# Yang–Mills and Mass Gap

Discussion in 'Physics & Math' started by Thales, Nov 29, 2017.

1. ### NotEinsteinValued Senior Member

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So you cannot point to the place in this thread where I have been defending GR. Glad we agree that you were misrepresenting my position!

So it's not a mass. Glad we agree!

I did, and that's my conclusion. Please show my your curve, then.

Please show that curve too, then.

Well, one of us clearly isn't able to judge the correctness of a simple equation. Heck, it's not even the correctness. One of us clearly isn't able to realize that multiplying positive numbers together results in a positive number!

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3. ### Kevin paul woodRegistered Member

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5. ### hansdaValued Senior Member

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Kevin, what are you trying to say?

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7. ### hansdaValued Senior Member

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Multiplying two negative numbers also results in positive number. Multiplying one positive number with a negative number results negative number.

8. ### NotEinsteinValued Senior Member

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Great! Progress!

Now let's again look at your equation:
$m=\frac{h}{c} \times \frac{1}{\lambda}$

You claim that the mass ($m$) is negative. So that means the right hand side is also negative. Please point me to the negative quantity/quantities on the right hand side.

9. ### hansdaValued Senior Member

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Plot$m$ in the y-axis and $\lambda$ in the x-axis. You will get the result.

10. ### NotEinsteinValued Senior Member

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Done!
http://www.wolframalpha.com/input/?i=y=1/x, x>0

At all valid values of $\lambda$, $m$ is positive. I have indeed gotten the result. QED

11. ### arfa branecall me arfValued Senior Member

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I think an important thing to consider is what is the motivation for taking a derivative like $\frac {dm} {d\lambda}$.
What does it mean physically, and what's the justification for any claim it has physical significance? Particularly since the derivative has a negative sign.

A derivative like that is meant to signify a small change in mass divided by a small change in wavelength, but this seems to not be connected to deBroglie's idea, which for one, requires that a particle with mass also has a velocity (related to its wavelength), what happened to the velocity? $E = mc^2$ is not related to particles with velocities because it defines rest mass and rest energy. QED.*

12. ### QuarkHeadRemedial Math StudentValued Senior Member

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I must say, I am amazed at the direction this thread has taken.

I remind you that the context of the opening post was gauge theory. Now you are all bickering about simple arithmetic - why?

Of course I know the answer - you don't understand the question.

I suggest we respond to posts in this subforum only if we have some knowledge of the the subject at hand. Is that at all unreasonable?

13. ### NotEinsteinValued Senior Member

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I fully agree; I've mentioned these points to hansda in this thread before, but I have yet to receive a response to those inquiries.

My involvement in the thread was prompted by the (quite common) mistake in post #24, which I merely pointed out. Then, in post #28, a claim was made (negative wavelengths?!), and I've since then been trying to get hansda to explain how (s)he got to that result.

You are absolutely right: I have no useful input on the original topic, which is why I haven't commented on it.

I suggest that the off-topic parts are split off into a separate thread, and placed in the appropriate subforum.

Edit: And I apologize for my part in pulling this thread so far off-topic. That was never my intention.

Last edited: Jan 25, 2018
14. ### QuarkHeadRemedial Math StudentValued Senior Member

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No! The apology should be mine - who am I to tell members what they should or shouldn't discuss?

Thing is, this is a subject that interests me, and it's possible I may have some minor contribution to make. Just possible.....

P.S. hansda is a well-known nut-job on this site, and as such is not worth the effort of arguing with.

15. ### NotEinsteinValued Senior Member

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No need to apologize. In Dutch, we have a saying, roughly translated: "if there are two people fighting, there are two people that have blame". I accept my part in that.

The first is true for me too. The latter: if only my knowledge of such things was that good.

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Which is why I'm totally fine with his/her and my posts being split off into a separate thread, to be relocated appropriately.

16. ### hansdaValued Senior Member

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Consider generation of mass from particle photon, as in pair production. Here energy is extracted from particle photon to generate mass. This can be considered as if propagation of photon is stopped under strong nuclear field. Perhaps this fact is highlighted by $d\lambda$ being negative.

17. ### hansdaValued Senior Member

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Consider $E=mc^2=hf=h\frac{c}{\lambda}$ and $c=f\lambda$ . Here minimum frequency(f) can be one and correspondingly maximum wavelength can be c. So that, $f=1$ and $\lambda=c$ . Considering these values, minimum mass(m) can be $m=\frac{h}{c^2}$ .

18. ### originHeading towards oblivionValued Senior Member

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The wavelength can be a 299792 km/sec? That is one fast length!
This has been a cold winter so far, as a matter of fact the temperature is 20 kg right now.

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19. ### QuarkHeadRemedial Math StudentValued Senior Member

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Duplicate. Sorry

20. ### QuarkHeadRemedial Math StudentValued Senior Member

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This is gibberish, as origin hinted.
Frequency times wavelength is NOT a speed. Furthermore, your claim above that "maximum wavelength can be c", is pure nonesense - even if it were not, it would conflict with your earlier erroneous claim.
How about if you give some thought to non-abelian gauge theories - i.e. stay on topic?
Just a thought......

21. ### NotEinsteinValued Senior Member

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Except that the strong nuclear field doesn't have anything to do with this. This can perfectly be described by QED only: electron pair production.

Except that $d\lambda$ isn't negative, nor is it the wavelength.

Please explain to me what is specifically interesting about a wave with $\lambda=c$. Also, $f$ can be lower than $1$; for example, it can be $0.5$. So your conclusion about a minimum mass is wrong.

22. ### arfa branecall me arfValued Senior Member

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One question some might be asking is: what exactly is a gauge?

I've seen it described as various things, including a choice of coordinates, a choice of global phase (as in quantum fields), and a choice of a global value for a potential such as a voltage.
Is it then, something quite general? And what exactly is the difference between a global and a local choice of a gauge?

ed: for instance, the "Coulomb gauge" appears to be the Coulomb potential with additional constraints on it (zero divergence, etc).

Last edited: Jan 26, 2018
23. ### QuarkHeadRemedial Math StudentValued Senior Member

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So - quick and dirty.

Suppose a field theory. Suppose further that at any chosen point this theory is invariant under a certain type of symmetry transformations - this is called "local invariance." Note it does imply our theory is globally invariant.

So if we assume this is true at each and every point, we may ask if this is true globally - Einstein called this the principle of General Covariance, and had trouble with it. In quantum physics this called "gauge invariance"

A gauge is a "choice" of symmetry transformations at each point such that our theory is globally invariant under any and all separate local trasformations

Of course there is serious mathematics involved. I think I can do it. Can you?