How Does a Space-Cruiser's Relative Speed Demonstrate Relativity Principles?

[Gregg]

1. Observer O sees a fire-engine leave its station $\frac{9}{\sqrt{2}}$ km due
north, where a super-shuttle had been launched $10^{-5}$ searlier. A space-cruiser flying north-east sees these two events also $10^{-5}$ s apart, but with the shuttle launch occurring after the fire-engine leaves the station.

Show that the speed of the space-cruiser relative to the Earth is $\frac{12c}{13}.$

2.

(I) $$x'=\gamma (x-ut)$$
(II) $$t'=\gamma (t-ux/c^2)$$

3.

In the stationary frame the events are $\Delta x$ distance apart, and they occur $\Delta t$ apart in time. The moving frame O' sees the first event happen before the second event so $\Delta t'=-\Delta t$. Now need to find how fast the O' frame is moving with respect to the Earth. Eq. II gives

$$\Delta t'=\gamma (\Delta t-ux/c^2)$$

move things around ...

$$u=\frac{2c^2 x t}{c^2 t^2+x^2} \ne 12c/13$$


What's the problem?

 

[vela]

The Lorentz transformation equations you're using are for when the moving frame moves with speed u in the +x direction. In this problem, the moving frame's velocity points along the line y=x.

 

[Gregg]

In the O frame the events are spaced $\Delta y$ apart. I thought that maybe $\Delta y'=\gamma (\Delta y- u_y t)$ and $\Delta x'=\gamma(\Delta x- u_x t)$ and $u_x=u_y$. But for this part of the problem they only ask about the time, and the co-ordinate transformations aren't really needed at this stage. So I look at $\Delta t'=\gamma (\Delta t-u \Delta y / c^2)$ ?

Or maybe I need to find
$$\frac{dx'}{dt'}=\frac{V_x-u}{1-u V_x/c^2}$$

$$\frac{dy'}{dt'}=\frac{V_y-u}{1-u V_y/c^2}$$

But these don't look useful now, they have velocities of the ship and of the moving frame with respect to the rest frame and give the velocity of what with respect to the moving frame? Very confusing

 

[Gregg]

I'm still unsure how to use the direction y=x to get 12c/13=u!

 

[vela]

Orient the coordinate system so the space cruiser's velocity points in the +x direction.

 

[Gregg]

So $$x \to \gamma (x/\sqrt{2} - ut) ?$$

Are the other equations right or relevant? I should be able to do this but I'm probably missing something ridiculously simple. It's really irritating to be on this same thing a day later! I must not understand something?

I rotate the co-ordinate axes which I imagine to look like a cross with event A at the origin and event B at a distance up the y-axis. There is a spaceship traveling NE with velocity u. After the rotation the the ship's velocity is directed along the x-axis and event A is still at the origin and event B is at the position (9/2,9/2). The positions don't seem to be of much importance, the transformation for time is the problem.

$$\Delta t'=\gamma (\Delta t-u x/c^2)$$

This can't be the right equation.

 

[vela]

In the rotated frame, you have event A occurring at the origin and event B occurring at (ct, x, y)=(3000 m, 4500 m, 4500 m). In the ship's frame, event A still occurs at the origin, and event B occurs at (ct', x', 4500 m) since y'=y. The usual Lorentz transformation equations relate the pair (ct, x) with the pair (ct', x').

 

[Gregg]

Oh ct=3000 x=4500. Turns out that

$$u=\frac{2c^2 x t}{c^2 t^2+x^2} = 12c/13$$

Thanks!

Also get,

$$x'=\gamma(x-ut)$$

$$\gamma =13/5$$

$$x=13/5(4500-12/13 \cdot 3000)=4500$$

I think the 5,12,13 triangle will appear somewhere, I think in

$$(ct)^2 + x^2 = (ct')^2 +x'^2$$