- #1
Maybe_Memorie
- 353
- 0
I have a few questions regarding the solution to this problem. First of all I have the Stress-Energy tensor for [itex]a[/itex] scalar fields [itex]\phi^a[/itex]
[tex]T_{Noether}^{\mu\nu} = \displaystyle \sum_a \frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial^{\nu}\phi^a - g^{\mu \nu}\mathcal{L}
[/tex]
To ensure symmetry I need to add a total divergence
[itex]\partial_{\lambda}K^{[\lambda\mu]\nu}[/itex] which is anti-symmetric in [itex]\mu[/itex] and [itex]\nu[/itex]
So I have a new tensor [tex]T^{\mu\nu}=T_{Noether}^{\mu\nu} + \partial_{\lambda}K^{[\lambda\mu]\nu}[/tex] I need to show that this thing is conserved.
[itex]\partial_{\mu}T^{\mu\nu}=\partial_{\mu}T_{Noether}^{\mu\nu}+\partial_{\mu}\partial_{\lambda}K^{[\lambda\mu]\nu}[/itex]
The solution then says that because [itex]\partial_{\mu}\partial_{\lambda}[/itex] is symmetric in the indicies and because [itex]K^{[\lambda\mu]\nu}[/itex] is anti-symmetric in those indicies then [itex]\partial_{\mu}\partial_{\lambda}K^{[\lambda\mu]\nu}=0[/itex]
I don't see why this is. I tried writing [itex]2K^{[\lambda\mu]\nu}=K^{[\lambda\mu]\nu}+K^{[\mu\lambda]\nu}[/itex] but that didn't get me anywhere. Now, after that, I have [itex]\partial_{\mu}T_{Noether}^{\mu\nu}=\partial_{\mu}\left[\displaystyle \sum_a \frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial^{\nu}\phi^a - g^{\mu \nu}\mathcal{L} \right]
= \displaystyle \sum_a \partial_{\mu}\left[\frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial^{\nu}\phi^a\right] - \partial^{\nu}\mathcal{L}
[/itex]
[itex] =\displaystyle \sum_a \left[\partial_{\mu}\frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial^{\nu}\phi^a + \frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial_{\mu}\partial^{\nu}\phi^a\right]- \partial^{\nu}\mathcal{L}[/itex]
[itex]=\displaystyle \sum_a \left[\frac{\partial \mathcal{L}}{(\partial\phi_a)}\partial^{\nu}\phi^a + \frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial_{\mu}\partial^{\nu}\phi^a\right]- \partial^{\nu}\mathcal{L}[/itex] after imposing the equations of motion. All fine up to here. Our solution then says that this thing is equal to
[itex]\displaystyle \sum_a \frac{\partial \mathcal{L}}{(\partial\phi_a)}\partial^{\nu}\phi^a- \partial^{\nu}\mathcal{L}[/itex]
I don't understand how this is arrived at. It then goes on to say that this is equal to [itex]\partial^{\nu}\mathcal{L}-\partial^{\nu}\mathcal{L}[/itex]. Again, I don't understand how this was achieved. Any pointers would be very appreciated. Thanks.
[tex]T_{Noether}^{\mu\nu} = \displaystyle \sum_a \frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial^{\nu}\phi^a - g^{\mu \nu}\mathcal{L}
[/tex]
To ensure symmetry I need to add a total divergence
[itex]\partial_{\lambda}K^{[\lambda\mu]\nu}[/itex] which is anti-symmetric in [itex]\mu[/itex] and [itex]\nu[/itex]
So I have a new tensor [tex]T^{\mu\nu}=T_{Noether}^{\mu\nu} + \partial_{\lambda}K^{[\lambda\mu]\nu}[/tex] I need to show that this thing is conserved.
[itex]\partial_{\mu}T^{\mu\nu}=\partial_{\mu}T_{Noether}^{\mu\nu}+\partial_{\mu}\partial_{\lambda}K^{[\lambda\mu]\nu}[/itex]
The solution then says that because [itex]\partial_{\mu}\partial_{\lambda}[/itex] is symmetric in the indicies and because [itex]K^{[\lambda\mu]\nu}[/itex] is anti-symmetric in those indicies then [itex]\partial_{\mu}\partial_{\lambda}K^{[\lambda\mu]\nu}=0[/itex]
I don't see why this is. I tried writing [itex]2K^{[\lambda\mu]\nu}=K^{[\lambda\mu]\nu}+K^{[\mu\lambda]\nu}[/itex] but that didn't get me anywhere. Now, after that, I have [itex]\partial_{\mu}T_{Noether}^{\mu\nu}=\partial_{\mu}\left[\displaystyle \sum_a \frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial^{\nu}\phi^a - g^{\mu \nu}\mathcal{L} \right]
= \displaystyle \sum_a \partial_{\mu}\left[\frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial^{\nu}\phi^a\right] - \partial^{\nu}\mathcal{L}
[/itex]
[itex] =\displaystyle \sum_a \left[\partial_{\mu}\frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial^{\nu}\phi^a + \frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial_{\mu}\partial^{\nu}\phi^a\right]- \partial^{\nu}\mathcal{L}[/itex]
[itex]=\displaystyle \sum_a \left[\frac{\partial \mathcal{L}}{(\partial\phi_a)}\partial^{\nu}\phi^a + \frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial_{\mu}\partial^{\nu}\phi^a\right]- \partial^{\nu}\mathcal{L}[/itex] after imposing the equations of motion. All fine up to here. Our solution then says that this thing is equal to
[itex]\displaystyle \sum_a \frac{\partial \mathcal{L}}{(\partial\phi_a)}\partial^{\nu}\phi^a- \partial^{\nu}\mathcal{L}[/itex]
I don't understand how this is arrived at. It then goes on to say that this is equal to [itex]\partial^{\nu}\mathcal{L}-\partial^{\nu}\mathcal{L}[/itex]. Again, I don't understand how this was achieved. Any pointers would be very appreciated. Thanks.