MTW Exercise 20.5 -- prove ∇2Φ=4πρ

[TerryW]

Homework Statement

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 $∇^2Φ = 4πρ$

Homework Equations

See below

The Attempt at a Solution

[/B]
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

 

[TSny]

actually $\Big(\frac{1}{r}\Big)_{,ii} = \frac {-3x^ix^i}{r^5} + \frac {1}{r^3}$

This is the correct expression before summing over $i$. What does it yield after summing?

 

[TerryW]

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 $∇^2φ = 0$RegardsTerryW

 

[TSny]

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

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.

 

[TerryW]

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

 

[TSny]

So if I start from:

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

$∇^2φ = 0$

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 $\rho = M \delta \left ( \vec r \right )$.

 

[TerryW]

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 supportRegards
TerryW

 

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

OK. That exercise looks like another tedious one!

 

[TerryW]

All I've done here is:

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

OK. That exercise looks like another tedious one!

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.RegardsTerryW

 

[TSny]

All I've done here is:

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

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.

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.

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.

 

[TerryW]

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?RegardsTerryW

 

[TSny]

Terry,
I'm traveling until Thursday. I'll take a look at it then.

 

[TSny]

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 $\varphi$.

 

[I like Serena]

Doesn't the problem statement imply that $\rho$ is constant throughout the volume?
That is, not a point mass?

 

[TSny]

Doesn't the problem statement imply that $\rho$ is constant throughout the volume?
That is, not a point mass?

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.

 

[TerryW]

Hi TNsy,

Something odd happened on this thread - I didn't get any notification of your post #13, I found it today when I checked on your recent posts!

So the conclusion is that I can't go down my initial route because:

"$T^{00}$ 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 $\varphi$."

(I tried to use QUOTE for the above but it didn't come out well!)

I did get the required results using 20.22, so I should be happy about that.

I've resolved my problem with part (b) so Exercise 20.5 is done!I've just reached the end of Chapter 20 and completed Exercise 20.9, getting the result φ,_αα = 0. Is this correct? Regards
TerryW

 

[TSny]

I did get the required results using 20.22, so I should be happy about that.

I've resolved my problem with part (b) so Exercise 20.5 is done!

OK, good! I think I also got it to work out, but it was very tedious for me.

I've just reached the end of Chapter 20 and completed Exercise 20.9, getting the result φ,_αα = 0. Is this correct?

I believe you should get $\phi^{\, , \alpha} \, _{;\alpha} = 0$. One index should be up and the other down so that there is a summation over $\alpha$. Also, the second derivative should be a covariant derivative. This makes $\phi^{\, , \alpha} \, _{;\alpha}$ a scalar quantity. So, if the equation of motion is satisfied in one frame, it will be satisfied in all frames.

$\phi^{\, , \alpha} \, _{;\alpha}$ is the d'Alembertian of $\phi$ and can be written as $\phi_{\, ; \alpha} \, ^{\alpha}$ or $\square \, \phi$. See MTW page 863, Exercise 32.10.

 

[TerryW]

Posted in error - finger trouble

 

[TerryW]

I originally worked through to get $\phi^{,α}~_{,α}$ and then had a mad moment and decided it looked nicer as $\phi^{,αα}$!

I also started out by looking at $T^{αβ}~_{;β}$ then decided, unnecessarily, to work in flat spacetime. I should have just carried on with the covariant derivative!

Might be looking at Exercise 32.10 in a few years time!RegardsTerryW