***Gauge theory, the Aharonov-Bohm effect, and Dirac monopoles

Posted by brandonrayhaun On 2014-09-21 2018-12-12 Filed under path integral formulaion, physics, quantum mechanics 4 Comments

Suggested background: modern physics and quantum mechanics, basic electricity and magnetism
I thought I’d kick off my blog with some beautiful physics that I learned in a quantum mechanics course during my junior year: Dirac’s argument that, if magnetic monopoles happen to exist, they must have quantized charge. Dirac’s idea was one of many insightful arguments (some of which we’ll see in future posts) motivating this so called quantization condition. I haven’t been able to find a resource which gives you Dirac’s story in its entirety, at least not in any accessible way. So this post will be a long, somewhat ridiculous journey during which we’ll learn some gauge theory and also the related Aharonov-Bohm effect, each jewels in their own right. You’ll probably laugh, you’ll probably cry, and hopefully by the end you’ll have a newfound appreciation for how ludicrous Dirac was.

Here’s the plan:

1) I’ll tell you what magnetic monopoles are.

2) I’ll introduce the relevant electricity & magnetism and gauge theory, as well as how the two manifest in quantum mechanics.

3) I’ll show how this manifestation leads to the Aharonov-Bohm effect, a quantum phenomenon in which the vector potential acquires a more physical status it did not enjoy in classical mechanics; we’ll see how the vector potential itself can physically impact a system through the wave function, even when the electric and magnetic fields are ${0}$ . To get a quantitative feel for this effect, we’ll study how a solenoid, whose fields vanish outside of it, can affect the interference pattern observed in a traditional two-slit experiment.

4) I’ll justify how an infinitesimally thin solenoid with one end at infinity and the other at the origin can reproduce the field of a magnetic monopole. We’ll derive the quantization condition by asserting that, in order for such a solenoid to be indistinguishable from a magnetic monopole, there should be no experiment we can perform that will allow us to detect the solenoid. Bearing in mind the way in which solenoids can impact the two slit experiment, we’ll show that the solenoid current, and thus the magnetic charge, must take on only discrete, quantized values if the length of the solenoid is to escape detection via the Aharonov-Bohm effect.

If the details don’t make sense to you yet, don’t worry. Just reread the agenda after every section to reorient yourself to the larger task at hand and the logical flow shoud become apparent to you.

Magnetic monopoles

Before we chase down this bizarre argument, it’d be nice to know what magnetic monopoles are. In short, magnetic monopoles are hypothetical particles that carry magnetic charge in the same way electrons carry electric charge. The main reasons they remain ‘hypothetical’ are that they are 1) supposedly massive enough that their production requires more energy than available in current particle accelerators, and 2) also rare enough in the universe that the probability of their detection is vanishingly small. How convenient!

Still, most physicists would bet money that they’re real. Relatively recently, work in grand unified theories and quantum gravity has produced evidence in favor of their existence, but we will content ourselves with a more obvious argument by beauty.

Let’s look at Maxwell’s equations in the presence of sources, as they’re traditionally written down. In some units, they’re

$\displaystyle \vec{\nabla}\cdot\vec{B} = 0, \ \vec{\nabla}\cdot\vec{E} = 4\pi\rho_e$

$\displaystyle \vec{\nabla}\times\vec{E} + \frac{1}{c}\frac{\partial\vec{B}}{\partial t} = 0, \ \vec{\nabla}\times\vec{B} - \frac{1}{c}\frac{\partial\vec{E}}{\partial t} = \frac{4\pi}{c}\vec{J}_e$

One might notice that the equations on the left look exactly like those on the right, except for the fact that ${0}$ on the left replaces ${4\pi\rho_e}$ and ${\frac{4\pi}{c}\vec{J}_e}$ on the right. This is precisely due to the fact that there is no magnetic charge. Just as the presence of electric charges gives us a non-zero divergence of the electric field, so too would the presence of magnetic charges lead to the non-zero divergence of the magnetic field. These equations are practically begging to be symmetrized, so let’s introduce some monopoles:

$\displaystyle \vec{\nabla}\cdot\vec{B} = 4\pi\rho_m, \ \vec{\nabla}\cdot\vec{E} = 4\pi\rho_e$

