Yang–Mills and Mass Gap

arfa brane said:
What I posted is about classical uncertainty, in the classical measurement of a classical wavelength, in the time domain.
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https://www.quora.com/How-does-the-uncertainty-principle-relate-to-Fourier-transforms
Well the link between Heisenberg and Fourier is just that - a classical wavepacket essentially exhibits that self same uncertainty relation! Amazing or not.
Can you show it doesn't follow from (the assumption) $$ mc^2 = hf $$?
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This article should help: http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html
The 2nd, full relativistic mass-energy-momentum expression there in (1) is needed to cover the general case; of both massless particles e.g. photon, and massive particles e.g. electron.
 
I give up. I was trying to demonstrate that hansda's assumption isn't just not well-defined, but clearly wrong. Now I'm being called on it.

Hey, I know it's wrong and I know why it's wrong, m'kay?

One reason is that if the photon had a rest mass, since it travels at c its energy would be infinite; one of the reasons $$ E = mc^2 = hf $$ is just wrong. The correct relation is $$ E = hf = hc/\lambda $$. But as Baez explains it, you can just assume "If we set the rest mass equal to zero (regardless of whether or not that's a reasonable thing to do), then E = pc."

Done, and done.
 
arfa brane said:
I give up. I was trying to demonstrate that hansda's assumption isn't just not well-defined, but clearly wrong. Now I'm being called on it.

Hey, I know it's wrong and I know why it's wrong, m'kay?...
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Well pardon me for thinking elsewise but that was not the impression gained from how you finished #79 with those last two lines. Why not there and then go on to provide the correct answer? Never mind.
One reason is that if the photon had a rest mass, since it travels at c its energy would be infinite; one of the reasons $$ E = mc^2 = hf $$ is just wrong. The correct relation is $$ E = hf = hc/\lambda $$. But as Baez explains it, you can just assume "If we set the rest mass equal to zero (regardless of whether or not that's a reasonable thing to do), then E = pc."

Done, and done.
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Indeed and indeed.
 
arfa brane said:
But you haven't defined a cycle.
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Consider the equation $$ w=2\pi f $$ . Here $$ 2\pi$$ is the cycle.

arfa brane said:
You haven't specified what is a "complete" cycle, or whether a cycle is a smooth function, etc.
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When $$ 2\pi $$ is completed, it is a complete cycle. It can be counted as a frequency, which can be repeated. $$2\pi $$ also can be considered as a sine wave, which is a smooth function. $$f(\theta)=\sin(2\pi ft) $$ .
 
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hansda said:
If the frequency is less than one, the cycle is incomplete.
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Can you now reconcile "the frequency is less than one" with $$ f(\theta) = sin (2\pi f t) $$?
Less than one what?

Also, is there a difference between the f on the LHS and the f on the RHS? Which is the function and which is frequency?
 
Q-reeus said:
Well the link between Heisenberg and Fourier is just that - a classical wavepacket essentially exhibits that self same uncertainty relation! Amazing or not.
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I was enrolled in a signal processing course once, and indeed it seems that although there's a classical interpretation for the uncertainties in frequency/time measurements, particularly in filter design, you can invoke quantum mechanics too. This shouldn't be all that surprising if you're dealing with optical signals (where they propagate through finite "time" apertures). And since "ordinary" electronic signals propagate at the speed of light (in say, a copper medium at < c ), it seems to apply there too.

But electronics engineers prefer the classical equations.
 
arfa brane said:
I was enrolled in a signal processing course once, and indeed it seems that although there's a classical interpretation for the uncertainties in frequency/time measurements, particularly in filter design, you can invoke quantum mechanics too. This shouldn't be all that surprising if you're dealing with optical signals (where they propagate through finite "time" apertures). And since "ordinary" electronic signals propagate at the speed of light (in say, a copper medium at < c ), it seems to apply there too.

But electronics engineers prefer the classical equations.
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Right. Just to clear up why I came down so hard in #80... The passage I objected to again:
"That said, hansda is assuming that E=mc2=hf holds. But this means there's a big problem with Einstein's 1905 paper on the photelectric effect, because it says photons have a rest energy and so a finite range."
I had read that as you claiming the photoelectric effect demanded photons have a finite rest energy, but only later realized the last part was actually connecting to the first sentence i.e. the E = mc^2 = hf claim of handsa. In short, I misinterpreted your intent - something that would have not happened with a differently structured sentence(s).
 
