# Notation question

1. Dec 14, 2006

### Jheriko

I have seen $\partial_{[\mu}\omega_{\nu]}$ this sort of notation ( the [ ] ) a few times now, what does it mean?

I am already familiar with $\partial_{\mu}$ as being the comma derivative, or partial derivative with respect to the coordinate with index $\mu$. Is this something similar but with respect to tensor components perhaps?

I have tried Google, but I think it ignores [ and ] in its search terms, and not knowing what the operation is actually called makes it harder to search too.

Any answer is appreciated, but preferably something short and sweet.

Last edited: Dec 14, 2006
2. Dec 14, 2006

### George Jones

Staff Emeritus
It means antisymmetrize:

$$\partial_{[\mu}\omega_{\nu]} = \frac{1}{2} \left( \partial_{\mu}\omega_{\nu} - \partial_{\nu}\omega_{\mu} \right).$$

3. Dec 14, 2006

### Jheriko

Thanks very much for the quick response. Short and sweet too. :)

4. Dec 14, 2006

### dextercioby

The 1/2 factor can very well be omitted. It's just a matter of convention.

Daniel.

5. Dec 14, 2006

### Jheriko

I would guess that the factor can be omitted since, if I am remembering correctly, $A_{\mu}B_{\nu} - B_{\mu}A_{\nu}$ is always anti-symmetric. I would also guess that the 1/2 is conventionally included to simplify some frequently occuring expression(s).

I suppose I can always just work out what factor is used in what text from whatever ends up as the result of expanding the [ ].

Thanks for the help guys. It is much appreciated.

6. Dec 14, 2006

### coalquay404

The factor of $1/2$ (or, more generally in the case of a $m$ index tensor, $1/m!$) can indeed be omitted. However, it's good practice to include it since it tends to simplify nasty combinatorial factors encountered in, for example, index calculations involving differential forms.

You may also find that the degree to which this particular convention is enforced depends very much on the context in which it is used. Relativists, for example, always seem to adhere to it whereas a lot of the string theory literature seems to ignore it.

7. Dec 19, 2006

### Chris Hillman

To halve or halve not?

Following up on coalquay's remark about context, choosing one convention might be convenient for some purposes, e.g. it might be nice to be able to write (using the 1/2 convention) simply $$T_{ab} = T_{(ab)} + T_{[ab]}$$, while other conventions might be more convenient for other purposes. Kinda like the multiple conventions for writing orthogonal polynomials; if you want your polynomials to form an orthonormal set, certain rational factors are required; if you only need orthogonality, these are simply a bother and you will probably adopt a convention which omits them.

8. Dec 19, 2006

### Jheriko

Now I am even more confused.

I assumed from the previous posts that $T_{[a}U_{b]}$ means $T_{a}U_{b} - T_{b}U_{a}$. So what does $T_{[ab]}$ mean? Or $T_{(ab)}$ for that matter?

I will guess based on my existing knowledge that () is "symmetrisation" as opposed to "anti-symmetrisation", which would fit with the statement that "any tensor can be expressed as the sum of a symmetric tensor and an anti-symmetric tensor".

In the case of $T_{(ab)}$ or $T_{[ab]}$. we would have to decompose the original tensor into lower rank tensors like $T_{ab} = V_{a}W_{b}$, then we would have $T_{(ab)} = V_{a}W_{b} + V_{b}W_{a}$ and $T_{[ab]} = V_{a}W_{b} - V_{b}W_{a}$.

Is this correct? Or am I groping in the dark still?

I suppose that really I need to get a decent textbook on tensors to work through...

Last edited: Dec 19, 2006
9. Dec 19, 2006

### dextercioby

$$T_{(ab)}=\left(\frac{1}{2}\right) \left(T_{ab}+T_{ba}\right)$$

where the 1/2 factor can be missing.

The same goes for the antisymmetrized tensor.

Daniel.

10. Dec 19, 2006

### Jheriko

That does seem quite obvious now that I have seen it. Don't know why I suddenly thought I would need to decompose the tensor to be able to do this.. I should have realised that the $V_{b}W_{a}$ in my expressions were equal to $T_{ba}$.

11. Dec 19, 2006

### robphy

When the combinatorial factor is included in the definition, one can write
$$A_{[ab]}=\frac{1}{2!}\left(A_{ab}-A_{ba}\right)$$
and refer to $$A_{[ab]}$$ as the "[totally] antisymmetric part of $$A_{ab}$$"... and similarly for $$A_{(ab)}$$

12. Dec 22, 2006

### Chris Hillman

Oh dear!

I am sorry that I confused you! The point is that different books/papers use different conventions. The factor 1/2 (for both symmetrization and antisymmetrization operators) allows you to write a second rank tensor as the sum of its symmetric part plus its antisymmetric part, which is often useful. If you don't need to do that, though, the factors are just a pain, so in other contexts they are often ommitted. You will often have to infer from context which definition is intended.

Right, that's the idea: you can symmetrize or antisymmetrize any pair of indices in any tensor and then by trivial algebra this will hold true (if you use the convention with factors of 1/2). Similar but more elaborate statements can be formulated for symmetrizing and so on over three or more indices, but these will usually require additional terms.

Probably you don't actually need to know very much at all about "tensor analysis" to jump right in and start reading about physical applications of tensor analysis. This is the approach advocated in MTW Gravitation, for example.

13. Jan 3, 2007

### Jheriko

I have been reading Dirac's "General Theory of Relativity" from Princeton Milestones in Physics (http://press.princeton.edu/titles/5813.html), I have been able to make my way through the book well enough so far. Basically that has been my reference for tensors, in combination with the internet.

I came across the [] and () notation for the first time in an internet textbook which I was reading to help "fill out" my background knowledge.

The only problem I have with the book is the small size, I think the author assumes that the reader is already familiar with Special Relativity and using tensors in that context, although he does cover the basics (very briefly) and in fairness it was almost enough for me to learn what I needed in order to get through the book. No exercises though...

I am thinking of buying MTW at some point, I have heard it mentioned and praised in many places (including here).

14. Jan 6, 2007

### Chris Hillman

Yes, Dirac's book is fun for those who already know gtr to see how he compresses the essence into such a short book (not one extraneous word is to be found!), but very few beginners will benefit from reading this book, IMO.

I highly recommend MTW, but some students complain that they find this tough going. Some easier textbooks which offer a fine introduction to modern notation are listed in http://www.math.ucr.edu/home/baez/RelWWW/reading.html#gtr. Enjoy!