Say im trying to prove $\frac{\partial \dot{T}}{\partial \dot{q}^i} - 2\frac{\partial {T}}{\partial {q^i}} = - \frac{\partial {V}}{\partial {q^i}}$ from the Lagrangian equation: $L = T - V$, and the euler-langrange equation, where $T$ is the kenetic energy and $V$ is the potential energy.

Below I have attached what I call my almost proof. I can't get past the last line.

I'm pretty sure I have done my math correct up until the last line.

enter image description here

My question is a conceptual one: how do total time derivatives of partial derivatives of functions work?

Taking $\frac{d}{dt}$$\frac{\partial T}{\partial \dot{q}^i}$ should give me $\frac{\partial \dot{T}}{\partial \dot{q}^i} - \frac{\partial {T}}{\partial {q^i}}$ but I'm not sure how to take the total time derivative of a partial derivative. How does this work?

  • $\begingroup$ Related: physics.stackexchange.com/q/428990/2451 $\endgroup$
    – Qmechanic
    Sep 28, 2020 at 4:04
  • $\begingroup$ The prerequisites of the proof are not clear. Is the potential velocity-independent ? Is the kinetic term $T$ coordinate-independent ? And actually, why do you want to show $$\frac{\partial\dot{ T}}{\partial \dot{q}^i }- 2\frac{\partialT}}{\partial q^i }=-\frac{\partial V}{\partial q^i}$$ $\endgroup$ Oct 1, 2020 at 12:53

1 Answer 1


This will be another entry in my long-running rant series which is (barely) hyperbolically titled "There's no such thing as a total derivative."

If you have a well-behaved function of two variables $f:\mathbb R\times \mathbb R\rightarrow \mathbb R$, then you can define the derivatives with respect to its first and second slots to be

$$\partial_1 f : (x,y) \mapsto \lim_{h\rightarrow 0}\frac{f(x+h,y)-f(x,y)}{h}$$ $$\partial_2f : (x,y) \mapsto \lim_{h\rightarrow 0}\frac{f(x,y+h)-f(x,y)}{h}$$

We call these functions the partial derivatives of $f$. In an abuse of notation, we often write them as $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$, where we are supposed to implicitly realize that $x$ and $y$ are the quantities we intend to plug into the first and second slots, respectively. However, this tradition can be ... problematic.

Now consider a function $\gamma:\mathbb R\rightarrow \mathbb R\times \mathbb R$ which eats some number $t$ and spits out the pair $\big(a(t),b(t)\big)$, where $a$ and $b$ are functions from $\mathbb R$ to $\mathbb R$. We can compose $f$ with $\gamma$ to obtain a single function from $\mathbb R$ to $\mathbb R$ as follows:

$$(f\circ \gamma) : \mathbb R\rightarrow \mathbb R$$

$$ t\mapsto f\bigg(a(t),b(t)\bigg)$$

Being a function from $\mathbb R$ to $\mathbb R$, we can take its regular, calculus 101 derivative:

$$(f\circ \gamma)'(t) = (\partial_1f)\bigg(a(t),b(t)\bigg) \cdot a'(t) + (\partial_2 f)\bigg(a(t),b(t)\bigg)\cdot b'(t)$$ which is usually written (in another abuse of notation) as

$$(f\circ \gamma)'(t) = \frac{\partial f}{\partial a} a'(t) + \frac{\partial f}{\partial b}b'(t) \equiv \frac{df}{dt}$$

It is this object which is called the "total derivative" of $f$, but that is a questionable description of it because it drops $\gamma$ completely out of the notation/phrasing. I would accept "the total derivative of $f$ along the path $\gamma$," but the president of calculus has stopped taking my calls on this matter.

With those preliminaries out of the way, the kinetic energy $T$ is a function of two variables, so we can define $\partial_1 T$ and $\partial_2T$ without issue. Given some function $q$, we can define $\gamma(t) = \bigg(q(t),\dot q(t)\bigg)$ and define the composite function $(T\circ \gamma)(t) = T\bigg(q(t),\dot q(t)\bigg)$. What you call $\dot T(t)$ is really $(T\circ \gamma)'(t)$.

At this point, you would be rightly confused as to how to take a partial derivative of $\dot T$. After all, $\dot T$ is a function from $\mathbb R$ to $\mathbb R$! The secret comes in two parts. First, you can specialize to $T$ which is quadratic in its second argument (the velocity); something of the form $T(x,v)=\frac{1}{2}g(x)v^2$ would do.

