Initial Value Equations from ADM Formalism - Trying again

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  • #1
TerryW
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Homework Statement:
I'm having another attempt to get a bit of help working out how MTW (21.115) (attached) can be derived from:

##I = \frac{1}{16π}\int[ \dot π^{ij} γ_{ij} - N\mathcal{H} -N_i\mathcal{H^i}]d^4x + \int\mathfrak{L}_{Field}d^4x \quad## MTW (21.95)
Relevant Equations:
In the hope of making progress, I'm going to present my working so far bit by bit to see if anyone can spot where I am going wrong. I'll start with a bit that seems to work OK:

I am first going to vary ## - N\mathcal{H} ## wrt ##g_{ij}##
##\delta(- N\mathcal{H}) = \delta(-N[\gamma^{-\frac{1}{2}}(Trπ ^2 - \frac{1}{2}(Trπ )^2 -\gamma^{\frac{1}{2}}R])##

and I am going to concentrate on the first part and find ##\delta(-N[\gamma^{-\frac{1}{2}}(Trπ ^2 - \frac{1}{2}(Trπ )^2]) ## wrt ##g_{ij}##.

##\delta(-N[\gamma^{-\frac{1}{2}}(Trπ ^2 - \frac{1}{2}(Tr π )^2]) = \delta(-N^2[(-g)^{\frac{1}{2}}]^{-1}(Trπ ^2 - \frac{1}{2}(Tr π )^2) ##

## = -N^2(-1)[(-g)^{\frac{1}{2}}]^{-2} \frac{1}{2}(-g)^{\frac{1}{2}}g^{ij} (Trπ ^2 - \frac{1}{2}(Tr π )^2)\delta g_{ij} -N\gamma^{-\frac{1}{2}}\delta(Trπ ^2 - \frac{1}{2}(Tr π )^2)##

## = \frac{1}{2}N^2(-g)^{-\frac{1}{2}} g^{ij} (Trπ ^2 - \frac{1}{2}(Tr π )^2)\delta g_{ij} -N\gamma^{-\frac{1}{2}}\delta(Trπ ^2 - \frac{1}{2}(Tr π )^2)##

## = \frac{1}{2}N\gamma^{-\frac{1}{2}} g^{ij} (Trπ ^2 - \frac{1}{2}(Tr π )^2)\delta g_{ij} -N\gamma^{-\frac{1}{2}}(\delta(g_{js}π^{sm}g_{mi}π^{ij}) - \frac{1}{2} 2(Trπ)π^{ij}\delta g_{ij}##

## = \frac{1}{2}N\gamma^{-\frac{1}{2}} g^{ij} (Trπ ^2 - \frac{1}{2}(Tr π )^2)\delta g_{ij} -N\gamma^{-\frac{1}{2}}(π^{im}g_{ms}π^{sj} +g_{ms}π^{sj} π^{im}- (Trπ)π^{ij})\delta g_{ij}##


## = [\frac{1}{2}N\gamma^{-\frac{1}{2}} g^{ij} (Trπ ^2 - \frac{1}{2}(Tr π )^2) -2N\gamma^{-\frac{1}{2}}(π^{im}π_m{}^j +g_{ms}π^{sj} π^{im}- \frac{1}{2}(Trπ)π^{ij})]\delta g_{ij}##

This has produced two of the terms in MTW's (21.115) which makes me feel that my process for finding the variation wrt ##g_{ij}## is probably sound.

Can anyone suggest where I might be missing something?


Regards


Terry W





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Answers and Replies

  • #2
TerryW
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My original post had quite a few looks but didn't get any replies. I am going to reply to myself and continue to set out my workings in the hope that someone else will join the conversation and help me resolve my problem.

In the last part, I managed to produce two of the terms in (21.115). This next part produces another bit of (21.115) but this time there is a discrepancy:

I am now going to look at the second part of N##\mathcal{H} ( = N\gamma^{\frac{1}{2}}R)## to find the variation wrt ##g_{ij}##

So ##\delta(N\gamma^{\frac{1}{2}}R) = \delta((-g)^{\frac{1}{2}}R)##

= ##\delta((-g)^{\frac{1}{2}})R + (-g)^{\frac{1}{2}}\delta R)##

= ##\frac{1}{2}(-g)^{\frac{1}{2}}g^{ij}\delta g_{ij} R + (-g)^{\frac{1}{2}}\delta (g_{ij}R^{ij})##

= ##(\frac{1}{2}N\gamma^{\frac{1}{2}}g^{ij}R + N\gamma^{\frac{1}{2}}R^{ij})\delta g_{ij}##

= ##-[N\gamma^{\frac{1}{2}}(R^{ij} -\frac{1}{2}g^{ij}R)]\delta g_{ij} + 2N\gamma^{\frac{1}{2}}R^{ij}\delta g_{ij}##


I now have another element of (21.115) with ##-[N\gamma^{\frac{1}{2}}(R^{ij} -\frac{1}{2}g^{ij}R)]## but to achieve this, I have had to add in an extra term ##2N\gamma^{\frac{1}{2}}R^{ij}##.

