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MTW Exercise 20.5 -- prove ∇2Φ=4πρ

  1. Sep 7, 2017 #1

    TerryW

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    1. The problem statement, all variables and given/known data
    I have a couple of issues with this problem, but first of all, I want to find out what I am missing in my attempt to prove ##∇^2Φ = 4πρ##

    2. Relevant equations
    See below

    3. The attempt at a solution

    If ##Φ = \frac {-M}{r}##

    Then ##Φ_{,i} = \frac {Mx^i}{r^3}##

    So ##Φ_{,ii} = \frac {-3Mx^ix^i}{r^5} + \frac {M}{r^3} = \frac {-2M}{r^3}##....(1)

    Taking ##ρ = \frac{M}{V} = \frac{3M}{4πr^3}##

    Gives ##Φ_{,ii} = \frac {-8πρ}{3}##

    Not what I wanted at all!

    I've looked through some of my solutions to previous exercises and found that I've used something like (1) before (actually ##\Big(\frac{1}{r}\Big)_{,ii} = \frac {-3x^ix^i}{r^5} + \frac {1}{r^3}##) and it produced the correct answer. Why doesn't it work here?


    TerryW
     
  2. jcsd
  3. Sep 7, 2017 #2

    TSny

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    This is the correct expression before summing over ##i##. What does it yield after summing?
     
  4. Sep 8, 2017 #3

    TerryW

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    Hi TSny - coming to my rescue again!

    Depending on the way I look at this I can give two answers!

    1. If I just follow the rules of tensor algebra, I would say that ##x^ix^i = r^2##, so ##\Big(\frac{1}{r}\Big)_{,ii} = \frac{-3r^2}{r^5} +\frac{1}{r^3} = \frac{-2}{r^3}##

    2. My alternative is to evaluate ##∇^2φ## explicitly for x,y & z:

    ##\frac {∂2φ}{∂x^2} = \frac{-3x^2}{r^5} + \frac{1}{r^3}##

    ##\frac {∂2φ}{∂y^2} = \frac{-3y^2}{r^5} + \frac{1}{r^3}##

    ##\frac {∂2φ}{∂z^2} = \frac{-3z^2}{r^5} + \frac{1}{r^3}##

    So ##∇^2φ = \frac{-3x^2-3y^2-3z^2}{r^5} + 3\frac{1}{r^3} = 0 ## !!!

    I also get ##∇^2φ = 0 ## if I used spherical polar coordinates.

    Looking at the 'tensor algebra' route, if I re-wrote my expression for ##\Big(\frac{1}{r}\Big)_{,ii}## as ##\Big(\frac{1}{r}\Big)_{,ii} = \frac{-3r^2}{r^5} +\frac{δ^i_i}{r^3} ## then ##\Big(\frac{1}{r}\Big)_{,ii} = 0##

    So I now have three consistent ways of showing ##∇^2φ = 0##


    Regards


    TerryW
     
  5. Sep 8, 2017 #4

    TSny

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    You can write this out as you did in your first way in 2.

    ##\Big(\frac{1}{r}\Big)_{,ii}## means ##\sum_{i=1}^{3} \Big(\frac{1}{r}\Big)_{,ii} = ## ##\sum_{i=1}^{3} \left (-3\frac{x^ix^i}{r^5} +\frac{1}{r^3} \right)##

    So, the ##\frac{1}{r^3}## part gets summed even though it has no ##i## indices.
     
  6. Sep 10, 2017 #5

    TerryW

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    So if I start from:

    ##φ_{,i} = \frac{-Mx^i}{r^3}## and follow the process above, I end up with

    ##∇^2φ = 0##

    I can however take a different route using the fact that ##φ_{,i} = \frac{-Mx^i}{r^3}## represents the gravitational field and then connect the field to its 'source" by

    ##\int φ_{,i}.dS = \int 4πρdV##

    ##\int φ_{,ii}dV= \int 4πρdV##

    ∴ ##∇^2φ = 4πρ##

    How do I reconcile this ambiguity?


    TerryW
     
  7. Sep 10, 2017 #6

    TSny

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    Here, you are assuming that you have a point source M. The expressions ##φ_{,i} = \frac{-Mx^i}{r^3}## and ##∇^2φ = 0## are valid everywhere except at ##r = 0##, where you have a singularity. This can be handled in terms of a delta function. [Edit: I don't think the minus sign should be in the expression ##φ_{,i} = \frac{-Mx^i}{r^3}##.]

    As shown in Jackson's EM book for example, you can take care of ##r = 0## by writing ##∇^2(\frac{1}{r} ) = -4 \pi \delta \left ( \vec r \right ) ##.
    ##\rho## for the point source can be taken to be ##\rho = M \delta \left ( \vec r \right ) ##.
     
