Energy in the Schwarzschild spacetime

[pervect]

I had a thought that I wanted to share in another thread, but it wandered way off track and quite properly was closed. But I thought the separate idea that I had spawned from the old thread was worthy of posting in a new thread. I do not want to re-open the old thread, though!

In flat space-time, the relation E=mc^2 is a special case of the more general relation

$$E^2 - (pc)^2 = (m\,c^2)^2$$

i.e. when p=0, $E=mc^2$. Here E is the energy, p is the momentum, c is the speed of light, and m is the invariant mass, which happens to be equal to the relativistic mass when p=0.

It will be convenient to re-write this so that everything is dimensionless by dividing both sided by $m^2 c^4$. This gives.

$$\left(\frac{E}{m c^2} \right)^2 - \left(\frac{p}{mc} \right)^2 = 1$$

There's a similar relationship in the curved space-time of the Schwarzschild geometry, which represents a single, massive, non-rotating gravitating body. In the same dimensionless form, and for purely radial motion, this is:

$$\left(\frac{E}{mc^2}\right)^2 - \left(\frac{dr}{d\tau} / c \right)^2 = 1-\frac{r_s}{r}$$

Here E is what's knowing as the "Energy at inifinity", a conserved constant for an object free-falling (i.e following a geodesic) in the Scwarzschild spacetime. $r_s$ is the Schwarzschild radius, and r is the Schwarzschild radial coordinate. This comes from "orbits in strongly curved spacetime", https://www.fourmilab.ch/gravitation/orbits/, which is basically a publically available summary of some results from MTW"s textbook "Gravitation". I've converted the original notation to replace $\tilde{E}$ with E/m, and made other changes as needed to deal with the issue of the original source using geometric units. I've also set the angular momentum L to zero.

If we set $dr/d\tau = 0$, we see that we now have

$$\left(\frac{E}{mc^2} \right)= 1-\frac{r_s}{r}$$

or

$$ E = \sqrt{1-\frac{r_s}{r}} \, m c^2$$

Note that as r approaches r_s, E , the "energy at infinity", approaches zero.

For the reasons stated, this could be regarded as a sort of curved space-time equivalent to "E=mc^2".

 

[jartsa]

In flat space-time the kinetic energy of an object is
$$E_{kin}=E-mc^2 $$An object dropped from infinity has kinetic energy
$$ E_{kin} = E - \sqrt{1-\frac{r_s}{r}} \, m c^2$$

Those two equations above look similar.At the event horizon an object dropped from infinity has kinetic energy
$$ E_{kin} = E $$

If a light pulse and an object, both far from event horizon, both have energy $E$ then both have kinetic energy $E$ at the event horizon, after freefalling there.

Somehow I feel that the light pulse and the object must have the same momentum at the event horizon.

 

[mitochan]

dr/d$\tau$=0 means no motion or $E_{kin}=0$.
$$E=mc^2\sqrt{1-\frac{r_s}{r}}$$
is interpreted as sum of rest energy and minus gravitational energy in Newtonian mechanics
$$mc^2-G\frac{Mm}{r}=mc^2(1-\frac{GM}{rc^2})$$ where M is mass at r=0. $$r_s=\frac{2GM}{c^2}$$ meets it.

 

[Ibix]

$$\left(\frac{E}{mc^2} \right)= 1-\frac{r}{r_s}$$
$$ E = \sqrt{1-\frac{r}{r_s}} \, m c^2$$

I believe you meant $r_s/r$ in both of these expressions.

 

[Ibix]

An object dropped from infinity has kinetic energy

Kinetic energy as measured by who?

At the event horizon an object dropped from infinity has kinetic energy

Again, kinetic energy as measured by who? Note that it doesn't make sense for an observer at infinity to discuss the kinetic energy of an object crossing the horizon, nor are there local hovering observers to measure the energy, so the obvious guesses don't work here.

Somehow I feel that the light pulse and the object must have the same momentum at the event horizon.

The light pulse has a null four-momentum and the massive object does not. If they have the same energies measured at a given event then they cannot have the same momenta since $E=p$ in one case and $E\neq p$ in the other.

