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Timeline for The usage of chain rule in physics

Current License: CC BY-SA 4.0

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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
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
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