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Why and how do magnetic monopoles and (elementary) electric dipoles break microscopic T-symmetry?

+ 2 like - 0 dislike
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I always thought that on the microscopic fundamental scale, T-symmetry of the laws of nature should hold and the fact that it is broken at macroscopic scales (arrow of time) can loosely speaking be seen as some kind of coarse-graining effect.

Now the (not online available) particle physics lecture notes I am reading say that magnetic monopoles and elementary electric dipoles (such as hypothetical electric dipole moment of the electron ?) would break T-symmetry at the microscopic scale. I dont understand this.

In particular concerning magnetic monopoles, a quantum field theory with monopoles taking the role of electrons and positrons is S-dual to conventional QED which conserves T-symmetry...

asked May 24, 2015 in Theoretical Physics by Dilaton (4,295 points) [ revision history ]

Perhaps you could provide some other way to identify the resource, like author, title, date of publication, publisher, etc.?

I always thought that on the microscopic fundamental scale, T-symmetry of the laws of nature should hold

According to the mainstream flow, there may not be "microscopic" (underlying) description, due to lack of the corresponding knowledge, so I do not understand how you dare to ask such questions. What do you mean by that?

@Dimension10, unfortunately the Prof. has written that part of the lecture notes himself in German, and I dont know where he has taken it from...
 

2 Answers

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Only the combination $CPT$ of the charge conjugation $C$, of the parity reversal $P$ and of the time reversal $T$ is expected to be fundamentally true and is true in any local Lorentz invariant quantum field theory (in the Euclidean theory $CPT$ becomes a rotation of angle $\pi$, which is continously connected to the identity in the rotation group and so is obviously a symmetry). In fact in the Standard Model $CP$ and so $T$ are violated. This was first experimentally observed in 1964 in the study of kaons, see http://en.wikipedia.org/wiki/CP_violation and is explained by the existence of three generations which allow for a non-trivial $CP$ violating phase in the $CKM$ matrix. Another possible source of $CP$ and so $T$ violation in the Standard Model comes from the theta angle of $QCD$ but experimental bounds show that the theta angle is very near to zero.

Obviously these microscopic violations of $T$ have nothing to do with the macroscopic thermodynamical arrow of time. Most of the macroscopic phenomena are controlled by $QED$ which is $T$ invariant.

A non-trivial electric dipole moment for the electron or the neutron would be a sign of $T$ and so $CP$ violation. Indeed by general principle the electric dipole moment $\vec{m_e}$ is expected to be proportional to the spin $\vec{S}$: $\vec{m_e}=\alpha \vec{S}$ for some $\alpha$. But under $T$, $\vec{S} \mapsto - \vec{S}$ and $\vec{m_e} \mapsto \vec{m_e}$ (to remember these transformations it is enough to look at some naive classical expressions) and the electron remains an electron and so if $T$ is a symmetry, $\alpha=-\alpha$ i.e. $\alpha=0$. Similarly one can show that a non-trivial electric dipole moment violates $P$ but preserves $C$. Remark that most electrically charged particles have a non-trivial magnetic dipole moment, proportional to the spin. This does not violate $T$ or $P$ because a magnetic dipole moment transforms under $P$ or $T$ in the opposite way of a electric dipole moment.The observed $CP$ violation in the Standard Model implies the existence of a non-trivial electric dipole moment for the electron but it is a very indirect effect and the expected contribution is so small that it is out of reach experimentally.

I think that a magnetic monopole does not violate $T$ or $CP$. Indeed by $S$ duality we expect a monopole to have a electric dipole moment and to preserve $T$ as an ordinary electron with an magnetic dipole moment preserves $T$. Similarly a magnetic monopole with a non-trivial magnetic dipole moment will violate $T$. A possible confusing point is that the $S$-duality does not commute with $P$ and $T$. Indeed $S$-duality is essentially applying the Hodge star to the 2-form field strength and the Hodge star depends on a choice of orientation which is reversed by $P$ and $T$.

A dyon, i.e a particle with both electric and magnetic charges, violates $CP$, and so $T$ if its electric charge is not appropriately quantized. In a theory with pure electric charges, it is known since Dirac that the magnetic charge  is quantized but one has to use $CP$ symmetry to also prove the quantization of electric charges of dyons. The fact that a non-trivial theta angle implies non-trivial electric charges for dyons is called the Witten effect.

answered May 25, 2015 by 40227 (4,660 points) [ revision history ]
edited May 25, 2015 by 40227
why would the electric dipole moment be proportional to spin?

Because from the analysis of the representations of the Poincaré group the spin is the only "intrinsic direction" attached to an "elementary" particle so any non-trivial vector property of such a particle has to be proportional to it. It is well-known that it is the case for the magnetic dipole moment of the electron for example.

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For micro/macroscopic T violation, just as 40227 said, $T$ is broken in the standard model, which means it is already broken at the microscopic level.

As for why an elementary electric dipole/magnetic monopole breaks T, here's my analysis:

(1) Electric dipole: Let's consider electro-magneto-statics in some reference frame. Your dipole interaction term must be an object that transforms like $e\vec{E}\cdot\vec{r}$. Now you change to a boosted frame, you will have terms consisting of $er_iB_j$, which really means your Lagrangian must contain this latter term in the first place, and this term is odd under $T$.

(2) Magnetic monopole: Again consider electro-magnetostatics with no current in some reference frame. Your monopole interaction term must transform like $qV(\vec{r})$, where $q$ is the magnetic charge (which is odd under $T$), and $ \nabla V(\vec{r})=\vec{B}$, this means $V$ is also odd under $T$. In a boosted frame, $V$ will receive a contribution from the electric field, hence you must include $q\int \vec{E}$ in your interaction, and this term is odd in $T$. 

Update: The whole (2) is quite naively wrong, see 40227's comment below.

answered May 25, 2015 by Jia Yiyang (2,465 points) [ revision history ]
edited May 26, 2015 by Jia Yiyang

I don't understand the last part of 2). Let us consider the usual case of an electric charge $e$ (which is even under $T$). Then without currents the interaction term is the usual $eV$ where $V$ is the electric potential, which is even under $T$. It seems to me that you would then conclude that in a boosted frame one has to include $e \int \vec{B}$ which is odd in $T$. But the correct additional term is not this one but $e \vec{A}.\vec{v}$ which is even in $T$. Similarly in the magnetic case the additional term is  not $q\int \vec{E}$ but $q \vec{A}.\vec{v}$ for some $\vec{A}$ such as $\vec{\nabla} \times \vec{A}=\vec{E}$. This $\vec{A}$ is even under $T$, $q$ is odd, $\vec{v}$ is odd so this term is even under $T$.

@40227, you are quite right, I retract the second part.

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