# Compton scattering; conservation of momentum violated?

While reading the derivation of the formula $\lambda' - \lambda = \frac{h}{ m_ec}(1-cos(\theta))$ on Wikipedia, they point out that the momentum gained by the electron is larger than the momentum lost by the photon:

$$p_e=\frac{\sqrt{h^2(\nu-\nu')^2 +2h(\nu-\nu')m_ec^2}}{c} > \frac{h(\nu-\nu')}{c} = p_{\gamma}-p_{\gamma'}$$

How is this not a violation of conservation of momentum?

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ShayanJ
Gold Member
I don't think that's the right of way talking about it! When we consider SR in analysing phenomena, we should keep in mind that it is no longer true that energy and momentum are separately conserved, but it is the total 4-momentum $P^\mu=(\frac E c, \vec p)$ which is conserved. So such a comparison has no meaning and violates nothing!

jtbell
Mentor
How is this not a violation of conservation of momentum?

Momentum is a vector quantity. The incoming and outgoing photons are in different directions. The difference in the magnitudes of their momenta is not a meaningful physical quantity, as far as I know.

• PeroK
Staff Emeritus
2021 Award
Shyan, what you wrote is wrong. It is not necessary to reply to every single question here, particularly if you don't know the answer. If the energy-momentum 4-vector (or any 4-vector) is conserved, each and every component is separately conserved.

jtbell - and Wikipedia - has it right. Comparing the magnitudes of two vectors is, in general (and in this particular case) comparing them along two different directions. You don't expect equality here.

• Matterwave and Nugatory
jtbell
Mentor
If the energy-momentum 4-vector (or any 4-vector) is conserved, each and every component is separately conserved.

...but the magnitude of the 3-momentum is not. (just to make this explicit)

Staff Emeritus