# In Newtonian pressure, what type of function is force?

This is pressure in Newtonian mechanics:

$$P=\frac {dF}{dA}.$$ What does this mean? (Doesn't it mean that force is a function of area?) What type of function is force?

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## 1 Answer

The ratio is meant to denote $$p = \lim_{\Delta A\to 0} \frac{\Delta F}{\Delta A}$$ where $\Delta A$ is the area of a particular piece of the surface whose magnitude we send to zero; and $\Delta F$ is the correspondingly small force that acts on this small area. In practice, it's enough to choose $\Delta A$ small enough so that the pressure is constant (the force is linear in the area) over that small area within the error margin. In principle, we want $\Delta A$ to be really infinitely small.

While the formula above still looks pretty much identical to the definition of a derivative, the ratio is not a derivative in any useful sense because $F$ isn't a natural function of area $A$. We're dealing with one particular physical situation in which the total area $A$ is fixed and it is not a variable at all.

But in fact, we may describe the situation in such a way that it becomes possible to interpret $dF/dA$ as a derivative. Just divide the total area $A$ to (almost) infinitely many (nearly) infinitesimal areas $\Delta A$. Now, the key unnatural point is to order these pieces of area $\Delta A$; choose an order or schedule how these areas are added one by one, starting from $0$ and finally reaching the total area $A$. For example, you divide the rectangular area $A$ to a grid and add the small squares one by one, a row after a row.

If $A$ is the symbol for the total area of the squares in the subset that have already been added and $F$ represents the force that acts on this subset of the (original, whole) $A$ composed of the areas $\Delta A$ that were already added, then $F=F(A)$ is a function of $A$ and $dF/dA$ evaluated at the value of $A$ corresponding to the point at which we added a particular $\Delta A$ (or $dA$, if you wish) is precisely the pressure at the same point of space (or point of the surface) where this $\Delta A$ is located.

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+1 for the inclusion and explanation of the limit –  user24901 Jun 17 '13 at 10:04
@Lubos Motl after going into the limits part, when you have written about adding the small areas, cant we call that integration? Moreover, would that be line integral, or surface integral? –  Saurabh Raje Jun 18 '13 at 4:07
Pleasure, @DamienIgoe. And Saurabh, no, we can't really write it as an ordinary integral because the integral for the force, $F = \int dA\,p$, is a two-dimensional integral, e.g. $\int dx\int dy\, p$ when the area is a rectangle in the $xy$-plane. On the other hand, to interpret $dF/dA$ as a derivative that may be inverted to get an integral expression for $F$ as an integral of $p$, we need to interpret the summation as a one-dimensional integral, over "area". There's no natural way to convert one-dimensional integrals to two-dimensional ones or vice versa. –  Luboš Motl Jun 18 '13 at 12:49
Well, perhaps except for integrating the 2D integral over 1 variable first, and then integrate these 1-dimensional integrals over another coordinate. This may be done but the definition of coordinates to cover the area $A$ is not unique so the function of one variable $F=F(A)$ won't be unique, either. –  Luboš Motl Jun 18 '13 at 12:51
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