# How do total time derivatives of partial derivatives of functions work?

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.

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?

• Sep 28, 2020 at 4:04
• 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 $$\frac{\partial\dot{ T}}{\partial \dot{q}^i }- 2\frac{\partialT}}{\partial q^i }=-\frac{\partial V}{\partial q^i}$$ Oct 1, 2020 at 12:53

## 1 Answer

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 $$\frac{\partial \mathcal{\dot \tau}}{\partial \dot r} = 2r\dot \theta^2$$ $$\frac{\partial T}{\partial r} = r\dot \theta^2$$ $$\frac{d}{dt}\frac{\partial T}{\partial \dot r}=\underbrace{\ddot r = r\dot \theta^2}_{\text{Only on-shell!}} = \frac{\partial \mathcal{\dot \tau}}{\partial \dot r} - \frac{\partial T}{\partial r}$$

• Thank you for this thorough reply! What do you mean when you say "on-shell"? Sep 28, 2020 at 12:57
• @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. Sep 28, 2020 at 14:42
• @dimes Can you be more specific with your objection? Sep 28, 2020 at 14:42
• 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 $(\dot{\tau} \circ \dot{T})(t) = \dot{\tau}(q(t), \dot{q}(t))$? Sep 28, 2020 at 14:47
• @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? Sep 28, 2020 at 15:05