Question about the weak field approximation

  • Thread starter Terilien
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  • #1
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How come in the weak field approximation, where the metric is equal to,
ds^2=-(1+2phi)dt^2 + dr^2(1-2phi). where of course dr is the three distance. why is phi multiplied by 2?

I have two more stupid question regarding a different approach. please just explain it to me as i want to to see this through to end(in understand all the other parts of the chapter).

first off
i don't understand some of the steps here.

http://www.mth.uct.ac.za/omei/gr/chap7/node3.html

How do we derive the inverse of the perturbed metric(i know its silly). More precisely why is the unverse of the perturbed metric the inverse of the minkowski metric minus the inverse of the perturbation?

Also why is the christoffel symbol in the weak field approximation equal to -1/2n^a0(h_00,0). where h is the perturbation and n is the minkowski metric. Seeing as how we are plugging in the metric itself, which is the sum of the perturbation and the minkowski metric, why isn't the christoffel symbol, equal to, -1/2g^a0(g_00,0).

Please help me with this and write it out. I know that at this point gr is probably beyond me(at least without help from a mentor), but i'd like to reach einstein's equationa t the very least, especially since i'm so close.

so please just explain this to me.
 
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Answers and Replies

  • #2
pervect
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"Why" is always a tricky question, but I'll point out that the above expression gives the correct value for gravitational time dilation:

If
[tex]
d \tau^2 = 1 + 2 \Phi dt^2
[/tex]

then
[tex]
d\tau \approx 1 + \Phi
[/tex]

becase
[tex]\sqrt{1 + x} \approx 1 + \frac{1}{2}x - \frac{1}{8} x^2 ...
[/tex]
via a taylor series expression around x=0. If you ignore the second order and higher terms, you get the result shown.

You can also get this expression from comparison with the Schwarzschild metric.

I think you can see why the inverse metric works out the way it does if you perform a matrix multipliation between the inverse metric and the regular metric. You should get the identity matrix to the first order, i.e. the errors should all be quadratic in terms of the pertubation, 1-x^2.

The Christoffel symbols are computed via the standard formula, for instance see the wikipedia article http://en.wikipedia.org/wiki/Christoffel_symbol

I'm not quite sure what level of explanation you're looking for there.
 
  • #3
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Wait can you explain how we get the formula for the inverse of the perturbed metric ? I mean derive it via the the expansion(seeing as how you said first order I'm assumign you used one).

Please i'm very close and would appreciate it.
 
  • #4
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The perturbed metric is:
[tex] g_{ \alpha \beta }= \eta_{\alpha \beta} + \epsilon\ h_{\alpha \beta} [/tex]

We guess, up to terms of order [tex] \epsilon [/tex] (dropping higher order terms), the inverse metric is:
[tex] g^{ \beta \gamma }= \eta^{\beta \gamma} - \epsilon\ h^{\beta \gamma} [/tex]

The guestion is what must be [tex] h^{\beta \gamma} [/tex]. We find out by multiplying the two matrices and demanding the result is the unit matrix up to first order in [tex] \epsilon [/tex] (dropping terms of higher order):

[tex] g_{ \alpha \beta }g^{ \beta \gamma } = \delta_\alpha^\gamma + \epsilon \ ( h_{\alpha \beta} \ \eta^{\beta \gamma} - \eta_{\alpha \beta} \ h^{\beta \gamma}) + O(\epsilon^2) [/tex]

From that we conclude that [tex] h_{\alpha \beta} \ \eta^{\beta \gamma} = \eta_{\alpha \beta}\ h^{\beta \gamma} [/tex]. To get rid of [tex] \eta [/tex] on the right hand side, we multiply both sides by [tex] \eta^{\alpha \delta} [/tex] (summation over repeated indices implied) and obtain:

[tex] h_{\alpha \beta}\ \eta^{\alpha \delta}\ \eta^{\beta \gamma} = h^{\delta \gamma} [/tex]

That shows h with upper indices is not the matrix inverse of h with lower indices but is obtained by 'raising the indices' using the non-perturbed metric [tex] \eta [/tex].
 
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  • #5
rbj
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"Why" is always a tricky question, but I'll point out that the above expression gives the correct value for gravitational time dilation:

If
[tex]
d \tau^2 = 1 + 2 \Phi dt^2
[/tex]

then
[tex]
d\tau \approx 1 + \Phi
[/tex]

becase
[tex]\sqrt{1 + x} \approx 1 + \frac{1}{2}x - \frac{1}{8} x^2 ...
[/tex]
via a taylor series expression around x=0. If you ignore the second order and higher terms, you get the result shown.

wouldn't that be

[tex]
d\tau \approx (1 + \Phi) dt
[/tex]

?

should also be

[tex]
d \tau^2 = (1 + 2 \Phi) dt^2
[/tex]

unless c is normalized to 1 or something like that. no?
 
