What part of physics describes what happens just outside of an electron?

[Anon42]

TL;DR Summary

What is the strength of the electric field of an electron just outside of the electron, and is it strong enough to enable pair production?

Coulomb's law for three dimensional space is an empirical law that describes the forces between two stationary point charges and is defined as:
$$\vec{F}=\frac{K q_1 q_2 (\vec{r}_1-\vec{r}_2)}{|\vec{r}_1-\vec{r}_2|^3}$$
From Coulomb's law, the magnitude and direction of an electric field produced by a point charge can be modeled. $q_2$ is removed from the original equation, as is ${r}_2$. The cartesian coordinate system is then defined so that the origin of the coordinate system is the point charge and the vector ${r}_1$ always originates from the origin. The equation then becomes:
$$\vec{E}=\frac{K q_1 \vec{r}_1}{|\vec{r}_1|^3}$$

Is it possible to define the magnitude of ${r}_1$ as the charge radius of an electron $(2.82 ⋅ 10^{-15} m)$ plus a couple Planck's constants$(6.62607004 ⋅ 10^{-34} \frac{m^2}{kg⋅s})$ and get a meaningful result? The Planck constants are added on to the electron radius to make sure that the electric field solved for is in fact outside of the electron. For good measure, we can let $|{r}_1|=3.00 ⋅ 10^{-15} m$

So for our equation, the variables are defined as:
$q_1$ is taken be the charge of an electron, $1.60217662 *10^{-19} C.$
${r}_1$ is taken to be the magnitude and direction of a position in space with the vector originating from the electron with: $|{r}_1|=3.00 ⋅ 10^{-15} m$
$K$ is taken to be the Coulomb Constant $8.9875517923 * 10^{9} \frac{kg⋅m^{3}}{s^{2}⋅C^{2}}.$

I tried solving this equation for electric field strength just outside an electron say in some arbitrary x-direction and got the following:

$$\vec{E}=\frac{8.9875517923⋅10^{9} \frac{kg⋅m^{3}}{s^{2}⋅C^{2}} 1.60217662⋅10^{-19} C⋅[3.00 ⋅10^{-15} m, 0 m, 0 m]}{|3.00 ⋅10^{-15}|^3}$$
$$\vec{E}=1.599960595 ⋅10^{20} \frac{V}{m}$$

If this calculation is done right, then an electron has an electric field strength of roughly $10^{20} \frac{V}{m}$ in the space just outside of it!

However, Wikipedia states that the Schwinger limit, the limit which the electromagnetic field becomes nonlinear occurs at roughly $10^{18} \frac{V}{m}.$ Electron-positron pair production happens at fields at and above the Schwinger limit. Every electrons in the universe doesn't spontaneously produce electron-positron pairs. That leads me to believe that the electric field strength just outside of an electron must somehow be less than $10^{20} \frac{V}{m},$ and closer to $10^{18} \frac{V}{m}.$. Either that or the field is in fact on the order of $10^{20} \frac{V}{m},$ and some quantum mechanism prevents pair production near electrons.

My question is this: Did I make a mistake in calculating the electric field strength just outside of an electron? If not, then why don't all electrons spontaneously generate electron-positron pairs? Or am I just misunderstanding the Schwinger limit in some way? The electric field strength just outside of an electron is enormous. How does our current understanding of QED describe this?

 

[PeterDonis]

the electric field of an electron just outside of the electron

An electron is not a little solid ball; it does not have a well-defined boundary. So "just outside of the electron" does not make sense.

For distance scales as small as you are talking about, classical electrodynamics doesn't work anyway. You need to use quantum electrodynamics. You correctly posted this question in the quantum physics forum, but you are not actually using quantum physics in your analysis.

Did I make a mistake in calculating the electric field strength just outside of an electron?

Your mistake was more basic: trying to use classical electrodynamics instead of quantum electrodynamics.

am I just misunderstanding the Schwinger limit in some way?