    Last edited: Sep 10, 2017
  8. Sep 13, 2017 #7

    TerryW

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    Hi TSny,

    This works fine for ##\int ∇^2φdV = \int 4πρdV## but taking the next step to assert that ##∇^2φ = 4πρ## leads one to assign a radial functionality, i.e. ##∇^2φ = \frac {3M}{r^3}## but is it legitimate to do this?

    I was going to go on to my original issues with Ex 20.5, but I've just spotted a mistake which I now need to work through to see if it clears one of my problems, so it may be a little while before I post again.

    Thank you for your continued support


    Regards



    TerryW
     

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  9. Sep 13, 2017 #8

    TSny

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    I'm not seeing the last part of this where you state ##∇^2φ = \frac {3M}{r^3}##. Can you elaborate on this?

    OK. That exercise looks like another tedious one!
     
  10. Sep 13, 2017 #9

    TerryW

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    All I've done here is:

    ##4πρ = 4π\frac{M}{V} = 4πM(\frac{4πr^3}{3})^{-1} = \frac {3M}{r^3}##

    I've already successfully worked out (20.29) but I did it using (20.22) which is a bit of a slog. I started off using (20.20) and (20.21). This produced the correct result for ##T^{00}## but the ##t^{μν}_{L-L}## results aren't correct. This is where I found an error earlier today so I'm reworking it.

    I've had a similar problem with part (b) - I get all of (20.30) except that I get ##\frac{1}{6}ρφ## instead of ##\frac{1}{2}ρφ##

    I'll post again when I've had a good further look.


    Regards


    TerryW
     
  11. Sep 13, 2017 #10

    TSny

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    I'm not sure what ##r## represents in this expression.

    If you write the mass density of the particle as ##\frac{M}{V}##, then ##V## would be the volume occupied by the point mass. So, ##V = 0## which makes ##\frac{M}{V}## indefinite. The way to handle this singular mass density is to use the delta function as ##\rho = M \delta \left (\vec r \right)## which expresses the fact that the mass density is zero everywhere except at ##r = 0##. At ##r = 0## the density is "infinite" as expressed by the delta function.

    In Poisson's equation ##\nabla ^2 \phi = 4 \pi \rho##, both sides are functions of position. For a point mass M located at the origin, ##\phi = -\frac{M}{r}##. If you evaluate ##\nabla ^2 \phi## at any point other than the origin, we know ##\nabla ^2 \phi = -M \nabla ^2 \frac{1}{r} = 0##. This is in agreement with the right hand side of Poisson's equation since ##\rho = 0 ## at any point other than the origin.

    At the origin, ##\nabla ^2 \phi## is singular. But ##\nabla ^2 \phi = -M \nabla^2 \frac{1}{r} = -M\left (-4 \pi \delta \left (\vec r \right) \right) = 4 \pi\ M \delta \left (\vec r \right)##. This agrees with the right hand side of Poisson's equation.

    OK, sounds good. I will be traveling for the next week, so I won't have much time until I get back. Hope the problem goes well for you.
     
  12. Sep 19, 2017 at 9:50 AM #11

    TerryW

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    Hi TSny,

    This is my attempt to derive (20.29) from (20.20) and 20.21):

    Starting with ##t^{00}_{L-L}##

    ##H^{0α0β}_{L-L} = \mathfrak {g}^{00}\mathfrak {g}^{αβ}-\mathfrak {g}^{α0}\mathfrak {g}^{β0}##
    This implies ##α≠0, β≠0##

    Giving us
    ##H^{0i0j}_{L-L} = \mathfrak {g}^{00}\mathfrak {g}^{ij}-\mathfrak {g}^{i0}\mathfrak {g}^{j0} = \mathfrak {g}^{00}\mathfrak {g}^{ij}##

    So
    ##H^{0i0j}_{L-L} = (-g)g^{00}g^{ij}##


    ##[H^{0i0j}_{L-L}]_{,ij} = [(-g)g^{00}g^{ij}]_{,ij}##

    ##= [(-g)_{,i}g^{00}g^{ij}+(-g)(g^{00})_{,i}g^{ij}+(-g)g^{00}(g^{ij})_{,i}]_{,j}##

    ##=[(-g)_{,ij}g^{00}g^{ij}+(-g)_{,i}(g^{00})_{,j}g^{ij}+(-g)_{,i}g^{00}(g^{ij})_{,j}]##

    ##+[(-g)_{,j}(g^{00})_{,i}g^{ij}+(-g)(g^{00})_{,ij}g^{ij}+(-g)(g^{00})_{,i}(g^{ij})_{,j}]##

    ##+[(-g)_{,j}g^{00}(g^{ij})_{,i}+(-g)(g^{00})_{,j}(g^{ij})_{,i}+(-g)g^{00}(g^{ij})_{,ij}]##

    To evaluate this expression, I've used:

