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Expressing density as dm/dv

by felipeek
Tags: density, dm or dv, expressing
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felipeek
#1
Mar27-14, 07:07 PM
P: 2
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!
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BiGyElLoWhAt
#2
Mar27-14, 07:30 PM
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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
BiGyElLoWhAt
#3
Mar27-14, 07:31 PM
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#itex+Droid=fml
sorry if its not rendering guys, I had to type it all out :/

Simon Bridge
#4
Mar27-14, 07:34 PM
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Expressing density as dm/dv

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.
Simon Bridge
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Mar27-14, 07:39 PM
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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}}##)
felipeek
#6
Mar28-14, 01:38 AM
P: 2
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!!!
Simon Bridge
#7
Mar28-14, 03:44 AM
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Quote Quote by felipeek View Post
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?
it's supposed to represent a mass distribution
http://en.wikipedia.org/wiki/Mass_distribution
... but you are right, I'm being sloppy.

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])?
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.

[quote]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?##
The resulting math need not lead to something physically realizable no.
Though it often does.

You have to ask yourself what it would mean to say such and such in the math.
... being careful about the definitions we made at the start.

I was hoping to show how the approach can be confusing.
HomogenousCow
#8
Mar28-14, 09:37 AM
P: 362
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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