abskebabs Posted May 27, 2006 Share Posted May 27, 2006 I remeber reading the book Quantum a guide for the perplexed, and in one section about Black holes, the contributer Paul Davies mentions negative energy, the Casimir effect, and the negative energy flux going into black holes; which on it's own I found intriguing and enthralling. Not only this, but he goes on to say;"The theoretical possibility of creating a flux of negative energy-in effect a beam of cold and dark, rather than a beam of heat and light- offered some bizarre and puzzling scenarios. Suppose such a beam were directed at a hot object rather than a black hole, such as an oven with an aperture protected by a shutter. It would seem the contents of the oven would lose energy and cool down. But this would be a clear breach of the celebrated second law of thermodynamics, by the loss of heat of the oven which would amount ot a loss of entropy, and the second law forbids the entropy of a closed system going down.(The beam itself has zero entropy.) The second law is the lynchpin of themrodynamics, and any violation would open the way to a perpetual motion machine, which is not thought possible." What are your opinions on this, and do you think a violation would ever represent a possibility? Link to comment Share on other sites More sharing options...
abskebabs Posted May 28, 2006 Author Share Posted May 28, 2006 I have another question now, as negative energy can exist, deos that mean negative mass can too(at least in relation to a vacuum)? Link to comment Share on other sites More sharing options...
swansont Posted May 28, 2006 Share Posted May 28, 2006 There are a lot of conceptual things that are impossible; the devil's in the details. The Casimir effect arises from changing the boundary conditions of the vacuum with a conductor, and it requires work to assemble. The question becomes: How would you make the beam? I suspect the answer to that will explain why entropy wouldn't decrease. Negative energy is a term that presupposes a zero-value, which is chosen for convenience. Link to comment Share on other sites More sharing options...
abskebabs Posted May 28, 2006 Author Share Posted May 28, 2006 Negative energy is a term that presupposes a zero-value, which is chosen for convenience. Ah, I see. It is only negative then if we presuppose the energy of a vacuum to be zero, even though it isn't(like using it as a standard). What is the energy of a vacuum by the way? Is such a thing known? Link to comment Share on other sites More sharing options...
Gilded Posted May 28, 2006 Share Posted May 28, 2006 If I remember correctly QM states a certain amount of energy associated with "each point of space". However, this would lead to having an infinite amount of energy within any volume and I don't think it's generally thought to be infinite, just damn big. Perhaps someone else knows better. Link to comment Share on other sites More sharing options...
swansont Posted May 28, 2006 Share Posted May 28, 2006 This is actually being discussed on a few other current threads (here's one). The value you get is infinite when you solve the QM equations. But that's inconvenient, so the infinity is generally discarded. However, the Casimir force shows that the solution is not entirely unphysical; you have to understand your boundary conditions to make sure you're doing things properly. Link to comment Share on other sites More sharing options...
abskebabs Posted April 19, 2007 Author Share Posted April 19, 2007 This is actually being discussed on a few other current threads (here's one). The value you get is infinite when you solve the QM equations. But that's inconvenient, so the infinity is generally discarded. However, the Casimir force shows that the solution is not entirely unphysical; you have to understand your boundary conditions to make sure you're doing things properly. I'm sorry if it seems innapropriate ressurecting this thread, but I was just having a look and realised that I failed to ask a pertinent question. Why are the values you get infinite and why is it mathematically legitimate to discard the infinities? I think this process is called renormalisation. Please could you shed some light on this? Link to comment Share on other sites More sharing options...
fredrik Posted April 20, 2007 Share Posted April 20, 2007 Why are the values you get infinite and why is it mathematically legitimate to discard the infinities? I consider it a sign of problems, and I don't consider it logically legitimate to just remove it. I think when the theories are better understood and improve it shouldn't have to be removed like that. It seems we can do it and not get punished but that does not justify it from a logical point of view. I think we're lucky, and there is a better, more proper resolution. It seems to be the fact that the background spacetime is assumed to be a rigid and fixed prior reference (=thus with "infinite" inertia) that implies these infinite energies. This is just an approximation anyway until we know better, it isn't logically satisfactory as is. /Fredrik Link to comment Share on other sites More sharing options...
swansont Posted April 20, 2007 Share Posted April 20, 2007 What we see are differences in energies between states. As I said before, what constitutes zero is often an arbitrary choice. You get infinity because the solution to the problem is [math]E_k = (n + \frac{1}{2})\hbar\omega_k[/math] for every mode, k. There are an infinite number of modes, so that extra 1/2 becomes infinite, even when there are no photons present (n=0), when you sum over all k. The presence of the Casimir force indicates that this is true — something is there. Empty space isn't really empty. But if it's always there, in empty space, and you're only interested in the energy difference between two states (when n changes), you are subtracting it off anyway. In a way it's just a matter of whether you are doing that first or last. Doing it first makes the math easier for many problems. Link to comment Share on other sites More sharing options...
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