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The contiunity equation and differentiation

  1. Dec 25, 2007 #1
    The contiunity equation is given as:

    [tex] ln(\rho) + ln(A) + ln(V)=constant[/tex]

    "Differentiating Eq. 11.43 we get:

    [tex]\frac{d\rho}{\rho}+ \frac{dA}{A} +\frac{dV}{V}=0[/tex]

    My question is, what is this differentiated with resepct to?

    Its clearly not [tex]\rho[/tex], A or V alone.
     
    Last edited: Dec 25, 2007
  2. jcsd
  3. Dec 25, 2007 #2

    Gokul43201

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    It depends on what your "constant" is actually independent of. For instance, if the top equation is a constant in space or time, then a derivative with repsect to space or time co-ordinates will give you the bottom equation.
     
  4. Dec 25, 2007 #3
    Then, shouldnt it be [tex]\frac{d\rho}{dx}\frac{1}{\rho}[/tex], (assuming d.w.r.t. x).

    [tex]d\rho[/tex] alone, is meaningless.
     
  5. Dec 25, 2007 #4

    Gokul43201

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    While in the above case, it may simply be laziness of notation, I don't believe it is, in general, meaningless. But I'm not qualified to talk about differential forms, so I'll let someone else do it.
     
  6. Dec 26, 2007 #5

    HallsofIvy

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    It's not, strictly speaking, a derivative, it is a total differential. Just as, with f(x,y), both x and y are functions of t, we have (chain rule)
    [tex]\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}[/tex]
    and the total differential is
    [tex]df= \frac{df}{dx}dx+ \frac{df}{dy}dy[/tex]
    with no mention of a specific variable, t.
     
  7. Dec 26, 2007 #6

    arildno

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    Go along the curve of constant value C, with an accretion [tex](d\rho,dA,dV)[/tex]
    Then, we have two equations:
    [tex]\ln(\rho)+\ln(A)+\ln(V)=C[/tex]
    [tex]\ln(\rho+d\rho)+\ln(A+dA)+\ln(V+dV)=C[/tex]
    Subtracting, we get:
    [tex](\ln(\rho+d\rho)-\ln(\rho))+(\ln(A+dA)-\ln(A))+(\ln(V+dV)-\ln(V))=0[/tex]
    an equation for the related accretions whose linear part is given by your second expression.
     
  8. Dec 26, 2007 #7

    Gib Z

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    I know this completely goes against everything I know about mathematical rigor, especially since I haven't learned about differential geometry yet, but...

    [tex](\ln(\rho+d\rho)-\ln(\rho))+(\ln(A+dA)-\ln(A))+(\ln(V+dV)-\ln(V))=0[/tex]

    Dividing through by [tex]dA\cdot dV \cdot d\rho[/tex] we get some nice difference quotients giving:

    [tex]\frac{1}{\rho \cdot dA \cdot dV} + \frac{1}{A\cdot d\rho \cdot dV}+\frac{1}{V\cdot dA \cdot d\rho} = 0[/tex]

    Then, multiplying by [tex]dA\cdot dV \cdot d\rho[/tex] yields the desired result. Would this be an acceptable thing to do, pretending I knew the rigorous theory of differentials?
     
  9. Dec 26, 2007 #8

    arildno

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    You have, for finite accretions:
    [tex]\ln(\rho+d\rho)-\ln(\rho)=\frac{d\rho}{\rho}+O(d\rho^{2})[/tex]
    and so on, by simple Taylor expansions.

    The retained LINEAR equation is that relation which must hold between the differential accretions.
     
  10. Dec 26, 2007 #9

    Gib Z

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    What is a finite accretion :( ? And doesn't that Taylor series only have a radius of convergence of 1?
     
  11. Dec 26, 2007 #10

    arildno

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    A triple of real numbers (dp,dA,dV), not all zero.
     
  12. Dec 26, 2007 #11

    Gib Z

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    Can those differentials really be considered real numbers? Even when they are considered numbers in non-standard analysis they had to construct the hyper reals to do so.
     
  13. Dec 26, 2007 #12

    arildno

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    Well, I made the notational sin using d's rather than [itex]\bigtriangleup[/itex]'s instead.
     
  14. Dec 26, 2007 #13
    Differentiating is the calculation of tangent planes to a surface. There is no need for "with respect to" to do that. The "wrt" has the meaning of picking a preferred parameterisation (of curves on the surface).
     
  15. Dec 26, 2007 #14
    Thanks HallyofIvy and arildno.

    Also, the equation above it says:

    [tex]dp + \frac{1}{2}d(V^2)+\gamma dz=0[/tex]

    Here, [tex]d(V^2)=2VdV[/tex]. But in this case isnt it d.w.r.t. V? I.e. [tex]\frac{d}{dV}(V^2)[/tex]

    Or is that ok in terms of differential notation? I suppose its just another application of the total differential on [tex]V^2[/tex].

