Does it require energy to remain at "rest"?

  • #1
stevendaryl
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The Schwarzschild metric (ignoring the angular parts) looks like this:

##ds^2 = (1 - \frac{2GM}{c^2 r}) c^2 dt^2 - \frac{1}{1 - \frac{2GM}{c^2 r}} dr^2##

The Rindler metric in 2 spacetime dimensions looks like this:

##ds^2 = (g r)^2 dt^2 - dr^2##

They are very different, but they have some similarities, namely that a test particle initially at "rest" (that is, ##\frac{dr}{dt} = 0##) that is unsupported will "fall" to lower values of ##r##.

One critical difference is that to keep something at rest in the Schwarzschild metric need not require any energy, while keeping something at rest in the Rindler metric does require energy.

To see that it's possible for something to be at rest in the Schwarzschild metric, imagine a spherical shell completely surrounding a black hole at a safe distance away. An object could then rest on that shell without requiring any energy.

In contrast, the Rindler metric is just flat spacetime in a different coordinate system. An object being at "rest" in the Rindler metric implies that it is accelerating in the usual inertial coordinates, which requires expending energy.

So thinking about what the two metrics mean, physically, allows you to see that it is possible in the one case to have an object at rest without expending energy in the one case, but not in the other. My question is that if you have an unfamiliar metric, is there some mathematical test for whether it is possible to have an object at rest that requires no energy?

Maybe it's simply a matter of whether there are orbits? (geodesics that return to the same spatial location after an amount of time)?
 
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  • #2
PeterDonis
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imagine a spherical shell completely surrounding a black at a safe distance away. An object could then rest on that shell without requiring any energy.
But the spherical shell itself has to also be at rest, and that's really what you need to look for. There is no analogue of the spherical shell at rest in Rindler coordinates in flat spacetime.

In more technical terms, I think the criterion would have to be something like:

(1) The spacetime has a timelike Killing vector field and is foliated by spacelike 3-surfaces such that each integral curve of the timelike KVF intersects each 3-surface exactly once. (Note that it is not required that the KVF be hypersurface orthogonal, although it is in the Schwarzschild case.)

(2) There exists a set of integral curves of the timelike KVF such that their intersection with each spacelike 3-surface forms a closed 2-surface with the same geometry (same area, same shape).
 
  • #3
pervect
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We don't seem to be talking about energy as being the metric at infinity. So , what do we mean by "energy" in this context?
 
  • #4
stevendaryl
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We don't seem to be talking about energy as being the metric at infinity. So , what do we mean by "energy" in this context?
Locally, you can use the special relativistic notion of energy. The original question is about a test mass.
 
  • #5
PeterDonis
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what do we mean by "energy" in this context?
I would say that "no energy required" means something like "static equilibrium is possible", and "energy required" means, correspondingly, "static equilibrium is not possible".
 
  • #6
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To see that it's possible for something to be at rest in the Schwarzschild metric, imagine a spherical shell completely surrounding a black hole at a safe distance away. An object could then rest on that shell without requiring any energy.
For a radially free falling observer the shell would be doing work on the object.

In contrast, the Rindler metric is just flat spacetime in a different coordinate system. An object being at "rest" in the Rindler metric implies that it is accelerating in the usual inertial coordinates, which requires expending energy.
It doesn't require energy in Rindler coordinates, just like it doesn't require energy in Schwarzschild coordinates. Comparing the energy in the non-inertial coordinates for the Schwarzschild case to energy in the inertial coordinates for the Rindler case is the wrong analogy.
 
  • #7
stevendaryl
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For a radially free falling observer the shell would be doing work on the object.
No, do you think it requires energy for something to sit at rest on the surface of the Earth?

It doesn't require energy in Rindler coordinates, just like it doesn't require energy in Schwarzschild coordinates. Comparing the energy in the non-inertial coordinates for the Schwarzschild case to energy in the inertial coordinates for the Rindler case is the wrong analogy.
I think you're confused. The comparison is: A rocket can rest on the surface of a planet without being turned on. A rocket cannot be at a constant Rindler coordinate without being turned on.
 
  • #8
Orodruin
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The comparison is: A rocket can rest on the surface of a planet without being turned on. A rocket cannot be at a constant Rindler coordinate without being turned on.
I think your simile is lacking. If there was an accelerating hard surface in Minkowski space, then it would be perfectly possible for the rocket to remain at rest in Rindler coordinates without being turned on. This is the simile you have to do. Locally, the two cases are indistinguishable.

