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by **Ben6993** » Wed Aug 10, 2016 5:55 am

Fred wrote:

I suppose the correct way of doing it is to get the action by taking the integral of the Lagrangian then using the principle of stationary action.

Agreed. One 'problem' with the lagrangian is that it contains potentials which are not unique: you can add any constant to a potential and it does not change the result of minimising the action (i.e gauge invariance). That is because one minimises by differentiating, and a constant term always goes to zero on differentiation. So if you are not minimising the action, the lagrangian could give you any answer you wanted by adding an arbitrary constant to the potential. Not sure that your X is a constant, but whatever it is it would need to preserve gauge invariance, and a constant would do that OK.

by **FrediFizzx** » Sun Aug 07, 2016 11:45 am

Of course the way around this is by renormalization.

http://isites.harvard.edu/fs/docs/icb.t ... Theory.pdf

See eqs. 12 and 13 where now they have a bunch of extra terms in the Lagrangian. But is this really "physical"?

by **FrediFizzx** » Sun Aug 07, 2016 10:53 am

Ben6993 wrote:Try a different way. Drop X as it isn't in the QED lagrangian that you originally presented.

Your earlier formula was m c c = e e / 4 π ε0 length
So 'm c c length' is a constant. I.e. 'm c c' is inversely proportional to length.

Classical electron length (radius) is approximately 10^-15 metres.
Planck length is approximately 10^-35 metres.
With a ratio of 10^20.

Electron mass is approximately 0.5 MeV, corresponding to a length about 10^-15 metres.
So [in order to try to answer your question as to what would be the effect of an electron having a radius of the Planck length]:
an electron mass of 0.5 *10^20 MeV would correspond to an electron radius with Planck length.

A Planck mass is about 10^21 MeV, so the electron mass would have to be of the order of a Planck mass.

Ben, that is not the point. I am saying that $X$ is needed in the Lagrangian to make it consistent with what we know from experiment that the electron in experiments show no substructure down to the order of 10^-22 m. Put 10^-22 m in for $r$ and you will still need $X$ to make the Lagrangian consistent with experiment.

by **Ben6993** » Sun Aug 07, 2016 1:46 am

Try a different way. Drop X as it isn't in the QED lagrangian that you originally presented.

Your earlier formula was m c c = e e / 4 π ε0 length
So 'm c c length' is a constant. I.e. 'm c c' is inversely proportional to length.

Classical electron length (radius) is approximately 10^-15 metres.
Planck length is approximately 10^-35 metres.
With a ratio of 10^20.

Electron mass is approximately 0.5 MeV, corresponding to a length about 10^-15 metres.
So [in order to try to answer your question as to what would be the effect of an electron having a radius of the Planck length]:
an electron mass of 0.5 *10^20 MeV would correspond to an electron radius with Planck length.

A Planck mass is about 10^21 MeV, so the electron mass would have to be of the order of a Planck mass.

by **FrediFizzx** » Sat Aug 06, 2016 4:16 pm

Ben6993 wrote:Fred wrote:
... what could X be if the effective radius is the Planck length?

But the KE and PE are still zero, irrespective of the value of X, so we have a stationary electron with no PE.

That is not true. The PE of the Lagrangian might be zero but that doesn't mean the PE of the electron is zero. You have to find a solution of the Lagrangian which I have shown you what the solution is. We know what the potential energy is for an electron; it is the energy due to the observed mass, $E = mc^2 \approx 0.511$ MeV.

by **Ben6993** » Sat Aug 06, 2016 3:10 pm

Fred wrote:

... what could X be if the effective radius is the Planck length?

But the KE and PE are still zero, irrespective of the value of X, so we have a stationary electron with no PE.

IMO such an electron, which is 'free and easy' and' laid back' under no external pressure, will not undergo collapse to a point particle but will remain an uncollapsed field with range necessarily greater than the Planck length. This is because an electron field will only collapse at an interaction with another particle/field ... which is not present under your constraint of zero PE.

by **FrediFizzx** » Sat Aug 06, 2016 12:30 pm

Ben6993 wrote:Fred

So have you answered your own question, now? To within 8% accuracy?
I am still not clear what your original question was. It seemed at first that you were hoping to calculate a very small (near zero) radius. But that is associated with a near infinite naked mass. Yet the electron mass plugged into the QED Lagrangian is a physical mass not a naked mass, so you can't get a naked radius outputted after inputting a physical mass ... no matter how much you amend the QED lagrangian.

