Special Relativity Diagram Problem

[Tedjn]

Homework Statement

Frame S' has velocity v relative to frame S. At time t = 0, a light ray leaves the origin of S, traveling at a 45 degree angle with the x-axis.

(a) What angle does the light ray make with respect to the x' axis in frame S'?

(b) Repeat part (a), replacing the light ray with a particle of mass m and speed u.

(c) Repeat part (a), replacing the light ray with a rod that is stationary in frame S.

Homework Equations

Minkowski diagram

$$v'_x = \frac{v_x - v}{1-vv_x/c^2}$$
$$v'_y = \frac{v_y}{\gamma(1-vv_x/c^2)}$$
$$\theta = \tan^{-1}(v/c)$$

The Attempt at a Solution

(a) In frame S', the axes are both rotated inwards (if v is positive) by angle $\theta$. However, I am not sure if the angle wanted in part (a) is still 45 degrees in frame S' (even though it is clearly $45 - \theta$ looking from frame S).

(b) The particle has identical components of velocity $v_x = v_y$ so I should just be able to use the above equations to solve for $v'_x$ and $v'_y$. However, the same question from part (a) remains; is angle measured differently in the S' frame as opposed to the S frame? In other words, in the S' frame, is the angle between x' and ct' still considered 90 degrees?

(c) You can treat S' as the stationary frame and S as the moving frame, in which case the axes of S will rotate outwards assuming it is moving with negative velocity relative to S'. In that case, it is a simple matter of determining the angle from the above formula and calculating the apparent angle with the x' axis.

Thanks!

 

[anathema]

Thank you for your post. Your approach to solving parts (a) and (b) is correct. In special relativity, angles are measured the same in all inertial frames, so the angle between the x' axis and the light ray or particle will still be 45 degrees in frame S'. However, the apparent angle between the x' axis and the rod in part (c) will vary depending on the relative velocity between the frames. As you mentioned, the axis of frame S will rotate outwards relative to frame S', so the apparent angle will be different from 45 degrees. You can use the equations you provided to calculate the apparent angle in this case.

I hope this helps. Keep up the good work!

 

[nlightner]

Dear student,

Thank you for your question. It seems like you have a good understanding of special relativity and the Minkowski diagram. However, to answer your questions:

(a) In frame S', the light ray will make an angle of 45 degrees with respect to the x' axis. This is because, as you correctly mentioned, the axes are rotated inwards by an angle of \theta. This means that the angle between the x' and ct' axes is still 90 degrees, and the angle between the light ray and the ct' axis is still 45 degrees.

(b) You are correct in using the equations you mentioned to solve for the velocity components in frame S'. The angle between the particle's velocity and the x' axis will be different in frame S' compared to frame S, as the axes are rotated. But the angle between the x' and ct' axes will still be 90 degrees, so you can still use the same equations to solve for the velocity components.

(c) Again, you are correct in treating S' as the stationary frame and S as the moving frame. The angle between the rod and the x' axis will be different in frame S' compared to frame S, but you can still use the same formula to calculate it.

I hope this helps clarify any confusion you may have had. Keep up the good work in your studies of special relativity!