *all* of the back end (assuming the ship is
> a cylinder and you are looking at it exactly side-on).
> That back end would look like an ellipse. The side
> "toward you" will be contracted and will fit exactly
> between your gun "markers".
Okay, then its settled. You can see lorentz contraction.
>
> Penrose-Terrell rotation is really a combination of
> contraction and skew. At sufficient distance (so that
> perspective isn't important) it looks exactly like a
> rotation. Closer up, it doesn't quite.
>
> You seem to have included the skew but forgotten the
> contraction.
I seem to have just been reporting what it says in Terrell's paper.
Russell Blackadar
First of all, we are talking about an image on (say) a
photographic plate. It is not rotated in spacetime, it
is (apparently!) rotated in space.
Secondly, if the object is sufficiently distant (as I
said) for perspective to be unnoticable, then there is
no way you can tell visually that the front of the ship
is at the electron gun, or further distant from it along
the same sight line. Surely you can see this.
>
> >
> > >
> > > If you set up a camera where both electron guns could be observed firing
> at
> > > the same time, the image of the space ship would be slightly skewed, so
> you
> > > could see a portion of the back end, but I can't understand how the ship
> > > could appear to be 2000 miles long when its ends are 1000 miles from
> each
> > > other.
> >
> > You would see *all* of the back end (assuming the ship is
> > a cylinder and you are looking at it exactly side-on).
> > That back end would look like an ellipse. The side
> > "toward you" will be contracted and will fit exactly
> > between your gun "markers".
>
> Okay, then its settled. You can see lorentz contraction.
Yes, definitely you can, if both of the following are
true:
(1) you know the true shape of the object and can in
some way visually distinguish its side from its rear
(spheres will be a problem!), and
(2) you have some way to perceive depth.
Another way to put condition #2 is to say that perspective
is visible -- exactly as I (and Terrell; I haven't read
Penrose) have said. At sufficient distance (e.g. through
the Hubble) it is not. Of course there may be other ways to
perceive depth, but we are talking about just the visual
appearance here.
Btw, condition #1 is by no means a triviality if we are
talking about images in the Hubble.
>
> >
> > Penrose-Terrell rotation is really a combination of
> > contraction and skew. At sufficient distance (so that
> > perspective isn't important) it looks exactly like a
> > rotation. Closer up, it doesn't quite.
> >
> > You seem to have included the skew but forgotten the
> > contraction.
>
> I seem to have just been reporting what it says in Terrell's paper.
No, you have ignored his final section where he talks
about perspective.
Jonathan Doolin
> Jonathan Doolin wrote:
> >
> > "Russell Blackadar" <rus...@mdli.com> wrote in message
> > news:3C58523D...@mdli.com...
> > > Russell Blackadar wrote:
> > >
> > > [snip my earlier posting]
> > >
> > > Jonathan, I am guessing that your lack of response means
> > > that you have retreated to your lair, to do some further
> > > pondering and calculation. That's a good thing. But let
> > > me stand at the entrance and shout a few more words of
> > > encouragement.
> > >
> > > First, please ignore the next-to-last paragraph of my
> > > last posting. I wrote it earlier than the rest -- before
> > > remembering that there are no tidal forces at all in this
> > > problem -- and I forgot to revise it later.
> >
> > Ahh, but the effect which streches and breaks the tether between
the
> > two space ships. What is this effect called?
>
> If you are thinking about how to describe it in
> (say) the rear rocket's accelerated coordinates,
> I would call that effect "front rocket is moving
> faster than I am". Or, "front rocket has been
> accelerating for a longer time than I have, and
> has done work on the tether, stretching it to its
> breaking point."
a) front rocket left before I did (during and after acceleration)
b) front rocket completed its acceleration before I did. (after
acceleration)
c) front rocket is moving faster than I am (during acceleration)
d) front rocket has been accelerating for a longer time than I have.
