Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Einstein Field Equations

  1. Jul 2, 2012 #1
    Hi all,

    I have been trying to solve the Einstein Field Equations for its (0,0) component. So I have got that (c=1)

    Einstein Tensor (upper,0,0)=8*pi*G*T(upper,0,0)

    Now, lets see what T (0,0) really is. It is energy density, right? So According to famous E=mc^2 the energy density is the same as mass density, assuming the speed of light to be equal to one. Therefore, the (0,0) components of the Stress-Energy Tensor is just the mass density. So we have that

    Einstein Tensor (upper,0,0)=8*pi*G*ρ

    Now, lets see what Einstein Tensor G(0,0) really is

    Ricci(0,0) - 1/2*g(upper 0,0)*Ricci scalar

    Ricci scalar is obtained by contracting it with metric tensor so we have that

    Ricci (upper,0,0) - 1/2*g(upper 0,0)*Ricci (upper0,0)*g(lower, 0,0)=8*pi*G*ρ

    So g(upper 0,0)*g(lower, 0,0) is 1 so we have that

    Ricci (upper 0,0) - 1/2*Ricci(upper 0,0)=8*pi*G*ρ

    1/2 Ricci (upper 0,0)=8*pi*G*ρ


    Ricci(0,0)=4*pi*G*ρ


    Now look carefully to the Right Hand Side of the Equation. It is the same from the Poisson's Equation where

    Set of second partial derivatives of the Gravitational potential=4*pi*G*ρ

    Therefore, is it true that the zero-zero component of the Einstein Tensor, and subsequently the Ricci tensor, is just the [Set of second partial derivatives of the Gravitational potential] or 4*pi*G*ρ

    Thanks!


    P.S I apologize for not using MathCodes--never used them before and would appreciate if someone will show me how to use them.
     
    Last edited: Jul 2, 2012
  2. jcsd
  3. Jul 2, 2012 #2

    Mentz114

    User Avatar
    Gold Member

    With c=G=1
    [tex]
    \begin{align}
    R^\mu_\mu - (1/2)g^\mu_\mu R &= 8\pi T^\mu_\mu\\
    R &= -8\pi \rho
    \end{align}
    [/tex]
    for a pressureless, static matter ( dust) solution.

    The Riemann and Ricci tensors contain first and second derivatives of the metric.
    [tex]
    R^{r}_{msq}=\Gamma ^{r}_{mq,s}-\Gamma ^{r}_{ms,q}+\Gamma ^{r}_{ns}\Gamma ^{n}_{mq}-\Gamma ^{r}_{nq}\Gamma ^{n}_{ms}
    [/tex]

    If you right click on a formula, you get a menu. Select 'Tex' to see the format.
     
    Last edited: Jul 2, 2012
  4. Jul 2, 2012 #3
    ^
    You did with one lower and one upper indexes. That's not what I meant. When I wrote (upper 0,0) it meant that BOTH zero and zero are upper indexes.

    Also, I would appreciate you solving Ricci tensor zero-zero (both upper!) with 4*pi*G*density on the Right Hand Sight.

    And why did you think that both indexes are equal? The are mu and nu; not mu and mu.

    And, I didn't ask about the Riemann tensor at all. There was also no point making G=1

    In general, I would like to see the proof that set of partial derivatives of the gravitational potential=ricci tensor with zero-zero
     
    Last edited: Jul 2, 2012
  5. Jul 2, 2012 #4

    Dale

    Staff: Mentor

    If you are going to be this picky then you really need to make the effort to learn LaTeX.
     
  6. Jul 2, 2012 #5
    ^
    Absolutely true! I just cannot find where I can learn it.
     
  7. Jul 2, 2012 #6

    Dale

    Staff: Mentor

  8. Jul 2, 2012 #7
    OK, my question is how to solve this equation, assuming c=1

    [itex]\mathrm R_{}{}^{00}[/itex]=[itex]\nabla^2 \phi[/itex]

    where [itex]\phi[/itex] is a gravitational potential from Poisson's equation.

