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covariant derivative of metric tensor


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#1 Off-diagonal

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Posted 20 November 2007 - 11:08 AM

why should covariant derivative of metric tensor have to be 0?


thanks,
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#2 timo

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Posted 20 November 2007 - 11:54 AM

I'm not exactly sure what you mean by "why"? Of course you can just plug it in and verify it:
 D_\mu g_{\alpha \beta} := \partial_\mu g_{\alpha \beta} - \Gamma^i{}_{\mu \alpha} g_{i \beta} - \Gamma^i{}_{\mu \beta} g_{\alpha i} = \dots = 0. (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.
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#3 ajb

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Posted 20 November 2007 - 12:22 PM

There is no reason at all why the covariant derivative (aka a connection) of the metric tensor should vanish. You can of course insist that this be the case and in doing so you have what we call a metric compatible connection. This will put some condition of the connection coefficients and furthermore insisting that they be symmetric in lower indices will produce the unique Christoffel symbols. As you know the Christoffel symbols can be expressed in terms of the metric and it's first derivatives.

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.
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#4 Norman Albers

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Posted 2 December 2007 - 11:47 PM

Doesn't inner product involve magnitudes and angle? Is the term 'conformal'?
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#5 ajb

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Posted 3 December 2007 - 11:25 AM

Doesn't inner product involve 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 (M,g) is a diffeomorphism which preserves the metric up to a scale.

More precisely,

Let f :M \rightarrow M be a diffeomorphism. Then f is said to be a conformal transformation iff

f^{*}g_{f(p)} = e^{2 \sigma}g_{p}

where \sigma is a function of M

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).
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#6 Norman Albers

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Posted 9 December 2007 - 05:54 PM

Solid math, ajb. So, then, two transplanted vectors could stretch by differing amounts but their angle, defined I think by: cos \theta=\frac{v\cdot u}{|v||u|} , stays the same?
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#7 ajb

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Posted 9 December 2007 - 06:02 PM

The angle between two vectors remains the same after a conformal transformation. You have used the metric to define what you mean by \cdot and ||

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.
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#8 Norman Albers

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Posted 9 December 2007 - 06:06 PM

I have never taken this sort of mathematics course, so I cannot yet follow the Wikipedia of 'diffeomorphism'. I learned all I know from the mathematically strong text, Introduction to General Relativity by Adler, Bazin, and Schiffer (McGraw-Hill). I am testing and seeking to expand my understandings. . . . . I can understand the first statement about a transform being differentiable...
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#9 ajb

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Posted 9 December 2007 - 06:18 PM

Have a look at the notes by Carroll. They are a good introduction, but are not really general enough if you want a good understanding of differential geometry.

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.
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#10 Norman Albers

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Posted 9 December 2007 - 06:23 PM

Thank you, 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.
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#11 scalbers

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Posted 9 December 2007 - 06:34 PM

A little off topic - as this reminds me of conformal map projections (e.g. Mercator, Lambert, Polar Stereographic). They will all change the map scale over different latitudes while preserving shapes and angles.
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#12 ajb

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Posted 9 December 2007 - 06:34 PM

I don't know the book by Adler, Bazin, and Schiffer. It seems quite old, but then by the 1960's most of the framework of general relativity was established. It is possible that notation has changed slightly.

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.
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#13 Norman Albers

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Posted 9 December 2007 - 07:05 PM

The last printing of my edition was 1975. I took the course in grad school at Stanford in 1972 from Schiffer. He was a short, punchy German, who would fill five blackboards, turn briefly to say, "Do you see?" and then continue full blast. There are many places I could quote, such as p.117: "When Minkowski first introduced the F_{ab} tensor into electrodynamics, he had in mind that it should transform as a tensor only under Lorentz transformations. However, as we see here, no such restriction appears necessary, for the Maxwell equations go over very easily indeed into a completely covariant form. The reader should not be blinded by our mathematical transformations into assuming the that the statement "F^{ab} is a tensor" is a purely mathematical one. It is a very important and far-reaching physical principle which can be motivated by mathematical elegance, but must also be tested by physical experiment. The fact that F^{ab} 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.
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#14 ajb

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Posted 10 December 2007 - 07:39 PM

Tensors, tensor densities and connections are simply fundamental in physics and differential geometry.

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.
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#15 Norman Albers

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Posted 13 December 2007 - 01:41 AM

So simple, as far as physics is differentiable.
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#16 CPL.Luke

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Posted 14 December 2007 - 11:43 PM

also to my knowledge when physicists use the statement 'is a tensor' they mean that it transforms by the summation partial x^a with respect to x^a
' 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.
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#17 ajb

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Posted 15 December 2007 - 01:07 PM

In algebra, you think of tensors as multilinear maps.

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.
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#18 Norman Albers

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Posted 15 December 2007 - 02:10 PM

I appreciate the depth of your math knowledge, ajb. My GR text spends much time developing the differential "transplanting of vectors". Here is another zinger from p.47: "The use of unsymmetric \Gamma^i_{ab} 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?
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#19 ajb

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Posted 15 December 2007 - 04:12 PM

You could have an antisymmetric part in general, although the definition of a metric is that it is symmetric.

An antisymmetric part would be more like a symplectic form. As you know F_{ab} is antisymmetric and you can use this as part of a symplectic form. I see no reason why you could not consider the object \hat{g}_{ab} = g_{ab} \pm \Omega_{ab}. Where  \Omega 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 M and build the symplectic manifold (T^{*}M, \Omega ). In local Darboux coordinates we have \Omega = dx^{a}\wedge dp_{a}. Then in these local coordinates you can build \hat{g}_{ab} 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).
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#20 Norman Albers

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Posted 15 December 2007 - 04:37 PM

Ouch! I do appreciate having my butt kicked, 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 (?)
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