# Why can different observers agree on their relative velocity

## Main Question or Discussion Point

I'm having trouble understanding how the lorentz transformations are posed intially.

Why can different observers agree on their relative velocity? I don't understand why this velocity is exempt from all of the other relativistic phenomena that occur.

Also, in the initial derivation, I'm told that one could make the following equations

$x'=G(x-ut)$ $x=G(x'+ut')$

I understand the reasoning behind the signs, but I think I dont see any clear reason for wanting to use different times here. Of course, the point of this is to conclude that time intervals aren't absolute, however, I don't get why, for example:

the term (x'+ut') is constructed that way because this coordinate is measured to be going away by O (unprimed frame). But why is the translation a function of the time measured by the O' frame? namely, why is it $ut'$ that is considered by the O frame. Why isn't it $ut$? That makes more sense to me.

Thanks.

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PeterDonis
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Why can different observers agree on their relative velocity? I don't understand why this velocity is exempt from all of the other relativistic phenomena that occur.
Relativity doesn't say that different observers must measure different values for everything. Relative velocity between two observers is symmetric: if you are moving at a certain speed relative to me, I must be moving at the same speed relative to you.

(Note that I said "speed" here; actually, your velocity relative to me will have the opposite sign from my velocity relative to you, because velocity includes direction, and the direction of your motion relative to me is opposite from the direction of my motion relative to you. Perhaps that is what was confusing you?)

in the initial derivation
Can you give a reference? We can't really help you understand what the derivation you're reading is saying if we can't see the context.

Hi. I'm following Serway's book on modern physics at the moment. He uses a similar derivation to the one I saw in class, and on both occassions I saw them saying the first step I included in my post is an intuitive set of equations.

How do they measure their relative velocities though? If O' measures it as $\frac{dx'}{dt'}$ then it shouldn't be the same as $\frac{dx}{dt}$ right? Since these are supposed to be transformed.

I'm not confused by the signs. I'm confused about the measurement of speed. I don't see how this is exempt from all the other conclusions of the theory.

Thanks.

PeterDonis
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How do they measure their relative velocities though? If O' measures it as $\frac{dx'}{dt'}$ then it shouldn't be the same as $\frac{dx}{dt}$ right? Since these are supposed to be transformed.
Lorentz transformations act on coordinates, yes, but you can't just assume that that changes the magnitude of all expressions involving those coordinates. To see if relative speeds change when you change frames, you have to transform the coordinates specifying the worldlines you are interested in, and then derive the speeds in the new frame from the transformed expressions for the worldline coordinates.

So, for example, in the unprimed frame the primed observer is moving along the worldline $x = u t$, by definition. That means $dx / dt = u$ for that worldline, whereas for the unprimed observer in this frame, he is at $x = 0$ for all $t$.

Now transform those two worldlines to the primed frame. The worldline $x = u t$ for the primed observer becomes the worldline $x' = 0$ for all $t$, by definition of the primed frame. We can show this by using the inverse transformation $x = G ( x' + u t' )$ and $t = G ( t' + u x' )$; we substitute into $x = u t$ to obtain $G ( x' + u t' ) = u G ( t' + u x' )$, which simplifies to $x' + u t' = u t' + u^2 x'$, or $(1 - u^2) x' = 0$, or $x' = 0$, as desired.

Now do the same substitution for the worldline $x = 0$. What do you obtain? And what does it give for $dx' / dt'$ for that worldline?

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I'm sorry, but I don't really follow you. I have only vague notions about what world lines are. If I'm understanding you correctly, then I think your argument is invalid because you are presupposing the validity of the lorentz transformations to prove that $u$ is invariant. I want to know from the beginning why $u$ is invariant.

Did I misunderstand you?

Could you maybe give me a basic argument as to why $u$is invariant from both observers?

Thanks.

Nugatory
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Could you maybe give me a basic argument as to why $u$is invariant from both observers?
That's an observational result. Every device that has been been built to measure relative speeds has produced that result to the limits of experimental accuracy whenever it is practical to do the experiment. Furthermore, when we assume that that result will hold even when it's not practical to do the experiment (there's no practical way of mounting a radar gun on a fast-moving subatomic particle) we come to conclusions that also agree with observations.

Thus, unless and until observations tell us otherwise (and I would bet very long odds against that happening), we are justified in assuming that A's velocity relative to B as measured by B will be the negative of B's velocity relative to A as measured by A.

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PeterDonis
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I have only vague notions about what world lines are.
A worldline is a curve in spacetime that describes a particular object--where it is in space at every instant of time. The description of a worldline in a specific coordinate chart will be a function; in the case we're discussing, the function we're using is the function $x(t)$ giving the $x$ coordinate of the object for each $t$ coordinate.

I think your argument is invalid because you are presupposing the validity of the lorentz transformations to prove that $u$ is invariant. I want to know from the beginning why $u$ is invariant.
As Nugatory said, it's an observational result. However, there are also good theoretical reasons for it, to do with the general properties that the Lorentz transformation must have. The key property for this discussion is that the transformation from the primed frame to the unprimed frame must be the inverse of the transformation from the unprimed frame to the primed frame. Working through the details of what that implies, IIRC, will turn out to require that $u' = -u$, i.e., that the velocity of the unprimed observer in the primed frame must be the negative of the velocity of the primed observer in the unprimed frame.

A.T.
I don't see how this is exempt from all the other conclusions of the theory.
It's not exempt, but follows directly from a main postulate of the theory: the equivalence of inertial frames of reference. If one inertial observer would measure another inertial observer to be faster than vice versa, there would be no symmetry and thus no equivalence.

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robphy
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The spacetime viewpoint of Minkowski (analogous to Euclidean geometry) might help here.

On a position-time diagram, plot "someone at rest" and "someone moving with constant velocity u".
You have just drawn the worldlines of two inertial observers.

In euclidean geometry, given two lines that cross at a point, consider the unit-vector along each line from that point.
(In special relativity, the intersection point is "the meeting event" and those unit vectors are analogous to the observers' 4-velocities [their t-hat-axes].)

In euclidean geometry, the angle between those unit-vectors is invariant... in the sense that you can rotate the diagram and the angle between is still the same.
(In special relativity, the analogous "angle" is called the rapidity.... the arc-length along the "unit-circle" [the invariant unit-hyperbola asymptotic with the light-cone of the meeting event].)

In euclidean geometry, each guy can break up the other guy's unit-vector into components with magnitudes cos(ang), sin(ang).... with ratio tan(ang).
(In special relativity, the analogous quantities are called the time-dilation factor $\gamma$, $\gamma(u/c)$, and the relative-speed/c $u/c$.)

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bcrowell
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