This is my first post in Physics Forums.(adsbygoogle = window.adsbygoogle || []).push({});

I am trying to a dimensional check of Einstein's field equations. Unfortunately, most books consider c = 1 or sometimes even G = c = 1, when presenting the field equations. This makes it very difficult to do a dimensional check.

In spite of this, I think that I have found the dimensions of the diagonal terms in the Ricci tensor. They all seem to be inverse length squared. The terms in the Ricci scalar too seem to have the same dimensions, inverse length squared.

This means that the Robertson Walker metric tensor's diagonal elements should be dimensionless. But the last 2 diagonal terms in the metric tensor are a^2*r^2 and a^2*r^2*sin^2(theta). Clearly, they have the dimension squared length, whereas the first diagonal term -1 has no units and the second diagonal term a^2/(1-k*r^2) too has no units. (a is dimensionless, k has the dimension inverse length squared and r has the dimensions of length).

Thanks in advance.

Sundar

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# I Dimensional check of Einstein's Field Equations

Have something to add?

Draft saved
Draft deleted

Loading...

Similar Threads for Dimensional check Einstein's |
---|

A Any 2-dimensional Lorentzian metric can be brought to this form? |

A Two Dimensional Ricci curvature |

A 2+2 dimensional spacetimes |

I 6-dimensional representation of Lorentz group |

**Physics Forums | Science Articles, Homework Help, Discussion**