As I understand it, for a tensor of any rank I can produce a corresponding scalar in the following way: Create an inverted form of the tensor by lowering its upper indices and raising its lower indices, and then taking the inner product of this tensor and the original one.
My only question is, is there a name for this result?
CompuChip
Jul20-08, 04:49 AM
Invertibility?
Metric?
Adjointness (or whatever you call "raising/lowering indices", i.e. going from co- to contravariant and vv)
robphy
Jul20-08, 01:54 PM
Loosely speaking, you can call it the "square-norm" since what you are really doing is analogous to g_{ap} A^a A^p .
(The quantity A_a= g_{ap} A^p is called the metric-dual of A^a .)
Given A^a{}_b{}^{cd}, you are forming the scalar using the metric [and its inverse]:
g_{ap} g^{bq} g_{cr} g_{ds} A^a{}_b{}^{cd} A^p{}_q{}^{rs}
snoopies622
Jul20-08, 02:28 PM
Thanks, robphy.
I was wondering because in another thread someone mentioned using R^{abcd}R_{abcd} (the "Kretschmann scalar"?) in order to show that the event horizon of a Schwarzschild black hole is not a real singularity, and it occurred to me that such a thing might be useful in other circumstances as well, so surely there must be a name for it...
By the way, am I correct in my belief that the "square-norm" of any metric tensor is the number of dimensions of its manifold?
CompuChip
Jul20-08, 02:53 PM
Yes, you can show that very easily. By definition, g^{ab} g_{bc} = \delta^a_c so if you contract a with c you get g^{ab} g_{ba} = \sum_{i = 1}^d 1 = d .
snoopies622
Jul20-08, 02:59 PM
By definition, g^{ab} g_{bc} = \delta^a_c so if you contract a with c you get g^{ab} g_{ba} = \sum_{i = 1}^d 1 = d .
Ah, yes, of course. Thanks, CompuChip.
robphy
Jul20-08, 04:55 PM
By the way, am I correct in my belief that the "square-norm" of any metric tensor is the number of dimensions of its manifold?
As seen from CompuChip's response, the metric tensor must have an inverse for that calculation. (Note: The metric of a Galilean spacetime is degenerate.)
snoopies622
Jul21-08, 04:49 PM
..The metric of a Galilean spacetime is degenerate.
I don't know what this means. Are you referring to a Euclidean metric? Minkowskian? And whether or not their matrix representations have inverses? Don't they?
robphy
Jul21-08, 06:15 PM
I don't know what this means. Are you referring to a Euclidean metric? Minkowskian? And whether or not their matrix representations have inverses? Don't they?
The nondegenerate (i.e. invertible) metric of an n-dimensional Euclidean space has
the diagonal form (+1,+1,...,+1,+1) in rectangular coordinates.
The nondegenerate (i.e. invertible) metric of an n-dimensional Minkowskian spacetime has
the diagonal form (-1,+1,...,+1,+1) in rectangular coordinates.
The degenerate (i.e. non-invertible) metrics of an n-dimensional Galilean spacetime has
the diagonal forms (0,+1,...,+1,+1) [for the spatial metric] and (+1,0,...,0,0) [for the temporal metric] in rectangular coordinates.