High School Velocity Addn Paradox: SR Postulate Explained

[jeremyfiennes]

TL;DR

Apparent velocity addition paradox for three observers.

Velocity addition paradox: In observer A's frame, observers B and C move away from him at the speed of light c, B to the left and C to the right. In B's, frame A and C are both moving away from him at c, i.e. at the same speed. In both A's and C's frames they are moving at different speeds, namely away from each other at c. Because all three are inertial, according to the first SR postulate all their views are correct. How is this apparent anomaly explained?

 

[Ibix]

You cannot define an inertial frame moving at the speed of light. Attempting to do so requires a timelike and a spacelike vector that are both also null. Thus your maths is based on a self-contradiction and the paradox stems from that.

 

[Nugatory]

In observer A's frame, observers B and C move away from him at the speed of light c, B to the left and C to the right.

Can't happen of course, because of the light-speed limit, but we can fix that annoyance by choosing some speed slightly less than $c$ and get to your question:

How is this apparent anomaly explained?

What anomaly? There's no reason why two speeds that are the same using one frame have to be the same using another frame.

 

[jeremyfiennes]

- 1) Right, thanks. Slightly less than the speed of light.
- 2) But doesn't SR say that no inertial frame is preferred, effectively that all are correct?

 

[Ibix]

- 1) Right, thanks. Slightly less than the speed of light.

There is no problem if you use "slightly less than the speed if light" as your frame velocity. Your problem, that multiple velocities transform to the same velocity, only occurs if you attempt to apply the transforms with a frame velocity of $c$. That is invalid, as noted above.

 

[vanhees71]

- 1) Right, thanks. Slightly less than the speed of light.
- 2) But doesn't SR say that no inertial frame is preferred, effectively that all are correct?

Sure all are correct, but the components of four-vectors depend on the reference frame. What's invariant are the four-vectors themselves, i.e.,
$$\underline{u}=u^{\mu} \underline{e}_{\mu} = u^{\prime \mu} \underline{e}_{\mu}'$$
for the four-velocities. The components change due to the change of the basis by a Lorentz transformation,
$$u^{\prime \mu}={\Lambda^{\mu}}_{\nu} u^{\nu}.$$
This implies that
$$\underline{u}={\Lambda^{\mu}}_{\nu} u^{\nu} \underline{e}_{\mu}'=u^{\nu} \underline{e}_{\nu}$$
and thus
$$e_{\nu} = {\Lambda^{\mu}}_{\nu} \underline{e}_{\mu}'.$$
So the basis vectors change covariantly (that's why they have lower indices) and the vector components contravariantly (that's why they have upper indices).

 

[jeremyfiennes]

Bit beyond me, but I'm sure you're right! Thanks.

 

[silverrahul]

- 2) But doesn't SR say that no inertial frame is preferred, effectively that all are correct?

Yes, all frames are correct. Different frames will see different things. None of them is more correct than the other. Each frame is correct in what it observes and it will be different that what other frames observe.

 

[Grasshopper]

- 2) But doesn't SR say that no inertial frame is preferred, effectively that all are correct?

My take: As I understand it, so does Galilean relativity, and in Galilean relativity, inertial frames can disagree on speeds. This is shown in the Galilean velocity transformation (such transformations wouldn’t have much meaning if speed were universal, I don’t think). Speed is coordinate system dependent. There is no universal reality to speed, except speed that is defined by a chosen coordinate system. I’m at rest on Earth. But I’m also spinning along the rotation. But I’m also orbiting the sun. But I’m also orbiting the center of the galaxy. But I’m also moving within the galaxy cluster. But I’m also moving with respect to the ISS. So what is my speed? My speed according to me is zero with respect to myself, because I reside at the origin of my coordinate system. Speed is coordinate system dependent. It only has value in comparison to some chosen frame of reference. And because you can choose any inertial frame of reference, you can choose any speed for any object.

Except the speed of light, of course.

 

[PeterDonis]

in Galilean relativity, inertial frames can disagree on speeds

More precisely, in Galilean relativity, different inertial frames will disagree on all finite speeds. Whereas, in SR, different inertial frames will disagree on speeds less than the (finite) speed of light, but will all agree on the speed of light, and speeds greater than the speed of light are impossible.

 

[Grasshopper]

More precisely, in Galilean relativity, different inertial frames will disagree on all finite speeds. Whereas, in SR, different inertial frames will disagree on speeds less than the (finite) speed of light, but will all agree on the speed of light, and speeds greater than the speed of light are impossible.

This is probably a pedantic question, but aren’t there trivially infinite frames that would agree on speeds since they are all at rest wrt each other? Or do they all count as one IRF in Galilean relativity?

 

[PeterDonis]

aren’t there trivially infinite frames that would agree on speeds since they are all at rest wrt each other?

I suppose you could consider different choices of spatial origin and orientation, or "zero" moment of time (i.e., space and time translations and spatial rotations), to be different frames still at rest wrt each other. If so, just interpret my "different inertial frames" to not do that. Or consider my statement to assume that we have already fixed the time and space origin and orientation of all frames to be the same, since those transformations are irrelevant to what we are currently discussing.