Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relativity and free energy

  1. May 10, 2005 #1
    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?
     
  2. jcsd
  3. May 10, 2005 #2

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  4. May 11, 2005 #3
    So wait, would the object lose energy and come to rest in the center then?
     
  5. May 11, 2005 #4

    Chronos

    User Avatar
    Science Advisor
    Gold Member
    2015 Award

    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.
     
  6. May 11, 2005 #5

    Ich

    User Avatar
    Science Advisor

    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.
     
  7. May 11, 2005 #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).
     
  8. May 11, 2005 #7

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
    Last edited: May 11, 2005
  9. May 12, 2005 #8

    Chronos

    User Avatar
    Science Advisor
    Gold Member
    2015 Award

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Relativity and free energy
  1. Free fall and relativity (Replies: 32)

Loading...