    Here is my logic:

    [itex]\mathrm G_{}{}^{00}[/itex][itex]=[/itex][itex]8[/itex][itex]\pi[/itex][itex]\mathrm G[/itex][itex]\rho[/itex]

    [itex]\mathrm R_{}{}^{00}[/itex][itex]-[/itex][itex]\frac{1}{2}[/itex][itex]g_{}{}^{00}[/itex][itex]\mathrm R_{}{}^{}[/itex][itex]=[/itex][itex]8[/itex][itex]\pi[/itex][itex]\mathrm G[/itex][itex]\rho[/itex]

    [itex]\mathrm R_{}{}^{00}[/itex][itex]-[/itex][itex]\frac{1}{2}[/itex][itex]g_{}{}^{00}[/itex][itex]\mathrm R_{}{}^{00}[/itex][itex]g_{00}{}^{}[/itex][itex]=[/itex][itex]8[/itex][itex]\pi[/itex][itex]\mathrm G[/itex][itex]\rho[/itex]

    [itex]\mathrm R_{}{}^{00}[/itex][itex]-[/itex][itex]\frac{1}{2}[/itex][itex]\mathrm R_{}{}^{00}[/itex][itex]=[/itex][itex]8[/itex][itex]\pi[/itex][itex]\mathrm G[/itex][itex]\rho[/itex]

    [itex]\frac{1}{2}[/itex][itex]\mathrm R_{}{}^{00}[/itex][itex]=[/itex][itex]8[/itex][itex]\pi[/itex][itex]\mathrm G[/itex][itex]\rho[/itex]

    [itex]4[/itex][itex]\mathrm R_{}{}^{00}[/itex][itex]=[/itex][itex]4[/itex][itex]\pi[/itex][itex]\mathrm G[/itex][itex]\rho[/itex]

    Poisson's equation tells us that

    [itex]\nabla^2 \phi[/itex][itex]=[/itex][itex]4[/itex][itex]\pi[/itex][itex]\mathrm G[/itex][itex]\rho[/itex]

    So it must be that
    [itex]4[/itex][itex]\mathrm R_{}{}^{00}[/itex][itex]=[/itex][itex]\nabla^2 \phi[/itex]

    So my question is what is its solution?

    Also [itex]\phi[/itex][itex]=[/itex][itex]\frac{GM}{R}[/itex], right?
     
    Last edited: Jul 2, 2012
  9. Jul 2, 2012 #8
    First off [itex]\frac{1}{2}R^{00}=8 \pi G \rho[/itex] is not equivalent to [itex]R^{00}=4 \pi G \rho[/itex]

    It's equivalent to [itex]R^{00}=16 \pi G \rho[/itex]

    And there is a limiting case somewhere to compare the metric with the Newtonian gravitational potential I believe. But I do not believe [itex]R^{00}= \nabla^2 \phi[/itex]
     
  10. Jul 2, 2012 #9
    I am confused...
     
  11. Jul 2, 2012 #10

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    This step is not correct. The Ricci scalar [itex]R[/itex] involves all of the components of the Ricci tensor:

    [tex]R = R^{ab} g_{ab} = R^{00} g_{00} + R^{11} g_{11} + R^{22} g_{22} + R^{33} g_{33}[/tex]

    In fact, even the above is only true in a coordinate chart where the metric is diagonal, i.e., where the only nonzero components have a = b. In a more general coordinate chart you will need to include terms like [itex]R^{01} g_{01}[/itex], etc.

    Also, EtherWind is correct that the equation you give relating [itex]R^{00}[/itex] to the Newtonian "potential", [itex]R^{00} = \nabla^{2} \phi[/itex], is not true in general. First, it is a limiting case where gravity is very weak; second, even that limiting case only applies in situations where the field is static, i.e., where gravity at a given radius does not change with time.

