Time dilation - No Need to Do the Math
Stephen Bint
newsgroups. I am grateful to those who discussed it, even though it got a
tiny bit personal at times :)
I was frustrated by the account of the twin paradox in the FAQ for
sci.physics (http://www.faqs.org/faqs/physics-faq/part4/), because it
resorts to saying that there is assymetry in the situation. One twin has
undergone an acceleration and it is that one which is younger.
That cannot be the case, because time dilation is supposed to explain why a
light beam, travelling from floor to ceiling of a moving spaceship will not
trace out its V-shaped path faster than c. Time must appear to be slowed
down or else the observer, looking through the window will see the beam
travel faster than c.
But if two people pass eachother on skateboards, watching eachother's
bouncing beams, they must both see time dilation in the other, or one of
them will see the beam going faster than c.
Claiming an assymetry based on the acceleration histories of the observer,
overlooks the fact that the time dilation must be real and identical in both
directions at all times or else someone will see the bouncing beam going
faster than c.
If only one skateboard rider is younger, what did this younger one see, the
other getting older? The beam going even faster than expected, much more
than c?
It baffles me, that intelligent people, capable of algebra, cannot see that
the assertion that time is slowed in a moving frame is a symmetrical
assertion, because frames move relative to eachother, symmetrically.
Besides which, the formula for time dilation is a function of the velocity
of the observed frame and makes no mention whatsoever of the acceleration of
either frame, before or while the calculation is made.
I was told again and again by several posters, "Do the math. Just plug the
numbers into the formulas and it all works out just fine." These were the
same people who want me to factor in acceleration, which is not mentioned in
the formula, because time dilation is purely a function of the relative
velocity, which is the same in both directions.
Plug the relative velocity of skateboarders A and B into the formula and
find out how much slower B's clock is, than A's. And how much slower A's is,
than B's. I don't need to "do the math" to see that this formula is telling
me two things which cannot both be true.
I suppose it depends on your definition of "working out fine".
Because I constructed my example with two moving twins, to force them to
face the symmetry, the defenders of relativity introduced a new twist:
acceleration causes time to speed up, cancelling out the age difference, in
both skateboarders. This is an new addition to relativity. Einstein never
said anything of the sort.
To say that acceleration makes time run faster, implies that if one of the
skateboarders is accelerating, the time contraction due to acceleration will
at least partially counteract the dilation caused by the velocity (predicted
by Einstein), allowing the other to see the bouncing beam move faster than
c.
Still, though frustrated, I thoroughly enjoyed the thread I started and I am
grateful to all of the people who had the patience to contribute. In return,
I can only assure you that I will continue to post under the name "Stephen
Bint", so as not to undermine your killfile strategies.
And may I say in conclusion, that trying to bring the sweet light of reason
to the proponents of relativity, is like trying to bring salvation to the
DAMNED.
Stephen Bint
Dan Bloomquist
Stephen Bint wrote:
>
> And may I say in conclusion, that trying to bring the sweet light of reason
> to the proponents of relativity, is like trying to bring salvation to the
> DAMNED.
>
Why not go back to Jon's quite eloquent explanation of the twins and
read it again. You seemed to get it then.
>
> Stephen Bint
>
Best, Dan.
--
http://lakeweb.net
http://ReserveAnalyst.com
dbAtLakewebDotCom
Stephen Bint
"Dan Bloomquist" <lak...@citlink.net> wrote in message
news:3FA081EF...@citlink.net...
>
>
> Stephen Bint wrote:
> >
> > And may I say in conclusion, that trying to bring the sweet light of
reason
> > to the proponents of relativity, is like trying to bring salvation to
the
> > DAMNED.
> >
>
> Why not go back to Jon's quite eloquent explanation of the twins and
> read it again. You seemed to get it then.
>
> >
> > Stephen Bint
> >
>
> Best, Dan.
>
thread. Here's a quote:
> By symmetry, he sees exactly the same things happening as he
> observes your clock during his journey. And *neither* of you
> actually *sees* the other's clock "run slow," because of the
> Doppler effect caused by your approaching each other.
And I am to blame for failing to bring the matter to a head. Because I
insisted in my example, to be told what the twins actually see, I have
demanded that Jon mask the time dilation with the doppler effect.
He has reiterated the explanation given in the FAQ, but done a better job of
it. The twins, by his reckoning, have aged less than an Earth-bound
observer. But the formula for time dilation is symmetrical, because the "v"
factor is identical for observers in both frames, looking at the other when
one frame is moving relative to the other. This cannot be reconciled with
the observers in one of the frames being younger .
The problem is not "deciding" which frame is the one running slow, by
clutching at an asymmetry. The problem is that time dilation fails to
prevent witnesses seeing the bouncing beam going faster than light, unless
it works both ways and the Earth-bound observer has also aged less than the
twins.
If you say either frame is the slow one, you deviate from what the formula
asserts, which is plainly that both are slow by the same factor.
When you say
> Why not go back to Jon's quite eloquent explanation of the twins and
> read it again. You seemed to get it then.
you are appealing to me to be open-minded enough to be able to believe it is
possible that I am wrong. I would ask the same of you.
Stephen Bint
dlzc@aol.com (formerly)
"Stephen Bint" <bi...@iname.com> wrote in message
news:3fa07dad$0$118$65c6...@mercury.nildram.net...
...
> I started a thread about the twin paradox and cross-posted it to three
> newsgroups. I am grateful to those who discussed it, even though it got a
> tiny bit personal at times :)
>
> I was frustrated by the account of the twin paradox in the FAQ for
> sci.physics (http://www.faqs.org/faqs/physics-faq/part4/), because it
> resorts to saying that there is assymetry in the situation. One twin has
> undergone an acceleration and it is that one which is younger.
>
> That cannot be the case, because time dilation is supposed to explain why
a
> light beam, travelling from floor to ceiling of a moving spaceship will
not
> trace out its V-shaped path faster than c. Time must appear to be slowed
> down or else the observer, looking through the window will see the beam
> travel faster than c.
>
> But if two people pass eachother on skateboards, watching eachother's
> bouncing beams, they must both see time dilation in the other, or one of
> them will see the beam going faster than c.
If you travel away from the Earth, you find that signals from the Earth are
slowed as well. Just not as slowed as the Earth determines your signals to
be.
> Claiming an assymetry based on the acceleration histories of the
observer,
> overlooks the fact that the time dilation must be real and identical in
both
> directions at all times or else someone will see the bouncing beam going
> faster than c.
>
> If only one skateboard rider is younger, what did this younger one see,
the
> other getting older? The beam going even faster than expected, much more
> than c?
If both frames are moving the same speed, but different directions, they
will both age the same.
> It baffles me, that intelligent people, capable of algebra, cannot see
that
> the assertion that time is slowed in a moving frame is a symmetrical
> assertion, because frames move relative to eachother, symmetrically.
Not true. The acceleration provides the asymmetry. The duration of the
velocity provides the difference. And the moving and non-moving frames do
not agree on the *amount* the other is affected.
> Besides which, the formula for time dilation is a function of the
velocity
> of the observed frame and makes no mention whatsoever of the acceleration
of
> either frame, before or while the calculation is made.
True. This *is* relativity. Absolutes are not at hand.
> I was told again and again by several posters, "Do the math. Just plug
the
> numbers into the formulas and it all works out just fine." These were the
> same people who want me to factor in acceleration, which is not mentioned
in
> the formula, because time dilation is purely a function of the relative
> velocity, which is the same in both directions.
>
> Plug the relative velocity of skateboarders A and B into the formula and
> find out how much slower B's clock is, than A's. And how much slower A's
is,
> than B's. I don't need to "do the math" to see that this formula is
telling
> me two things which cannot both be true.
>
> I suppose it depends on your definition of "working out fine".
>
> Because I constructed my example with two moving twins, to force them to
> face the symmetry, the defenders of relativity introduced a new twist:
> acceleration causes time to speed up, cancelling out the age difference,
in
> both skateboarders. This is an new addition to relativity. Einstein never
> said anything of the sort.
Time dilation doesn't much care about *direction*, only speed.
> To say that acceleration makes time run faster, implies that if one of
the
> skateboarders is accelerating, the time contraction due to acceleration
will
> at least partially counteract the dilation caused by the velocity
(predicted
> by Einstein), allowing the other to see the bouncing beam move faster
than
> c.
Acceleration does skew the results with SR. SR is for use with inertial
frames.
> Still, though frustrated, I thoroughly enjoyed the thread I started and I
am
> grateful to all of the people who had the patience to contribute. In
return,
> I can only assure you that I will continue to post under the name
"Stephen
> Bint", so as not to undermine your killfile strategies.
>
> And may I say in conclusion, that trying to bring the sweet light of
reason
> to the proponents of relativity, is like trying to bring salvation to the
> DAMNED.
Quite a lot of work you've cut for yourself. Not only do you know more
than anyone else, but you get to shine the light of revelation for us all.
Since you've committed the same sins as everyone else who has approached
relativity, including the big chip on your shoulder, my guess is there is a
plank in your eye.
Try analyzing your two skateboarders from a third stationary frame? Tried
using addition of velocities?
David A. Smith
Stephen Bint
You have said,
> If both frames are moving the same speed, but different directions, they
> will both age the same.
>
So, like the person who wrote the account of the twin paradox in the FAQ,
you do believe that time dilation affects the rate of aging.
The formula for time dilation is a function of velocity. If I am moving at a
given velocity with respect to you, then you are moving at the exact same
velocity with respect to me. To whatever extent there is time dilation and a
reduction of aging, it is predicted to be the same for you as for me,
because v is the same, and the formula is a function of v. You see me in
slow motion and I see you in slow motion, so both our ages are reduced by
the same amount.
Is that not the case?
Stephen
"dl...@aol.com (formerly)" <dlzc1.cox@net> wrote in message
news:uY%nb.123151$gv5.60210@fed1read05...
dlzc@aol.com (formerly)
"Stephen Bint" <bi...@iname.com> wrote in message
news:3fa09253$0$104$65c6...@mercury.nildram.net...
> Dear David Smith:
>
> You have said,
>
> > If both frames are moving the same speed, but different directions,
they
> > will both age the same.
> >
> So, like the person who wrote the account of the twin paradox in the FAQ,
> you do believe that time dilation affects the rate of aging.
Let me say yes, this way: If I travel between here and the store in my car
in different ways, do I show different mileage on my car for each trip?
Yes, the path you travel racks up different "mileage".
> The formula for time dilation is a function of velocity. If I am moving
at a
> given velocity with respect to you, then you are moving at the exact same
> velocity with respect to me.
Not quite correct.
> To whatever extent there is time dilation and a
> reduction of aging, it is predicted to be the same for you as for me,
> because v is the same, and the formula is a function of v.
Not necessarily true. Since one frame does not see time or length the same
as the other, inferred velocities are bumfuzzled. For example a moving
triplet and a stationary triplet will not agree how fast the moving triplet
is moving. (I added a third twin, to merge with your gedanken. I hope
their mom doesn't mind.)
> You see me in
> slow motion and I see you in slow motion, so both our ages are reduced by
> the same amount.
Not the same amount, no. The sign is correct ("slowed"), but the magnitude
is not.
> Is that not the case?
Answered. Look at the FAQ again.
David A. Smith
Dan Bloomquist
Stephen Bint wrote:
> "Dan Bloomquist" <lak...@citlink.net> wrote in message
>>Why not go back to Jon's quite eloquent explanation of the twins and
>>read it again. You seemed to get it then.
>
> Since you suggested it, I did. It stands out as the best post on this
> thread. Here's a quote:
>
>
Jon:
>>By symmetry, he sees exactly the same things happening as he
>>observes your clock during his journey. And *neither* of you
>>actually *sees* the other's clock "run slow," because of the
>>Doppler effect caused by your approaching each other.
>
> The problem is not "deciding" which frame is the one running slow, by
> clutching at an asymmetry. The problem is that time dilation fails to
> prevent witnesses seeing the bouncing beam going faster than light, unless
> it works both ways and the Earth-bound observer has also aged less than the
> twins.
>
> If you say either frame is the slow one, you deviate from what the formula
> asserts, which is plainly that both are slow by the same factor.
>
Do you see the *sees*? The FoR 'is slow', is not the same as the
observation 'is slow'. From the earth, the observation is that both
clocks are running at the same rate! I think Jon made the distinction
between observations and *is* very clear.
>
> you are appealing to me to be open-minded enough to be able to believe it is
> possible that I am wrong. I would ask the same of you.
>
It isn't about being 'open minded'. The math works. The observations are
consistent. Since 1905 there has been no hole found by minds very much
greater than ours. (I'm just a little mind. :) ) As you muddle through
all the 'paradoxes', (And I still do), learn the principles. When it is
all very clear in your head, then you can look beyond it for something
new. But without a contradictory observation to SR, that something new
will only be an extension of SR.
Stephen Bint
> > The formula for time dilation is a function of velocity. If I am moving
> at a
> > given velocity with respect to you, then you are moving at the exact
same
> > velocity with respect to me.
>
> Not quite correct.
>
In that case, if I am moving at 10kph with respect to you, at what velocity
will you be moving, with respect to me? Greater, or less?
Stephen
"dl...@aol.com (formerly)" <dlzc1.cox@net> wrote in message
news:sy0ob.123157$gv5.53477@fed1read05...
Bonnie Granat
> David,
>
> > > The formula for time dilation is a function of velocity. If I am
moving
> > at a
> > > given velocity with respect to you, then you are moving at the exact
> same
> > > velocity with respect to me.
> >
> > Not quite correct.
> >
> In that case, if I am moving at 10kph with respect to you, at what
velocity
> will you be moving, with respect to me? Greater, or less?
>
> Stephen
>
You haven't yet said what my velocity is. Am I stationary?
--
Bonnie Granat
Granat Technical Editing and Writing
http://www.editors-writers.info
Rick Sobie
The atoms in the one twin are moving faster and hence
due their moving fast, the function of aging,
is slowed. The normal interactions of the atoms
are slowed as opposed to the functions of atoms
in a normal state of rest.
The twin on earth, ages, the twin travelling
does not age at the same rate. Time is not
consistent between bodies at rest and bodies
in accelleration. Simply because atoms act
differently at rest, or under the influence
of accelleration.
Time is nothing in and of itself. If it were
then you would say, poppycock. They are the same age.
One just didn't age as well as the other.
Stephen Bint
"Bonnie Granat" <bgr...@editors-writers.info> wrote in message
news:3fa0...@andromeda.5sc.net...