 

[Ibix]

dr/d$\tau$=0 means no motion or $E_{kin}=0$.
$$E=mc^2\sqrt{1-\frac{r_s}{r}}$$
is interpreted as sum of rest energy and minus gravitational energy in Newtonian mechanics
$$mc^2-G\frac{Mm}{r}=mc^2(1-\frac{GM}{rc^2})$$ where M is mass at r=0. $$r_s=\frac{2GM}{c^2}$$ meets it.

I'd say this would be clearer written as$$\begin{eqnarray*}
E&=&mc^2\sqrt{1-\frac{r_s}{r}}\\
&\approx&mc^2\left(1-\frac{r_s}{2r}\right)\\
&=&mc^2-\frac{GMm}r
\end{eqnarray*}$$So what this is saying is that for an object of mass $m$ at rest at radial coordinate $r$, in order to reach infinity it needs to be given energy $GMm/r$ or, more precisely, $mc^2\left(\sqrt{1-\frac{r_s}r}-1\right)$ so that its energy when it reaches infinity (which is what $E$ is) is $mc^2$.

 

[PeterDonis]

An object dropped from infinity has kinetic energy...

Kinetic energy relative to a static observer.

At the event horizon an object dropped from infinity has kinetic energy...

No, because there are no static observers at the event horizon.

Somehow I feel that the light pulse and the object must have the same momentum at the event horizon.

Your feeling is based on a misconception. See above.

Also, "momentum at the event horizon" makes no sense.

 

[Ibix]

Kinetic energy relative to a static observer.

To be clear, I think that what @jartsa wrote is the kinetic energy relative to a static observer as redshifted by being measured by an observer at infinity.

A free-falling object has four momentum with t-component $P^t=E/(1−r_s/r)$ in Schwarzschild coordinates (in this case I'm interpreting $E$ as being an energy, not an energy per unit mass). A hovering object has four velocity with components $V^t=1/\sqrt{1−r_s/r}$ and $V^r=V^\theta=V^\phi=0$. I think that makes the total energy of the falling object as measured by the static observer $E_{\mathrm{local}}=g_{tt}P^tV^t=E/\sqrt{1−r_s/r}$, and hence the locally measured kinetic energy $E/\sqrt{1−r_s/r}−mc^2$. You need to multiply by the gravitational redshift, $\sqrt{1-r_s/r}$, to get the original expression. Or am I misunderstanding something?

 

[PeterDonis]

To be clear, I think that what @jartsa wrote is the kinetic energy relative to a static observer as redshifted by being measured by an observer at infinity.

Yes, you're right. Or you can think of it as the work that can in principle be extracted at infinity by slowly lowering the object from infinity to rest at a finite radial coordinate $r$ (as long as $r$ is above the horizon).

 

[jartsa]

Kinetic energy as measured by who?

All observers say that the energy at infinity of a motionless object at radius $2r_s$ is:
$mc^2* \sqrt{1/2}$. Right?

And all observers say that a free falling object always has the same energy at infinity:
$E=mc^2$

So, the energy at infinity difference between an object motionless at $2r_s$ and an identical object dropped from infinity to $2r_s$ is:
$mc^2* \sqrt{1/2} - mc^2$

So I thought that energy at infinity difference is caused by a difference of kinetic energy at infinity.Or how about this explanation:

I drop a potato into a black hole, at the event horizon the rest energy at infinity of the potato is zero - see post #1. So the energy at infinity of the potato is 100% kinetic energy at the event horizon. So the kinetic energy of the potato is $mc^2$ at the event horizon.

 

[PeterDonis]

I drop a potato into a black hole, at the event horizon the rest energy at infinity of the potato is zero

"Rest energy at infinity of the potato" is meaningless nonsense.

A correct statement is that the energy at infinity of a potato dropped from $r = 2 r_s$ will be larger than the energy at infinity of an object at rest at some $r$ close to $r_s$ that the potato passes as it falls.

I thought that energy at infinity difference is caused by a difference of kinetic energy at infinity.

"Kinetic energy at infinity" is meaningless nonsense for any object that is not at infinity.

"Energy at infinity" (without the "kinetic" qualifier or the "rest" qualifier) has a well-defined technical meaning, but you can't just wave your hands and add qualifiers to that term and expect it to still be meaningful.

