# Perpendicular relative velocities

## Homework Statement

Rocket A moves with speed .75c in a northerly direction relative to an origin. Rocket B moves west (relative to that origin) with speed .4c. As B moves, it emits radiation with a wavelength 100nm normal to its line of motion, in the northerly direction. What is the wavelength of the radiation as seen by A?

## Homework Equations

Doppler shift equations

## The Attempt at a Solution

So far I have two different approaches to a solution, the first approach finds the wavelength of the radiation as seen by a person at the origin (which is redshifted) because the source is moving away at .4c. Then use that wavelength to find the wavelength as seen by a person moving away at .75c (to represent the rocket A). I'm not so sure about this solution...

The second idea I had was to find the relative velocity of B as seen by A (since it would be a right triangle) and using those to find the angle between the light source and the observer. However, I am unsure how to find the separate components of the relative velocity. Do I keep the .75c speed in one direction and uy = uy' /(gamma(1-ux*v/c^2) to find the other direction, if that is so I have no idea what the relative velocity would be... This method is really tripping me up and I think it is the correct way. Could someone help me get on the right track?
All incite is appreciated

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The first way is incorrect. The observer should be at A position, not at the origin.

The second way is correct, but you will have to recalculate the emission direction.

I think the simplest way is to use Lorentz transformation to find photon's momentum-energy four-vector:
$$p^i = \left( \frac{h\nu}{c}, hk_x, hk_y, hk_z \right)$$
in origin's reference frame and then find it in A rocket's reference frame.

Darn, I am unfamiliar with this energy momentum four vector, my professor has just given us the velocity addition formulas and a few doppler shift formulas. Is it possible to find the "resultant" velocity of ship B according to ship A?

Yes, it is.

The corresponding formulas are
$$v^A_x = \frac{v^0_x - V_A}{1-(v^0_x V_A)/c^2}$$
$$v^A_y = \frac{v^0_y\sqrt{1-V_A^2/c^2}}{1-(v^0_x V_A)/c^2}$$

The "0" superscript means the origin's reference frame (RF) and "A" means A rocket's RF.

Remember that emission direction in RF A and RF B is different.

In fact, four-vectors is another way to write these formulas. I like it because it's easier for me to remember one matrix than four equations.

O, that does sound useful, only had minor exposure to matrices so far but I'm building my base knowledge. Another question: for VsubA do I use the speed of rocket ship A? I'm unsure as to what variable represents what speed. Sorry to be such a noob :/

Yes, V_A is the speed of rocket A.

Thank you very much Maxim, I think I've got it, have a good one