Why should gluons move at a speed determined by $\mu_0$ and $\varepsilon_0$?

Question

I understand that the speed of light can be derived from Maxwell's equations, giving $c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}$

I furthermore understand how the principle of invariance of laws w.r.t. inertial reference frames gives rise to special relativity in order to preserve the above equation.

I am aware that gluons are theoretically massless and also travel at $c$.

I am also aware that the speed of light is considered to be the "speed of massless particles" or "the speed of information", but I'll get to that in a moment.

My question is: why should the speed of gluons by given by the electric and magnetic constants $\mu_0$ and $\varepsilon_0$? This connection seems sensible in the case of the photon, an electromagnetic particle, but why should this apply to the gluon as well?

I reject the "all massless particles" and "speed of information" answers as an explanation because they don't actually explain anything -- the situation is just as mysterious after these "answers" are given as before.

• If "all massless particles" is really the answer, then $c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}$ needs to explain how $\mu_0$ and $\varepsilon_0$ are derived from $c$, not the other way round. This is two new mysteries: firstly, how do we obtain $\mu_0$ and $\varepsilon_0$ from $c$ in a philosophically sound manner, and secondly, why should the classical derivation coincidentally obtain the same answer?

• If "speed of information" is the answer, we need both to supply a sensible fundamental definition of "information" and furthermore show that photons and gluons actually satisfy that definition. Then we still have the problem in the bullet point above.

Can anyone shed some light (ha ha) on this? How can we present these results such that the speed of the gluon is naturally given by electronic and magnetic constants, or how do we derive $\mu_0$ and $\varepsilon_0$ from some common concept?

Answer 1

I think you are looking upon this from the wrong direction. The definition of $\mu_0$ and $\epsilon_0$ is a just convenient way of setting the scale and units of $E$ and $B$.

The fundamental object to consider here is indeed $c$. Which brings us to the question: What would happen if gluons had a different speed of propagation $c_g\neq c$?

The short answer is that special relativity does not support two different speeds of light, as it would lead to inconstancies.

For instance if gluons moved with $c_g<c$ then you could switch to a reference frame where gluons would be stationary, which is a bit problematic.

Also if you have massive particles with associated speed of light $c_g<c$ you could choose a reference frame where thse where moving at the speed of light or even faster. These particles would then appear to have infinite amount of energy or even appear to be tachyons, which is even more problematic.

Answer 2

I'm of the school "all massless particles" as you've identified as to why $c$ shows up in Maxwell's equations, so I'm going to derive here how to arrive at $\mu_0$ and $\epsilon_0$ from $c$.

It's a fairly common viewpoint that $c$, like many other fundamental constants, acts as a unit conversion factor, giving a relation between time units and space units. A lot of equations are much more clear when written in a way consistent with their units. Lorentz transformations for example are much clearer when written in a unit appropriate form:

$$ct'=\gamma (ct-\frac{v}{c}x)$$

$$x'=\gamma (x-\frac{v}{c} ct)$$

Time values are written with a $c$ to make them consistent with space values, and velocities should be written relative to $c$. I'm going to follow a similar approach with Maxwell's equations. In their usual form, Maxwell's equations are a mess with respect to units.

$$\nabla \cdot E = \frac{\rho}{\epsilon_0}$$

$$\nabla \cdot B=0$$

$$\nabla \times E = -\frac{\partial B}{\partial t}$$

$$\nabla \times B = \mu_0 J + \mu_0 \epsilon_0 \frac{\partial E}{\partial t}$$

We have a few hints as to how to put this in a more unit consistent form. In the wave equation derived from Maxwell's equation, $B$ and $E$ are related by a factor of $c$, suggesting that $cB$ is a better choice than $B$ for consistency with $E$. We also choose to write our time values as $ct$. We also note that $c\rho$ forms a 4-vector with $J$. Rewriting Maxwell's equations:

$$\nabla \cdot E = \frac{c\rho}{c \epsilon_0}$$

$$\nabla \cdot cB=0$$

$$\nabla \times E = -\frac{\partial cB}{\partial ct}$$

$$\nabla \times cB = c\mu_0 J + \frac{\partial E}{\partial ct}$$

In this form, the quantities $c\mu_0$ and $1/c \epsilon_0$ stand out. These are actually equal:

$$c \mu_0 = \frac{\mu_0}{\sqrt{\epsilon_0 \mu_0}} = \sqrt{\frac{\mu_0}{\epsilon_0}}$$

Similarly:

$$\frac{1}{c \epsilon_0} = \frac{\sqrt{\epsilon_0 \mu_0}}{\epsilon_0} = \sqrt{\frac{\mu_0}{\epsilon_0}}$$

These are identified as the wave impedance.

