Re: How real are the "Virtual" partticles?

On 2005-06-14, Eugene Stefanovich <eugenev@xxxxxxxxxxxx> wrote:
> Igor Khavkine wrote:

>> Once you express
>> everything in terms of physical renormalized particles, there is only
>> one kind of photon. My statement is simply that at finite times, the
>> state of the system will be in a superposition of states, some of which
>> will have non-zero photon number.
>
> When electric charges move slowly, the amount of energy converted to
> radiation is very small. If I remember correctly, the energy of
> radiation is proportional to the square of acceleration of charges.
> This is in disagreement with the strong Coulomb interaction between
> charges that is independent on charges' velocities and accelerations.

That is only the energy radiated to infinity. It does not take into
account energy stored in the E&M field. See more below.

> Emission of real photons by accelerated charges can explain such effects
> as radio transmissions. However, it cannot account for the
> static Coulomb or magnetic interactions.

That depends entirely on how you interpret the word "photon". A simple
true statement is that there is enough information stored in a Fock
state to account for any kind of E&M field configuration. To figure out
how, you may want to solve a simpler puzzle first.

The E&M field is really just a collection of oscillators, so consider a
simple harmonic oscillator with one degree of freedom as a toy model.
What is the equivalent of a field operators A(x) and E(x)? What is the
equivalent of a momentum mode field operators a_k and a*_k? What does a
state with a definite number of quanta look like? What is the
expectation value of the position observable in such a state? What does
a state with a non-trivial expectation value for the position observable
look like? Once you figure this out, trace back your steps and figure
out how a non-trivial E&M field configuration is described in Fock
space. Whether you want to relate this description to photons or not is
your business.

>> There is no approximation in the derivation of the Lienard-Wiechert
>> potentials. In particular, radiation is not neglected. The standard
>> formula for total radiated power is derived from the Lienard-Wiechert
>> potentials. If radiation were neglected, this formula would be utter
>> nonsense. Instead, it agrees quite well with experiment.
>
> Could you remind me which experiment can DIRECTLY measure the
> retardation of full electromagnetic interactions between charges?

All experiments that detect interactions between charges measure the
"full" interaction. Empirically it is not possible to select a part of
the E&M field at your leisure.

> To save time, I would like to note that experiments on radio
> transmissions can only probe the part of interaction transmitted by
> real photons. There is another (even stronger) part of interaction
> (Coulomb and magnetic) which does not depend on the acceleration of
> charges and does not involve emission/absorption of real photons.
> The speed of propagation of Coulomb and magnetic forces has not been
> unambiguously measured yet.

I've said it before, and I'll say it again. Empirically it is not
possible to select a part of the E&M field at your leisure. All that you
can see in experiment is that either your test charge moves or not.
And I'll repeat this again if necessary.

>> I don't know why you put scare quotes around the word derivation. To
>> remind you how it is obtained, here is a sketch.
>>
>> The system we are trying to solve involves particles coupled to
>> electromagnetic fields. The equations of motion are:
>>
>> div F(x) = j[x,x(t)], (*)
>> d[p(t)]/dt = j[x(t)].F[x(t)], (**)
>>
>> where F is the Faraday tensor, x(t) denotes the collective worldlines
>> and p(t) denotes the collective momenta of the particles. Note that
>> equation (*) is linear non-homogeneous, so if we know j[x(t)] for all
>> times, this equation can be eliminated by using a Green function
>> solution.
>>
>> A(y) = int D(y-x) j[x,x(t)] dx, (***)
>>
>> where A(y) is the vector potential, with F = curl A, and the Green
>> function D(x-y) represents the Lienard-Wiechert potentials. The
>> remaining equations are then encapsulated in
>>
>> d[p(t)]/dt = j[x(t)].curl A[x(t)], (****)
>>
>> where A is given by (***). Instead of a system of coupled PDE's and
>> ODE's, we now have a system of integro-differential equations. The
>> remaining unknowns are the collective world lines x(t). Nothing is
>> neglected, since once the world lines are solved for, the full E&M field
>> configuration is obtained.
>>
>> Once the solution is known, it can be checked that the total momentum
>> (carried by both particles and fields) is constant for any choice of
>> spacial slice. At the same time, the explicit form of equation (****) is
>> retarded and no superluminal propagation takes place.
>
> OK, I may agree with your derivation and accept that Lienard-Wiechert
> potential take into account (indirectly) the emission of radiation.
> However, Maxwell's theory is classical and approximate. It would be nice
> to have a proof of retardation in a full quantum theory.

