# Lorentz transformations (2nd year relativity)

1. Oct 11, 2010

### affans

1. The problem statement, all variables and given/known data

A light signal is sent from the origin of a system K at t = 0 to the point x = 1 m, y = 8 m, z = 13 m. a) At what time t is the signal received?
b) Find ( x', y', z', t' ) for the receipt of the signal in a frame K' that is moving along the x axis of K at a speed of 0.6c.

2. Relevant equations

The lorentz transformations:

x' = $$\gamma * (x - vt)$$
y' = y
z' = z

3. The attempt at a solution

Part a was easy. I got the right answer. I just took the length of the vector given by the co-ordinates and divided by the speed of light. The answer is $$5 * 10^8$$ I am having trouble with part b.

Ofcourse, y' and z' were easy to get. t' (i had 3 tries, and i used them all) so I lost a mark there. I have one try left on x'.

Using the equation, we first have to solve for $$\gamma$$. Plugging the numbers into the equation for gamma:
$$\frac{1}{\sqrt{1-(v/c)^2}}\; \text{yields} \; 1.25$$

Then using the lorentz transformation I have the following eqn:
x' = $$\gamma$$ (x - vt) . Plugging in the numbers yeilds
x' = 1.25(1 - 0.6 * c *(5E-8))

I get 10 as the answer. It is wrong. I also thought t = 0 could work since thats when the event happened. But the answer 1.25 is also wrong.

My third attempt yielded 11.25m however, I am scared to submit it. If anyone can please verify my number for me.

2. Oct 11, 2010

### diazona

The Lorentz transformations are more accurately written
$$\Delta t' = \gamma(\Delta t - v\Delta x/c^2)$$
$$\Delta x' = \gamma(\Delta x - v\Delta t)$$
The $\Delta$ indicates that the numbers to be plugged in should be the difference between two spacetime events. So putting in t=0 is a mistake that you should not make again.

Can you show your calculations for the other two attempts?