Tensorial Calculation and antisymmetric tensors

[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

 

[Dickfore]

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

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?

 

[vnikoofard]

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

 

[vnikoofard]

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

 

[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:
<br /> \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}<br />
<br /> W^{\mu \nu} = -\frac{1}{2} \, \epsilon^{\mu \nu \rho \pi} \, \tilde{W}_{\rho \pi}<br />

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

 

[Dickfore]

Then, you will have:
<br /> \partial_{\mu} V_{\nu} - \partial_{\nu} V_{\mu} = 0<br />

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:
<br /> \oint{V_{\mu} \, dx^{\mu}} = 0<br />

Do you know what this means?

 

[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?

 

[vnikoofard]

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

 

[Dickfore]

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?

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).

 

[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".

 

[Dickfore]

i don't know what to recommend, sorry.

 

[vnikoofard]

Thank you again for your help, my friend!

 

[haushofer]

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

 

[vnikoofard]

Thanks Haushofer! :)