I have a question relating to photons and momentum. I made two Feynman diagrams, I hope I used them right. (Time goes bottom->top, space goes left->right)
In the diagram on the left, a beam-riding/photon sail configuration is shown. A high-energy (high-frequency, low wavelength) photon comes in from the left, then is reflected by the atom, leaving the atom with some of the energy originally within the photon, and the photon with less energy as it leaves.
The diagram on the right depicts the situation I am asking about. A photon stimulates an atom to emit a photon of a similar frequency, giving the atom a kick but taking energy from its energy levels. Is this at all correct/possible? Or do I have everything wrong?
bump?
>>7701715
at first glance it doesn't look like momentum is conserved. Remember, light has momentum too!
>>7701798
Yeah, that's why beam-riding (on the left) works, right?
The photon transfers some of its momentum to the atom.
For things with mass, p=m*v, so it increases the atom's velocity.
For photons and other massless particles, p=E/c, and E=hv (where v is the frequency), so the photon's frequency decreases.
In my diagram on the right, I guess I'm asking if an atom emitting a photon gives it momentum, through a decrease in the energy of the atom.
Lasers work by using photons to cause excited electrons in atoms to release light of a similar phase and energy, right? When a laser releases a collimated beam, does it give it momentum because of the photons leaving it?
>>7702851
You don't just need to worry about conserving the magnitude of momentum, but the direction as well. So if you take out that bottom photon in the right picture it would work.
>In my diagram on the right, I guess I'm asking if an atom emitting a photon gives it momentum, through a decrease in the energy of the atom.
Lasers work by using photons to cause excited electrons in atoms to release light of a similar phase and energy, right? When a laser releases a collimated beam, does it give it momentum because of the photons leaving it?
yes.
>>7701798
Feynman diagrams don't conserve momentum.
Only the sum over all diagrams is physical.
Due to sentences like
>When a laser releases a collimated beam, does it give it momentum because of the photons leaving it?
with
>it gives it
your post is too much work to read and understand.
Let me just point out that the Feymann diagrams in QED which you probably took that kind of picture from are about interaction between photons and electrons. Here the left pic doesn't mate sense because it's always 2 wiggly and one straight line.
In the Lagrangian in my pic this corresponds to the term of the form [math]\psi A\psi[/math], A being the photon and [math]\psi[/math] the electron.
It's also not really necessary to speak of time and space for those diagrams (they represent just terms in some exponential function expansion and the rest is interpretation), what matters is to say which lines are ingoing and which are outgoing (this is what's done in drawing an imaginary "time"-line, horizontally, space doesn't matter, really). From this, and adopting time as you propose, the right hand picture shows two outgoing photons.
And if that helps: Energy and momentum will be conserved. If energy isn't changed internally (excited state of the atom), then yes the particles will have to make up for it by changing kinetic energy/momentum
>>7702891
Sorry about the Feynmann diagrams. I should have said something like "spacetime diagrams with particles". Looking it up, yeah they're about particle interactions, not particles in space over time.
I'm not exactly a professional physicist; I'm just trying to verify if an idea I had was correct.
>>7702959
>I should have said something like "spacetime diagrams with particles"
You can't describe quantum interactions (directly) in terms of spacetime trajectories. The only physical quantity is the sum over diagrams.
>>7702993
Really? Why not? They act like particles enough where the specifics don't really matter, right?
What do you mean by "sum over diagrams"? Do you mean quantities like charge and colour? Why not momentum?
>>7702994
(not him)
The Feynman diagrams each represent a mathematical expression, aking to the terms [math] \frac{x^n}{n!} [/math] in
[math] e^x = \frac{x} {1!} + \frac{x^2} {2!} + \frac{x^3} {3!} + \, ... [/math]
except the terms are much more complicated and the diagrams are used to find out what they really are (humans are better with pictures than with algebra).
If you look at observable in quantum mechanics (field strengths or combined momentum of a state of a bunch of particles), the theory will tell you which pictures to draw and thus which sums to compute and add.
Pic related I guess.
>>7702889
>Feynman diagrams don't conserve momentum
yes it does, each Feynman diagram conserves it separately.