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  • #6
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The perturbed metric is:
[tex] g_{ \alpha \beta }= \eta_{\alpa \beta} + \epsilon h_{\alpha \beta} [/tex]

We gess that up to first order in epsilon the inverse metric is:
[tex] g^{ \alpha \beta }= \eta^{\alpa \beta} + \epsilon h^{\alpha \beta} [/tex]

The question is what is [tex] h^{\alpha \beta} [/tex] ?

We find out by multiplying the two matrices and demanding the result is the unit matrix up to first order in epsilon (dropping terms of higher order):

my question is how do you derive the inverse from a series expansion(i'm assuming that the term first order means that there is a series expansion).

also:

http://www.mth.uct.ac.za/omei/gr/chap7/node3.html [Broken]

In step 23 they show the formula for the christoffel symbols, but in the part where they usually place the inverse metric, they've only placed the inverse minkowksi metric. What happened to the perturbation?
 
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  • #7
pervect
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wouldn't that be

[tex]
d\tau \approx (1 + \Phi) dt
[/tex]

?

Ooops, yep.
 
  • #8
nrqed
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my question is how do you derive the inverse from a series expansion(i'm assuming that the term first order means that there is a series expansion).
You impose that [itex] g_{\mu \alpha} g^{\mu \beta} = \delta_{\alpha}^{\beta} [/itex]

So you start with [itex] g_{\mu \alpha} \equiv \eta_{\mu \alpha} + h_{\mu \alpha} [/itex], and you try something of the form

[tex] g^{\mu \beta} = \eta^{\mu \beta} + c h^{\mu \beta} [/tex]

where "c" is a constant to be determined. You see quickly that, in first order in h, things work if you pick c=-1. But to second order in h, it does not work. So you must add a quadratic term in h to the inverse metric and fix the coefficient of that term and this goes on forever.
also:

http://www.mth.uct.ac.za/omei/gr/chap7/node3.html [Broken]

In step 23 they show the formula for the christoffel symbols, but in the part where they usually place the inverse metric, they've only placed the inverse minkowksi metric. What happened to the perturbation?

They work in first order in h. Since the derivatives are already all of first order in h, one must simply use the approximation of the flat inverse metric there.

Patrick
 
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  • #9
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The perturbed metric is:
[tex] g_{ \alpha \beta }= \eta_{\alpa \beta} + \epsilon h_{\alpha \beta} [/tex]

We guess, up to terms of order [tex] \epsilon [/tex] (dropping higher order terms), the inverse metric is:
[tex] g^{ \beta \gamma }= \eta^{\beta \gamma} + \epsilon h^{\beta \gamma} [/tex]

The guestion is what must be [tex] h^{\beta \gamma} [/tex]. We find out by multiplying the two matrices and demanding the result is the unit matrix up to first order in [tex] \epsilon [/tex] (dropping terms of higher order):

[tex] g_{ \alpha \beta }g^{ \beta \gamma } = \delta_\alpha^\gamma + \epsilon ( h_{\alpha \beta} \eta^{\beta \gamma} - \eta_{\alpa \beta} h^{\beta \gamma}) + O(\epsilon^2) [/tex]

From that we conclude that [tex] h_{\alpha \beta} \eta^{\beta \gamma} = \eta_{\alpha \beta} h^{\beta \gamma} [/tex]. To get rid of [tex] \eta [/tex] on the right hand side, we multiply both sides by [tex] \eta^{\alpha \delta} [/tex] and obtain:

[tex] h_{\alpha \beta} \eta^{\alpha \delta} \eta^{\beta \gamma} = h^{\delta \gamma} [/tex]

That shows h with upper indices is not the matrix inverse of h with lower indices but is obtained by 'raising the indices' using the non-perturbed metric [tex] \eta [/tex].


where do we get the negative sign? and aren't the two matrices multitplied by one another equal to the indeitity map and just that. Where do the higher order terms come from?
 
  • #10
pervect
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If we take
[tex]
\begin{array}{ccdc}a&0&0&0\\0&b&0&0\\0&0&1&0\\0&0&0&1\end{array}
[/tex]

it's inverse is

[tex]
\begin{array}{cdcc}\frac{1}{a}&0&0&0\\0&\frac{1}{b}&0&0\\0&0&1&0\\0&0&0&1\end{array}
[/tex]

We can taylor series expand 1/(1+x) and 1/(1-x), or we can observe that for instance

[tex]\frac{1}{1+x} = \frac{1-x}{1-x^2} \approx 1-x[/tex]
 
  • #11
nrqed
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where do we get the negative sign? and aren't the two matrices multitplied by one another equal to the indeitity map and just that. Where do the higher order terms come from?

