What is the Unit of the Metric Tensor?

In summary, the metric tensor g_{ij} has no units, as it is used to measure angles and distances in spacetime. The stress-energy tensor, which appears on the right side of Einstein's equation, has components with different units such as pressure and density, and is used to measure the flow of energy and momentum through spacetime.
  • #1
Antonio Lao
1,440
1
If the cosmological constant, [itex] \Lambda[/itex] has units of reciprocal time squared then what is the unit for the metric tensor, [itex]g_{ij}[/itex]?
 
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  • #2
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.
 
  • #3
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).
 
  • #4
Antonio Lao said:
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
 
  • #5
Russell,

Thanks. Will start studying this GR tutorial.
 

1. What is a metric tensor?

A metric tensor is a mathematical object that is used to define the distance or interval between points in a space. It is a generalization of the concept of distance in Euclidean geometry and is used in the study of differential geometry and general relativity.

2. What are the units of the metric tensor?

The units of the metric tensor depend on the specific space it is being used to define distances in. In general, it has units of length squared, such as meters squared in the case of a 3-dimensional Euclidean space.

3. How is the metric tensor related to the concept of curvature?

The metric tensor is essential in determining the curvature of a space. In a flat space, the metric tensor is constant and has a specific form, while in a curved space, the metric tensor varies from point to point, reflecting the curvature of the space.

4. Can the metric tensor be used in any dimension?

Yes, the metric tensor can be used in any dimension. In fact, it is a fundamental concept in higher-dimensional geometry and is used to define distances and angles in spaces with more than three dimensions.

5. How is the metric tensor used in general relativity?

In general relativity, the metric tensor is used to define the spacetime geometry, which is curved due to the presence of mass and energy. The curvature of spacetime is then related to the motion of objects in the presence of this curvature, giving rise to the effects of gravity.

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