# A name for this tensor ?

1. Aug 21, 2006

### dextercioby

Assume a torsionless space-time. Is there any name for this tensor, provided it exists?

$$G_{\alpha\beta} =\nabla_{\mu}\nabla_{\nu} T^{\mu\alpha|\nu\beta}$$

,where G is of course the Einstein tensor...?

My comment would be that it doesn't exist, since the algebra of the covariant derivatives implies that T=0.

Daniel.

Last edited: Aug 21, 2006
2. Aug 21, 2006

### pmb_phy

When you refer to "this tensor" it is unclear whether you're speaking of G or T. G is the Einstein tensor as you said. T is the stress-energy-momentum tensor and it is non-zero at points in spacetime where matter is present. And I don't see how you arrived at the assumption "My comment would be that it doesn't exist, since the algebra of the covariant derivatives implies that T=0." Covariance doesn't mean that T = 0 even when G = 0. Also, a tensor A = 0 can exist and have A = 0. That just means that A is the null tensor.

Pete

3. Aug 21, 2006

### coalquay404

First thing to note is that your indices are all wrong. I assume you meant to have

$$G_{\alpha\beta} = \nabla^\mu\nabla^\nu T_{\mu\alpha |\nu\beta}$$

The next question to ask is what you mean by having a line between $$\mu\alpha$$ and $$\nu\beta$$. Also, you don't indicate your reason for using early alphabet Greek script and late alphabet Greek. If one sees notation like this in the literature, for example, there is always a reason for using indices in that way.

Unfortunately, unless you explain exactly what $$T$$ is, I suspect the question doesn't have an answer.

4. Aug 21, 2006

### pervect

Staff Emeritus
I don't understand the bar notation either.

5. Aug 21, 2006

### coalquay404

Me neither. I've spent a bit of time trying to think of what it could be but can't come up with anything. In the physics literature, I've only ever seen vertical bars used in indices to indicate covariant differentiation with respect to a connection on a hypersurface (this is notation is usually used in tandem with the semi colon and comma notation for covariant and partial derivatives, respectively). I honestly can't think what else it could mean although it obviously doesn't make any sense in this case.

I'd appreciate if the OP could respond to the point about the bar so that we can help to clear up the confusion.

Last edited: Aug 21, 2006
6. Aug 23, 2006

### pmb_phy

If you study relativity, (journals, texts etc.) you will eventually see "|" in place of ";". They mean exactly the same thing. Its just a difference of notation.

Pete

7. Aug 24, 2006

### robphy

As mentioned above, the vertical bar can denote a covariant differentiation... However, with the derivative operators $$\nabla$$ explicitly present, that meaning is unlikely.

A pair of vertical bars is also used with a pair of parentheses or a pair of square brackets to exclude indices from symmetrization or antisymmetrization, respectively.

Looking over the original post again, it may be a (nonstandard) notation to indicate that the first pair of indices are to be grouped together and similarly for the last pair... with the grouping implying some kind of symmetry.

8. Aug 24, 2006

### coalquay404

They do not mean the same thing in the context of relativity. Vertical bars on indices can mean one of two things:

(a) That certain indices are to be excluded from tensor symmetrization or anti-symmetrization. In this case the vertical bars appear in pairs and the indices to which the parity operation do not apply then appear within the bars. This notation has been standard since at least Eisenhart's main book.

(b) Covariant differentiation with respect to an induced metric. If you use semi-colons for covariant differentiation with respect to a spacetime connection, vertical bars are taken to mean differentiation w.r.t. an induced connection on some submanifold. This use of bars is standard is the way they are used in every textbook I can think of. Neither can I think of a single paper where the notation has a different meaning to this.

9. Aug 24, 2006

### pmb_phy

I'm merely going by the literature that I've read in th4e past. It is the notation of tensor analysis and relativity uses tensor analysis and sometimes some author will use "|" rather than ";". If you have a text which uses the vertical bar elsewise then please post the name of the text.

Pete

10. Aug 24, 2006

### coalquay404

I'll post lots:

awking & Ellis,
:D'Eath's "Supersymmetric Quantum Cosmology",
:Wald - both "General Relativity" and "Quantum field theory in curved spacetimes and black hole thermodynamics",
:Misner, Thorne, & Wheeler (beginning in chapter 21 - Variational methods, and continuing throughout the rest of the book),
:Almost all of the Cambridge Monographs on Mathematical Physics books directly related to relativity.

That's pretty much all of the standard GR textbooks, and each of them uses the bar notation in the way I've described. To that list you can add pretty much every paper that's been published over the past forty years which deals with the ADM formulation, canonical GR more generally, analysis of the constraint equations, conformal decomposition of the constraints, studies of the Lichnerowicz and Lichnerowicz-York methods, several of Witten's papers on positive energy, the positive energy papers from Schoen & Yau (although they did also use D-notation), and lots more.

I'm not denying that you may have seen the notation used in the way you say. I'm just saying that that way is *not* the standard.

