Does Energy Gravitate? A Check of Understanding

[DrGreg]

How does one copy such diagrams from here to Microsoft WORD?? All I get with COPY is the text, not the illustrations?

First, right-click on the image and choose the appropriate copy option from the popup menu. In Firefox, that option is called "Copy Image"; it might be called something different in other browsers. Second, within Word, use "Paste Special" rather than "Paste", and choose "Device Independent Bitmap".

Note that if you're keeping things to quote in future messages, you should also keep the URL where you got it from so you can quote the URL too (like A.T. did).

In fact, if you're keeping something for a possible future quote on PF, your best option is to hit the "Quote" button and then copy-and-paste the code that appears in the editing window (which will include a link to the original post).

 

[Naty1]

Thanks for the advice DrGreg..I don't have the options you list ...but my daughter figured out an approach so I now have your charts in my notes...

 

[Infrared]

For example one can construct what is known as the Bel-Robinson tensor $T_{abcd} = C_{aecf}C_{b}{}{}^{e}{}{}_{d}{}{}^{f} -\frac{3}{2}g_{a[b}C_{jk]cf}C^{jk}{}{}_{d}{}{}^{f}$ where $C_{abcd}$ is the Weyl curvature tensor. One can show that $T_{abcd} = T_{(abcd)}$ and that $\nabla^{a}T_{abcd} = 0$.

Could you explain why $\nabla^{a}T_{abcd}$ necessarily vanishes? It's not completely clear to me.

 

[WannabeNewton]

Could you explain why $\nabla^{a}T_{abcd} = 0$ necessarily vanishes? It's not completely clear to me.

I should have mentioned that it only holds in vacuum space-time $R_{ab} = 0$. This is another reason why such a prescription for a local gravitational energy density would not be complete.

Anyways, taking the divergence and expanding out the expression we have $\nabla^{a}T_{abcd} = C_{aecf}\nabla^{a}C_{b}{}{}^{e}{}{}_{d}{}{}^{f} - \frac{1}{2}g_{ab}\nabla^{a}\{C_{jkcf}C^{jk}{}{}_{d}{}{}^{f}\}+ \frac{1}{2}g_{aj}\nabla^{a}\{C_{bkcf}C^{jk}{}{}_{d}{}{}^{f}\} - \frac{1}{2}g_{ak}\nabla^{a}\{C_{bjcf}C^{jk}{}{}_{d}{}{}^{f}\}$. Note that in vacuum space-time we have $\nabla^{a}C_{abcd} = 0$. Furthermore, the second Bianchi identity reads $\nabla_{[a}R_{bc]d}{}{}^{e} = 0$ and in vacuum space-time we have $R_{abcd} = C_{abcd}$ so we also have that $\nabla_{[a}C_{bc]d}{}{}^{e} = 0$ in vacuum. Using these two results, we can greatly simplify the divergence of the Bel-Robinson tensor to just $\nabla^{a}T_{abcd} = C_{aecf}\nabla^{a}C_{b}{}{}^{e}{}{}_{d}{}{}^{f} - \frac{1}{2}C_{aecf}\nabla_{b}C^{ae}{}{}_{d}{}{}^{f}$.

Applying the second Bianchi identity once more, we have that $\nabla_{b}C^{ae}{}{}_{d}{}{}^{f} = \nabla^{a}C_{b}{}{}^{e}{}{}_{d}{}{}^{f} - \nabla^{e}C_{b}{}{}^{a}{}{}_{d}{}{}^{f}$
thus $\nabla^{a}T_{abcd} = C_{aecf}\nabla^{a}C_{b}{}{}^{e}{}{}_{d}{}{}^{f} - \frac{1}{2}C_{aecf}(\nabla^{a}C_{b}{}{}^{e}{}{}_{d}{}{}^{f} - \nabla^{e}C_{b}{}{}^{a}{}{}_{d}{}{}^{f} )\\ = \frac{1}{2}C_{aecf}\nabla^{a}C_{b}{}{}^{e}{}{}_{d}{}{}^{f} + \frac{1}{2}C_{aecf}\nabla^{e}C_{b}{}{}^{a}{}{}_{d}{}{}^{f}\\ = \frac{1}{2}C_{eacf}\nabla^{e}C_{b}{}{}^{a}{}{}_{d}{}{}^{f} + \frac{1}{2}C_{aecf}\nabla^{e}C_{b}{}{}^{a}{}{}_{d}{}{}^{f}\\ = \frac{1}{2}(-C_{aecf} + C_{aecf})\nabla^{e}C_{b}{}{}^{a}{}{}_{d}{}{}^{f} = 0$