# Redshifting of forces in stationary space - times

1. Mar 18, 2013

### WannabeNewton

"Redshifting" of forces in stationary space - times

I'm trying to solve the following:

Let $(M,g_{ab})$ be a stationary spacetime with timelike killing field $\xi ^{a}$. Let $V^{2} = -\xi _{a}\xi ^{a}$ ($V$ is called the redshift factor).
(a) Show that the acceleration $a^{b} = u^{a}\triangledown _{a}u^{b}$ of a stationary observer is given by $a^{b} = \triangledown^{b}\ln V$.

(b) Suppose in addition that $(M,g_{ab})$ is asymptotically flat. Then, the "energy as measured at infinity" of a particle of mass $m$ and 4 - velocity $u^{a}$ is $E = - m\xi _{a}u^{a}$. Suppose a particle of mass $m$ is held stationary by a (massless) string, with the other end of the string being held by a stationary observer at infinity. Let $F$ denote the magnitude of the local force exerted by the string on on the particle. According to part (a), we have $F = mV^{-1}(\triangledown ^{a}V\triangledown _{a}V)^{1/2}$. Use conservation of energy arguments to show that the magnitude of the force exerted on the other end of the string by the observer at infinity, is $F_{\infty } = VF$.

So here's what I got:

(a) A stationary observer's 4 - velocity must be proportional to the time - like killing vector and it must be normalized to -1 so we find that for a stationary observer, $u^{a} = \frac{\xi ^{a}}{(-\xi ^{c}\xi _{c})^{1/2}}$. Now we compute, $\triangledown _{b}u^{a} = \frac{\triangledown _{b}\xi ^{a}}{(-\xi ^{c}\xi _{c})^{1/2}} + \frac{\xi ^{a}\xi ^{c}\triangledown _{b}\xi _{c}}{(-\xi ^{c}\xi _{c})^{3/2}}$ so $u^{b}\triangledown _{b}u^{a} = \frac{\xi^{b}\triangledown _{b}\xi ^{a}}{(-\xi ^{c}\xi _{c})} + \frac{\xi ^{a}\xi^{b}\xi ^{c}\triangledown _{b}\xi _{c}}{(-\xi ^{c}\xi _{c})^{5/2}} = -\frac{\xi^{b}\triangledown ^{a}\xi _{b}}{(-\xi ^{c}\xi _{c})}$ where the second term vanishes because it is a contraction of a symmetric tensor with an anti - symmetric one and I have swapped the indices in the first expression using killing's equation. Therefore, $a^{a} = u^{b}\triangledown _{b}u^{a} = \frac{1}{2}\frac{\triangledown ^{a}(-\xi^{b}\xi _{b})}{(-\xi ^{c}\xi _{c})} = \frac{1}{2}\triangledown ^{a}\ln V^{2} = \triangledown ^{a}\ln V$ as desired.

(b) This is where I'm totally stuck. As far as conservation of energy goes, we know that $E$, as defined above, is a conserved quantity along the worldline of the stationary particle; physically $E$ is the energy needed to bring in the particle from infinity to its orbit. Here we have a stationary observer at infinity holding this particle stationary by a long thread. There is a tension force at the end the observer holds and at the end the particle hangs by. Let's say the observer exerts the force $F_{\infty }$ at event $P_{1}$ and the particle feels the local force $F$ at event $P_{2}$. As far as I can tell, all we know is that in between these events, $E$ is constant. But how do I relate $E$ to $F_{\infty}$ and how do I do this using the conservation of energy explained above?

I should note that I tried something on a whim and looked at $\triangledown _{b}E$. We know that for the stationary particle hanging from the string, for which this total energy is attributed, $E = -m\xi _{a}u^{a} = -\frac{m\xi_{a}\xi^{a}}{(-\xi _{c}\xi^{c})^{1/2}} = m(-\xi^{c}\xi_{c})^{1/2}$ so if we compute the derivative we get $\triangledown _{b}E = mV\triangledown _{b}\ln V$ thus $(\triangledown ^{b}E\triangledown _{b}E)^{1/2} = mV(\triangledown ^{b}\ln V\triangledown _{b}\ln V)^{1/2} = VF$ but I have NO idea how this quantity is related to $F_{\infty }$, if at all. If it is related somehow I have no idea how one uses conservation of energy to arrive at the relation. Thanks in advance!

