# Yang–Mills and Mass Gap

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

1. ### hansdaValued Senior Member

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Consider pair production from light. Say, E=mc^2=hf. or dm/df=h/c^2. So mass generation will follow this curve.

3. ### NotEinsteinValued Senior Member

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That is incorrect. The proper formula for light (and other massless particles) is: E=pc=hf.

5. ### hansdaValued Senior Member

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Why? Are you aware of pair production from light.

How will you correlate this energy with the mass(m) of a massive particle.

7. ### NotEinsteinValued Senior Member

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Yes, I am aware of that, and I'm also aware that that completely irrelevant here.

https://en.wikipedia.org/wiki/Mass–...rict_mass–energy_equivalence_formula,_E_=_mc2

So the full formula is: $E^2=(pc)^2+m^2 c^4$
Set $m=0$ for massless particles, set $p=0$ for non-moving particles (in other words, to use the rest mass).

This is very basic physics stuff.

8. ### hansdaValued Senior Member

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Do you know, what exactly is the mass-gap problem? https://en.wikipedia.org/wiki/Mass_gap . So the problem is to find the mass of the lightest particle. If this lightest particle is converted into energy, the frequency we get will be the smallest interval of frequency. So, by knowing the smallest interval of frequency, the corresponding mass can be found out.

Irrelevant.

9. ### NotEinsteinValued Senior Member

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That is not relevant to whether $E=mc^2$ is the correct formula for massless particles or not.

So whether $E=mc^2$ is the right formula for massless particles or not is irrelevant to the question of whether $E=mc^2$ is the right formula for massless particles? Can you explain?

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11. ### hansdaValued Senior Member

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Consider the equation$E=mc^2=hf=\frac{hc}{\lambda}$ Or $m=\frac{h}{c} \times \frac{1}{\lambda}$.

Now $\frac{dm}{d\lambda}=-\frac{h}{c}\times \frac{1}{\lambda^2}$. From this equation we can see that, wavelength has to be negative to generate a positive mass.

Last edited: Jan 14, 2018
12. ### NotEinsteinValued Senior Member

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That conclusion is nonsense. You have shown how the wavelength associated to a particle (at rest) changes as the rest mass of a particle is changed. This doesn't demonstrate that negative wavelengths (which is a nonsensical concept anyway) is needed to have a positive mass.

Please only post these things in the appropriate subforums next time.

13. ### hansdaValued Senior Member

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Did you observe anything wrong with the math?

Correct. So what is your problem?

So what is your conclusion? I just made a logical conclusion based on the math.

Please clarify this statement with appropriate logic/arguments. Just making a statement like this, does not mean anything.

14. ### NotEinsteinValued Senior Member

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I don't think I said that?

My problem is with the conclusion you are drawing from said math.

You have a derivative of the mass of a particle at rest with respect to its wavelength. Please demonstrate how that last equation leads to your conclusion based on logic.

I think it will become clear what I meant by this over time...

15. ### ForcemanMay the force be with youRegistered Senior Member

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First thing about matter is that it can't be stretched without an appropriate spring force: a vector wouldn't be a v)9 without a proper Fnet. In net force there is always a stretch from gradient to bottom rest mass; a quotient in mass force such as antienergy not dark energy doesn't unequate matter from waves. A v(0) is the first wave in a cautious space such as TNT or bottom rest energy such as a sound wave.

16. ### NotEinsteinValued Senior Member

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(I assume "v)9" contains a typo; what did you mean to write?)

Why can't matter be stretched without a spring force?

How does one stretch something from a gradient to a bottom rest mass? What is a bottom rest mass?

What quotient in mass force? What is mass force? What is antienergy? Can you please rewrite your sentence as to not use a double negative (doesn't unequate)? How does one equate mass from waves?

How is v(0) a wave? What is cautious space? What does TNT (explosives) have to do with this? What is bottom rest energy? How is bottom rest energy a sound wave?

This all appears to be word salad?

17. ### hansdaValued Senior Member

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So, the math is perfect.

You have not explained your problem.

Can you see the RHS of the equation. It is negative. So, I made my logical conclusion.

So, you dont have explanations for your statement. You just made a blind statement. This is nonsense.

18. ### hansdaValued Senior Member

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If Fnet is zero, the mass will be subject to stress. The mass will have a strain. The stress-strain relationship is like a spring force.

Not able to understand this statement.

I think you are trying to explain mass to energy conversion.

19. ### NotEinsteinValued Senior Member

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I have made no claim either way.

I actually have explained my problem multiple times already: you have not explained how you reached your conclusion from that equation.

Ah, then your conclusion is wrong. The LHS of the equation is not the mass, but a derivative of the mass to its wavelength. There's no way to conclude the necessity of negative mass from that.

Also a wrong conclusion: me not giving you an explanation doesn't mean I don't have one. As I said, things are becoming clearer all the time...

20. ### hansdaValued Senior Member

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Seems you are not able to judge my simple equation. And you try to defend GR.

$\frac{dm}{d\lambda}=negative$ means as $dm$ becomes positive $d\lambda$ will be negative.

Negative wavelength also can be considered as reducing wavelength. This can generate a positive mass.

What is clearer?

21. ### NotEinsteinValued Senior Member

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Seems like one of us indeed isn't able to, yes.

But $dm$ is not a mass, it's an infinitesimal, so your conclusion is still unwarranted.

Reducing a wavelength traditionally means making the wavelength smaller. This does not imply a negative wavelength. I can reduce the volume on my television set without it becoming negative. So please explain what you mean by "reducing wavelength", as you appear to be using a different definition than the ones I'm familiar with.

Except a positive wavelength means a positive mass. Look at your own equation:
$m=\frac{h}{c} \times \frac{1}{\lambda}$.

$m$ (the mass) is positive if and only is $\lambda$ (the wavelength) is positive, because $h$, $c$, and $1$ are all positive by definition.

The meaning of the last sentence in my post #29.

22. ### NotEinsteinValued Senior Member

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It seems hansda has abandoned this thread?

23. ### hansdaValued Senior Member

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Seems you are still not able to judge.

GR equations are much more complicated.

$dm$ is infinitesimal change of mass.

You can draw a curve for $dm$ vs $d\lambda$ and check it out.

You can also check the graph for $m$ vs $\lambda$.

As you are not able to judge the correctness of a simple equation, this sort of prediction from you does not make any sense.