In Classical Mechanics by Goldstein it says:

$$ \sum \left\{ \left[ \frac{d}{dt} \left( \frac{\partial T}{\partial \dot q_j} \right) - \frac{\partial T}{\partial q_j} \right] - Q_j \right\} \delta q_j = 0. $$ 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 the partial derivative of $T$ with respect to an angle coordinate that the centripetal acceleration term appears. I don't understand what's said here that the partial derivative of $T$ vanishes when differentiating with respect to a Cartesian coordinate. How is that possible? Isn't $\dot x $ is a function of $x$?


1 Answer 1


This is a common point of confusion when one gets started in Lagrangian mechanics. The important thing to notice is that we are taking partial derivatives, not full derivatives with respect to $x$.

From the point of view of our partial derivatives, $x$ and $\dot{x}$ are completely separate variables with no relation to each other.

To give you a bit of intuition as to why this must be the case, consider a free particle in space (ie. no potential energy), no other particles to interact with.

Should it's kinetic energy depend on where it is in this space? No! So the derivative of the kinetic energy with respect to the position must vanish.

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    $\begingroup$ Your intuitive argument is misleading. The real reason is that the Lagrangian lives on the tangent bundle, not Galilean invariance. $\endgroup$ Commented Oct 10, 2016 at 2:11
  • $\begingroup$ The point of the intuitive argument I gave was solely to remind the asker that they already knew that $\frac{\partial \dot{x}}{\partial x}$ must vanish because they know that the form kinetic energy they are most familiar with, $\frac{1}{2} m \dot{x}^2$ does not depend on position. $\endgroup$
    – Mason
    Commented Oct 10, 2016 at 4:02
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    $\begingroup$ Do you think that bringing up the fact that the Lagrangian lives in the tangent bundle is really necessary to answer the above question? I believe it would it serve only to obfuscate a simple intuitive fact. The asker can learn about the tangent bundle on a later pass through mechanics. $\endgroup$
    – Mason
    Commented Oct 10, 2016 at 4:06

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