What really is the charge of the electron?

[Somali_Physicist]

In my study of Quantum physics i have arrived at the natural shielding caused by photon to electron/anti-electron pair fluctuations.The shielding reduces as you get closer the electron which naturally leads me to the question.What actually is the charge on an electron?

 

[PeterDonis]

What actually is the charge on an electron?

It depends on the energy at which you make the measurement of the charge. That is what the "shielding" is really telling you. A more technical description of what is going on is "renormalization group flow".

 

[ZapperZ]

In my study of Quantum physics i have arrived at the natural shielding caused by photon to electron/anti-electron pair fluctuations.The shielding reduces as you get closer the electron which naturally leads me to the question.What actually is the charge on an electron?

The charge of an electron is the published value in CODATA and PDG handbook. If this value is wrong, many particle accelerators, especially those synchrotron light sources and FELs, will be in deep doo doo, because they were built based on the physics that used that value.

We know about shielding. In fact, in condensed matter physics, this is an everyday issue. It still does not detract from us knowing its value, even considering the issue of quantum fluctuations that surround every particle that we know. Do the e/m measurement using Helmholtz coils as many times as you want, and you'll get the same answer.

Zz.

 

[PeterDonis]

The charge of an electron is the published value in CODATA and PDG handbook.

Which, as I understand it, is the low energy value--in renormalization group terms, it's the value that the coupling constant flows to in the limit of zero interaction energy. I agree that this value has been measured many, many times in experiments and is not in dispute.

 

[cosmik debris]

Which, as I understand it, is the low energy value--in renormalization group terms, it's the value that the coupling constant flows to in the limit of zero interaction energy. I agree that this value has been measured many, many times in experiments and is not in dispute.

Does the coupling constant change with self-energy? With that I mean if an electron has more kinetic energy does its self-energy increase and therefor the coupling constant?

Cheers

 

[PeterDonis]

Does the coupling constant change with self-energy?

"Self-energy" is a vague term. The coupling constant depends on the interaction energy--the total energy of all particles involved in the interaction, in the center of mass frame.

if an electron has more kinetic energy does its self-energy increase and therefor the coupling constant?

Kinetic energy is frame-dependent, so it can't be a factor in the coupling constant.

 

[cosmik debris]

Kinetic energy is frame-dependent, so it can't be a factor in the coupling constant.

Yes of course, thanks.

 

[vanhees71]

Does the coupling constant change with self-energy? With that I mean if an electron has more kinetic energy does its self-energy increase and therefor the coupling constant?

Cheers

By definition the change of the coupling constant comes from radiative vertex corrections, i.e., one-particle irreducible (1PI) Feynman diagrams with three external amputated legs (electron, positron, and photon lines) and loops.

However since QED is an Abelian gauge theory there are socalled Ward-Takahashi identities, which connect this tree-point vertex function with the electron self energy (1PI diagram with two external amputated electron-positron lines). This connects the wave-function renormalization of the electron/positron with the renormalization of the coupling constant.

In non-Abelian gauge theories you can play a similar trick using the socalled background-field gauge, where you get similarly simple Ward-Takahashi identities as in QED, because in this particular gauge the effective quantum action becomes gauge invariant itself. Then, in this gauge, this implies that the gauge-boson wave-function reormalization factor is related with the coupling-constant renormalization factor through a QED-like Ward identity, and thus you don't need to evaluate the vertex corrections to get the running of the coupling (and asymptotic freedom!) but the somewhat less complicated gauge-field self-energy.

For details, see my QFT manuscript:

https://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

 

[Vanadium 50]

B-level?

 

[vanhees71]

Argh. Sorry, I always overlook the levels.

 

[Somali_Physicist]

It depends on the energy at which you make the measurement of the charge. That is what the "shielding" is really telling you. A more technical description of what is going on is "renormalization group flow".

Charge measurement fluctuates with energy as this induces photons to spontaneously come into existence and thus increasing shielding (i'm assuming). What i am asking is that if you get rid of the fluff of vacuum shielding what is the charge left on the electron.After all as the shielding causes the electron to be a function of radius how could we even measure this?

The charge of an electron is the published value in CODATA and PDG handbook. If this value is wrong, many particle accelerators, especially those synchrotron light sources and FELs, will be in deep doo doo, because they were built based on the physics that used that value.

We know about shielding. In fact, in condensed matter physics, this is an everyday issue. It still does not detract from us knowing its value, even considering the issue of quantum fluctuations that surround every particle that we know. Do the e/m measurement using Helmholtz coils as many times as you want, and you'll get the same answer.

Zz.

