Imagine an electron and a positron, initially held stationary some distance apart at time $t=0$. There is an attractive force between them, so they will approach one another.

I am told that all the fundamental laws of physics are CPT-symmetric, so let's try that out.

First, a P-inversion. Reflect the universe in such a way that the two particles switch places.

Then, a C-inversion. The electron becomes a positron and vice-versa. In other words, we are back to where we started.

Now a T-inversion. Running the dynamical laws backwards in time, the attractive force becomes a repulsive one: the particles move further apart instead of approaching.

This is not CPT symmetry as I understand it. What am I doing wrong?


Two basic misunderstandings here: 1) under time reversal the velocity changes sign but not the acceleration or the force, which remains attractive, and 2) symmetry of the laws does not imply that a particular situation is symmetrical, only that the transformed situation (which may be different) is also a valid solution of the laws. Which it is - there is no reason electrons and positrons can't move apart.

  • $\begingroup$ "only that the transformed situation (which may be different) is also a valid solution of the laws" - does this mean the laws having the transformation applied as well? i came to a similar conclusion as the op but thinking about electron and a proton, in the proton's frame time reversal does nothing to it, but it makes the electron behave as a positive charge would if time were not reversed... $\endgroup$ – jheriko Jan 9 '14 at 17:52
  • $\begingroup$ now, if i use a neutron instead and pretend there are no residual forces at play and consider gravity I get a similar situation - except that the trajectory is not reversed under a charge or parity reversal. now i believe that this disobeys the laws of gravitation unless the law is transformed as well (there are no repulsive forces without negative mass afaik, and unlike signs will repel) - but then this seems to be nearly trivial - it will take exceptional cases to break that symmetry then, surely - like the values involved not being regular old real numbers? $\endgroup$ – jheriko Jan 9 '14 at 17:54

CPT symmetry (described as a dynamical symmetry) means that a dynamical evolution$s=s(t)$ is mapped into another possible dynamical evolution $s'=s'(t)$ if you transform under CPT every single state $s'(t) := CPTs(-t)$ connected by the initial dynamical evolution (reversing also the chronological order due to the presence of $T$).

In other words, $s'$ solves the equations of motion if $s$ does.

This is the case in the considered situation.

In $s'$ the particles moves further apart because you also changed the initial conditions: the particle attract each other, but the (new) initial relative velocity is positive. In fact it now decreases as $t$ increases...

In fact, we can think of $s$ as defined in $[0,T]$. At $t=0$ the particle have zero velocity (in the laboratory) and they reach a great velocity at $t=T$. During the motion, the velocity of the electron points towards the positron and vice versa.

The CPT reversed story $s'$ is instead defined over $[-T,0]$. The (swapped) particles have a great velocity at the initial instant $t=-T$. Differently from what happens along $s$, now the velocity of the electron diverges from the positrons and vice versa. Therefore the particles start their motion with divergent velocities though they attract each other. Due to this attractive force, the velocities decrease and vanish at $t=0$.


While it is pointing out the reader into some general directions, the general nature of answer https://physics.stackexchange.com/a/70858/139287 is in my taste too much handwaving. My personal perspective on this problem is as follows:

First, let us describe the particles move further apart by a formula. We can do this as follows: Let $l(t)$ designate the (non-zero, non-directional) distance between the two particles at time $t$. Now define the formula $B(t_1,t_2) :\Leftrightarrow t_1 > t_2 \Rightarrow l(t_1) > l(t_2)$. If $\forall t_1, t_2: B(t_1,t_2)$, ie. if $B(t_1,t_2)$ holds for all times $t_1$ and $t_2$, then we will call this particles move further apart.

In the experimental scenario before the C, P, T transformation, this is the case.

Now let us do the C, P, T transformation. It has no effect on the definition of $l$ but it transforms $B(t_1,t_2)$ into $B(-t_1, -t_2)$. So after the transformation we are interested whether $-t_1 > -t_2 \Rightarrow l(-t_1) > l(-t_2)$ holds for all $t_1, t_2$. This is the case. So after the T transformation the particles still move further apart.

To see the latter claim formally, you might want to apply some laws of logic. To see it informally, let us have a look at an example.

Let us chose in the situation before the T transformation the values $t_1 = 1$ and $t_2 = 0$. The precondition of the implication holds, $1 > 0 $, and the conclusion also holds, $l(1) > l(0)$. We know this, as in the situation before the T transformation the particles are moving further apart.

Now let us have a look at the situation after the CPT transformation, where the CPT theorem tells us that the transformed property agin holds. Here, let us pick $t_1 = -1$ and $t_2 = 0$. We obtain $1 > 0$ as precondition, which holds, and therefore also $l(1) > l(0)$.

The problem you fell victim of is, in my opinion, an incorrect application of the CPT theorem.

The same mistake has been made in https://arxiv.org/pdf/1103.4937.pdf, where the author argues that the CPT theorem produces repulsive gravitation between matter and anti-matter. In https://arxiv.org/pdf/1108.5117.pdf the problem associated with this line of reasoning is explained in detail. I guess the situation here is similar in nature, although easier to grasp.

Edit: Corrected a wrong use of a symbol.

  • $\begingroup$ Minor comment to the post (v3): In the future please link to abstract pages rather than pdf files. $\endgroup$ – Qmechanic Aug 13 '19 at 12:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.