# Partial Derivative vs Total Derivative: Function depending Implicitly and Explicitly on Variable

I have a quick question that I hope someone can shed some light on for me. Its to do with partial derivatives and total derivatives (In this case I was working on Lagrangian Formalism but its really a general question). Im a theoretical physics student so although I do study a good deal of mathematics, we dont really go into the same level of detail that say a mathematician would (Lets just say some serious abuse of notation does occur more than one would like)

So lets assume I have some function $L = L(q_i(t),q_i'(t))$.

If I take the partial derivative of $L(q_i(t),q_i'(t))$ then, through the chain rule, I get

$$\frac {\partial L}{\partial t} = \frac {\partial L}{\partial q_i} \frac {\partial q_i}{\partial t} + \frac {\partial L}{\partial q_i'} \frac {\partial q_i'}{\partial t}.$$

Sound so far.

Lets assume now that the function $L \to L(t,q_i(t),q_i'(t))$ (i.e an explicit time dependence is now introduced in addition to the implicit time dependence.) If I now take the total derivative of said function, then I get

$$\frac {dL}{dt} = \frac{\partial L}{\partial t} + \frac {\partial L}{\partial q_i} \frac {\partial q_i}{\partial t} + \frac {\partial L}{\partial q_i'} \frac {\partial q_i'}{\partial t}.$$

But how would I now compute $\frac {\partial L}{\partial t}$? If I just apply the same logic and the chain rule, I would say that

$$\frac {\partial L}{\partial t} = \frac{\partial L}{\partial t} + \frac {\partial L}{\partial q_i} \frac {\partial q_i}{\partial t} + \frac {\partial L}{\partial q_i'} \frac {\partial q_i'}{\partial t}.$$

However, here comes my issue with the whole thing. If the above is true, then that would mean that

$$\frac {dL}{dt} = \frac {\partial L}{\partial t}$$

and also that

$$\frac {\partial L}{\partial q_i} \frac {\partial q_i}{\partial t} + \frac {\partial L}{\partial q_i'} \frac {\partial q_i'}{\partial t}=0.$$

Is it just that for this particular scenario the total and partial derivatives are equal, or am I missing something here? I've heard that the Leibniz notation breaks down somewhat here and can lead to confusion (something to do with the fact that the $\frac{\partial L}{\partial t}$'s on either side of the equation mean two totally different things); is there something that Im doing wrong?

• But can I not then use the chain rule to say that $\frac {\partial L}{\partial t} = \frac {\partial L}{\partial q_i} \frac {\partial q_i}{\partial t} + \frac{\partial L}{\partial q_i'} \frac{\partial q_i'}{\partial t}$ ? – P.Sauerborn Jan 13 '18 at 12:23
• Above “Sound so far” you're computing the total derivative $dL/dt$ (just like below), not the partial derivative. – Hans Lundmark Jan 13 '18 at 14:17
• @Hans I guessed before that the partial derivative respect to $t$ of $L(q(t),\dot q(t))$ is zero, as $t$ is not an explicit variable of $L$, however I couldn't justify this approach. – Masacroso Jan 13 '18 at 16:17
• @Hans Yes, you are right. I managed to find a textbook that approaches the subject a bit better and more clearly. The the chain rule states that $$\frac{\partial L}{\partial t} = \frac{\partial L}{\partial q_i} \frac{\partial q_i}{\partial t} + \frac{\partial L}{\partial q_i'} \frac{\partial q_i'}{\partial t}$$ if $L(q_i(u,t),q_i'(v,t))$ i.e. if there are more than one parameter. In my case, $L(q_i(t),q_i'(t))$ is still ultimatly only a function of $t$ so the partial derivative doesnt really exist in that sense – P.Sauerborn Jan 13 '18 at 16:35
• @Masacroso: If you insert $q(t)$ and $\dot q(t)$ into the function $L$, then you get a composite function depending on only one variable ($t$, that is), so it doesn't make sense to talk about partial derivatives; the derivative $d/dt$ is the only one around. But for $L(t,q,v)$, viewed only as a functions of some (independent) variables $t$ and $q$ and $v$, it makes perfect sense to talk about partials, and if that function doesn't actually depend on $t$, well then obviously $\partial L/\partial t=0$ (so there's nothing at all to justify). – Hans Lundmark Jan 13 '18 at 16:35

I think the main point is to always ask "how do I currently look at the function". You can always break a function to different blocks and look at it differently.

