Having trouble connecting Lorentz transformations with my problem

[m00npirate]

Homework Statement

A ship is moving at 0.45c with respect to earth, and a beacon is fired perpendicular to the ship at 0.65c with respect to the ship. Find the velocity of the beacon with respect to earth.

Homework Equations

The Attempt at a Solution

My main problem here is seeing which numbers go where in the equations. My book basically just shows them, then has an example of adding velocities in one dimension >_>.

From what I gather, u_x and u_y are the components of the velocity of the beacon with respect to earth. u'_x and u'_y are the components of the velocity in the ship's frame, making u'_x = 0 and u'_y = 0.65c
This however makes the v's in the denominator cancel out, and I get u_x = v and u_y = 0.65\sqrt{1/ (v^{2}/c^{2}})
Plugging in my value of u_x for v I get u_y = 0.65\sqrt{1/ (u_{x}^{2}/c^{2}}) which doesn't lead me anywhere.
Its pretty clear that I'm doing something wrong, I just ned a nudge in the right direction. Here's my picture of the problem (not to scale =P)
http://www.geocities.com/zombierobopirate/relativity.png

Thanks in advance

 

[alphysicist]

Hi m00npirate,

What does v represent in this problem? What numerical value does it have?

 

[m00npirate]

v is the velocity of the beacon with respect to earth, which I am ultimately trying to find.

 

[alphysicist]

v is the velocity of the beacon with respect to earth, which I am ultimately trying to find.

No, I believe v is the velocity of the reference frame of the ship (relative to the earth). The primed velocities are the velocity of the beacon in the ship's reference frame; so what is v?

 

[ice ace]

Since the beacon is fired in perpendicular direction, v_{perpendicular}= 0.65c with respect to rocket, and since the rocket is moving at 0m/s in perpendicular axis with respect to earth, applying lorentz transformation in that perpendicular axis will still give beacon velocity of 0.65c

 

[alphysicist]

Hi ice ace,

Since the beacon is fired in perpendicular direction, v_{perpendicular}= 0.65c with respect to earth, since the rocket is moving at 0m/s in perpendicular axis with respect to earth. Applying lorentz transformation in that perpendicular axis with v= 0c (rocket v in that direction wrt earth), you'll obtain u'x=ux=0.65c

I don't think that is right. Those particular velocity equations were derived assuming that the reference frame moved with speed v along the x (and x') axis, so v would not be zero.

 

[ice ace]

Maybe I'm wrong, but I know that v= 0.45c in that axis that the Earth and spaceship travel in, but v in the perpendicular axis is 0, is it not?=>spaceship is not moving in the perpendicular axis with respect to earth.

 

[alphysicist]

Maybe I'm wrong, but I know that v= 0.45c in that axis that the Earth and spaceship travel in, but v in the perpendicular axis is 0, is it not?=>spaceship is not moving in the perpendicular axis with respect to earth.

I don't think that's the way those equations were derived. First you say that you have one frame move with velocity v relative to another. Then the direction of that velocity v defines the x (and x') axis, and the y and z (and y' and z') axes are perpendicular to the direction of v.

So the meaning of v in all the equations is the same; in both the u_x and u_y equations v is the velocity of the frame of reference in the x direction (because that's how the x direction was defined).