>
> "Stephen Bint" <bi...@iname.com> wrote in message
> news:3fa098d6$0$103$65c6...@mercury.nildram.net...
> > David,
> >
> > > > The formula for time dilation is a function of velocity. If I am
> moving
> > > at a
> > > > given velocity with respect to you, then you are moving at the exact
> > same
> > > > velocity with respect to me.
> > >
> > > Not quite correct.
> > >
> > In that case, if I am moving at 10kph with respect to you, at what
> velocity
> > will you be moving, with respect to me? Greater, or less?
> >
> > Stephen
> >
>
> You haven't yet said what my velocity is. Am I stationary?
>
anyone or anything else. I am moving at 10kph with respect to you. It could
be that you are travelling along a road at 45kph and I am overtaking you at
55kph. Or you could be standing at the side of the road while I pass you in
a car doing 10kph.
The "v" in the time dilation equation refers only to the observed object's
speed with respect to the observer. Our speeds with respect to everything
else have no effect on the outcome of the calculation of time dilation.
Jon Bell
Stephen Bint <bi...@iname.com> wrote:
>
>The problem is not "deciding" which frame is the one running slow, by
>clutching at an asymmetry. The problem is that time dilation fails to
>prevent witnesses seeing the bouncing beam going faster than light, unless
>it works both ways and the Earth-bound observer has also aged less than the
>twins.
>
>If you say either frame is the slow one, you deviate from what the formula
>asserts, which is plainly that both are slow by the same factor.
In your previous example, all factors (time dilation and kinematic) are
symmetrical between the two twins, so they experience the same elapsed
time. In the classic "twin paradox", the time dilation is symmetric, but
the kinematic factors aren't. Therefore the twins experience different
elapsed times.
Here's a description of a classic "twin paradox" situation, in the same
style as my description of your previous example:
The scenario: You stay behind on Earth while your twin goes on a space
journey. From your point of view he travels away from the Earth at a
speed of 0.8c for 5 years, covering a distance of 4 light-years in the
process, then immediately turns around and returns, again at a speed of
0.8c, taking another 5 years for the return trip. The total trip duration
is 10 years by your reckoning.
Relativity predicts that your twin experiences less elapsed time
because of time dilation:
(5 years) * sqrt (1 - 0.8^2) = 3 years
for each leg of the trip, and from his point of view the round trip lasts
only 6 years.
The relativistic time dilation equation predicts that each twin's clocks
"run slower" in the other twin's reference frame. So why can't your twin
conclude that the trip must be shorter for you, than it is for him? The
answer lies in the fact that your experiences are not symmetrical. Your
twin is at rest in two different inertial reference frames, one during the
outbound trip and another one during the inbound trip. You remain at rest
in a single inertial reference frame during the entire journey. Your twin
has to fire his spaceship's engines at the turnaround point. You do
nothing.
By examining what both of you actually *see* by watching each other's
clocks/calendars through telescopes, we can show that both of you must
come to the same conclusion: the trip lasts 10 years for you, and 6 years
for him.
First let's look at what's happening from your point of view.
To be specific, assume that your twin starts out on his journey at the
very beginning of the year 2004. You watch his clock (calendar?) through
your telescope as he recedes. The rate at which you receive the images of
each of his "new years" depends not only on his time dilation, but also on
the fact that he is moving away from you, so we have to use the
relativistic Doppler-effect equation. You see his calendar turn over to a
new year at intervals of
(1 year) * sqrt ((1 + 0.8) / (1 - 0.8)) = 3 years.
Therefore,
At the beginning of 2007, you see his calendar turn over to 2005.
At the beginning of 2010, you see his calendar turn over to 2006.
At the beginning of 2013, you see his calendar turn over to 2007.
This is the point at which you see him turn around and begin his return
trip. In your reference frame, he actually began the return trip at the
beginning of 2009 (2004 + 5), but you don't see it until 4 years
later because it happened 4 light-years away from you.
During his return trip, we have to switch the signs in the Doppler-shift
equation because now he's approaching, not receding. You now see his
calendar turn over to a new year at intervals of
(1 year) * sqrt ((1 - 0.8) / (1 + 0.8)) = 1/3 year.
Therefore,
At the beginning of May 2013, you see his calendar turn over to 2008.
At the beginning of September 2013, you see his calendar turn over to
2009.
At the beginning of January 2014, you see his calendar turn over to 2010,
and he arrives home. Ten years have elapsed on your calendar, and 6 years
have elapsed on his.
What does this look like from your twin's point of view, as he watches
your clock/calendar through his telescope?
As he is traveling away, he sees your calendar turn over a new year at
intervals of three years, just like you do his. Therefore,
At the beginning of 2007 (2004 + 3), he sees your calendar turn over to
2005.
But three years is the length of the outbound trip, according to him,
because of time dliation. So at this point he turns around and begins his
return trip. Now he sees your calendar turn over a new year at intervals
of 1/3 year, just like you do his. Therefore,
At the beginning of May 2007, he sees your calendar turn over to 2006.
At the beginning of September 2007, he sees your calendar turn over to
2007.
At the beginning of January 2008, he sees your calendar turn over to 2008.
At the beginning of May 2008, he sees your calendar turn over to 2009.
At the beginning of September 2008, he sees your calendar turn over to
2010.
At the beginning of January 2009, he sees your calendar turn over to 2011.
At the beginning of May 2009, he sees your calendar turn over to 2012.
At the beginning of September 2009, he sees your calendar turn over to
2013.
At the beginning of January 2010, he sees your calendar turn over to 2014.
But now three years have elapsed (on his calendar) since he turned around,
so he has now returned home. Six years have elapsed on his calendar, and
10 years have elapsed on yours, in agreement with what you observe.
Note that although we did not use the time dilation equation directly in
predicting what your twin sees your clock/calendar doing, it is implicit
in the relativistic Doppler-shift equation. In deriving the relativistic
Doppler shift equation, one must use the time-dilation equation.
--
Jon Bell <jtbe...@presby.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA
Stephen Bint
I see that you are asserting, that the accelerations undergone by one
observer affect time dilation. Yet the velocity, v, is the only variable in
the time dilation formula, is it not? Does that not imply, that the
calculation of the degree of time dilation for either party, is irrespective
of anything but their velocity, with respect to eachother?
Will you indulge me, by considering an example of mine, that illustrates the
symmetry implicit in the assertion of time dilation?
Two skateboards glide past eachother at speed.
On each skateboard is an observer and a device which sends a light pulse
from a source which is 1m above the base, vertically down to a mirror on the
base, which reflects it directly vertically back to the source. To one of
the observers (either one) the light pulse on the other board traces a
V-shaped path, which is longer than the vertical one. If there is no time
dilation, the observer will see the pulse travel a greater distance in the
same time, and therefore, see it travelling faster than c.
Time dilation slows the event on the other skateboard, the exact amount
necessary to make the light pulse travel this longer path at exactly c. The
length of the V-shaped path and consequently, the time dilation required, is
purely a function of the relative velocity of the two skateboards.
Furthermore, the path that A sees the pulse trace on B's skateboard is
exactly the same length as the path seen by B, taken by the light pulse on
A's skateboard.
So the time dilation must be identical in both directions, regardless of how
they acheived their relative velocity.
So what if your travelling twin was passing earth and viewing such an
experiment there. Either he will see time slowed down there, or he will see
the light pulse travelling faster than c there. Is that not the case?
"Jon Bell" <jtbe...@presby.edu> wrote in message
news:bnq8m5$i2i$1...@jtbell.presby.edu...
Bonnie Granat
> Jon,
>
> I see that you are asserting, that the accelerations undergone by one
> observer affect time dilation. Yet the velocity, v, is the only variable
in
> the time dilation formula, is it not? Does that not imply, that the
> calculation of the degree of time dilation for either party, is
irrespective
> of anything but their velocity, with respect to eachother?
>
> Will you indulge me, by considering an example of mine, that illustrates
the
> symmetry implicit in the assertion of time dilation?
>
> Two skateboards glide past eachother at speed.
>
> On each skateboard is an observer and a device which sends a light pulse
> from a source which is 1m above the base, vertically down to a mirror on
the
> base, which reflects it directly vertically back to the source. To one of
> the observers (either one) the light pulse on the other board traces a
> V-shaped path, which is longer than the vertical one. If there is no time
> dilation, the observer will see the pulse travel a greater distance in the
> same time, and therefore, see it travelling faster than c.
>
But if the assumption for this experiment is that c is the top speed and the
constant speed of light, the mere fact that the light path is longer tells
me that time is dilated.
Doesn't it, Bill Hobba???
Ron House
> Claiming an assymetry based on the acceleration histories of the observer,
> overlooks the fact that the time dilation must be real and identical in both
> directions at all times or else someone will see the bouncing beam going
> faster than c.
>...
> To say that acceleration makes time run faster, implies that if one of the
> skateboarders is accelerating, the time contraction due to acceleration will
> at least partially counteract the dilation caused by the velocity (predicted
> by Einstein), allowing the other to see the bouncing beam move faster than
> c.
There seem to be two misunderstandings here, one possibly yours and one
by those who (not having read the previous posts) I presume have been
talking about acceleration to you.
To put the record straight:
(1) An unaccelerated and an accelerated observer are NOT equivalent, but
(2) acceleration does NOT cause time dilation.
Suppose twin A stays at home and twin B travels out, accelerates towards
earth, and returns younger than twin A.
Acceleration has NOT (repeat, NOT) caused the time dilation of twin B.
In A's frame, B travelled faster than A at all times (except the small
reversal time) and so suffered a relative time dilation.
No equivalent argument can be made from twin B's point of view, because
B does not have a single reference frame. B on the outward journey will
see A suffering a dilation, but B's own return to earth later, from B's
own earlier frame, will seem like an even greater relative speed with an
even greater time dilation (twice A's) that will lead B to conclude that
B will suffer time dilation relative to A. That is, both A and B predict
that B will age less. This is because A has only one reference frame for
the whole time, but B has two different frames at different times, and
from either of these frames, B will conclude that B will age less.
But acceleration had nothing to do with it. If B were replaced by two
entities, B1, who makes the outward journey at constant speed, who
passes B2 moving in the reverse direction and who syncronises clocks
with B2 as they pass, then the time on B2's clock when he gets back will
show the same time dilation as the accelerated original B. Since neither
B1 nor B2 accelerated, this dilation is not due to the acceleration.
Any observer in a single reference frame, according to SR, will have an
interpretation of reality that is consistent with those of all other
observers, once allowance for the relative motion is made. But an
accelerated observer is not in a single reference frame, because the
acceleration changes his frame. That is the only relevance of
acceleration to the situation.
The simplest way I know to think of SR is as a geometric theory. The
formula for the length of the diagonal of a right triangle is the well
known Pythagoras' formula, but, on the time axis, bung in a minus sign.
This has the effect that, for space-like intervals, the shortest
distance between two points is a straight line, whereas for timelike
intervals, this is the longest distance. Bent timelike curves are
shorter than straight ones. Accelerated paths (by the definition of
acceleration) are bent ones, and therefore shorter.
To take this analogy a step further, suppose two points are marked on
the ground, P and Q. There is one piece of straight rope connecting P
and Q, and another curved piece doing likewise. You know just by looking
at it that the curved piece must be longer than the straight one. But it
is not that the curve in the rope has somehow stretched it. Now replace
P and Q by the timespace locations of A at the beginning and end of B's
journey, the straight rope by A's stationary (on earth) trajectory, the
curved rope by the path of the spaceship carrying B, "curve in the rope"
by "acceleration of B", and "longer" by "shorter". With these changes,
the situations are identical, and there is nothing baffling or
paradoxical about it at all.
--
Ron House ho...@usq.edu.au
http://www.sci.usq.edu.au/staff/house
Dirk Van de moortel
"Stephen Bint" <bi...@iname.com> wrote in message news:3fa07dad$0$118$65c6...@mercury.nildram.net...
> I started a thread about the twin paradox and cross-posted it to three
> newsgroups. I am grateful to those who discussed it, even though it got a
> tiny bit personal at times :)
Troll alert:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/AChance.html
Dirk Vdm
Martin Hogbin
"Stephen Bint" <bi...@iname.com> wrote in message news:3fa0893c$0$122$65c6...@mercury.nildram.net...
.>
> you are appealing to me to be open-minded enough to be able to believe it is
> possible that I am wrong. I would ask the same of you.
There is a difference. Dan's view is supported by thousands
of physicists throughout the world who have studied the subject
in great detail for nearly a century.
You should start by understanding how two inertial observers
in relative motion would measure one another's clocks. It
would then be useful to understand how they would _see_
one another's clocks, which is different.
You have a choice. You can either join the list of crackpots
who frequent this group, and be treated with scorn and derision
by the physicists here, or you can make an effort to understand
the subject.
Martin Hogbin
Dirk Van de moortel
"Martin Hogbin" <sp...@hogbin.org> wrote in message news:bnqmjp$3q7$1...@hercules.btinternet.com...
He has been here before with another name.
Same troll, new name.
Dirk Vdm
sal
understand the heart of the problem, which is clock synchronization.
(Well, that's _one_ heart of the problem, anyway...)
I posted some of the math for the twin paradox in your earlier thread,
but I only posted it on alt.sci.physics. So, I've taken this
opportunity to spruce it up by adding the affine transformations to
rezero the clocks.
Others have addressed the issue of what each twin sees through a
telescope, so I haven't done that here. I also didn't take the extra
step of filling space with "virtual observers" and reporting what they
see. That's also informative and is easier than dealing with telescopes
since it doesn't involve Doppler shift -- you can do it just from the
transforms as given here.
With no further introduction, here it is again.
(Pick holes, if you will -- I'm always willing to learn...)
===================================================================
The problem as stated was "two twins fly to Earth, each from a distance
of 1 ly, at a velocity of 0.75c. What happens?"
I worked this one through. There's no contradiction. It is, however,
fiendishly confusing.
Typing equations in Ascii is not a lot of fun so I'm going to try to
keep this brief. (Ha, ha.)
I'll try to describe what happens in English, with limited use of
equations. First, I made a few simplifications of the original
statement of the problem to keep the math tractable:
-- Set c=1
-- All dimensions are the same -- time and distance are both measured
as lengths (conversion factor = c, which we set equal to 1).
-- I used a velocity of 1/2 rather than 0.75
-- I assumed that A starts at location -1, B starts at location +1,
and there's an observer in the middle at location 0 who remains
stationary in the initial inertial frame (before anybody
accelerated). Call him "O".
