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This doubt is very silly, but anyway, I think it's worth asking. The problem is: when we work with mathematics, in many situations we want to consider sets $A$ and $B$ and functions $f : A \to B$. When we specify a function, the letter we use is irrelevant, so that $f : \Bbb R \to \Bbb R$ given by $f(x)=x^2$ is the same thing as $f(y)=y^2$ or $f(E)=E^2$. Also, we try to make a sharp distinction between the function $f$ and its value $f(a)$ for the point $a$.

Now, in Physics I'm a little confused. For instance, when we think about momentum, we have $p = mv$, but wait, there are some details that appear to be not clear at first. If we are to differentiate, integrate, or whatever, this thing must be a function. But, there are many ways in which this can be a function:

  1. We can think of $\mathbf{p} : \mathbb{R}^3 \to \mathbb{R}^3$ as a function of velocity: $\mathbf{p}(\mathbf{v})=m\mathbf{v}$ and think of $m \in \mathbb{R}$ fixed.
  2. We can think of $\mathbf{p} : \mathbb{R} \to \mathbb{R}^3$ as a function of mass: $\mathbf{p}(m)=m\mathbf{v}$ for $\mathbf{v}$ fixed.
  3. We can think of $\mathbf{p} : \mathbb{R}\times \mathbb{R}^3 \to \mathbb{R}^3$ as a function of both mass and velocity: $\mathbf{p}(m,\mathbf{v})=m\mathbf{v}$.
  4. We can think of $\mathbf{p} : \mathbb{R} \to \mathbb{R}^3$ as a function of time: $\mathbf{p}(t)=m\mathbf{v}(t)$ where we think of $\mathbf{v}$ as a function of time.
  5. We can think of $\mathbf{p} : \mathbb{R} \to \mathbb{R}^3$ as a funciton of time: $\mathbf{p}(t)=m(t)\mathbf{v}(t)$ where both mass and velocity are time dependent.

And these are only some of possible ways to think about it. Indeed, if we allow mass and velocity vary with even more parameters, then we can get $p$ as function as much more things. In that case, it's not clear at first what function $p = mv$ defines. There are tons of situations in physics where this kind of thing occur, and it becomes a little messy to deal with it this way. The books don't use the notation of of function, so we end up without knowing well what is happening.

So, how do we deal with this kind of situation? How to remove these ambiguities?

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There's no global answer. A physicist would say that you just have to be aware of context. In Newtonian mechanics, you can rarely go wrong thinking of your core dynamical quantity as a curve mapping the reals into 3-space, and with your background parameter being time, but even that depends on what you're doing. –  Jerry Schirmer Sep 25 '13 at 17:28
    
Are you familiar with the implicit function theorem? Many of the "functions" you describe are probably better understood by mathematicians as relations, which can be converted locally into functions via this if you need to do so. –  Logan M Sep 25 '13 at 18:54
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2 Answers

This is what physicists call abuse of notation. In mathematics, as you say, one function symbol, such as $f$, denotes one particular function $f: A \rightarrow B$ and the letter we use to denote its argument is irrelevant.

In physics, one could say that the letter used to denote a function's argument is also part of that function's name. This is especially rampant in thermodynamics. The inner energy of a system, for example, is denoted by $U$ and can be expressed as a function of entropy, volume and particle number, $U(S, V, N)$ or as a function of temperature, pressure and chemical potential, $U(T, p, \mu)$.

Technically, you could remove the ambiguity by using different symbols for them: $U_{SVN}$ would be one function and $U_{Tp\mu}$ another function. But as I said, by considering the argument list as part of the name we can get rid of the ambiguity.

So the "function" symbol alone is a placeholder for any number of functions that relate a set of physical quantities to another physical quantity, and what $p = mv$ defines is the relationship between all functions that give you mass, velocity and momentum.

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I would even say that most physicists would call this a mild abuse of notation - there's loads worse in physics. Picky mathematicians beware! –  Emilio Pisanty Sep 25 '13 at 17:07
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You will have no problem if you (correctly) use differentials.

Starting with $p=mv$, you have always: $dp = dm ~ v + m ~dv$.

Now, in most contexts, $dm = 0$, so $dp = m ~dv$.

With the above context, if you want to use a different parametrization, for instance time, you will have : $dp = (m \frac{dv}{dt}) dt$

Within a general context, you will have : $dp = (v\frac{dm}{dt} + m \frac{dv}{dt}) dt$

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Just my opinion, but I think that the OP is looking in the direction of mathematical rigor so, unless you are giving to symbols like $\text d x$ a precise meaning, this doesn't seem to me what the OP is looking for. –  pppqqq Sep 25 '13 at 18:06
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