Length contraction and time dilaton

  • #1
bernhard.rothenstein
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Please let me know if length contraction could be derived without using synchronized clocks.
Thanks in advance.
 

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  • #2
Dale
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Please let me know if length contraction could be derived without using synchronized clocks.
Thanks in advance.
I would try radar coordinates.
 
  • #3
bernhard.rothenstein
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radar coordinates and length contraction

I would try radar coordinates.
Thanks for your answer but as far as I know radar detection could lead to length dilation and to length dilation as well, Lorentz contraction being detected only under special experimental conditions.
 
  • #4
yuiop
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Please let me know if length contraction could be derived without using synchronized clocks.
Thanks in advance.

Assuming:
The speed of light is constant for any observer.
Time dilation is 1/sqrt(1-v^2/c^2)
Proper length of the rod in rest frame = Lo

The time taken for a signal to be sent from one end of the rod to a mirror at other end and return to the original end is 2Lo/c in the rest frame (frame S). Since there is only a single clock the round trip time in frame S' is 2Lo/(c sqrt(1-v^2/c^2)) = 2Lo/sqrt(c^2-v^2). The time in frame S can also be determined from the sum of the catch up time L/(c-v) and return time L/(c+v). Equating the two expressions for the total time in frame S' we get:

2Lo/sqrt(c^2-v^2) = L/(c-v) + L/(c+v)

2Lo/sqrt(c^2-v^2) = (L(c+v) + L(c-v))/((c-v)(c+v))

2Lo/sqrt(c^2-v^2) = 2Lc/(c^2-v^2)

L = Lo(c^2-v^2)/(c sqrt(c^2-v^2))

L = Lo sqrt(c^2-v^2)/c

L = Lo sqrt(1-v^2/c^2)

No clocks were harmed or synchronised during this derivation :P
 
  • #5
DrGreg
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Please let me know if length contraction could be derived without using synchronized clocks.
Thanks in advance.
No, it is impossible. Both length contraction and time dilation are related to the relativity of simultaneity. If you change the synchronisation convention, the amount of contraction or dilation may change too.

For example in Selleri coordinates, if an observer in the rest frame observes a moving object, the usual length contraction and time dilation applies because the rest frame uses Einstein sync. But if a moving observer observes a static object in the rest frame, because the moving observer uses the same defintion of simultaneity as the rest frame, the moving observer measures length dilation and time contraction. With absolute sync, dilation and contraction are also absolute. So a moving observer measures an object moving faster than herself (relative to the rest frame) as length-contracted & time-dilated, but an object moving slower as length-dilated & time-contracted.

Of course all of the above applies only to "one-way" time-dilation directly between two objects. It does not apply to the "two-way" round-trip cumulative time-dilation that you get in the twins paradox, which can be measured using proper-time clocks, with no sync convention, and can be explained using k-calculus.

The flaw in kev's argument (post #4) is his assumption of the time dilation formula, which depends on clock synchronisation.
 
  • #6
granpa
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So a moving observer measures an object moving faster than herself (relative to the rest frame) as length-contracted & time-dilated, but an object moving slower as length-dilated & time-contracted.

no. a moving object would measure all objects moving faster or slower than itself as being length contracted. thats what relativity is all about. every observer considers themselves to be stationary.
 
  • #7
Mentz114
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My understanding is that length contraction and time dilation both follow from the invariance of the proper interval. This is a postulate and doesn't mention synchronization explicitly but I suppose it might be hidden somewhere.
 
  • #8
granpa
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the length of an object is the distance between the position of the front and back of the object at one simultaneous moment.
 
  • #9
DrGreg
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So a moving observer measures an object moving faster than herself (relative to the rest frame) as length-contracted & time-dilated, but an object moving slower as length-dilated & time-contracted.

no. a moving object would measure all objects moving faster or slower than itself as being length contracted. thats what relativity is all about. every observer considers themselves to be stationary.
You are absolutely right if we are talking about clocks synchronised in the way specified by Einstein, which is the assumption we almost always make in Special Relativity. But I am talking about the weird consequences when you choose a non-standard method of clock synchronisation.
 
  • #10
DrGreg
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My understanding is that length contraction and time dilation both follow from the invariance of the proper interval. This is a postulate and doesn't mention synchronization explicitly but I suppose it might be hidden somewhere.
The equation

[tex]ds^2 = dx^2 + dy^2 + dz^2 - c^2 \, dt^2[/tex]​

is derived using Einstein-synced coordinates. In Selleri coordinates moving at speed v, the interval is

[tex]ds^2 = \frac{dx^2}{\gamma^2} + dy^2 + dz^2 - c^2 \, dt^2 + 2 \, v \, dx \, dt[/tex]​

(see this thread, in particular, post #7 and correction in post #10).
 