Call the inertial frame they all started in the "base frame".
Then at time 0 in the base frame, an alarm goes off in each spaceship
and at the origin, telling A, B, and O it's time to start. A and B
immediately accelerate to +1/2 and -1/2 respectively.
Gamma and beta are as usual: beta=velocity, gamma=1/sqrt(1-beta^2).
Let's start looking at things from the point of view of 'O'.
In this frame, A is moving at beta=1/2, gamma=2/sqrt(3). The Lorentz
transform for frame(O) -> frame(A) looks something like this, with a
unit width font and a 1-dimensional world (y and z components are
dropped out):
| 2/sqrt(3) -1/sqrt(3) |
| -1/sqrt(3) 2/sqrt(3) |
To get from frame(O) -> frame(B) it looks like this:
| 2/sqrt(3) 1/sqrt(3) |
| 1/sqrt(3) 2/sqrt(3) |
And the transform from A --> O is the same as the one from O --> B
(just reverse the velocity, of course).
Since A starts at X=(0,-1) in the base frame, we transform that to A's
(new) frame and we see that X' = (1/sqrt(3), -2/sqrt(3)).
The distance to the origin (X'=0) went down, 'cause it was contracted,
so it's just 2/sqrt(3) in A's frame. BUT the TIME changed, too --
it's 1/sqrt(3), _not_ 0. This is where the confusing part starts, and
it's worth a few words.
No time passed for A, and his clock still reads 0. However, if, after
accelerating, he _again_ synchronizes his clock with someone located
at X=0, the result won't be the same -- as far as he's concerned, O
and B have both now got clocks that read incorrectly. In simple
terms, he needs to use light-speed signals the sync up his clocks, and
the SOL delay messes up the operation. Since A's clock hasn't
actually changed, and still reads 0, we're going to just subtract
1/sqrt(3) from all times in A's frame from now on. (This makes things
a little messier and is somewhat error prone, unfortunately.)
Anyway, now we've got A with a clock that reads 0, and an apparent
distance to the origin of 2/sqrt(3). He's coming toward O at v=1/2,
so at time t=2 in O's frame, A arrives. In O's frame, the coordinates
are (2,0). We transform that to A's frame, and get
(4/sqrt(3),-2/sqrt(3)) -- BUT we need to subtract 1/sqrt(3) to get A's
actual clock reading. So A's clock must read 3/sqrt(3) = sqrt(3).
That's ~ 1.7, which is rather less than the 2 units that elapsed for
O. A aged less than O.
The behavior of B, from the point of view of O, is identical,
including the subtraction of 1/sqrt(3) to zero his clock. So, as far
as O can see, B also ages by 3/sqrt(3). Since the math is the same
I'm not including it here.
Now, the interesting part: We work it out again from A's point of
view. In A's frame, beta(O) = -1/2, gamma(O) = 2/sqrt(3), and the
transform from A to O is the same as the one from O to B. But we also
need the transform from A --> B, and for that we need beta(B) in A's
frame -- beta'(B), if you will.
B's velocity vector in B's rest frame is (1,0) (moving through time at
rate 1, stationary in space). We can transform that to O's frame, and
we get (2/sqrt(3),-1/sqrt(3)) (this is the 4-velocity with two terms
dropped out, of course). We can now transform that from O's frame to
A's frame using the transform we worked out back at the top, and we
get (5/3,-4/3). The "time" coordinate is just gamma, so gamma(B) in
A's frame is 5/3. The "x" coordinate is gamma*beta, so we divide it
by gamma and see that beta(B) in A's frame must be -4/5. If we plug
that back into the formula for gamma it works out to 5/3 again (which
is a relief): 1/sqrt(1-16/25) = 1/sqrt(9/25) = 5/3.
So the Lorentz transform to get from A's frame to B's frame is
| 5/3 4/3 |
| 4/3 5/3 |
Now, let's go back and figure out how long it takes A to get to the
origin; then we'll look at B again.
A finished accelerating at coordinates (0, -2/sqrt(3)) in his new
frame, including the clock correction. Now, we need to figure out
where the origin is in A's frame. It's _not_ 1 unit away, because
we're measuring the distance to travel in _O_'s frame (and that's the
fundamental asymmetry in this problem, by the way). Looking at O's
frame from A's frame, all distances are contracted, and the length A
needs to travel is 1/gamma = sqrt(3)/2, which is about 0.87.
Now, A sees O approaching at beta=-1/2, so the elapsed time is
(sqrt(3)/2)/(-beta) = sqrt(3). Let's look to see what's happening at
time sqrt(3) in O's frame.
We need to add back 1/sqrt(3) to A's clock and then transform it.
With the 1/sqrt(3) added in, A's coordinates at that point are
(4/sqrt(3),-2/sqrt(3)) (he's stationary in his own rest frame, of
course, so he's still sitting at -2/sqrt(3)). That transforms to
(2,0) in O's frame. Sigh of relief - A and O arrive at the same event
at the same time. Elapsed time for A = sqrt(3), just as we found
before, and elapsed time for O is 2, just as we found before.
Now, let's look at B from A's frame. In the base frame, B's starting
coordinates were (0,1). Transformed to A's frame, they're
(-1/sqrt(3),2/sqrt(3)) ... but with the clock correction of 1/sqrt(3)
that gives us -2/sqrt(3). It seems that B got a head start -- he
started his engine a EARLY from A's point of view. THIS IS IMPORTANT.
The "simultaneous" events in the base frame were NOT simultaneous in
the moving frames of A and B -- and that's where "intuition" totally
jumps the tracks.
If we transform that directly to B's frame using the matrix above
(after adding back that pesky 1/sqrt(3)), we get
(1/sqrt(3),2/sqrt(3)), which we're pleased to see agrees with earlier
results. Of course, we need to subtract 1/sqrt(3) from B's clock, too,
after which we see that B starts at time 0 by _his_ clock.
Now, at time 3/sqrt(3) in A's frame, we find
(3/sqrt(3)+1/sqrt(3),-2/sqrt(3)) maps to (4/sqrt(3),2/sqrt(3)) in B's
frame. Since 2/sqrt(3) is where B's been sitting still in his own
rest frame, A and B do indeed meet at that point. When we subtract
the clock correction of 1/sqrt(3) from B's time we see that his clock
reads 3/sqrt(3) = sqrt(3) -- the same as A's, and the same as what O
predicted. The twins aged the same amount.
Whew.
Again, the problem throughout is that once two frames are moving
relative to each other, they _cannot_ have synchronized clocks -- each
one sees that the other's clocks are skewed progressively with
distance. So, the result works out mathematically, it's supported
experimentally, but it doesn't make a whole lot of sense intuitively.
If you hate the clock corrections -- well, that's the point. The
skewed clocks are what make the math work, and they're what make the
whole thing so confusing.
=========================
To email directly replace nospam with foobox
dlzc@aol.com (formerly)
"Stephen Bint" <bi...@iname.com> wrote in message
news:3fa098d6$0$103$65c6...@mercury.nildram.net...
> David,
>
> > > The formula for time dilation is a function of velocity. If I am
moving
> > at a
> > > given velocity with respect to you, then you are moving at the exact
> same
> > > velocity with respect to me.
> >
> > Not quite correct.
> >
> In that case, if I am moving at 10kph with respect to you, at what
velocity
> will you be moving, with respect to me? Greater, or less?
I am assuming I am stationary...
Remember the title of the thread? Since I am disallowed from doing the
math, I can only answer approximately. A little less than 10kph.
Something around the 12th or 18th decimal place.
Relativity only shows up when you move *very* fast, or when you can measure
very precisely. Otherwise Newton (or another of history's geniuses) might
have spotted it.
David A. Smith
sal
math for the "symetric twin" paradox. Since this discussion is really
just the earlier twin paradox restated, the same analysis applies to both.
However, there's one specific point which should be addressed, and which
Steve Bint should try to grok. Until you get this you'll never see why
the math works out, let alone understand what "really" happens.
Stephen Bint wrote:
>
>
> It baffles me, that intelligent people, capable of algebra, cannot
> see that the assertion that time is slowed in a moving frame is a
> symmetrical assertion, because frames move relative to each other,
> symmetrically.
>
When two frames are moving relative to each other, their clocks can NOT
be in sync. What this means, in confusing English, is that when I move
through SPACE in a frame which is moving relative to me, I am also
moving through TIME in that frame! The X axes are not aligned --
they're skewed.
So, the time rates are slowed symetrically ... but the rate of time
change with X position is mirror-imaged between the frames.
If this makes no sense to you (and I expect it does) then you should
really try to work through the math to see why it's asserted as part of
SR. The math, though messy, is _just_ linear algebra -- nothing more.
--
To email me directly, take out nospam and put back foobox.
Harry
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:3fa0df79$1...@usenet01.boi.hp.com...
What was his old name, do you think?
Harald
Harry
"Stephen Bint" <bi...@iname.com> wrote in message
news:3fa07dad$0$118$65c6...@mercury.nildram.net...
> I started a thread about the twin paradox and cross-posted it to three
> newsgroups. I am grateful to those who discussed it, even though it got a
> tiny bit personal at times :)
>
> I was frustrated by the account of the twin paradox in the FAQ for
> sci.physics (http://www.faqs.org/faqs/physics-faq/part4/), because it
> resorts to saying that there is assymetry in the situation. One twin has
> undergone an acceleration and it is that one which is younger.
I agree, the faqs can be improved for sure.
Acceleration certainly isn't the official explanation; it's merely an
indicator of asymmetry.
(But you could have noticed, if you read it carefully: "and" instead of
"therefore"...)
SNIP
> It baffles me, that intelligent people, capable of algebra, cannot see
that
> the assertion that time is slowed in a moving frame is a symmetrical
> assertion, because frames move relative to eachother, symmetrically.
But everyone sees that!
> Besides which, the formula for time dilation is a function of the velocity
> of the observed frame and makes no mention whatsoever of the acceleration
of
> either frame, before or while the calculation is made.
True.
> I was told again and again by several posters, "Do the math. Just plug the
> numbers into the formulas and it all works out just fine." These were the
> same people who want me to factor in acceleration, which is not mentioned
in
> the formula, because time dilation is purely a function of the relative
> velocity, which is the same in both directions.
Acceleration of the ref. frame of observation is not part of SRT, and
therefore not included in the equations either.
However, it is acknowledged that acceleration has no relativistic effect on
the accelerated clock... and then it gets tricky! ;-)
I may post another time more about that point, which is one that made me
choose LET instead of SRT as standard theory.
In case you are really new, LET is the abbreviation used in this group for
Lorentz' interpretation of his transformations and SRT, which differed from
that of Einstein. Nowadays only Einstein's interpretation is taught at
school. No need to tell more, you can find many hours of reading in the
archives... but the same is true for the Twin paradox!
> Plug the relative velocity of skateboarders A and B into the formula and
> find out how much slower B's clock is, than A's. And how much slower A's
is,
> than B's. I don't need to "do the math" to see that this formula is
telling
> me two things which cannot both be true.
>
> I suppose it depends on your definition of "working out fine".
>
> Because I constructed my example with two moving twins, to force them to
> face the symmetry, the defenders of relativity introduced a new twist:
> acceleration causes time to speed up, cancelling out the age difference,
in
> both skateboarders. This is an new addition to relativity. Einstein never
> said anything of the sort.
Wrong, he did, in 1918; it was -as he put it himself- his clever way out of
the Twin problem (in German, sorry).
> To say that acceleration makes time run faster, implies that if one of the
> skateboarders is accelerating, the time contraction due to acceleration
will
> at least partially counteract the dilation caused by the velocity
(predicted
> by Einstein), allowing the other to see the bouncing beam move faster than
> c.
Really? Make sure that you know which acceleration has what effect on which
clock and according to who!
> Still, though frustrated, I thoroughly enjoyed the thread I started and I
am
> grateful to all of the people who had the patience to contribute. In
return,
> I can only assure you that I will continue to post under the name "Stephen
> Bint", so as not to undermine your killfile strategies.
>
> And may I say in conclusion, that trying to bring the sweet light of
reason
> to the proponents of relativity, is like trying to bring salvation to the
> DAMNED.
Don't overestimate yourself!
Harald
Paul Cardinale
[crap snipped]
I see that you are both a jackass and an idiot.
You believe that because you are too stupid/lazy/arogant to learn the
math, that all you need to do is wave your arms and try to apply your
own ignorant comic-book misconception of relativity. You are a
clueless ineducable blubbering baffoon.
Paul Cardinale
Dirk Van de moortel
"sal" <beli...@nospam.com> wrote in message news:XJ8ob.13181$lC5....@nntp-post.primus.ca...
> The problem as stated was "two twins fly to Earth, each from a distance
> of 1 ly, at a velocity of 0.75c. What happens?"
Earth uses coordinates (x,t)
Right twin uses coordinates (x',t')
Left twin uses coordinates (x",t")
World line Earth: x = 0 (t-axis)
World line Right twin: x' = 0 (t'-axis)
World line Left twin x" = 0 (t"-axis)
Right twin is present at event R: (x,t) = (D,0)
where/when his clock is set to t' = 0,
and he approaches Earth with relative speed v
Left twin is present at event L: (x,t) = (-D,0)
where/when his clock is set to t" = 0,
and he approaches Earth with relative speed v
[M]
/|\
/ | \
t"/ t \t'
/ | \
/ | \
/ | \
/ | \
-----[L]------|------[R]------x
Transformation Earth coord --> Right twin coord:
{ x' = g( (x-D) + vt )
{ t' = g( t + v(x-D)/c^2 )
Transformation Earth coord --> Left twin coord:
{ x" = g( (x+D) - vt )
{ t" = g( t - v(x+D)/c^2 )
where in both cases
g = 1/sqrt(1-v^2/c^2)
Meeting of the 3 world lines at event M:
{ x = 0
{ x' = 0
{ x" = 0
==> 2 lines of algebra ==>
(x,t) = ( 0, D/v )
(x',t') = ( 0, D/(gv) )
(x",t") = ( 0, D/(gv) )
According to Earth D/v seconds have passed.
According to both twins D/(vg) seconds have passed.
Dirk Vdm
Dirk Van de moortel
"Harry" <harald.v...@epfl.ch> wrote in message news:3fa12183$1...@epflnews.epfl.ch...
> See below.
>
> "Stephen Bint" <bi...@iname.com> wrote in message
> news:3fa07dad$0$118$65c6...@mercury.nildram.net...