$\displaystyle \vec{\nabla}\times\vec{E} + \frac{1}{c}\frac{\partial\vec{B}}{\partial t} = \frac{4\pi}{c}\vec{J}_m, \ \vec{\nabla}\times\vec{B} - \frac{1}{c}\frac{\partial\vec{E}}{\partial t} = \frac{4\pi}{c}\vec{J}_e$

One can easily check that making the field substitutions

$\displaystyle \vec{E}\rightarrow \vec{B}, \ \vec{B}\rightarrow -\vec{E}$

and also the source substitutions

$\displaystyle \rho_e\leftrightarrow \rho_m, \ \vec{J}_m\leftrightarrow \vec{J}_e$

leaves the modified Maxwell’s equations invariant; this is known as electric-magnetic duality. Since a symmetric theory is a more beautiful theory (at least to most), this is usually used as a simple amuse-bouche for why we should take magnetic monopoles seriously. But Dirac’s argument doesn’t have much to do with their existence–instead it is meant to illustrate how assuming their existence leads to their quantization, so let’s get into the machinery we need to understand this.

Electromagnetism preliminaries

First, some reminders: the magnetic field of an ideal, long solenoid with radius ${a}$ is given by $\vec{B}(\vec{r}) = \mu_0in$ when $|\vec{r}|\leq a$ , and $0$ otherwise. Here, ${\mu_0}$ is the vacuum permeability, ${i}$ is the current, and ${n}$ indicates how densely it is wound.

Once we know the fields, electricity and magnetism tells you how to calculate forces on particles from them via the force law:

$\displaystyle \vec{F} = q(\vec{E} + \vec{v}\times\vec{B})$

It is clear from this relation that there can be no electromagnetic influence in a region where ${\vec{B}}$ and ${\vec{E}}$ are identically 0 because the electromagnetic force vanishes then as well!

One often defines the electric scalar potential, ${\varphi}$ , and the magnetic vector potential, ${\vec{A}}$ , by the relations

$\displaystyle \vec{B}=\vec{\nabla}\times\vec{A}, \ \vec{E} = -\vec{\nabla}\varphi - \frac{\partial \vec{A}}{\partial t}.$

In classical electromagnetism, the potentials, strictly speaking, only serve as a useful tool used in place of fields. Potentials aren’t responsible for anything physical and can’t be measured; the fields do all the heavy lifting. In fact, the alternative description they afford us is ‘redundant’ in the sense that one has some freedom in choosing what potentials they want to work with. More concretely, if ${\varphi}$ and ${\vec{A}}$ describe the physics at hand, then so do ${\varphi'}$ and ${\vec{A}'}$ given by

$\displaystyle \varphi' = \varphi - \frac{\partial\chi}{\partial t}, \ \vec{A}' = \vec{A} + \nabla \chi$

where ${\chi}$ is an arbitrary function of position and time. You can verify this as an exercise by showing that

$\displaystyle \vec{B'} = \vec{\nabla}\times\vec{A'} = \vec{\nabla}\times\vec{A}=\vec{B}, \text{ and}$

$\displaystyle \vec{E}' = -\vec{\nabla}\varphi' - \frac{\partial\vec{A}'}{\partial t} =-\vec{\nabla}\varphi - \frac{\partial\vec{A}}{\partial t} = \vec{E}$

i.e., the fields described by the two sets of potentials are identical, and thus Maxwell’s equations are invariant under these transformations (hint: use the fact that ${\vec{\nabla}\times\vec{\nabla}\chi = 0}$ ). The changes ${\varphi\rightarrow\varphi'}$ and ${\vec{A} \rightarrow \vec{A}'}$ are referred to together as a gauge transformation. The statement that electromagnetism is a gauge invariant theory is precisely the statement that, if you tell me ${\varphi}$ and ${\vec{A}}$ describe the physics of a system, then I can pick any function ${\chi}$ I want and do all the computations using ${\varphi'}$ and ${\vec{A}'}$ perscribed above.

