What is the metric form ##g_{μν}## in this context?

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  • #1
hykhor0601
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Does anyone know how to solve this?
##ds^2=g_{μν}(x)dx^μdx^ν= -(r^6/l^6)[1-(Ml^2/r^2)]dt^2+{1/[(r^2/l^2)-M]}dr^2+r^2dΦ^2##

Does anyone know how to solve this?
 
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  • #2
Orodruin
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What do you mean by "solve"? There is nothing to "solve", there is just a line-element.
 
  • #3
Ibix
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Solve in what sense? It appears to be just a statement of a metric, missing a definition of ##l## and assuming ##d\Phi## is a solid angle element.
 
  • #4
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Can you provide some context here? Where di this equation come from? What book course etc...?

and what have you tried so far?

We had to modify your equation using Latex to make it readable:
ds^2=gμν(x)dx^μdx^ν= -(r^6/l^6)[1-(Ml^2/r^2)]dt^2+{1/[(r^2/l^2)-M]}dr^2+r^2dΦ^2

Basically the changes were to bracket the equation on both sides with double # and to replace the gμν with g_{μν} to make them into subscripts.
 
  • #5
hykhor0601
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Can you provide some context here? Where di this equation come from? What book course etc...?

and what have you tried so far?

We had to modify your equation using Latex to make it readable:


Basically the changes were to bracket the equation on both sides with double # and to replace the gμν with g_{μν} to make them into subscripts.
https://arxiv.org/pdf/0909.1347.pdf The metric is given in (9)
 
  • #6
hykhor0601
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Solve in what sense? It appears to be just a statement of a metric, missing a definition of ##l## and assuming ##d\Phi## is a solid angle element.
The conditions is in the article below, please help me
https://arxiv.org/pdf/0909.1347.pdf
 
  • #8
hykhor0601
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∂∨μ[(√-g)g∨μν∂∨ν(Φ)]-m^2√(-g)Φ=0
What is the meaning of ∂∨μ,√-g,∂∨ν(Φ) in this equation?

How to give the form of g^μν from the metric (9) in the article?(or how do i get the form of ∂∨μ[(√-g)g∨μν∂∨ν(Φ)]-m^2√(-g)Φ=0 )
https://arxiv.org/pdf/0909.1347.pdf
Please help me, thank you
 
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  • #10
Demystifier
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Why are you interested in this particular paper if you are not familiar with basics of general relativity?
 
  • #11
hykhor0601
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Again, what do you mean by "solve"? The metric is not a thing you solve. What are you trying to find out?
just the form of gμν that can be sub into the equation of motion
 
  • #12
Orodruin
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just the form of gμν that can be sub into the equation of motion
You already have ##g_{\nu\nu}##. What is your actual problem?
 
  • #13
hykhor0601
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You already have ##g_{\nu\nu}##. What is your actual problem?
can you write it in the form after substitude into
∂µ[√ (−g)gµν∂ν(ϕ) ]− m^2√ (−g)ϕ = 0
 
  • #14
Ibix
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I don't understand. You can read off the components of ##g_{\mu\nu}## from the line element and ##g## is just its determinant. Where's the difficulty?

It would also help immensely if you spent a few minutes reading the LaTeX Guide linked below the reply box. Then you could type maths in a way that we could actually read it.
 
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