Well how do we know that this shielding doesn't reduce the effective electric charge and thus all classical experiments? after all its only at small distances that this effect fluctuates.Similar how we can assume relativistic effects don't exist at Newtonian speeds.

 

[PeterDonis]

What i am asking is that if you get rid of the fluff of vacuum shielding what is the charge left on the electron

And the answer is that you can't get rid of the vacuum shielding without getting rid of the electron itself as well. The two are inseparable. There is no such thing as an electron without the vacuum shielding but with charge still left.

The best you can do is to probe the electron with a wide range of interaction energies and see how the measured charge varies.

as the shielding causes the electron to be a function of radius

I have no idea what you mean by this.

 

[ZapperZ]

Well how do we know that this shielding doesn't reduce the effective electric charge and thus all classical experiments? after all its only at small distances that this effect fluctuates.Similar how we can assume relativistic effects don't exist at Newtonian speeds.

So where is this reduction? Don't you think that if there is such a thing, it will be detected? And if it can't be detected, what difference does it make, because it is now a unicorn that you've made up that can't be measured?

Zz.

 

[PeterDonis]

how do we know that this shielding doesn't reduce the effective electric charge and thus all classical experiments?

You have it backwards. The "effective electric charge" is what we actually measure in experiments, and there are measurements from various high energy accelerators showing how the measured charge varies with interaction energy. The "shielding" is our best current theoretical explanation of why the measured charge varies with interaction energy (heuristically, because a higher energy probe can penetrate further into the shielding).

 

[Somali_Physicist]

You have it backwards. The "effective electric charge" is what we actually measure in experiments, and there are measurements from various high energy accelerators showing how the measured charge varies with interaction energy. The "shielding" is our best current theoretical explanation of why the measured charge varies with interaction energy (heuristically, because a higher energy probe can penetrate further into the shielding).

so the charge varies from a base point which we know of , okay.

And the answer is that you can't get rid of the vacuum shielding without getting rid of the electron itself as well. The two are inseparable. There is no such thing as an electron without the vacuum shielding but with charge still left.

The best you can do is to probe the electron with a wide range of interaction energies and see how the measured charge varies.
I have no idea what you mean by this.

Alright that makes sense, what i meant in the second part was referring to the fine constant changing the closer you get to an electron.

 

[vanhees71]

So where is this reduction? Don't you think that if there is such a thing, it will be detected? And if it can't be detected, what difference does it make, because it is now a unicorn that you've made up that can't be measured?

Zz.

But indeed it can be detected. The running of the em. coupling has been measured. At the Z peak it's about $\alpha_{\text{em}}(m_Z) \simeq 1/128$, compared to $\alpha_{\text{em}}(0) \simeq 1/137$. Here are just two examples found easily on Google Scholar:

https://arxiv.org/abs/hep-ex/0505072
https://arxiv.org/abs/hep-ph/9411377

 

[Somali_Physicist]

But indeed it can be detected. The running of the em. coupling has been measured. At the Z peak it's about $\alpha_{\text{em}}(m_Z) \simeq 1/128$, compared to $\alpha_{\text{em}}(0) \simeq 1/137$. Here are just two examples found easily on Google Scholar:

https://arxiv.org/abs/hep-ex/0505072
https://arxiv.org/abs/hep-ph/9411377

Still doesn't make sense.So charge increases the closer you get to the electron yet you are saying we definitely know the charge?

 

[vanhees71]

I don't understand where your problem is. The charge of the electron is well determined. It's -1 in natural units. The electromagnetic coupling constant $\alpha_s$ is "running" as is any coupling constant in QFT. There's nothing undefined or uncertain about the electron's charge at all.

 

[Demystifier]

I don't understand where your problem is. The charge of the electron is well determined. It's -1 in natural units. The electromagnetic coupling constant $\alpha_s$ is "running" as is any coupling constant in QFT. There's nothing undefined or uncertain about the electron's charge at all.

What can still be confusing to many is that the electron charge is a coupling constant itself, so if all coupling constants run, then why is the electron charge always exactly -1? (I know why, but I don't know how to explain it at a B-level.)

 

[Demystifier]

(I know why, but I don't know how to explain it at a B-level.)

Or let me try. Doing theoretical physics usually consists of two parts:
(i) Writing the equations.
(ii) Finding the solutions of those equations.
Usually the equations in (i) look simple, but the solutions in (ii) look very complicated. Very often we cannot find exact solutions, but only approximative ones.