And in another message:
> On a second (third?) thought, I remain unconvinced. If Lienard-Wiechert
> potentials take into account radiation, then they should describe the
> dissipation of energy (e.g., some part of energy of two colliding
> charges is converted to the energy of emitted radiation).
> I don't think Lienard-Wiechert potentials can do that. Probably, you
> need to add the Lorentz-Dirac equation for proper description of
> radiation reaction. Though Lorentz-Dirac equation has its own serious
> problems.

1. Yes, it would be nice to have a full quantum calculation. However, to
zeroth order in hbar what you should get is the above calculation. If
you expect different, check this expectation at the door before starting
the calculation.

2. The integro-differential formulation given above takes *all* E&M
fields into account. If two colliding charges loose energy into the
field, then that is described in the above formulation. If you don't
think so, please point out a flaw in going from (*,**) to (***,****).

3. The Lorentz-Dirac equation is a consequence of the above formulation
of the coupled field-particle system when a certain shape is assumed for
the electron. There are difficulties when the shape's size is taken to
be point-like. However this difficulty is resolved in the quantum theory
since the electron is not point-like, it is described by a wave packet
of non-singular charge density.

> Could you remind me how total radiated power is computed from
> Lienard-Wiechert potentials?

The details are in Section 14.2 of Jackson's E&M text. The assumption is
that we know the trajectory of the given particle for all times, then we
know the E&M field at all space-time points by making use of the L-W
potentials. The E and B fields are calculated by taking derivatives of
the L-W potentials. The fluence of the field is given by the Pointing
vector S=ExB. The absolute value of S is integrated over a sphere of
radius R centered at the particle. The result is the power flux through
the surface of the sphere due to the field.

However, looking at the expression for S, we see that it contains
several terms with different power law decays with respect to the
distance R. The leading term decays as 1/R^2, while higher order terms
are O(1/R^3). The sphere surface area grows as R^2. So after
integration, the leading term in the power flux is a constant, while
subleading terms are O(1/R). In other words, part of the power flux
escapes to infinity (this part is the famous Larmour formula), while
another part is of the energy lost by the particle gets trapped in the
E and B fields localized near the particle. The total energy, the
particle's kinetic energy and the integrated E^2+B^2 is conserved at all
times.

>> QED reduces to the above system in the classical limit. I expect no
>> different qualitative behavior in the quantum case.
>
> This is not obvious at all.

Mathematically, perhaps no. This is where physical intuition comes in.
If you can't convince yourself, you must check the mathematics yourself.
AFAIK so far, your have neither performed a fully relativistic
calculation nor have you checked the classical limit. Therefore,
whatever results you think you have can neither be taken seriously, nor
trusted to be correct.

> In QED, the "force" term
> j[x(t)].F[x(t)] corresponds to the interaction j.A in the Hamiltonian
> (or Lagrangian). As we discussed exhaustively, this term is trilinear
> in particle operators. You were right to point out that such
> interactions are characteristic for fictitious "bare" particles.
> There is a lot of additional work required to derive the
> interaction potentials
> between physical particles. This work has been done in RQD.
> The interaction potentials between physical particles appear to be
> instantaneous.

Every time you bring up the "they appear to be instantaneous" argument
you get the exact same reply. Shall we try again?

This is insufficient. Terms in the Hamiltonian (at finite order in 1/c
at that) are not measurable. You need to perform a dynamical calculation
and evaluate some observables, such as particle positions. And, I can't
stress this enough, it has to be done to all orders in 1/c, otherwise
you throw 100 years of relativity out the window.

Igor

.

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