# Justification of length contraction in light-clock presentation

1. Aug 5, 2012

### bcrowell

Staff Emeritus
Suppose you've just established the existence of time dilation using Einstein's 1905 postulates and the light clock. Is there any nice, easy way to go on and prove length contraction?

Here are a couple of incorrect arguments that I've been guilty of using in the past:

Symmetry: The Lorentz transformation treats time and space the same, so we must have length contraction by the same gamma factor. Wrong because the analogy ruler:distance::clock:time is wrong -- a clock's world-line is a line, a ruler's a strip. Also wrong because taking v<1 (c=1) introduces an asymmetry (interchange of x and t makes v>1). We can also tell it's wrong because the resulting effects should be in opposite directions (moving clock reads too low, contracted ruler measures things to be to big) and don't behave the same in terms of permanence.

Constancy of c: Under a Lorentz transformation, we must have $c=x/t\rightarrow (\gamma x)/(\gamma t)$ so that c stays the same. Wrong because $\gamma$'s interpretation as length contraction and time dilation refers to a rest frame, which doesn't exist for light; the actual factor of interest for something moving at c is the Doppler shift factor.

The Feynman Lectures have an argument in 15-5 based on the null result of the Michelson-Morley experiment. I'm sure this is fine logically, but the presentation is IMO not very good and overly abbreviated, and I think a full presentation of this argument would be rather long and complex. I think people like the light clock because it's a quick wham-bam demonstration of time dilation that works well for students with minimal math, and it doesn't require the full development of the Lorentz transformation. It seems unfortunate to have to follow that with a more complicated argument in which the full machinery of the LT is required.

Is there some other approach that is both correct and simple?

2. Aug 5, 2012

### Staff: Mentor

Just turn the light clock 90 degrees, so it is parallel to the direction of motion. If it doesn't length contract then you get the wrong period.

3. Aug 5, 2012

### Vorde

Depends how rigorous you want to be probably.

In my introductory relativity class we derived the length contraction law from looking at the Lorentz boosts of two nearby points (or at least I think that's how we did it, I can't remember the exact method), and that worked simply (it did need knowledge of the LTs though).

A non-rigorous way would be to say that if an object is time dilated it goes a certain distance $d$ in some time $t$ that you measure, but because to them time is moving slower they think they moved a smaller distance.
(I think this doesn't work upon close review but it gets the point across)

I think you could work out an argument based off of the consistency of the spacetime interval, but I'm not sure what that would be.

4. Aug 5, 2012

### bcrowell

Staff Emeritus
I think this is identical to Feynman's argument about the null result of the Michelson-Morley experiment, which, although correct, seems less than ideal for the reasons I gave in #1.

5. Aug 5, 2012

### Muphrid

Let me see if I understand this right:

To establish time dilation, say there are mirrors/detectors at $x = 0, d$, and a light beam bounces between them. We establish time dilation from the idea that, if a light beam starts at $0e_t + 0e_x$ and bounces between the two mirrors, it will hit the first mirror again at $2de_t$ (in $c=1$), but isotropy of spacetime implies that any other timelike vector can be used to measure the coordinate time between $0$ and $2d e_t$. All other coordinate times are at least as big or greater than $2d$, and this is time dilation.

The coordinate distance between the two mirrors is measured along a line of constant coordinate time cutting through the worldlines of the two mirrors. This is what breaks the symmetry between the two cases: in the first, we're considering the same interval in both frames and just resolving the components differently. In the second, we're finding the distance between two parallel worldlines as measured along differing third lines, and while in the Euclidean case the perpendicular is the shortest connecting line and all others are longer, in the Minkowski case the perpendicular is the longest and all others are shorter.

Maybe that's too mathy an explanation (I know we're all familiar with the math here), but I'm hoping it inspires some simple physical interpretation.

6. Aug 5, 2012

### zonde

Isotropy of c.
After boost speed of light is faster in direction perpendicular to direction of motion under Galilean transformation.

7. Aug 6, 2012

### Staff: Mentor

Yes, I saw those objections, but didn't understand them. It seems no more difficult to do a light clock in one orientation than in another, so if one is easy the other is easy too. The math is like those problems from when velocity was introduced where you would calculate how much time it takes for a bicyclist (or some other fast transport) to catch up with a jogger (or some other slow transport).