I think that your questions will be answered in my post. One point I did not emphasized, however, and an important point made by smallphi is that the raising and lowering of the indices on h is done with the unperturbed metric. I used that in my post, without putting the emphasis on that point.

Patrick
 
  • #12
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Terilien said:
where do we get the negative sign? and aren't the two matrices multitplied by one another equal to the indeitity map and just that. Where do the higher order terms come from?

If you don't put the negative sign by hand initially, at the end you will get [tex] h_{\alpha \beta}\ \eta^{\alpha \delta}\ \eta^{\beta \gamma} = - h^{\delta \gamma} [/tex] so it is equivalent. Read what nrqed said.

You get extra terms of order epsilon^2 in the unit matrix equation because you are approximating the inverse metric, not finding it exactly.
 
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  • #13
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My question is how to do we taylor series expand it. with respect to what
?

where do we get that c is equal to -1 and what do you mean by first order?

so essentially how do we approximate it?
 
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  • #14
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The small parameter is epsilon and you Taylor expand with respect to unperturbed metric eta. There was a sign typo in my post where I have outlined the calculation that is now corrected.
 
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  • #15
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so essentially what we're doing i we are guessing that that certain term is the inverse. and since epsilon is small we can ignore the higher order terms in epsilon without making too much of an error.

I still don't think I udnerstand it all that well, but I have a better idea.

Still why do we make that particular guess?

also as for the christoffel symbol, why do we not include the whole metric, instead of the jusepsilon times the minkowski metric?
 
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  • #16
441
2
Think of it this way. Given the intial perturbed metric, the inverse metric can be found simply by inverting the 4x4 matrix. The resulting formula will be hairy and will depend on epsilon and h_lower indices. We can Taylor expand the resulting expression for the inverse matrix around the Minkowski matrix using epsilon as the small parameter. We know there will be zero order term (Minkowski) and terms proportional to epsilon, epsilon^2 etc.

If we multiply the perturbed matrix by the full Taylor expansion of the inverted matrix you have to get the unit matrix exactly so no terms proportional to epsilon or higher power of epsilon should survive. The terms proportional to epsilon^1 are generated from the part of the Taylor expansion up to epsilon^1 (first order terms) no higher. I used that fact to consider only that part of the Taylor expansion and to cancel the first order term in the unit matrix. I haven't made an artificial choice because initially h with upper indices was completely unspecified and it was determined (not assumed) so that it cancels out the first order term in the unit matrix. Therefore IT IS the right coefficient of epsilon^1 in the actual Taylor expansion of the inverse metric. We derived it, did not assume it.

Later they are interested only in terms of first order in epsilon because Newtonian gravity is a weak perturbation of Minkowski. They raise indices with Minkowski because any term that is already first order in epsilon should be multiplied by zero order in epsilon (Minkowski part of the raising inverse metric) in order to get up to the first order in the result.
 
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  • #17
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ok the first part makes pefect sense. however i still don't quite understand what you said about the second part.

I:E why does the inverse matrix term equal epsilonN^munu rather tahn n^uv +h^uvepsilon

i think I understand now, but any extra explanation would be better.
 
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  • #18
nrqed
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ok the first part makes pefect sense. however i still don't quite understand what you said about the second part.

I:E why does the inverse matrix term equal epsilonN^munu rather tahn n^uv +h^uvepsilon

i think I understand now, but any extra explanation would be better.

Here is the way I see it.

First, for the flat metric we have [itex] \eta_{\mu \alpha} \eta^{\mu \beta} = \delta_{\alpha}^{\beta} [/itex]

Now, we impose that the full metric also obeys this condition,[itex] g_{\mu \alpha} g^{\mu \beta} = \delta_{\alpha}^{\beta} [/itex]

next, we define [itex] g_{\mu \alpha} \equiv \eta_{\mu \alpha} + h_{\mu \alpha} [/itex] where , by assumption, all the components of h are small compared to one so one may do an expansion in h.

So far, nothing was derived. These are all our initial assumptions. Now, imposing that the two conditions above must be compatible gives us the expansion for [itex] g^{\mu \alpha} [/itex] which may be worked out order by order in the small quantity h.

Hope this makes sense.
 
  • #19
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I already do understand this. This is actually the first time I've encountered taylor series(outside of derivation). Its cool. anyway I understand einstein's equation, thus I'm happy it was my original objective. I started studying mathematics last june(conics) and I'm so happy to have finally made it here!

Thank you for your help. Now I'm wondering where i should go from here....
 

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