Angle of reflection with moving mirror, SR

[SpaceTrekkie]

Homework Statement

A mirror moves perpendicular to its plane with speed (beta)c. A light ray is incident on the
mirror from the \forward" direction (i.e., vm dot vl < 0, where vm is the mirror's 3-velocity and vl is the light ray's 3-velocity) with incident angle µ (measured with respect to the mirror's normal vector).


find cos (phi) where phi is the reflected angle.

Homework Equations

Unsure, I believe the law of cosines is involved, but I am not sure how that would fit into any other equation.

The Attempt at a Solution

We had an example of when the mirror is moving parallel, using the 4-momentum. This is very similar (I think), but I can't seem to get my head around it. I figure for the mirror, the velocity in the x direction will be zero, because it is moving up the y with speed (beta)c. But that is where we get stuck...

can anyone point me in the right direction?

 

[tiny-tim]

A mirror moves perpendicular to its plane with speed (beta)c. A light ray is incident on the
mirror from the \forward" direction (i.e., vm dot vl < 0, where vm is the mirror's 3-velocity and vl is the light ray's 3-velocity) with incident angle µ (measured with respect to the mirror's normal vector).

find cos (phi) where phi is the reflected angle.

Hi SpaceTrekkie!

Just find the angle in the frame in which the mirror is stationary, take the opposite angle, and convert back to the original frame.

 

[SpaceTrekkie]

Okay, so in the rest frame the angle of reflection will be equal to the angle of incidence? So sin(theta_in) = -cos(theta_iin)?

Which would give: P_in = (E/c)(sin(theta_in), -cos(theta_in), 0, 1)
And thus P(prime)_in = (E/c) ( (gamma)sine(theta)in - B), -cos(theta_in), 0, (gamma)(1-B))
And then P(prime_out) = (E/c)(gamma)sine(theta_in), -

Is that the right approach at all? and if so, how does the Vm dot Vl come into play?

 

[tiny-tim]

Hi SpaceTrekkie!

(have a theta: θ and a phi: φ and a gamma: γ )

Okay, so in the rest frame the angle of reflection will be equal to the angle of incidence? So sin(theta_in) = -cos(theta_iin)?

uhh? you mean sinθ_in = sinθ_out.

Which would give: P_in = (E/c)(sin(theta_in), -cos(theta_in), 0, 1)
And thus P(prime)_in = (E/c) ( (gamma)sine(theta)in - B), -cos(theta_in), 0, (gamma)(1-B))
And then P(prime_out) = (E/c)(gamma)sine(theta_in), -

Is that the right approach at all? and if so, how does the Vm dot Vl come into play?

What's B?

v_m·v_l < 0 just means that the light is reflected off the "front" of the mirror rather than the back

 

[SpaceTrekkie]

sorry about the confusion of my notation. B = v/c, and yes, i did mean sin(theta in) = sin (theta_out).

ooo ok, the so the V_m and V_l are not relevant to getting the answer, aside from the fact that they are clarifying the direction of the light.

Was my approach of using the 4-momentum correct?

I think that using that method, when I take the transform, it should transform the y component, and not the x, since the mirror is moving along the y axis, but I am not sure if that is okay to do, since the formulas for the lorentz transformation hold y = y'

 

[tiny-tim]

Was my approach of using the 4-momentum correct?

I think that using that method, when I take the transform, it should transform the y component, and not the x, since the mirror is moving along the y axis, but I am not sure if that is okay to do, since the formulas for the lorentz transformation hold y = y'

Hi SpaceTrekkie!

Yes, using the 4-momentum and transforming it with the Lorentz transformation is exactly the right way to do it.

(the Lorentz transformation only works for 4-vectors such as 4-position and 4-momentum)

x and y are "dummy" indices … you can use any letters …

if you're using x and y, then you must make x the direction of travel.