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

Explanation of the following field equation

  1. Dec 1, 2014 #1
    9fe9108d287a6994c9f123644a43f3ba.png
    I have been learning GR since past few days and have come across the following problem. I just cant get what result exactly the equation wants to be deliver:olduhh:. Please help me out....
    Please a bit theoretical:)

    How do we describe Gravitational field through Einstein's field equations?? [i.e. the mathematical part of GR]?

    [Note: edited by mentor to merge two closely related threads]
     
    Last edited by a moderator: Dec 1, 2014
  2. jcsd
  3. Dec 1, 2014 #2

    Nugatory

    User Avatar

    Staff: Mentor

    That's the Einstein field equation. It gives the relationship between the curvature of spacetime at a given point (described by the two ##R## terms) and the amount of mass, energy, and pressure at that point (given by the ##T## term).

    Once the curvature is known, it is possible to calculate the path an object in free fall would follow through that curved spacetime, and good old ##F=ma## to calculate how forces other than gravity (whose effects are already in the curvature) will move the object off that free-fall path.

    There's a pretty decent introduction to all of this at http://preposterousuniverse.com/grnotes/grtinypdf.pdf [Broken]
     
    Last edited by a moderator: May 7, 2017
  4. Dec 1, 2014 #3

    ShayanJ

    User Avatar
    Gold Member

    I think you should replace all "curvature"s with "metric"!
    Because at the end, people don't get curvature from that PDE, but the metric!
    The geodesic equation is in terms of the Christoffel symbols which is in terms of metric, so the geodesic equation is written in terms of metric!
    Also those Rs are quantities which are calculated from Riemann curvature tensor by contraction and I think there may be different Riemann tensors with the same Ricci curvature and Ricci scalar which means (even if) Riemann curvature characterizes a particular solution completely, those Rs can't.
    So I would say its indeed the metric of the space-time which we get from EFE, not the curvature.
     
    Last edited by a moderator: May 7, 2017
  5. Dec 2, 2014 #4

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    But they get the metric by solving the PDE, after choosing some conditions that they want the metric to satisfy. (For example, if you want a spherically symmetric spacetime, you impose that condition on the metric and then solve the EFE to obtain relationships between the metric coefficients.) So it's not true that "people don't get curvature from that PDE"; they do, just with an intermediate step (deriving the metric).

    It may be that you were trying to say the same thing:

    But saying it this way seems like a quibble: you get the metric by solving the EFE, and you get the curvature from the metric. Why doesn't that count as getting the curvature by solving the EFE?

    (One can also, btw, solve the EFE without deriving an explicit expression for the metric; one way is to move the trace term to the RHS, obtaining ##R_{\mu \nu} = 8 \pi \left( T_{\mu \nu} - \frac{1}{2} g_{\mu \nu} T \right)##. This is an equation for the Ricci tensor, which has a direct physical interpretation, so you can use it to make predictions without going via the metric as an intermediate step.)
     
    Last edited: Dec 2, 2014
  6. Dec 2, 2014 #5

    ShayanJ

    User Avatar
    Gold Member

    Of course, you may get anything you want from the metric after calculating it using the PDE and some assumptions!
    My point is, Einstein Field Equations are PDEs with metric components as the unknown functions, not any kind of curvature.
    Also the essential quantity in GR is the metric tensor and so I think the main part of getting the gravitational field of an energy distribution, is calculating the metric of space-time in the presence of that energy distribution.
    I just wanted to emphasize that the metric tensor is the main thing and anything else is secondary.
     
  7. Dec 2, 2014 #6
    May anyone refer to any link that will help me to understand the concept better.
     
  8. Dec 2, 2014 #7

    ShayanJ

    User Avatar
    Gold Member

    It doesn't work that way. You should at first know Special Relativity(SR) well. Then, you should learn Tensor analysis and Riemannian geometry and then you're ready to learn General Relativity(GR) and understand the Field equations.
    There is another way too. After learning SR, you may study about GR without math but that will just teach you what GR is saying, not how you can use it.
    For learning GR fully with its math, you should learn Tensor analysis and Riemannian geometry and there is no escape.
    But I can give you a simple introduction(which is in no way enough). GR says that gravity is the curvature of space-time which is encoded in a quantity called Riemann curvature tensor([itex] R^\rho_{\ \sigma \mu \nu} [/itex]). Riemann tensor is calculated from Christoffel symbols which, in turn, are calculated form the metric tensor.The metric tensor of a "space" is a quantity which completely characterizes its "shape".(Words inside double quotations are not rigorous!). So GR says gravity is a consequence of the different shapes space-time takes in the presence of the different mass-energy distributions(which is encoded in the quantity stress-energy tensor [itex] T_{\mu \nu} [/itex].).
    From Riemann curvature tensor, we may calculate Ricci curvature ([itex] R_{\mu \nu}[/itex]) and Ricci scalar ([itex] R [/itex]) which appear in the Einstein Field Equations. So, the field equations are equations for finding the components of the metric tensor and so for finding the shape of the space-time when the mass-energy distribution is known.
    Knowing the shape of the space-time(the metric tensor), we can realize how the straight lines look like in that space-time(find geodesics). Then GR says that a freely-falling mass, traverses a geodesic of space-time. The fact that space-times with different metrics, have different geodesics, which are different from geodesics of Minkowski space-time(Space-time in the absence of gravitating mass and energy), results in the explanation of what we call gravity in terms of the geometry of space-time.
     
  9. Dec 2, 2014 #8

    Nugatory

    User Avatar

    Staff: Mentor

    There's a link to http://preposterousuniverse.com/grnotes/grtinypdf.pdf [Broken] in my post above. I don't think you'll find any more gentle introduction than that.
     
    Last edited by a moderator: May 7, 2017
  10. Dec 3, 2014 #9

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Um, you do realize that curvature is described by derivatives of the metric coefficients, right? If you have a PDE with the metric components as unknown functions, you have a PDE describing curvature. So it's not a question of choosing one or the other. Both are always involved.

    Depends on what you're trying to do. You can measure spacetime curvature directly by measuring tidal gravity. Depending on the spacetime and your choice of coordinates, you may not be able to measure any metric coefficients directly at all. So from a physical point of view, the curvature has a more direct meaning than the metric coefficients.
     
  11. Dec 3, 2014 #10

    ShayanJ

    User Avatar
    Gold Member

    I know that. But can you get some kind of curvature directly from the field equations? I guess the answer, in general, is no!
    That's right, but in such a situation, I would say that experimental methods give incomplete information!
     
  12. Dec 3, 2014 #11

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    What? The Einstein field equations are given explicitly in terms of curvature tensors. Getting the Einstein curvature tensor is far more direct than getting the metric coefficients. $$G_{ab}=8\pi T_{ab}$$ In a vacuum the even more direct implication of the Einstein field equations is that the Ricci curvature is 0. This is a direct result.

    I can't understand your insistence on obtaining the metric coefficients. In some instances, it might be more prudent to use tetrad methods, in which case the metric is simply the Minkowski metric, and all the physical quantities lie elsewhere (i.e. in the tetrads themselves, the connection forms, and curvature tensors).
     
  13. Dec 3, 2014 #12

    ShayanJ

    User Avatar
    Gold Member

    OK guys, I surrender!!! Actually because I lost track of things and got a little confused. I should go back and think!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Explanation of the following field equation
Loading...