Redshifting of forces in stationary space - times

[Passionflower]

If by "rotational acceleration" you mean a tangential \phi component to the proper acceleration, yes, I thought so too until I did the computation. See my previous posts.

I mean the rotation around the test mass' axis.

 

[PeterDonis]

I mean the rotation around the test mass' axis.

Meaning rotation relative to Fermi-Walker transport along the test mass' worldline? Yes, that's still present, but it's a whole other can of worms that I'm not trying to open at this time. I'm only looking at the proper acceleration of the trajectory, i.e., only at the 4-velocity vector, not the entire orthonormal tetrad for each observer.

 

[Passionflower]

Meaning rotation relative to Fermi-Walker transport along the test mass' worldline? Yes, that's still present, but it's a whole other can of worms that I'm not trying to open at this time. I'm only looking at the proper acceleration of the trajectory, i.e., only at the 4-velocity vector, not the entire orthonormal tetrad for each observer.

Seems to me that if it rotates fast it could become a factor in calculating the Doppler shift.

 

[WannabeNewton]

Very nice, thanks for the computations Peter! I'll take a look at the blog post regarding the centrifugal forces. I'm still trying to understand what you said regarding the centrifugal force at the very end of your post #44, the last paragraph that is, and trying to connect it to Pervect's analogy with the electromagnetic field.

Until then however, regarding the issue of why $F^{b} = -\nabla^{b} E$ worked in giving us $F_{\infty} = VF$: even for an arbitrary stationary axisymmetric space-time it is eluding me why this holds true (because so far all we have worked out is the time-like and tangential components vanish). Maybe it was just a coincidence?

 

[WannabeNewton]

I mean the rotation around the test mass' axis.

By this, do you mean precession of the spin axis due to frame dragging? This is non zero even for stationary observers. Here is a thread where I had to find the precession for an observer at rest, with respect to the background minkowski metric, at the center of a rotating shell in the linearized approximation: https://www.physicsforums.com/showthread.php?t=675475

If you'll notice, the precession $\frac{\mathrm{d} \vec{S}}{\mathrm{d} t} = \frac{4}{3}\frac{M}{R}\vec\omega \times \vec{S}$ is very similar in form to the expression relating the rate of change of a vector as seen in an inertial frame to its rate of change in a frame rotating with constant angular velocity $\Omega$ in the case where the vector is at rest in the rotating frame: $\frac{\mathrm{d} \vec{Q}}{\mathrm{d} t} = \vec\Omega \times \vec{Q}$

 

[PeterDonis]

Maybe it was just a coincidence?

I don't think so, but I am holding off on commenting further until I've done some more computations.

 

[PeterDonis]

if I'm right, the effect of frame dragging is to change how "centrifugal force" works; it changes the *magnitude* of proper acceleration required for a given equatorial trajectory, not its direction. What actually gets "dragged" tangentially is the "zero point" of centrifugal force: the closer to the rotating hole you are, the more the "zero point" of centrifugal force is biased in the direction of the hole's rotation.

The blog post is still not complete--in fact, it has now grown into two blog posts because the first one went over the 10000 character PF limit on blog posts! However, I can at least report enough results to address the topic of this thread. (The blog posts are also going to address some more general questions about the behavior of proper acceleration in Kerr spacetime, which I find interesting in their own right since there are some key differences from the non-rotating case that I didn't see coming. But more on some of that below.)

Regarding the intuitive guesses I made, quoted above, I was half right. The effect of frame dragging, at least in the equatorial plane of Kerr spacetime, does not affect the direction of proper acceleration, only its magnitude. However, the effect on the "zero point" of centrifugal force, i.e., the point of maximum proper acceleration required to maintain altitude, is the opposite of what I guessed: the maximum point moves "backwards" relative to the hole's rotation, i.e., the more the hole rotates, the more the maximum point moves in the retrograde direction--the direction of tangential motion opposite to the hole's rotation.

