Does Relativity Affect Energy Conservation in Gravity-Driven Systems?

In summary, the problem is that if you drill a hole through a massive body and drop something through, it will gain energy and come to rest in the center.
  • #1
gonzo
277
0
Okay, first I explain what I think I understand, then what my problem is.

As I understand it, if you have two massive bodies connected by a perfect frictionless spring, so that they fall towards each other by gravity and then bounce back from the spring, they will eventually come to rest with the energy carried off by gravity waves. This is because information can't travel faster than light, so that at any given moment while they are falling towards each other, they "feel" like they are farther apart and thus have less pull than they "should" (and so the spring stores up less energy) and as they are moving away, they "feel" like they are closer, and so the spring requires more energy to separate them, and all this lost energy is carried away as said by gravity waves.

Assuming I understand that much correctly, what happens then if you drill a hole through a massive body and drop something through. Without relativity you would think it would pendulum back and forth forever, being braked as it passes the center as much as it accelerates on the way down, reaching the same height on the other side that it was dropped from.

However, if we use relativity this is seems analogous yet opposite to the spring problem. As you approach the the center of the massive body, the gravitational force is constantly reduced. This means that at any given point, the object will "feel" more pull, which means it will arrive at the center will more speed than you would have assumed without relativity, and seemingly have gained energy. What's worse, the problem continues on the way past the center as it goes "up" the other side. At any given moment it will "feel" less gravity on and breaking it, so it seems it should reach a higher point on the other side than it was dropped from, and have gained energy, and this should continue on each swing.

What am I missing here?
 
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  • #2
What you are missing is that GR says that (ignoring gravitational radiation for the time being) that a body following a geodesic through the center of another body will have a conserved potential. It won't be the same potential as the Newtonian case, but it will have a conserved potential.

The exact solution including gravitational radiation would be extrordinarly messy, and would best be approached by assuming that the body follows a path that's almost a geodesic, and dealing with the departure of the body from this path as a small pertubation. (A body emitting gravitational radiation will not follow a geodesic exactly, only a body emitting no gravitational radiation will follow an exact geodesic. When the amount of radiation is very small, as is likely to be the case, pertubative methods can be used).

Assuming the usual conditions required for energy conservation in GR, i.e. an asymptotically flat space-time, the "Bondi mass" of the 2-body system would go down as the system radiated gravitatioanl energy, so that the total of the mass in gravitational radiation and in the system itself remained constant.
 
  • #3
So wait, would the object lose energy and come to rest in the center then?
 
  • #4
You are missing the patent on your free energy machine. I am still kicking myself for not getting a patent on the clapper. Have you checked your electric bill lately? You would think they would have been the first to jump at 'free energy'. Not that they would drop the rate, just increase the profit margin.
 
  • #5
You are using the stationary solution that all masses outside the current position of the falling object have zero effect. That is not necessarily true in the dynamic case - for example the Earth will appear as an ellipsoid rather than a sphere for a moving body.
I don´t know the result of exact calculations, but I think that´s where your assumptions are wrong.
 
  • #6
Thanks for the pointless sarcasm Chronos, it really helps. I obviously don't think this is really a source of free energy, which is why I'm asking this question. I want to see why my analysis is wrong. Idiocy like your reply don't really add much to my understanding of the situation. Thanks anyway.

Ich, an actually useful answer, and one which I was leaning towards on my own. I was thinking that in the static case each outer shell as you move towards the center exactly cancels. However, in the dynamic case, as you suggest, this seems unlikely to be the case.

I was trying to think of a rough analysis, and it seems to me on first glance that since the shell section behind you is closer than the "rest" on the other side, that it's information will reach you sooner, so that at any given time you will have different information about how deep you are relative to different parts of the body you are falling through, which would of course negate the static cancellation affect in Newtonian mechanics.

Assuming this means the portion behind/above you (closer to the surface) has a greater impact on you as you fall, then maybe this amount would exactly equal (or be slightly more than) the apparent excess you experience from my first analysis. That would be the simplest general analysis from my POV. I would guess that if the amount behind you breaks you more than you are losing energy each time (presumably in the form of gravity waves) and will eventually end up at the center, whereas if it cancels exactly, you return to the infinite pendulum scenario.

However, I really don't know nearly enough about this to have any idea is this is correct.

Unfortunately Pervects answer was a bit over my understanding. I was hoping for a simple (if not entirely accurate) analysis to get the general idea (like with two bodies falling towards each other on a perfect spring ... a not realistic scenario, but it serves to illustrate the idea in broad strokes that make it accessible to people like me).
 
  • #7
gonzo said:
So wait, would the object lose energy and come to rest in the center then?

Yes, at least that's what I think should happen. Consider the analogous case of a charge oscillating in a conservative electric potential, slowly emitting radiation in the process. Eventually one expects the charge to come to rest at the minimum of the potential.
 
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  • #8
The 'spring' is not totally efficient. A small amount of energy is lost due to gravitational radiation. Orbiting bodies will eventually merge. Fortunately, this takes a very long time - even for extremely dense objects [e.g., neutron stars] in tight orbits.
 

Related to Does Relativity Affect Energy Conservation in Gravity-Driven Systems?

1. What is the theory of relativity?

The theory of relativity, proposed by Albert Einstein in the early 20th century, is a concept in physics that explains the relationship between space and time. It suggests that the laws of physics are the same for all observers, regardless of their relative motion.

2. How does relativity relate to free energy?

The theory of relativity does not directly relate to free energy. Free energy is a term used in thermodynamics to describe a system that is able to perform work without any external energy input. However, some scientists have proposed theories that incorporate elements of relativity in their explanations of free energy.

3. Can relativity be used to create free energy?

No, relativity cannot be used to create free energy. The laws of thermodynamics, which are based on empirical evidence, state that energy cannot be created or destroyed, only transformed from one form to another. Therefore, it is not possible to create free energy using the principles of relativity.

4. What is the connection between relativity and the concept of perpetual motion?

Perpetual motion, or the idea of a machine that can run indefinitely without an external energy source, goes against the laws of thermodynamics. The theory of relativity does not offer any evidence or explanations for the possibility of perpetual motion. In fact, Einstein himself stated that perpetual motion machines are impossible.

5. Are there any practical applications of relativity in the field of free energy?

At this time, there are no practical applications of relativity in the field of free energy. While some scientists may incorporate elements of relativity in their theories of free energy, these theories have not been proven or applied in any practical way. The laws of thermodynamics remain the basis for understanding and utilizing energy in our daily lives.

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