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In the text books I have consulted about special relativity, it is always postulated that if frame A moves with speed v relative to frame B, then in frame B the other frame moves with speed -v. I mean this seems to make intuitive sense, but given that pretty much nothing else in special relativity makes intuitive sense, and that observers in the two frames will not agree on distance and time, why will they agree on the speed? I know this is very basic but since all the formulas in special relativity rest on this postulate, I would like to understand why this is so.

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The simple answer is: because the situation is completely symmetrical. This becomes even more clear when we choose the positive direction of $x'$ to be the negative direction of $x$. In that case $v' = v$.

Each of the reference frames sees another reference frame moving in its negative $x$ or $x'$ direction. Nothing can distinguish one of the frames from the other and they are each measuring the same quantity (their relative speed). According to the Principle of Relativity, the physics laws must be the same for each frame. And because there is no way to distinguish one frame from the other, they must therefore measure the same value for their relative speed.


Imagine two identical spaceships each with a measuring rod and a clock. We have assured beforehand that the rods are completely identical with the same length, and the clocks are identical as well, ticking at the same rate. You and your twin sister are piloting the spacecrafts. You both take off, and after a while you encounter each other in empty space. You see the other spacecraft approach and measure its speed with your rod and clock. Your sister sees the other spacecraft approach and measures its speed with her rod and clock. You each see exactly the same thing, and are measuring the same quantity with identical measuring instruments. The situation is completely symmetrical, with no way to distinguish one from the other. You and your sister could swap spaceships, and you would still experience the exact same thing. Due to that symmetry, you and your sister must measure the same value.

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It turns out that this is analogous to the relationship between the angles $\phi$ between two radii in Euclidean space. The pair of radii cuts a certain arc-length of the unit circle...and a certain area of a sector of the unit circle. Choosing one sense [once and for all] to be the positive direction, $\phi_{BA}=-\phi_{AB}$.

In relativity, the analogous quantity is called the rapidity $\theta$ between observer 4-velocities. It turns out that spatial velocity $v_{AB}=c\tanh\theta_{AB}$ (analogous to slope $m=\tan\phi$). So, $v_{BA}=-v_{AB}$ . We can generalize this to a vector relationship as spatial-velocity vectors in each frame.

(There is a subtlety however. This is talking about (say) the $x$-component of the velocities in the "space" of each observer. As 4-vector components of 4-velocities on a spacetime diagram, these spatial-component 4-vectors are not parallel... since each observer has a different sense of "space"... a different set of simultaneous events. Geometrically, using the tip of each 4-velocity, that observer's "space" is parallel to the tangent plane to the "circle" ["hyperbola"].)

robphy-circles


UPDATE

To amplify @AndrewSteane 's answer, with "ticks" and "space-ticks" ("sticks") and to help visualize Minkowski spacetime (a nonEuclidean geometry), I've drawn in what I call light-clock diamonds on a spacetime diagram drawn on "rotated graph paper".

You can now read off the tick marks for each observer.
Velocity is rise/run: $v=\Delta x/\Delta t$ in each frame, both agreeing on the forward direction.

robphy-RRGP-symmetry

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There are various ways to prove this, and depending on your taste you will find one way or another more basic or intuitive. But beware of one pitfall. This is the pitfall of asserting that since coordinate systems are arbitrary, you can come up with any value you like for the speed. Any assertion that it is just an arbitrary coordinate speed has failed to grasp what speed we are talking about. We are talking about speed as indicated by standard measuring devices such as standard metre-sticks and standard clocks. The claim is that if two inertial frames are in relative motion, then if a point fixed in frame B moves past $v t_A$ of frame A's metre-sticks in a time $t_A$ registered by clocks at rest in A, then a point fixed in frame A moves past $v t_B$ of frame B's metre-sticks in a time $t_B$ registered by clocks at rest in B. You see it is the same $v$ in these two assertions. The question is how to prove it.

One convenient approach is to consider a third frame, which I will call S. We pick S such that A and B both have the same speed relative to S, but in opposite directions. This helps to bring out the symmetry between A and B. You can say that whatever factor affects the clocks of B relative to S, that same factor affects of the clocks of A relative to S, because A and B have the same speed relative to S. A similar statement applies to the rods or metre-sticks or whatever you want to call the standard distance-measuring devices. Those of A and B are affected by the same factor relative to S.

Finally, then, we ask the question, "how many of A's sticks does a point fixed in B pass in the time $t_A = 1$ unit registered by A?" In principle S can figure this out, or observe it, just by looking at the A clocks and sticks and noting where B has got to when a nearby A-clock reads "1 unit". (By "looking at" here you can imagine that S can stand at any event and write down the value indicated by an A clock or stick as it passes by.) They will find some answer or other, let's call it $N$. S can also investigate the question "how many of B's sticks does a point fixed in A pass in the time $t_B = 1$ unit, registered by B?" Again, S can figure this out, or observe it just by looking at the B clocks and sticks and watching where A goes. But owing to the overall symmetry, S must get the same answer $N$. And this $N$ is the speed you asked about.

