# Having trouble connecting Lorentz transformations with my problem

#### m00npirate

1. 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.

2. Homework Equations

3. 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, ux and uy 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 ux = v and uy = 0.65$$\sqrt{1/ (v^{2}/c^{2}})$$
Plugging in my value of ux for v I get uy = 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

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#### alphysicist

Homework Helper
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

Homework Helper
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, vperpendicular= 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

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#### alphysicist

Homework Helper
Hi ice ace,

Since the beacon is fired in perpendicular direction, vperpendicular= 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

Homework Helper
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).

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