  • #11
yuiop
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No, it is impossible. .....
The flaw in kev's argument (post #4) is his assumption of the time dilation formula, which depends on clock synchronisation.

Here is an attempt without assuming time dilation...

Derivation based on a Michelson-Morely type experiment.

Assuming:
The speed of light is constant for any observer.
Proper length of the MM apparatus arms in rest frame (Frame S) = Lo
Frame S' has a velocity of v relative to frame S.
Length of the parallel arm in frame S' = Lp'
Length of the transverse arm in frame S' = L'

The time in frame S' can also be determined from the sum of the catch up time Lp'/(c-v) and return time Lp'/(c+v) for the parallel arm to give:

(1) t' = 2*Lp'*c/(c^2-v^2) .. See post#4

The length of the diagonal path of the photon moving along the transverse arm in frame S' is 2*sqrt((v*t'/2)^2 + L'^2) from Pythagorous theorem. The time taken is the same as for the parallel arm, from the null MM result. The light signal travel time for this arm is the diagonal path length devided by the speed of light to give

(2) t' = 2*sqrt((v*t'/2)^2 + L'^2)/c

Rearranging (2) to isolate t' on the left hand side:

(3) t'^2 = 4((v^2*t'^2/4 + L'^2)/c^2

(4) t'^2 = t'^2*v^2/c^2 + 4L'^2/c^2

(5) t'^2-t'^2*v^2/c^2 = 4L'^2/c^2

(6) t'^2(1-v^2/c^2) = 4L'^2/c^2

(7) t'^2 = 4L'^2/(c^2(1-v^2/c^2))

(8) t' = 2L'/(c*sqrt(1-v^2/c^2))

(9) t' = 2L'/sqrt(c^2-v^2)


Equating (1) and (9)

(10) 2*Lp'*c/(c^2-v^2) = 2L'/sqrt(c^2-v^2)

(11) Lp' = L'*sqrt(c^2-v^2)/c

(12) Lp' = L'*sqrt(1-v^2/c^2)


So far we have derived an expression for the ratio of the length L' of the transverse arm to the length Lp' of the parallel arm, both in Frame S'.

Now we have to use some logical deduction to relate L' and Lp' to Lo in the rest frame. The length of the transverse arm L' in S' has to be the same as Lo in S. Logical contradictions arise if the transverse arm length contracts. Imagine two hollow cylinders of the same radius orientated with their long axis parallel to the direction of motion. Transverse length contraction would have cylinder A passing inside cylinder B in one frame and cylinder B passing inside cylinder A in another, which would be inconsistent. Form this logical deduction we can conclude:

Transverse length L' = Lo
Parallel length Lp' = Lo*sqrt(1-v^2/c^2)

because we now know the relationship of L' to Lp'.

Only a single clock is used in the above derivation and an implicit assumption that the speed of light is the same in both directions.

--------------------------------------------------------------------------------------
[EDIT]

We can work out time dilation if we consider the length contraction above to be correct.
The time To in the rest frame for a photon to traverse either arm is:

(13) To = 2Lo/c
(14) To = 2L'/c
(15) L' = To*c/2

Taking equation (9) t' = 2L'/sqrt(c^2-v^2) and re-arranging:

(16) L' = t'*sqrt(c^2-v^2)/2

Equating (15) and (16)

(17) To*c/2 = t'*sqrt(c^2-v^2)/2

(18) t' = To*c/sqrt(c^2-v^2)

(19) t' = To/sqrt(1-v^2/c^2)
 
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  • #12
DrGreg
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...an implicit assumption that the speed of light is the same in both directions.
That assumption is precisely the assumption of Einstein clock synchronisation! :smile:
 
  • #13
yuiop
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That assumption is precisely the assumption of Einstein clock synchronisation! :smile:


Does not the postulate that the speed of light is a constant value for all observers in an inertial reference frame imply the speed of light is the same in all directions?
 
  • #14
Mentz114
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DrGreg,
thanks for revealing the implied the clock synch'ing in your post#10.
I found the posts you pointed to very interesting.

M
 
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  • #15
Dale
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Does not the postulate that the speed of light is a constant value for all observers in an inertial reference frame imply the speed of light is the same in all directions?
That would be a reasonable implication, but proponents of alternate synchronization conventions like to interpret that postulate to refer to the two-way speed of light, which remains constant and isotropic regardless of synchronization convention.
 
