Do the various propagators of the Klein-Gordon equation really produce a Green's function which is zero for times t <= 0, and also zero outside the light cone of the impulse?
A Green's function is an impulse response. The response should be after the response, not before it.
The Feynman propagator produces a function which is not zero for t < 0. Times t < t tell the probability amplitude for a particle traveling back in time. It is not a Green's function in the above sense. How is its use justified?
What about the retarded propagator? Wikipedia says that it is zero for t <= 0 and also outside the light cone.
Let us check what is known about various propagators.
Tim Evans (2018) explains why the retarded propagator has to be integrated with a contour at the infinity in the lower complex plane, if t > 0. That gives the sum of residues of the poles. If t < 0, then the contour includes no poles and the integral is zero.
Continuity of a Green's function is not well-defined because the Dirac delta source at t = 0 and x = 0 is infinite.
The contour integral in the retarded propagator goes to zero when t ---> +0. That is reasonable.
How to solve the response of a classical wave equation to an impulse?
__
| | hammer
| |=========
|__|
|
v
---------------------------------
drum skin
Consider the example where we hit a drum skin with a hammer whose head is not sharp. A drum skin approximately satisfies the massless Klein-Gordon equation (= the ordinary wave equation).
Let us try to solve the behavior of the drum skin with Green's functions. The response to a sharp Dirac delta impulse is the Green's function (given as its Fourier decomposition):
G(E, p) = 1 / (2 π)^4 * 1 / (E^2 - p^2 ),
where the wave is
exp(-i (E t - p • x)).
Above, E is the energy of the wave, p is the spatial momentum of the wave, t is the time, and x is the spatial location.
G(E, p) goes to infinity when E and p go to zero.
Question. If there is a "smooth" impulse to the massless Klein-Gordon equation, does there exist a solution? Equivalently, if we write a "smooth" source to an inhomogeneous massless Klein-Gordon equation, does there exist a solution?
Can we form a solution from the Green's function? If we set E = 0 in the Green's function, then in an n-dimensional space, let us look at the integral
∞
∫ 1 / r^2 * r^(n - 1) dr,
0
which decides absolute convergence.
If n >= 2, then the integral diverges for large r.
If n <= 2, then the integral diverges for small r.
The Green's function, when put to the massless Klein-Gordon equation, has a Dirac delta source. A Dirac delta source, and the associated solution with a Green's function, seem to have problems with divergence.
Let us put a unit ball source.
Daniel Fischer (2013) writes that the Fourier transform for n = 4 is
~ 1 / p^2 * J_2(p).
The Bessel function J_2 has an asymptotic behavior 1 / sqrt(p).
Let us put a gaussian source.
Daniel Fischer (2013) confirms that an n-dimensional gaussian has a Fourier transform of the form:
~ exp(-π p^2).
These sources look much better with respect to diverging behavior. Note that for small |p|, the Fourier transform is just like for the Dirac delta because exp(-π p^2) is then roughly 1.
The contour integration trick does not work for the gaussian, because
exp(-π E^2)
has a very large value if E is imaginary and |E| is large. The contour running at infinity should have very small function values. We can probably prove the convergence of the integral directly.
Time asymmetry
There is a serious problem in Green's functions. They are typically symmetric in time. But we want a solution where there is no wave in the past. That is, the initial value is a zero field in the past. The wave is created by the source. Can we still utilize Green's functions?
Maybe we can, for the following reason:
| impulse
v
|<---- Δx ----->|
---------●-------------------------------- tense string
support
Suppose that you press with your finger to make a wave in a tense string. The wave will spread to the left and the right. How to remove the left wave? Attach the string to a fixed support at the position of the left edge of the impulse area.
The wave which spreads to the right may be relatively unchanged if the impulse is very quick, despite the fixed point on the left. By a "quick impulse" here we mean that the wave can reach the ball during the impulse, but does not have time to proceed much farther. The width Δx of a quick impulse should be roughly
Δx = c Δt,
where c is the speed of the wave and Δt is the time of the impulse.
If the impulse is slow, then the wave to the right in the fixed case is much weaker than in the free case because the fixed point prevents the finger from pressing the string lower.
If we want the field to be zero in the past, we "fix" it to zero there. The source will create a wave to the future direction, and the wave might be somewhat similar to a Green's function.
We should calculate with a computer the waveform in various cases.
How does a string react to a Dirac delta impulse?
Does a Dirac delta impulse make sense at all? It contains an infinite force. The result may be strange.
Let us analyze the Dirac delta response in 1 + 1 dimensions.
d^2 ψ / dt^2 - d^2 ψ / dx^2 = δ(t , x).
Let us approximate the Dirac delta impulse with a box impulse that is Δx wide and Δt long.
Δx
____
__/ \__ tense string
^
| impulse
We assume that ψ is zero before the sudden impulse. The impulse forces the second time derivative to jump up suddenly in a narrow area. Around that area, ψ has to satisfy the homogeneous equation. The string is concave downward there.
Suppose that we have a solution with a box size A. We shrink both sizes of the box by a factor 1/2. We get a new solution by shrinking the width and the time of the string deformation by the same factor 1/2. Second derivatives grow 4-fold.
We conclude that nothing strange happens. The string behaves reasonably with a narrow box of impulse.
The wave going backward in time is useful in some configurations
Z+ -------------------------
| virtual photon p
e- -------------------------
\ /
~~~
virtual photon q
-----------------------------------> t
In the vertex correction diagram, when the electron absorbs the virtual photon q, we can also imagine that the electron emits it backward in time.
If there is input which matches the supposed emission backward in time, then emitting backward in time is no problem. This is probably the logic behind the Feynman propagator.
Conclusions
A major question of classical physics is why large amounts of very high-frequency or very low-frequency waves do not arise in wave phenomena. When calculating classical wave processes we instinctively apply cutoffs which remove such frequencies.
If lots of extreme frequencies would arise, then we would have infrared and ultraviolet divergences in classical physics, too.
There is probably no proof that extreme frequencies cannot arise. Coupled classical fields are nonlinear systems, and usually it is impossible to prove anything about nonlinear systems. The Clay Institute Millennium problem about the smoothness of Navier-Stokes is one example.
Above we gave some heuristic arguments why classical physics behaves well. Impulses are smooth, not Dirac delta impulses, and their Fourier decomposition contains very small amounts of extreme frequencies. The gaussian is a prime example of a smooth impulse.
Our analysis suggests that the divergences of Feynman integrals are an artifact from using Dirac delta impulses. The divergences are removed if we switch to more realistic impulse forms, for example, gaussians.
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