Does Energy Gravitate? A Check of Understanding

[PeterDonis]

yup. there is non gravitational spacetime curvature...
A simple example. A particle with rectilinear motion passes a stationary [or inertial] observer in their common frame. A coincident accelerating observer sees the same particle following a curved worldline...it appears spacetime is curved...that is a coordinate effect related to the acceleration, not a gravitational characteristic of the particle.

This isn't spacetime curvature. The Riemann curvature tensor is zero. Put another way, there is no tidal gravity present. Spacetime curvature is not a coordinate effect.

Lets draw the flat spacetime of an inertial observer on on a flat piece of graph paper. [ She sees a particle with rectilinear motion as a straight line.] If we switch to a non-inertial frame, an accelerated observer but still in the absence of gravitation, we are now drawing a curved grid, but still on the same flat sheet of paper. [He is accelerating and sees a curved trajectory of the same particle.]

If the sheet of paper is still flat, then there is no spacetime curvature. Spacetime is the sheet of paper, not the coordinate grid. I doubt if DrGreg intended any other interpretation.

GRAVITATIONAL "spacetime curvature" refers to the curvature of graph paper itself, regardless of observer

"Gravitational spacetime curvature" is the only kind of spacetime curvature there is. Whatever kind of curvature the curvature of the coordinate grid is, it isn't spacetime curvature. See above.

 

[Naty1]

Quote by Naty1

yup. there is non gravitational spacetime curvature...

PeterDonis

This isn't spacetime curvature.

we agree...and the rest of your post is of course correct...Thanks

my wording was misleading...I edited my earlier post to read 'apparent' curvature...

In that earlier discussion DrGreg also provided a related mathematical insight which further confirms your comments:

In Minkowski [flat] coordinates in Special Relativity, 4-acceleration is just the coordinate derivative of 4-velocity with respect to arc-length (proper time), and the 4-velocity is the unit tangent vector of the worldline. As the 4-velocity has a constant length its derivative must be orthogonal to it. The 4-acceleration is the curvature vector; orthogonal to the worldline and its length is the reciprocal of the worldline's "radius of curvature".

In non-inertial coordinates in GR, the 4-acceleration is defined as a covariant derivative. This takes into account (and removes) any curvature of spacetime or "apparent curvature" due to using a "non-square grid", and leaves us with curvature that is a property of the worldline itself, not the spacetime or the choice of coordinates. Then everything else I said in the last paragraph is still true in a coordinate-independent sense.

[I don't usually post mathematical descriptions because I know very well there are too many aspects, subtleties, that elude me.]

For those interested in a more detailed discussion, and a complete actual quote from DrGreg on the subject of 'apparent curvature' , try here:

space time curvature caused by fast electron [Nov 7, 2011]
https://www.physicsforums.com/showthread.php?t=548148

edit: One of the great things about these forums is that there are always experts ready to keep us non experts within proper bounds!

 

[WannabeNewton]

The extrinsic curvature of a smooth curve embedded in a Riemannian or semi-Riemannian manifold is in no way a coordinate dependent quantity. The only thing it depends on is how the embedding map is defined. A circle of radius $r$ naturally embedded in $\mathbb{R}^{2}$ has an extrinsic curvature of $1/r$; this is totally unrelated to whether we use polar, cartesian or w\e coordinates on $\mathbb{R}^{2}$.

The only "apparent" coordinate dependent behavior of curves I can possibly think of in relation to what is being talked about is the fact that the Christoffel symbols need not vanish identically for e.g. polar coordinates on $\mathbb{R}^{2}$ so when we write down the geodesic equation in polar coordinates we get some non trivial expression of the form $\frac{\mathrm{d} ^{2}x^{\mu} }{\mathrm{d} s^2} = f^{\mu}(x^{\nu})$ whereas in Cartesian coordinates we just get $\frac{\mathrm{d} ^{2}x^{\mu'} }{\mathrm{d} s^2} = 0$ but regardless the geodesics will be straight lines.

EDIT: Sort of unrelated but interesting nonetheless: http://www.phas.ubc.ca/~dwang/extrinsic.pdf

 

[A.T.]

Here is how DrGreg explained that to me in more general terms several years ago in these forums: [some restatements of the same points appear using different words]

Lets draw the flat spacetime of an inertial observer on on a flat piece of graph paper. [ She sees a particle with rectilinear motion as a straight line.] If we switch to a non-inertial frame, an accelerated observer but still in the absence of gravitation, we are now drawing a curved grid, but still on the same flat sheet of paper. [He is accelerating and sees a curved trajectory of the same particle.]

GRAVITATIONAL "spacetime curvature" refers to the curvature of graph paper itself, regardless of observer, whereas visible [apparent] curvature in space is related to the distorted, non-square grid lines drawn on the graph paper, and depends on the choice of observer. In the absence of gravity, spacetime [graph paper] is always "flat" whether you are an inertial observer or not; non-inertial observers will draw a curved grid on flat graph paper.

DrGreg:

Gravitational spacetime curvature is coordinate independent; the observer based curvature obviously is not.

Here is the picture by DrGreg:

Here the full post with more explanations:
https://www.physicsforums.com/showthread.php?p=4281670&postcount=20

 

[Naty1]

AT...thanks for the post from DrGreg...My discussion with him is now likely three years ago or so w.o any diagrams; had not seen those VERY nice illustrations from this year...Thanks..How does one copy such diagrams from here to Microsoft WORD?? All I get with COPY is the text,
not the illustrations?

 

[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 $\nabla^{a}T_{abcd} = C_{aecf}\nabla^{a}C_{b}{}{}^{e}{}{}_{d}{}{}^{f} - \frac{1}{2}C_{aecf}(\nabla^{a}C_{b}{}{}^{e}{}{}_{d}{}{}^{f} - \nabla^{e}C_{b}{}{}^{a}{}{}_{d}{}{}^{f} )\\ = \frac{1}{2}C_{aecf}\nabla^{a}C_{b}{}{}^{e}{}{}_{d}{}{}^{f} + \frac{1}{2}C_{aecf}\nabla^{e}C_{b}{}{}^{a}{}{}_{d}{}{}^{f}\\ = \frac{1}{2}C_{eacf}\nabla^{e}C_{b}{}{}^{a}{}{}_{d}{}{}^{f} + \frac{1}{2}C_{aecf}\nabla^{e}C_{b}{}{}^{a}{}{}_{d}{}{}^{f}\\ = \frac{1}{2}(-C_{aecf} + C_{aecf})\nabla^{e}C_{b}{}{}^{a}{}{}_{d}{}{}^{f} = 0$