Conservation of energy and momentum of a proton

[forrealfyziks]

Homework Statement

Consider a head-on, elastic collision between massless photon (momentum Pnot and energy Enot) and a stationary free electron. Assuming that the photon bounces directly back with momentum p (in the direction of -Pnot) and energy E, use conservation of energy and momentum to find p.

Homework Equations

massless: E=pc
E=$$\gamma$$mc²
p=$$\gamma$$mu
Maybe relevant...but probably not : E=hf

The Attempt at a Solution

I'm assuming that p is the momentum of the electron, because it is the only momentum not denoted in the problem. Note: Pnot is momentum of photon before collision, p is momentum of photon after collision, and m is the mass of the electron. I've set up this:

(Pnot)c + mc² = pc + $$\gamma$$mc²
Pnot=p_e - p

I remove a c from the top equation and isolate Pnot on the left side.
Pnot= p + $$\gamma$$mc - mc

I plug this into the second equation

p + $$\gamma$$mc - mc = p_e - p = $$\gamma$$mu - p

From here I try a couple different things, but my main method seems to be putting p on one side and pulling things out

2p= m($$\gamma$$u - $$\gamma$$c + c)

thanks for any help

 

[Jhero]

you can give me

Thank you for your post. I am a scientist who specializes in physics, and I would be happy to help you with this problem.

Firstly, I would like to clarify that p is indeed the momentum of the photon after the collision, not the electron. This is because the photon is the only particle that changes direction and momentum in this collision, while the electron remains stationary. Therefore, the momentum conservation law can be written as:

Pnot = p

Now, let's look at the energy conservation law. We know that the energy of a photon is given by E=pc, and the energy of an electron is given by E=\gamma mc^2, where \gamma is the Lorentz factor. In this case, the electron is initially at rest, so its energy is simply mc^2.

Using these equations, we can write the energy conservation law as:

Enot + mc^2 = E + mc^2

Note that the energy of the photon after the collision is the same as its energy before the collision, since the photon is massless and therefore its energy does not change.

Now, we can substitute E=pc into this equation and solve for p:

Enot + mc^2 = pc + mc^2
Enot = pc
p = \frac{Enot}{c}

Therefore, the momentum of the photon after the collision is equal to the initial energy of the photon divided by the speed of light.

I hope this helps you with your problem. If you have any further questions, please don't hesitate to ask.