Relativity - velocity of object travelling perpendicular to the observer

[struggles]

Homework Statement

1. Two light beams collide head on. Calculate their relative velocity.
2. (c) A particle moves north at speed 0.85c relative to an observer standing on the Earth. What is the velocity of this particle as observed by a fast ship traveling east on the Earth at speed 0.9c? Give the direction of travel with respect to a compass direction
   

Homework Equations

The Attempt at a Solution

1) A little confused by this question. I thought that speed of light is a constant no matter what reference frame it is viewed from so their relative velocity would be just c?

2) So speed between inertial frames u = 0.9c. Gamma γ = 1/√1-u²/c² = 1/√0.19
However I'm unfamiliar with working with the velocity of the object perpendicular to the relative velocity between the frames.
I've found V_y' = V_y/γ(1 - uV_x/c² where the dash frame is the frame of the ship. For this question would i take V_x = 0 and V_y = 0.85c and just end up with V_y' = V_y/γ? There would then be only 'north' velocity and none in the east direction?

 

[andrewkirk]

Two light beams collide head on. Calculate their relative velocity.

Your confusion is well justified. This is a weird question. Normally when we ask about relative velocity, we ask 'what is the velocity of A relative to B', which means 'in the inertial, momentarily comoving reference frame of B, what is the velocity of A'.

But light beams cannot have inertial reference frames (or any other sensible reference frames), so the question cannot be interpreted in the usual way.

We could interpret it as meaning this:
In the inertial reference frame of an observer standing near the point where the two light beams will meet, with the beams traveling along the x axis, let $x_a(t)$ and $x_b(t)$ be the x coordinates of the two wavefronts at time $t$. Then what is $\frac d{dt}(x_a-x_b)$?

The answer to this will be $2c$. That doesn't break any laws of physics because the $2c$ is not the velocity of any particle, massless or otherwise, or of any wave. It is a purely abstract quantity and hence not bound by relativistic speed limits.

 

[struggles]

Your confusion is well justified. This is a weird question. Normally when we ask about relative velocity, we ask 'what is the velocity of A relative to B', which means 'in the inertial, momentarily comoving reference frame of B, what is the velocity of A'.

But light beams cannot have inertial reference frames (or any other sensible reference frames), so the question cannot be interpreted in the usual way.

We could interpret it as meaning this:
In the inertial reference frame of an observer standing near the point where the two light beams will meet, with the beams traveling along the x axis, let $x_a(t)$ and $x_b(t)$ be the x coordinates of the two wavefronts at time $t$. Then what is $\frac d{dt}(x_a-x_b)$?

The answer to this will be $2c$. That doesn't break any laws of physics because the $2c$ is not the velocity of any particle, massless or otherwise, or of any wave. It is a purely abstract quantity and hence not bound by relativistic speed limits.

I think that kind of makes sense. So you can view the relative change in velocity as being greater than c but it doesn't actually go faster than c? Any chance of a hint of how to do the second part of my question?

 

[haruspex]

Two light beams collide head on. Calculate their relative velocity.

I would guess that the idea here is to apply the usual relativistic sum of velocities to show that the answer is still c.

For the second part, you could use time dilation to figure out the apparent velocity of the particle in the Northerly direction. Then recombine with the ship's velocity (I think Pythagoras applies here) to get the relative velocity. The result looks sensible anyway.

 

[andrewkirk]

you can view the relative change in velocity as being greater than c

It's a relative change in position (of the two wavefronts), not a relative change of velocity.

but it doesn't actually go faster than c?

One needs to choose one's words carefully in relativity. What is the 'it' referring to?