Can the kinetic energy be a function of the position vector?

,I got one confusion when reading Goldstein's Classical Mechanics (page 20, third edition). After getting the equation $$\sum \left\{\left[\frac{\mathrm{d}}{\mathrm{d}t}{\left(\frac{\partial T}{\partial \dot{q_{j}}} \right)} - \frac{\partial T}{\partial q_{j} } \right] - Q_{j} \right\} \ \delta q_{j} = 0$$ then it says that

Note that in a system of Cartesian coordinates the partial derivative of $T$ with respect to $q_{j}$ vanishes. Thus, speaking in the language of differential geometry, this term arises from the curvature of the coordinates $q_{j}$. In polar coordinates, e.g., it is in the partial derivative of $T$ with respect to an angle coordinate that the centripetal acceleration term appears.

My question is: Is the above statement general, i.e., that the kinetic energy $T$ does not depend on the position. I wonder why velocity can't depend on the particle's position vector. I mean, why couldn't we have cases where $\vec{v} = \vec{v}(\vec r ,t)$, so that the kinetic energy depends on $q_{j}$ or $\vec{r}$?

I will use the example of the second answer in this post What is the difference between implicit, explicit, and total time dependence, i.e. $\frac{\partial \rho}{\partial t}$ and $\frac{d \rho} {dt}$?

suppose we have a velocity distribution of gas whose function is $\vec {v}=\vec {v}(x,y,z,t)$

then the partial derivative with respect to $x$ is

$$\frac{\partial T} {\partial x}=mv\frac{\partial v} {\partial x}$$

since $v$ depends on $x$, I think in this case. $\frac{\partial T}{\partial q_{j} }$ does not vanish.

Note that you are calculating the PARTIAL derivative with respect to $q$, you must consider anything that is not $q$ ( $\dot q$ may implicitly depend on q and t but that doesn't matter when taking the partial derivative with respect to $q$) is just considered a constant, thus its partial derivative vanishes. I can't tell you more, it's just the properties of the partial derivatives, I suggest you get a good book on calculus and do a lot of exercises.

It is the same thing when you work out the Euler-Lagrange equations, $\dfrac{\partial L}{\partial q} = \mathrm{Force(q)}$, because you consider T a constant (as it doesn't explicitly depend on position but on velocity)

• Sorry for the mistakes, I did it fast, I hope they are now corrected – SaudiBombsYemen Jan 22 '16 at 20:29
• Thanks for replying! But I think the statement means that the kinetic energy does not depend on $\vec r$. That is where it puzzles me, not the partial derivatives. – FaDA Jan 22 '16 at 21:02
• BTW, sorry about the -1. But someone else did it, not me. – FaDA Jan 22 '16 at 21:04
• @FaDA I'll try to think if a can improve my answer. Kinetic energy does depend on r as you derive it to get velocity. – SaudiBombsYemen Jan 22 '16 at 21:06
• I have edited the post. I didn't mean $\vec{v}$ depends only on $\vec{r}$ instead of time. $\vec{v}$ can depend on both position and time, which means the partial derivative with respect to $q_j$ not equal to zero. – FaDA Jan 22 '16 at 21:07

The only way that quote makes sense to me is if he assumes his conclusion. He must have written some constraints on the allowed forms of kinetic energy.

• Thanks for reply! The statement seems to be quite general in the context. That is why I get puzzled. – FaDA Jan 23 '16 at 8:54

The author says, in Cartesian coordinates (denote them $x$, $\dot{x}$), kinetic energy does not depend on the second derivative of the coordinate. This is a general statement about Cartesian coordinates, not any coordinate system.

Thus, if $T$ depends on $q$, it is solely because it is, in a sense, non-Cartesian --- i.e., curvilinear. Hence this argument is made from the stance point of differential geometry, an unnecessary digression that the author made.