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

In summary, the effect of increasing Planck's constant on blackbody radiation is the red line below. Y-axis is frequency.
  • #36
em3ry said:
Perhaps this is why the physical significance of planks constant is obscure.
For the purpose of your question, have you decided how the other quantities in that equation will change to keep the equation true? Until you do so the question is incomplete.

Also, for your reference “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.626 070 15 x 10–34 when expressed in the unit J s, which is equal to kg m2 s–1, where the metre and the second are defined in terms of c and DeltanuCs.” So this is how Planck’s constant is tied to the SI units.
 
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  • #37
em3ry said:
There doesn't seem to be a Larmor formula for absorbance
The microscopic processes of Electromagnetism and Mechanics care not about the direction of time. If you run time (including the boundary conditions) backwards, Larmor will describe absorbance. This forms the basis for the principle of detailed balance and Einstein's A and B coefficients for light in equilibrium.

It might be a good idea to try to understand something before opining.

.
 
  • #38
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.
 
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  • #39
hutchphd said:
The microscopic processes of Electromagnetism and Mechanics care not about the direction of time. If you run time (including the boundary conditions) backwards, Larmor will describe absorbance. This forms the basis for the principle of detailed balance and Einstein's A and B coefficients for light in equilibrium.

It might be a good idea to try to understand something before opining.

I understood that perfectly but unlike the emission process the absorption process depends on chance encounters with other particles at exactly the right moment when the incoming light wave arrives.

Breaking something apart is the exact opposite of putting it together but putting it back together tends to be much harder
 
  • #40
Yes, and this is all in textbooks. So study them please. Your questions are not foolish but they are elementary. Do the work.
 
  • #41
That is what I am trying to do
 
  • #42
em3ry said:
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.
That is, I think, the best approach.

One of the most important effects of this is to change the stability of nuclei. The repulsive forces between protons will increase relative to the nuclear forces, so that heavier nuclei that are currently stable will become unstable. This will vastly change chemistry.

Also, nuclear fusion will be more difficult since the Coulombic barrier will be relatively higher. So stars and the stellar life cycle will change. There will be less synthesis of heavier elements, like carbon. That will have obvious implications for life.

I have a table that I made once with other effects. I will see if I can find that tomorrow. One thing that I know changed was the length of physical objects relative to electromagnetic radiation, but I cannot remember which way it changed.
 
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  • #43
em3ry said:
That is what I am trying to do
No you are expecting answers to questions you have not even thought about for a minute. Do the work. Why are you talking about ##\mu_0##? What happened to Planck?
 
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  • #44
Wow.
 
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  • #45
hutchphd said:
No you are expecting answers to questions you have not even thought about for a minute. Do the work. Why are you talking about ##\mu_0##? What happened to Planck?
I don't think his questions are so easy to answer. Furthermore the answers might not lie in his books, at least not explicitly.
 
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  • #46
I've not read the entire thread, but the answer was never as simple as today, because the International System of Units (SI units) have been finally defined by giving the fundamental constants a definite value. The only exception is the gravitational constant which cannot be measured accurately enough. That's why the only exception is the definition of the second, which still uses a material constant, namely the frequency of a certain hyperfine transition of the ##^{133}\text{Cs}## atom, i.e., one fixes this transition frequency ##\Delta \nu_{\text{Cs}}## to a certain value, defining the unit second for time.

All other units are defined by giving the fundamental constants ##c## (speed of light in vacuo), ##h=\hbar/(2 \pi)## (Planck's action), ##e## (elementary charge), ##N_{\text{A}}## (Avogadro number), and ##k_{\text{B}}## (Boltzmann constant) definite values. This shows that all the numbers in the SI units are just determined by defining the units.

In electromagnetism the constants ##\epsilon_0## and ##\mu_0## are now both derived quantities (in the SI before 2019 ##\mu_0## was fixed by the then valid definition of the Ampere, but now the Ampere is defined by the definition of the elementary charge and the second.

For details see

https://en.wikipedia.org/wiki/2019_redefinition_of_the_SI_base_units
 
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  • #47
This is why we should just use Lorentz-Heaviside units when doing EM 😜
 
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  • #48
Well, for electricians the HL units were pretty inconvenient ;-)). Of course, in theoretical physics HL units are the best choice.
 
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  • #49
@vanhees71 do you agree with what Dale says, that changing the dimensionful constants just changes the system of units we are using and to see physical changes we have to change the dimensionless constants?
 
  • #50
Yes, and it depends also on the system of units used, which "constants" you need. E.g., formulating classical (microscopic) electrodynamics in terms of the SI you apparently need two independent constants of this kind, for which usually ##\mu_0## and ##\epsilon_0##. Both have no deeper physical meaning than to introduce an extra unit for electric charge or electric current in addition to the three basic units for time, length, and mass which is all you need in classical physics (besides temperature for thermodynamics). These constants (permeability and permittivity of the vacuum or magnetic and electric field constant) are related to the speed of light by ##c=1/\sqrt{\epsilon_0 \mu_0}##.

Note that these constants have changed with the new system of units. This is due to the redefinition and the particular choice of the values for ##\nu_{\text{Cs}}##, ##c##, ##e##, and ##h##. These are, of course, chosen such that, as to the best of our abilities to measure them in terms of the older definitions of the SI units, these units change as little as possible. In fact afaik the largest change is indeed in the realm of the electromagnetic units, i.e., in the value for ##\mu_0## which was in the old system exactly ##4 \pi \cdot 10^{-7} \text{N}/\text{A}^2##. The relative change is ##\delta \mu_0/\mu_0 \simeq 1.5 \cdot 10^{-10}##.

From the point of view of relativity the older electromagnetic systems of units, the Gaussian system or its "rationalization" the Heaviside-Lorentz system, are more natural, and there you need only one fundamental constant. Here one usually uses the speed of light ##c##. It's more natural, because you measure electric and magnetic field components (which relativistically are combined to the Faraday or field-strength four-tensor ##F_{\mu \nu}##). The reason, why there's only one "conversion constant" is that one does not introduce an additional base unit for electric charge or current.
 
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  • #51
em3ry said:
. 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.
 
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  • #52
You seem to think that you are telling me something that I don't know

em3ry said:
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

latest


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:
$$F_\text{C} r^2 = constant = \alpha \hbar c$$
 
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  • #53
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.
 
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  • #54
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.
 
  • #55
em3ry said:
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.
 
  • #56
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
 
  • #57
em3ry said:
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.
 
  • #58
I specified it here:
em3ry said:
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.
 
  • #59
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
 
  • #60
em3ry said:
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.
 
  • #61
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.
 
  • #62
So did you read the list of things that the fine structure constant controls?
 
  • #63
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
 
  • #64
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.
 
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  • #65
em3ry said:
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.

em3ry said:
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##.

em3ry said:
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.
 
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  • #66
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?
 
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  • #67
em3ry said:
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
 
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  • #68
em3ry said:
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.

em3ry said:
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.

em3ry said:
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.
 
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  • #69
Ok. I agree with that. I guess I owe you an apology then.
 
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