Activity for Derek Elkinsâ€
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #292637 | Initial revision | — | 2 months ago |
| Answer | — |
A: Why are there infinitely many modes of electromagnetic radiation? In what should come as no surprise, you don't determine the number of normal modes by the number of $\cos$ subexpressions occurring in the solution. (I guess there's a sense where you could say that you're "counting the cosines", but it's not this simple syntactic sense.) It would help to have the... (more) |
— | 2 months ago |
| Edit | Post #292531 | Initial revision | — | 2 months ago |
| Answer | — |
A: Should (lone) black holes emit gravitational waves? There are no gravitons in General Relativity, and black holes don't evaporate in General Relativity (i.e. there is no Hawking radiation). The consequence of Hawking radiation, or something similar, being empirically verified, would be an empirical refutation of General Relativity, precisely because i... (more) |
— | 2 months ago |
| Comment | Post #288244 |
I believe it could have infinite variance in position in that case, which would be consistent with starting with two momentum eigenstates, i.e. plane waves, for the photons. Starting with two more localized photons would lead to uncertainty in their momenta and thus an uncertain final momentum of any... (more) |
— | 2 months ago |
| Comment | Post #288251 |
There is not a "singular" way of solving a boundary condition problem, so saying "solve this problem as a boundary condition problem" doesn't specify an approach. Nevertheless, I assume you mean something like [this](https://tutorial.math.lamar.edu/Classes/DE/BVPEvals.aspx). This corresponds to my se... (more) |
— | over 1 year ago |
| Comment | Post #288251 |
I'm not quite sure how you are getting an eigenfunction problem or what you expect the eigenfunctions look like. Can you elaborate? Ignoring the $x(c)=0$ condition momentarily, we can easily show that the solution to this ODE is $x(t) = a\sin(t\sqrt{\frac{k}{m}})$ with $a$ free. Adding the $x(c)=0$ c... (more) |
— | over 1 year ago |
| Edit | Post #288244 |
Post edited: |
— | over 1 year ago |
| Edit | Post #288244 | Initial revision | — | over 1 year ago |
| Answer | — |
A: Why is it forbidden for two photons to turn into one? Imagine two equivalent (e.g. same frequency) photons colliding with each other head-on. The linear momentum of the system is $0$ because each photon's momentum has the same magnitude but is pointing in opposite directions. If these two photons collide and form a single photon (and nothing else), t... (more) |
— | over 1 year ago |
| Comment | Post #286662 |
This is a question about opthamology, biology, childhood development, and/or medicine, not physics. (more) |
— | over 2 years ago |
| Edit | Post #285649 | Initial revision | — | almost 3 years ago |
| Answer | — |
A: How are the assumptions behind two ways of deriving the Rayleigh-Jeans law related? One way this can be explained is from the perspective of numerically approximating an integral. From this perspective, the concordance of "continuous" and "low frequency" has to do with the low frequency regime being where the approximation of the "continuous" integral is valid/best. If you wanted... (more) |
— | almost 3 years ago |
| Comment | Post #285373 |
Just because you get the correct result doesn't mean that your derivation is correct. At the time of writing the question you seemed uncertain as to how/why we could go from $\int_V \rho_f d\tau$ to $\oint_S \sigma_f da$. The way I read your concern(admittedly with a lot of guessing on my part to fil... (more) |
— | almost 3 years ago |
| Edit | Post #285373 |
Post edited: Improve formatting of implications |
— | almost 3 years ago |
| Suggested Edit | Post #285373 |
Suggested edit: Improve formatting of implications (more) |
helpful | almost 3 years ago |
| Comment | Post #285373 |
Where are you getting the $\sigma da = \rho d\tau$ equation? As you allude to, $da$ is referring to an area element and $d\tau$ to a volume element, so it's not clear what it means to equate these expressions. (more) |
— | almost 3 years ago |
| Edit | Post #285238 |
Post edited: |
— | almost 3 years ago |
| Edit | Post #285238 | Initial revision | — | almost 3 years ago |
| Answer | — |
A: How can the kinetic energy equation be intuitively understood? Your analysis seems pretty good. To take an equivalent but more traditional example, imagine we throw a ball upwards. Ignoring air resistance and approximating the gravitational force as constant, the ball will accelerate downward with constant acceleration. If the initial (upwards) velocity is $v0$ ... (more) |
— | almost 3 years ago |
| Comment | Post #285233 |
The OP knows what the difference between kinetic energy and momentum is. The question is how to intuitively understand why kinetic energy grows quadratically with velocity. Momentum was only brought up for contrast. (Also, kinetic energy is *not* conserved in most scenarios; total energy is.) (more) |
— | almost 3 years ago |
| Comment | Post #284761 |
It doesn't make sense to say the "Laplace operator represents waves" any more than it would make sense to say the second partial derivative of time does. You could say that the *solutions* to that differential equations] models waves, but this isn't a property of the operator. (Admittedly, there is a... (more) |
— | about 3 years ago |
| Edit | Post #284671 |
Post edited: |
— | about 3 years ago |
| Edit | Post #284671 |
Post edited: Point out the reparameterization motivation of the discussion in the middle. Some other changes. |
— | about 3 years ago |
| Edit | Post #284671 | Initial revision | — | about 3 years ago |
| Answer | — |
A: What is "order" and "disorder" in entropy? "Disorder" and "order" don't mean anything with regards to entropy. This is a common "science popularization" level description of entropy that is "not even wrong" in that, as you've seen, these terms are not defined, and especially not defined in terms of usual physical models. Now one could defi... (more) |
— | about 3 years ago |
| Comment | Post #284073 |
You have a definition of the Poisson bracket, so there's nothing that can be expressed with this notion of Poisson bracket that can't be expressed without by simply using its definition. Besides that, what answer are you expecting that [the Wikipedia page for Poisson bracket](https://en.wikipedia.org... (more) |
— | about 3 years ago |
| Comment | Post #283954 |
The Lagrangian doesn't minimizes the action. The Lagrangian is the integrand of the integral defining the action. The action is a function of trajectories. The trajectories that make the action stationary (not necessarily minimal) are the physically realizable trajectories. The Lagrangian isn't a tra... (more) |
— | about 3 years ago |
| Edit | Post #280620 | Initial revision | — | almost 4 years ago |
| Answer | — |
A: How do constraints work in Lagrangian systems? As apparently seems par for this book, the theorem statement you're paraphrasing involves vaguely defined or completely undefined terms. For example, it talks about "admissible (smooth) paths" but this is the first use of the term "admissible" in the text.^[The notation $f'q$ is bizarre. The authors ... (more) |
— | almost 4 years ago |
| Edit | Post #280558 | Initial revision | — | almost 4 years ago |
| Answer | — |
A: How does probability conservation work in Dirac's original formulation of relativistic QM? I recommend https://www.mat.univie.ac.at/neum/physfaq/topics/position.html which, while a bit hard to read, is more comprehensive and written by someone more authoritative than me. My research didn't start there, but I covered several of the same sources before finding it. The tl;dr is the "naive"... (more) |
— | almost 4 years ago |