Yang–Mills and Mass Gap

[hansda]

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

 

[NotEinstein]

Say, E=mc^2=hf.

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

 

[hansda]

That is incorrect.

Why? Are you aware of pair production from light.

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

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

 

[NotEinstein]

Why? Are you aware of pair production from light.

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

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

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.

 

[hansda]

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

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.

https://en.wikipedia.org/wiki/Mass–energy_equivalence#Applicability_of_the_strict_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.

Irrelevant.

 

[NotEinstein]

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.

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

Irrelevant.

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?

 

[Thales]

http://www.sciforums.com/threads/inertia-and-relativity.160396/

Reading it, hansda.

 

[hansda]

Thanks for reading my paper titled "Structure of a Particle" uploaded at academia.

Reading it, hansda.

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.

 

[NotEinstein]

Thanks for reading my paper titled "Structure of a Particle" uploaded at academia.



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.

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.

 

[hansda]

That conclusion is nonsense.

Did you observe anything wrong with the math?

You have shown how the wavelength associated to a particle (at rest) changes as the rest mass of a particle is changed.

Correct. So what is your problem?

This doesn't demonstrate that negative wavelengths (which is a nonsensical concept anyway) is needed to have a positive mass.

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

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

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

 

[NotEinstein]

Did you observe anything wrong with the math?

I don't think I said that?

Correct. So what is your problem?

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

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

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

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

 

[Forceman]

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.

 

[NotEinstein]

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.

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

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

In net force there is always a stretch from gradient to bottom rest mass;

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

a quotient in mass force such as antienergy not dark energy doesn't unequate matter from waves.

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?

A v(0) is the first wave in a cautious space such as TNT or bottom rest energy such as a sound wave.

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?

 

[hansda]

I don't think I said that?

So, the math is perfect.

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

You have not explained your problem.

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.

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

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

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

 

[hansda]

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.

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.

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.

Not able to understand this statement.

A v(0) is the first wave in a cautious space such as TNT or bottom rest energy such as a sound wave.

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

 

[NotEinstein]

So, the math is perfect.

I have made no claim either way.

You have not explained your problem.

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

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

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.

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

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...

 

[hansda]

I have made no claim either way.

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

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

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

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.

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

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...

What is clearer?

 

[NotEinstein]

Seems you are not able to judge my simple equation.

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

And you try to defend GR.

Please point out where I have done that in this thread.

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

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

Negative wavelength also can be considered as reducing wavelength.

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.

This can generate a positive mass.

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.

What is clearer?

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

 

[NotEinstein]

It seems hansda has abandoned this thread?

 

[hansda]

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

Seems you are still not able to judge.

Please point out where I have done that in this thread.

GR equations are much more complicated.

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

$$dm$$ is infinitesimal change of mass.

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.

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

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.

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

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

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