That is incorrect. The proper formula for light (and other massless particles) is: E=pc=hf.Say, E=mc^2=hf.
That is incorrect.
The proper formula for light (and other massless particles) is: E=pc=hf.
Yes, I am aware of that, and I'm also aware that that completely irrelevant here.Why? Are you aware of pair production from light.
https://en.wikipedia.org/wiki/Mass–...rict_mass–energy_equivalence_formula,_E_=_mc2How will you correlate this energy with the mass(m) of a massive particle.
Yes, I am aware of that, and I'm also aware that that completely irrelevant here.
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.
That is not relevant to whether $$E=mc^2$$ is the correct formula for massless particles or not.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.
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?Irrelevant.
Reading it, hansda.
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.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.
I don't think I said that?Did you observe anything wrong with the math?
My problem is with the conclusion you are drawing from said math.Correct. So what is 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.So what is your conclusion? I just made a logical conclusion based on the math.
I think it will become clear what I meant by this over time...Please clarify this statement with appropriate logic/arguments. Just making a statement like this, does not mean anything.
(I assume "v)9" contains a typo; what did you mean to write?)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.
How does one stretch something from a gradient to a bottom rest mass? What is a bottom rest mass?In net force there is always a stretch from gradient to 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?a quotient in mass force such as antienergy not dark energy doesn't unequate matter 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?A v(0) is the first wave in a cautious space such as TNT or bottom rest energy such as a sound wave.
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...
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.
I have made no claim either way.So, the math is perfect.
I actually have explained my problem multiple times already: you have not explained how you reached your conclusion from that equation.You have not explained your problem.
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.Can you see the RHS of the equation. It is negative. So, I made my logical conclusion.
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...So, you dont have explanations for your statement. You just made a blind statement. This is nonsense.
I have made no claim either way.
$$\frac{dm}{d\lambda}=negative $$ means as $$dm$$ becomes positive $$d\lambda$$ will be negative.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...
Seems like one of us indeed isn't able to, yes.Seems you are not able to judge my simple equation.
Please point out where I have done that in this thread.And you try to defend GR.
But $$dm$$ is not a mass, it's an infinitesimal, so your conclusion is still unwarranted.$$\frac{dm}{d\lambda}=negative $$ means as $$dm$$ becomes positive $$d\lambda$$ will be negative.
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.Negative wavelength also can be considered as reducing wavelength.
Except a positive wavelength means a positive mass. Look at your own equation:This can generate a positive mass.
The meaning of the last sentence in my post #29.What is clearer?
Seems like one of us indeed isn't able to, yes.
Please point out where I have done that in this thread.
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.