(during acceleration)
e) front rocket accelerated the same amount of time that I did (after
acceleration completed)
>In GR terms, that longer time is
> due to the gravitational blue shift of front ship
> as measured by the rear ship.
I'll just call it the "Stretchy effect" unless you have a better term for
it.
>
> In your other post, you talked about breaking any
> tether "no matter how strong", as if you think
> there is some magic going on here.
I was wrong. I admit this. If you had a tether that could strech to double
its length, it could easily handle two space-ships that duplicated the two
electron-gun experiment.
> The only magic
> is that we have set up the problem with the specific
> requirement that the acceleration be identical.
Right. If the accelerations are the same, the two ships will drift apart.
I agree with you now. I was wrong before.
> If something is impeding the front ship (tether
> pulling it back) or impelling the rear ship on
> (tether pulling it forward) then the problem setup
> requires that each ship adjust its thrust to
> compensate.
In cases of low acceleration (32feet/sec^2 or so) and short distance (300
yards or so) the process of getting to .866c would take so long that any
normal tether would have no trouble forcing the two ships to adjust. The
tether would overcome the "stretchy effect"
>So, by requirement of the problem,
> there is always sufficient force provided by the
> difference in rocket thrust to pull the tether
> apart.
Right. If the two accelerations are *exactly* the same, eventually the
tether has to strech.
>
> (And if it is a very very weak tether, even the
> *lack* of a slight difference in thrust in the
> other direction, is sufficient to break it.)
>
Yes. But if the tether is strong and not very stretchy, it will force the
rear ship to accelerate faster and the front ship to accelerate slower. In
the *incredibly* fast acceleration described in the two electron gun
experiment, a "born-rigid" tether would slow the front electron enormously,
and cause the rear electron to snap forward.
>
> When it came clear to me that
> > this actually happened, I thought it seemed very similar to a tidal
effect,
> > but I will admit that it is not a "tide" in the common
> > earth-ocean-moon-centripital-motion-gravity-sense of the word, as it is
a
> > streching due to linear acceleration instead of orbital acceleration.
>
> Tides are not due to orbital acceleration! That's
> a common myth. In Newtonian physics, they are due
> to nonuniformity of the gravitational field. The
> lunar tides on Earth as they are -- with Earth
> orbiting at constant distance from the moon -- are
> pretty much the same as they would be if the Earth
> were not orbiting, but rather falling directly
> toward the moon at its current distance away.
Right. Tides caused by gravitational gradient. Stretchy effect caused by
uniform acceleration. Is stretchy effect many orders of magnitude below
tidal force? Yes, stretchy effect barely noticeable at 1 gravity and earth
diameter. Or is it? Earth is a "born-rigid" body 7600 miles across. Water
is not as rigid. I will think more on this.
>
> In GR, tides are due to the curvature of spacetime,
> which causes neighboring geodesics to deviate along
> one spatial direction, and come together along the
> other.
Right. But I'm trying to understand it. Not memorize it.
Russell Blackadar
[snip points of agreement, down to discussion of tides]
> Right. Tides caused by gravitational gradient. Stretchy effect caused by
> uniform acceleration. Is stretchy effect many orders of magnitude below
> tidal force? Yes, stretchy effect barely noticeable at 1 gravity and earth
> diameter. Or is it? Earth is a "born-rigid" body 7600 miles across. Water
> is not as rigid. I will think more on this.
The magnitude of the effect is not the point. The point is
that tidal effects can act on a set of mutually noninteracting
objects each moving inertially (i.e. following a geodesic in
spacetime) and are (if large enough or measured for long
enough) observable in inertial coordinate systems moving
along with those particles.
The rocket scenario is an entirely different thing; the ships
themselves are not inertial. And for hypothetical inertial
bodies in our system, there is no effect at all on those
bodies that can be measured in their own frames. So, I
think this effect fails to qualify as a tide. OTOH a pair
of inertial bodies *will* come closer together over time in
our accelerated coords. But that is a coordinate effect; an
artifact of our using curvilinear (Rindler) coordinates in
flat spacetime. I think the correct term for such an effect
is "pseudoforce". Not tide. But as I said, I am open to
correction if some GR guru wants to weigh in on this subject.