EDIT: Just to clarify, what I'm saying is as follows. There are standard notations for a covariant derivative. Undoubtedly, the most common is $$\nabla$$ - everyone is familiar with that. However, sometimes you will also see a ';' placed next to an index to signify covariant differentiation. The point about the comma notation is that in the context of relativity it is always used to denote covariant differentiation with respect to a connection on a spacetime manifold.

The reason why the bar notation is used to signify covariant differentiation is quite different. Often, people are interested in imbedded submanifolds in a spacetime, $$M$$. Typically, you'll take a spacetime and imbed a spatial hypersurface $$\Sigma$$ in it by means of some imbedding $$\phi:\Sigma\to\phi(\Sigma)\subset M$$. This spatial hypersurface inherits an induced metric $$\phi^{*}g$$, where $$g$$ is a metric on $$M$$. You can then define a covariant derivative on the hypersurface in a straightforward way - it's the covariant derivative on the hypersurface (i.e., with respect to $$\phi^{*}g$$) for which the bar notation is used, not the spacetime covariant derivative.

Last edited: Aug 24, 2006
11. Aug 25, 2006

### pmb_phy

I wish you had given equation numbers as well as page numbers. I don't have the time or energy to look through every page to find what vertical bar you're looking for.

Pete

12. Aug 25, 2006

### coalquay404

That's what the index in a book is for. Look for any of the following:

First fundamental form/induced metric
Second fundamental form/extrinsic curvature
Hypersurfaces/foliations

and anything which deals with specific gravity of black holes and therefore black hole mechanics and thermodynamics. If you're not going to bother to look at the books, you'll just have to take my word for it.

13. Aug 25, 2006

### pmb_phy

I looked at the books you refered to before I asked for page/equation numbers but I didn't see that notation. I myself am in the habit of always giving page/equation numbers when I reference a book.

Then again I don't have a need to know what you're speaking about so don't worry about me not verifying your claim.

Best wishes

Pete

14. Aug 25, 2006

### George Jones

Staff Emeritus
In Adler Bazin, and Schiffer (possibly the most influential pre-MTW relativity book), a single vertical bar is used to denote an ordinary partial derivative derivative (like a comma), and a double vertical bar is used to denote a covariant derivative (like a semicolon).

15. Aug 25, 2006

### pervect

Staff Emeritus
I had only previously seen the first sort of usage you describe (symmetrization) where the bars occured in pairs inside either () to exclude indices from symmetrization, or in pairs inside [] to exclude indices from anti-symmetrization.

I don't see anything in the index of MTW or Wald on "induced metrics" or "first fundamental form", so I suspect that the reason I haven't run across the second sort of usage is that the topic isn't discussed in these textbooks (if you happen to know differently, I'd like to know). MTW in particular has some problems with its indexing sometimes.

16. Aug 25, 2006

### George Jones

Staff Emeritus
The notation is defined in equation (21.57) in MTW; I don't see this notation in Wald, but a different notation is used on page 257.

Last edited: Aug 25, 2006
17. Aug 26, 2006

### coalquay404

As I said earlier, MTW start using this notation in Chapter 21 when they discuss foliations of a spacetime (from memory, I think Eq. (21.105) defines the conjugate momenta to a three-metric and uses the bar notation). It is used several times in the later chapters. D'Eath starts using this notation in Section 2.4. Hawking & Ellis introduce the notation in section 2.8 and use it throughout the book, particularly in the chapters on the Cauchy problem and causal structure. I'm traveling at the moment and don't have my copy of Wald, but I'm sure it's used there also.

Last edited: Aug 26, 2006
18. Aug 26, 2006

### coalquay404

Well aren't you just so special! Good for you!

19. Aug 27, 2006

### pmb_phy

That was intended as a suggestion for your further postings habits for the benefit of the reader since what you refer to in a text may not be possible to find without skimming through the entire text. It was a suggestion not in need of a sarcastic response.

And yes. I am special in that sense.

Bye

Last edited: Aug 27, 2006
20. Sep 1, 2006

### dextercioby

It's interesting this simple wrongly stated question raised so much debate, even on a simple matter of notation.

G is the Einstein tensor in curved spacetime and T is an unknown tensor with mixed (2,2) symmetry (of the known Riemann tensor for a torsionless manifold) which is actually specified by that vertical bar: antisymmetry wrt to first & second, third & fourth independently and symmetrical wrt to changing the first group of 2 indices and the second one. This question arised naturally, since, in the linearized limit, the Einstein tensor (actually the Ricci tensor as well) can be put in this form

$$G_{\alpha\beta}=\partial^{\mu}\partial^{\nu} Z_{\mu\alpha|\nu\beta}$$

ALWAYS, and it's a simple matter of Young tableaux techinque to justify it. Z is to be determined (just like T in the posed question above; sorry about the confusion with the energy-momentum tensor, it has nothing to do with it).

And i thought that could be generalized in the curved case, keeping the same symmetries of the Z(or T) tensor, namely (2,2). Apparently it cannot and i had realized that only after posting the message & there was no way i could delete the thread afterwards.

Daniel.