    You might want to check out this page by John Baez on the meaning of Einstein's Equation:

    http://math.ucr.edu/home/baez/einstein/einstein.html
     
  12. Jul 2, 2012 #11
    I understand now...I will get back to my calculations and Reply to this thread in a couple of days. Because I am a first year undergrad, it is kinda difficult for me to grasp these concepts.
     
  13. Jul 2, 2012 #12

    WannabeNewton

    User Avatar
    Science Advisor

    Well if [itex]\frac{1}{2}x = 8 [/itex] then [itex]x = 16[/itex] but anyways what you are thinking of is the limiting case of the field equations. What you are trying to say would be valid for the linearized field equations where you consider a small perturbation on a background flat space -time. Anyways in this limit you can recast the field equations, with a bit of plugging in and simplifying etc. and using that for small velocities [itex]T^{00}[/itex] will dominate and [itex]\partial _{t}[/itex] will be higher order, and write it as [itex]\triangledown ^{2}\bar{h^{00}} = -16\pi \rho [/itex] ([itex]h[/itex] is the perturbation). You can compare that to the usual Poisson's equation to extract a solution which will eventually give you the weak field Newtonian metric.
     
    Last edited: Jul 2, 2012
  14. Jul 3, 2012 #13
    I reviewed my older calculations and I have that

    [itex]\mathrm T_{}{}^{00}[/itex][itex]=[/itex][itex]\rho[/itex][itex]c_{}{}^{2}[/itex]

    I mean that this fact is non-negotiable, right?

    This doesn't mean that all other components are zero--we are just considering its zero-zero component. (Right Hand Sight of the EFE has c^-4 term in original form, so having c^2 up and c^-4 down gives c^-2)

    [itex]\mathrm G_{}{}^{00}[/itex][itex]=[/itex][itex]8[/itex][itex]\pi[/itex][itex]\mathrm G_{}{}^{}[/itex][itex]\rho[/itex][itex]c_{}{}^{-2}[/itex]

    [itex]\frac{c_{}{}^{2}G_{}{}^{00}}{2}[/itex][itex]=[/itex][itex]4[/itex][itex]\pi[/itex][itex]\mathrm G_{}{}^{}[/itex][itex]\rho[/itex]

    if we set c=1, we have that

    [itex]\frac{1}{2}[/itex][itex]\mathrm G_{}{}^{00}[/itex][itex]=[/itex][itex]4[/itex][itex]\pi[/itex][itex]\mathrm G_{}{}^{}[/itex][itex]\rho[/itex]

    So we have that

    [itex]\frac{1}{2}[/itex][itex]\mathrm G_{}{}^{00}[/itex][itex]=[/itex][itex]\nabla^2 \phi[/itex]

    Please correct me if I am wrong.

    Also, [itex]\mathrm T_{}{}^{00}[/itex] of the Earth is [itex]\rho[/itex][itex]c^2[/itex][itex]=[/itex][itex]5520[/itex][itex]c^2[/itex], right?
     
    Last edited: Jul 3, 2012
  15. Jul 3, 2012 #14

    Dale

    Staff: Mentor

    Well, it isn't generally true, but it certainly can be taken as a given for a specific coordinate system that you wish to consider.

    All fine so far.

    This relies on the substitution: [itex]\nabla^2 \phi = 4 \pi \mathrm G \rho[/itex]. This substitution is generally true in Newtonian gravity, but it is only true in GR in the "weak field" limit.

    When solving the EFE in GR, the usual goal is to find a metric, [itex]g_{\mu\nu}[/itex], which satisfies the EFE for a given stress energy tensor. So you usually need to expand the Einstein tensor in terms of the metric.
     
  16. Jul 3, 2012 #15
    ^
    Thank you
     
  17. Jul 7, 2012 #16
    You wrote:

    "Ricci scalar is obtained by contracting it with metric tensor so we have that
    Ricci (upper,0,0) - 1/2*g(upper 0,0)*Ricci (upper0,0)*g(lower, 0,0)=8*pi*G*ρ"


    You did the contraction for getting the R scalar incorrectly. You did R= (g00)(R00), thinking that R00 was the only non-vanishing part of the Ruv tensor when Tuv has T00 as its only non-vanishing component. Surprisingly it turns out different than what you expected.

