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What is the density function?

  1. Jun 30, 2005 #1

    quasar987

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    I want a rigorous description of the density function (in cartesian coordinates) [itex]\rho(x,y,z)[/itex].

    I suggest that we define a function M(x,y,z,V'), where V' is a volume of any given shape centered on the point (x,y,z), giving the mass contained in that volume. Then define the density function as

    [tex]\rho(x,y,z) = \frac{\partial{M(x,y,z,V')}}{\partial{V'}}\vert_{V'=0}[/tex]

    I wrote the volume of the "enclosing volume" as V' to differentiate it from the "volume of integration" V such that

    [tex]M_{body} = \int_V \rho dV = \int_{x_1}^{x_2}\int_{y_1}^{y_2}\int_{z_1}^{z_2}\frac{\partial{M(x,y,z,V')}}{\partial{V'}}\vert_{V'=0}dzdydx[/tex]

    Any comments?
     
    Last edited: Jun 30, 2005
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  3. Jun 30, 2005 #2

    James R

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    Actually:

    [tex]M = \int_V \rho(x,y,z) dx\,dy\,dz[/tex]

    Your expression for [itex]\rho[/itex] doesn't quite match this.
     
  4. Jun 30, 2005 #3

    quasar987

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    It doesn't ?!
     
  5. Jun 30, 2005 #4
    it doesn't. coz it isn't complete differential. why you put |v' = 0?
     
  6. Jun 30, 2005 #5

    dextercioby

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    It makes no sense to write

    [tex]M=M (x,y,z,V) [/tex]

    How would you exemplify and justify such choice?

    Daniel.
     
  7. Jun 30, 2005 #6

    reilly

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    See any book on probability theory and/or measure theory-Lebesgue integration. A highly general density function is defined as a measureable (positive) function.

    For mass, any positive integrable point function will work as a density. For charge, any integrable function will do.

    Your definition makes no sense, as written. Why do you assume thaqt dx/dV is zero?
    If V is contained in W, what's the relation between the density with V vs. that of W? Etc., etc........ (In probability, you could have density conditional on a volume, but why?. Why mess up the notion that density is purely a point function. That is, if density is local, the local density does not depend upon far away charges, masses, or whatever.

    And, what's wrong with density=limit(V->0) deltaM/deltaV? This definition has survived for several centuries.

    Regards,
    Reilly Atkinson
     
  8. Jul 3, 2005 #7

    quasar987

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    I don't understand any of your objections so I won't reply to them.

    But reilly, how are defined M, V, [itex]\Delta M[/itex] and [itex]\Delta V[/itex] in

    [itex]\rho(x,y,z) = \lim_{\Delta V \rightarrow 0}\frac{\Delta M}{\Delta V}[/tex]?

    How could that limit span a function of x,y,z if M is not itself a function of x,y,z? and how could M be differentiated wrt V if M is not also a function of V?
     
  9. Jul 3, 2005 #8

    Galileo

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    Quasar, I really don't see a need for a mathematically rigorous definition of a mass-density function. The idea is physically clear as it is.

    What you WANT from the density function is that the integral of that function over some volume gives the mass enclosed by that volume. That is sufficient to completely characterize it. Take that as your definition.
     
  10. Jul 4, 2005 #9

    quasar987

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    Insightful post as always Galileo. So really there is no definition of M and V in

    [tex]\rho = \frac{dM}{dV}[/tex]

    other than that M is the mass of the object such that rho integrated over V equals M? Is this your opinion too reilly?
     
  11. Jul 4, 2005 #10

    Hurkyl

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    You were somewhat on the right track, quasar. The leap you were missing is that you want mass to be a set function. In particular, you wanted something like:

    For any region R, M(R) is the mass contained within R.

    This should be a special kind of set function called a measure.

    The function V(R) = volume of R is another measure.


    The thing we like to do with measures is to integrate (measurable) functions over regions. You're quite familiar with integrals such as [itex]\int_R f \, dV[/itex]. Well, volume isn't "special": we can do this with any measure.

    For example, we have the following identity:

    [tex]M(R) = \int_R \, dM[/tex]


    Now, back to density...

    It is often better to ask yourself, "What do I want a thing to do?" than to ask yourself "How do I construct this thing?" As Galileo said, then thing you really want to be able to say is:

    [tex]M(R) = \int_R \rho \, dV[/tex]

    In other words, the mass enclosed within a region is the integral of ρ with respect to volume within that region.


    And then we appeal to powerful mathematical machinery. (IOW the stuff that says we can do what we want to do) :biggrin:


    The Radon-Nikodym theorem says that, given two measures μ and ν satisfying some reasonable conditions, there is a function f such that:

    [tex]\nu(E) = \int_E \, d\nu = \int_E f \, d\mu[/tex]

    over any (Borel) set E.


    mass and volume ought to satisfy those reasonable conditions (I'll assume no point masses -- that surely messes things up), so this theorem guarantees the existance of a density function.


    There is a strong analogy with differentiation: in fact, we like to say that ρ is the Radon-Nikodym derivative of mass with respect to volume, and Royden writes it as:

    [tex]\rho := \left[ \frac{dM}{dV} \right][/tex]


    Voila, a rigorous existence proof.


    Anyways, now that you know that you need to think of mass and volume as being a function on sets, I'll let you work out a more "constructive" definition for ρ. Hint: you were with the right track thinking about regions enclosing a particular point.
     
  12. Jul 4, 2005 #11

    quasar987

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    Wow, that was sa-weet! :biggrin:


    Ok, err, so how about, for R centered on (x,y,z),

    [tex]\rho(x,y,z) = \lim_{\Delta R \rightarrow 0}\frac{M(R+\Delta R) - M(R)}{V(R+\Delta R) - V(R)}=\frac{dM}{dV}[/tex]

    ...and I can't help but add "evaluated at R = 0" *take cover*.
     
    Last edited: Jul 4, 2005
  13. Jul 5, 2005 #12

    Hurkyl

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    It's not that simple. What's a ΔR, for instance? How do you add sets?

    I can imagine three ways to proceed:

    Trying to figure out how to make everything make sense in the approach you're taking.

    Thinking geometrically about how the ε-δ definition of a limit for ordinary functions works.

    Thinking about the integrals.
     
  14. Jul 5, 2005 #13

    quasar987

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    I don't know man.

    This here

    [tex]M(R) = \int_R \rho \, dV[/tex]

    seems quite ok a definitnion and it involves integral, but that's probably not what you were thinking about.
     
  15. Jul 5, 2005 #14

    dextercioby

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    We usually call domains by [itex] D [/itex]. That [itex] R [/itex] can be mistaken with the set of reals.

    Daniel.
     
  16. Jul 11, 2005 #15

    quasar987

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    Can you just tell me what it is you had in mind Hurkyl?
     
  17. Jul 11, 2005 #16

    Hurkyl

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    The basic idea is that we want a small mass (ΔM) enclosed in a small volume (ΔV) containing the point in question, and take the ratio.

    So, what we want to do is something like:

    [tex]
    \rho(P) = \lim_{V(S) \rightarrow 0} \frac{M(S)}{V(S)}
    [/tex]

    where S ranges over all regions containing P. But there will be technical difficulties! (See if you can figure them out!)


    But, we can do this:

    [tex]
    \rho(P) = \lim_{\epsilon \rightarrow 0^+} \frac{M(B(P, \epsilon))}{V(B(P, \epsilon))}
    [/tex]

    where B(P, ε) is the open ball centered on P with radius ε.

    We could be somewhat more loose in what regions we allow the limit to range over, but I would expect any such workable variation to be equivalent to this one.
     
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