What is the law of reflection in special relativity?

[spaceofwaste]

Hi,

I'm working through an Excercise in Sean Carrol's spacetime and geometry book. The question asks you to consider an inertial frame S with coordinates $$x^\mu=(t,x,y,z)$$ and a frame S' with primed coordinates. Which is related to S by a boost v in the y direction. Imagine a wall( or mirror) lying along the line x'=-y'. From the point of view of S, what is the relationship between the angle of incidence (assuming ball travels in x-y plane only) and angle of reflection? Also what is velocity before and after?

Unfortunately there are not even one word answers let alone solutions in this book, so I don't even know if what I have is correct.

The way I proceeded, was firstly to consider what angle the mirror would appear at in S. Since the y-direction will be Lorentz contracted, but the x -direction will remain unchanged, the mirror should appear to be at a greater angle than 45 to someone in the S frame. This angle would precisely be:
$$\theta_M=arctan(\gamma \frac{\Delta x'}{\Delta y'})=arctan(\gamma)$$, the last equality following from the angle being 45 degrees in the primed frame meaning $$\Delta x'=\Delta y'$$

OK so that is my step 1. Now we need to work out how the incident velocity looks in S.

e.g. $$u_x(init)=\frac{\Delta x}{\Delta y}=\frac{\Delta x'}{\gamma (\Delta t'+v\Delta y')}= \frac{u_x'(init)}{\gamma(1+vu_y'(init))}$$ Similarly, $$u_y(init)=\frac{u_y'(init)+v}{(1+vu_y'(init))}$$.

From this we can work out the angle of the incident ray (wrt -x-axis) in the S frame, this is therefore $$\alpha=\frac{u_y(init)}{u_x(init)}$$ (which you can sub into from above). I work out then the actual incident angle, $$\theta_I=\alpha+\theta_M-90$$. (not 100% sure if this is correct)

Does this look like I'm going down the right path?

 

[spaceofwaste]

I come out with this monster finally anyway:

$$\theta_r-\theta_i=-2tan(\gamma)+arctan(\frac{u_(y')(i)+vu_(x')(i)u_(y')(i)}{\gamma(u_(x')(i)+v+v[u_(x')(i)]^2+v^2u_(x')(i)})+arctan(\frac{u_(x')(i)+vu_(x')(i)u_(y')(i)}{\gamma(u_(y')(i)+v+v[u_(y')(i)]^2+v^2u_(y')(i)})$$

where the $$u_(y')(i)$$ etc, represents the component of the incident velocity in the S', wrt to y' axis. $$\gamma$$ is just the Lorentz factor. v is the boost between S and S'.

Not sure if anyone knows what the expression shouled be for such a shift, and if this looks on the right track?

 

[Entropia]

Hi,

The law of reflection in special relativity is a bit different from the classical law of reflection in Newtonian physics. In special relativity, the law of reflection states that the angle of incidence is equal to the angle of reflection, but the angles are measured in different frames of reference. This means that the angle of incidence as seen in one frame may not be the same as the angle of incidence seen in another frame. In other words, the law of reflection in special relativity takes into account the effects of time dilation and length contraction on the angles of incidence and reflection.

In the scenario you described, the angle of incidence as seen in the frame S' (primed frame) would be 45 degrees, but as seen in the frame S, it would be arctan(γ) as you correctly calculated. This is because the angle of incidence is defined as the angle between the incident ray and the normal to the surface, and the normal vector is affected by the Lorentz contraction of the mirror.

As for the incident velocity, your calculations seem to be correct. The incident velocity in the S frame would be different from the incident velocity in the S' frame due to the effects of time dilation and length contraction. The incident angle, as you calculated, would be the arctan of the ratio of the y- and x- components of the incident velocity.

Overall, it seems like you are on the right track with your calculations and understanding of the law of reflection in special relativity. Keep up the good work!