    I'll modify what you were trying to do to it correctly. (I might make careless arithmetic mistakes, so check for them)

    Take Ruv - 1/2 guv R = 8 pi Tuv and do a scalar contraction. We get

    R - 2R = 8 pi T

    R = - 8 pi T where T is the contraction of the Tuv tensor.

    Now insert this back into the Einstein Equarions

    Ruv - (1/2)(- 8 pi guv T) = 8 pi Tuv

    Ruv = 8 pi [Tuv - 1/2 guv T]

    You wanted to look at the T00 equation. OK:

    R00 = 8 pi [ T00 - 1/2 g00 T]

    Note that it is not that only the T00 component of the stress energy tensor that is driving R00! The 1/2 g00 T term contains components in addition to T00!

    So if we have two otherwise identical mass distributions except one has more pressure, the one with more pressure (but the same mass!) will generate a stronger gravitational field. This might at first seem unphysical, but it actually makes sense. Faster particles are more strongly deflected by the gravitational field of a static mass distribution than are slower particles--this is because the static object actually turns out to generate velocity squared forces. So it makes sense that there is a recipricosity-- an object with velocity squared characteristics, being more affected by other gravitational bodies, also affects them more strongly than if it were stationary.

    Another place where the fact that the right-hand side of R00 = 8 pi [T00 - 1/2 g00 T] having the T term manifests itself is in the acceleration of the cosmological expansion. The [T00 - 1/2 g00 T] drives the acceleration of the expansion, and it contains pressure. Quantitatively both the mass density and the pressure in that term, if they are positive, produce deceleration of the expansion of the universe. The current view is that the observational evidence is that the Universe is expanding with positive acceleration. There is a reason (another relativistic cosmological equation) why the mass density appears to need to be positive...so this seems to leave us stuck with a negative pressure, a disturbing thought considering pressure goes as the square of the velocities. This is the "dark energy" problem.

    OK, so we need to add the T term to your equation. As you claimed there then is a Poisson Equation appearance. But you seem not aware that the Poisson nature is only approximate--the General Relativity equations have non-linear terms not appearing in the Poisson Equation. Also the General Relativity Equation has time derivatives.
     
  18. Jul 7, 2012 #17
    Thanks for your answer. Here I re-wrote what you tried to say in more clear way:

    [itex]R_{\mu\nu}-\dfrac{1}{2} g_{\mu\nu} R =8 \pi T_{\mu\nu}[/itex]

    [itex] R-2R=8\pi T[/itex] (how 1/2 turned into 2?)

    [itex]R=-8 \pi T [/itex]

    [itex]R_{\mu\nu}+\dfrac{1}{2} g_{\mu\nu}8 \pi T =8 \pi T_{\mu\nu}[/itex]

    [itex]R_{\mu\nu}= 8 \pi (T_{\mu\nu}-\dfrac{1}{2} g_{\mu\nu} T) [/itex]

    [itex]R_{00}= 8 \pi (T_{00}-\dfrac{1}{2} g_{00} T) [/itex]

    In any way, the T_{00} is the energy density. It would be much clearer if you make an example of this formula in use.
     
    Last edited: Jul 7, 2012
  19. Jul 7, 2012 #18
    GRstudent, I think you did not mean to type what you typed in your last post.
     
  20. Jul 7, 2012 #19
    ^
    Just look above--I re-edited my message. Thanks.
     
  21. Jul 7, 2012 #20
    "R−2R=8πT (how 1/2 turned into 2?)"

    When we did the contraction of the metric tensor we did (g00) squared + (g11) squared + (g22) squared + (g33) squared. That is (1) squared + (-1) squared + (-1) squared + (-1) squared. So the contraction of the metric tensor is 4. So the contraction of the -(1/2) guv R piece is -2R.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Einstein Field Equations
  1. Einstein field equation (Replies: 11)

Loading...