Question on weak field approximation

  • #1
WannabeNewton
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In A First Course in General Relativity, the use of the weak field approximation is confusing to me. I constantly get confused when the term "f(x) is only valid to first order in f..." for the Newtonian potential in the metric comes up. At a certain point the book states:
...1/2(-1/(1 + 2[tex]\phi[/tex]))(-2[tex]\phi[/tex]),0 (where ,0 is the time derivative)
and then the book simplifies this to:
[tex]\phi[/tex],0 + 0([tex]\phi[/tex][tex]^{2}[/tex])

I have NO IDEA how that step came about. Does this step and the whole "correct to first order" have to do with a Taylor expansion and if so how does one come to what the book came to? Thanks.
 

Answers and Replies

  • #2
Nabeshin
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[tex]\frac{1}{1+2\phi}} \approx 1 - 2 \phi + \mathcal{O}(\phi^2)[/tex]

So when you multiply it all out...
[tex]\partial_0 \left(\phi - 2 \phi ^2\right) = \partial_0 \phi - 4 \phi \partial_0 \phi [/tex]

Since we assume phi is small, phi dot is also small, so the product of two small quantities is really small! Specifically, of order phi^2. So

[tex] = \partial_0 \phi + \mathcal{O}(\phi^2) [/tex]
 
  • #3
PAllen
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[tex]\frac{1}{1+2\phi}} \approx 1 - 2 \phi + \mathcal{O}(\phi^2)[/tex]

So when you multiply it all out...
[tex]\partial_0 \left(\phi - 2 \phi ^2\right) = \partial_0 \phi - 4 \phi \partial_0 \phi [/tex]

Since we assume phi is small, phi dot is also small, so the product of two small quantities is really small! Specifically, of order phi^2. So

[tex] = \partial_0 \phi + \mathcal{O}(\phi^2) [/tex]

Is there something special about phi? Obviously, in general, smallness of phi implies nothing about its derivative.
 
  • #4
WannabeNewton
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If you don't mind, could you explain what big omicron of psi ^ 2 actually is? I had no idea it was that; I thought it was a zero being multiplied by the function (and I kept thinking why this was being showed on the text) =p.
 
  • #5
PAllen
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If you don't mind, could you explain what big omicron of psi ^ 2 actually is? I had no idea it was that; I thought it was a zero being multiplied by the function (and I kept thinking why this was being showed on the text) =p.

That's 'big O' notation. It means that the error is 'of the order' of the big O argument. Thus:

g(x) = f(x)+O(x^2)

means that g(x) = f(x) + <some function bounded by a constant times x ^2>
 
  • #6
WannabeNewton
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so what does it mean when psi is only correct to first order? Thanks.
 
  • #7
Nabeshin
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Is there something special about phi? Obviously, in general, smallness of phi implies nothing about its derivative.

In general, you are correct. But this is not something limited to phi, but rather a general feature of doing these "to order x^2" type taylor expansions. Generally, if phi is small, we want it to stay small so our approximation is always valid. The way we do this is by ensuring all derivatives of phi are equally small.

"Correct to first order" means just that: everything up to phi^1 is correct. You can think of the real function we are approximating as an infinitely long series expansion, so we're saying "what we've written down as the solution is correct for the first term of that expansion". Similarly, correct to phi^2 would be stating that we have the first two terms in the expansion correct.

Basically, the nth order comment is just a reference to how many terms in the taylor series we are interested in.
 
  • #8
WannabeNewton
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Ah ok I think I understand. I think the book should have had an omicron in front of the function and not a zero which was really confusing. If it isn't any trouble could you explain if the given first order approximation can be found using a taylor expansion (or am I missing something?).
 
  • #9
Nabeshin
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Ah ok I think I understand. I think the book should have had an omicron in front of the function and not a zero which was really confusing.
Yes, this probably would have helped! That notation is pretty standard, not sure where you're supposed to be first introduced to it...
If it isn't any trouble could you explain if the given first order approximation can be found using a taylor expansion (or am I missing something?).

Umm, is that not what I did in my first post? I'm not entirely sure what you're asking for here.
 
  • #10
WannabeNewton
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Just not used to seeing it put that way when using taylor series sorry.
 

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