There are four more terms to vary and I have found that surplus elements in one variation cancel surplus elements in another, but nowhere along the line can I find anything to cancel ##2N\gamma^{\frac{1}{2}}R^{ij}##.

Is there anyone out there able to join in this conversation?


Regards


Terry W
 
Last edited:
  • #3
MathematicalPhysicist
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Besides asking Kip Thorne or the other authors I don't have any other helpful advice.
Let us know how did you solve it in the end.
 
  • #4
TerryW
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Hi Mathematical Physicist,

I might try this at a later date when I have completed the presentation of my workings. I wonder if I am doing this all wrong but you haven't suggested that this is the case, so I'll carry on with the next bit, which is to find the variation of ##2N_i \pi^{ik}{}_{|k}##

So ##\delta(2N_i \pi^{ik}{}_{|k}) = \delta(2g_{ij}N^j \pi^{ik}{}_{|k})##

= ##\delta (g_{ij}N^j\pi^{ik}{}_{|k} + g_{ij}N^i\pi^{jk}{}_{|k})##

= ##\delta(g_{ij}(N^j\pi^{ik})_{|k} =g_{ij}(N^i\pi^{jk})_{|k} -g_{ij}N^j{}_{|k}\pi^{ik}-g_{ij}N^i{}_{|k}\pi^{jk})##

= ##(-N^j{}_{|k}\pi^{ik} -N^i{}_{|k}\pi^{jk})\delta g_{ij} = 2(N^i\pi ^{jk})_{|k}\delta g_{ij}##

This gives me another element of ( with 21.115) with ##(-N^j{}_{|k}\pi^{ik} -N^i{}_{|k}\pi^{jk})## but I have an unwanted term ##2(N^i\pi ^{jk})_{|k}##

In this case, the unwanted term turns out not to be a problem because it is cancelled out by a term which arises in the next chapter - which will appear when I get back from holiday!

TerryW
 
  • #5
MathematicalPhysicist
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Well I am no expert in GR, I haven't even finished reading Schutz's and the Solution manual I have for it.
Maybe I'll get back to it who knows when.
 
  • #6
TerryW
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For the next variation, I have to start working on the divergence part (so we can't just ignore it??) and take first of all:

##\delta[(-2\pi^{ij}N_j)_{,i}]##

= ##\delta[(-2\pi^{kj}N_j)_{,k}]##

= ##\delta[(-2\pi^{kj}g_{ij}N^i)_{,k}]##

=##\delta[-g_{ij}(2\pi^{kj}N^i)_{,k} - 2\pi^{kj}N^i g_{ij, k}]##

=## -2(\pi^{kj}N^i)_{,k}\delta g_{ij}##

=## -2(\pi^{kj})_{,k}N^i +\pi^{kj})N^i {}_{,k})\delta g_{ij}##

= ##[-2(\pi^{kj})_{|k}N^i +\pi^{kj})N^i {}_{|k}) +2(N^i(\Gamma ^k{}_{mk}\pi^{mj}+\Gamma ^j{}_{mk}\pi^{km}-\Gamma ^m{}_{mk}\pi^{kj}) +\pi^{kj}\Gamma^{i}{}_{mk}N^m)]\delta g_{ij}##

##= [-2(\pi^{kj}N^i)_{|k}) +2(N^i\Gamma ^k{}_{mk}\pi^{mj}+N^i \Gamma ^j{}_{mk}\pi^{km}-N^i\Gamma ^m{}_{mk}\pi^{kj} +N^m \pi^{kj}\Gamma^{i}{}_{mk})]\delta g_{ij}##

= ##= [-2(\pi^{kj}N^i)_{|k}) +2(N^i \Gamma ^j{}_{mk}\pi^{km} +N^m \pi^{kj}\Gamma^{i}{}_{mk})]\delta g_{ij}##
(Because ##N^i\Gamma ^k{}_{mk}\pi^{mj} = N^i\Gamma ^m{}_{mk}\pi^{kj}##)

My first term ##(-2(\pi^{kj}N^i)_{|k})## cancels the unwanted term from my last post, but now I am left with two more unwanted terms: ##2N^i \Gamma ^j{}_{mk}\pi^{km} +2N^m \pi^{kj}\Gamma^{i}{}_{mk}##

In my next post, I'll get rid of one of these two terms, but gain another unwanted term.

