# 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:
Pphoton,i = (Ei, Ei ni)
Pphoton,f = (Ef, Ef nf)
Pelectron,i = (me, 0)
Pelectron,f = (gamma me, gamma mevi)

## The Attempt at a Solution

Using conservation of momentum, we get this expression:
Pphoton,i + Pelectron,i = Pphoton,f + Pelectron,f

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

Homework Helper
If you first square, you get

|Pphoton,i + Pelectron,i|² = |Pphoton,f + Pelectron,f

How are you going to get |Pelectron,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 = |Pphoton,i + Pelectron,i|² = |Pphoton,f + Pelectron,f

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

Can anyone verify/refute this?

Staff Emeritus
Homework Helper
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:
Pphoton,i = (Ei, Ei ni)
Pphoton,f = (Ef, Ef nf)
Pelectron,i = (me, 0)
Pelectron,f = (gamma me, gamma mevi)

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

Pphoton,i = (Ei, Ei ni)
Pphoton,f = (Ef, Ef nf)
Pelectron,i = (mec2, 0)
Pelectron,f = (γmec2, γmecvf)

Squaring things loses information.

Isn't Pphoton,i + Pelectron,i = Pphoton,f + Pelectron,f
four linear equations in four unknowns, γ and γvf?

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

Last edited:
Staff Emeritus
Homework Helper
You will want the units of photons and electons to match. For n unitless, and all vector components in units of energy:

Pphoton,i = (Ei, Ei ni)
Pphoton,f = (Ef, Ef nf)
Pelectron,i = (mec2, 0)
Pelectron,f = (γmec2, γmecvf)
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 Pphoton,i + Pelectron,i = Pphoton,f + Pelectron,f
four linear equations in four unknowns, γ and γvf?

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