Yes, you are. A single electron cannot generate the kind of "strong electric field" (which is itself a somewhat misleading term since it suggests using classical electrodynamics as you have tried to do, which doesn't work--a better term would be "quantum field state") that is required to exceed the Schwinger limit.

 

[Anon42]

An electron is not a little solid ball; it does not have a well-defined boundary. So "just outside of the electron" does not make sense.

For distance scales as small as you are talking about, classical electrodynamics doesn't work anyway. You need to use quantum electrodynamics. You correctly posted this question in the quantum physics forum, but you are not actually using quantum physics in your analysis.Your mistake was more basic: trying to use classical electrodynamics instead of quantum electrodynamics.Yes, you are. A single electron cannot generate the kind of "strong electric field" (which is itself a somewhat misleading term since it suggests using classical electrodynamics as you have tried to do, which doesn't work--a better term would be "quantum field state") that is required to exceed the Schwinger limit.

I assumed this question would involved quantum electrodynamics, so that's why I posted it here.

Since classical electrodynamics does not accurately describe electric fields at small distances, is there any simple equation in QED that does, like there is in classical electrodynamics with (kq/r^2).

I don't have any background in QED, nor am I majoring in it. Is there a simple equation for me to understand how QED describes the field of an electron at small distances, or do I have to take a course in QED for this?

P.S. Thanks for the quick reply.

 

[PeterDonis]

Since classical electrodynamics does not accurately describe electric fields at small distances, is there any simple equation in QED that does

No, because the concept of "electric field" is not a basic concept in QED to begin with. It is only a name for certain particular kinds of states of the quantum electromagnetic field. Those states are not states that involve single electrons.

Is there a simple equation for me to understand how QED describes the field of an electron at small distances

There is no such thing in QED as "the field of the electron" the way you are thinking of it.

do I have to take a course in QED for this?

A course in QED would help you to understand how QED actually works, yes. But it won't tell you how to calculate "the field of the electron at small distances", since, as above, there is no such thing in QED.

 

[Anon42]

No, because the concept of "electric field" is not a basic concept in QED to begin with. It is only a name for certain particular kinds of states of the quantum electromagnetic field. Those states are not states that involve single electrons.There is no such thing in QED as "the field of the electron" the way you are thinking of it.A course in QED would help you to understand how QED actually works, yes. But it won't tell you how to calculate "the field of the electron at small distances", since, as above, there is no such thing in QED.

Alright, noted. Thank you.

 

[Dale]

Summary:: What is the strength of the electric field of an electron just outside of the electron, and is it strong enough to enable pair production?

Is it possible to define the magnitude of r1 as the charge radius of an electron (2.82⋅10−15m) plus a couple Planck's constants(6.62607004⋅10−34m2kg⋅s) and get a meaningful result?

No. The result is not meaningful. This is a classical computation assuming that the electron is a tiny classical ball. It is not.

To get something realistic then you will need to use QED, or at least non-relativistic QM. There is no classical calculation that can give a meaningful result in this regime.

Edit: never mind, I see I am hours late. I will just “second” @PeterDonis

 

[vanhees71]

No, because the concept of "electric field" is not a basic concept in QED to begin with. It is only a name for certain particular kinds of states of the quantum electromagnetic field. Those states are not states that involve single electrons.There is no such thing in QED as "the field of the electron" the way you are thinking of it.A course in QED would help you to understand how QED actually works, yes. But it won't tell you how to calculate "the field of the electron at small distances", since, as above, there is no such thing in QED.

Hopefully such a course will tell, how to calculate it. It's the self-energy (vacuum polarization) of the photon, from which you can infer also the vertex corrections and in some sense the "electrostatic field of an electron".

 

[PeterDonis]

in some sense the "electrostatic field of an electron"

Only "in some sense". That's the point. It does not, for example, imply that a single electron can trigger pair production, as the "naive" application of the Schwinger limit that the OP was attempting would.

 

[vanhees71]

Of course not. Schwinger pair production is the production of $\text{e}^+ \text{e}^-$ pairs due to a classical charge/classical field ("hemiclassical approximation").