In quantum mechanics, the Hamiltonian of a system is impacted by the presence of electromagnetic potentials:

$\displaystyle H(\vec{A},\varphi) = \frac{1}{2m}\left[\vec{p} - q\vec{A} \right]^2 + V+q\varphi.$

If you want a gauge invariant quantum theory of electromagnetism, you need to do more than just transform the potentials; you need to subject the wave function to a transformation as well. You should check that, if ${\psi}$ is a solution to the Schroedinger equation

$\displaystyle H(\vec{A},\varphi)\psi = E\psi$

then ${\psi'}$ is a solution to the Schroedinger equation

$\displaystyle H(\vec{A}',\varphi')\psi' = E\psi'$

where the gauge transformation is now

$\displaystyle \varphi' = \varphi - \frac{\partial\chi}{\partial t}, \ \vec{A}' = \vec{A} + \nabla \chi, \ \psi' = e^{iq\chi/\hbar}\psi.$

Of course, ${\psi'}$ differs from ${\psi}$ only by a phase factor and so represents the same physical state since all probabilities are conserved, so we’ve discovered a truly gauge invariant formulation.

To summarize the above, all we did was describe a way of changing problems in electricity and magnetism while retaining the physics–we have some freedom in choosing the potentials (the wavefunction must come along for the ride if we’re working in quantum mechanics) and we can exploit this as a computational tool.

As one final remark, I ask that you recall a theorem of vector calculus: if ${\vec{\nabla}\times \vec{A} = 0}$ in any simply connected region (i.e. regions where all loops can be continuously contracted to a point), then you can write ${\vec{A}}$ in that simply connected region as ${\vec{A} = \vec{\nabla}\chi}$ for some function ${\chi}$ .

If you’re not familiar with what it means for a space to be simply-connected, it’s ok. We only really care about a couple examples. Your run of the mill Euclidean space is pretty obviously simply-connected because you can take any loop anywhere and shrink it until it’s a point. However, if you consider all of Euclidean space, but then remove the ${\hat{z}}$ -axis, then a circle in the ${\hat{x}}$ – ${\hat{y}}$ plane cannot be contracted to a point continuously, because you would need to pull it through the missing ${\hat{z}}$ -axis.

This is actually all we need to start understanding Dirac’s argument.

Aharonov-Bohm effect

In 1959, Aharonov and Bohm demonstrated that, in contrast to classical electricity and magnetism, there actually can be quantum mechanical e&m effects on particles which never travel through a region where there are fields. Here, I’ll basically follow Griffiths approach (which you should take a look at), but in the interest of showing you something new, I’ll try to solve everything Griffiths solves in a different way. We’ll look at two systems: the quantum ring, for which I’ll flex gauge theory’s muscles a bit, and the modified two-slit experiment, for which I’ll use Feynman’s path integral formulation. For the sake of continuity, I recommend you look at these two methods only after you’ve read the rest of this post (or at least read through Griffiths approach).

To start with, consider an electron constrained to move on a ring of radius ${R}$ in the plane (I’ll commonly refer to this system as the quantum ring). This is a pretty easy system to solve the Schrodinger equation for, so I recommend you try it yourself. Here, I’ll just tell you that the eigenstates and eigenenergies are

$\displaystyle \psi_n(\theta) \propto e^{\pm in\theta}, \ E_n = \frac{\hbar^2 n^2}{2mR^2}, \ n\in\mathbb{Z}.$

Now imagine sticking a long solenoid with radius ${a<R}$ inside the ring (I’ll refer to this new system as the solenoidal quantum ring). If we followed classical intuition, we might argue that nothing should change, since the fields outside of the solenoid all vanish and the particle can never travel inside the solenoid where the fields are non-zero. However the eigenvalue problem does change:

$\displaystyle E_n = \frac{\hbar^2}{2mR^2}\left(n-\frac{q\Phi}{2\pi\hbar}\right)^2, \ n\in \mathbb{Z}.$

The free particle cares about the flux through the solenoid–the vector potential in this context can be interpretted as having some sort of physicality.

Gauge theoretic derivation of solenoidal quantum ring

The good news is that the magnetic field is zero outside of the solenoid, which is where our particle lives. This means that ${\vec{B} = \vec{\nabla}\times \vec{A} = 0}$ there as well. However we are not totally safe to write ${\vec{A} = \vec{\nabla}\chi}$ because the region in which ${\vec{\nabla}\times\vec{A}=0}$ is not simply-connected. The issue is that loop like paths that wind around the solenoid can’t be contracted into a point. We can eliminate these loopy paths from our region entirely by only considering the region with ${\theta\in(-\pi+\epsilon,\pi-\epsilon)}$ and taking the limit as ${\epsilon\rightarrow 0}$ . In other words, we are taking scissors to a point on the ring on which our particle lives and then bringing the ends closer and closer together until they’re basically connected.

