# Yang–Mills and Mass Gap

Where I have made this statement in post #148.
I never claimed that you did said that in post #148, so I don't know where you are getting that from?

Your post #149 does not say whether my post #148 is right or wrong. Anyway, you can give a fresh answer also.
Alright, let's take a fresh look.

Consider any time duration/interval as $$t$$units of time. Consider time-period of oscillation as $$T$$units of time. Here to get atleast one complete oscillation within the time interval $$t$$ units, $$T\leq t$$ .
I agree so far.

Here frequency $$f=\frac{t}{T}$$ or $$fT=t$$ .
This part is utter nonsense, because the units don't match.

I agree so far.

Good. You agree to this that $$T\leq t$$ as per my post #148.

This part is utter nonsense, because the units don't match.

Where the units are not matching.

Good. You agree to this that $$T\leq t$$ as per my post #148.
Only in order to get at least one complete oscillation, as you yourself state. My point is that this is not generally true.

Where the units are not matching.
You wrote: $$f=\frac{t}{T}$$
$$t$$ has units of time.
$$T$$ has units of time.
$$f$$ has units of inverse time.
The left-hand side thus has units of inverse time, while the right-hand side is unitless. This makes the equation nonsense.

Only in order to get at least one complete oscillation, as you yourself state.

Yes.

My point is that this is not generally true.

Why?

You wrote: $$f=\frac{t}{T}$$
$$t$$ has units of time.
$$T$$ has units of time.
$$f$$ has units of inverse time.
The left-hand side thus has units of inverse time, while the right-hand side is unitless. This makes the equation nonsense.

Seems you are not getting the point. To get the numerical value of rpm, you have to use this formula.

Yes.

Why?
Because it is physically possible to rotate an object 180 degrees, i.e. half of a full rotation.

Seems you are not getting the point. To get the numerical value of rpm, you have to use this formula.
But the formula is nonsense. It's equivalent to asking if you are 5 apples old, or if you are 6 waters high? You seem to not understand the importance of units. Without understanding how units work, one cannot do any meaningful maths in physics at all!

We know that massive, spinning particles are generated in pairs from particle photon through symmetry breaking, as in pair production. Here minimum energy of photon will be for one cycle. So, that minimum energy should correlate with minimum mass.

We know that massive, spinning particles are generated in pairs from particle photon through symmetry breaking,

as in pair production. Here minimum energy of photon will be for one cycle.

So, that minimum energy should correlate with minimum mass.
That's circular reasoning. You started out by saying that you are generating massive particles. Obviously there will be a minimum mass; their rest mass.

That's circular reasoning. You started out by saying that you are generating massive particles. Obviously there will be a minimum mass; their rest mass.

You can see wiki article on pair production.

Symmetry breaking is known from higgs mechanism
Which also isn't mentioned on that page. Are you even trying to back up your statements?

Which also isn't mentioned on that page. Are you even trying to back up your statements?

You dont know higgs mechanism?

You dont know higgs mechanism?
Please explain how it is relevant to your original statement, as it is not mentioned on that Wikipedia article.

Please explain how it is relevant to your original statement, as it is not mentioned on that Wikipedia article.

Particle photon is symmetrical. When its symmetry is broken, two assymetric massive particles are generated as in pair production.

Particle photon is symmetrical. When its symmetry is broken, two assymetric massive particles are generated as in pair production.

So, in pair production, the two particles generated can be considered as two half-photons.

So, in pair production, the two particles generated can be considered as two half-photons.

Thus minimum one photon cycle is necessary for minimum mass generation.

Thus minimum one photon cycle is necessary for minimum mass generation.

So, the minimum mass will be $$E=mc^2=h$$ or $$m=\frac{h}{c^2}$$ .

Particle photon is symmetrical.
Please provide a source that states that the photon is symmetrical in the same sense of the word as it is used in "symmetry breaking".

When its symmetry is broken, two assymetric massive particles are generated as in pair production.
In what sense are the resulting particle asymmetrical?

So, in pair production, the two particles generated can be considered as two half-photons.
False; electrons are not half-photons. Positrons are not half-photons. They are completely different types of particles.

Thus minimum one photon cycle is necessary for minimum mass generation.
Please define how you are using the word "cycle". Please define what you mean by "minimum mass generation".

So, the minimum mass will be $$E=mc^2=h$$ or $$m=\frac{h}{c^2}$$ .
The units don't match: the constant of Planck doesn't have units of energy, but energy time. Please correct your equations.

I see you still haven't managed to provide even a single source for your original statement. Why is that so difficult for you?

So, the minimum mass will be $$E=mc^2=h$$ or $$m=\frac{h}{c^2}$$ .

From my equations $$mr=\frac{4\hbar}{c}$$ or $$m=\frac{1}{r}\frac{4h}{2\pi c}$$ .

Comparing these two equations of mass, we can write maximum radius $$r$$ will be $$r=\frac{2c}{\pi}$$ .

From my equations $$mr=\frac{4\hbar}{c}$$ or $$m=\frac{1}{r}\frac{4h}{2\pi c}$$ .

Comparing these two equations of mass, we can write maximum radius $$r$$ will be $$r=\frac{2c}{\pi}$$ .
That last equation has a radius with units of length per time. A fine demonstration of "garbage in, garbage out".