Redshifting of forces in stationary space - times

[PeterDonis]

I will try to show that $\nabla_{a}E\nabla^{a}E = (F_{\infty})_{a}(F_{\infty})^{a}$ holds for all stationary (and if needed asymptotically flat) space-times or at least come up with some physical argument for it.

I think that both the stationary and the asymptotically flat assumptions are required: stationary so there is a well-defined notion of being "held at rest" (namely, following an orbit of the timelike KVF), and asymptotically flat so there is a well-defined notion of "infinity" so that the concepts of energy at infinity and force at infinity make sense. (Basically this amounts to requiring that there is an invariant way to normalize the timelike KVF so its length goes to 1 at infinity.)

 

[PeterDonis]

Working with the Kerr metric in Boyer-Lindquist coordinates, I got the acceleration of a hovering observer
in the local static frame basis to be $\frac{m\,r-{a}^{2}}{{r}^{2}\,\sqrt{{r}^{2}-2\,m\,r+{a}^{2}}}\ \partial_r$.

This doesn't seem to match what I posted in post #44. I posted:

$$a^a = \frac{M}{r^2} \frac{W^2}{V^2} \partial_r$$

where

$$V^2 = 1 - \frac{2M}{r}$$

$$W^2 = 1 - \frac{2M}{r} + \frac{a^2}{r^2} = V^2 + \frac{a^2}{r^2}$$

Substituting, this gives:

$$a^a = \frac{M}{r^2} \left( \frac{r^2 - 2 M r + a^2}{r^2 - 2M r} \right) \partial_r$$

Can you post some more details on how you arrived at your formula?

 

[WannabeNewton]

In Poisson's text, he does a similar thought experiment where he lowers the particle hanging from the string by a small amount. However he claims that the work done by the observer at infinity in moving the particle by a proper distance $\delta s$ is $\delta W = F_{\infty} \delta s$ where $F_{\infty} = [(F_{\infty})^{a}(F_{\infty})_{a}]^{(1/2)}$. I have no idea where he gets this formula from or what definition of work he is using. He then goes on to say that the energy extracted at infinity due to the lowering of the particle, $\delta E = (\nabla^{a}E \nabla_{a}E)^{(1/2)}\delta s$, should be equal to $\delta W$ via conservation of energy and equates to get $F_{\infty} \delta s = (\nabla^{a}E \nabla_{a}E)^{(1/2)}\delta s$ implying $F_{\infty} = (\nabla^{a}E \nabla_{a}E)^{(1/2)}$ but again I have no idea how he came up with those formulas.

 

[Mentz114]

This doesn't seem to match what I posted in post #44
...
...
Substituting, this gives:

$$a^a = \frac{M}{r^2} \left( \frac{r^2 - 2 M r + a^2}{r^2 - 2M r} \right) \partial_r$$

Can you post some more details on how you arrived at your formula?

I'm using the static frame field (in BL coordinates) from lecture notes ( see below)

The acceleration is $\nabla_\hat{\nu} u_\hat{\mu} u^\hat{\nu}$, calculated in the frame basis. For $a=0$ my value goes to the acceleration in the frame basis, for the Schwarzschild vacuum. I repeated your calculation and I get the same result ( as expected). This gives an $a=0$ value which is the coordinate basis value for the Schwarzschild.

I'm not 100% happy with the frame result although it gives the correct value when $a=0$.

The reference for the BL tetrad is http://casa.colorado.edu/~ajsh/phys5770_08/grtetrad.pdf section 5.22.

 

[WannabeNewton]

I hate to bring up an old thread but a more recent thread got me thinking about this again. It seems that the solution comes from the fact that for stationary, asymptotically flat space-times we can define a gravitational potential (analogous to the Newtonian potential) by $\varphi = \frac{1}{2}\ln (-\xi_{a}\xi^{a}) = \frac{1}{2}\ln V^{2} = \ln V$. But note that $\nabla^{b} \varphi$ will only be the local change in the potential; the change in the potential as measured at infinity will get red-shifted so $(\nabla^{b}\varphi)_{\infty} = V\nabla^{b}\varphi$. Now the net virtual work done on the particle, as measured at infinity, for a virtual displacement $\delta s^{b}$, will be $\delta W_{\infty} = (F_b)_{\infty}\delta s^{b} + m(\nabla_{b}\varphi)_{\infty} \delta s^{b}$ i.e. the work done by the observer at infinity plus the work done by the gravitational field as measured at infinity for this arbitrary virtual displacement. But the principle of virtual work says that for stationary particles, $(F_b)_{\infty}\delta s^{b} + m(\nabla_{b}\varphi)_{\infty} \delta s^{b} = 0$ for any virtual displacement $\delta s^{b}$ so this implies that $(F_b)_{\infty} + m(\nabla_{b}\varphi)_{\infty} = 0$ i.e. $(F_b)_{\infty} = -mV(\nabla_{b}\varphi) = -m\nabla_{b}V = -\nabla_{b}E$ because $E = -m\xi_{a}u^{a} = -mV^{-1}\xi_{a}\xi^{a} = mV$ for the stationary particle. So this seems to be why taking the gradient of the energy gives what we want. I haven't really been rigorous with this but it seems to be an ok argument.

 

[PeterDonis]

I haven't really been rigorous with this but it seems to be an ok argument.

This looks OK to me. The only comment I would make is that the argument works because force is taken to be a covector; i.e., force acts like a gradient, so it can be contracted directly with a virtual displacement. If force were a vector, you would need to bring in the metric to contract it with a virtual displacement; I think that would mess up the argument. I bring this up because IIRC the question of whether force is fundamentally a vector or a covector arose earlier in the thread.

 

[WannabeNewton]

Thanks for the help Peter! You're awesome