# Is there any derivation of length contraction through proper lenght?

1. Jul 17, 2009

### ELESSAR TELKONT

I'm doing some sort of notes on special relativity and i got this question, because proper time and time dilation have a relation. In fact we have that proper time is mathematically the arc length of a timelike curve. I will use geometrical units and the Minkowski Metric with the -+++ signature.

For example, we have two events $$A,B$$ separated timelike by a straight worldline which can be parametrized by (in the particular case of $$A$$ being at our origin of reference and at time $$x^{0}=0$$ and $$B$$ having only $$x^{1}$$ nonzero, the general case is obtained via Poincaré transformation).

$$\sigma^{\alpha}\left(x^{0}\right)=\left(x^{0},\frac{x_{B}^{1}}{x_{B}^{0}}x^{0},0,0\right)$$

Then

$$\frac{d\,\sigma^{\alpha}}{d\,x^{0}}=\left(1,\frac{x_{B}^{1}}{x_{B}^{0}},0,0\right)$$

and therefore proper time function is

$$\tau=\int\sqrt{\left\vert\eta_{\alpha\gamma}\frac{d\,\sigma^{\alpha}}{d\,x^{0}}\frac{d\,\sigma^{\gamma}}{d\,x^{0}}\right\vert}dx^{0}=\sqrt{1-\left(\frac{x_{B}^{1}}{x_{B}^{0}}\right)^{2}}x^{0}$$

but we have that

$$\frac{x_{B}^{1}}{x_{B}^{0}}=v$$

that is the speed with sign, therefore

$$\tau=\sqrt{1-v^{2}}x^{0}=\frac{1}{\gamma}x^{0}$$

or in differential form

$$dx^{0}=\gamma d\tau$$

For a general timelike curve parametrized by

$$\sigma^{\alpha}\left(x^{0}\right)=\left(x^{0},\sigma^{a}\left(x^{0}\right)\right)$$

we can write

$$\tau=\int\sqrt{\left\vert\eta_{\alpha\gamma}\frac{d\,\sigma^{\alpha}}{d\,x^{0}}\frac{d\,\sigma^{\gamma}}{d\,x^{0}}\right\vert}dx^{0}=\int\sqrt{\left\vert\delta_{ag}\frac{d\,\sigma^{a}}{d\,x^{0}}\frac{d\,\sigma^{g}}{d\,x^{0}}-1\right\vert}dx^{0}=\int\sqrt{1-\delta_{ag}\frac{d\,\sigma^{a}}{d\,x^{0}}\frac{d\,\sigma^{g}}{d\,x^{0}}}dx^{0}$$

but $$\delta_{ag}\frac{d\,\sigma^{a}}{d\,x^{0}}\frac{d\,\sigma^{g}}{d\,x^{0}}$$ is the squared norm of the instantaneous 3-velocity on the curve. Therefore we have in differential form

$$dx^{0}=\gamma(x^{0}) d\tau$$

and through this we have derived time dilation in general. In fact, for inertial observers (those that their frames of reference in spacetime are related through Poincaré transformations) the time that every observer actually measures is their proper time and the worldlines of the observers viewed by others are straight lines, then we can use

$$dx^{0}=\gamma d\tau$$

for derive

$$dx^{0}=\gamma dx^{\hat{0}}$$

or

$$dx^{\hat{0}}=\gamma dx^{0}$$

for two arbitrary inertial observers.

Therefore I formulate my original question: do you know similar derivation for the length contraction through proper length function (equivalent of the proper time for spacelike curves)?

2. Jul 17, 2009

### Mentz114

That looks most ingenious. But perhaps you can clear up a point for me.

You've parameterized the geodesic with coordinate time $x^0\equiv t$ which is fine because $t$ is affine if the velocity is constant. But there's a scale factor between proper length parameterized by $t$ and one parameterized by $\tau$ which seems to have been ignored.

I could have missed it, so I'm just asking for guidance.

Last edited: Jul 17, 2009
3. Jul 17, 2009

### clem

Proper time and proper mass are useful because they have physical meaning and their square is a Lorentz scalar. Each is the invariant length of a 4-vector.
In the rest system, each is the physical quantity.
This is not true for length. Length is one component, x_1-x_2 of a 4-vector.
The invariant scalar ds^2=dt^2-dx^2-dy^2-dz^2 is not easily related to measured length.
There is a discussion of this in <arXiv:0906.1919>.

4. Jul 17, 2009

### ELESSAR TELKONT

Ok Mentz. The derivation I have done is the one for time dilation and proper time. My question is if anyone has done already something similar to derive length contraction from proper length (i.e. proper time for spacelike curves) and not to derive it from the classic ends of a rod.