 

[Ibix]

All observers say that the energy at infinity of a motionless object at radius $2r_s$ is:
$mc^2* \sqrt{1/2}$. Right?

Yes

And all observers say that a free falling object always has the same energy at infinity:
$E=mc^2$

An object that was initially at rest at infinity, yes.

So, the energy at infinity difference between an object motionless at $2r_s$ and an identical object dropped from infinity to $2r_s$ is:
$mc^2* \sqrt{1/2} - mc^2$

Yes, and that means what I said in the part of my post that Peter quoted, or Peter's alternative description. But you didn't specify that you were talking about the energy difference at infinity - you just said "the energy", but energy is not an invariant quantity. Energy at infinity is invariant, though. So you've now specified who is measuring the energy and that's fine.

I drop a potato into a black hole, at the event horizon the rest energy at infinity of the potato is zero - see post #1. So the energy at infinity of the potato is 100% kinetic energy at the event horizon. So the kinetic energy of the potato is $mc^2$ at the event horizon.

No. The potential/kinetic energy division only makes sense where there's a way to define "not moving" (or, more formally, where there exists a timelike Killing vector). You can't define "not moving" when you are "at" the horizon because the horizon isn't a place in the relevant sense of the word and you can't be at rest "there".

If you said that as your potato dropped from rest at infinity approaches the horizon its kinetic energy as measured at infinity approaches $mc^2$, that would be correct. It's essentially a restatement of Peter's #9.

 

[mitochan]

So what this is saying is that for an object of mass at rest at radial coordinate , in order to reach infinity it needs to be given energy or, more precisely, so that its energy when it reaches infinity (which is what is) is .

As for the questions of "Given energy to what ? Who holds E ?", if the answer is "the object", it seems that the object of mass m would hold gravitational energy between itself and the centered object of mass M, no sharing.

 

[PeterDonis]

For the questions of "Given energy to what ? Who holds E ?", the answer seems to be "the object".

You can take this view if you take "energy" to mean "energy at infinity". That is unproblematically a property of the object--but it is only well-defined in stationary spacetimes.

You can also, with some caveats, split up energy at infinity into "kinetic" and "gravitational potential" energy; the latter is then best viewed, not as localized in the object itself, but as a property of the object and the gravitating mass $M$ as a system.

The real issue comes when you have more than one object that is a source of gravity in the system--for example, if you try to model "assembling" a planet or star from a lot of small pieces of matter. For cases like that, there is no way to localize the "gravitational potential energy" that makes a (negative) contribution to the mass of the planet or star after it is assembled (the contribution is negative because the mass after assembly is less than the total mass of the small pieces of matter the planet or star was assembled from).

Also, once you are dealing with spacetimes that are not stationary, all of the above goes out the window and there is no good notion of "energy" except the locally conserved stress-energy tensor.

 

[pervect]

I believe you meant $r_s/r$ in both of these expressions.

Yeah, let me fix that.

 

[mitochan]

Say the object fall from the position where $r>>r_s$ so
$$E=mc^2\sqrt{1-\frac{r_s}{r}} \approx mc^2$$
to the position $r = r_s$ where $E=0$, not freely but in quasi-static process with extracting (kinetic) energy continuously. it would allow us to transform all the rest energy $mc^2$ ,that is far away from the gravity source, to usable one.

 

[vanhees71]

The question is what you call "energy". Since the Schwarzschild spacetime is static, there's a conserved quantity for the motion of a test particle due to time-translation invariance of the action, and that quantity you can call "energy" in honour of Noether.

 

[jartsa]

Say the object fall from the position where $r>>r_s$ so
$$E=mc^2\sqrt{1-\frac{r_s}{r}} \approx mc^2$$
to the position $r = r_s$ where $E=0$, not freely but in quasi-static process with extracting (kinetic) energy continuously. it would allow us to transform all the rest energy $mc^2$ ,that is far away from the gravity source, to usable one.

Yes, like for example a vehicle with regenerative braking descends carrying one potato, at the lowest point the potato is unloaded and the vehicle starts climbing up, using the thousands of cubic meters of matter-anti matter fuel that was generated during the descend.