The charge $e=-1$ is a parameter that appears in (i). But that's not something that we can measure directly. What we directly measure is some physical measurable quantity $M$. To find how $M$ depends on $e$, we must do (ii), that is find the solution of (i). We don't know the exact solution, but we know an approximative solution that has the form
$$M\approx ae^2$$
where $a$ is constant. Another way to write this is
$$e^2\approx \frac{M}{a} \;\;\;\; (1)$$
This is a very simple expression, but unfortunately it is only approximately true. It would be nice if something as simple as that would be exactly true, so we ask the following question: Is it possible to find a similar relation that is both simple and exactly true at the same time?

It turns out that it is possible, by performing a trick that is called "renormalization". Instead of dealing with the charge $e$, we introduce a new quantity $e_R$ defined by the formula
$$e^2_R\equiv \frac{M}{a} \;\;\;\; (2)$$
The quantity $e_R$ is called renormalized charge. It does not appear in simple equations in (i), but it is useful because the solution (ii) looks simple when expressed in terms of $e_R$, rather than $e$. Formula (2) looks very similar to the formula (1), except that formula (2) is exactly true, by definition.

Since $M$ is something that we actually measure, it should not be surprising that it depends on details how the measurement is performed. In particular, it depends on energies of the particles, that is on how close the particles approach each other. Hence $M$ is really a function of energy, so we write $M=M(E)$. But then (2) implies that $e_R$ is also a function of energy, so (2) must be written as
$$e^2_R(E) = \frac{M(E)}{a} \;\;\;\; (3)$$
Thus we see that the renormalized charge $e_R$ is not a constant, but a function of energy. This dependence of $e_R$ on energy $E$ is called "running". At the same time, $e=-1$ is a true constant. The parameter $e$ is useful because it makes the equations (i) simple, while the $e_R(E)$ is useful because it makes the solution (ii) simple.

The final and perhaps most difficult question is what is the "true" charge? Is it $e$ or is it $e_R$? Well, the question is more philosophical than physical. It depends on what do you mean by "true". The $e$ is "true" because it appears in the fundamental equations in (i), while $e_R$ is "true" because it is directly related to something measurable.

 

[Somali_Physicist]

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Or let me try. Doing theoretical physics usually consists of two parts:
(i) Writing the equations.
(ii) Finding the solutions of those equations.
Usually the equations in (i) look simple, but the solutions in (ii) look very complicated. Very often we cannot find exact solutions, but only approximative ones.

The charge $e=-1$ is a parameter that appears in (i). But that's not something that we can measure directly. What we directly measure is some physical measurable quantity $M$. To find how $M$ depends on $e$, we must do (ii), that is find the solution of (i). We don't know the exact solution, but we know an approximate solution that has the form
$$M\approx ae^2$$
where $a$ is constant. Another way to write this is
$$e^2\approx \frac{M}{a} \;\;\;\; (1)$$
This is a very simple expression, but unfortunately it is only approximately true. It would be nice if something as simple as that would be exactly true, so we ask the following question: Is it possible to find a similar relation that is both simple and exactly true at the same time?

It turns out that it is possible, by performing a trick that is called "renormalization". Instead of dealing with the charge $e$, we introduce a new quantity $e_R$ defined by the formula
$$e^2_R\equiv \frac{M}{a} \;\;\;\; (2)$$
The quantity $e_R$ is called renormalized charge. It does not appear in simple equations in (i), but it is useful because the solution (ii) looks simple when expressed in terms of $e_R$, rather that $e$. Formula (2) looks very similar to the formula (1), except that formula (2) is exactly true, by definition.

Since $M$ is something that we actually measure, it should not be surprising that it depends on details how the measurement is performed. In particular, it depends on energies of the particles, that is on how close the particles approach each other. Hence $M$ is really a function of energy, so we write $M=M(E)$. But then (2) implies that $e_R$ is also a function of energy, so (2) must be written as
$$e^2_R(E) = \frac{M(E)}{a} \;\;\;\; (3)$$
Thus we see that the renormalized charge $e_R$ is not a constant, but a function of energy. This dependence of $e_R$ on energy $E$ is called "running". At the same time, $e=-1$ is a true constant. The parameter $e$ is useful because it makes the equations (i) simple, while the $e_R(E)$ is useful because it makes the solution (ii) simple.

Basically my question is if charge is changing due to screening and all our measurments are altered by this.Is it not fine to say that the electron charge is unknown due to this inherent uncertainty.Or is it such that the electron hits a value at a distance which stays constant from then on?

 

[Demystifier]

Basically my question is if charge is changing due to screening and all our measurments are altered by this.

If by charge you mean my $e_R$, then the answer is - yes.

 

[Somali_Physicist]

If by charge you mean my $e_R$, then the answer is - yes.

Ok so then how do we know what the "charge of an electron" really is.

 

[Demystifier]

Ok so then how do we know what the "charge of an electron" really is.

See the last paragraph in my post #20.