I What happens if you increase μ0 and decrease ϵ0? Or vice versa

[Vanadium 50]

. If a car is going 100 mph and I change it so that it is moving 50 mph then that is not just a change of units. I didnt redefine 100 mph to be 50 mph. I actually changed the cars motion.

If you said you were changing μ (e.g. by picking another material) that would be an OK analogy. But you didn't. You said you were changing μ₀. That's not changing the speed of one car, that's changing the entire universe. And what people are telling you repeatedly is that this has to be done consistently. Furthermore, when you do it consistently you redefine your system of units, and in many cases you end up right where you started.

 

[em3ry]

You seem to think that you are telling me something that I don't know

I have been thinking about the physical significance of Planck's constant.

The effect of increasing Planck's constant on blackbody radiation is the red line below. Y-axis is frequency

Apparently if Planck's constant were infinite then there would be no blackbody radiation. But we know that as long as the speed of light is finite then accelerating charges will emit light.

Wikipedia says:
So that got me thinking. The speed of light is $$\sqrt{\frac{1}{μ0 ϵ0}}$$. So that means that if you increase μ0 and decrease ϵ0 by the same amount then the speed of light would be the same but surely something in the universe would change.

So what would change if you increased μ0 and decreased ϵ0 by the same amount so that the speed of light was the same as before? Surely this would have some effect on the universe.

My question is about μ0 and ϵ0 but if you have any insight into Planck's constant then I will be glad to hear it too.

edit: I see that the Larmor formula depends on ϵ0 and c but not μ0. Decreasing ϵ0 increases the energy radiated by an accelerating charge.

$$P = {2 \over 3} \frac{q^2 a^2}{ 4 \pi \varepsilon_0 c^3}$$

edit2: Dale mentioned the Fine structure constant in post 1 below. It is:
$$\alpha = \frac{1}{4 \pi \varepsilon_0} \frac{e^2}{\hbar c} = \frac{\mu_0}{4 \pi} \frac{e^2 c}{\hbar}$$

edit3: The force between two separated electric charges with spherical symmetry (in the vacuum of classical electromagnetism) is given by Coulomb's law:
$$F_\text{C} = \frac{1} {4 \pi \varepsilon_0} \frac{q_1 q_2} {r^2}$$

comparing to the equation for the Fine structure constant we get:
$$F_\text{C} r^2 = constant = \alpha \hbar c$$

 

[em3ry]

I can see what's going on here. You people aren't even discussing my thread anymore. You are just continuing some argument that you were having between yourselves before i got here. Maybe its time for me to leave.

 

[etotheipi]

No need to throw a hissy fit, please. I think @Dale's post #2 and @Vanadium 50's post #7 sum up pretty well the right response to your question, and it's not really clear what you're trying to achieve.

Maybe, it would be worth getting hold of a copy of Sommerfeld's lectures on E&M (I think Vol. 3), where there is some nice discussion of electromagnetic units, in order to clear up some misconceptions.

 

[Ibix]

I can see what's going on here. You people aren't even discussing my thread anymore. You are just continuing some argument that you were having between yourselves before i got here. Maybe its time for me to leave.

Everyone seems to me to be pointing out that you can't just change $\mu_0$ without changing at least one other constant. Depending which other constant you change you may see genuinely different physics (if it's a dimensionless constant) or just unit redefinition (if it's another dimensionful constant). It isn't clear to me, at least, that you've accepted that point.

 

[em3ry]

so your answer to my question is that nothing will change?

if that is your answer then why don't you just say that? Like I say it's clear that you are just continuing some argument that you were having before I got here

 

[Ibix]

so your answer to my question is that nothing will change?

No. The answer is "maybe". As I just said, it depends what else you change, and I don't believe you've yet specified that.

 

[em3ry]

I specified it here:

I guess the place to start would be to increase $\mu$ while leaving everything else the same. $\alpha$ would increase the same amount. If c is unchanged then $\epsilon$ would have to decrease

Of course that might not be possible if the other factors are themselves functions of $\mu$. I think some of them might be.

 

[em3ry]

I probably shouldn't say this since it's irrelevant to my question and it's just going to get me sucked into your arguing but it occurs to me that any equation can be turned into a dimensionless constant k

a = b becomes:
k = b/a where k=1

A less trivial equation is the Larmor formula

$$P = {2 \over 3} \frac{q^2 a^2}{ c^3}$$

$$ 1.5 = \frac{q^2 a^2}{ P c^3}$$

That particular constant might not be variable though

so I guess the defining feature of a dimensionless constant is not just that it is constant and dimensionless but also that it's particular value is arbitrary in the sense that it's value can't be determined by the other values in the equation

 

[Ibix]

I specified it here:

Ok - I missed that this morning, apologies.

If you vary $\alpha$ then this does have real physical effects. The fine structure constant governs how strong the EM field is compared to other forces, so you'd see different spectral lines in atomic emissions among other things. There's a (probably non-exhaustive) list on the fine structure constant Wikipedia page.

 

[em3ry]

I asked what would change if the value of mu change. You people have spent 3 pages hammering away at me to get me to say "ok let's change the value of alpha too" as though that were somehow the most important thing in the universe. I don't care about alpha. If need to change alpha to change mu then change alpha. If you need to change the price of yoyos in china to change the value of mu then change the price of yo-yo's in china. I don't care. I just want to know the answer to my question. Lol.

