Activity for Derek Elkinsâ€
Type | On... | Excerpt | Status | Date |
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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) |
— | 10 months 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) |
— | 10 months ago |
Comment | Post #286662 |
This is a question about opthamology, biology, childhood development, and/or medicine, not physics. (more) |
— | over 1 year 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) |
— | about 2 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) |
— | over 2 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) |
— | over 2 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) |
— | over 2 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) |
— | over 2 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) |
— | over 2 years ago |