Special Relativity- Photon/Mirror Collision

In summary, a photon with frequency ν and angle of incidence θ is reflected from a mirror without changing its frequency. The momentum transferred to the mirror can be calculated using the conservation of momentum law, resulting in a final momentum of 2hνcosθ. The mirror's mass and original 4-momentum do not need to be known.
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Homework Statement


A photon of frequency ν is reflected without change of frequency from a mirror, with an angle of incidence θ. Calculate the momentum transferred to the mirror.

Homework Equations


E= hν
Conservation of four-momentum

The Attempt at a Solution


If the mirror is in the x-y plane, the incident photon will have initial 4-momentum
Pγ = (hν , hνsinθ , 0 , -hνcosθ) as the sum of the squares of the spatial momenta must be equal to the square of the time coordinate momentum.
The initial 4-momentum of the mirror:
Pm = (m , 0 , 0 , 0) as it is initially at rest in the chosen frame of reference and m is the mass of the mirror.

After the collision:
Pγ' = ( hν , hνsinθ, 0 , hνcosθ)
Pm' = (E , p)

Applying conservation law:
Pm + Pγ = Pγ' + Pm'
(Pm')^2 = (Pm)^2 + (Pγ)^2 + (Pγ')^2 + 2Pm (Pγ-Pγ') - 2Pγ⋅Pγ'
-E^2 + p^2 = -m^2 + 0 + 0 +2⋅0 -2{ -(hν)^2 + (hνsinθ)^2 - (hνcosθ)^2} (p^2 is the square of the norm of the spatial momenta components)
-E^2 + p^2 = -m^2 + 4(hνcosθ)^2

This is where I don't get what to do. I know that E^2 = m^2 + p^2 but if I substitute that in, the momentum term disappears and I'm left with 4(hνcosθ)^2 = 0.

I also know I get the right answer if E=m, then p= 2hνcosθ , but I don't see how I am able to say that. Is it because the mirror is not a particle in its own right, but rather a system? Or does I have to change to a different reference frame where the mirror will be stationary relative to it?
 
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  • #2
You do not need to know the mirror's mass or original 4-momentum. You also do not need to square any 4-momentum. All you need to do is to apply conservation of momentum.
 

Related to Special Relativity- Photon/Mirror Collision

1. What is special relativity?

Special relativity is a fundamental theory in physics that explains the relationship between space and time. It was developed by Albert Einstein in the early 20th century and is based on two main principles: the laws of physics are the same for all observers in uniform motion, and the speed of light is constant for all observers regardless of their relative motion.

2. What is a photon/mirror collision in special relativity?

A photon/mirror collision is a thought experiment often used to explain the concept of time dilation in special relativity. It involves a photon of light being emitted from a source and bouncing back and forth between two mirrors at the speed of light. According to special relativity, the observer of this experiment will measure the photon's travel time as longer than the observer on the moving photon.

3. How does special relativity affect the laws of physics?

Special relativity does not change the fundamental laws of physics, but it does change our understanding of space and time. It introduces the concept of time dilation, where time appears to pass slower for objects in motion relative to an observer. It also explains the phenomenon of length contraction, where objects in motion appear shorter in the direction of motion. These effects are only noticeable at high speeds, close to the speed of light.

4. What are the implications of special relativity in everyday life?

Special relativity has many practical applications in modern technology, such as in GPS systems and particle accelerators. It also plays a crucial role in our understanding of the universe, including the behavior of objects moving at high speeds and the concept of space-time. However, its effects are only noticeable at very high speeds and do not significantly impact our daily lives.

5. Are there any limitations to special relativity?

Special relativity is a well-tested and highly accurate theory, but it does have limitations. It only applies to objects in constant, uniform motion and does not take into account the effects of gravity. It also does not provide a complete explanation for the behavior of very small particles, which require the more comprehensive theory of general relativity. However, for most everyday situations, special relativity is a reliable and essential tool for understanding the physical world.

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