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In the context of special relativity:

I know how to transform proper length into moving length (so transform length from a frame in which the measured object is at rest into one in which it moves at some constant velocity), or the other way around. I also know how to derive it (the length contraction formula, that is) and feel mostly comfortable with the derivation & the logic behind it: we define "length" to be the distance between the two ends of the object at the same time, and agree that if the object is at rest then this time requirement is unnecessary, and use that to transform from the rest system (proper length) to another, moving at a constant relative velocity (moving length).

However, I struggle, for some reason, when I try to transform lengths between two systems in which the object isn't at rest – but directly. That is to say: the method I employ that seems to work just fine is to use the fact that I know the object's velocity and just directly calculate its proper length in its rest system; and then find the relative velocity of the object and any other system, and use length contraction ($L=\frac{L_0}{\gamma}$) to calculate its moving length in that system.

But what I was wondering is this:
How could I go about doing this without going through calculating its proper length as well? Say I wish to transform from system A to system B, neither of which is the object's rest frame, but without first calculating its proper length: I just want to use Lorentz transformations of time and space directly to obtain this result.
To emphasize: I could always derive a formula by using the various Lorentz factors (i.e calculate the proper length without admitting that this is what I'm doing. So first multiply by the appropriate Lorentz factor to transform to proper length, and then divide by the appropriate Lorentz factor to calculate the needed length through length contraction) but I was wondering if there's another way to do this that doesn't involve the proper length stop even implicitly: just direct time/space Lorentz transformations instead.

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3 Answers 3

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This spacetime diagram and some geometric thinking might help you see what is happening.

robphy-LengthContraction-A

I've drawn the worldlines of the endpoints of Carol's meterstick.

The proper length is the "distance between parallel timelike lines (in Minkowski Spacetime Geometry)”, which is along a spacelike segment Minkowski-perpendicular to the worldlines (analogous to what happens in Euclidean (and Galilean Spacetime) geometry.)

This segment is the "base" of a Minkowski-right-triangle
whose "hypotenuse" (the triangle side opposite the right angle) is along the spaceline of an observer (Alice) measuring the length of Carol's meterstick as the segment on Alice's spaceline cut by the parallel worldlines of Carol's meterstick.

The "base" is the side adjacent to the "Minkowski-angle" between spacelines that is equal to the relative-rapidity $\theta_{C,wrtA}$ from Alice to Carol, where $$V_{C,wrtA}=\tanh\theta_{C,wrtA} \mbox{ and } \gamma_{C,wrtA}=\cosh\theta_{C,wrtA}.$$

So, the length contraction formula is the expression of
the hypotenuse (the apparent length measured by Alice)
as the ratio of
the (base) "adjacent side" (the proper length of Carol's meterstick)
divided by [hyperbolic] cosine:

$$L_{C,wrtA}=\frac{L_{C,wrtC}}{\cosh\theta_{C,wrtA}}=\frac{L_{proper}}{\gamma}.$$

Thus, the proper length of Carol's meterstick can be written as $$L_{C,wrtC}=L_{C,wrt A}\cosh\theta_{C,wrtA},$$ in terms of Alice's apparent-length measurement and the relative-rapidity.


Now consider another observer Bob, who will obtain a similar expression for the same base of another right-triangle, this time with hypotenuse along Bob's spaceline: $$L_{C,wrtC}=L_{C,wrt B}\cosh\theta_{C,wrtB},$$ in terms of Bob's apparent-length measurement and the relative-rapidity from Bob to Carol.

robphy-LengthContraction-2

So, one can now eliminate the proper-length of Carol's meterstick (along the common base of their right-triangles) and write $$L_{C,wrt A}\cosh\theta_{C,wrtA}=L_{C,wrt B}\cosh\theta_{C,wrtB}$$ which relates their apparent-lengths of Carol's meterstick in terms of their relative rapidities to Carol.