Last edited by a moderator: Mar 19, 2013
2. Mar 19, 2013

### WannabeNewton

Actually, I came up with an argument that didn't use $E$ at all so now I'm wondering why Wald even gave that piece of information.

Anyways, here's the argument: We know that the observer at infinity is doing work in keeping the particle stationary. The amount of work $dW_{\infty}$ done by the observer in an infinitesimal space - time displacement $ds$ of the particle (i.e. the proper length between the infinitesimally separated events, which for the stationary particle is just displacement in time) is given by $dW_{\infty} = F_{\infty}ds$. In the local frame of the particle, it only feels the local force $F$ for, of course, the same displacement so the work done locally would be $dW = Fds$ giving us the relationship $F_{\infty} = F\frac{dW_{\infty}}{dW}$. To get a relationship between the two work done quantities, note that if all of $dW$ were converted into electromagnetic radiation and received by the observer at infinity, the associated energy of the radiation would of course get redshifted i.e. $dE_{received} = VdW$ but by conservation of energy the energy so received must equal the original work put in by the observer so $VdW = dW_{\infty}$ giving us $F_{\infty} = VF$.

Moreover, I'm still wondering why the $\triangledown_{b}E$ method actually gave the right answer..., I still can't find a physical justification for it. Does anyone know why it worked or was it just a coincidence?

3. Mar 19, 2013

### pervect

Staff Emeritus
I don't know if this will help any, but the way I view the end result of that section is that the integral of the force-at-infinity normal to any closed 3-surface is the amount of mass (Komar mass) that the three surface encloses.

This is obviously related to Gauss' law, but also different, due to the use of "force-at-infinity" rather than force.

4. Mar 19, 2013

### WannabeNewton

Hi pervect, thanks for the response. Actually this exercise was before the chapter on the komar mass and asymptotic flatness (where he makes use of the force at infinity to talk about the komar mass in exactly the way you described). It was at the end of the chapter on the schwarzschild solution actually. I was just curious because as the aforementioned calculation showed $(\triangledown _{b}E\triangledown^{b}E)^{1/2} = VF$ for the energy $E$ of the particle hanging from the string, as measured by the observer at infinity. I'm just wracking my head trying to see if there actually is a connection between $F_{\infty}$, the force exerted by the observer at infinity on his end of the string, and $\triangledown _{b}E = \partial _{b}E$ or if this was merely a fluke / coincidence.

The argument I gave in the second paragraph above, using work, didn't use the form of $E$ given by Wald (i.e. $E = -\xi _{a}u^{a}$) at all so it seemed like a red herring or at the least superfluous information but I'm also very uncomfortable with the notion of work, even if infinitesimal work, in GR. How is it defined, even if infinitesimally, in general in GR? I just hand - waved it.

Thanks again!

5. Mar 19, 2013

### Ben Niehoff

I have a feeling this exact same problem gets assigned a lot, because I've also done it. Let me dig through my old GR homework and see if I can decipher my solution...

6. Mar 19, 2013

### Ben Niehoff

OK, part (a) is very easy, but it looks like you got it already...

As for part (b): I took the expression for the "energy at infinity", and used the principle of virtual work. It turns out

$$E_\infty = m V$$
and hence

$$(F_\infty)_\mu = -m \nabla_\mu V$$
My version of this problem had a third part, which was to find the surface gravity of a black hole (defined as the force-at-infinity of a unit test mass sitting at the horizon). This involves just plugging in the appropriate expression for V and evaluating it at $r=2m$.

