B What would happen if the speed of light were different?

[PeterDonis]

Where can I see the high energy SM Lagrangian?

See, for example, here:

https://en.wikipedia.org/wiki/Electroweak_interaction#Lagrangian

A quadratic term of Higgs field is allowed.

As I understand it, it isn't, because before EW symmetry breaking, the Higgs field is not a single scalar field. It's a complex SU(2) doublet, and a direct quadratic mass term would break SU(2) gauge invariance.

 

[nikkkom]

See, for example, here:

https://en.wikipedia.org/wiki/Electroweak_interaction#Lagrangian

As I understand it, it isn't, because before EW symmetry breaking, the Higgs field is not a single scalar field. It's a complex SU(2) doublet, and a direct quadratic mass term would break SU(2) gauge invariance.

In that article, L_h actually has a term quadratic in Higgs field h, with $\lambda v^2$ dimensionful coefficient. THAT is exactly the coefficient I was talking about, (88.45 GeV)².

 

[PeterDonis]

In that article, Lh actually has a term quadratic in Higgs field h, with λv2\lambda v^2 dimensionful coefficient.

No, $\lambda v^2$ is not a Higgs quadratic mass term. $v$ is not the Higgs scalar, it's a constant in the potential for the Higgs field. The value of $\lambda$ is not the Higgs mass squared; the "after electroweak symmetry breaking" shows the Higgs mass term with $m_H$.

 

[nikkkom]

No, $\lambda v^2$ is not a Higgs quadratic mass term. $v$ is not the Higgs scalar, it's a constant in the potential for the Higgs field.

Indeed, $\lambda v^2$ is not the full Higgs quadratic mass term. It's a part of it - its constant coefficient.
Please do follow the link in question and see how L_h is defined there:

https://en.wikipedia.org/wiki/Electroweak_interaction#Lagrangian

Expand the squared parentheses, and you'll get a quartic term, a quadratic term, and a constant one. Quadratic one ends up with dimensionful coefficient.

 

[PeterDonis]

Quadratic one ends up with dimensionful coefficient.

Ah, I see. Yes, this is correct. This dimensionful coefficient ends up determining the vacuum expectation value of the Higgs field after spontaneous symmetry breaking. But the value of this coefficient is not the same as the mass $m_H$ that we observe for the Higgs boson now; the field $h$ in the high energy Lagrangian is not the same as the field $H$ in the low energy Lagrangian.

 

[nikkkom]

This dimensionful coefficient ends up determining the vacuum expectation value of the Higgs field after spontaneous symmetry breaking. But the value of this coefficient is not the same as the mass $m_H$ that we observe for the Higgs boson now

Sure. They are related via $\mu^2 = m_H^2/2 = \lambda v^2$. My point is that SM has one *dimensionful* parameter (while other 18+ parameters are all dimensionless).

 

[PeterDonis]

My point is that SM has one *dimensionful* parameter

Yes, I agree. How it appears in the Lagrangian differs depending on whether you are in the high energy or low energy regime. I'm not sure I would call it a "coupling constant" in either regime, since $v$ is a term in the potential and $m_H$ is a mass, which appears in a quadratic term that involves only one field, not multiple fields. But there is a dimensionful parameter, yes. By the reasoning I gave earlier in the thread, this would be an indication that the SM is not fundamental, but emergent from a lower level theory.

 

[nikkkom]

That's different from this post:

The currently accepted explanation for this is that the Higgs mass is emergent from spontaneous electroweak symmetry breaking; at high enough energies (as in the very early universe, before electroweak symmetry breaking occurred), all of the Standard Model fields are massless.

Here, you said that Higgs mass is emergent *in SM*. It is not. Higgs mass is a direct consequence of $\mu^2$ coefficient.

I agree that SM may be supplanted by an extended theory without dimensionful parameters. Koide rule and other mass relations are signs of it.

 

[Buzz Bloom]

Higgs mass is a direct consequence of μ² coefficient.

Hi nikkkom:

I would much appreciate seeing a generally accessible reference which explains the above quote.

The closest discussion I can find on the Internet regarding μ² is

https://en.wikipedia.org/wiki/Coefficient_of_variation .​

However, this seems to be completely unrelated to the Higgs mass.

Regards,
Buzz

 

[PeterDonis]

Here, you said that Higgs mass is emergent *in SM*. It is not.

I think this is a matter of terminology. The dimensionful coefficient that appears in the SM Lagrangian at high energy is not called the "Higgs mass" and is not described as a "mass" of the Higgs field; descriptions of the SM at those energies typically say, as I said, that all of the fields are massless there. But you are correct that there is still a dimensionful coefficient at those energies, with units of mass, that does appear in a quadratic term, so the usual terminology is sloppy.

 

[nikkkom]

Hi nikkkom:

I would much appreciate seeing a generally accessible reference which explains the above quote.

The closest discussion I can find on the Internet regarding μ² is

https://en.wikipedia.org/wiki/Coefficient_of_variation .​

However, this seems to be completely unrelated to the Higgs mass.

See $\mu^2$ in the definition of L_H here in section "The Higgs mechanism":

https://en.wikipedia.org/wiki/Mathematical_formulation_of_the_Standard_Model#The_Higgs_mechanism

 

[nikkkom]

I think this is a matter of terminology. The dimensionful coefficient that appears in the SM Lagrangian at high energy is not called the "Higgs mass" and is not described as a "mass" of the Higgs field; descriptions of the SM at those energies typically say, as I said, that all of the fields are massless there. But you are correct that there is still a dimensionful coefficient at those energies, with units of mass, that does appear in a quadratic term, so the usual terminology is sloppy.

You are right. $\mu$ is not the Higgs mass. Higgs mass is $\sqrt {2}\mu$.