We can write this in terms of the relative rapidity from Alice to Bob: $$\begin{align*} \theta_{B,wrtA} &=\theta_{B,wrtC} - \theta_{A,wrtC}\\ &= -(\theta_{C,wrtB}-\theta_{C,wrtA}) \end{align*}, $$ which is, of course, just a rewriting of the addition of rapidities: $$\theta_{C,wrtA}=\theta_{C,wrtB}+\theta_{B,wrtA} .$$

So, we have $$ \begin{align} L_{C,wrt B}\cosh\theta_{C,wrtB} &= L_{C,wrt A}\cosh\theta_{C,wrtA}\\ &=L_{C,wrt A}\cosh( \theta_{C,wrtB}+ \theta_{B,wrtA} )\\ &=L_{C,wrt A}\left( \cosh\theta_{C,wrtB} \cosh\theta_{B,wrtA} +\sinh\theta_{C,wrtB} \sinh\theta_{B,wrtA} \right)\\ &=L_{C,wrt A} \cosh\theta_{C,wrtB} \cosh\theta_{B,wrtA} \left( 1+\tanh\theta_{C,wrtB} \tanh\theta_{B,wrtA} \right)\\ L_{C,wrt B} &=L_{C,wrt A}\cosh\theta_{B,wrtA} \left( 1+\tanh\theta_{C,wrtB} \tanh\theta_{B,wrtA} \right)\\ &=L_{C,wrt A}\ \gamma_{B,wrtA} \left( 1+V_{C,wrtB} V_{B,wrtA} \right), \end{align} $$ which seems to agree with the result by @benrg .


postscript:
So, geometrically, this relates these two segments bounded by a pair of parallel lines in terms of the Minkowski-angle between the segments and a Minkowski angle from one segment to the Minkowski-perpendicular segment.

One could probably try to write things in terms of a Lorentz boost transformation. But note the two segments are not related by a boost (the endpoints of the corresponding segments have unequal square-interval).

In the Euclidean version of this problem, I don't think anyone would solve it with rotation matrices.

(There is likely a solution using boosts and projections (dot-products), or just vector-algebra and dot-products.)

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    $\begingroup$ Thank you very much for this incredibly detailed answer! It was super helpful. $\endgroup$
    – Shay
    Sep 28, 2021 at 15:18
  • $\begingroup$ I still don't understand this PSE site. There are so many questions and answers that are not even voted on or considered properly. $\endgroup$
    – Sebastiano
    Oct 15, 2021 at 20:26
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Write the worldlines of the endpoints of the rod as functions of an arbitrary parameter that isn't a coordinate, e.g.

$$\begin{eqnarray} x_k(q) &=& uq+kL \\ t_k(q) &=& q \end{eqnarray} \qquad \text{where } k\in\{0,1\}$$

Then use a Lorentz transformation to find $x'_k(q)$ and $t'_k(q)$, set $t'_k(q)$ to a constant (probably $0$) to find $q$ (with a dependence on $k$), plug it into $x'_k(q)$ to find the endpoints, and subtract. I get $L' = L/γ(1{+}uv)$.

You may be getting into trouble because you're writing the worldlines as functions $x(t)$. This will work if you carefully keep the parameter $t$ distinct from the coordinate $t$, but it's easy to get confused.

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  • $\begingroup$ Thank you very much for your help :) To be honest, this is the first time I've heard the term "worldlines": so far it was just an event or two and some simple transformations concerning them. Got me into a lot of very interesting reading and I'm looking forward to studying more :) $\endgroup$
    – Shay
    Sep 28, 2021 at 15:18
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You can't just use the Lorenz transforms on two endpoint measurements of the object in Frame A, because you would then end up with two non-simultaneous measurement events in Frame B, which would give the wrong answer.

What you need to do is to find two simultaneous events in Frame B that correspond to the endpoints of the object.

To simplify the maths, define the origins of Frames A and B to coincide with one end of the object at some arbitrary time, and assume that Frame A is at rest, and that both the object and frame B are moving in the x direction at speeds Vo and Vb.

The far end of the object in Frame A will be at x=l, where l is the contracted length of the object in Frame A. The wordline of the far end of the object will be a straight line through x=l, with a slope that is given by Vo. Where the line meets the x axis of Frame B is the length of the object in that frame.

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  • $\begingroup$ Thanks a lot for your help :) This explanation was very simple and intuitive and that was definitely a push in the right direction for me. $\endgroup$
    – Shay
    Sep 28, 2021 at 15:19

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