7. Mar 19, 2013

### WannabeNewton

Thanks Ben, I'm very thankful you've done it before. I feel like I'm going to go insane if I don't sort this out :[

EDIT: I typed this LITERALLY as you typed in your most recent response so pretend this message doesn't exist until I finish sorting this out xD

8. Mar 19, 2013

### WannabeNewton

Ben, I always thought that even though $E = mV$ was the energy "as measured at infinity" it was still the total energy of the particle, in this case the particle hanging off the string. I'm still having trouble seeing how to use virtual work to show the total energy of the stationary particle hanging off one end of the string (the total energy being measured at infinity) is related to the force exerted by the observer on the other end of said string via $\triangledown _bE = (F_{\infty})_b$ (what's extra unsettling is that $\triangledown _b$ is a 4 - gradient). Any hint as to where to get that? Thanks again.

9. Mar 19, 2013

### Ben Niehoff

Doesn't matter if $\nabla_\mu$ is a 4-gradient, since V is constant along the Killing direction.

$E=mV$ is the potential energy (measured at infinity) that the particle has in relation to the black hole. It's the gravitational potential energy. If you imagine displacing the particle by a small amount, $E$ will change. Therefore there must be a force holding the particle in place, and the "force at infinity" must be minus the gradient of the "energy at infinity".

You might find it odd that we're measuring potential energy at infinity when the particle is not at infinity. We can get away with this because we have a timelike Killing vector and spacetime is asymptotically flat. So there is a sensible notion of what it means to suspend a particle on a string from infinity.

10. Mar 19, 2013

### WannabeNewton

The reason I was hesitant to conclude $E$ would be the potential energy, as measured at infinity, of the dangling particle is because in general for asymptotic static space - times in which such an energy is definable, we say $E$ is the TOTAL energy which constitutes gravitational potential energy and energies that arise due to stress / pressure etc. Why, in this case for the stationary particle, can we claim $E$ must be just the potential energy?

11. Mar 19, 2013

### Ben Niehoff

Where is the stress, pressure, etc.? There is only a single particle.

12. Mar 19, 2013

### WannabeNewton

Ok but I'm just having trouble accepting it because we don't have an explicit form for the total energy in terms of the constituent energies and this is an arbitrary stationary space - time so I don't really know what to expect in terms of the different energies that could be attributed to a single particle hanging stationary in an arbitrary stationary space - time. For example we know that a stationary particle in a stationary axisymmetric space - time starts gaining coordinate angular velocity due to frame dragging. Won't this contribute to its total energy on top of the grav. potential energy?

Also, I'm also confused as to why the gradient of the potential energy due to the gravitational field is in any way related to the force applied by the observer at infinity; why should this force be derived from the total energy (or grav. potential energy) of the particle? I think I'm misunderstanding what the "energy as measured at infinity" of said particle actually is and how virtual work is applied to the stationary particle. Sorry for the questions Ben, thanks again.

Last edited: Mar 19, 2013
13. Mar 30, 2013

### Staff: Mentor

You don't need it because it's assumed that you know the Killing vector field, and the Killing vector field already contains all the information about how "local" energy varies relative to energy at infinity; that's what the "redshift factor" V represents. That seems to me to be the point of the problem.

Not if you hold it steady with a rope. In this case the local force felt by the rope would not be purely radial, but it would still be perfectly possible to hold an object steady at a constant energy at infinity. As far as I can see, this would also require holding the object at zero angular momentum, which may help with how you are visualizing the axisymmetric case (i.e., the spacetime is stationary but not static); see further comments below.

Because the object, by hypothesis, has no energy other than gravitational potential energy. Here's why:

By hypothesis, the object is following an orbit of the timelike KVF. This means, as far as I can see, that the object can't have any angular momentum; if it did, it would not be following an orbit of the timelike KVF. (Note that in the stationary axisymmetric case, these orbits have a nonzero "angular" component, with reference to asymptotic "non-rotating" observers at infinity; but they still have zero angular momentum. I believe there is a discussion of these "zero angular momentum observers" in Kerr spacetime in Carroll's online lecture notes.)

Since the object has zero angular momentum, and since it is also not moving radially (obviously, since it's following an orbit of the timelike KVF and those have no radial component), it has no energy other than gravitational potential energy, because it is not moving and so has no kinetic energy. (A better way to say this would be that the timelike KVF defines what "not moving" means.) Since the timelike KVF is an invariant feature of the spacetime, this sense of "not moving" is well-defined; and so the statement that the object has no energy other than gravitational potential energy is also well-defined.