For a more algebraic method, you can write down a derivation of the Lorentz transformation. The way it is normally done is to assume the relative velocities are equal and opposite, and then one shows that the transformation obtained under that assumption is consistent with the principles so one has an acceptable theory. But your question is whether this assumption has to be made: whether it is not just sufficient but also necessary. I think if you don't make it then the transformation you obtain will concern some sort of coordinate-choice which does not map very closely onto the behaviour of standard rods and clocks. If one then adapts the coordinates, rescaling them to match the standard measures, then one will find that the relative velocities are equal and opposite after all, unless there is a special case which I will present next.

I think the only way to avoid the conclusion of equal and opposite relative velocities is to assert from the outset that there is some sort of special direction, so that when one examines how the frames A,B, relate to the frame S which I introduced above, one would then claim that the situation is not necessarily symmetric because it depends on how the directions of relative motion relate to the special direction. That situation is called non-isotropy or anisotropy of space. You then have some definitions of terms to worry about. Physicists tend to feel intuitively that when we assert the Principle of Relativity we are including the assertion of isotropy, but anyone who wishes to carefully tease apart the assumptions will want to consider the isotropy assumption as separate. This is the kind of thing which philosophers like to consider. So the final answer to your question is that the equality of relative speeds for a pair of inertial frames turns on the assumption of isotropy of space. In this sense it is not self-evident but it follows from a very natural assumption.

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  • $\begingroup$ Would you agree that this only shows that $\|v\|=\|v'\|$, not necessarely that $v'=-v$? Indeed I was wondering whether the directions of $v$ and $v'$ can be compared in a meaningful way. I have posted a question to discuss this, an answer would be much appreciated. $\endgroup$
    – Filippo
    Sep 22, 2022 at 9:38
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You say that the moving observers don't agree on distance and time, but in an important sense that isn't exactly true. All the relativistic effects that result from two inertial frames moving relative to each other are symmetric. If, for example, in my frame 5 seconds pass on my clocks, while in your frame 4 seconds pass on your watch, it is also true that 4 seconds pass on my watch while 5 seconds pass on your clocks, so we each agree that the watch of the other is time dilated by the same amount.

Imagine, then, that we start off at rest relative to each other some measured distance apart with synchronised watches. At an agreed time we each accelerate at the same rate toward the other momentarily, then coast at a constant speed until we meet. If each of us does not see the other moving towards them at the same speed, then when we meet our watches must show different times (since speed is change of distance over time, and we both have to agree that the original distance between us has shrunk to zero). If the time differences we experience are not identical, when we have each been subjected to exactly the same conditions from the start of the experiment, then the principle of relativity could not apply. So asserting that all inertial reference frames are equivalent does demand, at least in the arrangement I described, that each frame moves at the same speed relative to the other.

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  1. A Lorentz transformation has to preserve the Minkowski metric. That means it has to be of the form $(t,x)\mapsto (t',x')=(At+Bx,Ct+Dx)$ where, for all $t$ and $x$, we have $t^2-x^2=(t')^2-(x')^2$. This in turn requires that $A^2-C^2=B^2-D^2=1$ and $AB-CD=0$, which in turn requires that for some $w$, we have $$A=D=1/\sqrt{1-w^2}\qquad\qquad B=C=-w/\sqrt{1-w^2}$$ Let's call this particular Lorentz transform $L(w)$.

  2. If you are traveling with respect to me at some velocity $v$, and if you pass me at $(t=0,x=0)$, then I will say that you are present at the event $t=1,x=v$. By point 1), the coordinates that you assign to that event must be $$t'=(1-vw)/\sqrt{1-w^2}\qquad\qquad x'=(v-w)/\sqrt{1-w^2}$$ for some value of $w$. But you are going to say you're standing still, so $x'$ must equal zero. Therefore $w=v$.

  3. Conclusion so far: If you are traveling with respect to me at (according to me) velocity $v$, then the Lorentz transform from my coordinates to yours must be $L(v)$. And conversely, if the Lorentz transform is $L(v)$ then I must say you are traveling at velocity $v$.

  4. If we transform from my coordinates to yours and then back again, we surely must get back the coordinates we started with. So if I say you are traveling at velocity $v$ and you say I am traveling at velocity $v'$, then $L(v')L(v)$ must be the identity.

  5. Solving $L(v')L(v)=I$ gives $v'=-v$.

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  • $\begingroup$ The first sentence in 4. is particularly nice. $\endgroup$ Jan 19, 2022 at 13:23
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It's true for the Lorentz transformations; it's true for the Galilean transformations. But it doesn't always have to be true. You need to assume isotropy of space along with some other assumptions. See: P. Moylan, “Velocity Reciprocity and the Relativity Principle” A.J.P., 90, 126 (2022). (I'm the author of that paper.)

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We can imagine a situation where the frames don't agree with the magnitudes of their realtive speeds.

Suppose 2 approaching stars. Each of them has inteligent life monitoring and sending the information of the relative velocity to the other, measured by the blue shift of the H spectrum.

If the results show the same blue shift, then $v_A = -v_B$ as stated by SR.

But suppose that they are consistently different, above the experimental error. The conclusion could be that the spacetime between them is not flat, there is a non symmetrical gravitational potential for some reason, that should be investigated.

Answering the question, the conclusion is that $v_A = -v_B$ is a requirement for a spacetime be Minkowskian.

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