[snip]
> > I was told again and again by several posters, "Do the math. Just plug the
> > numbers into the formulas and it all works out just fine." These were the
> > same people who want me to factor in acceleration, which is not mentioned
> in
> > the formula, because time dilation is purely a function of the relative
> > velocity, which is the same in both directions.
>
> Acceleration of the ref. frame of observation is not part of SRT,
Ha, but it *is* part of SRT:
http://groups.google.com/groups?&threadm=a3M1b.82977$F92....@afrodite.telenet-ops.be
> and
> therefore not included in the equations either.
I think it is not included in the standard twin paradox
treatment because it is irrelevant.
When the accelerations are taken large enough, such
that the cruising velocities are reached in a short time,
then what happens during these accelerating phases
can be ignored when compared with what happens
during the cruising phases.
Dirk Vdm
Dirk Van de moortel
"Harry" <harald.v...@epfl.ch> wrote in message news:3fa12123$1...@epflnews.epfl.ch...
[snip]
> > He has been here before with another name.
> > Same troll, new name.
> >
> > Dirk Vdm
>
> What was his old name, do you think?
Not sure yet. His style suggests a few candidates.
But I could be wrong of course, this one could be
brand-new after-all :-)
Dirk Vdm
Dirk Van de moortel
"Paul Cardinale" <pcard...@volcanomail.com> wrote in message news:64050551.03103...@posting.google.com...
You are too kind, sparing him like that ;-)
Dirk Vdm
Uncle Al
[snip}
No need to do the math, eh? Crackpot. Special Relativity is a set of
hyperbolic rotations in 4-space. Do those matrix multiplications
intuitively, git.
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/gratim.html
http://www.eftaylor.com/pub/projecta.pdf
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
Stephen Bint
When you warned me that G=EMC^2 Glazier <herbert...@webtv.net> is a
troll, I didn't like the way you elected yourself judge and jury over who is
a troll and who isn't. Nor did I like the way you saw fit to speak on behalf
of the others on this newsgroup.
Dirk:
"I think you don't want to know what the Bert believes.
You might want to know that in the beginning the Bert
used to be a Herb. We all wonder whether next year
the Bert will call himself a Terp or a Terk, and the year
after, whether the Terp will call himself a Perd or a Perl,
or perhaps whether the Terk will become a Kerl or a
Kerm. To survive on this forum one needs patience.
But you'll find that out soon enough."
I will be the judge of whether I want to know what Bert believes.
Now you are accusing me of being a troll, I like it even less. I have
started one thread on alt.sci.physics.new-theories about my cranky
strobing electron theory and contributed to a couple of other threads.
I have started two threads about the twin paradox, which I have
cross posted to three groups. I have confined myself almost entirely
to discussing the topics of these threads with those adults who have
chosen to discuss them with me.
Who the hell are you to declare me a troll, or any sort of pest? The threads
I have started concern subjects of great interest to me and I have every
right to discuss them here. I suppose you think the people contributing to
these discussions are all fools who have fallen into my trap. They come
across as a good deal more intelligent than you.
What I really take exception to is this:
> He has been here before with another name.
> Same troll, new name.
>
> Dirk Vdm
This is a lie. You are lying about me, though I have done nothing to you,
nor have I been rude to anyone on this list, except Uncle Al, who was much
ruder to me first.
I think you owe me an apology.
Stephen Bint
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:3fa0df79$1...@usenet01.boi.hp.com...
Dan Bloomquist
Stephen Bint wrote:
> Jon,
>
>
> So the time dilation must be identical in both directions, regardless of how
> they acheived their relative velocity.
>
> So what if your travelling twin was passing earth and viewing such an
> experiment there. Either he will see time slowed down there, or he will see
> the light pulse travelling faster than c there. Is that not the case?
>
This seems to be the most difficult hurdle in SR. It happens when you
try to make sense of the clocks from some god like frame. But this is
not just a time like phenomena. It is a spacetime phenomena.
You can not be in multiple frames at once and expect the clocks to make
sense with each other. This is because you are taking the privilege of
discounting the velocity which changes the distance between A and B.
When it is said to 'do the math', it means look at it one frame at a
time and don't drag the perceptions of one frame to another. It also
means don't create a special frame that sees something else. This
special frame is not in the theory so creating it creates a paradox in
your mind not the theory.
Stephen Bint
"Paul Cardinale" <pcard...@volcanomail.com> wrote in message
news:64050551.03103...@posting.google.com...
You can only resort to personal attacks. What you cannot do is face the
issue, which I keep bringing up for you to ignore. You call me names, you
bleat "do the math" and you run like a cur from the question at hand.
This is the question I have asked Jon Bell. You won't be able to answer it
and I suspect you won't even understand it:
Two skateboards glide past eachother at speed.
On each skateboard is an observer and a device which sends a light pulse
from a source which is 1m above the base, vertically down to a mirror on the
base, which reflects it directly vertically back to the source. To one of
the observers (either one) the light pulse on the other board traces a
V-shaped path, which is longer than the vertical one. If there is no time
dilation, the observer will see the pulse travel a greater distance in the
same time, and therefore, see it travelling faster than c.
Time dilation slows the event on the other skateboard, the exact amount
necessary to make the light pulse travel this longer path at exactly c. The
length of the V-shaped path and consequently, the time dilation required, is
purely a function of the relative velocity of the two skateboards.
Furthermore, the path that A sees the pulse trace on B's skateboard is
exactly the same length as the path seen by B, taken by the light pulse on
A's skateboard.
So the time dilation must be identical in both directions, regardless of how
they acheived their relative velocity.
So what if your travelling twin was passing earth and viewing such an
experiment there. Either he will see time slowed down there, or he will see
the light pulse travelling faster than c there. Is that not the case?
So there it is. Don't worry if you can't see the point or understand the
question. Just call me few names and bleat "Do the math!" over and over and
everyone will think you are mature and intelligent.
Stephen
Stephen Bint
actually copied from a british Open University program, where it is used as
an illustration of time dilation. Because time dilation prevents observers
from seeing the light pulse travel faster than c, it must apply equally to
both observers. If it doesn't one of them will see the light pulse move
faster than c.
I know I am repeating myself a lot, but there are so many posts since I gave
that example and not one of them has referred to it. Surely, the velocity
determines the length of the path and time dilation must be employed to slow
it - BOTH ways.
> When it is said to 'do the math', it means look at it one frame at a
> time and don't drag the perceptions of one frame to another. It also
> means don't create a special frame that sees something else. This
> special frame is not in the theory so creating it creates a paradox in
> your mind not the theory.
Yes, but time dilation mustr have a meaning. It must have an effect. It must
reconcile classical expectations of seeing things go faster than c, by
stretching the time frame of anything with a velocity, so that you see slow
motion in that moving frame. Literally.
Stephen
"Dan Bloomquist" <lak...@citlink.net> wrote in message
news:3FA1414F...@citlink.net...
Harry
>
> "Harry" <harald.v...@epfl.ch> wrote in message
news:3fa12183$1...@epflnews.epfl.ch...
> > See below.
> >
> > "Stephen Bint" <bi...@iname.com> wrote in message
> > news:3fa07dad$0$118$65c6...@mercury.nildram.net...
>
> [snip]
>
> > > I was told again and again by several posters, "Do the math. Just plug
the
> > > numbers into the formulas and it all works out just fine." These were
the
> > > same people who want me to factor in acceleration, which is not
mentioned
> > in
> > > the formula, because time dilation is purely a function of the
relative
> > > velocity, which is the same in both directions.
> >
> > Acceleration of the ref. frame of observation is not part of SRT,
>
> Ha, but it *is* part of SRT:
>
http://groups.google.com/groups?&threadm=a3M1b.82977$F92....@afrodite.telen
et-ops.be
Oops, I'll have a look! :-)
Einstein was certainly not aware of that in 1918, he simply declared that
SRT is for Gallilean systems and thought he needed his GRT for the answer...
So it's no part of the original SRT.. but of course, thinking has advanced
since then!
BTW, your second reference (Baez) from the above claims:
"
[TOTAL BALONEY FOLLOWS:]
"Einstein came up with special relativity, which handles
only unaccelerated frames of reference [...]
Then he realized that an accelerated frame of reference
is indistinguishable from a gravitational field - his famous
`elevator' thought experiment. This led him to invent general
relativity, which handles general frames of reference and gravity."
"
To the contrary, I can find no error, I'm sure it is historically accurate!
> > and
> > therefore not included in the equations either.
>
> I think it is not included in the standard twin paradox
> treatment because it is irrelevant.
But it is included in elaborate treatments.
Einstein and later Moller understood the necessity to explain the time
acceleration of the stay-at-home according to the traveller.
> When the accelerations are taken large enough, such
> that the cruising velocities are reached in a short time,
> then what happens during these accelerating phases
> can be ignored when compared with what happens
> during the cruising phases.
No it can't - you still need an explanation for the time jump as "seen" by
the traveller -see Einstein's paper of 1918 or Moller's book. (I can find
the refs if anyone is interested).
Harald
stmx3
> Dear David Smith:
>
> You have said,
>
>
>>If both frames are moving the same speed, but different directions, they
>>will both age the same.
>>
>
> So, like the person who wrote the account of the twin paradox in the FAQ,
> you do believe that time dilation affects the rate of aging.
>
> The formula for time dilation is a function of velocity. If I am moving at a
> given velocity with respect to you, then you are moving at the exact same
> reduction of aging, it is predicted to be the same for you as for me,
> because v is the same, and the formula is a function of v. You see me in
> slow motion and I see you in slow motion, so both our ages are reduced by
> the same amount.
>
> Is that not the case?
>
>
>
[snip]
The twin paradox arose specifically to point out the inconsistency of
SR. So, way back in the early 1900's, intelligent people like you were
finding it hard to come to grips with SR because of things like the
situation you bring up. If velocities are relative, how can it be wrong
to say that, from each twin's perspective, the other twin ages less?
Take a look at
http://www.phys.unsw.edu.au/~jw/twin.html
They have nice graphics and a decent explanation. You should be able to
convince yourself of the following scenarios:
1) Given twins, A & B,
--A accelerates left to relativistic speeds
--B stays put
--A reverses course and returns
**We find age(A) < age(B)
2) Given triplets, A, B & C,
--A accelerates left to relativistic speeds
--B accelerates right to relativistic speeds
--C stays put
--A & B have the same acceleration/deceleration profiles
--A & B reverse course and return
**We find age(A) = age(B) and both are younger than C
There is enough literature to support (1) and by symmetry you can glean
(2).
Dan Bloomquist
Stephen Bint wrote:
> Dan, what do you think of this bouncing beam device I described? It's
> actually copied from a british Open University program, where it is used as
> an illustration of time dilation. Because time dilation prevents observers
> from seeing the light pulse travel faster than c, it must apply equally to
> both observers. If it doesn't one of them will see the light pulse move
> faster than c.
>
Yes.
> I know I am repeating myself a lot, but there are so many posts since I gave
> that example and not one of them has referred to it. Surely, the velocity
> determines the length of the path and time dilation must be employed to slow
> it - BOTH ways.
>
Yes.
>
>>When it is said to 'do the math', it means look at it one frame at a
>>time and don't drag the perceptions of one frame to another. It also
>>means don't create a special frame that sees something else. This
>>special frame is not in the theory so creating it creates a paradox in
>>your mind not the theory.
>
>
> Yes, but time dilation mustr have a meaning. It must have an effect.
It doesn't have to have meaning. It is just the way nature works. The
effect is that c is the speed limit of nature.
> It must
> reconcile classical expectations of seeing things go faster than c, by
> stretching the time frame of anything with a velocity, so that you see slow
> motion in that moving frame. Literally.
>
We have never seen anything go faster than light so this 'classical
expectation' is only conjecture. 'It', SR, doesn't say stuff. It is just
a model of what nature says.
So, if you argue with SR you are arguing with nature.
> Stephen
Greg Neill
[snip]
Why do you top-post *and* put quoted text inline,
thus unnecessarily duplicating content? Do you
work for the department of redundancy department?
Dirk Van de moortel
"Harry" <harald.v...@epfl.ch> wrote in message news:3fa14846$1...@epflnews.epfl.ch...
>
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
> in message news:3fa12a93$1...@usenet01.boi.hp.com...
> >
> > "Harry" <harald.v...@epfl.ch> wrote in message
> news:3fa12183$1...@epflnews.epfl.ch...
[snip]
> > > Acceleration of the ref. frame of observation is not part of SRT,
> >
> > Ha, but it *is* part of SRT:
> >
> http://groups.google.com/groups?&threadm=a3M1b.82977$F92....@afrodite.telenet-ops.be
>
> Oops, I'll have a look! :-)
> Einstein was certainly not aware of that in 1918, he simply declared that
> SRT is for Gallilean systems and thought he needed his GRT for the answer...
> So it's no part of the original SRT.. but of course, thinking has advanced
> since then!
>
> BTW, your second reference (Baez) from the above claims:
> "
> [TOTAL BALONEY FOLLOWS:]
> "Einstein came up with special relativity, which handles
> only unaccelerated frames of reference [...]
> Then he realized that an accelerated frame of reference
> is indistinguishable from a gravitational field - his famous
> `elevator' thought experiment. This led him to invent general
> relativity, which handles general frames of reference and gravity."
> "
(This was an extract from
http://groups.google.com/groups?&threadm=bi75vd$qbm$1...@glue.ucr.edu )
> To the contrary, I can find no error, I'm sure it is historically accurate!
The first line is obviously wrong already :-)
And I don't have much to add to John's surrounding remarks.
> > > and
> > > therefore not included in the equations either.
> >
> > I think it is not included in the standard twin paradox
> > treatment because it is irrelevant.
>
> But it is included in elaborate treatments.
> Einstein and later Moller understood the necessity to explain the time
> acceleration of the stay-at-home according to the traveller.
Of course, it can be a nice exercise to really take accelerations
into account to describe the twins paradox. You can use my
calculations for it :-)
>
> > When the accelerations are taken large enough, such
> > that the cruising velocities are reached in a short time,
> > then what happens during these accelerating phases
> > can be ignored when compared with what happens
> > during the cruising phases.
>
> No it can't - you still need an explanation for the time
> jump as "seen" by the traveller -see Einstein's paper of
> 1918 or Moller's book. (I can find the refs if anyone is
> interested).
Sure it can. Just look at the equations at the bottom of my
treatment at
http://groups.google.com/groups?&threadm=a3M1b.82977$F92....@afrodite.telenet-ops.be
Take for instance the equation that gives proper time T as
a function of coordinate time t and take the limit for
a --> infinity:
T(t) = c/a argsinh(at/c)
You clearly see that for all values of t (!)
