# Expressing density as dm/dv

1. Mar 27, 2014

### felipeek

Is it correct to write this:

$\rho=\frac{dm}{dv}$

where $\rho$ is density, $dm$ is a differential of mass and $dv$ a differential of volume?

We know that $\rho=\frac{m}{v}$ when $m/v$ is constant. But, if density is not constant, or, in other words, $m/v$ changes, could we express the variation of $m/v$ as $dm/dv$ to calculate the density at one point?

I'm asking it because I've never seen density written in that way, however we can express mass as $m=\int \rho dv$. Is there some physical mistake in the first expression?

Thanks!

2. Mar 27, 2014

### BiGyElLoWhAt

Moments of inertia?
This is more a math question, but anyway, $\rho[\itex]is a function, yes? The real question is what is rho a function of? Probably either x,y, and/or z. [itex]\rho=\frac{m}{v}[\itex] now simply differentiate with respect to your independent variable. Example:[itex]\rho=\frac{m}{v}[\itex] where m = kx. So in this case [itex]\frac{d\rho}{dx}=\frac{1}{v}[\itex] make the substitution x=m therefore [itex]dx=dm d\rho=\frac{dm}{v}[\itex] So to give a direct answer, it could be legit, but you would have to have a function for v that you're differentiating. Hope that helps 3. Mar 27, 2014 ### BiGyElLoWhAt #itex+Droid=fml sorry if its not rendering guys, I had to type it all out :/ 4. Mar 27, 2014 ### Simon Bridge Welcome to PF; Sure - we would say that the amount of mass in volume element $dV$ is $dm : dm=\rho(\vec{r})\;dV$ ... where $\vec r$ is because density may vary with position. There is not so much a physical mistake in saying \rho(x,y,z)=\frac{dm}{dV} ... that just says that the density is the way mass varies with volume. Which would be the definition. It's just not usually a terribly useful way of putting it. Certainly if you have the mass function $m(\vec r)$ then you also have the density function. 5. Mar 27, 2014 ### Simon Bridge @BiGyElLoWhAt: Good LaTeX ...only ...it's forward slashes for the tags and backslashes for the LaTeX. So you wanted: Probably either x,y, and/or z. [itex]\rho=\frac{m}{v}$ now simply differentiate with respect to your independent variable.
Example:$\rho=\frac{m}{v}$ where m = kx. So in this case $\frac{d\rho}{dx}=\frac{1}{v}$ make the substitution x=m therefore $dx=dm d\rho=\frac{dm}{v}$​

Works better to use double-hash instead of itex and double-dollar instead of tex tags.

Nitpick: density may vary with all three Cartesian coordinates at the same time.

i.e. $\rho(x,y,x)=\rho_0e^{-(x^2+y^2+z^2)}$ ... would be a spherical cloud of matter centered on the origin.

You wanted to do: $\rho=m/V \implies \rho V=m$ - differentiate both sides wrt V.

$$V\frac{d\rho}{dV}+\rho = \frac{dm}{dV}$$... does the density depend on the volume?
Maybe increasing the volume of a container also sucks more material into it?
When we are playing with these definitions, we have to be careful about what the system is.

This is why I was careful to define dm as the amount of mass in volume dV (at some position $\small{\vec{r}}$)

Last edited: Mar 27, 2014
6. Mar 28, 2014

### felipeek

Guys, thank you so much for your answers. It really helped me a lot.

@SimonBridge: I understood all you have said and it really makes sense. In fact, the density must be a function of a 3D vector, since every volume element $dV$ on the xyz graph has their own density. However, it is still hard to me to understand some things. I don't know why but when Physics is mixed with calculus it gets really hard to understand :P. For example, I can't understand the physical meaning of this function $m(\vec r)$. What is m? Is m the mass in the position $\vec r$? But shouldn't the mass of the exact point $(\vec r)$ be a differential of mass?

Other thing that I would like to ask: In your second post you made a nice demonstration about density varying with all coordinates at the same time. But then you took the original expression $\rho=m/V$ and did a "pure-mathematical" thing - putting in that way - and transformed the original equation using implict differentiation into $V\frac{d\rho}{dV}+\rho = \frac{dm}{dV}$.Mathematically, it's easy to understand what you did, but this new expression is not the same as $\rho=dm/dV$, I guess. At least it is not the same when the density depends on the volume. So here's the question: Is the last expression a particular version of this general expression (Particularly, when $\rho$ does not depends on $V$)? Actually, this is other thing that I have troubles with. Sometimes the demonstration of physics expressions are done taking a known-expression and changing it using math, like you did. If the math is correct, can I be sure that the resultant expression is also physically correct?

I'm sorry with the size of my text, I tried to explain my doubts in the best way possible :P. And unfortunately english isn't my first language, so it makes even harder to me to write (if you find grammar mistakes this is why :P)

Thank you very much everyone!!!

7. Mar 28, 2014

### Simon Bridge

it's supposed to represent a mass distribution
http://en.wikipedia.org/wiki/Mass_distribution
... but you are right, I'm being sloppy.

Look what happens if the density does not vary with volume.

Note - if the total mass is a constant but the volume changes then $d\rho/dV = -m/V^2$
This may happen for a gas in a closed syringe.

8. Mar 28, 2014

### HomogenousCow

Point is, you need densities to formulate useful differential equations.
The various field equations in integral form, are not very useful for computational purposes.