  • #16
Dale
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bernhard: although Mentz did not mention it here, he came up with a really good idea for addressing this problem in another thread:
Another way of measuring the length would be to measure the time taken for the rod to pass a fixed point in your frame.
This would only require a single clock and would not require any assumptions about the 1-way speed of light.
 
  • #17
yuiop
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bernhard: although Mentz did not mention it here, he came up with a really good idea for addressing this problem in another thread:

"Another way of measuring the length would be to measure the time taken for the rod to pass a fixed point in your frame."

This would only require a single clock and would not require any assumptions about the 1-way speed of light.

That would be a good way to measure length contraction using a single clock but it does not derive length contraction. Another possible difficulty, is that in order to determine the velocity of the rod you would require two spatially seperated clcoks. Possibly you could doppler radar and compare the emitted frequency of the radar pulses to the returned frequency of the echoed pulses to determine the velocity and use that measure the length of the passing rod and that in turn has an assumption of isotropic speed of light. (You can not measure the length with a single clock without knowing the velocity). That still leaves the problem of deriving as opposed to measuring the length contraction.
 
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  • #18
Hans de Vries
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Please let me know if length contraction could be derived without using synchronized clocks.
Thanks in advance.

Length contraction can be derived from the basic wave-equations: The
equations for the electromagnetic potentials and the Klein Gordon equation.

See sections 2.1 http://physics-quest.org/Book_Chapter_EM_LorentzContr.pdf" [Broken].

The simplest way I can think of to measure length contraction in a given
reference frame is this:

Send out two laser beams under a certain angle. If the length contraction of
an object flying by is enough, then it will fit in between the two beams and
you'll never see the reflection of both beams at the same time.


Regards, Hans.
 
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  • #19
Mentz114
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Kev is right. It doesn't matter which method is used, you need two clocks.

But I also like this, from an article on the web by J. L. Safko -

Associated with this synchronization problem is the question of measuring lengths. To properly measure the length of a moving object, we must measure the position of both ends at the same time in our inertial frame. However, an observer at rest on the moving object would not agree that the measurements were made at the same time. The observer at rest with respect to the moving object, using her own clocks, would say that the position of the front end was measured at an earlier time than the position of the back end. So both agree that a measurement of length of a moving rod yields a shorter length than the measurement made in the frame of the rod. This is called the Lorentz contraction. This contraction is real in any sense of the word.
It seems to need synchronisation.

Maybe Hans has it. I'll read the cited chapters.
 
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  • #20
Dale
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Possibly you could doppler radar and compare the emitted frequency of the radar pulses to the returned frequency of the echoed pulses to determine the velocity and use that measure the length of the passing rod and that in turn has an assumption of isotropic speed of light.
Right, but it assumes only an isotropic 2-way speed of light, which can be measured with a single clock and therefore requires no synchronization. I think you are selling yourself short here, I am pretty sure that your idea would work and would answer the OP's question. Also, if you can measure it surely you can work backwards to derive it (although I will leave that as an exercise for the interested reader or OP :smile:)
 
  • #21
bernhard.rothenstein
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length contraction and time dilation

No, it is impossible. Both length contraction and time dilation are related to the relativity of simultaneity. If you change the synchronisation convention, the amount of contraction or dilation may change too.

For example in Selleri coordinates, if an observer in the rest frame observes a moving object, the usual length contraction and time dilation applies because the rest frame uses Einstein sync. But if a moving observer observes a static object in the rest frame, because the moving observer uses the same defintion of simultaneity as the rest frame, the moving observer measures length dilation and time contraction. With absolute sync, dilation and contraction are also absolute. So a moving observer measures an object moving faster than herself (relative to the rest frame) as length-contracted & time-dilated, but an object moving slower as length-dilated & time-contracted.

Of course all of the above applies only to "one-way" time-dilation directly between two objects. It does not apply to the "two-way" round-trip cumulative time-dilation that you get in the twins paradox, which can be measured using proper-time clocks, with no sync convention, and can be explained using k-calculus.

The flaw in kev's argument (post #4) is his assumption of the time dilation formula, which depends on clock synchronisation.
Please explain me what is the difference between Selleri and Tangherlini approaches.
 
  • #22
yuiop
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Please explain me what is the difference between Selleri and Tangherlini approaches.