Another point is that the electron can be a field as well as a mass. If you set the lagrangian equal to zero and then let KE equal zero.
Then that forces PE to be zero. So you have an electron at rest with no potential acting on it. Under those conditions the electron will be a field not a particle. This is probably just my non-standard opinion as I believe in particle --> field --> particle sequentiality, where the electron only becomes a particle at an interaction. And in your scenario there will be no interaction. So IMO your test electron will be a field. So what is the effective radius of the electron field?

Well I suppose I am assuming an electron to be both a particle and a field. There is an interaction; self-interaction as I described above. The plugged in mass in the Lagrangian is an observed mass ~= 0.511 MeV for an electron. I have no clue as to what a "naked" mass might be that you are referring to. It is OK that PE becomes zero in the Lagrangian when KE is zero because then we have,

$0 = \frac{e^2}{4\pi \varepsilon_0 l_P} - X - mc^2$ ,

which then gives,

$\frac{e^2}{4\pi \varepsilon_0 l_P} - X = mc^2$ .

But yes, we are trying to determine the effective radius $r$ of the electron field via $\bar{\psi} \psi=1/r^3$ . Obviously it can't be done with the QED Lagrangian that I originally posted. So the question remains; what could $X$ be if the effective radius is the Planck length?
...

by **Ben6993** » Sat Aug 06, 2016 1:05 am

Fred

So have you answered your own question, now? To within 8% accuracy?
I am still not clear what your original question was. It seemed at first that you were hoping to calculate a very small (near zero) radius. But that is associated with a near infinite naked mass. Yet the electron mass plugged into the QED Lagrangian is a physical mass not a naked mass, so you can't get a naked radius outputted after inputting a physical mass ... no matter how much you amend the QED lagrangian.

Another point is that the electron can be a field as well as a mass. If you set the lagrangian equal to zero and then let KE equal zero.
Then that forces PE to be zero. So you have an electron at rest with no potential acting on it. Under those conditions the electron will be a field not a particle. This is probably just my non-standard opinion as I believe in particle --> field --> particle sequentiality, where the electron only becomes a particle at an interaction. And in your scenario there will be no interaction. So IMO your test electron will be a field. So what is the effective radius of the electron field?

by **FrediFizzx** » Fri Aug 05, 2016 3:43 pm

thray wrote:Fred, I think of it this way:

A ball thrown perpendicular to a gravitational plane, at less than escape velocity, continuously sheds kinetic energy until it reaches its original rest energy -- before descending and accumulating kinetic energy by acceleration in the gravity field and returning to rest. Independent of the field, in free fall and noninteracting, its Lagrangian is zero, and as the song goes, "nothin' from nothin' leaves nothin'".

Introducing an interaction (observation) forces us to choose an arbitrary zero boundary condition. In other words, the Lagrangian is not absolute; special relativity is.

Well, I was just considering the rest frame so SR doesn't really apply. The Lagrangian I posted above is supposed to be invariant under local gauge condition and I am interested in when it is zero for the rest frame scenario. I suppose the correct way of doing it is to get the action by taking the integral of the Lagrangian then using the principle of stationary action. But I like short-cuts.

by **FrediFizzx** » Fri Aug 05, 2016 2:16 pm

Ben6993 wrote:Fred

Page 1 of http://www.theoretical-physics.net/dev/quantum/qed.html
says that QED works fine down to about 10^-13 cm .

Yes, I noticed that and they say that without giving any kind of reference. I've never seen that before but it makes some sense.

AFAIK, renormalisation works by settling on a 'smallest' scale of length above which the equations will work. And relying on cancellation to zero of counterbalancing balancing terms below that threshold length. Below the threshold length, the energies get bigger and there is a reliance on many very large energies cancelling out to zero. Setting a threshold smallest length (maximum energy) involves removing complicated vertices/interaction events from the calculations. However, if you want a new QED equation to be valid right down to the Planck length then you need to cater in the new QED lagrangian equation for all possible loop interactions, no matter how intricate the interaction vertices are. However, you need to derive formulae for them. I believe this would be done by adding a(n infinite) number of extra PE terms each of which would represent a different interaction vertex. And you need guage invariance too.