Limit{ a->Inf; T(t) } = 0
which says that the accumulated proper times can be made
arbitrarily small by taking the acceleration large enough.
Some more limits at
http://groups.google.com/groups?&threadm=3f150372$1...@usenet01.boi.hp.com
Cheers,
Dirk Vdm
Dirk Van de moortel
"Stephen Bint" <bi...@iname.com> wrote in message news:3fa13add$0$107$65c6...@mercury.nildram.net...
[snip]
> > He has been here before with another name.
> > Same troll, new name.
> >
> > Dirk Vdm
>
> This is a lie.
No, it is an educated guess, based on style and content.
> You are lying about me, though I have done nothing to you,
> nor have I been rude to anyone on this list, except Uncle Al, who was much
> ruder to me first.
>
> I think you owe me an apology.
With your combination of ignorance, arrogance, contempt
and stupidity, you might consider apologizing to your parents
for having wasted their time.
Dirk Vdm
Martin Hogbin
"Stephen Bint" <bi...@iname.com> wrote in message news:3fa13add$0$107$65c6...@mercury.nildram.net...
> I have started two threads about the twin paradox, which I have
> cross posted to three groups. I have confined myself almost entirely
> to discussing the topics of these threads with those adults who have
> chosen to discuss them with me.
So, are you going to learn some physics?
Martin Hogbin
Stephen Bint
"Greg Neill" <gnei...@OVE.netcom.ca> wrote in message
news:b1dob.15950$zH1.1...@weber.videotron.net...
with nested quotes. I like to be very specific about what I am talking
about, but I don't want to misprepresent people by editing what they have
said. And people who download "headers only" lose the beginning of the
thread and so the context, if they log on three days later. I have also been
called up for not quoting the whole damn thing, for that reason.
Am I spineless? I just want everybody to be happy. Is that bad?
Stephen
Stephen Bint
"stmx3" <stmx3N...@NOSPAMM.netscape.net> wrote in message
news:3FA15037...@NOSPAMM.netscape.net...
travel left at half the speed of A, staying central. Their relative
velocities could be identical, to the triplets in example two.
So the examples seem to contradict eachother as if they are not observing
the same laws. In the first, the ones at either end are different ages, in
the second they are the same age, simply because an extra traveller happened
by.
Stephen
Paul Cardinale
I know. I'm trying to reform.
Paul Cardinale
Stephen Bint
that velocity in a universe without a universal frame of reference, is not a
property of an object but a relationship between two objects. If I don't see
the fundamental symmetry implicit in the concept of relative velocity, or
even have an inkling why that is important, I don't belong here, do I,
Martin?*
Even less so, if I were to talk down to others as if they don't know shit.
Stephen
*irony
stmx3
>>
>>1) Given twins, A & B,
>>--A accelerates left to relativistic speeds
>>--B stays put
>>--A reverses course and returns
>>**We find age(A) < age(B)
>>
>>2) Given triplets, A, B & C,
>>--A accelerates left to relativistic speeds
>>--B accelerates right to relativistic speeds
>>--C stays put
>>--A & B have the same acceleration/deceleration profiles
>>--A & B reverse course and return
>>**We find age(A) = age(B) and both are younger than C
>>
>>There is enough literature to support (1) and by symmetry you can glean
>>(2).
>>
>
> But that is logically flawed. I could simulate example 2 by having a triplet
> travel left at half the speed of A, staying central. Their relative
> velocities could be identical, to the triplets in example two.
>
Sorry...I don't follow your scenario. I don't see how their relative
velocities will be the same as in example 2) or how their acceleration/
deceleration profiles will remain the same. I have been following your
thread fairly closely, so perhaps you can be more specific.
But, are you convinced that example 1) is ok? Do you follow the "logic"
that leads to its ultimate conclusion?
stmx3
[snip]
> Whew.
>
> Again, the problem throughout is that once two frames are moving
> relative to each other, they _cannot_ have synchronized clocks -- each
> one sees that the other's clocks are skewed progressively with
> distance. So, the result works out mathematically, it's supported
> experimentally, but it doesn't make a whole lot of sense intuitively.
>
> If you hate the clock corrections -- well, that's the point. The
> skewed clocks are what make the math work, and they're what make the
> whole thing so confusing.
>
> =========================
> To email directly replace nospam with foobox
>
>
Excellent post. You put a lot of time into it and I, for one,
appreciate your effort.
Stephen Bint
but a relationship between two objects. So if you stood at the roadside
while I drive past at 40kph, my velocity relative to you is 40kph and your
velocity, relative to me is also 40kph.
It would also be 40, if you were driving at 30 and I overtook you doing 70,
or if I was driving at 20 and you drove past in the opposite direction at
20.
The only difference between these three examples of our RV being 40, is what
the other objects (road, world, etc.) are doing. But the formula for time
dilation makes no reference to any other objects - only our relative
velocity, v.
So, if the relativistic speed of A in example (1) was 0.8c, and I keep a
triplet at the midpoint between them, travelling at 0.4c, he is at rest from
his POV, with ships travelling away from him at 0.4c to the left and the
right.
Which is example two, if the velocities of A and B wrt C are both 0.4.
Without reference to external objects or an absolute frame of reference,
they are identical, velocity-wise, regardless of their acceleration
histories.
Time dilation cannot depend on acceleration history, because if it did, it
would not be a function of velocity, which it most certainly is.
So I'm preaching now and let's assume you know a damn sight more about it
than I do, but I am just trying to explain why I think your examples
contradict. The A and B should be the same age in both, or neither. They
shouldn't be different in the first and the same in the second.
Remember, no external objects or acceleration histories play any part in the
calculation of time dilation.
Stephen
Bonnie Granat
Please explain to me how these two things could be true:
- B is traveling at half the speed of A
- A and B are traveling at identical velocities
I'd really like to know.
> So the examples seem to contradict eachother as if they are not observing
> the same laws. In the first, the ones at either end are different ages, in
> the second they are the same age, simply because an extra traveller
happened
> by.
>
The second example is the same as your original example, except you had
nobody left behind and stationary.
If I am all wrong, I want a noncrackpot (you know who you are) to tell me.
--
Bonnie Granat
Granat Technical Editing and Writing
http://www.editors-writers.info
Bonnie Granat
That's right. You came here with "There's no defense..." I think that's the
only reason people are annoyed. After all, they tolerate me -- they
sometimes ignore me, but they tolerate me. LOL.
Bonnie Granat
Yes. So you agree A and B in example 2 will be younger when they return to
C?
> Without reference to external objects or an absolute frame of reference,
> they are identical, velocity-wise, regardless of their acceleration
> histories.
>
> Time dilation cannot depend on acceleration history, because if it did, it
> would not be a function of velocity, which it most certainly is.
>
> So I'm preaching now and let's assume you know a damn sight more about it
> than I do, but I am just trying to explain why I think your examples
> contradict. The A and B should be the same age in both, or neither.
Why? In example 1, only one of them travels very fast, while in example 2,
they both do.
They
> shouldn't be different in the first and the same in the second.
Again, why? Why would you expect them to be the same when they do very
different things in example 1?
sal
Concise. (Something I'm not...)
For pedantic reasons we might also add:
at event x=0, t=0, (Earth frame)
x" = gD, t" = -gvD/c^2 (left twin)
x' = -gD, t' = -gvD/c^2 (right twin)
at event x"=0, t"=0, (left twin)
x = -D, t = 0 (earth frame)
x' = -2gD, t' = -2gvD/c^2 (right twin)
at event x'=0, t'=0, (right twin)
x = D, t = 0 (earth frame)
x" = 2gD, t' = -2gvD/c^2 (left twin)
The point is that the clocks are already highly skewed relative to each
other at the starting events. If one hoped to get a classical result,
he lost before he began. That's not obvious as the problem was stated
because each of the three starting events was stated as only taking
place in a _single_ reference frame -- and here's where a lot of the
confusion comes from: The use of human twins in the problem forces us
to think of "reference frames" as single observers, which is incomplete.
IOO the problem statement forces a flawed point of view on the reader.
Dirk Van de moortel wrote:
> "sal" <beli...@nospam.com> wrote in message news:XJ8ob.13181$lC5....@nntp-post.primus.ca...
>
>
>>The problem as stated was "two twins fly to Earth, each from a distance
>>of 1 ly, at a velocity of 0.75c. What happens?"
>
>
> Earth uses coordinates (x,t)
> Right twin uses coordinates (x',t')
> Left twin uses coordinates (x",t")
> World line Earth: x = 0 (t-axis)
> World line Right twin: x' = 0 (t'-axis)
> World line Left twin x" = 0 (t"-axis)
>
> Right twin is present at event R: (x,t) = (D,0)
> where/when his clock is set to t' = 0,
> and he approaches Earth with relative speed v
> Left twin is present at event L: (x,t) = (-D,0)
> where/when his clock is set to t" = 0,
> and he approaches Earth with relative speed v
>
> [M]
> /|\
> / | \
> t"/ t \t'
> / | \
> / | \
> / | \
> / | \
> -----[L]------|------[R]------x
>
> Transformation Earth coord --> Right twin coord:
> { x' = g( (x-D) + vt )
> { t' = g( t + v(x-D)/c^2 )
> Transformation Earth coord --> Left twin coord:
> { x" = g( (x+D) - vt )
> { t" = g( t - v(x+D)/c^2 )
> where in both cases
> g = 1/sqrt(1-v^2/c^2)
>
> Meeting of the 3 world lines at event M:
> { x = 0
> { x' = 0
> { x" = 0
> ==> 2 lines of algebra ==>
> (x,t) = ( 0, D/v )
> (x',t') = ( 0, D/(gv) )
> (x",t") = ( 0, D/(gv) )
> According to Earth D/v seconds have passed.
> According to both twins D/(vg) seconds have passed.
>
> Dirk Vdm
>
>
--
To email me directly, take out nospam and put back foobox.
Martin Hogbin
"Stephen Bint" <bi...@iname.com> wrote in message news:3fa1828d$0$118$65c6...@mercury.nildram.net...
>
> "Martin Hogbin" <sp...@hogbin.org> wrote in message
> news:bnrpsg$q5h$1...@sparta.btinternet.com...
> >
> > "Stephen Bint" <bi...@iname.com> wrote in message
> news:3fa13add$0$107$65c6...@mercury.nildram.net...
> > So, are you going to learn some physics?
> That's solid advice. I need to learn the most basic concepts, like learning
> that velocity in a universe without a universal frame of reference, is not a
> property of an object but a relationship between two objects.
That is actually a rather philosophical issue.
What you actually need to know what an inertial frame is and how
to make measurements in it.
> If I don't see
> the fundamental symmetry implicit in the concept of relative velocity, or
> even have an inkling why that is important, I don't belong here, do I,
> Martin?*
>
> Even less so, if I were to talk down to others as if they don't know shit.
You did not demonstrate a great knowledge of the principles
of the subject in your original question.
You referred to triangulation as a means of measuring the
distance between two moving objects. Perhaps you could
say precisely what you mean by that.
Martin Hogbin
stmx3
[snip]
> The only difference between these three examples of our RV being 40, is what
> the other objects (road, world, etc.) are doing. But the formula for time
> dilation makes no reference to any other objects - only our relative
> velocity, v.
>
> So, if the relativistic speed of A in example (1) was 0.8c, and I keep a
> triplet at the midpoint between them, travelling at 0.4c, he is at rest from
> his POV, with ships travelling away from him at 0.4c to the left and the
> right.
>
> Which is example two, if the velocities of A and B wrt C are both 0.4.
I hear what you're saying. Even so, the outcome remains the same. When
A & B arrive back at C, they will both be the same age and both younger
than C.
>
> Without reference to external objects or an absolute frame of reference,
> they are identical, velocity-wise, regardless of their acceleration
> histories.
>
> Time dilation cannot depend on acceleration history, because if it did, it
> would not be a function of velocity, which it most certainly is.
Yes...time dilation is a function of velocity...or shall we say,
relative velocities. We can neglect the effects of acceleration.
>
> So I'm preaching now and let's assume you know a damn sight more about it
> than I do, but I am just trying to explain why I think your examples
> contradict. The A and B should be the same age in both, or neither. They
> shouldn't be different in the first and the same in the second.
>
I think sal gave a really good treatment of this. It was very well put
and used simple math. I think, in your scenario, as long as symmetry is
maintained, both twins will have the same age. Once you break the
symmetry (one undergoing acceleration/deceleration), then that skews the
respective space-time grid and changes the final outcome.
stmx3
[snip]
He did respond back to me and it basically amounts to performing a
velocity translation of the A, B, C triplets to the left, such that C is
now travelling at some speed (e.g. 0.4c). It doesn't make a difference
as long as the relative velocities remain the same.
I'm a certified noncrackpot and you are not wrong.
The twin paradox and all of relativity is fun and appeals to the senses,
so everyone has an opinion. Typically, there's a fundamental
misunderstanding that should be addressed before jumping straight into
the twin paradox.
Dirk Van de moortel
"sal" <beli...@nospam.com> wrote in message news:Gdgob.451$Dz6...@nntp-post.primus.ca...
> Nice.
>
> Concise. (Something I'm not...)
>
> For pedantic reasons we might also add:
>
> at event x=0, t=0, (Earth frame)
> x" = gD, t" = -gvD/c^2 (left twin)
> x' = -gD, t' = -gvD/c^2 (right twin)
>
> at event x"=0, t"=0, (left twin)
> x = -D, t = 0 (earth frame)
> x' = -2gD, t' = -2gvD/c^2 (right twin)
>
> at event x'=0, t'=0, (right twin)
> x = D, t = 0 (earth frame)
> x" = 2gD, t' = -2gvD/c^2 (left twin)
Yes, these are the so-called starting events as viewed
in the three frames.
>
> The point is that the clocks are already highly skewed relative to each
> other at the starting events. If one hoped to get a classical result,
> he lost before he began. That's not obvious as the problem was stated
> because each of the three starting events was stated as only taking
> place in a _single_ reference frame -- and here's where a lot of the
> confusion comes from: The use of human twins in the problem forces us
> to think of "reference frames" as single observers, which is incomplete.
> IOO the problem statement forces a flawed point of view on the reader.
If you allow me to return the compliment: concise.