In essence I am saying that tidal effects are the same
thing as spacetime curvature.
>
> >
> > In GR, tides are due to the curvature of spacetime,
> > which causes neighboring geodesics to deviate along
> > one spatial direction, and come together along the
> > other.
>
> Right. But I'm trying to understand it. Not memorize it.
Good; it's just that I think an essential part of the
"tide" concept is that it involves geodesics. Not
(or not essentially) accelerated motion. You won't
understand it unless (at least) you understand that
tides can occur where there is no proper acceleration
at all.
Russell Blackadar
I wrote:
[re Rindler coords]
a pair
> of inertial bodies *will* come closer together over time in
> our accelerated coords.
Or get further apart; it all depends on whether those
inertial bodies are moving with our direction of
acceleration or against it.
Paul B. Andersen
"Jonathan Doolin" <spence...@earthlink.net> wrote in message news:vy_58.8188$By6.9...@newsread2.prod.itd.earthlink.net...
>
> Incidentally, if this interpretation of the two electron gun experiment is
> correct, the two electrons are (or will be) separated by 2000 miles in their
> reference frame, but we "see" them separated only by 1000 miles in our
> reference frame.
>
> If these two electrons are matching pace with a 2000 mile long space-ship
> traveling by, it would see the front electron fired as its front end was
> passing the front electron gun, and it would see the back electron fired as
> its back end were passing by.
>
> From earth, the space ship would appear to be only 1000 miles long--it's
> front end would pass the front electron gun at the same moment that the back
> end passed the rear electron gun.
>
> However, Penrose and Terrel proved in 1959 that I am in error. The ship
> would appear to be 2000 miles long.
It might, but that would be a coincidence. The ship may appear to
have a length anywhere in the range 549 miles to 7280 miles.
You are now talking about visual observation, and visually observed
apparent length. This length is obviously (no?) highly dependent
on your vantage point. If the ship is approaching you along your
line of sight, it will appear lengthened by the factor sqrt((c+v)/(c-v)),
if it is receding from you, it appears shortened by sqrt((c-v)/(c+v)).
(If you think this looks a lot like Doppler shift, it is no coincidence,
it is basically the same phenomenon.)
If the ship is approaching, it will have the apparent speed v/(1-v/c)
or ca. 6c, if it is receding the apparent speed will be 0.46c.
If the ship passes you right in front of your nose, it will at
the moment you are adjacent to middle of the ship seem distorted;
the front end is shortened, the aft end is lengthened.
>
> If you set up a camera where both electron guns could be observed firing at
> the same time, the image of the space ship would be slightly skewed, so you
> could see a portion of the back end,
If you look at the ship from a point where the length axis is not
along your line of sight, it will look very strange. It will be bent
and rotated. Penrose rotation.
> but I can't understand how the ship
> could appear to be 2000 miles long when its ends are 1000 miles from each
> other.
It can look a lot longer, and the reason is simple.
You see the different part of the ship where it was when the light
that hits your eye was emitted.
So when the ship is approaching, and you NOW receive light simultaneously
from both ends of the ship, then the light from its aft end was emitted
when the ship was a lot father away than it was when the light from its
front end was emitted.
BTW, apparently superluminal velocities are actually observed
in jets from active galaxies. "Superluminal jets."
Paul
Russell Blackadar
>
> "Jonathan Doolin" <spence...@earthlink.net> wrote in message news:vy_58.8188$By6.9...@newsread2.prod.itd.earthlink.net...
[snip Paul's correct analysis]
> >
> > If you set up a camera where both electron guns could be observed firing at
> > the same time, the image of the space ship would be slightly skewed, so you
> > could see a portion of the back end,
>
> If you look at the ship from a point where the length axis is not
> along your line of sight, it will look very strange. It will be bent
> and rotated. Penrose rotation.
