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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.


13

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 ...


7

This creates a point of extremely focused energy at the middle point where the bubble collapses. In theory, this point focuses enough energy to trigger nuclear fusion. It is not currently accepted mainstream science to say that collapsing bubbles focus energy enough to cause nuclear fusion. Temperatures over 10,000K can be acheived, but are still well ...


3

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, ...


3

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 ...


3

Why can't fermions have a non-zero vacuum expectation value (VEV)? Lorentz invariance. If anything other than a Lorentz scalar has a non-zero VEV, Lorentz invariance would be spontaneously broken. For example, suppose we have a Lorentz invariant term in a Lagrangian for a vector $$ \mathcal{L} \supset m^2 A_\mu A^\mu. $$ Now suppose the vector obtains a ...


1

from a, b and c: An electromagnetic field is propagating by changing the field that's generated by electrically charged particles pass through the air and the space that is devoid of particles of space. An electromagnetic wave propagates, not an electromagnetic field. An EM wave is a propagating disturbance in the existing electromagnetic ...



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