Each of these extra PE terms represents a different type of QED virtual interaction vertex around the electron eg complicated emissions, absorptions of photons, positrons and electrons. These interactions would describe and mathematically account for the virtual cloud of photons etc surrounding the bare electron.

We are just talking about electron self-energy here; the charge of the electron is interacting with the electromagnetic field that it is creating. So I think there isn't a whole bunch of different vertex terms to consider here. They can all be lumped together anyways into the Coulomb field as a good approximation. And then the Coulomb field can be corrected for vacuum polarization.

https://en.wikipedia.org/wiki/Vacuum_po ... tic_fields

If $r$ is the Planck length then that is about an 8 percent correction to the Coulomb field. So the Coulomb field is about 8 percent stronger at Planck length for the electron. For the muon it is about 7 percent stronger. Provided this formula on there is correct which I believe it is.
...

by **thray** » Fri Aug 05, 2016 8:11 am

Oh, hell. You're right, Ben. I should have E^2 in the unreduced equation. E_0 in the final. My apologies.

by **thray** » Fri Aug 05, 2016 7:43 am

Ben6993 wrote:Tom

Agree that one shouldn't divide by zero.
How do you get negative mass from the unreduced equation? Your particle has zero momentum. So p=0. So you immediately get the reduced equation E=mcc. BTW should your unreduced equation have E_0 squared on the LHS?

You've made a math error, Ben. Instead of squared, the absolute rest state E_0 would become sqrtE, if we divide both sides by c^2

Mass and energy have a quadratic relation. Sqrt(E) = sqrt(m) times c; however, we need square c to represent light in terms of speed, E_0 = km with k = c^2

by **thray** » Fri Aug 05, 2016 7:17 am

FrediFizzx wrote:
thray wrote:So the QED electron radius can't go to zero, to avoid dividing by zero.

Tom

Right but say we stop the radius getting smaller at Planck length the theoretical smallest length so that,

$\frac{e^2}{4\pi \varepsilon_0 l_P} = mc^2$ ,

then the Lagrangian doesn't work since the LHS is a huge amount of energy. So there ought to be a term to counteract that like,

$\frac{e^2}{4\pi \varepsilon_0 l_P} - X = mc^2$ ,

so that the Lagrangian produces the correct observed mass-energy. What could $X$ possibly be?

Fred, I think of it this way:

A ball thrown perpendicular to a gravitational plane, at less than escape velocity, continuously sheds kinetic energy until it reaches its original rest energy -- before descending and accumulating kinetic energy by acceleration in the gravity field and returning to rest. Independent of the field, in free fall and noninteracting, its Lagrangian is zero, and as the song goes, "nothin' from nothin' leaves nothin'".

Introducing an interaction (observation) forces us to choose an arbitrary zero boundary condition. In other words, the Lagrangian is not absolute; special relativity is.

by **Ben6993** » Fri Aug 05, 2016 2:36 am

Fred

Page 1 of http://www.theoretical-physics.net/dev/quantum/qed.html
says that QED works fine down to about 10^-13 cm .

AFAIK, renormalisation works by settling on a 'smallest' scale of length above which the equations will work. And relying on cancellation to zero of counterbalancing balancing terms below that threshold length. Below the threshold length, the energies get bigger and there is a reliance on many very large energies cancelling out to zero. Setting a threshold smallest length (maximum energy) involves removing complicated vertices/interaction events from the calculations. However, if you want a new QED equation to be valid right down to the Planck length then you need to cater in the new QED lagrangian equation for all possible loop interactions, no matter how intricate the interaction vertices are. However, you need to derive formulae for them. I believe this would be done by adding a(n infinite) number of extra PE terms each of which would represent a different interaction vertex. And you need guage invariance too.

Each of these extra PE terms represents a different type of QED virtual interaction vertex around the electron eg complicated emissions, absorptions of photons, positrons and electrons. These interactions would describe and mathematically account for the virtual cloud of photons etc surrounding the bare electron.

Tom

Agree that one shouldn't divide by zero.
How do you get negative mass from the unreduced equation? Your particle has zero momentum. So p=0. So you immediately get the reduced equation E=mcc. BTW should your unreduced equation have E_0 squared on the LHS?

by **FrediFizzx** » Thu Aug 04, 2016 9:49 pm

thray wrote:So the QED electron radius can't go to zero, to avoid dividing by zero.

