- #1

Coldie

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Our textbook gives us equations to find the speed of objects in relation to others in the x, y, and z planes. These are:

[tex]v^'_x = \frac{v_x - u}{1 - uv_x/c^2}[/tex]

[tex]v^'_y = \frac{v_y}{\gamma(1 - uv_x/c^2)}[/tex]

[tex]v^'_z = \frac{v_z}{\gamma(1 - uv_x/c^2)}[/tex]

My first question is why is the equation for velocity in the x direction different from those in the y and z directions? Since all directions are relative, that would mean that simply turning 90º in any direction would mean that you'd have to use a different equation to find the velocity of something in a given axis?

My next question is from one of the homework problems, which says:

A and B are trains on perpendicular tracks. The velocities are in the station frame (S frame).

a) Find [tex]v_AB[/tex], the velocity of train B with respect to train A.

b) Find [tex]v_BA[/tex], the velocity of train A with respect to train B.

The picture shows that train A is going directly upwards from the train station at .8c, and train B is moving directly to the right from the station, also at .8c.

Now, just looking at the equations to finding the answer to part a you can tell that something's not right. The equation to find [tex]v^'_y[/tex] has [v_y] on the top, which in this case is 0 since train B is not moving vertically, so following the given equations, the vertical speed of train B relative to A is 0, which is not correct. The answer in the back is arrived at by using the equation for [tex]v^'_x[/tex] to find [tex]v^'_y[/tex] and using the equation for [tex]v^'_y[/tex] to find [tex]v^'_x[/tex]. Can someone tell me how you're supposed to know when to switch the x and y axes to use the proper equations to find the answer?

I hope I've made everything clear, thanks for the help.