So in the region outside the solenoid where ${\theta\in(-\pi+\epsilon,\pi-\epsilon)}$ , we can write ${\vec{A} = \vec{\nabla}\chi}$ for some function ${\chi}$ . Here, the line integral between any two points ${a}$ and ${b}$ will be given by

$\displaystyle \int_{a}^b\vec{A}\cdot d\vec{l} = \chi(b)-\chi(a).$

Using Stoke’s theorem, we also know that

$\displaystyle \oint\vec{A}\cdot d\vec{l} = \iint_{S}\vec{B}\cdot d\vec{A} = \Phi$

where ${\Phi}$ is the magnetic flux through the solenoid. Consider the line integral around our ring, and break it up into two integrals–one restricted to the region we defined above, and the other integrating over what remains. Then we have that

$\displaystyle \Phi = \oint\vec{A}\cdot d\vec{l} = \int_{-\pi+\epsilon}^{\pi-\epsilon}RA_{\theta}d\theta + \int_{\pi-\epsilon}^{-\pi+\epsilon}RA_{\theta}d\theta.$

If we take the limit as ${\epsilon\rightarrow 0}$ , the second term will vanish because the limits of integration become closer and closer together. Rewriting the left hand term gives us

$\displaystyle \Phi = \lim_{\epsilon\rightarrow 0}\chi(\pi-\epsilon)-\chi(-\pi+\epsilon)$

and so we can write some ${\chi}$ which satisfies this constraint, the simplest choice being

$\displaystyle \chi(\theta) = \frac{\theta}{2\pi}\Phi.$

Keeping all that in mind, we can choose a gauge transformation which turns the solenoidal quantum ring into the traditional quantum ring. We’ll let

$\displaystyle \vec{A}\rightarrow\vec{A}'= \vec{A}-\vec{\nabla}\chi = 0, \ \psi\rightarrow\psi' = e^{-iq\chi/\hbar}\psi.$

Now ${\psi'}$ will obey the free particle on a ring equation. The caveat is that since ${\chi}$ is not single-valued at ${\theta=\pi}$ , ${\psi'}$ will not be either, so we have to come up with alternative boundary conditions to enforce. Let’s derive those boundary conditions quickly before we get into solving the eigenenergy problem.

Using the definition of ${\psi'}$ , we get that

$\displaystyle \psi'(\pi-\epsilon) = e^{iq\Phi(\pi-\epsilon)/(2\pi\hbar)}\psi(\pi-\epsilon),\text{ and}$

$\displaystyle \psi'(\epsilon-\pi) = e^{iq\Phi(\pi-\epsilon)/(2\pi\hbar)}\psi(\epsilon-\pi)$

so taking the limit as ${\epsilon\rightarrow 0}$ and requiring that the original wave function we had before the gauge transformation be single-valued, we get

$\displaystyle \psi'(\pi) = e^{-iq\Phi/\hbar}\psi(-\pi)$

Now we’re ready to solve the particle on a ring problem. Using the Wikipedia formula

$\displaystyle \nabla^2f = \frac{1}{r}\frac{\partial}{\partial r}\left( r \frac{\partial f}{\partial r}\right) + \frac{1}{r^2}\frac{\partial^2 f}{\partial\theta^2} + \frac{\partial^2 f}{\partial z^2}$

and recognizing that the wavefunction of our particle can be parametrized by only ${\theta}$ , (i.e. ${\frac{\partial \psi}{\partial r} = \frac{\partial \psi}{\partial z} = 0}$ ) the Schrodinger equation reduces to

$\displaystyle -\frac{\hbar^2}{2m}\frac{1}{R^2}\frac{\partial^2}{\partial \theta^2}\psi(\theta) = E\psi(\theta).$

This is a pretty standard differential equation that most of you have probably seen in introductory quantum mechanics courses. Making the ansatz ${\psi'(\theta) \propto e^{i\alpha\theta}}$ gives us

$\displaystyle \frac{\hbar^2\alpha^2}{2mR^2}\psi' = E\psi'$

and enforcing the boundary conditions we derived above gives us (after a little algebra) ${\alpha = \frac{-q\Phi}{2\pi\hbar} + n}$ for ${n\in \mathbb{Z}}$ . We can rearrange this to get the allowed energy values for this system:

$\displaystyle E_n = \frac{\hbar^2}{2mR^2}\left(n - \frac{q\Phi}{2\pi\hbar}\right)^2, \ n\in\mathbb{Z}$

which is the same result derived in Griffiths.

