i have a question about compton scattering. so i understand that in compton scattering the angle of deflection of the photon is related to the difference in the wavelength of the photon before and after. my question is why, physically, the photon gets deflected? i mean in the cases of deflection what is the physical mechanism?

If you imagine watching the interaction in the centre of momentum frame then we have an incoming photon and electron interacting with each other. The interaction entangles the two wave functions and we get a new state that is in effect an excited state of the two objects i.e. it is no longer separable into an electron and photon but rather it's a combination of the two.

Then this new state can decay in a variety of ways subject only to the requirements that the total momentum remains zero and the total energy is conserved.

Hmm, on second thoughts I'm not sure if this approach leads anywhere interesting ...

@Relativisticcucumber squack

@Relativisticcucumber Oh wait, yes it does work.

The entangled state has to evolve back into a photon and electron with their energies unchanged, but since the only restriction on momentum is that it sums to zero the directions of the outgoing photon and electron can be different from the incoming directions.

That is, in the COM frame the interaction could look like this:

In the COM frame the energies and speeds of the two particles are the same before and after, but if you now observe this from a frame moving wrt the COM frame the speed of the electron and the doppler shift of the photon will be different before and after due to the different directions of the velocities.

How does one integrate the RHS?

what does $\mathrm{d}g_{ab}$ mean? What are you integrating over?

@ACuriousMind an attempt to do the inverse of $$- 2 \frac{\partial (L_M \sqrt{-g})}{\partial g^{\mu \nu}} \frac{1}{\sqrt{-g}}$$

yeah, that's not how the math works

ah ...

whats the inverse of $\partial / \partial g^{ab}$

ill-defined

partial derivatives don't have "inverses"

if you want to invert some sort of differential operator the general method is via Green's functions, but that's really not what you should be doing here

but you don't need to do anything inverting here, really, you just need to write down the correct Lagrangian :P

see e.g physics.stackexchange.com/q/91116/50583 for discussions of the correct Lagrangian density for a point particle in curved spacetime

When is the distribution function route a valid method to define the stress energy tensor?

what is "the distribution function"?

$$T^{\mu\nu} = \int \dfrac{\text{d}^3p}{E} F(\vec{x},\vec{p})p^{\mu}p^{\nu}$$

oh, you mean a phase space distribution

ah yes my bad

Probably valid for dust?

e.g. for a perfect fluid you'd also need something that produces a pressure term

Should your $T^{\mu \nu}(x)$ not be $T^{\mu \nu}(x) = \frac{1}{\sqrt{-g(x)}} \sum_a m_a \int d \tau_a \frac{dX_a^{\mu}}{d \tau_a} \frac{dX_a^{\nu}}{d \tau_a} \delta^4(x - X_a(\tau_a))$, coming from $T^{\mu \nu} = \frac{2}{\sqrt{-g(x)}}\frac{\delta}{\delta g_{\mu \nu}(x)} S$ where $S = - \sum_a m_a \int d \tau_a \sqrt{-g_{\rho \sigma}(X_a) \frac{dX_a^{\rho}}{d \tau_a} \frac{dX_a^{\sigma}}{d \tau_a} }$ and using $\dot{X}^2 = - 1$

Good luck thinking of all this backwards without knowing it forwards first

(yes, it should)

@ACuriousMind Why do more symmetries mean less divergences in view of renormalisaton?

@RyderRude where did you read that and what is that supposed to mean? :P

@ACuriousMind I read this a while back on the wikipedia of quantum gravity. It was something like "If a theory has additional symmetries, there are higher chances that it is renormalisable"

I don't think that's true

for example, supersymmetry improves a theory in view of renormalisability

oh, i must have misunderstood :P

yes, but supersymmetry is extremely special in that regard

oh

so it's just not any symmetries

in general I can see a vague idea here where symmetries constrain the admissible Feynman diagrams and if you have "fewer" diagrams altogether you of course have fewer diagrams that can diverge

but you can write down plenty of non-renormalizable terms that don't break symmetries of otherwise renormalizable theories - just add some high $\phi^{4n}$ terms to some renormalizable $\phi^4$ theory, for instance

@ACuriousMind Yeah. So symmetries just help us constrain the terms allowed in the lagrangian. But there is not much correlation between these constraints and divergences and renormalisation.

@ i read something like "When you have more symmetries, you can absorb infinities more effectively. I often read, there may be hidden non-trivial symmetries in the perturbation series of GR, which may lead to the absorption of all infinities"

that doesn't really make any sense to me

the idea is really just that symmetries can "protect" certain things from having to be renormalized

like gauge bosons can't acquire masses during renormalization due to gauge symmetry

yeah, this much makes sense. I don't know what they were talking about when they said the infinities in the GR perturbation series may get non-trivially absorbed

as if, the proof that that perturbation series is infinite is somehow an insufficient proof. I guess they meant substituting for the coupling constants some non-trivial energy dependent formula, which would lead to the absorption of all infinities

I think someone saw what supersymmetry was doing and generalized this to "symmetries" without basis :P

because what happens in supersymmetry is that roughly the bosonic and fermionic contributions cancel each other

oh

how does this help gravity? Does the gravitino term get added in the Einstein Hilbert Lagrangian?

i.e. you get a "bosonic divergence" and via supersymmetry you get a "fermionic divergence" with the opposite sign and they "absorb" each other

oh. this makes sense But this is not because some hidden non-trivial symmetry like they were saying :P. This must happen there we modified the Einstein Hilbert Lagrangian to include Gravitino terms?

so there's nothing hidden. we explicitly modified the theory :P

and the infinities still don't completely cancel, I guess. otherwise, we wouldn't look into string theory

10D SUGRAS are non renormalizable

physicsforums.com had a homework section

you can try there too

@RyderRude yes, it's not that supersymmetry magically makes all theories renormalizable

it's just that it render a bunch of theories renormalizable you wouldn't expect to be, or makes parameters immune to renormalization that otherwise would change

cf. "non-renormalization theorems"

@tush A chat that accepts questions about exercises is Problem Solving Strategies; also, see this meta post for alternatives to physics.SE.

thanks

hi -- say you have operators A,B,C and you take (1) [A,C] and (2) [B,C] and say A and B are compatible. then why if (1) is zero, then (2) is not zero

or why does (1) being zero not necessarily mean (2) must be zero

also honk

@Relativisticcucumber why would it mean that?

whenever you have questions like this, it's important to remember the identity commutes with everything

so if you choose $A$ as the identity operator and just any two non-commuting operators $B$ and $C$, you have $[A,B] = 0$ but $[B,C]\neq 0$

wait where does the identity come into play

oh i was too hasty with my reply sorry

reading now

well if we can measure A and B simultaneously then i feel their properties should be somewhat shared but hm i do follow ur counter example

i shall try to think of a more involved counter example. i think that will help me

that $A$ and $B$ commute just means that there is a joint eigenbasis

that $A$ and $C$ commute means that there is a joint eigenbasis for them, too

@Relativisticcucumber another counterexample is rotation operators - all the $L_i$ commute with $L^2$, but none of the $L_i$ commute with each other

LOL yes im reading that section

you have found the culprit of my confusion

wait actually no that is not my confusion and i just realized my issue i think

thank you

sorry for chaoticness i am in dire need of rest

@SillyGoose quack

I had to do that once in a lifetime

honk