# Tensorial Calculation and antisymmetric tensors

1. Jun 27, 2012

### vnikoofard

Hi Friends
I am reading the following paper
http://arxiv.org/abs/hep-th/9705122
In the page 4 he says that

$\tilde{W}_{\mu\nu}=0\Rightarrow V_{\mu}=\partial_{\mu}\lambda$

Where $\tilde{W}^{\mu\nu}\equiv\frac{1}{2}\epsilon^{\mu \nu\rho\sigma}W_{\rho\sigma}$ and $W_{\mu\nu}\equiv\partial_{[\mu}V_{\nu]}$ and $\epsilon$ is antisymmetric Levi-Civita tensor.

The above expression is a general argument and it is not related to the paper. I can not understand how can we drive $V_{\mu}=\partial_{\mu}\lambda$ from $\tilde{W}_{\mu\nu}=0$
Would someone please explain it for me

2. Jun 27, 2012

### Dickfore

Are you sure it doesn't say
$$\partial_{\nu} \tilde{W}^{\mu \nu} = 0$$
?

EDIT:

Oh, sorry, I saw that there is a V and a W, and the W is the anti-symmetrized derivative.

Do you know Stokes' theorem in 4-dimensional space-time?

Last edited: Jun 27, 2012
3. Jun 27, 2012

### vnikoofard

Yes, I am sure. :( You can check it in the mentioned paper.

4. Jun 27, 2012

### vnikoofard

Unfortunately I do not know. Is it related to Stokes's theorem?

5. Jun 27, 2012

### Dickfore

First of all, there is a one-to-one correspondence between $\tilde{W}_{\mu \nu}$, and $W_{\mu \nu}$. You just showed how to find $\tilde{W}$ if you know W. But:
$$\epsilon^{\mu \nu \rho \pi} \, \tilde{W}_{\rho \pi} = \frac{1}{2} \epsilon^{\mu \nu \rho \pi} \, \epsilon_{\rho \pi \sigma \tau} \, W^{\sigma \tau} = -\left(\delta^{\mu}_{\sigma} \, \delta^{\nu}_{\tau} - \delta^{\mu}_{\tau} \, \delta^{\nu}_{\sigma} \right) \, W^{\sigma \tau} = -W^{\mu \nu} + W^{\nu \mu} = -2 \, W^{\mu \nu}$$
$$W^{\mu \nu} = -\frac{1}{2} \, \epsilon^{\mu \nu \rho \pi} \, \tilde{W}_{\rho \pi}$$

Therefore, if you say $\tilde{W}_{\mu \nu} = 0$, then, so is $W_{\mu \nu} = 0$.

6. Jun 27, 2012

### Dickfore

Then, you will have:
$$\partial_{\mu} V_{\nu} - \partial_{\nu} V_{\mu} = 0$$

Integrate this over an arbitrary 2-dimensional surface with an element $df^{\mu \nu} = -df^{\nu \mu}$, and convert it to a line integral over the boundary of the surface. You should get:
$$\oint{V_{\mu} \, dx^{\mu}} = 0$$

Do you know what this means?

7. Jun 27, 2012

### vnikoofard

Thanks! Now I got it. When $W_{\mu \nu} =0$ means $\partial_\mu V_{\nu} -\partial_{\nu}V_{\mu}=0$. So for having this expression we must suppose that $V_{\mu} =\partial_\mu\lambda$ where $\lambda$ is a scalar. Because we can change order of derivations $\partial_{\mu}, \partial_{\nu}$. Is it correct?

8. Jun 27, 2012

### vnikoofard

Would you please explain more about the integral? It seems interesting.

9. Jun 27, 2012

### Dickfore

No, what you are proving is that $W_{\mu \nu} = 0$ is a necessary condition for $V_{\mu} = \partial_{\mu} \lambda$, which I though is trivial to show (because derivatives commute). But, I was trying to point out that it is also a sufficient condition. Well, locally at least (see Poincare's Lemma).

10. Jun 27, 2012

### vnikoofard

Dear Dickfore, I am really poor on Topology and such kinds of mathematics. Recently I decided to begin studying this topics. Can you please suggest me some good textbooks for self-study. I am thinking about 3rd edition of Frankle's book: "Geometry of Physics".

11. Jun 27, 2012

### Dickfore

i don't know what to recommend, sorry.

12. Jun 27, 2012

### vnikoofard

Thank you again for your help, my friend!

13. Jun 28, 2012

### haushofer

Frankle is nice, Nakahara is also very good for selfstudy; at least the chapters untill Fibre Bundles. After that it becomes a bit wuzzy.

14. Jun 28, 2012

### vnikoofard

Thanks Haushofer! :)

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook