The Possibility of Faster-Than-Light Communication?

[Vanadium 50]

Are you familiar with the modern definition of μ0\mu_0 and e0e_0? Both have a fixed relationship to the fine-structure constant

I don't think they do, since the fine structure constant is quantum mechanical, and c is purely classical:

μ₀ is just the definition of the ampere. If one measures space and time in the same units (just as one measures northness and eastness in the same units), then ε₀ = 1/μ₀.

 

[Dale]

c0 has a specific value. It doesn't matter what units you use to describe it.

Actually, if you dig down into this concept you will find that what you actually mean is not related to the speed of light but to the fine structure constant. Think, what do you mean by the specific value of the speed of light if you don’t use units to describe it.

 

[ohwilleke]

I don't think they do, since the fine structure constant is quantum mechanical, and c is purely classical

The fine structure constant is quantum mechanical. c is just as much a part of quantum physics as it is a part of classical physics. Special relativity is observed in both Maxwell's equations and in QED.

 

[Nugatory]

Are you familiar with the modern definition of $\mu_0$ and $\epsilon_0$? Both have a fixed relationship to the fine-structure constant, so any “measurement” of either can serve only to refine our best known value for that quantity.

I don't think they do, since the fine structure constant is quantum mechanical, and c is purely classical:

I'm using (or misunderstanding?) the 2019 definitions. We have $\alpha=\frac{\mu_0}{2}\frac{e^2c}{h}$ with $e$, $c$, and $h$ all fixed, and we have $\epsilon_0=\frac{1}{\mu_0c^2}$. That leaves $\mu_0$ and $\alpha$ proportional to one another with a fixed constant of proportionality, and $\epsilon_0$ in a fixed relationship with $\mu_0$. Thus there is no freedom to vary them independently.

 

[Vanadium 50]

Sorry this took so long. By the time I thought it through, I couldn't find the message any more.

I think we can agree that the equation alone is a necessary but not sufficient condition for a physical relationship in it. In the case of alpha, we don't want to imply that there is a relationship between c and the number 2 for instance. So we need to understand where each term comes from.

The h comes in because there is no classical way to measure alpha: it is purely quantum-mechanical.

The first complication is that one can "hide" the h. In units where h is 1, one can replace the h with a plain old one. In other units, you can absorb it in the definition of μ₀. So one can write down what looks like a purely classical α, even though it is really quantum mechanical.

Are there any parts that can't be hidden? Yes, the e². Not only is it present no matter what units you are using, every measurement of α on objects of charge q₁ and q₂ has a q₁ and q₂ in it. So that's real. And, to jump to A-level for one line, if you are discussing different fundamental forces, they have different coupling constants (α is the coupling constant for electromagnetism) because they have different elementary charges. So α is essentially the electric charge of the electron (squared).

Let me slightly rewrite what you wrote: \alpha \sim \frac{e^2}{\epsilon_0}. This is a force times a distance squared. You can sort of see where this is going to cause you trouble relativistically, because distance is not an invariant, and α, being just a number, must be. So the reason the c is there comes into the fact that by convention we measure space and time in different units. (To jump to an I.5 level for one line, this is actually wrapped up in the definition of ε₀; it comes from Coloumb's Law, where you have a Force (which has a time derivative) on one side of the equation and something purely spatial on the other)

Had we worked in conventional units for airplane flight, where horizontal distances are measured in nautical miles and vertical distances in 100's of feet, we'd see the constant 60.76 appearing in these definitions as well.

So that's why I don't consider α a measure of c.

 

[Dale]

So that's why I don't consider α a measure of c.

I agree that $\alpha$ is not a measure of c, but when people actually express what they mean by statements like “c has a specific value regardless of the units used to describe it” what they are looking at is in fact the fine structure constant.

 

[Vanadium 50]

when people actually express what they mean by statements like “c has a specific value regardless of the units used to describe it” what they are looking at is in fact the fine structure constant.

I think that when people write that, they either don't believe in relativity, or they are completely confused.

Historically we measured time in seconds and length in meters. Also, for airplanes, we measure horizontal distances in nautical miles and vertical distances in 100-feet increments. In one case, you need c's and in the other you need 60.76's. They serve exactly the same function.

 

[Meir Achuz]

Let's start with permeability. Do you think the 4π is a) an artifact of the definition of units or b) a physical constant that just accidentally happens to be 4π?

Muzero is just a change of units like the fundamental constant 5280.
muzero/4pi=10^-7 to change the units from the original emu system to SI.
10^-5 comes from changing from cgs to MKS units. The 10^-2 comes from the change (in 1880 something) from the Absolute Ampere (defined by the force between two wires) to the SI Ampere. I think that was done because the telegraphers defined the ohm for their convenience.
If you changed 5280, an American could run faster.

 

[NBrown]

Somehow I stumbled across this question, that was posted 3 years ago. It is a brilliant, clear and simple question in search of an equally straightforward answer. Most of the answers show that there are many people around that have some level of grasp of modern physics, and that are willing to talk at length around it, but I found no good, direct, clear, simple and plausible answer. I do not believe that there is in principle no simple and elegant answer. I also don’t have it - and I don’t know whether science has it (yet). @TerranIV, if you have in the meantime received a satisfying answer, I would be intrigued to know what it is.

 

[PeroK]

Somehow I stumbled across this question, that was posted 3 years ago. It is a brilliant, clear and simple question in search of an equally straightforward answer. Most of the answers show that there are many people around that have some level of grasp of modern physics, and that are willing to talk at length around it, but I found no good, direct, clear, simple and plausible answer. I do not believe that there is in principle no simple and elegant answer. I also don’t have it - and I don’t know whether science has it (yet). @TerranIV, if you have in the meantime received a satisfying answer, I would be intrigued to know what it is.

I think @TerranIV was last seen looking for the company that built Nigel Tufnell's amplifier.

 

[Dale]

It is a brilliant, clear and simple question in search of an equally straightforward answer.

The straightforward answer is “no”.

 

[NBrown]

The straightforward answer is “no”.

Thanks, Dale, I already know your thoughts on the topic, but I was asking TerranIV if he had received a satisfying answer (meaning satisfying to him). I don’t expect this to be answered by anyone other than TerranIV.

 

[Dale]

Then I recommend you PM them. The public forums are for public commentary, not for a one on one discussion.

 

[Vanadium 50]

but I was asking TerranIV if he had received a satisfying answer

Then a PM would possibly be appropriate, but he hasn't been here in 3-1/2 years. But it sounds like you dislike the answers so far. As they say "Two plus two continue to make four, despite the whine of the amateur for three or the cry for the critic for five."

 

[gmax137]

I think @TerranIV was last seen looking for the company that built Nigel Tufnell's amplifier.

OMG I had to google that.