- #1

TheMercury79

- 24

- 5

- Homework Statement
- What is the squared 4-momentum transfer between the particles

- Relevant Equations
- 4-momentum vector P = (E/c, ##\vec p##)

In a head-on collision between the proton and electron, what is the squared 4-momentum transfer between the two particles.

Starting with the difference in momentum of the electron with the 4-vectors before and after the event: $$(P-P')^2=P^2+P'^2-2P\cdot P'$$

The circumstances are such that the energies are large enough for the masses of the electron and proton to be ignored, thus ##E\sim pc## and ##E'\sim p'c##. Where ##E## is the energy of the electron before collision, ##E'## the energy of the electron after collision and ##\theta## is the angle between the the electron before and after the collision. Then:$$(P-P')^2=m_e^2c^2+m_e^2c^2-2(\frac{EE'}{c^2}-\vec p \cdot \vec p')\sim -2\frac{EE'}{c^2}(1-cos\theta)$$

The squared momentum transer would accordingly be ##Q^2=-2\frac{EE'}{c^2}(1-cos\theta)##

But I was told it is ##Q^2=-2\frac{EE'}{c^2}(1+cos\theta)##. These differ only by a plus and minus sign.

I think it is a bit weird that in my calculation above, I didn't have to include the proton at all. But then again I figure the momentum transfer is what the electron transfers to the electron (correct me if I'm not using the term momentum transfer correctly).

I am thinking I could perhaps use Mandelstam variables and include the proton in the calculation and see if I somehow end up with a positive cos-term instead of the negative.

Or is my calculation ##Q^2=-2\frac{EE'}{c^2}(1-cos\theta)## correct?

Starting with the difference in momentum of the electron with the 4-vectors before and after the event: $$(P-P')^2=P^2+P'^2-2P\cdot P'$$

The circumstances are such that the energies are large enough for the masses of the electron and proton to be ignored, thus ##E\sim pc## and ##E'\sim p'c##. Where ##E## is the energy of the electron before collision, ##E'## the energy of the electron after collision and ##\theta## is the angle between the the electron before and after the collision. Then:$$(P-P')^2=m_e^2c^2+m_e^2c^2-2(\frac{EE'}{c^2}-\vec p \cdot \vec p')\sim -2\frac{EE'}{c^2}(1-cos\theta)$$

The squared momentum transer would accordingly be ##Q^2=-2\frac{EE'}{c^2}(1-cos\theta)##

But I was told it is ##Q^2=-2\frac{EE'}{c^2}(1+cos\theta)##. These differ only by a plus and minus sign.

I think it is a bit weird that in my calculation above, I didn't have to include the proton at all. But then again I figure the momentum transfer is what the electron transfers to the electron (correct me if I'm not using the term momentum transfer correctly).

I am thinking I could perhaps use Mandelstam variables and include the proton in the calculation and see if I somehow end up with a positive cos-term instead of the negative.

Or is my calculation ##Q^2=-2\frac{EE'}{c^2}(1-cos\theta)## correct?

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