    Edit: Ah, it cant be [tex]\frac{d}{dV}(V^2)[/tex] because that would just yield 2V, and no dV term.
     
  16. Dec 26, 2007 #15
    Another thing I was not sure about:

    If you follow the derivation of the product rule it says:

    [tex] \Delta (uv)=u \Delta v +v\Delta u + \Delta u \Delta v[/tex]

    Am I not allowed to change the [tex]\Delta[/tex]'s into d's if I consider [tex]\Delta[/tex] very small? In which case this could also be viewed as an application of the product rule?

    I.e, use the product rule on [tex]\Delta (\rho VA)=0[/tex] and divide by the accretion [tex](d\rho,dA,dV)[/tex]
    to get the same result.
     
    Last edited: Dec 26, 2007
  17. Dec 26, 2007 #16

    Hurkyl

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    It's meaningful, you just haven't been taught the meaning.


    You are already encouraged to think of things like [itex]\rho[/itex], P, and V as 'numbers'. You know that they're really functions of your state, but you manipulate them as if they were ordinary numbers. For example, you write [itex]\ln \rho[/itex] when you really mean function composition; it is another function of your state, and evaluating it on a state [itex]\xi[/itex] gives you the number [itex](\ln \rho)(\xi) := \ln(\rho(\xi))[/itex]. You could view this as sloppiness, but it also turns out that there is a deep reason that suggests this really is the right way to be treating these things.


    Expressions like [itex]d\rho[/itex] are differential forms. But to keep in spirit, you simply think of a new kind of object called a 'differential', and it turns out that 'numbers' and 'differentials' relate in all the ways you would like them to relate, except for one -- you you can't divide 'differentials' to get a derivative1. In order to get derivatives, you have to pair a 'differential' with a 'vector'. (really, a vector field. But we call it a 'vector' to keep in the right spirit!) e.g. if you've parametrized your state space with the variables (s, t), then you have the 'vectors' [itex]\partial/\partial s[/itex] and [itex]\partial / \partial t[/itex]... and now everything works out:

    [tex]\left( d\rho \cdot \frac{\partial}{\partial s} \right) (\xi(s, t))
    =
    \frac{\partial}{\partial s} \left( \rho(\xi(s, t)) \right)
    [/tex]

    where the right hand side is an ordinary partial derivative that you learned in your calculus courses. As you might imagine, the left hand side is usually given the same notation: [itex]\partial \rho / \partial s := d\rho \cdot \partial/\partial s[/itex].


    1: Well, you probably can make some sense of division, but I won't try to do it here
     
    Last edited: Dec 26, 2007
  18. Dec 26, 2007 #17
    Why not? [tex] \frac{dy}{dx} = _{lim \Delta x \rightarrow 0} \frac{\Delta y}{\Delta x} [/tex]

    This is the differential, dy, divided by the differential dx.

    I.e. [tex] dy=f'(x)dx[/tex]

    Its exactly the division of dy by dx.
     
  19. Dec 26, 2007 #18

    Hurkyl

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    Division does make sense in some cases, but it doesn't work out so nicely in others. One of matt grime's favorite counterexamples is the identity

    [tex]
    \frac{\partial x}{\partial y} \frac{\partial y}{\partial z} \frac{\partial z}{\partial x} = -1
    [/tex]

    :smile:


    For example, if our state space is the plane (parametrized by (x, y)), and we define the scalar z by

    z = x + y

    then we also have

    x = z - y
    y = z - x

    and we compute

    [tex]
    \frac{\partial x}{\partial y} \frac{\partial y}{\partial z} \frac{\partial z}{\partial x}
    =
    (-1) \cdot 1 \cdot 1 = -1[/tex]


    The tricky point here is that the vector [itex]\partial / \partial y[/itex] is not well-defined on its own -- here, it was used to denote the directional derivative in the direction in which "y increases but z remains fixed". Similarly for [itex]\partial / \partial z[/itex] and [itex]\partial / \partial x[/itex].

    In other words, you have to have chosen a parametrization before these partial differentiation operators make sense, and [itex]\partial / \partial y[/itex] can mean different things depending on the parametrization. (e.g. the direction in which "y increases but z remains fixed" is different from the direction in which "y increases but x remains fixed", but alas, in the simplest notation, both would be notated [itex]\partial / \partial y[/itex])

    On the other hand, you need no such thing to make sense of differentials.


    Roughly speaking, differentials are to functions as vectors are to parametrizations. And to compute a derivative, you need both a differential and a vector.
     
    Last edited: Dec 26, 2007
  20. Dec 27, 2007 #19

    arildno

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    Not sure what you are saying here.
    By, definition, we have:
    [tex]\bigtriangleup(uv)=(u+\bigtriangleup{u})(v+\bigtriangleup{v})-uv[/tex]
    By merely using the arithemical law of distributivity, you gain your result.
     
  21. Dec 27, 2007 #20
    Im saying is it incorrect to replace the deltas with d's?
     
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