Specially your choice of "IOO" ;-)
Dirk Vdm
Bonnie Granat
Glad to know I'm not totally misunderstanding. I'm incapable of
understanding the math, because I barely know algebra, so I have to try to
graze here and there, reading material that makes my ears twitch, and try to
learn something. I am working on the math, but it's next to impossible to
catch up more than 40 years after I should've learned this stuff. Funny
thing is that now I *like* math. LOL. Back when I was a child, and even now,
though, it's very hard to get comprehensible definitions and beginning
concepts. I've found nothing at all on the Internet that gives basic math
definitions. Most places tell you to click on "flomdor" for an explanation.
When you get to the destination, you're told that a flomdor is an "akerdee"
as if you know what an akerdee is. If you click "akerdee," you're returned
to "flomdor" ad nauseam. Even the FAQ for this newsgroup is way beyond my
level of comprehension. But, I shall keep trying. Thank you for letting me
know I wasn't wrong.
Jeff Root
> Here's a description of a classic "twin paradox" situation, in
> the same style as my description of your previous example:
Jon, that was great! Simple, clear, and concise!
-- Jeff, in Minneapolis
.
Jeff Root
>> you are appealing to me to be open-minded enough to be able to
>> believe it is possible that I am wrong. I would ask the same of you.
>
> There is a difference. Dan's view is supported by thousands
> of physicists throughout the world who have studied the subject
> in great detail for nearly a century.
Plus hundreds of thousands of advanced physics students and
millions of undergraduate students who did not go on to become
physicists. Some of whom *did* go on to become engineers and
technicians who actually use and depend on the accuracy of the
calculations of relativity in their work. Plus many millions
of people who have read books or articles by Einstein, Arthur
Eddington, Stephen Hawking, George Gamow, Lillian R. Lieber,
Isaac Asimov, Kip Thorne, John D. Barrow, and a host of others
who found relativity to be fascinating, not flawed.
xxein
xxein: Don't worry about it. Solve it. Ignore the ng for a while
and put ten+ self-honest years into tying up loose ends. Question
yourself, not at the bidding of this ng, but from within and what you
can accept as rock hard logic (but start from scratch, not the
assumptions of others).
Still, in the end, even with supporters, all will remain inconclusive.
But you might be able to show that past theories are less conclusive
and may be subset subjectivities in an ever increasing awareness of an
objectivity in this universe.
Just do it without peer pressure and without letting your panties get
into an uproar.
If you get lucky enough to find a good cause of/for gravity, it will
help a lot (more like indispensable). I got lucky (with effort) and
almost any "new discovery" of this universe is just a 2 to 10 minute
assimilation, but I still learn more (not much thanks to this ng).
Don't worry about it. Solve it. If you are good enough to think in
the objective sense, you will not worry about anything but the
solution.
That is what it is all about. Good luck.
Stephen Bint
"Bonnie Granat" <bgr...@editors-writers.info> wrote in message
news:3fa1...@andromeda.5sc.net...
people on this newsgroup is down to a difference in what we mean by the word
"velocity". The meaning of "velocity" comes into it before any formulas are
employed, or diagrams are drawn, because before you can do either, you need
to decide which of the observers is moving and which is not. If you try to
say both observers are moving equally, time dilation occurs for both and
they are both younger than eachother at the end, which is impossible. So let
me explain what I mean when I use the word "velocity".
If you and I were in space and floating towards eachother, which of us would
you say is the stationary one and which would you say is the moving one?
The answer, according to Newton, is "we are both moving with respect to the
other". The laws of physics require velocity to be a pure measure of how
much we are moving with respect to eachother. If the gap between us is
getting smaller by ten metres every second, our relative velocity is ten
metres a second. Physics formulas do not make any reference to which of us
is moving. They tell us how fast we will fly apart when we bounce off
eachother, if we can tell the formula how fast we were approaching eachother
and how much mass we are each and how bouncy we both are.
The value of velocity we put in will be ten metres a second. You could say I
am the stationary one, or that you are, it doesn't matter. The formulas have
no slot for us to plug that information in. The formulas only ask for our
relative velocity, or how much we are moving from eachother's point of view.
Velocity means something weird to common sense. From the day we are born, we
habitually think of the ground as "not moving". When things move relative to
the ground, we say they are moving and when they stay still with respect to
the ground, we say they are stationary.
Physicists never do this, except by accident. When a physicist tells you the
velocity of something, they always mean "with respect to" something else.
They never say something is moving, or stationary, with respect to nothing.
Velocity is always a comparison of two objects, or could be described as a
relationship between two objects.
So, in answer to your query, because the time dilation formula is concerned
with this relative velocity I have described, there is no difference in
three ships relationship to eachother, or effect on eachother, whether they
choose situation 1 or situation 2:
1. A stays still
C flies left at 0.4c
B flies left at 0.8c
2. A flies right at 0.4c
C stays still
B flies left at 0.4c
They are the same because the relative velocities of all pairs of objects in
the experiment are the same in both cases. It's like playing billiards on a
train with really good suspension. If the train stays at the same speed, the
balls behave exactly the same as they would on a concrete floor.
Now the way this strikes at the heart of the matter goes beyond what you
asked. If you want to see how the meaning of velocity brings on the paradox,
read on.
The fact that the amount of time dilation predicted by Einstein depends on
the velocity of the object relative to the observer, implies a symmetry
which is difficult to explain, so I will ask you to imagine something.
Imagine you and I are in glass cubes travelling past eachother in space so
that I see you passing me at 0.2c. Our cubes are filled with smoke and be
both have lasers at the back of our cubes, pointing forwards. We both turn
our lasers on around the time we pass eachother. Because of the smoke, we
can both watch the progress of the laser beams on eachother's cubes.
Einstein says this, very boldly and clearly: The speed of light in a vacuum
is the same for all observers in all frames of reference. This means that,
though I am travelling at 0.2c from your point of view, you will not see my
laser beam travel at 1.2c, but at 1.0c, the constant, universal, speed of
light for all observers. It also means that I will not see your beam travel
at 1.2c, but at 1.0c, the same as you.
Einstein says that two things happen to make the light travel at 1c on the
other ship. The ship is shorter because we have relative velocity (so it
doesn't travel as far) and time is slower on the other ship, so the light is
effectively moving in slow motion. So a certain amount of time dilation must
occur so you will see my beam go at 1.0c. It depends entirely on our
relative velocity.
It does not matter which one of us has accelerated. The amount of time
dilation must be enough to shrink 1.2c, to 1.0c. If I was passing you at
0.3c, more time dilation would be needed. The amount needed depends
entirely on how fast we are going past eachother. That is why the formula
for time dilation asks us one thing only: the relative velocity. The amount
of time dilation required is unaffected by how we came to be passing
eachother. It only cares whether it's 0.2 or 0.3c.
So the paradox is this: To see my beam going at 1c, you must see me in slow
motion. But if I am to see your beam going at 1.0c, I must see you going in
slow motion. So if we are twins, we will both be younger than eachother.
Einstein's theory is telling us that twins that move with respect to
eachother will be younger than eachother.
If the people on this newsgroup who say, one of us has accelerated in the
past and that one appears slow, but the other doesn't, they overlook
something. If one of us does not see the other slowed, then instead we will
see the laser on the other's ship going faster than c.
But the whole point of relativity is to explain how "The speed of light in a
vacuum is the same for all observers in all frames of reference".
And thank you for reading all that.
Stephen
Richard
Don't be discouraged by the crackpots Stephen, you're argument was not
only correct, but very well constructed. OTOH, there's nothing like
math, and I already proved in another thread that the clocks will be
ticking at the same rate. For another form of the argument: As the two
identical ships pass each other, each will perceive the other to be
contracted by a factor 1/gamma. Since both perceive their speed wrt the
other to be v, then the time that one ship takes to pass the other is
just
delta t = (L + L/gamma)/v
Since this is true for both observers, because of the symmetry involved,
then
delta t = delta t'
Which is to say that if the regular passage of regular intervals marked
along the hulls of the ships are used to keep time, then the clocks are
ticking at the same rate.
This is nothing more than the formalization of your argument, thus no
need to simply speculate, the math is right there in front of you.
Mind you, the clocks are not necessarily ticking equally though, because
the ships are not necessarily moving at the same rate wrt local space,
nor are they necessarily at the same potential. Thus in order to provide
a symmetrical arrangement in the real universe, the ships must be moving
equally wrt local space and be at the same potential (gravitational and
electromagnetic). The Hafele-Keating planes do not provide such a
symmetry because they were moving relative to, and within, the Earth/Sun
gravitational and electromagnetic fields. As a nuclear engineer in a
similar group once said "I don't need SR, I only need the factor gamma,
because the remainder cancels out it is irrelevant to the prediction."
Richard Perry
Richard Perry
Stephen Bint
You speak of local space, electromagnetic and gravitational fields affecting
clocks. Since the mechanism of the clock is not specified, you are talking
about the passage of time being variable and possibly more. If you can avoid
using quadratic equations (I'm quite good at English) I would like to be
told more.
And what the hell! I'll try and understand that formula you just gave me.
How hard can it be?
delta t is the time taken to pass the starship 1, measured from starship 0
delta t is the time taken to pass the starship 0, measured from starship 1
v is relative velocity
L is what? c?
gamma ...is a strange enigma
Stephen
"Richard" <no_mail...@yahoo.com> wrote in message
news:3FA1CA5F...@yahoo.com...
Paul Cardinale
> "Paul Cardinale" <pcard...@volcanomail.com> wrote in message
> news:64050551.03103...@posting.google.com...
> >
> > [crap snipped]
> >
> > I see that you are both a jackass and an idiot.
> > You believe that because you are too stupid/lazy/arogant to learn the
> > math, that all you need to do is wave your arms and try to apply your
> > own ignorant comic-book misconception of relativity. You are a
> > clueless ineducable blubbering baffoon.
> >
>
> You can only resort to personal attacks. What you cannot do is face the
> issue, which I keep bringing up for you to ignore. You call me names,
You deserved every bit of it by making asinine assertions about a
theory without bothering to learn the theory. Moreoverm your premise
that you needn't bother to calculate what a theory predicts in order
to claim that it's predictions are false goes beyond idiotic.
> you
> bleat "do the math" and you run like a cur from the question at hand.
>
> This is the question I have asked Jon Bell. You won't be able to answer it
> and I suspect you won't even understand it:
>
"Ah. Arogance and stupidity all in one package. How efficient of
you."
> Two skateboards glide past eachother at speed.
>
> On each skateboard is an observer and a device which sends a light pulse
> from a source which is 1m above the base, vertically down to a mirror on the
> base, which reflects it directly vertically back to the source. To one of
> the observers (either one) the light pulse on the other board traces a
> V-shaped path, which is longer than the vertical one. If there is no time
> dilation, the observer will see the pulse travel a greater distance in the
> same time, and therefore, see it travelling faster than c.
>
> Time dilation slows the event on the other skateboard, the exact amount
> necessary to make the light pulse travel this longer path at exactly c. The
> length of the V-shaped path and consequently, the time dilation required, is
> purely a function of the relative velocity of the two skateboards.
>
> Furthermore, the path that A sees the pulse trace on B's skateboard is
> exactly the same length as the path seen by B, taken by the light pulse on
> A's skateboard.
>
> So the time dilation must be identical in both directions, regardless of how
> they acheived their relative velocity.
>
> So what if your travelling twin was passing earth and viewing such an
> experiment there. Either he will see time slowed down there, or he will see
> the light pulse travelling faster than c there. Is that not the case?
>
Each twin will observe the other's clock to be running slow.
Note however that the formula for time dilation, by itself, is
insufficient for resolving the twin paradox.
Paul Cardinale
Richard
Stephen Bint wrote:
>
> Richard,
>
> You speak of local space, electromagnetic and gravitational fields affecting
> clocks. Since the mechanism of the clock is not specified, you are talking
> about the passage of time being variable and possibly more. If you can avoid
> using quadratic equations (I'm quite good at English) I would like to be
> told more.
>
> And what the hell! I'll try and understand that formula you just gave me.
> How hard can it be?
>
> delta t is the time taken to pass the starship 1, measured from starship 0
> delta t is the time taken to pass the starship 0, measured from starship 1
> v is relative velocity
> L is what? c?
> gamma ...is a strange enigma
gamma is 1/sqrt(1-(v/c)^2)
L is the length of your ship, and L/gamma is the length of the other
ship. Thus, since the other ship is moving wrt you at v, then the time
required for the ships to pass each other is just the length of both
ships divided by the speed, or
delta t = (L + L/gamma)/v
OTOH, the other ship also thinks that the speed is v and that the total
distance is L + L/gamma, so
delta t = delta t'
I.e. the clocks are ticking at the same rate.
The effect of the local field on clocks is evident when the clock is a
"rod clock" which is just another name for a "light clock" that keeps
time by measuring the period of a light beam (photon) bouncing back and
forth between parallel mirrors. As you've already noted the light clock
presents a special sort of conundrum for the relativists. OTOH the
behavior of the light clock provides us with the factor gamma when the
clock is moving wrt Earth, in that it will slow down simply because the
beam must travel farther between the mirrors when the clock is in motion
wrt Earth. The beam is then taking a zig-zag path. Ordinary Euclid says
that the delay is described by a factor gamma. IOW gamma is a strictly
Galilean factor, borrowed by the relativists and elevated to the status
'invariant', but of course it isn't any such thing, it's just plain old
Euclidean gamma.
The local field 'is' local space, that simple, thus the success of GR
despite the insanity of SR. GR in fact prevents the testability of SR,
simply because SR assumes a local isotropic space, and of course our
local space is instead anisotropic. Thus no test of SR can even be
conceivable performed within the confines of our solar system. OTOH, as
you've correctly noted, there is no need to do the math, SR is self
contradictory.
Richard Perry
Stephen Bint
Initially I thought you said:
> 1) Given twins, A & B,
> --A accelerates left to relativistic speeds
> --B stays put
> --A reverses course and returns
> **We find age(A) < age(B)
>
> 2) Given triplets, A, B & C,
> --A accelerates left to relativistic speeds
> --B accelerates right to relativistic speeds
> --C stays put
> --A & B have the same acceleration/deceleration profiles
> --A & B reverse course and return
> **We find age(A) = age(B) and both are younger than C
>
> There is enough literature to support (1) and by symmetry you can glean
> (2).
Then later, after we agreed that by changing our frame of reference we could
see these situations as identical, simply by adding a moving oveserver into
situation 1), you say:
> I hear what you're saying. Even so, the outcome remains the same. When
> A & B arrive back at C, they will both be the same age and both younger
> than C.
>
But in the first description, you said that A is younger than B and now what
are you saying? If the situations are identical, then if C is always the one
in the middle, the relationship of A's age to B's age must be the same in
both examples. I have obviously misunderstood something.
