Expressing 4-potential as a Fourier integral

  • Context: Graduate 
  • Thread starter Thread starter dnquark
  • Start date Start date
  • Tags Tags
    Fourier Integral
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
7 replies · 2K views
dnquark
Messages
13
Reaction score
0
The following formula is taken from a book by Rubakov:

[tex]A_\mu = \int_{k_0 \ge 0} [e^{ikx}a_\mu(k) + {\rm c.c.}] d^4k[/tex]

I am trying to understand why it makes sense to integrate only over the positive values of the frequency k0. (Pretty much any time I write down Fourier integrals, I integrate over the entire frequency space). Does it have something to do with the A four-vector being real valued?.. Something having to do with causality?..
 
Physics news on Phys.org
I'm not sure what the formula refers to, but since quantum mechanically, frequency is energy, restricting to positive frequencies means restricting to positive energies, which is generally accepted to be a good thing to do.
 
dnquark said:
The following formula is taken from a book by Rubakov:

[tex]A_\mu = \int_{k_0 \ge 0} [e^{ikx}a_\mu(k) + {\rm c.c.}] d^4k[/tex]

I am trying to understand why it makes sense to integrate only over the positive values of the frequency k0. (Pretty much any time I write down Fourier integrals, I integrate over the entire frequency space). Does it have something to do with the A four-vector being real valued?.. Something having to do with causality?..

It is because the four-vector A is real-valued. Therefore the Fourier coefficients must satisfy

[tex]a_\mu (-k) = a_\mu^*(k)[/tex]

and hence, the second term with the complex conjugate effectively includes all the negative-frequency components.

Although, it appears the integral is only restricting the time component of k to be greater than zero, so perhaps the reality condition I gave is too strict. However, the correct answer will be something similar. Probably to do with unitarity.
 
Ben Niehoff said:
Although, it appears the integral is only restricting the time component of k to be greater than zero, so perhaps the reality condition I gave is too strict. However, the correct answer will be something similar.

Yes, and that's what bugs me about this... I can't figure out whether this is something trivial not worth thinking about, or has some depth behind it. The discussion of positive/negative frequencies/energies is reminiscent of this discussion (courtesy of John Baez): http://math.ucr.edu/home/baez/photon/fourier.htm" Unfortunately, there's too little context for me to fully understand the argument presented.

Note, by the way, that if we were to assume that the solutions live on a light cone, k0 = |k|, then k0 > 0 would also imply that all k components are > 0. This would be true for Fourier components of the fields. However, the gauge term of the vector potential does not have to respect k0 = |k| (I guess it's analogous to being "off-shell" when talking about particles). I wonder if causality would force us to draw the conclusion "k0 > 0 implies k1,k2,k3 > 0" regardless...
 
Last edited by a moderator:
PeterDonis said:
I'm not sure what the formula refers to, but since quantum mechanically, frequency is energy, restricting to positive frequencies means restricting to positive energies, which is generally accepted to be a good thing to do.

Right, see page 14 of Rubakov. Many quantum field theory books (e.g., Kaku, Ryder, Ticciati) explicitly write the integration measure as [itex]\delta \left(k^\nu k_\nu - m^2 \lright) \theta\left(k^0\right) d^4 k[/itex], where [itex]\theta[/itex] is the Heaviside step function. With [itex]m=0[/itex]. The delta function ensures that support is only on the ligh tcone, and the step function ensures that support is on the future light cone. See the first integral on page 14. It is interesting to do the [itex]k^0[/itex] integration.
dnquark said:
Note, by the way, that if we were to assume that the solutions live on a light cone, k0 = |k|, then k0 > 0 would also imply that all k components are > 0.

Why?

Suppose 1 = k0 = -k1 and 0 = k2 = k3.
 
George Jones said:
Right, see page 14 of Rubakov. Many quantum field theory books (e.g., Kaku, Ryder, Ticciati) explicitly write the integration measure as [itex]\delta \left(k^\nu k_\nu - m^2 \lright) \theta\left(k^0\right) d^4 k[/itex], where [itex]\theta[/itex] is the Heaviside step function. With [itex]m=0[/itex]. The delta function ensures that support is only on the ligh tcone, and the step function ensures that support is on the future light cone. See the first integral on page 14. It is interesting to do the [itex]k^0[/itex] integration.
We might have different editions; are you talking about the Fourier representation of Klein-Gordon field?

George Jones said:
The delta function ensures that support is only on the ligh tcone, and the step function ensures that support is on the future light cone. See the first integral on page 14. It is interesting to do the [itex]k^0[/itex] integration.

This I don't quite understand -- referring to [itex]k^\nu k_\nu = 0[/itex] as "light cone". To me, this is a relation between frequency and wavenumber, or energy and momentum. A "light cone" is an object in the spacetime domain. For instance, the (causal) wave equation Green's function in the time domain (something like [itex]\delta(t-|x|/c)/|x|[/itex] represents a future light cone, and the anti-causal branch represents the past light cone. But in the frequency domain, what's the motivation for associating positive frequencies with future and negative ones with the past?..

George Jones said:
...
Why?

Suppose 1 = k0 = -k1 and 0 = k2 = k3.
Because when it's late I make bizarre statements like that :) (you are obviously correct, that conclusion didn't make any sense)
 
Let me make my question more concrete: why does [itex]\omega > 0[/itex] refer to the future light cone?.. What is the relationship between this cone (a dispresion relation, really) in frequency-momentum space and the time dynamics?
 
dnquark said:
Let me make my question more concrete: why does [itex]\omega > 0[/itex] refer to the future light cone?.. What is the relationship between this cone (a dispresion relation, really) in frequency-momentum space and the time dynamics?

I'm guessing that that is done to ensure parity consistency. If that's the case then you could choose either positive [itex]\omega[/itex] or negative. But positive sounds more conventional (right hand rule?)

A positron is a reverse parity image of an electron with respect to time, isn't it? Electrons are traditionally of more interest - they are understood to have positive energy or positive motion or positive momentum.
 
Last edited: