# Understanding length contraction in SR

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1. Mar 30, 2015

### vinven7

I seem to be missing something fundamental about length contraction in SR. I would be grateful if someone can point out the error in the following logic.

Suppose I have a method to measure lengths of objects in the following manner: I place a mirror at the right end of the object, send a light pulse from the left end (assume it's a one dimensional rod that i am measuring) calculate the time taken for the round trip, length is of course simply given by:

L = cT/2

Now, I see someone do the same measurement in another frame of reference that is moving at speed v wrt to my frame and this person calculates the time as T'.

Since T' = Tγ
shouldn't i conclude that the length as measured by the other person is

L' = cTγ/2 = Lγ ?

In other words, the length should increase and not contract as we now understand it? Where did I go wrong? I'm pretty stupid so any help is great help. Thanks!

2. Mar 30, 2015

### Orodruin

Staff Emeritus
You have the wrong thinking about length contraction and time dilation. They are both based on very precise statements within special relativity.

The length of an object is the difference between the x coordinates at a given fixed time (in the case of one-dimensional space). As such, this is not really what you are measuring when you perform your procedure. Your procedure only works because the object you are measuring does not move and so you can conclude that whatever the x-coordinates were at one time, they will be at other times as well.

3. Mar 30, 2015

### vinven7

Hi Thank you for your reply - I do understand that I am missing something basic. Would you please explain a bit, and perhaps use my specific example to point out where I am wrong? I would be so grateful, thank you

4. Mar 30, 2015

### bcrowell

Staff Emeritus
I have a philosophical objection to Orodruin's #2, which is that coordinates are not fundamental in relativity, and therefore we don't need to define anything fundamentally in terms of them. It is in fact nicer to define things using operational definitions, as vinven7 has attempted to do. But this is just a matter of taste, not a factual disagreement.

Vinven7, how about this as an explanation. Consider *Galilean* relativity for a moment rather than SR. A car is driving by while I stand on the sidewalk. The front bumper passes by me at event P, and the back bumper passes by me at event Q. The spatial distance between P and Q is zero. Can I then infer that the car is infinitely foreshortened in Galilean relativity? I would say not, because P and Q are not simultaneous.

A nice treatment of SR and GR in the operational style you were attempting is given in Geroch, Relativity from A to B.

5. Mar 30, 2015

### PAllen

One glaring error is that T' = T/γ, γ > 1 of course. Time is dilated not expanded. Moving clocks are found to run slow, not fast.

6. Mar 30, 2015

### vinven7

γ = 1/√(1-v∧2/c∧2) in the equation that i wrote. So that T' = T/√(1-v∧2/c∧2) = Tγ... Did i go wrong somewhere?

7. Mar 30, 2015

### vinven7

I'm sorry but I don't really see how this applies to my logic - could you please tell me where I have assumed simultaneity in the argument i put forth? (thanks for the book recommendation btw. I will look into it)

8. Mar 30, 2015

### PAllen

You seem to taking part the leading part of the Lorentz transform. That is for transforming coordinates, not intervals. A muon (treated as clock) elapses e.g. < 1 microsecond from upper atomsphere to ground (else it would decay). Our ground clocks measure 1 microsecond elapsed per 300 meters of traval. Thus, an interval measured on a moving clock is 'our interval' divided by γ. You have defined T and T' as as interval measured in one frame, versus some equivalent interval measured in frame moving relative to the former. Thus, for this purpose, you have T' = T/γ : 1 microsecond = 10 microseconds /10, for example.

9. Mar 30, 2015

### vinven7

If I understand you correctly: Suppose in one frame of reference, and event begins at T1 and ends at T2.

Interval ΔT = T2 - T1.

In a moving frame, T2' = T2γ and T1' = T1γ (as T1 and T2 are simply co-ordinates of time in this frame)

ΔT' = T2' - T1' = T2γ - T1γ = ΔTγ ..? But you are saying ΔT' = ΔT/γ?