As far as I can tell, none of the alternative coordinate systems uphold the postulate that the speed of light is constant for all inertial observers and that the laws of physics are the same in all inertial reference frames.
 
  • #23
bernhard.rothenstein
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selleri interval

The equation

[tex]ds^2 = dx^2 + dy^2 + dz^2 - c^2 \, dt^2[/tex]​

is derived using Einstein-synced coordinates. In Selleri coordinates moving at speed v, the interval is

[tex]ds^2 = \frac{dx^2}{\gamma^2} + dy^2 + dz^2 - c^2 \, dt^2 + 2 \, v \, dx \, dt[/tex]​

(see this thread, in particular, post #7 and correction in post #10).


is it correct to present selleri interval as
x^2(1-V^2/c^2)-2Vt(a)x-c^2t(a)^2=x^2-c^2t(E)^2
in an one space dimensions approach where t(a) represents the reading of the nonstandard synchronnized clock whereas t(E) represent the reading of the standard synchronized clock.
I think that working with standard and nonstandard synchonized cloks it is essential to mention which clock measures the time which appears in the presented equations.
in understanding Selleri's approach it is hard for me to understand why in the equations he proposes it is not correct to present the first one as
x'=x(1-VV/cc)^1/2-Vt(a)/(1-VV/cc)^1/2
the inverse of which is
x=(x'+Vt'(E))/(1-VV/cc)^1/2
 
  • #24
DrGreg
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Does not the postulate that the speed of light is a constant value for all observers in an inertial reference frame imply the speed of light is the same in all directions?
If you choose to word the postulate in that way, then, yes it does. The effect of the postulate, under that interpretation, is to specify how you synchronise clocks. "Einstein synchronisation" and "one-way light speed isotropy" are essentially the same thing. Either you have both or you have neither.

However there is a weaker version of the postulate, that doesn't mandate a synchronisation convention, which says no more that the speed of light does not depend on the motion of its emitter. In other words, one photon cannot overtake another travelling in the same direction. Under this interpretation, the fact that the "two-way speed of light" is isotropic is a consequence of the other postulate, that the laws of physics look the same to all inertial observers.

It may be a useful experience (for readers who have advanced beyond beginner level and who are comfortable with algebra) to see what happens when you deliberately "mis-synchronise" clocks as explained in this post which I mentioned before. With mis-synced clocks, time dilation and length contraction need not take the usual Lorentz forms; so any proof of time dilation or length contraction by the Lorentz factor must somewhere (directly or indirectly) be making use of Einstein-sync along the way.
 
  • #25
DrGreg
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Right, but it assumes only an isotropic 2-way speed of light, which can be measured with a single clock and therefore requires no synchronization. I think you are selling yourself short here, I am pretty sure that your idea would work and would answer the OP's question. Also, if you can measure it surely you can work backwards to derive it (although I will leave that as an exercise for the interested reader or OP :smile:)
The problem here is converting doppler factor to velocity. Coordinate velocity is coordinate distance divided by coordinate time (not proper time); you need to know how to sync clocks before you can even define what velocity is!
 
  • #26
DrGreg
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Please explain me what is the difference between Selleri and Tangherlini approaches.
When I refer to "Selleri coords" or "Tangherlini coords", I mean the same thing -- just a convenient label to apply to the coords defined here. I haven't read either Selleri's or Tangherlini's work on this subject; if you can point me to a website I would be interested.

is it correct to present selleri interval as
x^2(1-V^2/c^2)-2Vt(a)x-c^2t(a)^2=x^2-c^2t(E)^2
in an one space dimensions approach where t(a) represents the reading of the nonstandard synchronnized clock whereas t(E) represent the reading of the standard synchronized clock.
I think that working with standard and nonstandard synchonized cloks it is essential to mention which clock measures the time which appears in the presented equations.
Your equation follows from my two equations when you make the distinction between [itex]t_E[/itex] in the first and [itex]t_a[/itex] in the second. You also have -V instead of v which means you are measuring velocity the other way round (E relative to a instead of a relative to E, I think).

in understanding Selleri's approach it is hard for me to understand why in the equations he proposes it is not correct to present the first one as
x'=x(1-VV/cc)^1/2-Vt(a)/(1-VV/cc)^1/2
the inverse of which is
x=(x'+Vt'(E))/(1-VV/cc)^1/2
Is this a reference to Selleri's own publication?
 
  • #27
bernhard.rothenstein
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When I refer to "Selleri coords" or "Tangherlini coords", I mean the same thing -- just a convenient label to apply to the coords defined here. I haven't read either Selleri's or Tangherlini's work on this subject; if you can point me to a website I would be interested.

