# covariant derivative of metric tensor

### #1

Posted 20 November 2007 - 11:08 AM

thanks,

### #2

Posted 20 November 2007 - 11:54 AM

(note that I have implicitely used the symmetry of g and the two last indices of the Christoffel symbols - iow: I have not bothered on the exact position of the indices).

If "why?" was meant in the sense of "what is the physical meaning?": I dunno at the moment. It would probably be something like that relations (scalar products) between vectors remain unchanged under parallel transport.

"Ocean madness is no excuse for ocean rudeness" -- H.J. Farnsworth

### #3

Posted 20 November 2007 - 12:22 PM

As Atheist said, insisting that the connection be metric compatible does indeed mean that the length of vectors is unchanged under parallel transport with respect to the connection. In other words, it respects the inner product.

Mathematical Ramblings.

### #4

Posted 2 December 2007 - 11:47 PM

*inner product*involve magnitudes and angle? Is the term 'conformal'?

**we specialize in the elimination of**

*Singularities-R-Us,**and*

**embarrassing orders of infinity****. http://laps.noaa.gov/albers/physics/na**

*pesky asymptotic dependencies*### #5

Posted 3 December 2007 - 11:25 AM

Doesn't

inner productinvolve magnitudes and angle? Is the term 'conformal'?

The inner product is really a map that takes a vector and a dual vector and spits out a number (or whatever your field is). Generally, you can think of it as a generalisation of the dot product. You can use the inner product to define what you mean by the angle between vectors. I don't think this concept is always useful.

A conformal transformation on (pseudo-)Riemannian manifold is a diffeomorphism which preserves the metric up to a scale.

More precisely,

Let be a diffeomorphism. Then is said to be a conformal transformation iff

where is a function of

Now, conformal transformation on "stretch" they don't "twist". That is they do not preserve lengths but preserve angles (as defined via the inner product/metric).

Mathematical Ramblings.

### #6

Posted 9 December 2007 - 05:54 PM

**ajb**. So, then, two transplanted vectors could stretch by differing amounts but their angle, defined I think by: , stays the same?

**we specialize in the elimination of**

*Singularities-R-Us,**and*

**embarrassing orders of infinity****. http://laps.noaa.gov/albers/physics/na**

*pesky asymptotic dependencies*### #7

Posted 9 December 2007 - 06:02 PM

This is all very standard stuff Albers, as you often talk about aspects of general relativity I thought you would be quite familiar with this.

Mathematical Ramblings.

### #8

Posted 9 December 2007 - 06:06 PM

**we specialize in the elimination of**

*Singularities-R-Us,**and*

**embarrassing orders of infinity****. http://laps.noaa.gov/albers/physics/na**

*pesky asymptotic dependencies*### #9

Posted 9 December 2007 - 06:18 PM

There are some notes via my website that may also be useful, but I know there are some typos. Really the only way to learn differential geometry is to do differential geometry. Most physics books on general relativity are poor and lack the mathematical background needed.

Mathematical Ramblings.

### #10

Posted 9 December 2007 - 06:23 PM

**ajb**. If you can, check out that book. I find it to be written with a clear mathematician's sense. This is most excellent, because one must understand all the assumptions made in creating mathematical physics. To my mind the best authors are those who show clearly their assumptions. They are leaving you the freedom, at a future date, to go differently.

**we specialize in the elimination of**

*Singularities-R-Us,**and*

**embarrassing orders of infinity****. http://laps.noaa.gov/albers/physics/na**

*pesky asymptotic dependencies*### #11

Posted 9 December 2007 - 06:34 PM

### #12

Posted 9 December 2007 - 06:34 PM

Wald I would also recommend.

I think that the books by Nakahara and Bertlmann give the best introduction to differential geometry as needed in physics.

Mathematical Ramblings.

