# Why does flowing current break time reversal symmetry?

I am reading about the Quantum Hall Effect. In a course, they wrote that

And I don't know what it means. About TR symmetry means that the same physical laws describe a considered system with $$t \to -t$$. Also, I know that with this transformation $$E \to E$$, which means that TR symmetry doesn't change the sign of the electric field. But why does flowing current break this symmetry?

• Well time reversal will reverse the direction of the current, right? Because the charge carriers making up the current will move in the opposite direction. Hence $\mathbf J \to - \mathbf J$ under time reversal, while $\mathbf E \to \mathbf E$. I think that is all they are saying.
– ummg
Commented Nov 15, 2023 at 15:23

The article claims time reversal is broken by dissipation, that is: it's a Second Law of Thermodynamics thing. I disagree with this usage of time reversal.

The root of the Hall effect is the Lorentz Force Law:

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

under time reversal, that becomes:

$$\vec F' = q(\vec E + (-\vec v) \times (-\vec B)) = \vec F$$

Velocity is obviously time-odd, and the magnetic field depends on current, which is also time-odd.

Since you didn't include any formulae in your question, I had to look it up, and just got:

$$R = V/I$$

i.e., Ohms law. $$R$$ is clearly time even, while $$I$$ is time odd. Meanwhile, since

$$\vec E = -\nabla \Phi$$

is time even, one might suspect $$V$$ is also time even...with the distressing conclusion that Ohm's Law breaks time reversal invariance.

But: the voltage in question is defined as:

$$V = \Phi_{in} - \Phi_{out}$$

where "in" and "out" are defined by direction of current flow, so the labels are swaped

$$R = \frac{\Phi_{in} - \Phi_{out}}I \rightarrow \frac{\Phi_{out} - \Phi_{in}}{-I} = R$$

restoring faith in Ohm.

• Imagine a battery connected to a resistor. This device flows current according to ohm's law, but if you time-reverse this process it doesn't follow Ohm's law. Of course microscopically it is still a valid solution of the equations of motion, but it doesn't make sense from an entropic point of view (like many other time reversed examples of irreversible processes in thermodynamics). So I think the article is correct that Ohm's law apparently violating time reversal symmetry is related to dissipation. Commented Nov 16, 2023 at 17:32
• Do newtons laws violate time reversal? No. Is a billiard break reversible, or dropping a jar of marbles? No. It is a confusing misuse of the term.
– JEB
Commented Mar 1 at 21:28
• I agree that the microscopic laws do not break time reversal symmetry in the case of Ohm’s law. I think what I am saying is clear if you think about what happens microscopically for one particle performing ohmic conduction. Electrons get accelerated due to the E field, and collide with impurities and get scattered, imparting energy into impurities as heat. Then their velocity becomes zero (on average) and gets accelerated again due to the E field, and so on. Time reversal does not change the sign of E field. If you look at it microscopically, the time reversed path of the electrons is a… Commented Mar 2 at 22:56
• …valid solution to the equations of motion. But it is a process that violates the second law of thermodynamics since it absorbs heat from the system to move against the electric field. So, ohms law violates time reversal in the “thermodynamically irreversible” sense, even though the microscopic laws are time reversal symmetric all along. Commented Mar 2 at 22:57
• I don’t think the “in” and “out” discrepancy upon time reversal makes Ohm’s law TR symmetric. Consider the local version of Ohm’s law $\vec{j} = \sigma \vec{E}$. Upon time reversal, $E$ doesn’t change but the current density reverses. Ohm’s law is clearly violated. Commented Mar 2 at 23:07