In retrospect, of course, this is obvious. What frame dragging does is to make it easier (i.e., to require less force) to maintain altitude if you are circling the hole along with the hole's rotation, and to make it harder if you are circling against the hole's rotation. This can be interpreted as "adding some centrifugal force" due to the hole's rotation, so of course a ZAMO, who is rotating with the hole, will find it easier to maintain altitude than a static observer. (The part that isn't quite obvious is that even a static observer gets some "help" from the hole's rotation, since the maximum is always at some nonzero retrograde angular velocity.)

This means that, in the equatorial plane, the force required to hold an object static is indeed purely radial (as I posted before), and so is the gradient of the energy at infinity (since in the equatorial plane we have E = \sqrt{g_{tt}} = \sqrt{1 - 2M/r}). It's then straightforward to show (as I think we've already done in this thread) that the force at infinity is equal to minus the gradient of energy at infinity.

However, for objects held static at a point that is not in the equatorial plane, both the force and the gradient of energy at infinity will have a non-radial component (a \theta component in the Boyer-Lindquist chart for Kerr spacetime). This part will not be in my forthcoming blog post, but it looks like a computation of the components will show that, although the individual components are not identical, the ratio of the \theta component to the r component is the same for both (so both vectors point in the same direction--or opposite directions if we take the minus sign into account), and the magnitude of the gradient of the energy at infinity is indeed the "redshift factor" times the magnitude of the local force vector (rest mass times proper acceleration). In other words, the equation (force at infinity = - gradient of energy at infinity) does not hold at the component level, but it *does* hold as an invariant equation summed over all nonzero components. The computations are rather messy so I won't post them unless someone asks, but this at least suggests that the same should be true for a generic stationary spacetime.

 

[Mentz114]

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In other words, if I'm right, the effect of frame dragging is to change how "centrifugal force" works; it changes the *magnitude* of proper acceleration required for a given equatorial trajectory, not its direction. What actually gets "dragged" tangentially is the "zero point" of centrifugal force: the closer to the rotating hole you are, the more the "zero point" of centrifugal force is biased in the direction of the hole's rotation. We'll see how it pans out; I plan to put the details in a PF blog post, and I'll post a link and a quick summary here when I do.

I read this post with interest ( although I cut it here for brevity).

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 $\frac{m\,r-{a}^{2}}{{r}^{2}\,\sqrt{{r}^{2}-2\,m\,r+{a}^{2}}}\ \partial_r$.

For a test body in orbit an constant r with angular velocity $\dot{\phi}$, the expression for the radial acceleration is a bit long, but the Taylor expansion in $a$ around zero is

$-\frac{\left( {r}^{3}-3\,m\,{r}^{2}\right) \,{\dot{\phi}}^{2}-m}{{r}^{2}-2\,m\,r} + \frac{2\,a\,m\,\dot{\phi}\,\sqrt{{r}^{2}\,{\dot{\phi}}^{2}+1}}{\sqrt{r}\,{\left( r-2\,m\right) }^{\frac{3}{2}}}+\frac{2\,{a}^{2}\,m\,{\dot{\phi}}^{2}}{{\left( r-2\,m\right) }^{2}} + ...$
which shows the contribution of $a$ to the acceleration clearly. This expression depends on the signs of $a$ and $\dot{\phi}$, so there are different forces required in the spin and anti-spin directions.

In the equatorial plane only the r-component of proper acceleration is non-zero.

 

[WannabeNewton]

So what you are saying is that based on your calculations for kerr space-time in the usual coordinates, $\nabla_{\mu}E\nabla^{\mu}E = (F_{\infty})_{\mu}(F_{\infty})^{\mu}$ held true? While it isn't immediately physically obvious to me, 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.

 

[PeterDonis]

So what you are saying is that based on your calculations for kerr space-time in the usual coordinates, $\nabla_{\mu}E\nabla^{\mu}E = (F_{\infty})_{\mu}(F_{\infty})^{\mu}$ held true?

With a minus sign on the LHS, yes.

 

[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
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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