### #13

Posted 9 December 2007 - 07:05 PM

*motivated*by mathematical elegance, but must also be

*tested*by physical experiment. The fact that is a tensor under Lorentz transformation embodies a large part of special relativity theory. Our assumption that it is a tensor in a general Riemannian space-time leads to important consequences for the electrodynamics of accelerated systems of reference. The methodological principle that laws of nature which appear in tensor form in a particular coordinate system should be interpreted as valid in every system is called the

*principle of covariance*. Its philosophic motivation is the postulate that no coordinate system should be distinguished in the formation of physical laws. It is, however, mathematically somewhat ambiguous and comes in practice to the old principle that we should try to explain facts with the simplest and most aesthetically satisfactory theory." Indeed, I have suggested that perhaps inside an event horizon, physics is not at all the same.

**we specialize in the elimination of**

*Singularities-R-Us,**and*

**embarrassing orders of infinity****. http://laps.noaa.gov/albers/physics/na**

*pesky asymptotic dependencies*### #14

Posted 10 December 2007 - 07:39 PM

As the Lorentz transformations are in fact linear infinitesimal diffeomorphisms in hindsight it is not surprising that tensor objects under Lorentz transformations are tensors in the general sense.

Mathematical Ramblings.

### #15

Posted 13 December 2007 - 01:41 AM

**we specialize in the elimination of**

*Singularities-R-Us,**and*

**embarrassing orders of infinity****. http://laps.noaa.gov/albers/physics/na**

*pesky asymptotic dependencies*### #16

Posted 14 December 2007 - 11:43 PM

' for covariant indices and opposite that for contravariant indices.

this is just the general transformation law or tensors, although when mathematicians say that something is a tensor I believe it means that "something is linear with respect to more than 1 argument, hence why the dot product is a tensor mathematically.

### #17

Posted 15 December 2007 - 01:07 PM

In geometry, it is more useful to think of tensors as objects who's components transform in a certain way under diffeomorphisms (or maybe some subgroup of the diffeomorphism group). There is no need for me to be any more specific here as a quick google search will show you the details.

The point here is that the components of the tensor transform as to compensate for how the basis changes leaving the entire object invariant. (From this point of view tensors are

*just*scalars on a larger space!) I think this fact is not stressed enough in physics texts.

Mathematical Ramblings.

### #18

Posted 15 December 2007 - 02:10 PM

**ajb**. My GR text spends much time developing the differential "transplanting of vectors". Here is another zinger from p.47: "The use of unsymmetric coefficients has been considered only in later developments of the theory of GR, in the attempt to unite electromagnetic theory and gravitation theory..." QUESTION: What about the metric tensor itself? Could it be useful to consider an antisymmetric part?

**we specialize in the elimination of**

*Singularities-R-Us,**and*

**embarrassing orders of infinity****. http://laps.noaa.gov/albers/physics/na**

*pesky asymptotic dependencies*### #19

Posted 15 December 2007 - 04:12 PM

An antisymmetric part would be more like a symplectic form. As you know is antisymmetric and you can use this as part of a symplectic form. I see no reason why you could not consider the object . Where is the symplectic two form. (You may want some factors in there etc... ). I would suggest starting with a symplectic form as it is non-degenerate. Maybe it has been done already?

What you could do is start from a manifold and build the symplectic manifold . In local Darboux coordinates we have . Then in these local coordinates you can build and then start asking about it properties. One natural question is if there exists a connection which preserves this. It would be a "mix" of the Levi-Civita and Fedosov connections some how. (If it exists).

Another extension is super Riemannian geometry. The metric for the odd part is antisymmetric (really it is all supersymmetric).

Mathematical Ramblings.

### #20

Posted 15 December 2007 - 04:37 PM

**ajb**. Have you read much of Doug Sweetser's GEM presentation (on a far-away forum in a distant galaxy)? I lost it at the point where he deals with an exterior derivative of the vector potential; I think he is constructing a rank two object (?)

**we specialize in the elimination of**

*Singularities-R-Us,**and*

**embarrassing orders of infinity****. http://laps.noaa.gov/albers/physics/na**

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