 

[Buckethead]

if the fine structure constant changes, it is not "due to" a change in c, e, or h. Which of c, e, and h change if the fine structure constant changes is a matter of choice of units, not physics. The physics is all in the fine structure constant.

I would very much like to understand the meaning of this quote. I do not want to introduce philosophy, so I will just mention briefly what I perceive to be the problem with my mental ability to understand this quote. It seems to have logical implications that contradict my philosophical view of reality.

I assume for the purpose of discussing the OP's question that that the speed of light changes, and also the fine structure constant changes correspondingly based on
α = 2π e2 / h c,that is, c and α vary reciprocally assuming no changes in e and h.

I think I might be able to help here. (or I must just screw things up...we'll see)

The secret to resolving this difficulty is to keep in mind is that the fine structure constant being dimensionless, is a specific value in all units. i.e. it is 1/137 in CGS and in SI units. Therefore your selection of units will change the value in different ways if you change the value of one single constant (such as c). In other words, in one system if you double the value of c then the fine structure constant would let's say double, but in another system if you double c then the fine constant value might triple because it has no units.

To make this even clearer. Image you have the equation c=c in SI units. Now in this case both the left side and the right side have units. c is not dimensionless. So if you change the units from SI to CGS then c=c is still true. now take another equation, one where the left side has no units (just like a dimensionless constant) such as 1=c. Now you have a problem. if you use units of c=1 light second per second, then the equation is true. However if you use CGS units for the value of c, then the equation is no longer true since 1 does not equal 299,792,458.

(this statement might need correcting) What this implies is that in a dimensionless constant, changing one value would give different results depending on the units of measurement when measuring that one constant such as c or h or e. Since any units of measurement is arbitrary (you can make up any system of units you want just by changing the definition of a second or a meter for example) you can arbitrarily change the value of the fine structure constant for any given change in one of the constant values rendering the exercise completely useless.

I hope that helps. And if so I'm glad to be able to help instead of always being the one to ask the question.

 

[PeterDonis]

The secret to resolving this difficulty is to keep in mind is that the fine structure constant being dimensionless, is a specific value in all units. i.e. it is 1/137 in CGS and in SI units.

Yes.

Therefore your selection of units will change the value in different ways

You realize this contradicts what you just said, right? If the value is the same in all units, your selection of units can't change the value.

The correct thing to say is that, because the fine structure constant is 1/137 in any system of units, if you change units such that the value of $c$ changes (say from 299,792,458 to 1, because you're switching from SI units to "natural" relativity units), the values of $e$ and/or $h$ must also change, so as to keep the fine structure constant's value the same, 1/137. In other words, it's impossible to take some system of units, and from it construct another system of units where the value of $c$ is different but everything else is the same.

now take another equation, one where the left side has no units (just like a dimensionless constant) such as 1=c

This is not correct. $c$ is a speed, not a dimensionless number. The fact that we can choose units where $c = 1$ does not mean $c$ is dimensionless in those units. A dimensionless number, like the fine structure constant, is dimensionless in all systems of units.

this statement might need correcting

Your instincts here are sound, unlike in the rest of your post. See above.

I hope that helps. And if so I'm glad to be able to help instead of always being the one to ask the question.

Unfortunately, I don't think your comments are helping because you're confused about the actual issue. See above.

 

[Buckethead]

You realize this contradicts what you just said, right? If the value is the same in all units, your selection of units can't change the value.

That was sloppy of me. I meant to say, "Therefore your selection of units will change the value (of the fine structure constant) if you change the value of a constant on the right side of the equation". But your way of saying it is much better.

The correct thing to say is that, because the fine structure constant is 1/137 in any system of units, if you change units such that the value of cc changes (say from 299,792,458 to 1, because you're switching from SI units to "natural" relativity units), the values of ee and/or hh must also change, so as to keep the fine structure constant's value the same, 1/137. In other words, it's impossible to take some system of units, and from it construct another system of units where the value of cc is different but everything else is the same.

now take another equation, one where the left side has no units (just like a dimensionless constant) such as 1=c.

This is not correct. c is a speed, not a dimensionless number. The fact that we can choose units where c=1 does not mean c is dimensionless in those units. A dimensionless number, like the fine structure constant, is dimensionless in all systems of units.

I said "one where the left side has no units (just like a dimensionless constant) ". I did not say the equation was a "dimensionless number". I admit this was a bad example because the equation is not dimensionless on both sides. But I was making clear the fact that if an equation results in a dimensionless number then changing units can make the equation invalid if one were to change the value of one of the constants that is not dimensionless, and this is where I think Buzz was finding difficulty.

Unfortunately, I don't think your comments are helping because you're confused about the actual issue.

Back to the drawing board.

 

[PeterDonis]

I meant to say, "Therefore your selection of units will change the value (of the fine structure constant)

Still wrong. Go back and read what I said again, carefully.

one where the left side has no units (just like a dimensionless constant)

The number "1" does not necessarily have no units. Go back and read what I said again, carefully.

the equation is not dimensionless on both sides

An equation has to be either dimensionless on both sides or have the same dimensions on both sides. Otherwise it's not a valid equation.

Back to the drawing board.

Yes. And I strongly suggest not posting again until you have improved your understanding.

 

[Dale]

The number "1" does not necessarily have no units. Go back and read what I said again, carefully.

I disagree with this (I agree with the rest). The number 1 does not have units, but some people may be sloppy and not write the units explicitly if they are understood.

This can be important in understanding e.g. the difference between Planck units and geometrized units. In both sets of units c is unity, but in Planck units it should be a dimensionful 1 Lp/Tp and in geometrized units it is a dimensionless 1.