Now that we have a feel for what is happening, we can start thinking about the more traditional Aharonov-Bohm effect, which is studied in the context of electron interference.

Recall how an interference pattern is created in the ordinary two slit experiment from accumulated phase due to electron propagation. The idea is that placing a solenoid near the two slits (and making sure that electrons never pass through it) provides a new way for the two separate beams to acquire a new relative phase, and thus changes the interference pattern. If we look at a point directly on the opposite side of the solenoid, by symmetry, the phase the beams from the two different slits accumulate via the paths they take is the same, and so can be ignored. The Ahoronov-Bohm effect predicts that the solenoid will contribute a relative phase of

$\displaystyle \Delta \phi = \frac{q\Phi}{\hbar}$

between paths from the two slits so that the interference pattern that’s observed is different from that in the original two slit experiment. As in the case of the solenoidal quantum ring, placing a solenoid in between the two slits of a two slit experiment modifies the physics via the vector potential.

Path integral formulation of the modified two-slit experiment

To me, the most intuitive way to see how the Aharonov-Bohm effect arises is to use path-integrals. To appreciate the real solution, we’d need to use a bit of homotopy (more specifically, winding numbers to analyze all possible paths around the solenoid) but we’ll content ourselves with a bit of a simplification.

If you’re familiar with the Lagrangian formalism, you’ll remember that

$\displaystyle \mathscr{L}'(\vec{r},\vec{v})=\mathscr{L}(\vec{r},\vec{v}) - q\vec{v}\cdot\vec{A}(\vec{r})$

where ${\mathscr{L}}$ is the Lagrangian of the free particle. The action of a path ${\vec{r}}$ for this Lagrangian is given by

$\displaystyle S'[\vec{r}(t)] = \int\mathscr{L}'[\vec{r},\vec{v}; t]dt=S[\vec{r}(t)]-q\int\frac{d \vec{r}}{d t}\cdot\vec{A}(\vec{r})dt = S[\vec{r}(t)]-q\int \vec{A}(\vec{r})\cdot d\vec{r}$

where ${S}$ is the action without the solenoid present, and ${S'}$ is the action with the solenoid present.

Now, we want to determine the propagation amplitude from the electron source to the point directly opposite the solenoid. Of course, in Feynman’s formalism, we determine the propagator by summing a phase factor related to the action over all paths, namely

$\displaystyle K' = \int \mathscr{D}[\vec{r}(t)]\exp\left(\frac{i}{\hbar}S'[\vec{r}(t)] \right)=\int\mathscr{D}[\vec{r}(t)]\exp\left(\frac{i}{\hbar}S[\vec{r}(t)] \right)\exp\left(-\frac{iq}{\hbar}\int\vec{A}(\vec{r})\cdot d\vec{r} \right).$

We can define ${K'_i}$ by the same integral as ${K'}$ , except restricting the integral to be over paths which travel through slit ${i}$ . Since each path either travels through slit 1 or 2, then we would have ${K' = K'_1 + K'_2}$ .

Now, because ${\vec{\nabla}\times\vec{A} = 0}$ outside the solenoid, we know that the line integral ${\int \vec{A}(\vec{r})\cdot d\vec{r}}$ is the same for all paths which go through the same slit. In other words, if two different paths travel through the same slit ${i}$ , then their line integrals will be identical. (If we were being very rigorous, we would need to to justify this fact by considering only simply-connected regions). We can use this fact to pull the second exponential factor in the definition of ${K'_i}$ outside of the path integral,

$\displaystyle K'_i =\exp\left( -\frac{iq}{\hbar}\int_{C_i} \vec{A}(\vec{r})\cdot d\vec{r}\right) \int_{\text{slit i}} \mathscr{D}[\vec{r}(t)] \exp\left(\frac{i}{\hbar}S[\vec{r}(t)] \right) = \exp\left(i\phi_i\right)K_i$

where ${C_i}$ is any path which travels through slit ${i}$ , ${K_i}$ is the path integral for the system without the solenoid, and ${\phi_i := -\frac{q}{\hbar} \int_{C_i}\vec{A}(\vec{r})\cdot d\vec{r}}$ . Finally, we’re ready to derive the Aharonov-Bohm effect. The ordinary amplitude is given by

