$dT/dx=0$ always true?

Question

In a Classical Mechanics book I found the assumption that for an arbitrary particle with constant mass in the Real line $dT/dx=0$, with T the Kinetic Energy i.e. $T=(m·\dot x^2)/2$

My hypothesis is that the author used the following 'identity' $$d\dot x^2/dx=0$$

But solving the differential equation (correct me if I am wrong please) I get to $\dot x=f(t)$

Which I think could wrong because it could be that $\dot x=f(x,t)$ couldn't it?

Could it be saying the particle is at constant velocity? Without the "dot" superscript, it may be no velocity (momentum mv). — Robert DiGiovanni, Nov 27 '20 at 14:16

score 4 · Accepted Answer · answered Nov 27 '20 at 14:27

4

In classical Mechanics, where we work with Lagrangian $\mathcal{L}(q,\dot{q},t)$ which is a function of generalizing coordinate and velocity, We take velocity and coordinate to be independent variables. Why? Look here.

So kinetic energy which is $$T=f(\dot{q})$$

When we take its derivative with respect to $q$, it's turn out to be zero. $$\frac{dT}{dq}=0$$

As it doesn't depend on the generalized coordinate.

answered Nov 27 '20 at 14:27

Young Kindaichi

11,746

2

Nice answer, I believe this reasoning holds true even if $T$ allows for relativistic effects – TheVal Nov 27 '20 at 14:59

$dT/dx=0$ always true?

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