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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?

Thanks very much for any help you can give!

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  • $\begingroup$ 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} $ ? $\endgroup$ – P.Sauerborn Jan 13 '18 at 12:23
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    $\begingroup$ Above “Sound so far” you're computing the total derivative $dL/dt$ (just like below), not the partial derivative. $\endgroup$ – Hans Lundmark Jan 13 '18 at 14:17
  • $\begingroup$ @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. $\endgroup$ – Masacroso Jan 13 '18 at 16:17
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    $\begingroup$ @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 $\endgroup$ – P.Sauerborn Jan 13 '18 at 16:35
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    $\begingroup$ @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). $\endgroup$ – Hans Lundmark Jan 13 '18 at 16:35
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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.

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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 $${\tt L}'(\tau)={\partial L\over\partial t}\cdot1+\sum_i{\partial L\over\partial q_i}q_i'(\tau)+\sum_i{\partial L\over\partial q_i'}q_i''(\tau)\ .$$

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