Timeline for Derivation of $E=pc$ for a massless particle?
Current License: CC BY-SA 3.0
8 events
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| Jun 23, 2020 at 15:35 | comment | added | Charles Hudgins | Wish I could upvote this answer more than once. Too many leave undergrad without a good understanding of Noether's theorem and related concepts. | |
| Nov 6, 2015 at 0:36 | comment | added | Robin Ekman | Consider Hamiltonian mechanics in canonical coordinates $(x,p)$. Then translations map $(x,p) \mapsto (x+a,p)$. Clearly the vector field that generates this is $\partial/\partial q$. But in canonical coordinates the symplectic form is $dq\wedge dp$, so $\partial/\partial q \mapsto dp$. Thus while it's strictly speaking $\partial/\partial q$ that is the generator of translations, we can associate it with the coordinate p in this way. | |
| Nov 6, 2015 at 0:25 | comment | added | Robin Ekman | @fffred I think the analogy is somewhat a "deep" insight that you get after much reading from several sources, but I would recommend Armold's Mathematical Methods of Classical Mechanics, because he talks a lot about the Lie algebra structure in Hamiltonian mechanics. | |
| Nov 6, 2015 at 0:22 | comment | added | fffred | Could you explain a bit more? or point to places where to read about it? I have some trouble finding it. | |
| Nov 6, 2015 at 0:21 | comment | added | Robin Ekman | The concept of generator of translations applies also to classical mechanics in the Hamiltonian formulation. The difference is that classically the generator is a vector field, whereas the quantum generator is an operator. But they are analogous because vector fields and operators are both Lie algebras. | |
| Nov 6, 2015 at 0:20 | comment | added | fffred | I'm confused: definition of momentum as generator of translations appears, to my knowledge, in quantum mechanics. Is there a way to describe the same thing using only special relativity? without waves? | |
| Nov 5, 2015 at 23:24 | history | edited | Robin Ekman | CC BY-SA 3.0 |
added 402 characters in body
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| Nov 5, 2015 at 23:16 | history | answered | Robin Ekman | CC BY-SA 3.0 |