If you do this, you can massage the expression for $\dot T$ into the form $$\dot T = \dot q\left[ (\ldots) \dot q^2 + (\ldots) \ddot q\right]$$

The problem with this is that there's no way to interpret $\dot T$ as a function of $q$ and $\dot q$, because there's a $\ddot q$ sitting there. We now come to the second subtlety: if you use the Euler-Lagrange equations, you can show that when the equations of motion are satisfied, the term in square brackets above depends only on $q$. As a result, there exists some function $\mathcal{\dot \tau}(a,b)$ such that when the equations of motion are satisfied,

$$\mathcal{\dot \tau}\bigg(q(t),\dot q(t)\bigg) = \dot T(t) = \bigg[ \text{terms depending only on }q(t),\dot q(t) \bigg]$$ $$\implies (\partial_2\mathcal{\dot \tau})\bigg(q(t),\dot q(t)\bigg) \equiv \frac{\partial \mathcal{\dot \tau}}{\partial \dot q} = \bigg[ \text{other terms depending only on }q(t),\dot q(t) \bigg]$$

and furthermore, you will find that on-shell,

$$\frac{d}{dt}\bigg[(\partial_2 T)\bigg(q(t),\dot q(t)\bigg)\bigg] = \frac{\partial \mathcal{\dot \tau}}{\partial \dot q} - (\partial_1 T)\bigg(q(t),\dot q(t)\bigg)$$

which would typically be written

$$\frac{d}{dt} \frac{\partial T}{\partial \dot q} = \frac{\partial \dot T}{\partial \dot q} - \frac{\partial T}{\partial q}$$

To recap, the notation $\frac{\partial \dot T}{\partial \dot q}$ belies quite a bit of subtlety. The "total derivative" $\dot T \equiv \frac{d}{dt}T\big(q(t),\dot q(t)\big)$ has a $\ddot q$ in it, so it's not at all clear how to interpret it as a function of $q$ and $\dot q$, which would be necessary to find the partial derivative.

However, if you enforce the equations of motion, you can find a function $\mathcal{\dot \tau}\bigg(q(t),\dot q(t)\bigg)$ which will agree with the function $\dot T(t)$ (on-shell), and it is this object which you can differentiate with respect to its second argument ($\dot q$). Note that despite the notation, $\dot\tau$ is not actually the time derivative of some other function $\tau$; I added the dot to suggest that you should interpret $\dot \tau$ as a function of $q$ and $\dot q$ whose value is the rate of change of the kinetic energy along a physical trajectory.

As a concrete example in more compact notation (and in 2D), consider $T = \frac{1}{2} (\dot r^2 + r^2\dot \theta^2)$ and $V=0$. The equations of motion are

$$\ddot r = r\dot \theta^2$$ $$\ddot \theta = 0$$

Differentiating with respect to time, $$\dot T = \dot r\ddot r + r\dot r \dot \theta^2 + r^2\dot \theta \ddot \theta$$

We now need to use the equations of motion to get rid of the second derivatives, and we find

$$\dot T = r\dot r \dot \theta^2 + r\dot r\dot \theta^2 = 2r\dot r\dot \theta^2$$ therefore, we define the function $$\mathcal{\dot \tau}(r,\dot r,\theta,\dot \theta) = 2r\dot r \dot\theta^2$$ and note that $$\frac{\partial \mathcal{\dot \tau}}{\partial \dot r} = 2r\dot \theta^2$$ $$\frac{\partial T}{\partial r} = r\dot \theta^2$$ $$\frac{d}{dt}\frac{\partial T}{\partial \dot r}=\underbrace{\ddot r = r\dot \theta^2}_{\text{Only on-shell!}} = \frac{\partial \mathcal{\dot \tau}}{\partial \dot r} - \frac{\partial T}{\partial r}$$

  • $\begingroup$ Thank you for this thorough reply! What do you mean when you say "on-shell"? $\endgroup$
    – dimes
    Sep 28, 2020 at 12:57
  • 1
    $\begingroup$ @dimes The phrase on-shell refers to the fact that we are imposing the equations of motion as constraints. In my example at the end, $\frac{d}{dt} \frac{\partial T}{\partial \dot r} = \ddot r$ which is only equal to $r\dot\theta^2$ if $r(t)$ satisfies the Euler-Lagrange equations. $\endgroup$
    – J. Murray
    Sep 28, 2020 at 14:42
  • 1
    $\begingroup$ @dimes Can you be more specific with your objection? $\endgroup$
    – J. Murray
    Sep 28, 2020 at 14:42
  • $\begingroup$ I agree with your general proof up until where you write $\dot{\tau}(q(t), \dot{q}(t)) = \dot{T}(t)$. Why do we say $\dot{\tau}$ is equal to $\dot{T}$ if we have already defined $\dot{T}$ in $\dot{\tau}$? Isnt $(\dot{\tau} \circ \dot{T})(t) = \dot{\tau}(q(t), \dot{q}(t)) $? $\endgroup$
    – dimes
    Sep 28, 2020 at 14:47
  • 1
    $\begingroup$ @dimes No. $\dot \tau$ is a function of $q$ and $\dot q$, while $\dot T$ is a function of time. They agree when the equations of motion are satisfied, but they are different functions. Do you understand my concrete example? $\endgroup$
    – J. Murray
    Sep 28, 2020 at 15:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.