Light Reflection: Energy and Momentum Conservation
Most recent answer: 02/16/2013
Q:
We know that the wall has some coefficient of restitution n shows inelastic collision with every object.But when the particles of light collide with the wall its velocity is not changed it remains const i.e. 3x10^8 m/s.what is the reason behind the const velocity of light even after undergoing inelastic collision??
- Kushmeet singh saluja (age 17)
Agra,India
- Kushmeet singh saluja (age 17)
Agra,India
A:
There are a couple different questions here.
1. Light can bounce elastically off a surface, with a small transfer of energy.
2. There can be inelastic events, such as when the light is absorbed by the material.
Let's look at case (1), where the light has momentum p0 and bounces directly off something of mass M. After the collision, the mass will recoil a little, picking up some momentum and a tiny bit of kinetic energy. The final momentum of the light is not quite -p0but rather -p0Mc/(2p0+Mc). Assuming M is reasonably big, that's not far from -p0.
The energy lost by the photon is then 2c(p0)2/(Mc+2p0), which is very close to 2(p0)2/M. That's just what would be expected classically when the photon transfers momentum 2p0 to the object of mass M. It all hangs together,
For case (2) we have to think about quantum events in the wall, where lumps of energy get absorbed. Then almost all the energy (p0c) goes into internal modes of the wall, and only a tiny bit into the wall recoil.
The reason you rarely get partially inelastic collisions, like those where a ball bounces of a wall weakly, has to do with the quantum states available to take up the energy in the wall. There are types of scattering (Raman) and fluorescence that are partially inelastic, but they're not really typical reflections.
Mike W.
1. Light can bounce elastically off a surface, with a small transfer of energy.
2. There can be inelastic events, such as when the light is absorbed by the material.
Let's look at case (1), where the light has momentum p0 and bounces directly off something of mass M. After the collision, the mass will recoil a little, picking up some momentum and a tiny bit of kinetic energy. The final momentum of the light is not quite -p0but rather -p0Mc/(2p0+Mc). Assuming M is reasonably big, that's not far from -p0.
The energy lost by the photon is then 2c(p0)2/(Mc+2p0), which is very close to 2(p0)2/M. That's just what would be expected classically when the photon transfers momentum 2p0 to the object of mass M. It all hangs together,
For case (2) we have to think about quantum events in the wall, where lumps of energy get absorbed. Then almost all the energy (p0c) goes into internal modes of the wall, and only a tiny bit into the wall recoil.
The reason you rarely get partially inelastic collisions, like those where a ball bounces of a wall weakly, has to do with the quantum states available to take up the energy in the wall. There are types of scattering (Raman) and fluorescence that are partially inelastic, but they're not really typical reflections.
Mike W.
(published on 02/16/2013)
Follow-Up #1: constant light speed
Q:
Hey mike if we focus on the second possibility that some particles may absorb in the wall this implies that the intensity of the light will decrease so this may lead to decrease in velocity of light but still it is taken as const. so plzz clear me...
- Kushmeet singh saluja (age 17)
Agra,India
- Kushmeet singh saluja (age 17)
Agra,India
A:
Kushmeet- Whoops, I didn't really answer your previous question directly. I showed that conservation of momentum and energy are consistent with a constant light speed, given special relativity, but that doesn't really say why the speed is constant, including when the intensity changes.
Here's two ways to think of it.
1. Classically, light is just a wave solution to Maxwell's equation for electricity and magtnetism. The speed of the wave comes directly out of those equations. The equations are linear, which means that if you increase or decrease the field strengths the properties (like speed) of the wave don't change.
2. You can think of light as made of photons, which have zero rest mass. A zero rest-mass particle can only have energy and momentum if it travels at the speed of light.
Mike W.
Here's two ways to think of it.
1. Classically, light is just a wave solution to Maxwell's equation for electricity and magtnetism. The speed of the wave comes directly out of those equations. The equations are linear, which means that if you increase or decrease the field strengths the properties (like speed) of the wave don't change.
2. You can think of light as made of photons, which have zero rest mass. A zero rest-mass particle can only have energy and momentum if it travels at the speed of light.
Mike W.
(published on 02/23/2013)