Why partial derivative in Lagrange is partial?

Question

When trying to arrive to Euler-Lagrange equation, Susskind does terrific job but I have one problem. Let's consider motion in only $x$ direction with respect to time.

$$x(t) = \hat x(t) + \epsilon f(t)$$ (variation where $\hat x(t)$ is the true trajectory and x(t) the varied one)

$$\frac {dx}{d\epsilon} = f(t)$$

Then we differentiate action with respect to $\epsilon$ to see how action changes when $\epsilon$ changes:

$$\frac{dS}{d\epsilon} = \int \frac{\partial L}{\partial x} \frac{\partial x}{\partial \epsilon}$$ (note that i didn't fully write the second part for $\dot x$ but it should be included.

I wonder now, why did he bring partial derivative for $L$ in $\frac{dS}{d\epsilon}$ but not for $\frac{dx}{d\epsilon}$.

I understand that he does chain rule for $L$ as it's a function of $x$ which is function of $t$, but i never knew that doing chain rule transforms derivative notation from full to partial. would appreciate some help.

Answer 1

The Lagrangian is a function of $x$ and $\dot{x}$: $$L = L(x, \dot{x})$$ In Lagrangian mechanics, you have to think of position and velocity as independent variables. Forget that one is the derivative of the other and simply think of them in the same way you think of the $x$ and $y$ coordinates in an ordinary coordinate system.

When you do a transformation $$x(t) = \hat{x}(t) + \epsilon f(t),$$ then both $x$ and $\dot{x}$ will depend on $\epsilon$: $$L(x(\epsilon) , \dot{x}(\epsilon))$$ Now, you want to know the change of the Lagrangian $dL$ when you change $\epsilon$ by a little amount $d\epsilon$. This change $d\epsilon$ will cause a small change in $x$ given by $$dx = \frac{\partial{x}}{\partial{\epsilon}} d\epsilon.$$ Why do we use a partial derivative here? Because in general, $x$ might not only depend on $\epsilon$. In fact, in our example, we know that $x$ also depends on time. But we only want to know what happens to $x$ when we change $\epsilon$ a little bit, keeping everything else fixed. This is exactly what the partial derivative is designed to do. Now, similarly, the change in $\dot{x}$ will be $$d\dot{x} = \frac{\partial \dot{x}}{\partial\epsilon} d\epsilon.$$ Now, how does $L$ change when we change both $x$ and $\dot{x}$ by the amount we just calculated? We simply add up the change caused by $dx$ and the change caused by $d\dot{x}$ (Again, think of $x$ and $\dot{x}$ like $x$ and $y$ coordinates. Then you can draw $dx$ and $d\dot{x}$ as little displacement vectors.): $$dL = \frac{\partial{L}}{\partial \dot{x}}d\dot{x} + \frac{\partial{L}}{\partial x} dx = \frac{\partial L}{\partial \dot{x}} \frac{\partial \dot{x}}{\partial \epsilon} d \epsilon + \frac{\partial L}{\partial x}\frac{\partial x}{\partial \epsilon} d\epsilon $$ The right hand side is really the total change of $L$ due to $\epsilon$, so the corresponding rate of change $dL/d\epsilon$ is a total derivative To determine it, we now simply divide both sides by $d\epsilon$ (I know this isn't rigerous, but I think it helps to build an intuitive unterstanding): $$\frac{dL}{d\epsilon} = \frac{\partial L}{\partial \dot{x}} \frac{\partial \dot{x}}{\partial \epsilon}+\frac{\partial L}{\partial x}\frac{\partial x}{\partial \epsilon}$$ And there you have it.

Answer 2

Then we differentiate action with respect to $\epsilon$ to see how action changes when $\epsilon$ changes:

$$\frac{dS}{d\epsilon} = \int \frac{\partial L}{\partial x} \frac{\partial x}{\partial \epsilon}$$ (note that i didn't fully write the second part for $\dot x$ but it should be included.

I wonder now, why did he bring partial derivative for $L$ in $\frac{dS}{d\epsilon}$ but not for $\frac{dx}{d\epsilon}$.

Presumably he is trying to emphasize that $x$ is a function of both $\epsilon$ and $t$, whereas $S$ is not.

However, if you want to be really pedantic you could argue that $S$ is also a function of the endpoints, so really the derivative on $S$ should be partial as well.

We are usually not so pedantic in physics because there are usually way worse notational issues to deal with.

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To explain further, consider the action: $$ S(\epsilon, t_1, t_2, q) = \int_{t_1}^{t_2}dt L(\hat x(t)+\epsilon f(t), \dot {\hat x}+\epsilon\dot f(t), t, q)\;, $$ where $q$ represents all the parameters we don't usually write explicitly.

To first order in $\epsilon$ the action is: $$ S(\epsilon, t_1, t_2, q) = \int_{t_1}^{t_2}dt \left(L(\hat x(t), \dot {\hat x}, t, q) + \epsilon f(t)\frac{\partial L}{\partial x}(\hat x,\dot {\hat x}, t, q) + \epsilon \dot f(t)\frac{\partial L}{\partial \dot x}(\hat x,\dot {\hat x}, t, q) \right)\;. $$

Therefore, by definition of the derivative, we have $$ \frac{dS}{d\epsilon}(t_1,t_2,q) = \int_{t_1}^{t_2}dt \left( f(t)\frac{\partial L}{\partial x}(\hat x,\dot {\hat x}, t, q) + \dot f(t)\frac{\partial L}{\partial \dot x}(\hat x,\dot {\hat x}, t, q) \right)\;. $$

If you would prefer to rename: $$ f(t) \to \frac{\partial x}{\partial \epsilon} $$ go for it.

If you would prefer to rename: $$ \frac{dS}{d\epsilon} \to \frac{\partial S}{\partial \epsilon} $$ go for it.

Regardless of how you use the notation, the result is the same.