For clem. Proper Lenght is an invariant under Poincaré transformations since is derived from the metric, in fact is the same as proper time only that is for spacelike curves (both are scalars). Remember that two events that are spacelike separated in certain frame reference are simultaneous and in that reference frame (and by invariance in all reference frames) the spatial length of the straight segment between events must coincide with the proper length of the segment in every reference frame.

5. Jul 17, 2009

### Mentz114

My question is answered. What you've done is calculate the scale factor between parameterizations

$$\tau=\int\sqrt{\left\vert\eta_{\alpha\gamma}\frac{ d\,\sigma^{\alpha}}{d\,x^{0}}\frac{d\,\sigma^{\gam ma}}{d\,x^{0}}\right\vert}dx^{0}=\int\sqrt{\left\v ert\delta_{ag}\frac{d\,\sigma^{a}}{d\,x^{0}}\frac{ d\,\sigma^{g}}{d\,x^{0}}-1\right\vert}dx^{0}=\int\sqrt{1-\delta_{ag}\frac{d\,\sigma^{a}}{d\,x^{0}}\frac{d\, \sigma^{g}}{d\,x^{0}}}dx^{0}$$

from which one can generalize to all frames.

I don't think this can be done for lengths because $t\equiv x^0$ has a special role. But I could be wrong.

6. Jul 18, 2009

### clem

Your first sentence agrees with my interpretation.
The problem is that "proper length" defined that way means you have to know at what times to measure the length in any frame to get the "proper length".
What is your definition of "spatial length"?

7. Jul 18, 2009

### ELESSAR TELKONT

well clem and all others, let me put my point explicitly (I'm working on it right now):

You have two events that you can connect by a spacelike straight line (or by a spacelike curve). Then in spacetime you can obtain the arc length of the curve between this events by means of the line integral

$$\lambda=\int\sqrt{\eta_{\alpha\gamma}dx^{\alpha}dx^{\gamma}}$$

or in differential form

$$d\lambda^{2}=\eta_{\alpha\gamma}dx^{\alpha}dx^{\gamma}$$

where I will take for the metric the -+++ signature. Since arc length (which is not other thing that the application of the metric), not only in relativity but in every riemmanian or semi-riemmanian manifold, is an invariant when you apply the isometries of that manifold, in this case the Poincaré Transformations.

In fact, this $$d\lambda$$ is the proper length between two infinitesimally spacelike separated events which coincides with the euclidean length that someone actually measures when is in the frame where both are simultaneous.

For the sake of simplicity suppose that the arc between these two events is a straight line and that the event $$A$$ is at origin in frame of reference and time and let $$B$$ be at $$\left(x^{0}_{B},x^{1}_{B},0,0\right)$$, then we can parametrize this arc by

$$\sigma^{\alpha}(x^{1})=\left(\frac{x^{0}_{B}}{x^{1}_{B}}x^{1},x^{1},0,0\right)$$

Differentiating with respect the parameter

$$\frac{d\,\sigma^{\alpha}}{d\,x^{1}}=\left(\frac{x^{0}_{B}}{x^{1}_{B}},1,0,0\right)$$

and applying the metric

$$\eta_{\alpha\gamma}\frac{d\,\sigma^{\alpha}}{d\,x^{1}}\frac{d\,\sigma^{\gamma}}{d\,x^{1}}=-\left(\frac{x^{0}_{B}}{x^{1}_{B}}\right)^{2}+1$$

then we can obtain $$\lambda$$ by means of the line integral which is trivial

$$\lambda=\sqrt{-\left(\frac{x^{0}_{B}}{x^{1}_{B}}\right)^{2}+1}x^{1}$$

And I got stuck here since I don't see a $$\gamma$$ out there since $$\frac{x^{0}_{B}}{x^{1}_{B}}$$ is, strictly speaking, not really speed although it's less than 1 and is in fact the inverse of the superluminal velocity to make travel beetween those events AB. However if I take that as velocity I have in differential form

$$d\lambda=\frac{1}{\gamma}dx^{1}$$

expression that's telling us exactly the opposite than lenght contraction. Then I assume there is something missing that I haven't seen already that can make $$d\lambda=\gamma dx^{1}$$ expression that is correct for lenght contraction.