Stephen
"stmx3" <stmx3N...@NOSPAMM.netscape.net> wrote in message
news:3FA1967...@NOSPAMM.netscape.net...
Richard
Of course it isn't, what is required in addition is to note that it
isn't a paradox, it's an outright contradiction.
>
> Paul Cardinale
Richard
And of course the studies haven't yet been done to determine the
detrimental psychological effect of gaining an understanding of
something that isn't understandable. If you ever get to the point that
you understand it, then do me a favor, don't try to explain it to me, I
already understand that it is inconsistent. I have no desire to define
time as having a rate, when in fact a rate requires a time interval to
define. Time progresses at 1 second per second regardless of your frame
of reference, and a second isn't what a clock says it is if clocks don't
agree. In this case you have changed standards but kept the same name
for the respective units. It's a bit like calling a meter a millimeter,
but retaining the millimeter that was already called a millimeter. Just
stupidity personified.
Richard Perry
Big Bird
> It baffles me, that intelligent people, capable of algebra, cannot see that
> the assertion that time is slowed in a moving frame is a symmetrical
> assertion, because frames move relative to eachother, symmetrically.
The reason is that all these intelligent people capable of algebra
have themselves been where you are now: ignorant of special
relativity, only in posession of some snippets and chunks gleaned from
the pop-literature.
Every physics freshman or highschool student who first learns SR
encounters exactly the same conceptual problems that you are
encountering -- that's why the twin-"paradoxon" and the various other
well-known riddles around relativity are such clichees.
Those other people then decided to learn what SR actually says and how
one is actually supposed to understand these things. Contrary to what
you falsely assert in the subject-line, there is very much the need to
do the math.
Until you have computed for yourself what SR actually has to say about
a given situation, you CANNOT POSSIBLY proclaim that its statements
are in conflict with observation. Because you have no idea what its
statements ARE.
> Besides which, the formula for time dilation is a function of the velocity
> of the observed frame and makes no mention whatsoever of the acceleration of
> either frame, before or while the calculation is made.
You will forever be puzzled by relativity, as long as you think in
terms of a "formula for time dilation". You might as well give up
right here and now and either join the kooks or shut up. If you want
to *understand* special relativity, then you'll have to bite the
bullet and sit down and figure out by yourself for yourself (with the
aid of a good book, one might hope) what kind of mathematical
statements follow from which postulates in what order under what
assumptions.
Only if you have some mathematical grasp of WHY the Loretz transforms
look the way they do (and you have probably never actually seen the
form they take when they're used by actual physicists) can you
appreciate what might happen if you tweak this piece here or poke into
that end there.
You CANNOT understand SR if you think it is a bunch of formulas. Each
of the mathematical entities you may have encountered MEANS something.
And all your problems stem from a lack of grasp of that *meaning*.
Eyesee
well so maybe I'll
throw my own scenario in for you guys to help:
Two trains, the Mayflower and the
Yang-Mill, are on a head on collision on a one-way track. They are moving
towards each other
at the relative velocity V. Along the course, the Mayflower turns on its
headlights. Now,
1. whose reference frame is the light beam in after it is emitted?
2. does the recipient of the light beam see the light doppler shifted or
not?
3. does the sender of the beam see the light as doppler shifted or not?
After some thought, I've concluded the following:
1. Without a preferred frame, the Lorentz Transformation results in a
contradiction- both
observers observe time dilation and length contraction of the other guy's
frame. But
seems this is merely because the events aren't simultaneous in both frames.
So,
the time dilation and length contraction are a relefact of the math. To
claim the length
contraction (and time dilation)
as real effects of relative velocity would be like me measuring a 1 meter
ruler you're
holding with my 1 meter ruler from the 1 cm mark of my ruler while sneezing,
then declaring
that mine sneezing has caused the ruler in your non-sneezing frame to
lengthen.
2. If the velocity of light is independent of source velocity then the
Doppler Shift is not
a relativistic effect since after the light leaves the source, there can't
be a doppler shift
if there is no preferred frame.
Bruce Richmond
What is acceleration but a change in velocity.
While traveling along a straight road a car veers off to the left and
travels in a straight line. At some later point it veers back and
returns to the original road. In doing this the car will have
traveled more distance than it would have if it had continued on its
original path.
What caused the greater distance? Was it the turn? If immediately
after making the first turn the car had turned again, to travel
parallel to the original path, there would have been almost no change
in distance. The change in distance was not caused by the turn, or
angular acceleration, but by the time spent traveling in a different
direction.
In a similar manner, the change in time is not caused by the
acceleration, but by the time spent traveling at the resulting
velocity. So if it makes you feel better, forget about the
accelerations, it is the one that changes velocities that ages the
least ;-)
In the examples above, the difference is that in (1) "B stays put",
whereas in (2) "B accelerates".
Jon Bell
Paul Cardinale <pcard...@volcanomail.com> wrote:
>
>Each twin will observe the other's clock to be running slow.
>Note however that the formula for time dilation, by itself, is
>insufficient for resolving the twin paradox.
Yes. The traveling twin (in the usual formulation of the twin paradox)
must be at rest in two different inertial reference frames, one during his
outbound trip, the other during his inbound trip. During both outbound
and inbound trips, the stay-at-home twin's clock runs more slowly in each
of those two reference frames. However, "between" the two reference
frames (i.e. when the traveling twin switches from one frame to another)
something happens that does not apply to the stay-at-home twin (who
remains in the same frame throughout, without switching frames).
I hope to post a more detailed description in the next few days. I've had
to deal with little things like faculty meetings, grading homework,
getting my eyeglass prescription fixed... and tomorrow night we have to
deal with ghosts and goblins. ;-)
--
Jon Bell <jtbe...@presby.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA
Dirk Van de moortel
"Jeff Root" <je...@freemars.org> wrote in message news:964c68dd.0310...@posting.google.com...
> Jon Bell (jtbe...@presby.edu) wrote:
>
> > Here's a description of a classic "twin paradox" situation, in
> > the same style as my description of your previous example:
>
> Jon, that was great! Simple, clear, and concise!
I agree.
Shame that Bint decided to toss it away without even looking
at it, which he proves with the first line of his reply:
| "I see that you are asserting, that the accelerations
| undergone by one observer affect time dilation".
Genuine pearls for a genuine swine.
Dirk Vdm
Dirk Van de moortel
"Eyesee" <Eye...@here.com> wrote in message news:A_lob.8877$811.59...@newssvr21.news.prodigy.com...
> Hello everyone, interesting discussion. I'm a little confused about SR as
> well so maybe I'll
> throw my own scenario in for you guys to help:
>
> Two trains, the Mayflower and the
> Yang-Mill, are on a head on collision on a one-way track. They are moving
> towards each other
> at the relative velocity V. Along the course, the Mayflower turns on its
> headlights. Now,
>
> 1. whose reference frame is the light beam in after it is emitted?
It is in space-time, which is defined as the set of all events
of the universe. A reference frame is something that is
associated with an observer who measures or thinks about
events. So the light is in everyone's frame. Not everyone
might be able to see it. But it "is in" everyone's frame.
That means that your question is in fact meaningless.
Saying 'after the beam is emitted' means that the beam stopped
shining after a while. A beam is a collection of emitted waves or
photons.
> 2. does the recipient of the light beam see the light doppler shifted or
> not?
Yes, if they move w.r.t. each other, with a factor
sqrt( (c+V)/(c-V) ).
> 3. does the sender of the beam see the light as doppler shifted or not?
Seeing is photons hitting your retina. He does not see the
light since the light is moving away from him. Normally you
don't get hit by your own bullets.
>
> After some thought, I've concluded the following:
At this stage, I'm afraid that thought is the last thing you need.
You need to study first. Try for instance:
http://www.physicsguy.com/ftl/html/FTL_part1.html
http://www.physicsguy.com/ftl/html/FTL_part2.html
When you have had a very careful look at those, do some
thinking again... Maybe you will be able to answer your
questions yourself already... Maybe you will decide that
it is time to read a good book. See Physics FAQ for an
excellent list.
Enjoy...
Dirk Vdm
Harry
>
> "Greg Neill" <gnei...@OVE.netcom.ca> wrote in message
> news:b1dob.15950$zH1.1...@weber.videotron.net...> > news:3fa14402$0$116$65c6...@mercury.nildram.net...
> >
> > [snip]
> >
> > Why do you top-post *and* put quoted text inline,
> > thus unnecessarily duplicating content? Do you
> > work for the department of redundancy department?
> >
> >
> I wish there was a guide book. I find it difficult to read text
interspersed
> with nested quotes. I like to be very specific about what I am talking
> about, but I don't want to misprepresent people by editing what they have
> said. And people who download "headers only" lose the beginning of the
> thread and so the context, if they log on three days later. I have also
been
> called up for not quoting the whole damn thing, for that reason.
>
> Am I spineless? I just want everybody to be happy. Is that bad?
>
> Stephen
>
I agree, I started to do the same myself recently - without any duplication!
It can be most efficient and reader friendly.
Harald
Harry
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:PEdob.115592$cN4.5...@phobos.telenet-ops.be...
>
> "Harry" <harald.v...@epfl.ch> wrote in message
news:3fa14846$1...@epflnews.epfl.ch...
> >
> > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
wrote
> > in message news:3fa12a93$1...@usenet01.boi.hp.com...
> > >
> > > "Harry" <harald.v...@epfl.ch> wrote in message
> > news:3fa12183$1...@epflnews.epfl.ch...
>
> [snip]
>
> > > > Acceleration of the ref. frame of observation is not part of SRT,
> > >
> > > Ha, but it *is* part of SRT:
> > >
> >
http://groups.google.com/groups?&threadm=a3M1b.82977$F92....@afrodite.telen
et-ops.be
> >
> > Oops, I'll have a look! :-)
> > Einstein was certainly not aware of that in 1918, he simply declared
that
> > SRT is for Gallilean systems and thought he needed his GRT for the
answer...
> > So it's no part of the original SRT.. but of course, thinking has
advanced
> > since then!
> >
> > BTW, your second reference (Baez) from the above claims:
> > "
> > [TOTAL BALONEY FOLLOWS:]
> > "Einstein came up with special relativity, which handles
> > only unaccelerated frames of reference [...]
> > Then he realized that an accelerated frame of reference
> > is indistinguishable from a gravitational field - his famous
> > `elevator' thought experiment. This led him to invent general
> > relativity, which handles general frames of reference and gravity."
> > "
>
> (This was an extract from
> http://groups.google.com/groups?&threadm=bi75vd$qbm$1...@glue.ucr.edu )
>
> > To the contrary, I can find no error, I'm sure it is historically
accurate!
>
> The first line is obviously wrong already :-)
> And I don't have much to add to John's surrounding remarks.
OK, you are right, I overlooked the glitch: it should be "handled" instead
of "handles". ;-)
Harald
stmx3
Sorry for the confusion. Let's add a third example for clarity.
1) Given twins, A & B,
--A accelerates left to relativistic speeds
--B does not accelerate or decelerate
--A reverses course and returns
**We find age(A) < age(B)
2) Given triplets, A, B & C,
--A accelerates left to relativistic speeds
--B accelerates right to relativistic speeds
--C does not accelerate or decelerate
--A & B have the same acceleration/deceleration profiles
--A & B reverse course and return
**We find age(A) = age(B) and both are younger than C
3) Given triplets, A, B & C, and an observer, O
--A,B,C are moving left at 0.4c relative to O
--Now, forget about O.
--A accelerates left to 0.4c relative to C (the 3rd triplet)
--B accelerates right to 0.4c relative to C (the 3rd triplet)
--C does not accelerate or decelerate
--A & B have the same acceleration/deceleration profiles
--A & B reverse course and return to C
**We find age(A) = age(B) and both are younger than C
But all this presupposes that you concur with example 1).
Richard
You are presupposing that the mere effect of acceleration causes a clock
to tick slower, when in fact it is only the relative motion of the clock
that matters. It doesn't matter who accelerated, each will regard the
other's clock to be ticking slower than his own. What you are spouting
is not SR, nor is it consistent.
If the initial acceleration of one of the twins causes a reduction in
clock rate, then the second acceleration (which you are calling a
'deceleration') should reduce that rate even further, and yet though
accelerated in both instances the second acceleration increases clock
rate rather than reduces it as in the first case, then having passed
through the K frame the same continuous acceleration suddenly decides to
decrease the clock speed again. Oddly enough this occurs exactly when
the clock comes to a perfect stop wrt K in its turnaround. Now certainly
there are other frames in which the clock comes to a stop at some other
point during the acceleration, in fact there are an infinite number of
frames in which the acceleration never even brings the clock to a stop,
but only either speeds it up more or slows it down in its motion.
Let's suppose for instance a frame K'' moving at greater than the speed
of A, wrt C, at all times: From this frame K'', the first acceleration
of A would cause the A clock to increase in ticking rate, since this
clock is losing speed wrt K''. The second acceleration of A (turn
around) would cause the clock to reduce ticking rate, since it causes
the clock to gain speed wrt K''. Thus, wrt K'', the A clock which was
already ticking slower, first ticks less slow, then slows back down to
its initial rate. Thus wrt K'' this clock has ticked more than the C
clock. OTOH, the B clock first slows down in ticking rate', since it
gains speed, and then increases in ticking rate. Thus wrt K'' the B
clock has ticked less than C. Since A has ticked more than C, and B
ticked less than C, then they (A and B) cannot have possibly ticked the
same wrt C, regardless of what C perceived. And thus a direct
contradiction in your acceleration argument emerges with little effort.
Richard Perry
Paul Cardinale
No. It would only be a contradiction if SR predicted two or more
different values for the same measurement; which it doesn't. But
since you can't do the math either, you'll never realize the truth.
Paul Cardinale
Stephen Bint
"Paul Cardinale" <pcard...@volcanomail.com> wrote in message
news:64050551.03103...@posting.google.com...
> "Stephen Bint" <bi...@iname.com> wrote in message
> > So what if your travelling twin was passing earth and viewing such an
> > experiment there. Either he will see time slowed down there, or he will
see
> > the light pulse travelling faster than c there. Is that not the case?
> >
>
> Each twin will observe the other's clock to be running slow.
So, will they both be younger than eachother, afterwards? They have
experienced the same degree of time dilation, have they not?
G. Orme
> Stephen Bint wrote:
>>
>> And may I say in conclusion, that trying to bring the sweet light of
>> reason to the proponents of relativity, is like trying to bring
>> salvation to the DAMNED.