Since the equation for alpha contains mu I would have thought it rather trivial that we must change alpha too. You people are the ones making a big deal out of nothing. You are clearly continuing some bizarre argument that you have been having between yourselves.

 

[Ibix]

So did you read the list of things that the fine structure constant controls?

 

[em3ry]

A very interesting list indeed

Alpha is related to the probability that an electron will emit or absorb a photon

the fine-structure constant is one fourth the product of the characteristic impedance of free space, Z0 = μ0c, and the conductance quantum

 

[em3ry]

I tend to think of $\epsilon$ and $\mu$ as defining the strength of the electromagnetic force. Alpha then relates that to various other things

$$\\ \alpha = \frac{1}{4 \pi \varepsilon_0} \frac{e^2}{\hbar c} = \frac{\mu_0}{4 \pi} \frac{e^2 c}{\hbar} = \frac{k_e e^2}{\hbar c} = \frac{e^2}{2\varepsilon_0 c h} = \frac{c \mu_0}{2 R_K} = \frac{e^2 Z_0}{2h} = \frac{e^2 Z_0}{4\pi \hbar}$$

I tend to think of Alpha as the speed of the electron in the Bohr model. But since Planck's constant exists outside of the Bohr model then perhaps alpha does too.

 

[Dale]

I tend to think of ϵ and μ as defining the strength of the electromagnetic force.

That is actually a better way to think of $\alpha$.

So I dug up my old table. If $\alpha$ increases then material objects will get longer when measured in meters. In other words, the length of a metal bar of a certain number of atomic lengths as measured using an atomic clock and a pulse of light will increase with increasing $\alpha$. Similarly, the duration of a pendulum tick will increase when measured by an atomic clock if $\alpha$ increases.

I asked what would change if the value of mu change. You people have spent 3 pages hammering away at me to get me to say "ok let's change the value of alpha too" as though that were somehow the most important thing in the universe. I don't care about alpha. If need to change alpha to change mu then change alpha.

Then you have indeed missed the point. Only changing $\alpha$ changes the physics.

If you change only $\mu_0$ and $\hbar$ leaving everything else constant then there is no physical impact at all. You have merely changed your units away from SI. If you change only $\mu_0$ and $c$ you have again only changed units with no physical changes.

If you change only $\alpha$ and $\mu_0$ then you have kept your SI units the same but you have made real physical changes as I described above. If you change only $\alpha$ and $\hbar$ then you have both changed the physics and your units. Furthermore, if the change in $\alpha$ is the same as before then the physical changes will be the same. The physical changes are purely determined by $\alpha$, not by $\mu_0$.

You people aren't even discussing my thread anymore. ...
I don't care. I just want to know the answer to my question. Lol.

I am a little unclear why you are saying this. I have answered your question. I have talked about how the change you asked about would alter nuclear stability, fusion, length measurements, and time measurements. It is certainly not an exhaustive list, but it is a good start.

Your question required some clarification. I think that you still are missing the importance of that point and that you don’t really see why it was important to clarify. But once it was clarified we have indeed answered your question too.

 

[em3ry]

I asked what changes when you change mu. You keep asking me over and over what else I want to change.

If something depends on mu then I want to change it.
If something doesn't depend on mu then I don't want to change it.

If alpha depends on mu then change it too.
If alpha doesn't depend on mu then don't change it.

If nothing in the universe changes then the answer to my question is "nothing changes"

You want to make this a philosophical discussion about dimensionless constants.
I DONT CARE

I just want to know what changes if you change mu. Why is this so hard for you to understant?

 

[hutchphd]

You want to make this a philosophical discussion about dimensionless constants.
I DONT CARE

I will now retract my previous statement about your questions not being foolish

 

[Dale]

You keep asking me over and over what else I want to change.

Yes, because something else must change. And the result of changing $\mu_0$ depends entirely on what else changes.

If something depends on mu then I want to change it.
If something doesn't depend on mu then I don't want to change it.

Unfortunately the universe doesn’t work the way you want it to. None of these things depend on $\mu_0$.

You can change $\mu_0$ without changing $\alpha$. You can change $\mu_0$ without changing $\hbar$. You can change $\mu_0$ without changing $c$. You can change $\mu_0$ without changing $e$. So none of these quantities actually depend on $\mu_0$.

But you cannot change $\mu_0$ without changing at least one of $\alpha$, $\hbar$, $c$, or $e$. So you must specify what else changes. Nothing else automatically changes but something else must change. This is not a part of the question that can be avoided.

You have chosen to also change only $\alpha$, which is a good choice since it is the simplest choice with some physical consequences. And I have described the result of that. So I have indeed answered your question fully.

You want to make this a philosophical discussion about dimensionless constants.
I DONT CARE

You entirely misunderstand. I dislike philosophical discussions. This is not philosophical at all, it is purely physical. You want the physical world to behave in a way that it simply does not, and no amount of all-caps complaints will change that.

Your question requires a specification of what else happens. If you are unwilling to make that specification then you simply do not have a well-formed question. Such a question cannot be answered, not even with “nothing happens”. This is not a philosophical issue, it is a physical requirement.

“What happens if $\mu_0$ increases?” is not a well formed physical question. It has no answer at all because it is incomplete.

“What happens if $\alpha$ and $\mu_0$ both increase together and no other constants change?” is a well formed physical question with the answer I have given. As near as I can tell this is the question you have asked, and I have answered.

 

[em3ry]

Ok. I agree with that. I guess I owe you an apology then.