Timeline for When can one write $a=v \cdot dv/dx$?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 4, 2015 at 7:19 | comment | added | Martín-Blas Pérez Pinilla | "... if one knows what he's doing..." is the crux of the matter. | |
| Mar 3, 2015 at 17:45 | comment | added | joshphysics | @Martín-BlasPérezPinilla I'm somewhat inclined to agree, although I think the abuse has at least two possibly advantages: (1) it's easier and simpler to write and therefore a bit more readable (2) I think it can benefit one's intuition if one knows what he's doing. Having said this, I completely agree that it's often extremely confusing to students, along with a lot of other physics derivative conventions. | |
| Mar 3, 2015 at 8:29 | comment | added | Martín-Blas Pérez Pinilla | "It is common to abuse notation here and use $v$..." Is an obnoxious custom that causes many confusions to the students. | |
| Feb 4, 2015 at 17:10 | comment | added | joshphysics | @KDN Nowhere do I claim that a function must be invertible to be differentiable, nor is that implied by my answer. In order for one to define velocity as a function of the position in some neighborhood of a point $x_0$ along the trajectory of a particle, the position must be an invertible function of time in some neighborhood of $t_0$ such that $x(t_0) = x_0$. If not, the trajectory could intersect itself at say $x_*$, and the velocity at $x_*$ would be ambiguous. | |
| Feb 4, 2015 at 16:30 | comment | added | KDN | This implies that a function must be invertible in order to have a well defined spatial derivative, but that isn't true. Many functions must restrict their inverses to a particular domain, while still having well defined derivatives outside that domain (e.g., sine and cosine). Even if the velocity cannot be expressed analytically as a function of position, the spatial derivative may still be well defined, and thus the OPs original integral may still be valid, even if no clean expression for v(x) can be found. | |
| Feb 12, 2014 at 12:58 | vote | accept | pppqqq | ||
| Feb 22, 2013 at 1:53 | history | answered | joshphysics | CC BY-SA 3.0 |