When the vehicle is back at the starting point there is 0.1 liters of fuel left.

(I cleverly used volume to tell the amount of fuel, so that I don't have to deal with such things as mass at infinity)

 

[PeterDonis]

a vehicle with regenerative braking descends carrying one potato, at the lowest point the potato is unloaded and the vehicle starts climbing up, using the thousands of cubic meters of matter-anti matter fuel that was generated during the descend

The potato is superfluous in your example. If your regenerative braking scheme works at all, it works just fine with just the vehicle itself: as it descends, it transfers energy at infinity to the battery, or whatever your regenerative braking scheme is using to store energy, and then as it ascends, it transfers energy at infinity back out. There is no need to unload anything at the bottom, because by hypothesis nothing has any energy at infinity at the bottom anyway, except the battery. All of the energy at infinity, which at the bottom is, by hypothesis, the full rest energy of the vehicle, is stored in the battery at the bottom.

Your remark that there will be 0.1 liters of fuel left at the top illustrates this last point: you could have saved that fuel by not carrying the potato at all, and not having to unload anything at the bottom.

Also, your remark about "thousands of cubic meters" of fuel collected is a vast understatement. At the bottom, the energy at infinity of the vehicle (equal to the vehicle's rest energy, or close to it) is blueshifted by an amount that approaches infinity as the bottom point approaches the horizon. So the fuel storage required will also approach infinity.

 

[pervect]

As is often the case, the coordinate singularity of the Schwarzschild coordinates can make simplistic analysis confusing.

There aren't any particular pathologies in the equations I wrote - if we consider an object falling from a large height (infinity), we can use the fact that $dr/d\tau$ = 0 at infinity to say that it's constant energy-at-infinty is equal to E=mc^2 as it falls. We also can say that $dr/d\tau$ starts out as zero at infinty, and remains finite, decreasing to a value of -c (as $dr/d\tau$ is negative for an infalling object) when r=$r_s$.

Where things start to go wonky, to use the not-so-technical term, is if one tries to interpreting the math in terms of a static observer. The basic issue is static observers simply don't exist at the event horizon.

There are various other simplifications that have been made along the route, some of which have been commented on by other posters, but it's the best I can do at I-level.

Trying to split the total energy into a kinetic part and a potential part is not very useful, and is usually not done in the textbooks. It's also inhernetly frame dependent. However, as long as one chooses a frame that actually exists (which excludes a stationary observer at the event horizon), it can technically be done, I suppose.

 

[PeterDonis]

Trying to split the total energy into a kinetic part and a potential part is not very useful, and is usually not done in the textbooks. It's also inhernetly frame dependent.

While it's true that this split is not often done in the textbooks, it can be done in a way that is not frame-dependent, using the timelike Killing vector field of the spacetime. In other words, the kinetic part is kinetic energy relative to stationary observers--the ones whose worldlines are orbits of the timelike KVF. Which observers those are is not frame-dependent.

 

[jartsa]

There is no need to unload anything at the bottom, because by hypothesis nothing has any energy at infinity at the bottom anyway

Unloading a zero mass_at_infinity potato buys us the future absence of a non-zero mass_at_infinity potato. I mean unloading the load saves fuel, not immediately, but at later time.

I think I got it right. A potato disappeared, some matter - anti-matter fuel appeared at the place where the potato was. No conservation laws were broken.

 

[PeterDonis]

A potato disappeared, some matter - anti-matter fuel appeared at the place where the potato was

In your potato scenario, that's the net effect of the whole operation, comparing the starting and ending states at infinity, yes. But if you have some "regenerative braking" scheme that can convert the captured energy into matter-antimatter fuel, you could just use it at infinity to convert mechanical energy into matter-antimatter fuel. You don't need to have a black hole handy.

 

[mitochan]

Say the object fall from the position where so

to the position where , not freely but in quasi-static process with extracting (kinetic) energy continuously. it would allow us to transform all the rest energy ,that is far away from the gravity source, to usable one.

Say we would like to use thus generated energy at the original position far away from the gravity source, we should carry energy by lifting it up against gravity force so we have to consume part of the energy. It would claim to the statement that all the rest energy is transformed to usable one.