It's true that since $$L = L(q_i(t),q_i'(t))$$ only really depends on $$t$$, you can't do partial derivative on $$t$$, only total derivative. But you could take a partial derivative with regard to $$q_i$$ or $$q'_i$$. So the total derivative with regards to $$t$$ would look:

$$\frac {dL}{dt} = \frac {\partial L}{\partial q_i} \frac {dq_i}{dt} + \frac {\partial L}{\partial q_i'} \frac {d q_i'}{d t}.$$

And $$\frac {\partial L}{\partial q_i}$$ exist - e.g. if $$L(t^2, sin(t)) = t^2sin(t)$$ then $$\frac{\partial L}{\partial q_i} = \frac{\partial L}{\partial t^2} = sin(t)$$.

This way of mixing total and partial derivatives is used for example in the Method of Characteristics to solve the (Partial Differential) transport equation.

Now, you could decide that $$q_i = t$$, then you would get $$L = L(t,q_i'(t))$$. Now you can take the partial derivative w.r.t. $$t$$. It would be equal to whatever the function is, keeping $$q_i'(t)$$ as a constant.

e.g. if $$L(t, sin(t)) = t^2sin(t)$$ then $$\frac{\partial L}{\partial t} = 2tsin(t)$$. I.e. you keep the $$sin(t)$$ as a constant.

SO - your mistake is twofold:

1. as mentioned, in the first case you computed the total derivative, not the partial (which doesn't exist w.r.t. $$t$$); and

2. in the 2nd case, the partial derivative (w.r.t. $$t$$) is NOT equal to the total derivative (w.r.t. $$t$$). i.e.

$$\mathbf {\frac {\partial L}{\partial t} \neq \frac{dL}{dt}} = \frac{\partial L}{\partial t} + \frac {\partial L}{\partial q_i} \frac {d q_i}{d t} + \frac {\partial L}{\partial q_i'} \frac {d q_i'}{d t}.$$

$$\frac {\partial L}{\partial t}$$ in the 2nd case would mean you keep $$q_i, q_i'$$ as constants, and there's no further derivation for it unless you have the actual function $$L$$.

So, going back to my 1st point - the total derivative of $$t^2 sin(t)$$ will always be the same, but you may decide to break it to partial derivatives of $$t$$ and $$sin(t)$$, or $$t^2$$ and $$sin(t)$$ or something else. And sometimes these partial derivatives actually have some meanings.

Your problem comes from an abuse of notation (that is very common in this field), namely that different things are denoted with the same letters. You have a function $$L$$ of $$2n+1$$ variables, namely $$(t,q_1,\ldots, q_n,q_1',\ldots, q_n')\mapsto L(t,q_1,\ldots, q_n,q_1',\ldots, q_n')\ .\tag{1}$$ This function is given a priori, even before you know that $$t$$ should be time, and that, later on, the $$q_i$$ and $$q_i'$$ will be functions of $$t$$, whereby $$q_i'(t)={d\over dt}q_i(t)$$. This function $$L$$ has partial derivatives $${\partial L\over\partial t},\quad {\partial L\over\partial q_i},\quad {\partial L\over\partial q_i'}\ ,$$ all of them functions of the same variables given in $$(1)$$.

Now you start doing physics, with the already sketched interpretation of $$t$$. In order to do this properly we denote the "running time" with $$\tau$$. We then have the functions $$\tau\mapsto t(\tau)=\tau, \quad \tau\mapsto q_i(\tau),\quad \tau\mapsto q_i'(\tau):={d\over d\tau}q_i(\tau)\ .$$ Plugging these functions into $$L$$ we obtain a new function $$\tau\mapsto {\tt L}(\tau):=L\bigl(\tau,q_1(\tau),\ldots, q_n(\tau),q_1'(\tau),\ldots, q_n'(\tau)\bigr)\ .$$ This function $${\tt L}$$ of one variable $$\tau$$ has a derivative $${\tt L}'(\tau)={\partial L\over\partial t}\cdot1+\sum_i{\partial L\over\partial q_i}q_i'(\tau)+\sum_i{\partial L\over\partial q_i'}q_i''(\tau)\ .$$