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bernhard.rothenstein
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Please let me know if length contraction could be derived without using synchronized clocks.
Thanks in advance.
Thanks in advance.
I would try radar coordinates.Please let me know if length contraction could be derived without using synchronized clocks.
Thanks in advance.
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.I would try radar coordinates.
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.Please let me know if length contraction could be derived without using synchronized clocks.
Thanks in advance.
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.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.
The equationMy 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.
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.
That assumption is precisely the assumption of Einstein clock synchronisation!![]()
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.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?
This would only require a single clock and would not require any assumptions about the 1-way speed of light.Another way of measuring the length would be to measure the time taken for the rod to pass a fixed point in your frame.
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.
Please let me know if length contraction could be derived without using synchronized clocks.
Thanks in advance.
It seems to need synchronisation.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.
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 OPPossibly 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.
Please explain me what is the difference between Selleri and Tangherlini approaches.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.
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).
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.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?
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!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)
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.Please explain me what is the difference between Selleri and Tangherlini approaches.
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 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.
Is this a reference to Selleri's own publication?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
Thanks for your help.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?
When you say "(x,0)", what is the "0"? Is it y? Is it t?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.
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. ThanksWhen you say "(x,0)", what is the "0"? Is it y? Is it t?
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.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.It seems to me that Selleri coordinates would not avoid synchronized clocks, they would just use a non-standard convention of synchronization.
Then the simple answer to the OP:Therefore any proof of Lorentz contraction must make use of a synchronisation convention somewhere, either explicitly or by quoting some other sync-dependent result.
Is "no".Please let me know if length contraction could be derived without using synchronized clocks.
Sorry for the delay in replying. I am not able to log in to this forum every day.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