What is the Unit of the Metric Tensor?

[Antonio Lao]

If the cosmological constant, \Lambda has units of reciprocal time squared then what is the unit for the metric tensor, g_{ij}?

 

[franznietzsche]

It has no units. A metric tensor in any coordinate can have no units, just look at how the components of a diagonal metric tensor fits into the distance formula, if it had units the units on the left would not match the units on teh right.

 

[Antonio Lao]

franznietzsche,

Thanks. But I'm still not clear why the righthand side of Einstein's field equations, the momentum-energy tensor has components of different units (units in pressures, and units in densities).

 

[PFanalog57]

I'm still not clear why the righthand side of Einstein's field equations, the momentum-energy tensor has components of different units (units in pressures, and units in densities).

Here is an excellent GR tutorial:

http://math.ucr.edu/home/baez/gr/outline2.html

The METRIC is the star of general relativity. It describes everything about the geometry of spacetime, since it let's us measure angles and distances. Einstein's equation describes how the flow of energy and momentum through spacetime affects the metric. What it affects is something about the metric called the "curvature". The biggest job in learning general relativity is learning to understand curvature!

Mathematically, the metric g is a tensor of rank (0,2). It eats two tangent vectors v,w and spits out a number g(v,w), which we think of as the "dot product" or "inner product" of the vectors v and w. This let's us compute the length of any tangent vector, or the angle between two tangent vectors. Since we are talking about spacetime, the metric need not satisfy g(v,v) > 0 for all nonzero v. A vector v is SPACELIKE if g(v,v) > 0, TIMELIKE if g(v,v) < 0, and LIGHTLIKE if g(v,v) = 0.

The STRESS-ENERGY TENSOR. The stress-energy is what appears on the right side of Einstein's equation. It is a tensor of rank (0,2), and it defined as follows: given any two tangent vectors u and v at a point p, the number T(u,v) says how much momentum in the u direction is flowing through the point p in the v direction. Writing it out in terms of components in any coordinates, we have

T(u,v) = T_ab u^a v^b

 

[Antonio Lao]

Russell,

Thanks. Will start studying this GR tutorial.