Why can’t you solve 2 dimension conservation of momentum questions using the linear momentum equation, e.g. m1v1+m2v2= new m1V1+ new m2V2?

This subtle question is one of the best questions that combines physics and math.

m1v1 + m2 v2 = m1 u1 + m2u2 in vector notation.

In fact, it has two main difficulties:

First: We have 4 unknowns, u1x, u1y, u2x and u2y and only 3 equations., Conservation of vector momentum and conservation of scalar energy.

This means that there is no unique deterministic solution. There are an infinite number of solutions.

Second: in the general case, the speed of particles falls within the relativistic range where Einstein's special relativity is a must.

But don't worry because the solution can be found if we treat the problem statistically by assuming that a statistical variable is known i.e. the scattering angle of one of the two particles Phi or Theta [1,2[.

We now have 4 equations in 4 unknowns and the solution is simple and straightforward.

The most striking example is the so-called Compton scattering which is explained as In Compton scattering the photon changes wavelength and decays into a lighter mass.

Here we apply Compton's relativistic scattering model, which is the only known case where the photon loses only part of its energy and part of its momentum to the electron.

It is known as the elastic collision model of the classic billiard balls shown in Fig. 1.

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Fig. 1 Classic model of Compton scattering billiard balls.

The increase in the wavelength of the photons is calculated from the conservation E and P before and after collision, Lambda and Lambda *, and is given by,

Lambda * - Lambda = h (1-cos phi) / m (0) C . . . . . . . . . (1)

Eq.1 is the statistical solution to the question which is tested experimentally and has been found to be correct.

We all know that this suggestion that the photon has a moving mass m = hf / c ^ 2 brings many problems from our colleagues in the field of QM where they assume m to be literally zero.

1- Can a photon have an infinite lifetime? Questions and Answers, Quora, 2022.

2- Compton scattering, ResearchGate, Q&A, 2021.

The short answer to the question is that it is ill-defined due to the infrared divergences of the electromagnetic field. How to deal with these, however, is known and is presented in any textbook on the subject.

The statement that the photon is a massive particle is, simply, wrong.

The longer answer, that adresses the issues raised in the accompanying text, is, now, also, a standard exercise in relativistic kinematics-though its experimental verification was the reason Arthur Compton was awarded the Nobel Prize in Physics.

It goes as follows:

Electromagnetic waves are a superposition of a macroscopic number of photons; one photon isn't an electromagnetic wave. Electromagnetic waves are classical objects, they're solutions to Maxwell's equations; photons are quantum objects, they're the 1-particle excitations of the electromagnetic field.

In particular, photons are massless particles, so the non-relativistic approximation doesn't make sense for them. It is possible to check whether it does make sense for electrons.

In Compton scattering it is possible to study the approximation where the process is electron + photon -> electron + photon. What's conserved is the total energy, momentum and angular momentum, as well as electric charge and lepton number.

So the energy of the initial state, hv + sqrt(p^2c^2+m^2c^4), is equal to the energy of the final state,

hv' + sqrt(p'^2c^2 + m^2c^4).

So the initial state contains one photon and one electron and the final state contains one photon and one electron. That is the simplest realization.

In the frame where the electron of the initial state is at rest (which always exists, since the electron is a massive particle), it's possible to compute the shift in frequency between the photon of the initial state and the photon of the final state

h(v-v') = sqrt(p'^2c^2+m^2c^4)

The (modulus of the) momentum of the initial photon, q, is related to its energy by E = hv = qc; similarly the (modulus of the) momentum of the final photon, q', is related to its energy by E' = hv' = q'c since the photon is a massless particle.

Momentum conservation then implies that

q = p'cosθ + q'cosθ'

0 = p'sinθ - q' sinθ'

where the angles are the angles that the momenta make with the direction defined by the direction of the initial photon.

These equations are necessary and sufficient for determining v', θ and θ' and hence p' and q', in terms of q.

So there's no problem, the equations are consistent, so, indeed, there's no issue of fractionalization. Energy and momentum of the electron and the photon take continuous values because spacetime isn't compact.

It was the experiment-carried out by Compton-that verified this calculation and measured the frequency shift and the angle of the final photon-that showed that conservation of energy and momentum does indeed hold for processes that involve quantum particles (whose properties depend on Planck's constant not being negligible) and, thereby, showed that Bohr, Kramers and Slater were wrong-and, in particular-that it ought to be expected to hold in beta decay, also.

Normally, I don't comment on answers, but if the answer came from Tours, France (a country I love and consider my second home), it would be different.

The answer itself is interesting, but unfortunately wrong.

I sense a certain presence of the iron guards of Schrödinger's PDEs, speaking in the name of God without reasoning!

Shut up and calculate, for these are the orders from heaven.

The answer I quoted, taken from a Google search and Wikipedia, is:

(Yes, in the Compton effect, an electron can only absorb part of the energy of the incident photon, not all of it. The absorbed energy is transferred to the recoiling electron, and the photon loses energy and changes direction.)

The meaning or conclusion is that the hypothesis of the infinite life of the photon and perhaps the whole theory of QM collapses.