Tom

Right but say we stop the radius getting smaller at Planck length the theoretical smallest length so that,

$\frac{e^2}{4\pi \varepsilon_0 l_P} = mc^2$ ,

then the Lagrangian doesn't work since the LHS is a huge amount of energy. So there ought to be a term to counteract that like,

$\frac{e^2}{4\pi \varepsilon_0 l_P} - X = mc^2$ ,

so that the Lagrangian produces the correct observed mass-energy. What could $X$ possibly be?

by **thray** » Thu Aug 04, 2016 4:24 pm

FrediFizzx wrote:The first two terms are KE and we go to the rest frame so those are zero. That just leaves the interaction term and the mass-energy term. Since we are in the rest frame, we are only interested in the zeroth component of A which results in a self-energy term for the electron.

Good question, Fred!

Einstein's original equation pointing to mass-energy equivalence, m = L/c^2, is a measure of the energy content for a given state of mass, L being the Lagrangian. The general equation E = mc^2 is an equation of state for the entire universe at rest.

The unreduced equation, E_0 = m^2c^4 + (pc)^2 for a particle with mass yet with zero momentum, implies negative mass (an unphysical state).

So the QED electron radius can't go to zero, to avoid dividing by zero.

Tom

by **FrediFizzx** » Thu Aug 04, 2016 3:54 pm

The first two terms are KE and we go to the rest frame so those are zero. That just leaves the interaction term and the mass-energy term. Since we are in the rest frame, we are only interested in the zeroth component of A which results in a self-energy term for the electron.

by **Ben6993** » Thu Aug 04, 2016 3:05 pm

I have followed Susskind's online course but am not fluent in this.
On the scale: first exposure ---> familiar -----> fluent usage ... I am on the extreme left.

If you look at the terminology in the following powerpoint presentation:
http://www.tech.plym.ac.uk/Research/app ... ldQED2.pdf
then your four component terms (of ℒ) in order break down into:
1. first part of the Dirac lagrangian
2. Maxwell lagrangian
3. interaction Lagrangian
4. second term in Dirac Lagrangian

You have equated 'the time term within the third part' with 'the fourth part'.
Ie 'the time term within the interaction Lagrangian' with 'the second term of the Dirac lagrangian'.
Ie you are assuming that the KE parts of ℒ are:
Part 1, Part 2 and the spatial parts of Part 3 and then you are setting the sum of these to zero.

I think that Part 1 is KE and Part 4 is PE, but I am not sure how to separate KE from PE in Parts 2 and 3.

[Nagging away at my mind is a point in Jay's recent thread wrt Equation 2.10 where you can use an energy of the form of a KE of a PE.
Where would that go in a Lagrangian? I.e. is 'the KE of a PE' a KE or a PE?]

by **FrediFizzx** » Thu Aug 04, 2016 11:11 am

Obviously the Lagrangian was not equal to zero so there must be a term missing that would take it to zero. At least that is what I surmise.

by **Ben6993** » Thu Aug 04, 2016 5:22 am

Let me have a quick amateur reply before the experts step in.
Your classical radius agrees with http://scienceworld.wolfram.com/physics ... adius.html
But I imagine your question is "why is the QED lagrangian giving a classical radius"?

You say you have set ℒ equal to zero and then let KE equal zero.
As the ℒ equals KE - PE, setting ℒ =0 is equivalent to setting KE = PE.
Then you set KE to zero which implies PE equals zero also.
So I imagine you could even find 2=4 under those conditions.

To use ℒ, you plug ℒ into the Euler-Lagrangian equations eg here http://www.theoretical-physics.net/dev/quantum/qed.html
(Maybe you did this but it is not what you wrote?)
But I am sure you know all this, and I am not clear what question you are asking.

Perhaps you are querying why the QED Lagrangian does not use:
naked mass = infinite
naked charge = infinite
But these naked values are not observed in nature and can't be inputted to calculate a physical r. Can't even use them theoretically to calculate naked r = 0.

A maybe related question of mine follows. The physical electron is a combination of LH and RH non-physical electrons and gets its mass from interacting with the higgs field.
But IMO the LH and RH electrons are physical and maybe have different masses (perhaps very similar masses). [IMO this is closely related to electrons having hidden variables.]
Similarly the LH neutrino gets its mass from the higgs field. Presumably the LH neutrino is accepted as a "physical" particle. The RH particle may perhaps not be observed/produced if it has a much higher mass than that of the LH version.

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