B Time-symmetry in electromagnetism: a simple puzzle

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ConradDJ

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All basic laws of physics are said to respect CPT symmetry, and Maxwell's equations in particular are time-symmetric. But here's a simple scenario I find very puzzling:

Two particles of opposite charge attract each other. In the time-reversed picture, they'd repel each other, no? But they remain opposite charges, whether under time-reversal or CPT-reversal. So it seems that the laws of electromagnetism aren't time-symmetric after all.

There has to be an obvious answer, but my aging brain won't come up with it.
 

.Scott

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They will attract each other in both cases.
In both cases, they will accelerate away from each other.
In one case, you start with them at a distance heading straight for each other - and they are accelerating as they approach each other.
In the other case, they start close to each other - but with velocity vectors pointing away from each other. They decelerate as they move away from each other.

Let's use gravity as the force. It will pull whether time goes forward or backward. Something can be in orbit and if you reverse time, it is still in orbit - but going in the same direction. In both cases, gravity is attracting them it towards the planet. Toss a rock into the air. In reverse, it bounces from the ground goes up to a highest point, and then falls into your hand.
 
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Two particles of opposite charge attract each other. In the time-reversed picture, they'd repel each other, no?
No, the time reverse of an attractive force is still an attractive force. For example, a parabolic path pointing towards one direction is still a parabolic path pointing in the same direction when time reversed.
 

ConradDJ

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Thanks for the clear and timely responses -- very helpful. (Though I assume .Scott meant to say, "In both cases, they will accelerate towards each other."
 

vanhees71

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It's easily seen from Newton's 2nd law (you can of course also argue within SRT): ##\vec{F}=m \vec{a}##. Since under time reversal ##t \rightarrow -t##, ##\vec{x} \rightarrow \vec{x}##, and ##m \rightarrow m##. Thus ##\vec{a}=\ddot{\vec{x}} \rightarrow \ddot{\vec{x}}##, and thus also ##\vec{F} \rightarrow \vec{F}##.

You can also argue directly with Coulomb's Law. Of course here you need in addition ##q \rightarrow +q## under time reversal. This also implies that ##\vec{E} \rightarrow \vec{E}## and ##\vec{B} \rightarrow -\vec{B}##.
 

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