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K^2 is offline
Nov9-12, 04:55 PM
Sci Advisor
P: 2,470
Quote Quote by Studiot View Post
Would you like to put up some maths to back up this proposal?
So dp/dt = F isn't enough? I understand that you might be tempted to point out that dp/dt = ƩF, and ƩF=0. But do keep in mind that d/dt is a linear operator, and we need to decompose it if we want to look at individual flows rather than the net change.

Or do you want me to go deeper and write out some Feynman diagrams for a virtual photon exchange, showing that any application of force results in momentum flow?

Or do you need me to go a step deeper and derive relation between stress tensor and momentum flow? If you want me to be creative, I can do that using the fact that stress tensor is a conserved quantity under coordinate transformation, while angular momentum is a conserved quantity under rotation specifically.

When you are asking about something so fundamental as momentum flow due to an interaction, I really don't know what sort of level of math you are looking for.