Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

It is known that from the vacuum of a quantum field theory, virtual particle pairs are created and destroyed; is it possible to capture these particles thus obtaining free energy from the vacuum?

share|improve this question
3  
We earnestly hope not. If so, this would imply the "vacuums" of our universe are not true vacuums, and the universe could spontaneously tunnel to its "true", lower energy vacuum, wiping out everything in the process, at any time! –  WetSavannaAnimal aka Rod Vance Mar 31 at 12:13
2  
See also, for interest and amusement google.com/patents/US7379286 although mainstream physics with considerable justification does not take such patents seriously. This particular patent has "slicked" itself up over the years and looks respectable: the original provisional patent bore a photograph (very seldom in patents) of a putative vacuum energy extraction device (a small, plastic box) with a regular, everyday 240V mains lit lightglobe dangling from the box by a cable! It's one thing to be genuinely sucked in by misguided physics, but such a photo borders on outright fraud. –  WetSavannaAnimal aka Rod Vance Mar 31 at 12:21
1  
@WetSavannaAnimalakaRodVance not slick enough. There is a typo in the abstract. These sorts of patents and claims are harmful to the less educated that take them as evidence that this sort of physics is real. –  Brandon Enright Mar 31 at 15:55
    
Related: physics.stackexchange.com/q/11800/2451 and links therein. –  Qmechanic Mar 31 at 16:01
    
@BrandonEnright You're right. I must say I never looked at this one closely after the box and lightbulb - I stopped looking at the physics when the question of the writer's honesty was decisively answered for me by the box and lightbulb. It seems someone's hoping to build a pyramid on this one - even the name "Jovion" seems to be making a joke at the expense of anyone who puts money into this (it reminds me of Enron at the height of its integrity - they had shelf companies and fictitious individuals to hide debt in and they named them things like M. Yass (permute trivially to read "My Ass"). –  WetSavannaAnimal aka Rod Vance Mar 31 at 21:30

4 Answers 4

up vote 20 down vote accepted

The energy is borrowed from the Heisenberg Uncertainty Principle to create virtual particles and has to be paid back in a very short time.

$\Delta{t} \geq \frac{\hbar}{2\Delta{E}}$

This is why virtual particles live for very short times (i.e pop in and out of existence). We cannot manipulate this energy.

share|improve this answer
2  
For additional information, I recommend Prof. Tong's notes: damtp.cam.ac.uk/user/tong/qft/qft.pdf. The very first section on "why quantum field theory?" discusses virtual particles. –  JamalS Mar 31 at 12:14
    
These notes are good QFT notes but I think they are better suited for grad students! I'm wondering are there any undergrad level QFT notes? –  PhotonicBoom Mar 31 at 12:28
    
Actually they don't look as bad as I remembered them to be.. –  PhotonicBoom Mar 31 at 12:29
    
A. Zee's "Quantum Field Theory in a Nutshell" is perhaps better suited for undergraduates, as it focuses on the physical intuition/understanding. –  JamalS Mar 31 at 12:30

Whether you can extract energy from this or not (and I strongly suspect not) the Casimir effect is a consequence of vacuum fluctuations.

Essentially when two metallic plates are very close to each other, the wavelengths of virtual particles that can be created between the plates is restricted and hence there are fewer particles between the plates and outside, where no restriction occurs. This creates a pressure on the plates and pushes them together. Not a source of free energy, but an interesting (and experimentally verified) result of vacuum fluctuations.

Another well-known effect due to virtual particles is Hawking radiation. This says that when virtual particles created across the event horizon of a black hole, one can escape and the other fall in, essentially turning a virtual particle into a real particle, since it's antiparticle is inside the black hole. This is not free energy however, as the energy required comes from the mass of the black hole, causing it to (very, very) slowly evaporate over time.

So in short, no, we can't get free energy from vacuum fluctuations, but that's not to say it it doesn't have some very interesting effects.

share|improve this answer

The answer kinda is "You can, but why would you".

It is indeed possible to extract energy from the vacuum. It has been studied, both theoretically and experimentally, using a variety of metal plates and other Casimiresque gizmos.

The problem is just that it basically acts like a spring. To put the Casimir effect in action, you must first approach together the two metal plates, working against the pressure of the Casimir effect. You then let the plates go, giving back said energy (minus whatever loss in the process).

There's a number of other methods you can use for equivalent effects, such as oscillating metal plates to emit a (very weak) radiation from the Casimir effect. But none of them break energy conservation in the end.

Although I am sure that such devices will be plenty useful for various nanotechnology things.

Edit : Oh, by the way :

1) While the quantum vacuum is a nice explanation for it, it is (probably) not true. You can derive Casimir effects just from higher order QED effects (you can't really have a Casimir effect with metal plates because they are not perfect conductors. Though you do have the topological Casimir effect in compact spaces).

2) For some sources on various experimental gadgets using the Casimir effects, you can try "Frontiers in Propulsion Physics". It is quite a neat book that contains a lot of experimental results on such crazy topics (usually negative unfortunately!), and it contains quite a lot of references on applications of the Casimir effect.

share|improve this answer

No.

Just like in Chemistry and Thermodynamics, we never get anything for free.

On a mechanistic level, it's important to recognize that zero-point (vacuum) energy represents the lowest energy state waveform. I remember thinking that because the EM fields are everywhere and quantized, that there was some sort of magic taking place. Realistically, zero-point energy is more like a spring between objects (very good analogy Slereah) than something unknown. It is a mathematical consequence of very well understood relations.

The second matter of virtual particles requires a little more thought. When we expand (any) Schroedinger equation into its particle form (using the number operator (two opposing ladder operators) which changes the differential equation to a sum of a series of integrals), we get an infinite number of these terms (the virtual particles). They are a distraction at this level.

Finally, Nuclear_Wizard explained the Casimir effect brilliantly, and I won't add to that except to say that this effect represents the primary experiment validating the theory. To my knowledge and at this time, the other experiments have had a great deal of background noise to wade through (they're looking for the LOWEST energy level) and have not given nearly the experimental validation that the two plates experiment has.

One critical point that is fairly unrelated to your question (but somehow the accepted "right" answer), there is no time operator. The derivation of that relationship depends on somewhat specific circumstances relating to the decay of wavestates. If we look at a graph of the energies of incoming particles, there will be some sort of distribution - the particles don't all have the exact same energy level. If we repeat the experiment and instead look at the time of arrival (which gets us to the lifetime of the wavestate by subtraction) of particles (different setup, identically prepared source), I will also get a distribution. The Heisenberg equation (in that form) relates the uncertainty (width) of those distributions where the location of the peak represents the quantity of those distributions. There is no borrowing energy but only for a short time - the Heisenberg inequality represents a fundamental property of the Fourier Transform and NOT a vacuum energy lender.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.