I don't follow you

10. Mar 30, 2015

### PAllen

Are you familiar with time dilation? I assumed you were. It seems you are not grasping this aspect of SR. In the time a muon 'clock' elapses 1 microsecond, clocks in the ground frame elapse 10 microseconds. Forget primes, γ, etc., and confirm whether you understand/ agree with this. If not, that is where we need to start the clarification.

11. Mar 30, 2015

### vinven7

I would say yes - but this might also be where I might have some basic misunderstanding. But I do agree that 1 microsecond in the muon clock is 10 microseconds for us and vice versa.

12. Mar 30, 2015

### Staff: Mentor

Meaning, that person is at rest with respect to the clock? Then the time they should calculate is T, not T'. If the two clocks are identical, then people at rest with respect to each clock should measure the same elapsed time for one tick.

This is true, but it only seems backwards because you've described the scenario backwards. The clock is moving in your frame, so the coordinate time in your frame that it takes the clock to make one tick (which is T'), is longer than the time it takes for one tick to someone who is at rest relative to the clock (which is T). That's because the clock is time dilated relative to you, since it's moving relative to you, and you're describing everything in your rest frame. In the rest frame of the clock, the time is just T, as above.

No, you should conclude that the length measured by the person at rest relative to the clock is L. See above.

What length this corresponds to for you is a bit more complicated, because the clock is moving relative to you, so it's not as clear what the quantity L' that you have calculated means.

13. Mar 30, 2015

### Staff: Mentor

No; see my previous post. T' = Tγ is correct; it just doesn't mean what the OP thinks it means.

14. Mar 30, 2015

### PAllen

I'm trying to go with the OP definition, as best as I understand it. Moving clock measures 1 second (T')' ground clocks it passes find 10 seconds (T). Thus T' = T/γ in this sense.

15. Mar 30, 2015

### Staff: Mentor

Yes, but I don't think that understanding is correct. I don't think the OP has the correct understanding either. He has described the scenario backwards, but doesn't realize it, so the latter part of his specification in the OP is incorrect (more precisely, it's not consistent with the first part).

Yes, but that's not what the OP is actually describing. He's describing two different clocks. One is at rest (in his frame), and measures time T. The other is moving (in his frame), but T'= Tγ, according to the OP, is the time measured by a person moving with the clock (at least, that's how it seems to me). But that is wrong: the time measured by the person moving with the clock is T, since by hypothesis this clock is identical to the stationary clock (the OP says "the same measurement" is being made). The time T' = Tγ should be the coordinate time measured for the moving clock by a stationary person, and that time is Tγ (because dividing it by γ, the time dilation factor, must give us the time measured by the person moving with the clock, which is T).

Last edited: Mar 30, 2015
16. Mar 30, 2015

### PAllen

I hope this doesn't confuse the issue, but I am going to try to work out, with more precise definitions, what I think the OP intends, and show how it is consistent with length contraction.

1) We start with the definition that length is defined by half round trip time (times c), measured by a standard clock, and take this to be valid in any inertial frame.

2) Train lab measures some distance as cT' /2 which we call L'.

3) Ground frame clocks measure γT' as the time interval of this train measurement. At first glance, this 'suggests' that ground frame corresponding length would be γL'. But this is not correct.

4) Ground frame is observing a moving measurement, and you have to first understand what ground frame would expect using ground measurements. For some moving ruler considered by the ground frame to be length L, using an array of ground clocks to time the emission, reflection, and receipt of light by one end of the moving ruler, would expect a round trip time of L/(c+v) + L/(c-v) = (2L/c)γ2.

5) Thus what we have (2L/c)γ2 = γT' = γ (2 L'/c), implying L = L'/γ . That is, a ground frame observer would consider that a rod the train measures as, e.g. 10 meters would be one meter. This is length contraction.