14. Mar 30, 2013

### WannabeNewton

So basically, as long as we can make the point particle have zero translational (radial) kinetic energy and zero rotational kinetic energy (i.e. zero angular momentum) then we can be sure the energy should be purely potential (this also relates to your later passage below). As for keeping it with zero angular momentum, this is indeed achieved immediately if the object is following orbits of the time - like killing field which I will refer to below.

I agree that if it is following an orbit of the time - like KVF that it cannot have any angular momentum. I actually talked about this in a different thread. Here was the relevant post: https://www.physicsforums.com/showpost.php?p=4325477&postcount=67 (also a problem from Wald actually).

I don't disagree with any of this and I can agree that the particle has no energy other than the gravitational potential (could you elaborate on what you mean by well - defined? I am aware that the concept is not always well defined but under what circumstances it it and why?). In this case we can conclude that the total energy as measured by the observer at infinity, $E = -m\xi _{a}u^{a}$, of the point particle being held at the other end of the string is simply the gravitational potential energy of the particle.

However, why is the gradient (or negative gradient) of this gravitational potential energy related to any kind of force as per GR and moreover why is it related to the force exerted by the observer at infinity on his end of the string?

Thanks Peter.

15. Mar 30, 2013

### Staff: Mentor

Perhaps "invariant" would have been a better word. The point I was trying to make is that you can only construct an invariant notion of "not moving" that works globally in a spacetime with a timelike KVF.

In any stationary spacetime, i.e., any spacetime with a timelike KVF.

Suppose we take an object of rest mass m and lower it from infinity to a point with a "redshift factor" V. Then, once the object is lowered, we convert its entire rest mass into radiation and send the radiation back out to infinity. We expect that when the radiation reaches infinity it will deliver energy $Vm$, because of gravitational redshift. This is one way of stating what the "energy at infinity" of the object means. (By conservation of energy, this also tells us that we can extract total work equal to $(1 - V) m$ by lowering the object.)

Now, what does the *gradient* of the energy at infinity mean? Obviously it means how much the energy at infinity *changes* if we lower the object by a very small additional amount (before converting its rest mass into radiation and sending the radiation back out to infinity). But this change in energy must go somewhere. Where does it go? It gets transmitted up the rope and extracted as work at the top. The work extracted will be equal to the force exerted at the top, i.e., "at infinity", times the distance traveled; but this is equivalent to saying that the force at infinity is minus the gradient of energy at infinity. (It's minus the gradient because the sign of the work extracted at infinity is opposite to the sign of the change in energy at infinity of the object. Compare with the formula $(1 - V) m$ above for the total work extracted.)

16. Mar 30, 2013

### WannabeNewton

Sorry Peter when I was talking about being well - defined I meant the notion of gravitational potential energy, not being stationary (that I am fine with). Wald points out that the notion of gravitational potential energy is not always well defined which is why I was being cautious.

Yeah this is the argument I gave in post #2. I was content with it but it still intrigued me to look for something more mathematical.

But how is work actually defined in GR? For example if we do indeed lower the particle by a small amount via the string then the only invariant measure of displacement is the space - time displacement and it of course won't be simply a spatial displacement because in physically lowering the particle via the string, between two events, there is some temporal displacement. So how is work actually defined in GR considering the classical definition?

Now back to the point at hand. Let's say we lower the string radially (and hence the particle) by some infinitesimal spatial amount $\delta r$. Of course $\triangledown _{r}E$ will give us the rate of change in the energy as measured at infinity of the particle per unit length as the particle is being lowered so the total change, as measured at infinity, over the infinitesimal displacement should be $\delta r\triangledown _{r}E$. Now let's say we convert this change in energy (as measured at infinity) into radiation and extract it back at infinity. By conservation of energy this should exactly equal negative the work done by the observer at infinity in radially lowering the string, call it $\delta W$. Using the classical definition of work, this should just be the force exerted by the observer at infinity, in radially lowering the string, times the displacement i.e. $\delta W = (F_{\infty})_{r}\delta r$ so the aforementioned equality argument based on conservation of energy tells us that $(F_{\infty})_{r}\delta r = -\delta r\triangledown _{r}E\Rightarrow -\triangledown _{r}E = (F_{\infty})_{r}$ (and of course the other components would be trivially equal since they will all vanish in the case of lowering the string / particle radially so we may as well say $(F_{\infty})_{b} = -\triangledown _{b}E$).

