Expressing density as dm/dv

  1. Is it correct to write this:

    [itex]\rho=\frac{dm}{dv}[/itex]

    where [itex]\rho[/itex] is density, [itex]dm[/itex] is a differential of mass and [itex]dv[/itex] a differential of volume?

    We know that [itex]\rho=\frac{m}{v}[/itex] when [itex]m/v[/itex] is constant. But, if density is not constant, or, in other words, [itex]m/v[/itex] changes, could we express the variation of [itex]m/v[/itex] as [itex]dm/dv[/itex] 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 [itex]m=\int \rho dv[/itex]. Is there some physical mistake in the first expression?

    Thanks!
     
  2. jcsd
  3. BiGyElLoWhAt

    BiGyElLoWhAt 1,057
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    Moments of inertia?
    This is more a math question, but anyway, [itex]\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
     
  4. BiGyElLoWhAt

    BiGyElLoWhAt 1,057
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    #itex+Droid=fml
    sorry if its not rendering guys, I had to type it all out :/
     
  5. Simon Bridge

    Simon Bridge 15,261
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    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.
     
  6. Simon Bridge

    Simon Bridge 15,261
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    @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}[/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]​

    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
  7. 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 [itex]dV[/itex] 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 [itex]\rho=m/V[/itex] 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 [itex]\rho=dm/dV[/itex], 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 [itex]\rho[/itex] does not depends on [itex]V[/itex])? 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!!!
     
  8. Simon Bridge

    Simon Bridge 15,261
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    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.

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