Last edited: Mar 30, 2015
17. Mar 31, 2015

### Staff: Mentor

Your notation is much clearer since you are putting the prime on the moving frame coordinates. Part of the confusion in the OP may be due to the fact that the prime does not appear to be used consistently in this way. In your notation, the time T' measured in the moving train lab will be the same as the time $\tau$ (I'll use a different letter to avoid any ambiguity--the OP calls this T) measured in an identical lab that is stationary (because in both cases the measuring apparatus is at rest relative to what is being measured). I'm still not sure the OP grasps that.

18. Mar 31, 2015

### PWiz

Looks like there is a conceptual confusion here. I will try to explain why lengths contract and time dilates without using too much math. To simplify the problem, let's consider the second postulate of special relativity:
The speed of light in a vacuum is observed to be invariant in all inertial frames/non-accelerating coordinate systems.

Well, this has enormous consequences. Since $c=\frac{x}{t}$, it means that the spatial coordinates and the temporal coordinates change upon a coordinate transformation to ensure the invariance of $c$. This is why time is also included when describing an event in a frame.

Let's say that A and B (non primed variables are A's coordinates and primed variables are B's coordinates) are two coordinate systems which had their origins coincide at $t=0$. Let's say that B moves at a constant velocity $v$ in the direction parallel to $x$ (ignore the $y$ and $z$ coordinates here). Now for this moment, let's use gallilean transformations and see where it takes us. A calculates the x coordinates of B by the equation $x'=x-vt$ and $t'=t$. Now if a torch is flashed in B from B's origin in the same direction as the $x'$ axis, it will travel at a speed $c=\frac{x'}{t'}$. If however, A tries to calculate the speed of the light using these coordinate transformations, he will get a value (for a non-zero value of $v$) which is lower than $c$. This cannot happen.
If the speed of light is to be the same for A as well (A sees it move at the same speed but our calculation is not agreeing with the observation at this point), then that means that the coordinates of B perceived by A will be "warped" in such a way to ensure this.

I want to talk about covariant vectors briefly over here. Let's say two people draw up coordinate systems to map the same arrow seen by them with different scales. It is important to realize that the numbers in the coordinate systems themselves will have no meaning unless you assign basis vectors to them. The object described is independent of the coordinate system one chooses. The actual "length" of the arrow won't change. We describe the actual length to be covariant. If the position vector of light in B is described by a function in terms of the parameter $t'$, then the tangent vector of this function will be invariant under all linear coordinate transformations.

OK, so what does this imply? It means that to ensure the constancy of $c$, the $x'$ coordinates seen by A will be slightly more than the $x-vt$ and $t'$ coordinates for A will be slightly less than $t$ (note that we can immediately establish that $t$ does not equal $t'$). So this means that the unit vector for the spatial direction shrinks, whereas for the temporal dimension expands (read about contravariant vectors) for the same point which is compared in A and B. It's like trying to map a point on a graph with $x=5$ onto another graph where $x=50$ for the same position on the graph paper. I can only do this if I divide my original scaling factor by 10. I strongly recommend you read about contravariant vectors and then try applying that concept to this problem (draw the the position time graph for light in both frames and see what must happen to the axes to ensure constant gradient [aka spacetime diagrams]). The scaling factor is a function of $v$, commonly known as $\gamma$, and it has a value of $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ which is calculated using spacetime intervals. Remember that using the logic I have so far described, a person who is at rest always perceives a moving clock to run slower than his clock (the basis vector for $t'$ is smaller than the basis vector $t$ by a factor of $\gamma$), and sees a moving rod to be shorter than its actual length. $t'=\gamma t$ where t is always the time measured by the clock in the frame (the stationary person) in which it is at rest (time dilation). This also yields the lorentz transformation for spatial coordinates, $x'=\gamma (x-vt)$.

I hope my post wasn't too long and boring :) (or full of errors for that matter)

Last edited: Mar 31, 2015
19. Mar 31, 2015

### vinven7

Thank you so much! I think I understand where I went wrong now.

20. Mar 31, 2015

### vinven7

Thank you for taking the time to write this description. Im very grateful for your help.