Is this a valid argument? I am not sure how one would derive this equality in full generality, using just this radiation argument if the force was not completely radial (i.e. the string was not lowered just radially) because of the usual dot product in the definition of work; two equal dot products doesn't necessarily imply the arguments are equal in all components. AFAIK there is no operational definition of components of force in terms of gradient of some kind of energy in GR, in general, in the sense of Newtonian mechanics.

If it is then I'm confused as to why the force exerted by the observer at infinity in lowering the string / particle should equal the force exerted by the observer in keeping the particle stationary (which is what the problem wants)?

Last edited: Mar 31, 2013
17. Mar 31, 2013

### Staff: Mentor

Yes, and what he means is that it is not well-defined in a spacetime that is not stationary. See further comments below.

As force times displacement, just as you would expect. (More precisely, as the inner product of the force 4-vector and the displacement 4-vector.) See further comments below.

This is true, but it doesn't matter, because the force 4-vector has no timelike component. More precisely, it has no component which is parallel to the timelike KVF, so when we take the inner product of force and spacetime displacement, only the spacelike terms in the displacement come into play. The reason the force has no timelike component is that such a component would change the rest mass of the object to which the force is applied, and that's not happening here. (It does happen in the case of non-conservative forces like friction, but in such cases you can't express the force as the gradient of a potential anyway, even in Newtonian mechanics.)

You can't do that; you already extracted that change in energy at infinity during the lowering process. That's what the change in energy at infinity *is*: it's the work you extract as you lower the object; i.e., as I said in my previous post, it's $(1 - V)m$ for an object of rest mass m lowered to a point where the redshift factor is V. The energy at infinity that is left to be converted into radiation is the energy at infinity that you *haven't* extracted yet, i.e., $Vm$.

This is true if you take out the part about turning the difference in energy at infinity into radiation; see above.

I'm not sure I see why the dot product is an issue; computing dot products is easy as long as you have an expression for the metric in whatever coordinate chart you're using. The key requirement for the conservation of energy argument to work is that the spacetime is stationary; see below.

You're right, there isn't, because Newtonian mechanics assumes that you can always define a potential for the force to be the gradient of, and in GR you can't always do that. You can only do it in the case of a stationary spacetime. Basically, Newtonian mechanics assumes "absolute space", so that a potential energy, which requires an absolute sense of "position", makes sense. In general there is no absolute sense of "position" in GR, but if a spacetime is stationary, the spacelike hypersurfaces orthogonal to the timelike KVF are all identical, so they serve as an "absolute space" in the necessary sense, and you can use "position relative to the timelike KVF" to define a potential energy.

For a small enough displacement they're the same because the redshift factor doesn't change measurably. If you look at the whole process of lowering the object from infinity to some redshift factor V, then during the process the force exerted at infinity has to change, yes. But you can make the conservation of energy argument using only infinitesimal displacements.