It also gives a chance to introduce spacetime diagrams, which you need to do anyway.

8. Aug 6, 2012

### bcrowell

Staff Emeritus
Well, the audience I have in mind is liberal arts students at a community college, who probably have nearly zero algebra skills -- the same audience as the intended readership of a book like Hewitt. The time dilation is extremely simple to derive in the light clock approach:

$$(vT)^2+L^2=c^2T^2$$
$$L^2=(c^2-v^2)T^2$$
$$T=\frac{L}{\sqrt{c^2-v^2}}$$

This differs from L/c by a factor of gamma. You're done.

In the longitudinal direction, let the unknown contraction factor be x:

$$2T=\frac{xL}{c-v}+\frac{xL}{c+v}=\frac{2\gamma L}{c}$$
$$x\left(\frac{1}{c-v}+\frac{1}{c+v}\right)=\frac{2\gamma}{c}$$
$$x\left(\frac{c+v}{(c-v)(c+v)}+\frac{c-v}{(c+v)(c-v)}\right)=\frac{2\gamma}{c}$$
$$x\left(\frac{c+v+c-v}{c^2-v^2}\right)=\frac{2\gamma}{c}$$
$$\frac{xc}{c^2-v^2}=\frac{\gamma}{c}$$
$$x=\gamma(1-v^2/c^2)$$
$$x=\sqrt{1-v^2/c^2}=1/\gamma$$

The first one is three lines, and the hardest algebra is arranging like terms on the same side of the equation.

The second one is seven lines, and it involves a lot of algebra that will seem like total voodoo to this audience. Bringing the two fractions over a common denominator is something they won't know/remember, and probably couldn't even do in a concrete case like 1/3+1/2. They won't know/remember the identity $(a+b)(a-b)=a^2-b^2$. Not only is it more lines, but the expressions are considerably more complicated.

I wish I had my copy of Hewitt handy -- it would be interesting to see how he does it.

9. Aug 6, 2012

### ghwellsjr

"...liberal arts students at a community college, who probably have nearly zero algebra skills" and you want to show them equations with square roots in them? I wouldn't show them even one equation, I would just show graphics.

A couple years ago, I attempted to explain these things graphically in this thread starting at post #78:

Unfortunately, since I'm mobile right now, I can't even watch my own videos, but as I remember, I never made it to the end in that thread. There was also the confusion that I said the mirrors in the animations were brown but they are really yellow (I'm partially color blind. In any case, as I remember, there were three more videos to get to the final one which I think I show in this post:

https://www.physicsforums.com/showpost.php?p=3803335&postcount=19

Anyway, if this sort of approach appeals to you, when I get to a real computer in a couple days I can clean things up.

10. Aug 6, 2012

### bcrowell

Staff Emeritus
I guess I'm particularly interested in a level somewhere above the one you're thinking of ("nearly zero," not zero) and below the one dalespam has in mind. Basically the same level that Hewitt is aimed at. The light clock does work well at that level for demonstrating time dilation and deriving the equation for gamma.

11. Aug 6, 2012

### grav-universe

Well, if you're wanting to keep it simple, then once you have the time dilation, you could use muon decay to demonstrate length contraction. Or to make it even simpler and perhaps more precise, in case muon decay has a half life or something and to help visualize better as well and to make things more interesting, you could set the timer on a bomb to explode in a proper time t.

The timer on a bomb at rest is set to explode in a time t, so at a speed v relative to an observer O, would normally have travelled a distance v t before exploding. But since it is moving, then due to time dilation, the timer on the bomb is observed to tick at a rate 1/γ slower, so instead is observed to travel a distance v t γ before it explodes. If it starts at one end of a ruler and just reaches the other end when it explodes, then v t γ is the proper length of the ruler in frame O.

From the perspective of an observer O' travelling with the bomb, the bomb is at rest, so explodes in a time t. Observer O is travelling past at a speed v, so will have travelled a distance v t when the bomb explodes according to observer O', so that is the distance travelled along the ruler of observer O according to observer O' as well, the length of the ruler as observer O' measures it. But observer O measured his ruler to be v t γ long while observer O' measures it to be 1/γ shorter, so that is the length contraction observed of a ruler travelling at v.