$$\eta = \sqrt{\frac{\mu_0}{\epsilon_0}} \approx 377 \Omega$$

A final form of Maxwell's equations:

$$\nabla \cdot E = \eta (c\rho)$$

$$\nabla \cdot cB=0$$

$$\nabla \times E + \frac{\partial cB}{\partial ct}= 0$$

$$\nabla \times cB - \frac{\partial E}{\partial ct} = \eta J $$

In this form, it's clear that $\eta$ acts like a conversion factor between the charge quantities and variations in the fields, which the charge quantities generate. The constant $\eta$ is closely associated with electromagnetism, while $c$ is a more generic relation between the much more common time and space quantities. It's also possible to rewrite Maxwell's equations in terms of the less insightful quantities:

$$\epsilon_0 = \frac{1}{c\eta}$$

$$\mu_0 = \frac{\eta}{c}$$

Answer 3

I suggest that you write Maxwell's equations in (the physicists') Gaussian units, rather than (engineers') SI:

$$\nabla \cdot \mathbf{E} = 4\pi\rho$$ $$\nabla \cdot \mathbf{B} = 0$$ $$\nabla \times \mathbf{E} = -\frac{1}{c}\frac{\partial \mathbf{B}} {\partial t}$$ $$\nabla \times \mathbf{H} = \frac{4\pi}{c}\mathbf{J}_\text{f} + \frac{1}{c}\frac{\partial \mathbf{D}} {\partial t}$$

I think this should shed some light on your question.

Answer 4

Why should gluons move at a speed determined by μ₀ and ε₀?

There’s a bit of an issue with that. Have you ever seen a gluon? Do you know anybody who has? Take a look at the confinement section of the Wikipedia gluon article:

"Although in the normal phase of QCD single gluons may not travel freely, it is predicted that there exist hadrons that are formed entirely of gluons - called glueballs. There are also conjectures about other exotic hadrons in which real gluons (as opposed to virtual ones found in ordinary hadrons) would be primary constituents. "

Nobody has ever seen a gluon. The gluons in ordinary hadrons are virtual. And virtual particles "only exist in the mathematics of the model”. That's not to say there is no underlying reality. There is "gluonic field", and it is real. It's the thing that's there that keeps the proton in one piece. But it doesn't consist of a swarm of gluons popping in and out of existence and flying around inside a proton. This picture is the wrong picture:

A proton is not full of quarks and gluons ready to spill out like beans from a bag. Take a look at proton-antiproton annihilation. What you get is things like pions, which soon decay to things like muons, then electrons and neutrinos, and photons. We don't see gluons, we see photons. Keep that thought in mind:

_{Image credit CSIRO, see The Big Bang & the Standard Model of the Universe}

I understand that the speed of light can be derived from Maxwell's equations, giving $c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}$. I furthermore understand how the principle of invariance of laws w.r.t. inertial reference frames gives rise to special relativity in order to preserve the above equation.

The expression is not totally unlike the mechanics expression for the speed of a shear wave: v = √(G/ρ) where G is the shear modulus of elasticity and ρ is density. Some people will tell you that space is not a medium, but it is. Google on Einstein elastic. Note the shear stress term in the stress-energy-momentum tensor. Check out LIGO for gravitational waves. Whilst gravitational waves aren’t the same thing as electromagnetic waves, they are “waves of distorted space”, and that plus things like this should be enough to persuade you that space is genuinely modelled as an elastic continuum. On that basis permittivity is the reciprocal of G and something like “how easy is it to deform space” whilst permeability is something like “how well does it spring back”.

My question is: why should the speed of gluons by given by the electric and magnetic constants μ₀ and ε₀? This connection seems sensible in the case of the photon, an electromagnetic particle, but why should this apply to the gluon as well?

Because the photon has "gluonic attributes". Think about the bag model. Note that “one visualization is that of an elastic bag”. A proton stays in one piece because of the elastic gluonic field. When you annihilate it with the antiproton, there is no magic. There is surely no way in which one fundamental field or excitation magically switches off whilst another one magically switches on. IMHO it must be something like the electric field and the magnetic field, which are “are better thought of as two parts of a greater whole”. Think about a wave in a rubber mat. Shake a rubber mat, and a shear wave propagates because of an elastic tension that opposes the stress. This is like the elastic tension of the bag model. See page 5 of http://arxiv.org/abs/0912.2678 where Milgrom mentions the strength of space. Where does the strong force go in low-energy proton-antiproton annihilation to gamma photons? Nowhere. Because that’s the thing that makes those photons go at c. The photon field is elastic and gluonic. The energy that was a gluon is now a photon, and photons move at a speed determined by μ₀ and ε₀. Or put more simply: gluons move at c because they are massless bosons, and massless bosons are waves in space, and that's what waves in space do. But when they do, we don't call them gluons any more. We call them photons.