Angle of reflection with moving mirror, SR

In summary, the conversation discusses the problem of finding the cosine of the reflected angle of a light ray incident on a moving mirror. The approach involves using the 4-momentum and transforming it with the Lorentz transformation. The direction of the light and the mirror's movement are clarified by the use of vm dot vl < 0.
  • #1
SpaceTrekkie
24
0

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?
 
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  • #2
SpaceTrekkie said:
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! :smile:

Just find the angle in the frame in which the mirror is stationary, take the opposite angle, and convert back to the original frame. :wink:
 
  • #3
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?
 
  • #4
Hi SpaceTrekkie! :smile:

(have a theta: θ and a phi: φ and a gamma: γ :wink:)
SpaceTrekkie said:
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 sinθin = sinθout.

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? :confused:

vm·vl < 0 just means that the light is reflected off the "front" of the mirror rather than the back
 
  • #5
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 Vm and Vl 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'
 
  • #6
SpaceTrekkie said:
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! :smile:

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. :wink:
 

1. What is the angle of reflection when a mirror is moving at relativistic speeds?

The angle of reflection is determined by the law of reflection, which states that the angle of incidence (the angle between the incident light ray and the normal to the mirror's surface) is equal to the angle of reflection (the angle between the reflected light ray and the normal to the mirror's surface). This law still applies when the mirror is moving at relativistic speeds.

2. How does the angle of reflection change when the mirror's speed approaches the speed of light?

As the mirror's speed approaches the speed of light, the angle of reflection will approach 90 degrees. This is because at relativistic speeds, space and time become distorted, causing the angle of incidence to approach 0 degrees. Therefore, the angle of reflection will approach 90 degrees in order to maintain the law of reflection.

3. Does the angle of reflection change if the mirror is moving towards or away from the incident light?

No, the angle of reflection will not change if the mirror is moving towards or away from the incident light. This is because the law of reflection is independent of the motion of the mirror or the incident light. The angle of reflection is only dependent on the angle of incidence and the properties of the mirror's surface.

4. Can the angle of reflection ever exceed 90 degrees with a moving mirror?

No, the angle of reflection cannot exceed 90 degrees regardless of the mirror's speed. This is because the law of reflection states that the angle of reflection must always be equal to the angle of incidence, and both angles cannot be greater than 90 degrees. Additionally, at relativistic speeds, the angle of incidence will always approach 0 degrees, making it impossible for the angle of reflection to exceed 90 degrees.

5. How does the angle of reflection with a moving mirror differ from a stationary mirror?

The angle of reflection will be the same for a moving mirror as it is for a stationary mirror, as long as the mirror's speed is not approaching the speed of light. However, at relativistic speeds, the angle of reflection will approach 90 degrees, whereas with a stationary mirror it will remain constant at the angle of incidence. This is due to the distortion of space and time at high speeds.

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