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- Thread starter CosmologyHobbyist
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Remember the uncertainty principle.

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- #4

vanesch

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This is a question that comes back regularly. The answer is: depends on exactly what one understands by "virtual particles". In a commonly used version of it, there is, at each interaction, perfect conservation of energy and momentum. Only, the "virtual particles" are not exactly like their real counter parts. One sometimes says that they are "off-shell", and they can carry imaginary mass for instance. It is only when one insists on them to ressemble a bit more to their real counterparts than they formally are, that one seems to run into a problem of conservation of energy.

Just let us give an example. Imagine a photon, with energy-momentum (k,0,0,k) (in units where c = 1). Now, imagine that this photon "splits" into a virtual electron-positron pair.

Well, the electron will have energy-momentum (k-e1, l1, l2, k+l3) and the positron will have energy-momentum (e1,-l1,-l2,-l3).

Normally, for a real electron, we have the on-shell condition that

(k-e1)^2 = m^2 + l1^2 +l2^2 +(k+l3)^2, but in this case, this will not hold. What will hold is that the sum of the 4-momentum of the virtual positron and electron will equal the 4-momentum (k,0,0,k), so there will be perfect conservation of energy and momentum!

But the particles will not have the right "mass" (can even have imaginary mass).

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Thanks guys! That clears things up. I think I better locate an easy read on the basics!

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Remember the uncertainty principle.

can you please explain this a little more?

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Nothing is real until it is "measured". So if the virtual particles are created and annihilate each other prior to any type of measurement (interaction with other particles). Then there is no net change in energy and thus no violation of the conversation of energy law.

Someone correct me if I am wrong.

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Energy conservation can be violated during short times.

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Isn't it more like

[tex]\Delta E \Delta A}{\left | \frac{\mbox{d}t}{\mbox{d}\langle A \rangle}\right |} \ge \frac{\hbar}{2}[/tex]

Where [itex]\Delta A = \sqrt{\langle A^2 \rangle - \langle A \rangle^2}[/itex] where [itex]\langle A \rangle = \langle \psi | A | \psi \rangle[/itex]

[tex]\Delta E \Delta A}{\left | \frac{\mbox{d}t}{\mbox{d}\langle A \rangle}\right |} \ge \frac{\hbar}{2}[/tex]

Where [itex]\Delta A = \sqrt{\langle A^2 \rangle - \langle A \rangle^2}[/itex] where [itex]\langle A \rangle = \langle \psi | A | \psi \rangle[/itex]

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We know that they are never detected: Whether or not they actually exist is up for debate. In the strictest Popperian approach, they clearly do not exist on a par with ordinary particles.

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Next, virtual particles do not exist. They are not even a part of the exact theory. They are nothing but a pictorial name for some mathematical terms appearing in certain approximative methods of solving otherwise difficult and abstract quantum equations.

Finally, energy of a closed system is always conserved, even during a short time. Only the energies of certain subsystems may be uncertain, but this is always compensated by energies of other subsystems, such that the total energy is conserved exactly.

- #14

reilly

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Regards,

Reilly Atkinson

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Think of the interaction between two electrons, which occures when one electron emits a virtual photon that is then absorbed by the other electron. To understand this idea of a virtual photon, you must understand that an electron can't emit a real photon and just remain an electron, because this would violate the conservation of energy and momentum. This is easy to see if you think of the initial electron at rest. The final state, after emitting a real photon, would consist of an electron and a photon moving off back-to-back, and that configuration necessarily has more energy than the initial at-rest electron.

With the uncertainty principle, this is easy to resolve. The uncertianty realtion, dEdt>h-bar, allows for deviations from energy conservation, but only for time dt determined by the degree of energy uncertianty. During the time period dt the photon is said to be virtual, and as long as the photon is reabsored by the other electron quickly enough, there is no measurable violation of conservation of energy.

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can you please explain this a little more?

The photon decays into a virtual electron (mass m) and a virtual positron (mass M) for a certain time t.

(k-a+ E1)²= m² + 1² + 2² + (k+3)²

(a+ E2)²= M² + (-1)² + (-2)² + (-3)²

E1 and E2 are the uncertainties in energy

t1 und t2 are the uncertainties in time

The process only is allowed if t1 > h/E1 and t2> h/E2

that means if for example t1>t2, the two particles must anihilate after at least t=t2 into a photon again.

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