    ##(-g) = (1-4φ), \text { }g^{00} = -(1-2φ) \text { and }g^{ij} = δ_{ij}(1+2φ)##

    ##\begin{align}[H^{0i0j}_{L-L}]_{,ij} = &-4φ_{,ij}(-1+2φ)(δ_{ij})(1+2φ) -4φ_{,i}(2φ_{,j})(δ_{ij})(1+2φ)-4φ_{,i}(-1+2φ)(δ_{ij})2φ_{,j}\nonumber \\ &-4φ_{,j}(2φ_{,i})(δ_{ij})(1+2φ)+(1-4φ)(2φ_{,ij})(δ_{ij})(1+2φ)+(1-4φ)(2φ)_{,i}(δ_{ij})2φ_{,j} \nonumber \\ &-4φ_{,j}(-1+2φ)(δ_{ij})(2φ)_{,i}+(1-4φ)(2φ)_{,j})(δ_{ij})(2φ)_{,i} +(1-4φ)(-1+2φ)(δ_{ij})2φ_{,ij}\nonumber \end{align}##

    ##\begin{align}[H^{0i0j}_{L-L}]_{,ij} = &-4φ_{,ii}(4φ^2-1) -8φ_{,i}φ_{,i}(1+2φ)-8φ_{,i}φ_{,i}(-1+2φ)\nonumber \\ &-8φ_{,i}φ_{,i}(1+2φ)+(1-4φ)2φ_{,ii}(1+2φ)+(1-4φ)2φ_{,i}2φ_{,i} \nonumber \\ &-8φ_{,i}(-1+2φ)φ_{,i}+(1-4φ)4φ_{,i}φ_{,i} +(1-4φ)(-1+2φ)2φ_{,ii}\nonumber \end{align}##

    So, ignoring terms ##φ_{,i}φ_{,i}φ## but not ##φ_{,ii}φ## (see below for reason*)

    ##\begin{align}[H^{0i0j}_{L-L}]_{,ij} = &4φ_{,ii} -8φ_{,i}φ_{,i}+8φ_{,i}φ_{,i}\nonumber \\ &-8φ_{,i}φ_{,i}+2φ_{,ii}-4φ_{,ii}φ+4φ_{,i}φ_{,i} \nonumber \\ &+8φ_{,i}φ_{,i}+4φ_{,i}φ_{,i} -2φ_{,ii}+12φ_{,ii}φ\nonumber \end{align}##

    ##[H^{0i0j}_{L-L}]_{,ij} = 4φ_{,ii} +8φ_{,i}φ_{,i}+8φ_{,ii}φ##

    So

    ##16π(-g)(T^{00} + t^{00}_{L-L}) = 4φ_{,ii} +8φ_{,i}φ_{,i}+8φ_{,ii}φ##

    ##16π(T^{00} + t^{00}_{L-L}) = (1+4φ)(4φ_{,ii} +8φ_{,i}φ_{,i}+8φ_{,ii}φ)##

    ##16π(T^{00} + t^{00}_{L-L}) = 4φ_{,ii} +8φ_{,i}φ_{,i}+24φ_{,ii}φ## ................(1)

    This gives me ##T^{00}## but the residue is a problem. *It's the point at which I started to use ##4πρ = \frac{3M}{r^3}## so that I could recast ##φ_{,ii}φ## as ##-3φ_{,i}φ_{,i}## but this gives me ##-64φ_{,i}φ_{,i}## in (1), leading to ##t^{00}_{L-L} = \frac {8}{π}φ_{,i}φ_{,i}##

    I tried an alternative route in which I used ##(-g) = (1-4φ), \text { }g^{00} = -(1-2φ+4φ^2) \text { and }g^{ij} = δ_{ij}(1+2φ+4φ^2)## as the second derivatives produce some more ##φ_{,ii}φ## terms, but it didn't bring me any closer to the desired result.

    Can you see where I've gone wrong?


    Regards


    TerryW
     
  13. Sep 19, 2017 at 5:50 PM #12

    TSny

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    Terry,
    I'm traveling until Thursday. I'll take a look at it then.
     
  14. Sep 20, 2017 at 5:32 PM #13

    TSny

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    Terry,

    I believe MTW expect you to work part (a) using (20.22), which you were able to do according to your post #9. The pseudotensor ##t^{00}_{L-L}## is defined by (20.22).

    Note that ##T^{00}## and ##\varphi## for this part of the problem have not been specified. In particular, ##\varphi## is not necessarily equal to ##-M/r##. As I understand it, part (a) is dealing with a general "nearly Newtonian metric" (discussed in section 18.4) with a general Newtonian potential ##\varphi##.
     

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  15. Sep 20, 2017 at 5:52 PM #14

    I like Serena

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    Doesn't the problem statement imply that ##\rho## is constant throughout the volume?
    That is, not a point mass?
     
  16. Sep 20, 2017 at 6:06 PM #15

    TSny

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    Hello, ILS.
    Yes, I agree that the problem is not dealing with a point mass. I don't think that ##\rho## is assumed to be constant throughout some volume. ##\rho## can vary with position and time.
     
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