# Length contraction and time dilaton

1. May 15, 2008

### bernhard.rothenstein

Please let me know if length contraction could be derived without using synchronized clocks.

2. May 15, 2008

### Staff: Mentor

3. May 15, 2008

### bernhard.rothenstein

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. May 15, 2008

### yuiop

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. May 16, 2008

### DrGreg

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. May 16, 2008

### granpa

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. May 16, 2008

### Mentz114

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. May 16, 2008

### granpa

the length of an object is the distance between the position of the front and back of the object at one simultaneous moment.

9. May 16, 2008

### DrGreg

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. May 16, 2008

### DrGreg

The equation

$$ds^2 = dx^2 + dy^2 + dz^2 - c^2 \, dt^2$$​

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

$$ds^2 = \frac{dx^2}{\gamma^2} + dy^2 + dz^2 - c^2 \, dt^2 + 2 \, v \, dx \, dt$$​

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

11. May 16, 2008

### yuiop

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)

Last edited: May 16, 2008
12. May 16, 2008

### DrGreg

That assumption is precisely the assumption of Einstein clock synchronisation!

13. May 16, 2008

### yuiop

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. May 16, 2008

### Mentz114

DrGreg,
thanks for revealing the implied the clock synch'ing in your post#10.
I found the posts you pointed to very interesting.

M

Last edited: May 16, 2008
15. May 16, 2008

### Staff: Mentor

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. May 16, 2008

### Staff: Mentor

bernhard: although Mentz did not mention it here, he came up with a really good idea for addressing this problem in another thread:
This would only require a single clock and would not require any assumptions about the 1-way speed of light.

17. May 16, 2008

### yuiop

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.

Last edited: May 16, 2008
18. May 16, 2008

### Hans de Vries

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.

Last edited by a moderator: May 3, 2017
19. May 16, 2008

### Mentz114

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 -

It seems to need synchronisation.

Maybe Hans has it. I'll read the cited chapters.

Last edited: May 16, 2008
20. May 16, 2008

### Staff: Mentor

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 )