The better question is if you can construct a global equilibrium situation where everything is static. This will depend on the global matter distribution and space-time geometry.
 
  • #9
stevendaryl
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I think your simile is lacking. If there was an accelerating hard surface in Minkowski space, then it would be perfectly possible for the rocket to remain at rest in Rindler coordinates without being turned on.
Sure. I guess the question then becomes: there are stationary rigid surfaces that have nonzero proper acceleration in Schwarzschild spacetime, but not in Minkowski spacetime.
 
  • #10
A.T.
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For a radially free falling observer the shell would be doing work on the object.
No
Just apply the definition of work in the free falling frame.

do you think it requires energy for something to sit at rest on the surface of the Earth?
Energy is frame dependent. In some frames the surface is doing positive work on the object.
 
  • #11
stevendaryl
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Energy is frame dependent. In some frames the surface is doing positive work on the object.
But whether or not a rocket is firing is certainly not frame-dependent. You're being distinctly unhelpful.

Okay, so a better description of the distinction between Schwarzschild spacetime and Minkowski spacetime is this: In the former, there exists static configurations of rigid objects where the surface of the objects have a nonzero proper acceleration. In the latter, this is not possible.
 
  • #12
Orodruin
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Sure. I guess the question then becomes: there are stationary rigid surfaces that have nonzero proper acceleration in Schwarzschild spacetime, but not in Minkowski spacetime.
Yes, but I do not see a contradiction here. It is a global property and Minkowski space is not globally equivalent to Schwarzschild spacetime.
 
  • #13
stevendaryl
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Yes, but I do not see a contradiction here. It is a global property and Minkowski space is not globally equivalent to Schwarzschild spacetime.
I didn't say there was a contradiction. I was interested in characterizing the difference between spacetimes where it is possible for a test mass to accelerate forever without expending energy and one where it is not possible.
 
  • #14
Orodruin
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I didn't say there was a contradiction. I was interested in characterizing the difference between spacetimes where it is possible for a test mass to accelerate forever without expending energy and one where it is not possible.
I am still not sure that this is a meaningful distinction. In particular, the "expending energy" part is not well defined. Obviously the force necessary to accelerate forever must come from somewhere, it is just a matter of what mechanisms you allow in order not to "count" as "the test mass expending energy".
 
  • #15
stevendaryl
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I am still not sure that this is a meaningful distinction. In particular, the "expending energy" part is not well defined.
Well, the rephrasing in terms of static structures seems like it covers it.
 
  • #16
Orodruin
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Well, the rephrasing in terms of static structures seems like it covers it.
But structures are extended objects. In themselves they do not have proper acceleration although their different parts may. To me it seems like asking for the difference between putting a glass on a book shelf and putting it on top of a quadcopter drone.
 
  • #17
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But whether or not a rocket is firing is certainly not frame-dependent.
We have a book sitting on a table. The table may be sitting on the surface of the earth or in the cabin of an accelerating spaceship. In both cases the table is exerting a force on the book and as a result the book is experiencing proper acceleration; these are frame-invariant facts.

However, the work done on the book by that force is frame-dependent. Choose coordinates in which the table is at rest and the work done is zero: it takes no energy to keep the book at rest. So what do we make of the coordinate-independent fact that the rocket is firing and generating energy? If we've chosen coordinates in which the table is at rest, then all the energy goes into increasing the kinetic energy of the rocket exhaust and none goes into increasing the kinetic energy of the book.

I don't see how it's helpful to think of this situation as using the energy of the rocket to keep the book at rest.
 
  • #18
pervect
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I would say that "no energy required" means something like "static equilibrium is possible", and "energy required" means, correspondingly, "static equilibrium is not possible".
OK, with that approach, an elevator is static only if you ignore the forces that are either pushing the elevator from behind, or pulling it from the front ( or some combination of the two) to make it accelerate.

If you do ignore those forces, the elevator is in a static equilibrium, but I think the OP objects to ignoring those forces, therefore it's not static.

It doesn't seem to me to be a great mystery then that an elevator can not accelerate unless there are unbalanced forces on it, though. Actually, with the right approach it could be made a non-trivial problem, perhaps. But if we seek the simplest answer, it's not too surprising.

Meanwhile, a planet or star, or other massive body, as a whole, without any exterior forces on it, doesn't accelerate, so it can be in a static equilibrium.
 
  • #19
PeterDonis
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The better question is if you can construct a global equilibrium situation where everything is static.
This is how I was approaching it in my previous posts. In the Schwarzschild case, you can do this. In the Rindler case, you can't.
 

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