Reference Frames with Different Origins

[mailman123]

Homework Statement

Alex is born in a spaceship and Bill is born on Earth just as Alex’s spaceship passes Earth at 0.90c
when t=t’=0 (event A). Planet Z is at rest in Bill’s reference frame (x,ct), where it is 45 light-years (ly) away. Alex, whose spaceship is at rest in the frame (x’,ct’), flies directly to and past Planet Z. Event B is when he passes the planet. In the meantime, Carl, who was born at the same time t=t’=t”=0 in his spaceship in the frame (x”,ct”), has been flying toward Earth at 0.9c and passes Planet Z just as Alex does (event B). Carl passes the Earth at event C.
Find the co-ordinates of events A, B, and C in the rest frames of Alex, Bill and Carl.

Homework Equations

$t'=\gamma(t-\frac{vx}{c^2})$
$x'=\gamma(x-vt)$

The Attempt at a Solution

I started by finding Event A in every reference frame, for Bill it is obviously x=0 ct=0 (just based on the definition), same for Alex. I was a bit unsure what to do for Carl here since his origin is shifted and his axes are pointing the other direction. I decided to trust the formulas, so I tried $x''=\gamma(x-vt)$ using $x=90ly$ (since he should be twice as far away as planet Z if him and Alex travel the same amount of time at the same speed and meet at planet Z), $v=-0.9c$ and $t=0$, this got me that according to Carl event A happened 206ly away, this seems rather ridiculous and wrong. I'm assuming my approach must be wrong, because they also said t''=0 but with my diagram (picture attached, black is x-ct, red is x'-ct', grey is x''-ct'') the t'' is clearly not 0... I think my confusion comes from the fact that the origins aren't the same, I haven't really had a question like that before and I'm not sure exactly what to do.

 

[Fredrik]

When the two coordinate systems don't have the same origin, you have to add a constant to each of the two transformation equations. You should be able to figure out what constants to use.

 

[mailman123]

So I should have something like $x''=\gamma(x-vt)-90ly$? is that right or does the constant need to go somewhere else? Also, does my diagram look correct? I'm not sure I'm interpreting the situation correctly... Does t=t'=t''=0 mean my x'' axis has to pass through event A at the other two frames origins? If not does that the ct'' coordinates of event A would not be 0?

 

[Fredrik]

Something like that yes, but not that. (The constant will be the x'' coordinate of event A, and it's not just -1 times the x coordinate of the t=0 event on Carl's world line). There will also be a constant on the right-hand side of the equation expressing t'' in terms of t and x. It seems to be more difficult than I thought at first to determine the constants, so right now I'm thinking about whether you really have to. You can use the fact that the slope of the world line of an object with velocity v is 1/v, and the slope of the corresponding simultaneity line is v. Then you may be able to treat this as a geometry problem, without worrying about the details of Poincaré transformations.

The diagram is correct, assuming that you just drew a rough picture and wasn't trying to get the slopes and distances exactly right.

There's an issue with the problem statement. It says that Carl was born at t=t'=t''=0. I assume that this is supposed to be t=t''=0, because the event where Carl was born (the point where the grey lines intersect the horizontal black line) doesn't have t'=0.

 

[Fredrik]

Unless you can find a simpler geometric solution (I haven't tried to find one), you will have to do something like this:

You know that there's a Lorentz transformation $\Lambda$ and real numbers a,b such that
$$\begin{pmatrix}t''\\ x''\end{pmatrix}=\Lambda\begin{pmatrix}t\\ x\end{pmatrix}+\begin{pmatrix}a\\ b\end{pmatrix}.$$ The constants can be determined by plugging in t=0 and x=0 (the unprimed coordinates of event A). "Determined" is perhaps too suggestive a word. In this case, I just mean that you will find a better notation to replace a,b.

You also know that there exist real numbers c,d such that
$$\begin{pmatrix}t\\ x\end{pmatrix}=\Lambda^{-1}\begin{pmatrix}t''\\ x''\end{pmatrix}+\begin{pmatrix}c\\ d\end{pmatrix}.$$ The constants are determined by plugging in t''=0 and x''=0. In this case, you can really determine c and d.

Now you can do things like solve the first equation for $\begin{pmatrix}t\\ x\end{pmatrix}$ and compare the result to the second equation. This approach should give you the values of a,b.