Last edited: Aug 6, 2012
12. Aug 6, 2012

### bcrowell

Staff Emeritus
Grav-universe, that's a really sweet argument. It's perfect for the level of presentation we're talking about, very simple.

In addition to the light clock, the other approach I've used in teaching this subject is a geometrical one, in which you prove that the Lorentz transformation must preserve area in the x-t plane, and then everything else just falls out. It occurred to me today that in this approach, it's possible to get the result for length contraction very straightforwardly, so I put that argument in this book:

http://www.lightandmatter.com/html_books/lm/ch23/ch23.html#Section23.1 [Broken]

See figures o and p. (In the html version of the book these two figures are awkwardly separated. The pdf version http://www.lightandmatter.com/lm/ may be easier to follow if you don't mind the large size of the file.)

Last edited by a moderator: May 6, 2017
13. Aug 6, 2012

### Staff: Mentor

You can do the whole Lorentz transform at once graphically without any algebra, but you have to have introduced them to spacetime diagrams and reference frames and the postulates.

Once you have done that you talk about how to relate two reference frames moving uniformly wrt each other. Because the moving one is moving uniformly you know that the time axis is a slanted line. Then you show them how the speed of light is wrong in the moving frame, so you have to tilt the space axis to satisfy the second postulate. That gets you to the Lorentz transform to a scale factor, which is determined by the first postulate, the stationary frame must look the same to the moving frame as the moving frame looks to the stationary.

Once you have that drawn you can show length contraction and time dilation geometrically. No equations needed, but it is quite abstract.

14. Aug 6, 2012

### Mark M

If you've already introduced time dilation, why not just use that to derive length contraction? For example, an observer measures the length of a steel rod to be $L$. In order to measure $L$, he shines a light from one end of the rod to the other, and measures how long it takes. So, the length of the rod is given by $L = ct$. An observer moving relative to him measures it to be $L'$. Using the same light signal to determine the length of the rod, he finds the length to be $L' = ct'$. Because of time dilation, $$t' = \frac {t}{\sqrt {1 - v^{2}/c^{2}}}$$ So, substituting for $t'$, you get $$L' = \frac {ct}{\sqrt {1 - v^{2}/c^{2}}}$$ Since $ct$ is the length measured by the original observer, we can re-write the above as $$L' = \frac {L}{\sqrt {1 - v^{2}/c^{2}}}$$ or, $L' = \frac {L}{\gamma}$

15. Aug 6, 2012

### bcrowell

Staff Emeritus
Mark M, your derivation doesn't seem quite right to me. You say "Using the same light signal to determine the length of the rod..." This seems to imply that we have an event A where the signal is emitted from one end of the rod and an event B where the signal is received at the other end, and both observers are basing everything on these same two events. There is a time interval t between them in the original (unprimed) frame. In the new frame, the interval is not $t'=\gamma t$. Under a Lorentz transformation, the time interval between two events that are lightlike in relation to one another changes by the Doppler shift factor, which isn't equal to $\gamma$. Also, at the end, you have this:

But this isn't right, because in the first expression you have division by the square root factor, which is a factor of $\gamma$, not $1/\gamma$.

16. Aug 6, 2012

### bcrowell

Staff Emeritus
This sounds interesting. Do you know of a presentation along these lines that I could look at?

17. Aug 6, 2012

### Muphrid

Yeah, Mark, that's not $L' = L/\gamma$, that's $L' = L\gamma$. You get this result because the light beam does not directly measure that distance. You're finding the components of a vector along the first mirror's worldline and resolving those components differently.

For any vector that's purely spatial, a boost will increase the spatial component, just as a for a vector that's purely in time direction, a boost will increase the time component. It's a basic property of hyperbolic geometry. The key difference between time dilation and length contraction is that in length contraction, we're not just resolving the components of a vector in a different basis. We're trying to find the distance between two worldlines along a certain spatial dimension. The cases are totally different.

Let $A(s) = s e_t$ and $B(w) = w e_t + d e_x$. Boosting these worldlines gives

$$A'(s) = s \gamma (e_t + \beta e_x) \quad B'(w) = w \gamma( e_t + \beta e_x) + d \gamma (e_x + \beta e_t)$$

We want to choose the $e_t$ components to be both the same (I'll go with zero). Clearly $s = 0$, and $w \gamma + d \gamma \beta \implies w = - d \beta$. The resulting x distance is $-d \gamma \beta^2 + d \gamma = d/\gamma$ exactly like you expect.

Now, just for fun, let's put those parameters back into our unboosted worldlines.

$$A(0) = 0 \quad B(-d \beta) = - d \beta e_t + d e_x$$

Clearly, the Minkowski length of $B(-d\beta)$ is $d/\gamma$.

So, what are we doing when we measure length contraction? We're looking at the different spacetime distances between the two worldlines according to different spacelike vectors.

So, what are we doing when we measure time dilation? We're looking at the components of a fixed vector according to different timelike vectors.

18. Aug 6, 2012

### Mark M

Oh, I totally forgot that the proper length should be multiplied by the square root factor, not divided by (which ends up giving you longer rods in moving frames.). And sorry about messing up the Lorentz factor.

Well, sorry for that, I didn't notice all of the errors. Thanks for the correction.

Muphrid, thanks for the correction, too.

Here is a derivation of the effects of special relativity, using Minkowski diagrams, as DaleSpam is describing:

http://www.physics.byu.edu/faculty/allred/222 11/minkowski 11.pdf

19. Aug 6, 2012

### bcrowell

Staff Emeritus
20. Aug 7, 2012

### zonde

But you use Pythagoras' theorem there.

You can single out that most difficult part in a single row like that:
$$\frac{1}{c-v}+\frac{1}{c+v}\equiv \frac{1}{c-v}\frac{c+v}{c+v}+\frac{1}{c+v}\frac{c-v}{c-v}\equiv \frac{c+v}{(c-v)(c+v)}+\frac{c-v}{(c+v)(c-v)}\equiv \frac{c+v+c-v}{(c-v)(c+v)}\equiv \frac{2c}{c^2-vc+cv-v^2}\equiv \frac{2c}{c^2-v^2}$$
you can test this part afterwards using sample values to convince that everything is correct - plug in values in first expression, calculate; plug in the same values in last expression, calculate; compare results. Imagine that you would have to convince them that Pythagoras' theorem is correct.

You can calculate length without unknown and afterwards just guess length contraction factor from result.

You can use more colorful illustration than light clock.
Lets say we have formation of military spaceships somewhere in deep space. Ships maintain their formation by sending light pulses to nearest ships and measuring round trip time of pulses i.e. it should remain constant. Now for a few seconds all the spaceships in formation accelerate and after that within some time they settle down in the same formation as before acceleration.
What is the length of formation from perspective of commanding ship that does not accelerate?
Basically it's the same light clock but it is easier to imagine how formation of ships length contracts.

What do you think?

21. Aug 9, 2012

### ghwellsjr

I decided to clean up my previous explanation and finish it since it never got finished on line before:

A graphical explanation of Special Relativity

22. Aug 9, 2012

### Austin0

hi

Very nice .
I do have a question. It captures the ellipsoidal geometry of the points of reflection as charted in the frame you are doing the animation from ,but does not incorporate the ellipsoidal profile of the contracted sphere . Kinematically it seems impossible for all the reflections to arrive back at the source point in the moving frame simultaneously, given a spherical geometry in the observation frame.
Do you disagree???

23. Aug 9, 2012

### ghwellsjr

If you imagine a horizontal line passing through the heads of the observers and rotate the entire graphic around that line, it will sweep out a perfect sphere for Homer's mirrors and an oblate sphere (or ellipsoidal sphere or whatever you call it) for Rover's mirrors. There is a single expanding sphere of light prior to it hitting either set of mirrors and there are two collapsing spheres of light reflecting separately off of each set of mirrors. Expanding or contracting spheres of light are always perfect spheres in any frame, they never have an ellipsoidal profile. It's only a moving set of mirrors that has an ellipsoidal profile.

(Maybe you should move your question to my thread and I'll move this response. No point in cluttering this thread.)

Last edited: Aug 9, 2012
24. Aug 10, 2012

### Austin0

No I think i will just retract the whole post. My eyes must be going. Having looked again I see that it is an ellipsoidal shape, just with only slight eccentricity, which looked spherical when I checked it out before. Having looked closer I see now that the velocity is only about .5 c with a contraction of approx, 0.86 so just scratch everything I said.
Thanks for doing the videos