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We often discuss mass-density, charge-density, and other such functions: $\rho(x,y)$ for ultra-thin plates and $\rho(x,y,z)$ for 3-d objects. The units for the output of these functions say mass or charge/unit volume like $\frac{kg}{m^3}$ or $\frac{C}{m^3}$. I cannot comprehend a density at a single point.

I suspect that when we say mass per unit volume "at a point", we mean the mass contained in the volume immediately around the point. Here, we say "immediately around" in a calculus sense so that the volume is approaching zero. This way the set of all points together is still the full object.

Is this correct? Or have I missed the mark?

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When we say the mass density is $\rho(x,y,z)$, we mean that the mass within any finite region $R$ is given by $$ M(R) = \int_R \rho(x,y,z)\ dx\,dy\,dz. $$ In other words, specifying the mass density $\rho(x,y,z)$ is a concise way of describing the function that takes a region $R$ as input and returns the mass $M(R)$ in that region as output.

The region $R$ can be arbitrarily small, so your intuition is on the right track. If we take $R$ to be a point, then the mass $M(R)$ is zero, no matter how large the mass density may be (as long as it's finite).

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  • $\begingroup$ Density is (M(R))/R. Is it the case that as R approaches a zero, M approaches a value, and the value that M approaches is the output of the mass-density function and the density at that "point" (i.e the density of the infinitesimal R)? $\endgroup$ – novawarrior77 Aug 17 '20 at 13:30
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    $\begingroup$ @novawarrior77 If I understand what you're asking, then yes: $M(R)/|R|$ approaches $\rho(x,y,z)$ as the size of $R$ goes to zero, where $(x,y,z)$ is the point that $R$ shrinks to and $|R|$ is the volume of $R$. $\endgroup$ – Chiral Anomaly Aug 17 '20 at 14:04
  • $\begingroup$ That is what I meant, stated more clearly! Thank you so much, I am satisfied. $\endgroup$ – novawarrior77 Aug 17 '20 at 14:20
  • $\begingroup$ @ChiralAnomaly Could the mass desnity function also be a function of temperature, i.e. $ρ(x, y, z, T)$? Should we still integrate only over the $3$ spatial coordinates? $\endgroup$ – Antonios Sarikas Feb 26 at 16:49
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Substance (that makes up the mass) is discrete. We have molecules, atoms, smaller particles, etc, ...

There are hints that the space itself is discrete, too (see about the Planck length), but we don't know for sure.

Then again, sometimes (almost always, in fact) it is useful to approximate the substance as smooth and homogenous on small enough scales and use the whole calculus aparatus we have available that uses real numbers.

That's how density becomes a scalar field.

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    $\begingroup$ What are the hints that space is discrete? $\endgroup$ – Deschele Schilder Aug 17 '20 at 8:59
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    $\begingroup$ @descheleschilder Certain theories work better if space is discrete I think is what's being aimed at. pbs.org/wgbh/nova/article/… has some insight into some of the high level stuff. $\endgroup$ – MichaelS Aug 17 '20 at 9:12
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Basically, you are correct. The mass contained in a point (when we speak of continuous materials) is zero.
However, we can indeed take a small amount of length, area, or volume, mathematically described as $dx$, $dA$, or $dV$ approaching zero. These are called length-, area-, or volume elements. To find the entire mass one has to sum up all the products of all infinitely small mass densities with the length, area, or volume elements at all points in the mass in the 1-, 2-, or 3d case. This summation becomes an integral of the products of the densities $\rho$ with the three different elements (assuming $\rho$ is independent of the position in $x$, $A$, or $V$):

$$m_{tot}=\int _x\rho dx,$$

for a mass on a line,

$$m_{tot}=\int _A\rho dA,$$

for a mass on a surface, and

$$m_{tot}=\int _V\rho dV,$$

for a mass in a volume.

If the mass density is dependent on the position in the mass, just replace $\rho$ by $\rho (x)$, $\rho (A)$, and $\rho (V)$.

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Mass density at a point is defined two ways:

  • the limit of the average mass density in a volume containing the point as the volume decreases to zero, and
  • as a field which is integrated to give mass.

Understanding how and when these two definitions are the same thing requires some measure theory -- at which time you learn how they are not the same thing.

Example of how they are the same thing. Suppose that the mass density (field) is a constant $1\, \mathrm{mg}/\mathrm{cm}^3$ at each point under consideration. Let $x$ be such a point. Let us calculate the limit of (for simplicity) spherical volume average densities for spheres centered at $x$. Let $r$ be the radius in $\mathrm{cm}$. The volume, $V$, and mass, $m$, are \begin{align*} V(r) &= \frac{4}{3} \pi r^3 \\ m(r) &= \int_{-r}^{r} \int_{-\sqrt{r^2 - z^2}}^{\sqrt{r^2 - z^2}} \int_{-\sqrt{r^2 - z^2 - y^2}}^{\sqrt{r^2 - z^2 - y^2}} 1\, \mathrm{mg}/\mathrm{cm}^3 \,\mathrm{d}x \,\mathrm{d}y \,\mathrm{d}z \\ &= \frac{4}{3} \pi r^3 \,\mathrm{mg}/\mathrm{cm}^3 \text{.} \end{align*}

(The explicit units might make this mass look like a density. Recall that "$r$" in "$r^3$" has distance units which cancel the distance units in the denominator of the explicit units.)

Then the mass density at $x$ is $\lim_{r \rightarrow 0} \frac{\frac{4}{3} \pi r^3 \,\mathrm{mg}/\mathrm{cm}^3}{\frac{4}{3} \pi r^3} = 1 \,\mathrm{mg}/\mathrm{cm}^3$. Notice that we must take the limit as $r \rightarrow 0$. We cannot evaluate the ratio of mass to volume at $r = 0$ since that involves division by zero. Now a graph of the function we are taking a limit of. From the algebraic cancellation (permissible under the limit, but not outside this limit), we expect to see a constant function.

Mathematica graphics

The point $(0,1)$ is omitted, because division by zero is undefined. To sneak up on the value there, we use a limit. Note that if the density field varied (small fluctuations around a mean density and/or a trend to higher or lower densities away from $x$) we would see these variations in the curve. This very simple model doesn't have such features.

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I'll add another point of view, since the question only seems like something that's highly advanced or that which only comes up in that area of physics: What you are asking is precisely akin to the Zeno's arrow paradox: https://en.wikipedia.org/wiki/Zeno's_paradoxes#Arrow_paradox

Basically, I'm sure you are familiar with derivatives, but they are not intuitive when applied to arbitrary quantities. Certainly we can talk about an average speed over some duration ∆t, and reason that when restricting the duration towards a single instant of time, we get the instantaneous speed at a given moment – a useful quantity which we know is well-defined.

"But in order to have a speed, you'd need to travel, and you can't travel if time doesn't go on!" Yeah, it's the same deal with there not being an intuitive "instantaneous" density (dm/dV) if you look at a point of mass, but nevertheless we work with derivatives and they work. :)

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