- #1

- 124

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter professor
- Start date

- #1

- 124

- 0

- #2

- 124

- 0

- #3

- 1,250

- 2

Einstein's General Relativity is the theory of the interaction between matter and spacetime. There is a system of equations, the Einstein Field Equations, that uniquely determines the 4-dimensional curvature in a region given the distribution of matter and energy.

The EFE are a system of 16 highly nonlinear coupled partial differential equations in 16 unknowns. Unless you have a truly exceptional backround in physics, it is unlikely that you have encountered anything like this. Without the aid of specail notations, it is essentially impossible to express the tens of thousands of individual terms (all quadratic differentials). Here is an example of one of the 16 components of the Reimann curvature tensor (which is highly simplified due to symmetry) which is just one a part of EFE:

The EFE are a system of 16 highly nonlinear coupled partial differential equations in 16 unknowns. Unless you have a truly exceptional backround in physics, it is unlikely that you have encountered anything like this. Without the aid of specail notations, it is essentially impossible to express the tens of thousands of individual terms (all quadratic differentials). Here is an example of one of the 16 components of the Reimann curvature tensor (which is highly simplified due to symmetry) which is just one a part of EFE:

- #4

Danger

Gold Member

- 9,607

- 248

That would make a great wallpaper pattern! The whole set would probably be enough to do my living room.Crosson said:Here is an example of one of the 16 components of the Reimann curvature tensor (which is highly simplified due to symmetry) which is just one a part of EFE:

- #5

- 124

- 0

- #6

- 124

- 0

- #7

- 1,250

- 2

Long before Einstein, Reimann found that distance on any curved (or not) space could be expressed by a metric. Here are some examples:

[tex] ds^2=dx^2+dy^2[/tex]

[tex] ds^2 = dr^2 +r^2 d\theta ^2 [/tex]

Hopefully, these two metrics look familiar to you Professor. They both represent flat space, one is in cartesian coordinates and one is in polar coordinates. It is noteworthy that although the two metrics look different, they both represent the same thing. Here is a metric that expresses distance on the surface of a sphere:

[tex]ds^2 = R^2d\phi ^2 + R^2 Sin^2(\phi) d\theta ^2 [/tex]

This surface now has a curvature (the first two metrics haves zero curvature they are flat). It is certainly not obvious by looking that this metric belongs to a curved surface where as the flat metric in polar coordinates does not. How can we tell the curvature of a space using the metric? Reimann and Gauss found that all you have to do is take a lot of derivatives!

Here is Reimann's general form of the metric:

[tex]ds^2 = \sum_{\mu,\nu} g_{\mu,\nu} dx^{\mu}dx^{\nu}[/tex]

Hopefully you can see that this is:

[tex] ds^2 = g_{0,0} (dx^0)^2 + g_{1,0}dx^1 dx^0 +....[/tex]

Notice that the metric coeffecients [itex]g_{i,j} [/itex] (there are 16 of them in 4-d space time because we have four coordinates [itex] x^i where i = 0,1,2,3[/itex]) contain all the information, for the reason we put them in 4x4 matrix called the metric tensor. Now like I said, once we have the metric tensor, we are only a few thousand derivatives away from finding the curvature.

From the metric tensor we define the affine connection:

[tex] \Gamma_{\sigma,\mu,\nu} = \frac{1}{2}\sum_{\mu,\nu} g^{\mu,\nu}(\partial _{\nu} g_{\mu,\sigma}-\partial_{\sigma}g_{\mu,\nu} + \partial_{\mu}g_{\sigma,\nu})[/tex]

As you can see, any particular affine connection (Ex. [itex]\Gamma_{1,1,1}[/itex]) contains 48 terms. So we already have 64 affine connections with 48 terms each.

From the affine connection we find the rank 4 reimann curvature tensor:

[tex]Reimann_{\lambda,\mu,\nu,\kappa} = \frac{1}{2}(\partial _{\kappa,\mu} g_{\lambda,\nu} +\partial_{\kappa,\lambda}g_{\mu,\nu} +\partial_{\nu,\lambda} g_{\kappa,\mu} + \sum _{\sigma,\epsilon} g_{\sigma,\epsilon}(\Gamma_{\epsilon, \nu, \kappa}\Gamma{\sigma,\mu,\kappa}-\Gamma_{\epsilon,\kappa,\lambda}\Gamma_{\sigma,\mu,\nu}[/tex]

For General Relativity we contract this to the Ricci Tensor:

[tex] R_{\mu,\kappa} = \sum_{\nu,\lambda} g^{\lambda,\nu} Reimann_{\lambda, \nu,\mu,\kappa} [/tex]

And the Ricci Scalar:

[tex] R = \sum_{\nu,\lambda} g^{\lambda,\nu} R_{\lambda, \nu} [/tex]

In the most general case, computation of the Ricci Scalar involves over 15,000 terms. But at least now we have reached a point where we can express the EFE:

[tex] R_{\mu,\nu} -\frac{1}{2} g_{\mu,\nu}R = \frac{8 \pi G}{c^2} T_{\mu,\nu}[/tex]

You should be able to interpret the left side of the equation as the curvature of space time. The right side is called the stress energy tensor, and it is basically energy density (and momentum densities in different directions).