$\displaystyle K \propto K_1 +K_2$

whereas the solenoid modifies it like

$\displaystyle \begin{array}{rcl} K' &\propto& K_1' + K_2' \\ &=& K_1e^{i\phi_1} + K_2e^{i\phi_2} \\ &=& e^{i\phi_1}\left(K_1 + e^{i(\phi_2-\phi_1)}K_2 \right). \end{array}$

The exponent in the front is an overall phase factor, and so doesn’t affect the physics. However, the exponent in front of ${K_2}$ induces a relative phase given by

$\displaystyle \Delta\phi = \phi_2-\phi_1 = -\frac{q}{\hbar}\int_{C_2}\vec{A}(\vec{r})\cdot d\vec{r} + \frac{q}{\hbar}\int_{C_1}\vec{A}(\vec{r})\cdot d\vec{r}=\frac{q}{\hbar}\oint\vec{A}(\vec{r})\cdot d\vec{r}$

where of course we can apply Stoke’s theorem to the closed line integral to convert it into a surface integral over ${\vec{\nabla}\times\vec{A} = \vec{B}}$ , i.e.

$\displaystyle \Delta \phi = \frac{q}{\hbar} \iint_S\vec{B}\cdot d\vec{a} = \frac{q}{\hbar}\Phi$

This of course is the same answer we presented above.

Thin solenoid as a model for magnetic monopoles

You may be wondering how the Aharonov-Bohm effect relates to magnetic monopoles. I’ll try to justify the notion that we can think of a magnetic monopole as special kind of solenoid–once we’ve established this equivalence, we’ll be able to derive the quantization condition by studying monopoles within the context of the Aharonov-Bohm effect. I may address the equivalence in more detail in future blog posts, however, for our purposes we can intuit and then assume the answer.

If you image an ordinary solenoid and stare at the field lines, you’ll notice that it forms a dipole, each end constituting one of the poles. That’s all well and good, but we need one less pole if we want a monopole, so what’s the solution? Just toss one of the poles out of course!

To be less coy, what we’re proposing is that taking one end of the solenoid (one of the poles) and dragging it off to infinity makes the end that we have near us exhibit the field of a solenoid,

$\displaystyle \vec{B} = \frac{g \mu_0\hat{r}}{4\pi r^2}$

where ${g}$ is the magnetic charge.

The quantization condition

Here’s where the miracle happens. To recap, so far what we’ve argued is that in the two split experiment, if one includes a solenoid in the setup, electron beams from the two different slits arrive at the wall with a phase difference proportional to the magnetic flux through the solenoid–this causes a change in the interference pattern. We’ve also motivated how a solenoid might serve as a useful way of thinking about magnetic monopoles.

From here, Dirac asserted that, if solenoids were really a good model for magnetic monopoles, then there should be no experiment whatsoever that would allow us to determine that we have a solenoid, and not a monopole; the special Dirac solenoid and magnetic monopole should be indistinguishable. To undermine the equivalence, we’ve actually already shown how one might distinguish between a solenoid and a monopole–simply perform the two slit experiment along where you think the solenoid lies and check whether or not the interference pattern changes. Does this mean that we really can’t use Dirac solenoids and monopoles interchangably?

Not quite. In fact, we may be able to use this to our advantage. Let’s look more closely–if the accumulated phase difference is a multiple of ${2\pi}$ , then ${e^{i\Delta\phi}=1}$ and the solenoid makes no detectable impact on the physics. This occurs when (in units ${\hbar=1}$ )

$\displaystyle e\Phi = 2\pi n$

which we can rewrite by using the fact that the flux is related to the magnetic charge via ${\Phi = 4\pi g}$ ,

$\displaystyle e g = \frac{n}{2}.$

This is the quantization condition. We used the Aharonov-Bohm effect to constrain the possible values of magnetic charge by demanding that the modified two-slit experiment should not be able to distinguish between a long solenoid and a monopole. Pretty clever, and sort of bizarre.

