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$$-\sum_i \frac{dx_i}{dt} \frac{\partial V(\{x\})}{\partial x_i} = - \frac{dV}{dt}$$

where $x_i$ is the coordinate of a configuration space and $V({x})$ is the potential energy function. I have only started learning multivariable calculus and don't understand why this is true.

244529
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2 Answers2

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This expression is nothing more than the chain rule written for multiple variables. Namely, if I have some function $f$ that is a function of many variables, I can express, \begin{gather*} \frac{d f}{d y} = \sum_j \frac{\partial f}{\partial x_j} \frac{\partial x_j}{\partial y} \end{gather*} Where the sum runs over all of the variables that $f$ depends on. The only difference is the names of the variables and functions and the fact that there is a negative sign multiplied on both sides. Also make sure that you are being careful with total and partial derivatives.

Matt Hanson
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  • oh but the book I got this from did multiply the total derivative of x_i with respect to t with the partial derivative of potential energy with respect to x_i. I think having the total derivative of x_i(the velocity) is important because it is required to show the conservation of energy in a system. Is the equation false if the total derivative of x_i with t is used? – 244529 Jan 15 '24 at 01:50
  • That would mean the $x_j$ are just functions of $t$, which is perfectly possible. I just wanted to point out that care with partial and total derivatives is necessary! – Matt Hanson Jan 15 '24 at 01:52
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What's the mathematical/physical reason this equation is true?...

$$-\sum_i \frac{dx_i}{dt} \frac{\partial V(\{x\})}{\partial x_i} = - \frac{dV}{dt}$$

Consider $V=V(a, b)$ to be a function of just two arguments. The generalization from 2 to $N$ is straightforward.

Now, let each argument $a$ and $b$ be evaluated at a point that depends on time, $x_1(t)$ and $x_2(t)$, respectively.

Physicists, somewhat confusingly, often use the same symbol for two different functions. For example, a physicist might write $V(t)$ to mean the function that results if $V(a, b)$ is evaluated at $V(x_1(t), x_2(t))$. Clearly, the function $V(t)$ can not be the same function as $V(a,b)$. To avoid this potential confusion, I'll write $$ \tilde V(t) \equiv V(x_1(t), x_2(t))\;. $$

Now, let $\tilde V(t)$ be evaluated at $t_0+\delta t$, where $\delta t$ is "small." (Meaning we can neglect all but the first order change with respect to $\delta t$.) $$ \tilde V(t_0 + \delta t) = V(x_1(t+\delta t), x_2(t+\delta t))\tag{A}\;. $$

The LHS of Eq. (A), to first order in $\delta t$, is: $$ \tilde V(t_0) + \delta t\left.\frac{d\tilde V}{dt}\right|_{t_0} $$

The RHS of Eq. (A), to first order in $\delta t$, is: $$ V(x_1(t_0), x_2(t_0)) + \delta t\left.\frac{dx_1}{dt}\right|_{t_0}\left.\frac{\partial V}{\partial a}\right|_{(a,b)=(x_1(t_0),x_2(t_0))} + \delta t\left.\frac{dx_2}{dt}\right|_{t_0}\left.\frac{\partial V}{\partial b}\right|_{(a,b)=(x_1(t_0),x_2(t_0))}\;. $$

The first order changes in $\delta t$ are equal on both sides of the equation, so we see: $$ \left.\frac{d\tilde V}{dt}\right|_{t_0} = \left.\frac{dx_1}{dt}\right|_{t_0}\left.\frac{\partial V}{\partial a}\right|_{(a,b)=(x_1(t_0),x_2(t_0))} +\left.\frac{dx_2}{dt}\right|_{t_0}\left.\frac{\partial V}{\partial b}\right|_{(a,b)=(x_1(t_0),x_2(t_0))}\;. $$

Or, reverting to a more condensed notation, writing $t$ instead of $t_0$, and using the symbol $V$ in place of $\tilde V$, we have: $$ \frac{dV}{dt} = \frac{dx_1}{dt}\frac{\partial V}{\partial x_1} + \frac{dx_2}{dt}\frac{\partial V}{\partial x_2}\;. $$

hft
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    Thank you for the bit about physicists using the same symbol $V$ for two different functions. I believe this a source of confusion for a great many students. – WillO Jan 15 '24 at 06:00