Relativistic momentum and energy

[bhsmith]

Homework Statement

A gamma ray photon with wavelength=2.0 * 10^-15m is traveling in the positive x direction undergoes a one dimensional collision with a stationary proton.
a.) find the relativistic momentum and energy of the photon and proton before the collision.
b.) assuming the momentum is exchanges between photon and proton, find the relativistic momentum and energy of the photon and proton after the collision.

Homework Equations

p(momentum)= mv(gamma)

Energy= M(gamma)

Where Gamma= 1/(sqrt(1-v^2))

The Attempt at a Solution

I'm not sure how to use these equations because i don't have the velocity of either the photon or the proton.

 

[fzero]

The speed of the photon is always the same, so if you think about it, you do know what it is! You're also given the initial velocity of the proton, just reread the problem if you missed it.

Note, the photon has zero mass, so it's momentum isn't given by the formula above. However, the equation

E^2 - (pc)^2 = (mc^2)^2

is always valid. In order to solve the problem you also need to use conservation of energy and momentum.

 

[bhsmith]

Wow, your exactly right. I don't know how i overlooked that.
But i don't understand how to use the equation for the conservation of momentum. The mas is zero, but i still have to have the momentum to find the energy or vice versa.

 

[fzero]

The momentum of a photon is inversely related to its wavelength. The constant of proportionality is one of the famous constants.

 

[bhsmith]

I think i remember something like that. Isn't it the debroglie wavelength equation where the momentum=h/wavelength?

 

[fzero]

Yes, that's the one. The equation for photons was known before de Broglie, but he was the one that applied it to the wave interpretation of massive objects.

 

[bhsmith]

Ok great! i got the first part down. Now for after the collision, I think i would use the
conservation equation where the total energy would be the same after as before and same goes for momentum.
But would the momentum of the photon be the same still? since it travels at the speed of light and has zero mass or would it change?

 

[fzero]

The final momentum of the photon will differ from the initial momentum both in direction and magnitude. But its energy also changes according to the equation in post #2.

 

[bhsmith]

I'm not quite getting this. Thanks for getting me this far though.
But i would add the energy and momentum of the photon and proton before the collision to get the total energy and momentum. Correct? then i know that has to be equal to the momentum and energy after the collision, but i don't understand how to find the separate momentum and energy for the two particles?

 

[fzero]

Write down the equations that you get. Since momentum conservation is a vector equation, it's really 2 conditions on the final momentum components. Energy gives a 3rd constraint. There are 3 unknowns: the magnitudes of the final momenta and the angle between the initial and final photon.

 

[bhsmith]

Ah ok, I'm really trying to understand this.

I would have P(total)= sqrt(E(total)^2 - m(total)^2)
But shouldn't i already know the final momentum because it should be equal to the momentum before the collision.

But Ptotal=P'total=P'(photon)+P'(proton) and the same thing goes for energy.
But so would if the speed for the photon is still the speed of light and that should be equal to Energy since it's massless then that would give me P(photon)=0... because P=mv(gamma) and mass is zero. I am so confused.

 

[fzero]

Ah ok, I'm really trying to understand this.

I would have P(total)= sqrt(E(total)^2 - m(total)^2)

This equation doesn't hold in general. For each particle we have

E_i^2 - |\vec{p}_i c|^2 = (m_i c^2)^2 ,

but

\left( \sum_i E_i \right)^2 - \left|\sum_i\vec{p_i} c\right|^2 \neq \left( \sum_i m_i c^2\right)^2,
What is true is that

\left( \sum_i E_i \right)^2 - \left|\sum_i\vec{p_i} c\right|^2 = \left( \sum_i E'_i \right)^2 - \left|\sum_i\vec{p'}_i c\right|^2.

where initial quantities are on the LHS and the primed quantities on the RHS are from the final state. This equation just follows from energy and momentum conservation.

But shouldn't i already know the final momentum because it should be equal to the momentum before the collision.

But Ptotal=P'total=P'(photon)+P'(proton) and the same thing goes for energy.
But so would if the speed for the photon is still the speed of light and that should be equal to Energy since it's massless then that would give me P(photon)=0... because P=mv(gamma) and mass is zero. I am so confused.

P=mv(gamma) is not valid for a photon exactly because the photon is massless. Let the initial photon momentum be \vec{k}. The relation to the wavelength is through the magnitude

| \vec{k}| = \frac{h}{\lambda}.

You will also need to find a convenient way to parametrize the directions of the particles to solve the problem.

The energy-momentum relation for the photon is

E_{\tex{photon}}^2 - |\vec{k} c|^2 = 0,

which is consistent with the relation

E_{\tex{photon}} = h f,

where f is the frequency of the photon.

 

[bhsmith]

OK, so say my total Energy is 1.24E-33 before the collision. I would have the same after the collision of course.
So i would have an equation for
1.24E-33= E'photon + E' proton = hf + m(gamma)
But by doing this then i can solve for photon energy and i have the mass of the proton, so i can solve for gamma which will give me the velocity. is this right?

 

[bhsmith]

I did that calculation, and the Final energy was not equal to the initial Energy. That must not be right.

 

[fzero]

OK, so say my total Energy is 1.24E-33 before the collision. I would have the same after the collision of course.
So i would have an equation for
1.24E-33= E'photon + E' proton = hf + m(gamma)
But by doing this then i can solve for photon energy and i have the mass of the proton, so i can solve for gamma which will give me the velocity. is this right?

I wouldn't worry about velocity and \gamma, since we're expected to solve for the momenta and energies. We can therefore work entirely in terms of the momenta, using

E'_{\tex{proton}} = \sqrt{ (mc^2) + |\vec{p'} c|^2 }.

Also, try to avoid substituting numerical values for symbols until you're at your final result. Otherwise it just makes the algebra harder.

 

[fzero]

I did that calculation, and the Final energy was not equal to the initial Energy. That must not be right.

Use k for the photon momentum and p for the proton, then the initial total energy is

|\vec{k}| c + m c^2 .

The final total energy is

|\vec{k'}| c + \sqrt{ (m c^2)^2 + |\vec{p'}c|^2} .

 

[bhsmith]

I think i have gotten it, turns out i converted something wrong last time i tried. It worked out just fine the second time around. Thank you so much for all of your help!