Timeline for The usage of chain rule in physics
Current License: CC BY-SA 4.0
28 events
when toggle format | what | by | license | comment | |
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Mar 15, 2022 at 19:32 | comment | added | David White | @MarkH, in my opinion, the mathematicians view math as the end-all and be-all, while physicists view math as a tool to accomplish a goal. If some "weird" math trick better represents reality or allows a physics solution that can't be achieved by any other method, so be it, even if such a math trick does not strictly follow the wishes of the "purists". | |
Mar 15, 2022 at 15:16 | answer | added | Brian | timeline score: 0 | |
S Oct 4, 2021 at 2:27 | history | suggested | gmz | CC BY-SA 4.0 |
fixed grammar
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Oct 4, 2021 at 2:18 | review | Suggested edits | |||
S Oct 4, 2021 at 2:27 | |||||
Dec 11, 2020 at 18:26 | comment | added | jnez71 | $\frac{dv}{dx}$ isn't some imprecise idea that physics textbooks use without checking validity of. Nor does it need the implicit function theorem to define. It is a perfectly fine mathematical construct when $v(x)$ is a velocity field. For discrete / finite particle systems, there is no concept of "infinitesimally close neighbor" so you won't see $\frac{dv}{dx}$ in those contexts. You will see it when describing flow though! Remember, $\frac{dm}{dt}$ is not the rate-of-change of a single particle's mass (Newtonian particles don't change mass). It is a flow of particles through a volume. | |
Dec 11, 2020 at 18:18 | comment | added | jnez71 | To some extent it does. I guess what it's missing is saying that, especially in continuum / field theories, $v(x,t)$ means "what is the velocity of the particle that happens to currently be at position $x$" (if $x$ is spatial coordinates) or "what is now the velocity of the particle that was at $x$ initially" (when $x$ is material coordinates). Thus $\frac{dv}{dx}(x,t)$ means "how different right now is the velocity of the particle at $x$ (spatial) or named $x$ (material) from its infinitesimally close neighbor?" | |
Dec 11, 2020 at 17:58 | comment | added | Brian | I think John Alexiou's post addresses that particular point | |
Dec 11, 2020 at 17:26 | comment | added | jnez71 | I usually see quantities like $\frac{dv}{dx}$ in the context of continuum mechanics, where $v(x)$ is a velocity field and $x$ is interpreted as indexing a specific particle in the continuum ("material coordinates") or specifying a location in space ("spatial coordinates"). The $v \cdot \nabla v$ nonlinear term in the Navier-Stokes is a famous instance of this. The answers below seem to be hung up on questions of differentiability, but I think the real question is about the physical meaning of $x$ in such derivatives. | |
Nov 17, 2020 at 19:52 | answer | added | AccidentalTaylorExpansion | timeline score: 2 | |
Nov 17, 2020 at 18:06 | history | edited | Brian | CC BY-SA 4.0 |
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Sep 13, 2020 at 12:08 | history | edited | Brian | CC BY-SA 4.0 |
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Aug 20, 2020 at 19:39 | vote | accept | Brian | ||
Aug 14, 2020 at 19:37 | comment | added | Mark H | @JohnRennie Mathematicians hate this one weird trick! | |
Aug 14, 2020 at 19:16 | answer | added | user541686 | timeline score: 2 | |
Aug 14, 2020 at 14:42 | comment | added | Koustubh Jain | dv/dx might mean the implicit differentiation of v and x, both as functions of time. Look up 3Blue1Brown's Essence Series on Implicit Differentiation, it might be helpful :-) | |
Aug 14, 2020 at 14:17 | answer | added | Brick | timeline score: 4 | |
Aug 13, 2020 at 21:22 | vote | accept | Brian | ||
Aug 14, 2020 at 0:11 | |||||
S Aug 13, 2020 at 21:19 | history | suggested | user1079505 | CC BY-SA 4.0 |
* → · in the equation
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Aug 13, 2020 at 20:33 | review | Suggested edits | |||
S Aug 13, 2020 at 21:19 | |||||
Aug 13, 2020 at 18:00 | history | tweeted | twitter.com/StackPhysics/status/1293970576562847745 | ||
Aug 13, 2020 at 16:54 | answer | added | WillO | timeline score: 20 | |
Aug 13, 2020 at 15:40 | history | became hot network question | |||
Aug 13, 2020 at 12:48 | answer | added | John Alexiou | timeline score: 8 | |
Aug 13, 2020 at 11:11 | comment | added | Syed Jaffri | If velocity=v(x(t),t), means velocity is an explicit function of both 'x' and 't' - then for same x , at two different times - it will have same veocity. It v=v(x(t))- then same as in v(x(t),x). If v=v(t), then v can have different velocity at same x at two different times.(since v is no longer a function of x- v has to have a unique value corresponding to a x,to be a function).- in this case, we can't write v as a function of x. | |
Aug 13, 2020 at 8:46 | comment | added | John Rennie | Welcome to life as a physicist :-) | |
Aug 13, 2020 at 7:59 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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Aug 13, 2020 at 7:56 | answer | added | Vid | timeline score: 15 | |
Aug 13, 2020 at 7:39 | history | asked | Brian | CC BY-SA 4.0 |