Why does massive carrier implies finite range?

In summary: I don't know how to calculate the range of interaction for the case of massive bosons.In summary, the professor explained that the original Weinberg-Salam-Glashow theory had to be augmented with the Higgs mechanism because it didn't give rise to massive gauge fields, yet since weak interaction had a finite range they knew it had to have a massive carrier. However, he was not able to provide a satisfactory explanation as to why the massless carrier implies a finite range.
  • #1
tomkeus
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Back on lecture about Standard model, professor said that original Weinberg-Salam-Glashow theory had to be augmented with Higgs mechanism because it didnt give rise to massive gauge fields, yet since weak interaction had finite range they knew it had to have a massive carrier, well three of them due to three SU(2) generators. When I asked him why massive carrier implies finite range, and massless infinite, I didnt get satisfactory (any) explanation.

So, can anyone please shed any light on this subject?
 
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  • #2
tomkeus said:
When I asked him why massive carrier implies finite range, and massless infinite, I didnt get satisfactory (any) explanation.
This is a very important fact. You are right to search for a rigorous answer.

The "bad reason" why this is important is technical. You should be able to calculate the associated potential using Cauchy's theorem.

The "good reason" is conceptual. Gauge invariance imposes the photon to be massless, but this is quite annoying to fix gauge invariance. A useful trick is to set a finite mass, that physically you claim is to small to be observable. Do your calculations, and set m=0 at the very end. It is a quite non-trivial property that this procedure will still respect gauge invariance at the end of the day. But many people use it.

Do you have Zee's "QFT in a nutshell" ? He discusses about all this at the conceptual level. This is a very good introductory book.

So the thing is something like : calculation of the massive scalar potential with a point source at the origin [tex]\rho=g\delta(\vec{x})[/tex]
Klein-Gordon : [tex]\partial_{\mu}\partial^{\mu}U+m^{2}U=\rho[/tex]
Time independence : [tex](-\Delta^{2}+m^2)U=g\delta(\vec{x})[/tex]
Fourier transform : [tex]U(\vec{x})=\frac{g}{2\pi^3}\int d^3k\frac{e^{i\vec{k}\vec{x}}}{\vec{k}^2+m^2}[/tex]
Go to polar coordinate, trivial angular integration, radial integration done be closing the contour (say) in the positive imaginary part you pick up the pole at [tex]k=im[/tex]. This is the important technical part. Details upon request (try it before !).
Finally
[tex]U(\vec{x})=\frac{g}{4\pi}\frac{e^{-mr}}{r}[/tex]
This is what we mean by finite range, because exponential is decreasing fast. Setting a vanishing mass, you recover the Coulomb potential :smile:
 
  • #3
I guess what you meant is that if the intermedia boson is massive, such as W-boson, then the related interaction should be short-ranged. For W,Z+,Z-, it is weak interaction.
Well~~why~~I am not sure. We can calculate the propogators in coordinate-space, and use the C-S equation to get the details.
May the reason is that these massive bosons decay during the propogating process due to their lower velocities than that of Light. As we know, "All three particles(W,Z+,Z-) are very short-lived with a mean life of about 3×10^-25 s".
 
  • #4
tomkeus said:
Back on lecture about Standard model, professor said that original Weinberg-Salam-Glashow theory had to be augmented with Higgs mechanism because it didnt give rise to massive gauge fields, yet since weak interaction had finite range they knew it had to have a massive carrier, well three of them due to three SU(2) generators. When I asked him why massive carrier implies finite range, and massless infinite, I didnt get satisfactory (any) explanation.

So, can anyone please shed any light on this subject?

A conceptual argument is from uncertainty principle. [tex] \Delta E \Delta t \sim \hbar [/tex]
Like what we often said, the higher the energy of the accelerator, the shorter the distance one can probe.
So, if an intermediate boson has exactly zero energy, then the range of the interaction should be infinite, this is the case of QED. However, if the intermediate boson has finite mass, we take the deviation [tex] \Delta E [/tex] to be the mass of the boson, then we get a finite range of the interaction.
However, although the intermediate bosons, gluons, are massless, the range of QCD is still short-ranged due to confinement.
I tried to look at the range of interaction from the propagator in coordinate space. For the massless case, I can work out the Fourier transform and the result is 1/x^2. But for the case of massive boson, the propagator of real space is kinda complicated, I can't work out the integral.
 

FAQ: Why does massive carrier implies finite range?

1. Why does a massive carrier have a finite range?

A massive carrier refers to a particle that has a rest mass, such as an electron or a proton. Due to the inherent mass of these particles, they are subject to the laws of physics that govern motion and energy. One of these laws is the conservation of energy, which states that energy cannot be created or destroyed, only transferred or converted. In order for a massive carrier to travel a certain distance, it requires a certain amount of energy. However, as the carrier moves further away from its source, it loses energy due to factors such as resistance and interactions with other particles. Eventually, the carrier will run out of energy and come to a stop, resulting in a finite range.

2. Can a massive carrier have an infinite range?

No, a massive carrier cannot have an infinite range. As explained in the previous question, the inherent mass of the carrier limits its ability to travel a certain distance. Even if the carrier were to travel in a vacuum with no external forces, it would eventually come to a stop due to the conservation of energy.

3. How does the mass of a carrier affect its range?

The mass of a carrier directly affects its range. The more massive the carrier is, the more energy it requires to travel a certain distance. This means that a carrier with a larger mass will have a shorter range compared to a carrier with a smaller mass. This is evident in the case of particles such as electrons and protons, where the more massive proton has a shorter range compared to the less massive electron.

4. Are there any exceptions to the finite range of a massive carrier?

Yes, there are exceptions to the finite range of a massive carrier. One example is a photon, which is a massless particle. Being massless means that a photon is not subject to the laws of energy conservation and can travel an infinite distance without losing energy. However, it should be noted that even though a photon's range is theoretically infinite, it can still interact with other particles and be absorbed or scattered, thus limiting its effective range.

5. What are the implications of a massive carrier having a finite range?

The finite range of a massive carrier has numerous implications in various fields of science. In particle physics, it limits the range of interactions between particles, thus affecting the fundamental forces that govern the universe. In communication and technology, it affects the range and efficiency of wireless signals. In astrophysics, it plays a crucial role in understanding the behavior and interactions of massive objects such as stars and galaxies. Overall, the finite range of a massive carrier is a fundamental concept that has implications in many areas of study.

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