Another way I have tried is not parametrizing with x^{1}, instead with time x^{0}. We have that

$$d\lambda^{2}=\eta_{\alpha\gamma}\frac{d\,\sigma^{\alpha}}{d\,x^{0}}\frac{d\,\sigma^{\gamma}}{d\,x^{0}}\left(dx^{0}\right)^{2}$$

but from the invariance of $$d\lambda$$ we have that

$$d\lambda^{2}=ds^{2}=-\left(dx^{0}\right)^{2}+d\ell^{2}$$

where $$d\ell^{2}=\left(dx^{1}\right)^{2}+\left(dx^{2}\right)^{2}+\left(dx^{3}\right)^{2}$$

therefore

$$\left(dx^{0}\right)^{2}=d\ell^{2}-d\lambda^{2}$$

so

$$d\lambda^{2}=\left(-1+\delta_{ag}\frac{d\,\sigma^{a}}{d\,x^{0}}\frac{d\,\sigma^{g}}{d\,x^{0}}\right)\left(d\ell^{2}-d\lambda^{2}\right)$$

or

$$\delta_{ag}\frac{d\,\sigma^{a}}{d\,x^{0}}\frac{d\,\sigma^{g}}{d\,x^{0}}d\lambda^{2}=\left(-1+\delta_{ag}\frac{d\,\sigma^{a}}{d\,x^{0}}\frac{d\,\sigma^{g}}{d\,x^{0}}\right)d\ell^{2}$$

$$d\lambda^{2}=\left(1-\frac{1}{\delta_{ag}\frac{d\,\sigma^{a}}{d\,x^{0}}\frac{d\,\sigma^{g}}{d\,x^{0}}}\right)d\ell^{2}$$

Obviously in this case $$\delta_{ag}\frac{d\,\sigma^{a}}{d\,x^{0}}\frac{d\,\sigma^{g}}{d\,x^{0}}>1$$ since the curve is spacelike and then its multiplicative inverse is less than one. However, again, I dont see how to make it through to obtain

$$d\lambda^{2}=\gamma^{2}d\ell^{2}$$

and therefore the desired expression

$$d\lambda=\gamma d\ell$$

which is valid in all frames of reference and could take a familiar form of the lenght contraction

$$d\ell=\frac{1}{\gamma} d\lambda$$

or if you want, as in introductory modern physics texts like Beiser,

$$dL=\frac{1}{\gamma} dL_{0}$$

I wonder it's sufficient explicit for you clem to understand finally what I'm talking about.

8. Jul 20, 2009

### Ich

The point is: "length" is not a vector. It is the distance between two timelike (hopefully straight and parallel) lines. In each coordinate system, the events that delimit "length" are different.

9. Jul 20, 2009

### Saw

In the thread "The show of the duel" (last part) I had a discussion with JesseM more or less on this issue. I argued that "proper time" and what sometimes is called "proper length" or "rest length" are not at the same level. "Proper time" between two timelike events is "the invariant interval", i.e., all observers agree on its value, because it is a simultaneity-free concept, while length, either coordinate or rest length, is not: no matter whether you measure the length of a rod in its rest frame or in a frame where it is moving, you assume that the two ends of the rod are measured simultaneously and that means a different thing for each frame. That's obvious for coordinate length but it's also true for rest length: even if you put one stick after the other to measure length, the value of those (say 1-metre) sticks has been measured with the Einstein convention and so the simultaneity version of the measurer taints the measurement. Contrary to that, "proper time" is a bunch of events (the clock ticks), each of which takes place at a given time and a given place, so all observers agree that they happen and how many of them happen between two markers: the passing of the clock by two references of another frame.

Well, I don´t know if I explained it well (I never do, the idea is not so clear to me), but maybe it has to do with what has been commented here: you cannot derive proper length as you do with proper time, because the concepts are not analogous, they do not have the same rank. Proper time is something that can be derived as universal because it is in the essence of the theory that its scalar aspect (number of ticks) is so, since (i) number of ticks equates to number of events and (ii) events are universal. Instead, the magnitude of length would not benefit from such advantage...

10. Jul 20, 2009

### Ich

Well, the "same place" version of the measurer taints proper time measurements also. But I concede that it's relatively easy to build devices that stay at the same place in one frame.

11. Jul 20, 2009

### Saw

The ticks of a clock moving in my frame happen in different places of my frame. But when I describe how many times the ticker of that clock ticks, I don't care where it ticks in my frame. My description is identical as the one given in the clock's frame: 100 ticks or whatever. But if you want to establish a comparison between the rate of those ticks and the rate of ticks of another clock at rest in my frame, then it's not more difficult, it's utterly impossible that I agree with the other frame. That's what relativity of simultaneity means, after all. Since the measurement of lengths is conditioned by that relativity of simultaneity, it seems also impossible that, in terms of length, two observers agree on a description in the same manner as they agree when they talk about proper time. I thought the mathematical difficulties expressed by Elessar might have something to do with that. But I do not have the math skills to follow that reasoning and make the link. Anyhow, I do not want to disturb the discussion. Please continue where it was.:shy:

12. Jul 20, 2009

### Meir Achuz

arXiv:0906.1919