In my parallel reply, I assumed that such a positioning
is what Jonathan had in mind, since the subject was
Penrose-Terrell rotation.
If, as you described earlier, the length axis *is* on the
line of sight, and the ship is next to you, it's certainly
not far enough away for the notion of Penrose-Terrell
rotation to be applicable. Even if it's at a greater
distance, Penrose-Terrell rotation is zero for
line-of-sight motion.)
>
> > but I can't understand how the ship
> > could appear to be 2000 miles long when its ends are 1000 miles from each
> > other.
>
> It can look a lot longer, and the reason is simple.
Yes it can, *if* you have a way to perceive depth. That
might not be the case if you are looking at a single 2D
image of a distant object. In such cases, the Doppler-
lengthened image (assuming the object is approaching)
would be indistinguishable from an 9mage of the object
at its normal length but rotated. (Of course neglecting
other effects, such as change of the object's color.)
In other words, Jonathan should not interpret your answer
as a refutation of Penrose and Terrell's predictions,
within their stated range of applicability.
Also, in my reply I said that Penrose-Terrell rotation is
a combination of contraction and skew, but I see that's
an oversimplification except in cases of the purely
transverse line of sight that I had in mind.
[snip Paul's good explanation, which btw I think Jonathan
already knew]
Paul B. Andersen
> "Paul B. Andersen" wrote:
> >
> > "Jonathan Doolin" <spence...@earthlink.net> wrote in message news:vy_58.8188$By6.9...@newsread2.prod.itd.earthlink.net...
>
> [snip Paul's correct analysis]
>
> > >
> > > If you set up a camera where both electron guns could be observed firing at
> > > the same time, the image of the space ship would be slightly skewed, so you
> > > could see a portion of the back end,
> >
> > If you look at the ship from a point where the length axis is not
> > along your line of sight, it will look very strange. It will be bent
> > and rotated. Penrose rotation.
>
> In my parallel reply, I assumed that such a positioning
> is what Jonathan had in mind, since the subject was
> Penrose-Terrell rotation.
Well.
He said, after correctly describing what would be _measured_
in "our reference frame", with no qualifications:
"However, Penrose and Terrel proved in 1959 that I am in error.
The ship would appear to be 2000 miles long."
My point was rather that what will be _visually_ observed
highly depend on our position in "our reference frame",
and that the above statement is a gross oversimplification.
> If, as you described earlier, the length axis *is* on the
> line of sight, and the ship is next to you, it's certainly
> not far enough away for the notion of Penrose-Terrell
> rotation to be applicable. Even if it's at a greater
> distance, Penrose-Terrell rotation is zero for
> line-of-sight motion.)
Sure.
> > > but I can't understand how the ship
> > > could appear to be 2000 miles long when its ends are 1000 miles from each
> > > other.
> >
> > It can look a lot longer, and the reason is simple.
>
> Yes it can, *if* you have a way to perceive depth. That
> might not be the case if you are looking at a single 2D
> image of a distant object. In such cases, the Doppler-
> lengthened image (assuming the object is approaching)
> would be indistinguishable from an 9mage of the object
> at its normal length but rotated. (Of course neglecting
> other effects, such as change of the object's color.)
Of course, if the ship came directly at you, you wouldn't see
anything but it's front.
But this is rather theoretical, don't you think?
> In other words, Jonathan should not interpret your answer
> as a refutation of Penrose and Terrell's predictions,
> within their stated range of applicability.
It wasn't meant as a refutation of anything.
> Also, in my reply I said that Penrose-Terrell rotation is
> a combination of contraction and skew, but I see that's
> an oversimplification except in cases of the purely
> transverse line of sight that I had in mind.
>
> [snip Paul's good explanation, which btw I think Jonathan
> already knew]
Maybe.
Paul