Four-Momenta derivation: is this mathematical step even valid?

[whyamihere]

Homework Statement

Hey all, this is my first post to the forum, but I've been reading for years. Anyways, I was reviewing a solution to a Compton Scattering derivation we learned in class, and am still trying to make sense out of it.

Homework Equations

Given, we have the initial and final four-vectors of the photon and electron:
P_photon,i = (E_i, E_i n_i)
P_photon,f = (E_f, E_f n_f)
P_electron,i = (m_e, 0)
P_electron,f = (gamma m_e, gamma m_ev_i)

The Attempt at a Solution

Using conservation of momentum, we get this expression:
P_photon,i + P_electron,i = P_photon,f + P_electron,f

This is where I get confused. Our tutor solved for the final electron term and then squared both sides like so:
|P_electron,i|² = |P_photon,f + P_electron,f - P_photon,i|²

Is this action legal? Don't you have to square them first, then solve for the final electron term? Please, someone help clarify this. Everyone I ask seems to not even think twice about it. Is it really just a trivial step?

 

[CompuChip]

If you first square, you get


|P_photon,i + P_electron,i|² = |P_photon,f + P_electron,f|²

How are you going to get |P_electron,f|² from that?

(Note that |a + b|² = |a|² + |b|² + (some very unpleasant cross terms) )

 

[whyamihere]

Thanks CompuChip.

Yes, that makes sense mathematically. It's the physics that I still am not getting. What is the reason for squaring both sides? Does that give us the square of the rest mass?

 

[whyamihere]

Actually, now I am convinced that I am right and my tutor made a mistake.

In order for this to be Lorentz invariant, the threshold energy s = |P_photon,i + P_electron,i|² = |P_photon,f + P_electron,f|²

So we MUST square each side first, otherwise it is not invariant.

Can anyone verify/refute this?

 

[vela]

The square of any four-vector is an invariant, and more generally, the product of any two four-vectors is an invariant.

 

[whyamihere]

That really clarifies everything. Thanks vela!

 

[Phrak]

Homework Equations

Given, we have the initial and final four-vectors of the photon and electron:
P_photon,i = (E_i, E_i n_i)
P_photon,f = (E_f, E_f n_f)
P_electron,i = (m_e, 0)
P_electron,f = (gamma m_e, gamma m_ev_i)

You will want the units of photons and electons to match. For n unitless, and all vector components in units of energy:

P_photon,i = (E_i, E_i n_i)
P_photon,f = (E_f, E_f n_f)
P_electron,i = (m_ec², 0)
P_electron,f = (γm_ec², γm_ecv_f)

Squaring things loses information.

Isn't P_photon,i + P_electron,i = P_photon,f + P_electron,f
four linear equations in four unknowns, γ and γv_f?

However, you haven't actually stated what your givens and unknowns are.

 

[vela]

You will want the units of photons and electons to match. For n unitless, and all vector components in units of energy:

P_photon,i = (E_i, E_i n_i)
P_photon,f = (E_f, E_f n_f)
P_electron,i = (m_ec², 0)
P_electron,f = (γm_ec², γm_ecv_f)

It's common to use units where c=1 and drop the explicit factors of c from expressions. For example

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

becomes

m^2 = E^2 - p^2.

Squaring things loses information.

Sometimes you don't care about that information. A common trick in relativity calculations is to arrange the four-vectors so that a well-placed 0 will wipe out terms you don't really care about when you square the equation.

Isn't P_photon,i + P_electron,i = P_photon,f + P_electron,f
four linear equations in four unknowns, γ and γv_f?

However, you haven't actually stated what your givens and unknowns are.