TerryW
 
  • #7
TerryW
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Now I am going to look at the variation ##\delta((N^i Tr\pi)_{,i})##

##\delta((N^i Tr\pi)_{,i})## = ##\delta((N^k g_{ij} \pi^{ij})_{,k})##

## = \delta((g_{ij}(N^k \pi^{ij})_{,k} + (N^k \pi^{ij})g_{ij, k})##

##= (N^k \pi^{ij})_{,k}\delta g_{ij}##

##= (N^k{}_{|k} - \Gamma^k{}_{mk} N^m)\pi^{ij} +N^k(\pi^{ij}{}_{|k} - (\Gamma^i{}_{mk}\pi^{mj} + \Gamma^j{}_{mk}pi^{mi} - \Gamma^m{}_{mk}\pi^{ij}))\delta g_{ij}##

##= ((N^k \pi^{ij})_{|k} - N^k(\Gamma^i{}_{mk}\pi^{mj} + \Gamma^j{}_{mk}\pi^{mi}) - N^m \pi^{ij}\Gamma^k{}_{mk} + N^k \pi^{ij} \Gamma^m{}_{mk})\delta g_{ij}##

Which, with a bit of index manipulation, becomes

##= ((N^k \pi^{ij})_{|k} -2N^m \Gamma^i{}_{mk}\pi^{kj})\delta g_{ij}##

The first term is an element of (21.115) and the second term cancels one of the unwanted terms in my last post (without adding a further unwanted term as stated previously).

Now there is just one more post to go.
 
  • #8
TerryW
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Here is the final piece of my 'derivation', taking the variation of ##-(2N^{|i}\gamma^{\frac{1}{2}})_i##

##-\delta(-(2N^{|i}\gamma^{\frac{1}{2}})_{,i }) = \delta(-(2N^{|i}\frac{(-g)^{\frac{1}{2}}}{N})_{,i})##

##= \delta(+2N^{|i}(-g)^{\frac{1}{2}}\frac{N_{,i}}{N^2} - 2(\frac{(-g)^{\frac{1}{2}}}{N}N^{|i})_{,i})##

Then using (21.85)

##= \delta(+2N^{|i}(-g)^{\frac{1}{2}}\frac{N_{,i}}{N^2} - 2\frac{(-g)^{\frac{1}{2}}}{N}N^{|i}{}_{|i})##

##= 2(\frac{1}{2}\frac{(-g)^{\frac{1}{2}}}{N^2}g^{ij}\delta g_{ij})N^{|i}N_{|i} - 2(\frac{1}{2}(-g)^{\frac{1}{2}}g^{ij}\delta g_{ij})\frac{N^{|i}{}_{|i}}{N}##

##= \gamma^{\frac{1}{2}}g^{ij}\frac{N^{|i}N_{|i}}{N} \delta g_{ij} - \gamma^{\frac{1}{2}}g^{ij}N^{|m}{}_{|m}\delta g_{ij}##

##= \gamma^{\frac{1}{2}}\frac{N^{|i}N^{|j}}{N} \delta g_{ij} - \gamma^{\frac{1}{2}}g^{ij}N^{|m}{}_{|m}\delta g_{ij}##

##\Big[ \Big( \frac{N^{|i}}{N} \Big) ^{|j} = \frac{N^{|ij}}{N} - \frac{N^{|i}N^{|j}}{N^2} \Rightarrow \frac{N^{|i}N^{|j}}{N^2} = -N \Big( \frac{N^{|i}}{N} \Big) ^{|j} +N^{|ij} \Big]##

Leading to

##\gamma^{\frac{1}{2}}\frac{N^{|i}N^{|j}}{N} \delta g_{ij} - \gamma^{\frac{1}{2}}g^{ij}N^{|m}{}_{|m}\delta g_{ij} = \Big(- \gamma^{\frac{1}{2}}g^{ij}N^{|m}{}_{|m} + \gamma^{\frac{1}{2}}N^{|ij} - \gamma^{\frac{1}{2}}N\Big( \frac{N^{|i}}{N} \Big) ^{|j}\Big)\delta g_{ij}##

To summarise:

I have now derived all the elements of 21.115 but I have three unwanted extras which are:

## 2N\gamma^{\frac{1}{2}}R^{ij} + 2\Gamma^j{}_{mk}\pi^{mj}N^i - \gamma^{\frac{1}{2}}N\Big( \frac{N^{|i}}{N} \Big) ^{|j}##

I can't see any prospect of these terms cancelling each other out and I don't think they qualify for the description of "source terms arising from fields other than geometry"

So unless someone can wade in and show me where I've gone wrong, I am stuck!!


Terry W
 

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