The dependent variables in the system of PDEs are the metric coefficients, the independent variables are the four coordinates of spacetime; the components of the stress energy tensor are the inhomogeneous terms. A straightfoward method of approach would be to choose a stress energy tensor and solve the system of PDEs for the metric coefficients; but, in all but the simplest cases, the equations are so ultra-complex that nothing on earth is capable of this method. A more much more tractable approach is to input a particular metric and stress energy tensor, and solve the EFE for the time evolution of that metric.

[tex] ds^2=dx^2+dy^2[/tex]

[tex] ds^2 = dr^2 +r^2 d\theta ^2 [/tex]

Hopefully, these two metrics look familiar to you Professor. They both represent flat space, one is in cartesian coordinates and one is in polar coordinates. It is noteworthy that although the two metrics look different, they both represent the same thing. Here is a metric that expresses distance on the surface of a sphere:

[tex]ds^2 = R^2d\phi ^2 + R^2 Sin^2(\phi) d\theta ^2 [/tex]

This surface now has a curvature (the first two metrics haves zero curvature they are flat). It is certainly not obvious by looking that this metric belongs to a curved surface where as the flat metric in polar coordinates does not. How can we tell the curvature of a space using the metric? Reimann and Gauss found that all you have to do is take a lot of derivatives!

Here is Reimann's general form of the metric:

[tex]ds^2 = \sum_{\mu,\nu} g_{\mu,\nu} dx^{\mu}dx^{\nu}[/tex]

Hopefully you can see that this is:

[tex] ds^2 = g_{0,0} (dx^0)^2 + g_{1,0}dx^1 dx^0 +....[/tex]

Notice that the metric coeffecients [itex]g_{i,j} [/itex] (there are 16 of them in 4-d space time because we have four coordinates [itex] x^i where i = 0,1,2,3[/itex]) contain all the information, for the reason we put them in 4x4 matrix called the metric tensor. Now like I said, once we have the metric tensor, we are only a few thousand derivatives away from finding the curvature.

From the metric tensor we define the affine connection:

[tex] \Gamma_{\sigma,\mu,\nu} = \frac{1}{2}\sum_{\mu,\nu} g^{\mu,\nu}(\partial _{\nu} g_{\mu,\sigma}-\partial_{\sigma}g_{\mu,\nu} + \partial_{\mu}g_{\sigma,\nu})[/tex]

As you can see, any particular affine connection (Ex. [itex]\Gamma_{1,1,1}[/itex]) contains 48 terms. So we already have 64 affine connections with 48 terms each.

From the affine connection we find the rank 4 reimann curvature tensor:

[tex]Reimann_{\lambda,\mu,\nu,\kappa} = \frac{1}{2}(\partial _{\kappa,\mu} g_{\lambda,\nu} +\partial_{\kappa,\lambda}g_{\mu,\nu} +\partial_{\nu,\lambda} g_{\kappa,\mu} + \sum _{\sigma,\epsilon} g_{\sigma,\epsilon}(\Gamma_{\epsilon, \nu, \kappa}\Gamma{\sigma,\mu,\kappa}-\Gamma_{\epsilon,\kappa,\lambda}\Gamma_{\sigma,\mu,\nu}[/tex]

For General Relativity we contract this to the Ricci Tensor:

[tex] R_{\mu,\kappa} = \sum_{\nu,\lambda} g^{\lambda,\nu} Reimann_{\lambda, \nu,\mu,\kappa} [/tex]

And the Ricci Scalar:

[tex] R = \sum_{\nu,\lambda} g^{\lambda,\nu} R_{\lambda, \nu} [/tex]

In the most general case, computation of the Ricci Scalar involves over 15,000 terms. But at least now we have reached a point where we can express the EFE:

[tex] R_{\mu,\nu} -\frac{1}{2} g_{\mu,\nu}R = \frac{8 \pi G}{c^2} T_{\mu,\nu}[/tex]

You should be able to interpret the left side of the equation as the curvature of space time. The right side is called the stress energy tensor, and it is basically energy density (and momentum densities in different directions).

The dependent variables in the system of PDEs are the metric coefficients, the independent variables are the four coordinates of spacetime; the components of the stress energy tensor are the inhomogeneous terms. A straightfoward method of approach would be to choose a stress energy tensor and solve the system of PDEs for the metric coefficients; but, in all but the simplest cases, the equations are so ultra-complex that nothing on earth is capable of this method. A more much more tractable approach is to input a particular metric and stress energy tensor, and solve the EFE for the time evolution of that metric.

Last edited:

- #8

- 124

- 0

- #9

- 13,055

- 604

Daniel.

Share:

- Replies
- 4

- Views
- 3K

- Replies
- 8

- Views
- 2K