Your equation follows from my two equations when you make the distinction between [itex]t_E[/itex] in the first and [itex]t_a[/itex] in the second. You also have -V instead of v which means you are measuring velocity the other way round (E relative to a instead of a relative to E, I think).

Is this a reference to Selleri's own publication?
Thanks for your help.
Please follow the following approach to Selleri.
Consider the inertial reference frames I and I' in the standard arrangement, I' moving with constant speed V relative to I in the positive direction of the overlapped OX(O'X') axes. At a point M(x,0) we find the clocks C1(x,0) and C2(x,0) at rest in I both synchronized with clock C0(0,0) located at the origin O(0,0) of I. Clock C1(x,0) is sinchronized using an isotropic light signal that propagates with speed C the second with an anisotropic light signal propagationg in the positive directioon of the OX axis with speed c, both emitted from O when C0 reads t=0. When the corresponding light signal arrives at C1 it reads t(E)=x/C whereas clock C2 reads t(a)=x/c, t(E) and t(a) being related by
t(E)=t(a)+(x/C)(1-C/c). (1)
Performing the Lorentz transformations to I' the result is
x'=gx[1-V/C(1-C/c)-gVt(a) (2)
t'(E)=gt(a)+g(x/C)(1-C/c-V/c) (3)
Imposing the condition of absolute simultaneity (dt(a)=0 implies dt'(E)=0 we impose the condition
1-C/c-V/C=0 (4)
and so
c=C/(1-V/C) (5)
with which (2) becomes
x'=x/g-gVt(a) (6)
(3) becoming
t'(E)=gt(a) (7)
recovering the Selleri transformation equations presented by the author as
x'=g(t-x/C) (8)
t'=gt (9)
the weak point consisting in the fact that they do not mention the clocks which display the involved times.
Do you find flows in the lines above which are only an attempt to put your help in my own way of thinking.
Thanks in advance.
t'=gt
 
  • #28
DrGreg
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At a point M(x,0) we find the clocks C1(x,0) and C2(x,0) at rest in I both synchronized with clock C0(0,0) located at the origin O(0,0) of I.
When you say "(x,0)", what is the "0"? Is it y? Is it t?
 
  • #29
bernhard.rothenstein
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absolute simultaneity

When you say "(x,0)", what is the "0"? Is it y? Is it t?
In a two space dimensions approach I characterize postion of a point or of a clock located at that point as M(x,y), C(x,y) and so M(x,0) stays for y=0. Thanks
 
  • #30
Dale
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It seems to me that Selleri coordinates would not avoid synchronized clocks, they would just use a non-standard convention of synchronization.
 
  • #31
bernhard.rothenstein
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selleri coordinates

It seems to me that Selleri coordinates would not avoid synchronized clocks, they would just use a non-standard convention of synchronization.
Not only Selleri but Abreu and Guerra (see arxiv and European Journal Physics) as well, use Einstein synchronization in I and nonstandard clock synchronization in I.
 
  • #32
DrGreg
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It seems to me that Selleri coordinates would not avoid synchronized clocks, they would just use a non-standard convention of synchronization.
I'm not sure whether you aimed that comment at me or Bernhard.

The only reason I mentioned Selleri coordinates in the first place was to provide a counterexample to the proposition that you could derive the standard formula for Lorentz contraction without a clock synchronisation convention (explicit or implied). If you don't use Einstein-synced coordinates, Lorentz contraction need not be true; e.g. between pairs of Selleri coord systems it's possible to have length dilation instead of contraction. Therefore any proof of Lorentz contraction must make use of a synchronisation convention somewhere, either explicitly or by quoting some other sync-dependent result.
 
  • #33
Dale
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Therefore any proof of Lorentz contraction must make use of a synchronisation convention somewhere, either explicitly or by quoting some other sync-dependent result.
Then the simple answer to the OP:
Please let me know if length contraction could be derived without using synchronized clocks.
Is "no".
 
