Relativity - Lorentz Transformation

[missmaths.]

Homework Statement

A is at the base station and given in K co-ordinates
B is on a spacecraft and given in K' co-ordinates.
The velocity of the spacecraft is v=0.8c

Question 1
After t = 2y (y = years) A sends a message by radio to B demanding a picture. Which time t' does B have when the signal arrives and what is B's distance x to the base station then?

Question 2
It took B t' = 1y time to take the picture and get the radio transmission on its
way. What time t is it for A and where (x) is the spacecraft when B launches
his message?

Question 3
What time t is it for A when the message from B arrives and where (x) is
B when this happens?

Homework Equations

\gamma = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}
so in this question \gamma = \frac{1}{0.6}

Lorentz's Transformations
t' = \gamma(t-\frac{vx}{c^{2}}),
x' = \gamma(x-vt),
t = \gamma(t'+\frac{vx'}{c^{2}}),
x = \gamma(x'+vt').

The Attempt at a Solution

For 1 I have done the following

x_{B}=vt,
x_{message}=c(t-2y)

So I need to find when these to meet i.e.
vt=c(t-2y)
which leads to
t=\frac{2y}{1-\frac{v}{c}} = \frac{2y}{1-0.8} = 10y
therefore
x_{B}=0.8c * 10y = 8Ly

then from the Lorentz t' equation I get 6y (just substituting values above in).


For 2 I have done the following

From 1 I saw that t' = 6y when B got the message therefore when B sends the message it must be 7y as tooks 1y to sort it.

Now I need to find t from this which is where I am stuck.

Obviously I use the t Lorentz equation however it is the entry for x' I am confused about.
As I am calculating t for A must I put in x'_{A} = -vt' or should I put in x' = 0 as the photo is at x' = 0.

I have calculated both ways and for x' = -vt' I got 4.2y and for x' = 0, I got 11.67.
The second seems more correct as it is greater then 10 the t found in part 1, but this is relativity so I am confused.

Once I have sorted this bit I am sure I can do part 3.

Any help on my problem would be much appreciated.
Thank You.

 

[vela]

Your instinct and reasoning that x' = 0 is correct. The primed coordinates have nothing to do with observer A; they are the coordinates B uses. You calculate for A by using the Lorentz transformations with the relative velocity of the two frames.