# Understanding the total derivative in terms of analysis

1. Jul 15, 2009

### Tac-Tics

I'm trying to work through an introduction to Lagrangian mechanics. I get the idea. You have a particle at point A traveling to point B. You have a functional which maps every path between those points to a scalar called the action of the path. The universe then does some number crunching, figures out which path has attributed to it the smallest amount of action, and the particle takes that path. Conceptually, not hard.

Physicst's calculus notation left over from the 17- and 1800s, on the other hand, is very hard. For me at least. I'm hoping to get a slightly more rigorous understanding of how a total derivative works.

So, let's say I have my functional defined by $$F[x] = \int_A^B L(x, \dot{x}, t)dt$$ where L is the Legrangian. We all learn in calculus that $$\frac{d}{dt}f = \frac{\partial f}{\partial t} + \Sigma_i \frac{\partial f}{\partial x_i}\frac{dx_i}{dt}$$. But the terms $$\frac{dx_i}{dt}$$ seem nonsensical to me. The $$x_i$$'s aren't functions, and you can't take the derivative of them! The $$\partial x_i$$'s in the denominators are a shrothand notation to indicate the direction of the derivative, or which variable you are taking the derivative with, holding the rest constant. How the heck should I interpret those $$\frac{dx_i}{dt}$$'s?

I know, of course, that when working with this kind of problem, the parameters to the function aren't totally independent in the problem. Once we admit that, though, L(x, y, z) is no longer a legitimate function following the definition of a function described by set theory. (And yes, I do realize that the theory of calculus came well before set theory came to dominate mathematics).

So what I'm looking for is a better "framework" for interpreting the total derivative. Clearly, the dependencies granted to the Legrangian by the definition of the functional must be taken into account. The total derivative would seem not to be an operation on the Legrangian itself, then, but rather, it would be an operation on the expression of the functional which resides under the integral sign.

Any thoughts on this would be greatly appreciated!

2. Jul 15, 2009

### Pinu7

Who says the Xs aren't functions? There is no reason we can't say they depend on other variables. Suppose that we have a function f(x,y), why can't we have x=h(u,v) and y=g(u,v)

Here's a specific example: Potential energy.
Potential energy is a function of N x's where N is the number of degrees of freedom of a system. The Xs are functions themselves X(t)s. This is because POSITION MAY CHANGE WITH TIME. So x is definitely a function of time. Therefore, potential energy is a function composition.

For example, let's say a particle is falling from rest in a uniform gravitational field. It's position(from high school physics) is x(t)=-(1/2)gt^2, while it's potential energy is V(x(t))=mgx, we could calculate dV/dt by plugging the first equation into the second.
i.e. V(x(t))=-(1/2)m(gt)^2 is is a function of t and we could easily see that dV/dt= -mgt
which we could directly calculate by using the chain rule(try it yourself).

Wiki Articles-
Function composition: http://en.wikipedia.org/wiki/Function_composition" [Broken]
Chain Rule proof: http://en.wikipedia.org/wiki/Chain_rule#Proof_of_the_chain_rule"

Last edited by a moderator: May 4, 2017
3. Jul 15, 2009

### Tac-Tics

The $$x_i$$'s may appear to vary in the context of the problem, but this is just a notational phenomenon. You are appealing to a physical explanation, as if each variable was a bead on a abacus which you could freely slide back and forth. But variables in a rigorous proof never vary. Once you say x is a real number, it can't be a function too. The real story of what's going on is deeper than function composition.

4. Jul 15, 2009

### sir_manning

Isn't x still a function even if it is a real number? Say x is some function x(t) (where t is just a dummy variable, but could represent time). Then x(t)=5 is a function that says x is constant and will always be equal to the real number 5. With this in mind, $$\frac{dx_i}{dt}$$ is always valid and may be 0.

5. Jul 15, 2009

### Tac-Tics

It isn't. It might be a real-valued function, but that is a different kind of beast than a real number. In short, you have to apply an argument to it to get a real out. The value of f, by itself, is a function. The value of f(x) is a real. Most importantly, $$f \neq f(x)$$!

You may call it splitting hairs, but I call it rigor ;-) When you work out a problem in physics, you can fudge the math and produce the correct answer. But I'm not trying to do a homework problem here, so I can afford to be as pedantic as I want in my analysis, as long as it helps my understanding.

(By the way, I want to say thanks to you guys for responding to my post at all. I have a somewhat unique background and I approach these problems with an entirely different toolset than someone going to school for physics or engineering. I know I'm a pain in the butt for the regulars. They like questions to be asked the same way as they are written in textbooks, but I enjoy taking my own approach to things).

6. Jul 15, 2009

### Office_Shredder

Staff Emeritus
The position vector x is assumed to be a function of t, with each coordinate being a function. Why do you question the existence of $$\frac{dx_i}{dt}$$ when differentiating f with respect to t, but let it slide when in the integral above you have the derivative with respect to time of the position vector already?

Last edited: Jul 15, 2009
7. Jul 15, 2009

### Tac-Tics

Perhaps I have missed something. The definition of the total derivative seems to type properly as long as the respected variable is a scalar and the remaining variables are all functions.

I will sit and think about this problem for a bit. That may have been the insight I needed. Thanks a lot!

8. Jul 16, 2009

### Pinu7

Yes, x is a "number" but it may change with time if it is moving.
For example,
at t=0, x=1
at t=1, x=4
at t=2, x=16
at t=3, x=64
(Exercise: Find the probable equation of position. And then find the velocity function. Then determine the force acting on the particle with mass m.)

As you see, for a particle, with each instant of time, there is a single position corresponding to it. This is the definition of a function.

9. Jul 16, 2009

### dx

There's nothing deep here. Let's consider the Lagrangian, which is supposed to be a function of two variables q and q'. What does it mean to say that the action S is the integral of the lagrangian over time? L is not a function of time, so how do we integrate it over time? If we have a path γ : R → C, where C is the configuration space, then at each instant of time t, we have a position γ(t), and a velocity γ'(t). Now we create a new function of time as follows: M(t) = L(γ(t), γ'(t)). The action for the path γ is now defined as S = ∫ M(t) dt. This is where mathematicians may feel a little uneasy when reading physics books, because physicists use the same name for the Lagrangian function L and what we have called M, and they also use q(t) for γ(t) and q'(t) for γ'(t) . So they write S = ∫ L(q,q') dt. The way we have introduced a time dependence into the Lagrangian function is called 'implicit' time dependence, and a total derivative is simply differentiation with respect to an implicit variable. So the total derivative Dt of the Lagrangian function L(q,q') with respect to t is

Dt L = dM/dt

where M is defined as above.

In general, if we have a time dependent lagrangian L(t,q,q'), then the associated 'M function' will be M(t) = L(t,γ(t),γ'(t)), and by the usual calculus rules, we get

Dt L = dM/dt = ∂L/∂t + (dγ/dt)(∂L/∂q) + (dγ'/dt)(∂L/∂q')

and switching to physicist notation (q for γ, L for M etc.), this becomes

Dt L = dL/dt = ∂L/∂t + (dq/dt)(∂L/∂q) + (dq'/dt)(∂L/∂q')

So, to summarize, you are right in saying q and q' are variables, not functions, but it is just convenient notation to use the same letters for varables q, q' and the functions γ(t), γ'(t), and the functions L(q,q') and M(t) = L(γ(t), γ'(t)).

Last edited: Jul 16, 2009