O . . . k. I suppose I should have written "But that would mean . . ." instead. Sorry for skipping over the bleeding obvious (need for a subjunctive) there.
 
arfa brane said:
Can you now reconcile "the frequency is less than one" with $$ f(\theta) = sin (2\pi f t) $$?
Less than one what?
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Say frequency is $$f=1 $$, it will become $$ g(\theta)=sin(2\pi t) $$. If $$ f=\frac{1}{2}$$, the function will be $$g_1(\theta)=sin(\pi t) $$. To complete one circle $$2\pi $$ or $$360^0 $$ is required for one wavelength. In the case of $$g_1(\theta) $$, wavelength will correspond to $$ \pi$$. This may not be true for particle photon.

arfa brane said:
Also, is there a difference between the f on the LHS and the f on the RHS? Which is the function and which is frequency?
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You can consider the function as $$ g(\theta)=sin (2\pi ft)$$.
 
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hansda said:
Say frequency is $$f=1 $$, it will become $$ g(\theta)=sin(2\pi t) $$. If $$ f=\frac{1}{2}$$, the function will be $$g_1(\theta)=sin(\pi t) $$. To complete one circle $$2\pi $$ or $$360^0 $$ is required for one wavelength. In the case of $$g_1(\theta) $$, wavelength will correspond to $$ \pi$$. This may not be true for particle photon.
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Right, so please point out the problem you see with setting $$f<1$$.
 
The one dimensional wavepacket model may be nice (actually I hate it but that's just me) but in reality we have three dimensions. The one dimensional wave packet doesn't address the properties of (say) a photon uncovered by the double slit experiment.
 
hansda said:
Try to calculate wavelength. You will observe the difficulty.
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Alright, let's do it. $$\lambda=\frac{c}{f}$$
Let's use units where $$c=1$$. Set $$f<1$$, for example, $$f=0.5$$
So now we get: $$\lambda=\frac{1}{0.5}=2$$

Where is the difficulty?
 
Would it be blasphemous to ask members to respond to the opening post?

I note that the thread starter has given up. I do not blame him (or her).

But gauge theory, in particular the non-abelian variety, is quite interesting and I believe important in physics.
 
QuarkHead said:
But gauge theory, in particular the non-abelian variety, is quite interesting and I believe important in physics.
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Where to start though? With continuous symmetry groups? With Wikipedia's page on that subject?

viz:
Continuous symmetry has a basic role in Noether's theorem in theoretical physics, in the derivation of conservation laws from symmetry principles, specifically for continuous symmetries. The search for continuous symmetries only intensified with the further developments of quantum field theory.
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Or with the special unitary groups, maybe why is U(1) important although it's abelian, then why SU(n) is important but non-abelian? What about what t'Hooft mentions in the article I keep referring to, about the two gauge theories, the Weinberg-Salam-Ward model and the Georgi-Glashow model, one of which doesn't fly because it predicts the decay of the proton? Is isospin a gauge, for instance, or a continuous symmetry which the Higgs field "gauges" by breaking it?
 
NotEinstein said:
Alright, let's do it. $$\lambda=\frac{c}{f}$$
Let's use units where $$c=1$$. Set $$f<1$$, for example, $$f=0.5$$
So now we get: $$\lambda=\frac{1}{0.5}=2$$

Where is the difficulty?
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Did you observe, here $$\lambda $$ is becoming $$\lambda=2c $$. You can consider $$fT=1 $$, where $$ T$$ is the time period. If $$f=\frac{1}{2} $$, $$T $$ becomes $$T=2 $$. So with half cycle it will travel $$c $$. This is not compatible with particle photon. You can check the equations for photon.
 
hansda said:
Did you observe, here $$\lambda $$ is becoming $$\lambda=2c $$.
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No, I didn't observe that, because that's wrong. Wavelength is a length (the hint is in the name), not a speed.

hansda said:
You can consider $$fT=1 $$, where $$ T$$ is the time period.
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This is wrong for similar reasons. Frequency has units of inverse time, not time.

hansda said:
If $$f=\frac{1}{2} $$, $$T $$ becomes $$T=2 $$.
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Exactly.

hansda said:
So with half cycle it will travel $$c $$.
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Do you mean: travel a distance of $$cT$$, or travel at the speed $$c$$? And I think both are false conclusions: nowhere did I specify the group velocity of the phase velocity of the wave.

hansda said:
This is not compatible with particle photon.
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Please explain why.

hansda said:
You can check the equations for photon.
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No, you are making the claim. Please give these equations and calculations yourself. In fact, I don't even know what specific equations you are referring to.
 
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