# Confusing Total Derivative and Partial Derivative in Classical Field Theory - Noether Theorem

I'm really confused about total derivatives and partial derivatives.

My multivariable calculus book (Guidorizzi vol 2 Um Curso de Calculo) says that if I have a function like $$f(a(u,v),b(u,v))$$ then the following is true:

$$\frac{\partial f}{\partial u}=\frac{\partial f}{\partial a}\frac{\partial a}{\partial u}+\frac{\partial f}{\partial b}\frac{\partial b}{\partial u}\tag{1}$$

But studying classical field theory, precisely in Noether's theorem for fields, I came across that

$$\frac{d\mathcal{L}}{dx^{\mu}}=\frac{\partial \mathcal{L}}{\partial\phi_{\alpha}}\frac{\partial\phi_{\alpha}}{\partial x^{\mu}}+\frac{\partial \mathcal{L}}{\partial \phi_{\alpha,\mu}}\frac{\partial \phi_{\alpha,\mu}}{\partial x^{\mu}}+\frac{\partial \mathcal{L}}{\partial x^{\mu}}\tag{2}$$ with $$\mathcal{L}=\mathcal{L}(\phi_{\alpha}(x^{\mu}),\phi_{\alpha,\mu}(x^{\mu}),x^{\mu})$$

Note: $$\alpha$$ fields $$\phi$$ for $$x^\mu$$ arguments ($$\mu=0,1,2,3$$) in lagrangian density $$\mathcal{L}$$

This is called total space-time derivative (also someone can give me a reference than explain and define formally ?, for bibliography reference purpose please!)

The problem is, I can understand why equation (2) is true and make sense. After all, we are varying in all $$\mathcal{L}$$ arguments. It seems like the (1) equation doesn't make sense if equation (2) is true. I'm thinking that equation (1) is a notation abuse and in fact should be a total derivative

$$\frac{\text d f}{\text d u}=\frac{\partial f}{\partial a}\frac{\partial a}{\partial u}+\frac{\partial f}{\partial b}\frac{\partial b}{\partial u}$$

Am I right?

I have a third question: If $$f$$ depends explicitly on $$u$$ then $$f(a(u,v),b(u,v),u)$$ and equation (1) doesn't make sense:

$$\frac{\partial f}{\partial u}=\frac{\partial f}{\partial a}\frac{\partial a}{\partial u}+\frac{\partial f}{\partial b}\frac{\partial b}{\partial u}+\frac{\partial f}{\partial u}$$

All of my reasoning makes me believe that in fact that is $$\frac{\text d f}{\text d u}=\frac{\partial f}{\partial a}\frac{\partial a}{\partial u}+\frac{\partial f}{\partial b}\frac{\partial b}{\partial u}+\frac{\partial f}{\partial u}$$

Is my calculus book wrong along with several other references I had studied?

In math a total derivative is simply the full derivative of a (possibly multivariate) function, and can be represented by a matrix at each point. In math, the notation $$\partial f/\partial x$$ vs $$df/dx$$ is simply that in the first case $$f$$ may depend on more variables, where in the second or is emphasized to only depend on a single argument. There is no difference in how the chain rule applies.

There is a difference in notation between math an physics: in $$f(x)$$, $$f(r)$$, $$f(p)$$ and $$f(t)$$ means the same function $$f$$ evaluated in $$x$$, $$r$$, $$p$$ or $$t$$, and it is clear what is meant by $$f(4)$$ when 4 is in the domain. In other words, there is a functional relationship between the arguments and the function value, and this is independent of the names given to the arguments.

In physics, $$f(x)$$ often means the value of some quantity in the point $$x$$, $$f(r)$$ the value of the same function or quantity after a change to polar coordinates, $$f(p)$$ the value of the Fourier transform of the original quantity in the point $$p$$, and $$f(t)$$ the composition of $$f$$ with some other function that depends on time. This is some kind of a conceptual relationship between an underlying physical quantity, and some other physical quantities, between which there does exist a functional relationship but that is not explicit in the notation.

When dealing with a Lagrangian density $$\mathcal L(\phi_\alpha, \phi_{\alpha,\mu}, x^\mu)$$, in math there is no such thing as $$\frac{d\mathcal L}{dx^\mu}$$, because $$\mathcal L$$ is not a function of a single variable $$x^\mu$$. $$\frac{\partial\mathcal L}{\partial x^\mu}$$ does of course exist, and is the same as in physics. A mathematically correct notation for the former would be

$$\frac{\partial}{\partial x^\mu}\left(\mathcal L\circ\Phi\right)(x^\mu)$$

where $$\Phi(x^\mu) = \left(\phi_{\alpha}\left(x^{\mu}\right), \phi_{\alpha, \mu}\left(x^{\mu}\right), x^{\mu}\right)$$.

Neither of your books are incorrect. For the calculus book you have two independent variables $$u$$ and $$v$$. Therefore, you need to use the partial derivative $$\partial$$.

In your physics book all functions depend on only one independent variable$$^*$$ $$x^\mu$$. Therefore, you use the total derivative $$\text d$$.

So, in your first case $$\text df/\text du$$ doesn't make sense, unless $$v$$ also depends on $$u$$.

For your "third" case, to avoid confusion it would be best to replace the $$u$$ dependence with some function say $$c(u)$$. Then you have $$f(a(u,v),b(u,v),c(u))$$. Then your partial derivative $$\partial f/\partial u$$ will have a term $$\frac{\partial f}{\partial c}\frac{\text dc}{\text d u}$$ Therefore we end up with

$$\frac{\partial f}{\partial u}=\frac{\partial f}{\partial a}\frac{\partial a}{\partial u}+\frac{\partial f}{\partial b}\frac{\partial b}{\partial u}+\frac{\partial f}{\partial c}\frac{\text dc}{\text du}$$

i.e. the term you have confusion about really should just include any $$u$$ dependence that has not been considered as a part of $$a$$ or $$b$$ already.

And this brings up a point that is worth mentioning. The break up of $$f$$ into $$a$$, $$b$$, and $$c$$ is completely subjective. In reality we just have $$f(u,v)$$. But sometimes it's more useful to specify composite functions to do our math. Note that this exactly what we do in calculus 1 for single-variable functions. For example, to find the derivative of $$f(x)=(x+1)^2$$, we first define a new function $$g(x)=x+1$$. Then we rewrite $$f$$ as $$f(g(x))=\left[g(x)\right]^2$$. Then we use the chain rule to find $$\text df/\text dx$$.

$$^*$$ I know this isn't actually a single variable, but in this case it is being treated as one. If you wanted to break it up into each part, then each part would involve a partial derivative.