 

[jartsa]

Say we would like to use thus generated energy at the original position far away from the gravity source, we should carry energy by lifting it up against gravity force so we have to consume part of the energy. It would claim to the statement that all the rest energy is transformed to usable one.

If you have one Joule at infinity hanging at low altitude, and you use 100 Joules at infinity to winch it up, then the 1 Joule at infinity transforms to 101 Joules at infinity.

Here are two alternative ways to phrase the above thing:

1. Vague and understandable: Energy gains energy when you lift it.
2. Less vague and less understandable: Energy at infinity gains energy at infinity when you lift it.

 

[PeterDonis]

Say the object fall from the position where $r>>r_s$ so
$$E=mc^2\sqrt{1-\frac{r_s}{r}} \approx mc^2$$
to the position $r = r_s$ where $E=0$, not freely but in quasi-static process with extracting (kinetic) energy continuously. it would allow us to transform all the rest energy $mc^2$ ,that is far away from the gravity source, to usable one.

Yes, because you are not allowing the object to fall freely, but are lowering it very slowly, so its energy at infinity continuously decreases during the process, and the extracted energy at infinity is captured at infinity and is usable there.

Say we would like to use thus generated energy at the original position far away from the gravity source, we should carry energy by lifting it up against gravity force so we have to consume part of the energy. It would claim to the statement that all the rest energy is transformed to usable one.

No, you don't have to lift the extracted energy up. You extracted it at infinity and can use it there.

 

[PeterDonis]

If you have one Joule at infinity hanging at low altitude, and you use 100 Joules at infinity to winch it up, then the 1 Joule at infinity transforms to 101 Joules at infinity.

This is irrelevant to the scenario @mitochan is describing. He's talking about taking 100 Joules at infinity (or whatever number you want to use) and winching it down, extracting energy from the winching process. Nothing ever has to get winched back up.

 

[pervect]

A somewhat related observation is that if you suspend an object in a static gravitational field by a (massless) string, the tension in the string varies with height.

This is discussed in Wald as a way to motivate a further discussion of energy in GR.

Thus, the local force, which must be exerted on a unit test mass to hold it in place is given by (equation snipped). However, if we choose to calculate the force which must be applied by a distant observer at infinity (e.g. by means of a long string), we find that the force differes from the local force by a factor of V.

V here is the "redshift" factor, formally the "length" of the Killing vector associated with the static gravitational field. This works out to be from memory $\sqrt{|g_{00}|}$ in the usual coordinates. The remarks can be extended to stationary fields if one wishes, though the discussion is clearer with a static field, such as the Schwarzschild metric.

The releavant derivation in Wald is left as an exercise, though I have a few comments on how I approached said exercise if anyone wants to get into that level of detail.

Less obvious is what the notion of a "massless" string is. Wald is not as clear as he could be, but I assume he means that $T^{00}$ has only tension components. If the string is not massless, the tension varies because of the weight of the string. The notion of "massless" is mildly coordinate depedent, the coordinates used are the coordinates that are natural for a static observer.,

 

[PeterDonis]

Wald is not as clear as he could be, but I assume he means that $T^{00}$ has only tension components.

That's not my understanding. My understanding is that he means that $T^{00}$ (the energy density) is zero, and the only nonzero component is the tension, which would be a single diagonal space-space component in the radial direction, $T^{rr}$, in Schwarzschild coordinates. Of course this means the material of the string violates energy conditions and is therefore not physically realistic; but it simplifies the math.

 

[pervect]

That's not my understanding. My understanding is that he means that $T^{00}$ (the energy density) is zero, and the only nonzero component is the tension, which would be a single diagonal space-space component in the radial direction, $T^{rr}$, in Schwarzschild coordinates. Of course this means the material of the string violates energy conditions and is therefore not physically realistic; but it simplifies the math.

I thought that's what I said? Pure tension in the rest frame, the density is zero.

 

[PeterDonis]

Pure tension in the rest frame, the density is zero.

You said

I assume he means that $T^{00}$ has only tension components.

Perhaps you intended to say just $T$ instead of $T^{00}$?