18. Mar 31, 2013

### WannabeNewton

I think I can argue it mathematically as follows. We know that $F_{b} = m\nabla _{b}\ln V = \frac{1}{2}m\nabla_{b}\ln V^{2} = \frac{1}{2}mV^{-2}\nabla_{b}V^{2}$. Consider the covariant expression $F_{a}\xi ^{a} = \frac{1}{2}mV^{-2}\xi ^{a}\nabla _{a}V^{-2}$ (here $\xi^{a}$ is the time - like KVF). If we can show this is zero in one coordinate system then it must be zero in all coordinate systems.
We know there always exist coordinates $(t,x^{1},x^{2},x^{3})$ such that $\xi ^{a} = (\frac{\partial }{\partial t})^{a},\partial _{t}g_{ab} = 0$. In particular, in the coordinate basis we just have that $\xi ^{\mu } = \delta ^{\mu }_{t}$. Evaluating our expression from before, this gives us $F_{\mu }\xi ^{\mu } = \frac{1}{2}mV^{-2}\xi ^{\mu}\nabla_{\mu}V^{2} = -\frac{1}{2}mV^{-2}\partial _{t}g_{tt} = 0$ hence $F_{a}\xi ^{a} = 0$ for all coordinate systems, which of course implies $(F_{\infty})_{a}\xi ^{a} = 0$. This codifies what you said about the force at infinity being completely orthogonal to the time - like KVF for any coordinate system. Note also that for the special case of the coordinate system adapted to the time - like KVF used in the computation above, $F_{t} = 0$. Seems to work, would you agree with this?

So by definition, the change in energy as measured at infinity IS the work extracted in lowering the string? I guess that makes sense if I think about it: we are measuring the change in energy *at infinity* but to actually measure this change we would have to first convert the change in energy into radiation and extract it back at infinity to measure it *at infinity* in the first place; the work done by the observer would be the negative of this since we do negative work in lowering a particle in a gravitational field but the work we extract via the radiation will of course be a positive addition. Is that correct to say?

My point was that if we use the conservation of energy argument for the infinitesimal displacement then we end up with something along the lines of $g_{ab}((F_{\infty}))^{a}(\delta s)^{b} = -g_{ab}(\triangledown^{a}E) (\delta s^{b})$ but why would this necessarily imply $(F_{\infty})^{a}=-\triangledown^{a}E$?

I'll respond to this after everything else is cleared up because I don't want to bombard you with too many questions at a single time.

So in summary we are saying the force used to lower the particle $F_{\infty}$ is being held constant and equal to the force used to hold the particle stationary since the lowering is only happening over an infinitesimal displacement; we don't need to change the force because $V$ is essentially constant over the infinitesimal displacement. In this sense this argument only works for infinitesimal displacements.

Last edited: Mar 31, 2013
19. Mar 31, 2013

### Staff: Mentor

This is all fine, but note that at the outset you assumed $F^b = m \nabla^b ln V$, which assumes that $m$ is constant. That's really the key assumption; the rest is just computing the gory details of how a timelike KVF works.

If you're lowering the string, yes. The more general definition is that if the object undergoes *any* process which starts with it at rest at infinity and ends with it at rest at some redshift factor V, the change in the object's energy at infinity is equal to the total energy that can be captured at infinity during the process, regardless of how it is captured.

For example, suppose we try a different process: we drop the object in free fall from infinity to some redshift factor V, and at that redshift factor we convert all of its kinetic energy into radiation, bringing it to rest ("rest" of course means "rest with respect to the timelike KVF"), and send the radiation back out to infinity, and capture it in a collector that stores it in a huge battery bank. The energy added to the battery bank will be equal to the change in the object's energy at infinity; we could verify that by attaching a rope to the object and using the energy stored in the battery bank to drive a winch that pulls the object back up to infinity; in other words, the inverse of lowering the object by a rope. The energy in the battery bank will be just enough to complete the pulling up process, with none left over. (This is all assuming no friction or other losses, of course.)

(It looks like the above is more or less what you are getting at with your next remarks, but perhaps the above will help to clarify things further.)

Note also that the above implies that, if we drop the object in free fall, its energy at infinity is *constant*; it doesn't change until we do something to capture its kinetic energy. This is the original reason that the term "energy at infinity" was invented, btw; it captures a key constant of free-fall motion.

Because you can "divide out" $g_{ab}$ and $(\delta s)^b$ on both sides of the equation, since they're the same on both sides.

Which is all you need to show that the force is the negative of the gradient of potential energy, since the gradient is defined using infinitesimal displacements. The total work done during the entire lowering process would require integrating over a range of small displacements, which would cover a range of redshift factors, of course; but you don't need that to show the relationship between the local force and the force at infinity at a single redshift factor.