Phew… it took us quite a bit of effort to get there, but this is a great starting point for learning more about this far-reaching topic. In future posts, I’ll try to talk more about why we can or can’t trust this result and then go through the ‘t Hooft-Polyakov monopole.

4 Comments for “***Gauge theory, the Aharonov-Bohm effect, and Dirac monopoles”

Your Math Student

says:

Whhhhooooooaa man! That was totally wiiiickkkkeeeddd!

[ペットキャリー]?在庫限り★?トップランナー Mサイズブラック×グリーン[O

says:

最近、私は、サイトのページの記事にコメントを残すに配慮の多くを与えるdidntの、さらにはるかに少ないのコメントを配置しています。あなたの素敵なポストを介して読書、時々そうするように、私を支援します。

Dreama

says:

The accident of finding this post has brehetingd my day

Wes Hansen

says:

Wow! This is crazy! I just left a comment on your Moonshine post, but this is much better.

William Tiller has demonstrated the existence of magnetic monopoles with some pH experiments – hold on now, I know!

To begin with, in my previous comment I linked to some papers by and about David Hestenes. In the paper of Hestenes I link to in that previous comment, Electron Time, Mass, and Zitter, Hestenes describes the Zitter model of the electron and tells us precisely what spin represents. The electron is a point charge which maps out a helical motion in spacetime and spin is a measure of helical orientation. Tiller’s magnetic monopoles are the source of this helical motion; the pilot wave has a net magnetic charge. There is a tremendous amount of synergy between Tiller’s model and that of Hestenes. I briefly write about this and link to the relevant papers in a Quora answer: https://www.quora.com/What-s-the-benefit-of-using-complex-numbers-instead-of-real-valued-vectors-in-quantum-mechanics/answer/Wes-Hansen-1.

I wish you would take a moment.

Okay, with regards to the magnetic monopoles:

“Three key experimental signatures heralded the onset of what we called “space conditioning”. One of these, in particular, was very important in our postulation that a partial SU(2) EM gauge symmetry state had been established in the experimental space. These three signatures were(18):
(1) A D.C. magnetic field-polarity effect wherein the south-labeled magnetic pole pointing upwards into a pH-measuring water vessel produced a large increase in the pH-measurement (more alkaline) while (ii) the reverse polarity north-labeled pole pointing upward into the water, led to a substantial decrease in the pH-measurement (more acidic). It is not possible for this to occur in a normal U(1) gauge state space because only magnetic dipoles are present in such a Gauge state, both the magnetic force and magnetic energy are proportional to H2 (so “sign” effect is not relevant) and no geometrical orientation can occur.
(2) The development of very low frequency oscillations, in the 10-1 to 10-4 hertz range, for air temperature, water temperature, pH and electrical conductivity of the water and
(3) The development of macroscopic coherence effects in both the air and the water over distances of ~10 feet wherein, for all four time-dependent property measurements, their Fourier spectra completely nested with each other.”
That’s from his https://tillerinstitute.com/pdf/White%20Paper%20XIX.pdf, page 20. See also https://tillerinstitute.com/pdf/White%20Paper%20XXIII.pdf.

Okay, in addition to this, we have the recent Pre-stimulus experiments conducted at UC Santa Barbara; those experiments used quantum random event generators and are backed up by more than 1500 samples from, I think, 46 or 48 different experiments using psuedo and truly random event generators. In fact, it was a meta-analysis published in Frontiers of Psychology in 2012 which motivated the Santa Barbara experiments; an update to that meta-analysis was published in SSRN in 2017. These experiments show that the human heart and brain respond in a meaningful way to an event 2 to 18 seconds before that event occurs in spacetime! Tiller can explain those experimental results. In addition to that, we have the thermodynamically inexplicable g-tummo experiments published in the PLOS/ONE journal in 2015: https://journals.plos.org/plosone/article?id=10.1371/journal.pone.0058244.
Tiller can explain those as well.

I don’t know if you will care about this stuff, being all into that stringy craze, but in the event . . .

Magnetic monopoles

Electromagnetism preliminaries

Aharonov-Bohm effect

Thin solenoid as a model for magnetic monopoles

The quantization condition

4 Comments for “***Gauge theory, the Aharonov-Bohm effect, and Dirac monopoles”

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