  • #34
DrGreg
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Thanks for your help.
Please follow the following approach to Selleri.
Consider the inertial reference frames I and I' in the standard arrangement, I' moving with constant speed V relative to I in the positive direction of the overlapped OX(O'X') axes. At a point M(x,0) we find the clocks C1(x,0) and C2(x,0) at rest in I both synchronized with clock C0(0,0) located at the origin O(0,0) of I. Clock C1(x,0) is sinchronized using an isotropic light signal that propagates with speed C the second with an anisotropic light signal propagationg in the positive directioon of the OX axis with speed c, both emitted from O when C0 reads t=0. When the corresponding light signal arrives at C1 it reads t(E)=x/C whereas clock C2 reads t(a)=x/c, t(E) and t(a) being related by
t(E)=t(a)+(x/C)(1-C/c). (1)
Performing the Lorentz transformations to I' the result is
x'=gx[1-V/C(1-C/c)-gVt(a) (2)
t'(E)=gt(a)+g(x/C)(1-C/c-V/c) (3)
Imposing the condition of absolute simultaneity (dt(a)=0 implies dt'(E)=0 we impose the condition
1-C/c-V/C=0 (4)
and so
c=C/(1-V/C) (5)
with which (2) becomes
x'=x/g-gVt(a) (6)
(3) becoming
t'(E)=gt(a) (7)
recovering the Selleri transformation equations presented by the author as
x'=g(t-x/C) (8)
t'=gt (9)
the weak point consisting in the fact that they do not mention the clocks which display the involved times.
Do you find flows in the lines above which are only an attempt to put your help in my own way of thinking.
Thanks in advance.
t'=gt
Sorry for the delay in replying. I am not able to log in to this forum every day.

For the benefit of other readers I should say that this seems to have no connection with the original question of this thread.

Let me try and reword this, as it's currently rather confusing.

Consider the inertial reference frames I and I' in the standard arrangement, I' moving with constant speed V relative to I in the positive direction of the overlapped OX & O'X' axes. At a point M(x,0,0) we find the clocks C1 and C2, and at the origin of I we find clock C0, all three at rest in I.

Clock C1 is synchronised to C0 in such a way as to make the speed of light isotropic when measured using C0 and C1,

[tex]t_E = x/c[/tex]​

being the time measured by C1 when a photon is received at M that was emitted from the origin when C0 time was zero.

Clock C2 is synchronised to C0 in such a way as to make the speed of light from O to M anisotropically equal to [itex]c_f[/itex] when measured using C0 and C2,

[tex]t_a = x/c_f[/tex]​

being the time measured by C2 when a photon is received at M that was emitted from the origin when C0 time was zero.

[itex]t_E[/itex] and [itex]t_a[/itex] are therefore related by

[tex]t_E = t_a + \frac{x}{c} \left( 1 - \frac{c}{c_f} \right)[/tex]. (1)​


At this point I get lost: I can't see how you got (2) and (3). There is a missing intermediate step that you will have to supply.
 
  • #35
DrGreg
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Thanks to a private message from Bernhard I can continue from where my previous post stopped.

(I measure V in the direction implied by post #27, opposite to the private message. Just a change of sign.)

Now consider Einstein-synced coordinates [itex](t'_E, x')[/itex] in the I' frame. These are related to the Einstein-synced coordinates [itex](t_E, x)[/itex] in the I frame by the Lorentz transform

[tex]x' = \gamma (x - V t_E) = \gamma \left(1 - \frac{V}{c}(1 - c/c_f) \right) x - \gamma V t_a [/tex] (2)
[tex]t'_E = \gamma (t_E - V x / c^2) = \gamma t_a + \gamma \left( 1 - \frac{c}{c_f} - \frac {V}{c} \right) \frac {x} {c} [/tex] (3)​

Now choose [itex]c_f[/itex] to be the value which makes equation (3) independent of [itex]x[/itex], i.e.

[tex] 1 - \frac{c}{c_f} - \frac{V}{c} = 0 [/tex] (4)​

which solves as

[tex] c_f = \frac{c}{1 - V/c} [/tex] (5)​

Substitution in (2) and (3) gives:

[tex] x' = \frac{x}{\gamma} - \gamma V t_a [/tex] (6)
[tex] t'_E = \gamma t_a [/tex] (7)​

That all seems correct to me.

If anyone other than Bernhard is following this, what we have done is define an alternative coordinate system [itex](t_a, x)[/itex] in the I frame which is related to the I'-frame coordinates [itex](t'_E, x')[/itex] via equations (6) and (7) instead of the usual Lorentz Transform. Equation (7) shows that both coordinate systems agree in their definitions of simultaneity -- if two events share the same [itex]t'_E[/itex] coord, they must also share the same [itex]t_a[/itex] coord.

Putting [itex]t_a = 0[/itex], equation (6) shows that a rod of length [itex]x'[/itex] that is stationary in the I' frame is longer in [itex](t_a, x)[/itex] coordinates, which is the counterexample to the original implied proposition of post #1.
 

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