20. Mar 31, 2013

### WannabeNewton

Cool. And don't get me wrong I was perfectly content with the physical argument you gave I just wanted to see for fun if it could be shown using a little tensor calculus :D.

Makes perfect sense, thanks Peter and the work done in pulling it back up would be negative the change in energy as the change in energy was stored exactly in the battery bank and the work done would simply deplete all this energy in essence yes?

Much better put by yourself however, if I do say so myself.

Unless I am missing something, I have to disagree here Peter. The equality is a sum over the indices so how can we just divide out the two recurring terms $g_{ab}$ and $(\delta s)^{b}$ as they are being summed over their respective indices? IF the force used to hold and lower the string was purely radial then the only surviving term of the summation would be $g_{rr}(F_{\infty}^{r})(\delta s)^{r}$ and since the rate of change of energy is only along the radial direction the other side of the equality would only have the surviving term $-g_{rr}(\triangledown ^{r}E)(\delta s)^{r}$ and in this case we can just divide out the two recurring terms to get $F_{\infty}^{r} = -\triangledown ^{r}E$ but this is again the example I gave in one of the previous posts. If the force was not purely radial then I can't see how we can just "divide" out the terms in the summation like that.

Sure, I agree with all of this

21. Mar 31, 2013

### Staff: Mentor

[Edit: I see what you're saying, I'll have to think about this some more.]

22. Mar 31, 2013

But all $h_{a}F^{a} = -h_{a}\triangledown ^{a}E$ tells us is that $h_{t}F^{t} + h_{1}F^{1} + h_{2}F^{2} + h_{3}F^{3} = -(h_{t}\triangledown ^{t}E + h_{1}\triangledown ^{1}E + h_{2}\triangledown ^{2}E + h_{3}\triangledown ^{3}E)$. If instead we had say $h_{b}F^{a} = -h_{b}\triangledown ^{a}E$ for all choices of the index $b$ (so no summation over a common index) then I would totally and completely agree with you that we could just say $F^{a} = -\triangledown ^{a}E#3 but since it IS a sum I am still not getting why we can just equate each component separately. 23. Mar 31, 2013 ### WannabeNewton I just saw this after I posted my latest reply so you can ignore that reply for now. I'll try to think about it too, thank you so much thus far! 24. Mar 31, 2013 ### PeterDonis ### Staff: Mentor I think the answer is that $g_{ab} F^a ( \delta s )^b = g_{ab} ( - \nabla^a E ) ( \delta s )^b$ is correct, but because the spacetime is stationary, we can always adopt a coordinate chart in which the equation only has one component. (And we can only define $E$, the energy at infinity, in a stationary, asymptotically flat spacetime, so there is no real loss of generality by assuming such a coordinate chart.) In the chart we adopted, $\nabla^a E$ has only one component, and $g_{ab}$ is diagonal, so the whole equation collapses to just one component. But you could adopt another coordinate chart in which $\nabla^a E$ had more than one component; if you do that, then you need to write the equation in the more general form. 25. Mar 31, 2013 ### WannabeNewton I'm trying to see why this would be true Peter but I can't seem to see it. nor can I see why this would be true. As far as I can see, the most natural coordinate chart to choose would be the one adapted to the killing vector field i.e. the one where$\xi ^{a} = (\frac{\partial }{\partial t})^{a}$and$\partial _{t}g_{ab} = 0$. In these coordinates,$F_{t} = mV^{-1}\nabla _{t}V = -mV^{-1}\partial _{t}g_{tt} = 0$implying$(F_{\infty})_{t} = 0$and$\nabla _{t}E = m\partial _{t}V = 0$so that$(F_{\infty})_{\mu}(\delta s)^{\mu} = (F_{\infty})_{i}(\delta s)^{i} = -(\nabla _{i}E)(\delta s)^{i}$, where the$i##'s run over the spatial components in these coordinates.

However, I cannot see another choice of coordinates for which $g_{ab} F^a ( \delta s )^b = g_{ab} ( - \nabla^a E ) ( \delta s )^b$ would reduce to a single component. How do we know such coordinates exist? Thanks Peter.