>>
>
> Why not go back to Jon's quite eloquent explanation of the twins and
> read it again. You seemed to get it then.
>
>>
>> Stephen Bint
>>
>
> Best, Dan.
Can I suggest an explanation of this effect, perhaps someone can tell me if
I have it right or not.
The problem with the twin paradox is based on frames of reference. Say
for example there is a planet A with an observer on it and a rocketship
close to it we'll call B. They are in a place where nothing else can be
seen, like a dust c,oud or if it's easier that they aren't aloud to look for
their position relative to any stars or objects.
Now they decide that either the planet A will go on a journey near the
speed of light and B will stay where he is. Either that or B will go for a
jounrye near the speed of light and A will stay where he is. Neither A nor B
know which will make the journey and they aren't allowed to look at any
stars to check.
All they will see is the other will move away from them and after a long
time will return. At that time one of them will see the other has aged less
than them.
From this the older one will deduce he must have stayed at rest and the
other moved away. I think though moving away is not a good way to describe
this as compared to each other, each moved away and returned. From observing
this motion A or B would be unable to tell which had actually moved.
The difference seems to be that if either A or B measured that they
underwent acceleration and decceleration then they would know they would
expect to be younger than the other.
Stephen Bint
"G. Orme" <newsg...@harmakhiss.org> wrote in message
news:3fa36fb9$0$28123$afc3...@news.optusnet.com.au...
Stephen
Dan Bloomquist
Stephen Bint wrote:
> "Paul Cardinale" <pcard...@volcanomail.com> wrote in message
>>
>>Each twin will observe the other's clock to be running slow.
>
> So, will they both be younger than eachother, afterwards? They have
> experienced the same degree of time dilation, have they not?
>
You just took the word 'observe' and turned it into 'be'. The twins
don't 'experience' time dilation. There is no time dilation _in_ the
observer's frame. The 'laws' of physics are identical for both twins in
their own frames.
As this thread seems to go in circles, I'll guess your next question is,
'How can they observe each other getting younger but they are not?'
Because you are assuming some master clock that says how old they are
without relevance to the distance between them. SR is a space-time
theory, it is not an absolute time theory. As long as you drag stuff
into the theory that isn't part of the theory you are not addressing the
theory.
Best, Dan.
--
http://lakeweb.net
http://ReserveAnalyst.com
dbAtLakewebDotCom
Stephen Bint
"Dan Bloomquist" <lak...@citlink.net> wrote in message
news:3FA3E579...@citlink.net...
Am I to blame for the thread going in circles? The FAQ for this newsgroup
and many posters who have berated me for my ignorance/stupidity/whatever,
have said that time dilation slows aging. When that becomes embarassing,
they switch to saying it doesn't. I go in circles, chasing them.
Stephen
Dan Bloomquist
Here it is, apples:
dilation slows aging for _one_(A) twin _relative_ to the other twin(B)
_if_ one twin(A) travels in multiple frames of reference. Time dilation
did not occur in either frame on it's own. The twins traveled in
different spacetimes _and_ returned to the same frame.
Oranges:
You said, 'So, will they both be younger than eachother, afterwards?'
I haven't seen anyone use the term 'afterwards' in the context of
inertial frames. I haven't seen anyone say 'they will _be_ younger than
each other.'
You said, 'They have experienced the same degree of time dilation.' SR
didn't say this, you did. I pointed out that you took the word 'observe'
and turned it into 'be'. You are still claiming a circle without
addressing the contraction you are creating.
Now, you can not point out a contradiction unless you say SR says
something it doesn't, it is that simple. You are inventing
contradictions in your mind and to support those contradiction you are
misrepresenting the theory.
G=EMC^2 Glazier
99.999999999 of"c' It is foreshortened by 80% in the direction it is
going. It it is a radio active particle its rate of decay will be
lengthen by 80% Bert
Stephen Bint
Dan,
You say that I cannot point out a contradiction. The FAQ says that a twin
flies away from earth and back again and has aged less. It says the reason
they haven't both aged less is because the has undergone several
accelerations.
The contradiction is this. If time dilation occurs, it happens both ways.
The effects of relative velocity must work two ways, because relative
velocity is a two-way relationship.
Because when earth looks at the travelling twin they must not see the light
in his cabin go faster than c, the twin must be dilated wrt earth.
Because when the travelling twin looks at the earth the twin must not see
the light on earth go faster than c, so the earth must be dilated, wrt twin.
The contradiction is, that time dilation occurs for both parties, but only
causes aging in one party.
It suggests that time dilation causes aging, sometimes.
Stephen
G. Orme
And the acceleration and decceleration A or B feels would relate directly to
the velocity he reaches so mathematically it should be equivalent.
Dan Bloomquist
Stephen Bint wrote:
>
> Dan,
>
> You say that I cannot point out a contradiction. The FAQ says that a twin
> flies away from earth and back again and has aged less. It says the reason
> they haven't both aged less is because the has undergone several
> accelerations.
>
In case you missed the page:
http://www.phys.unsw.edu.au/~jw/twin.html
> The contradiction is this. If time dilation occurs, it happens both ways.
> The effects of relative velocity must work two ways, because relative
> velocity is a two-way relationship.
Only if the frames _stay_ inertial.
>
> Because when earth looks at the travelling twin they must not see the light
> in his cabin go faster than c, the twin must be dilated wrt earth.
>
Not sure why the 'must' baggage, but yes.
> Because when the travelling twin looks at the earth the twin must not see
> the light on earth go faster than c, so the earth must be dilated, wrt twin.
>
yes.
> The contradiction is, that time dilation occurs for both parties, but only
> causes aging in one party.
>
Point out the contradiction in the link, explicitly. Quit pretending
twin B did _not_ change frames.
> It suggests that time dilation causes aging, sometimes.
>
There is no suggestion, there is no paradox. The math has been tested
against almost 100 years of observations. The math is explicit, there is
no 'sometimes'.
You are claiming a contradiction yet you can not point it out. You are
saying it doesn't make sense to you therefore it is wrong. And what it
amounts to is to say nature doesn't suit your your senses so nature is
wrong. Well, here is the real clue, nature could care less what you
think. Science is about building models around observations. SR is
observationally consistent. Consider the particle in an accelerator,
(among other experiments), as twin B.
But, if you'd like, continue to say the model is wrong without pointing
to the observation that backs your claim. And don't use the observation
you invent in your mind.
And, if you are sincere about getting this, it is about 'spacetime',
'spacetime', 'spacetime'.
T^2 = t^2 - s^2
There really is no more to it. Quite eloquent, IMHO.
> Stephen
Richard
Truth is only what can be empirically confirmed, and even then it is
suspect. SR is an untestable theory, OTOH GR and GR-like theories are
testable. No, despite what you are probably thinking right now, GR
doesn't in any way imply SR, nor can it, since GR takes as premise that
space is nowhere isotropic as is required of SR.
Richard Perry
Sam Wormley
>
> Truth is only what can be empirically confirmed, and even then it is
> suspect. SR is an untestable theory, OTOH GR and GR-like theories are
> testable. No, despite what you are probably thinking right now, GR
> doesn't in any way imply SR, nor can it, since GR takes as premise that
> space is nowhere isotropic as is required of SR.
>
> Richard Perry
Observational and Experimental Evidence Bearing on General Relativity
http://math.ucr.edu/home/baez/RelWWW/tests.html
General Relativity Tutorial
John Baez
http://math.ucr.edu/home/baez/gr/gr.html
Relativity on the World Wide Web
http://math.ucr.edu/home/baez/relativity.html
General Relativity and Cosmology FAQs
http://math.ucr.edu/home/baez/physics/
Developments in General Relativity: Black Hole Singularity and Beyond
http://arxiv.org/abs/gr-qc/0304052
Improved Test of General Relativity with Radio Doppler Data from the Cassini Spacecraft
http://arxiv.org/abs/gr-qc/0308010
What is the experimental basis of Special Relativity?
http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html
Physics is an experimental science, and as such the experimental basis for
any physical theory is extremely important. The relationship between
theory and experiments in modern science is a multi-edged sword:
1.It is required that the theory not be refuted by any experiment within
the theory's domain of applicability.
2.It is expected that the theory be confirmed by a number of
experiments which cover a significant fraction of the theory's
domain of applicability.
3.It is expected that the theory be confirmed by a number of
experiments which examine a significant fraction of the theory's
predictions.
Special Relativity (SR) meets all of these requirements and expectations.
There are literally hundreds of experiments which have tested SR, with
an enormous range and diversity, and the agreement between theory and
experiment is excellent. There is a lot of redundancy in these experimental
tests. There are also a lot of indirect tests of SR which are not included
here. This list of experiments is by no means complete!
Other than their sheer numbers, the most striking thing about these
experimental tests of SR is their remarkable breadth and diversity. An
important aspect of SR is its universality - it applies to all known physical
phenomena and not just to the electromagnetic phenomena it was
originally invented to explain. In these experiments you will find tests
using electromagnetic and nuclear measurements (including both strong
and weak interactions); gravitational tests are the province of General
Relativity, and are not considered here, see Experimental Tests of GR.
There are several useful surveys of the experimental basis of SR:
Y.Z.Zhang, Special Relativity and its Experimental Foundations,
World Scientific (1997).
G.Holton, "Resource Letter SRT-1 on Special Relativity Theory",
Am. J. Phys., 30 (1962), p462.
D.I.Blotkhintsev, "Basis for Special Relativity Theory Provided by
Experiments in High Energy Physics", Sov. Phys. Uspekhi, 9 (1966),
p405.
Newman et al. Phys. Rev. Lett. 40 no. 21 (1978), p1355.
Zhang's book is especially comprehensive.
Textbooks which have good summaries of the experimental basis of
relativity are:
M.Born, Einstein's theory of Relativity.
Bergmann, Introduction to the Theory of Relativity.
Moller, The Theory of Relativity.
M. von Laue, Die relativitätstheorie (in German).
Crank Information
http://www.google.com/search?hl=en&lr=&ie=ISO-8859-1&q=%22Richard+perry%22+site%3Ausers.pandora.be
http://groups.google.com/groups?q=group%3Asci.physics+author%3APerry
Richard
Which of these experiments were conducted in the absence of a net
gravitational potential? How many turn to gibberish when the Earth is
removed out from under them. You're an idiot and a crank, and performing
no service by countering sound reason.
Richard Perry
Stephen Bint
You were right to direct me back to the link and I found that I had
misunderstood it. The contradiction is that it says explicitly that Jane
observes Joe's clock going more slowly on both legs of the journey, then the
analysis and diagram show her receiving his anniversary signals more than
once a year on the return journey.
First it says:
"On the outward leg Jane observes Joe's clock to run slowly, and she
observes that it ticks slowly on the return run. So will Jane conclude that
Joe will have aged less?"
Then it says:
"In the second of these inertial frames, she receives a lot of anniversary
messages from Joe. If she pretended that she had been in this frame when Joe
sent them (the dashed line extrapolation of her returning world line, i.e.
if she had been travelling towards Earth at constant v for six of her
years), she would conclude that Joe had been sending them for eight of his
years."
In other words, Jane observes, on Joe's clock, eight years of Joe history in
six years. She watches the clock go slowly on both legs of the journey,
slower than hers, yet she sees eight anniversaries tick past in six of her
years.
On the diagram, she is receiving anniversary messages more than once a year,
so she must surely be seeing his clock go more quickly than her own on the
way back.
Why is that not a contradiction of "she observes that it ticks slowly on the
return run"?
Stephen
Stephen Bint
Sam Wormley
Bilge
>Which of these experiments were conducted in the absence of a net
>gravitational potential?
Which experiments that support your point of view were conducted in
the absence of nuclear forces? The strong force is 10^39 times the
strength of the gravity and even the weak force is 10^32 times the
strength of gravity. If gravity is a problem for testing special
relativity at any level here on earth, nuclear forces are a much
more serious problem for you.
Sam Wormley
Perhaps these numbers for relative strength of forces
http://scienceworld.wolfram.com/physics/FundamentalForces.html
Bilge
Their number for the weak force is way off.
That 10^-16 for the weak interaction should be 10^-6.
kenseto
> Besides which, the formula for time dilation is a function of the velocity
> of the observed frame and makes no mention whatsoever of the acceleration
of
> either frame, before or while the calculation is made.
>
> I was told again and again by several posters, "Do the math. Just plug the
> numbers into the formulas and it all works out just fine." These were the
> same people who want me to factor in acceleration, which is not mentioned
in
> the formula, because time dilation is purely a function of the relative
> velocity, which is the same in both directions.
Relative motion (velocity v in the time dilation equation) between A and B
is the vector difference of their absolute motions. Whoever has a higher
state of absolute motion his clock will run slower than the other clock.
This means that if A has a higher state of absolute motion then his clock
will run slower than B's clock and from B's point of view his clock will run
faster than A's clock. This concept destroys the SR concept of mutual time
dilation and illustrates the incompleteness of SR.
Ken Seto
>
> Plug the relative velocity of skateboarders A and B into the formula and
> find out how much slower B's clock is, than A's. And how much slower A's
is,
> than B's. I don't need to "do the math" to see that this formula is
telling
> me two things which cannot both be true.
>
> I suppose it depends on your definition of "working out fine".
>
> Because I constructed my example with two moving twins, to force them to
> face the symmetry, the defenders of relativity introduced a new twist:
> acceleration causes time to speed up, cancelling out the age difference,
in
> both skateboarders. This is an new addition to relativity. Einstein never
> said anything of the sort.
>
> To say that acceleration makes time run faster, implies that if one of the
> skateboarders is accelerating, the time contraction due to acceleration
will
> at least partially counteract the dilation caused by the velocity
(predicted
> by Einstein), allowing the other to see the bouncing beam move faster than
> c.
>
> Still, though frustrated, I thoroughly enjoyed the thread I started and I
am
> grateful to all of the people who had the patience to contribute. In
return,
> I can only assure you that I will continue to post under the name "Stephen
> Bint", so as not to undermine your killfile strategies.
>
> And may I say in conclusion, that trying to bring the sweet light of
reason
> to the proponents of relativity, is like trying to bring salvation to the
> DAMNED.
>
>
> Stephen Bint
>
>
>
Sam Wormley
Thanks
FrediFizzx
news:slrnbq9lun....@radioactivex.lebesque-al.net...
Isn't the one for the electromagnetic interaction off also? e =
sqrt(alpha) is about 0.0854 which is close to being 10^-1 and not
10^-3.
FrediFizzx