How to calculate the time until charged particles collide

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How to calculate time until charged particles (Electron/Positron for example) collide starting at velocity= 0 , distance x apart:
Using coulombs law, how do I get the velocity and position equation for a pair of unit charged particles?
A simple reference link would be great as I know this is simple and must be readily available, but couldn't find it...
 
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BvU said:
If you want to do this classically, use the SUVAT equations.
The SUVAT equations are for constant acceleration. But this case does not involve constant acceleration. So one has to deal with some calculus.

The direct, brute-force approach would be to write [coupled] differential equations for acceleration as a function of position for both particles and solve. *shudder*.

But one could note some symmetry about the center of mass and write one differential equation relating the rate of change of distance between the particles to the distance between the particles. Without writing this down, I suspect that it would still be a bit of a nasty equation to solve.

However, one can use conservation of energy to solve for closing velocity as a function of distance. Velocity is distance traveled per unit time. Invert it and you get time elapsed per unit distance. Now what you have to do is integrate this over distance to get elapsed time.
 
This may not be soluble analytically. One runs into problems quite rapidly trying to write position versus time for a small mass falling in Earth's gravitational field - a similar problem except that one can treat the Earth as stationary because its mass is so huge.
 
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This is not that hard I think. It is just the Kepler problem (with appropriate substitutions). The collision will occur after one quarter of the period.
 
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Here is the numerical solution:

1607621319935.png


Orange assumes constant acceleration, blue assumes it goes as 1/r2.

As you see, the collision happens ~20% sooner if the change in acceleration is considered.

I don't know if it is analytically solvable. If it is, the solution probably involves the equal areas in equal times rule, applied to a circular orbit and again to one where the semiminor axis goes to zero.
 
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Thanks for replies. I thought I was missing a simple solution to a simple problem, that seems not to be the case.

I was looking for a solution to discuss with my kids over dinner. As the discussion is about the Physics not the math, I use the approximate solution of constant acceleration shown above to be close enough.

Just for my knowledge, can the Acc =1/x^2 be entered into Wolframalpha or similar program to get a graph of distance/time? If so, what would I type in?
 
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Ibix said:
This may not be soluble analytically
I seem to remember someone (maybe @Redbelly98?) working out an analytical solution here a while back. The integral was not trivial.
 
Nugatory said:
I seem to remember someone (maybe @Redbelly98?) working out an analytical solution here a while back. The integral was not trivial.
I was thinking of @haushofer, who managed to get ##t(r)## for the gravitational case I mentioned (here) but couldn't invert it to get ##r(t)## in the general case. The integration was indeed non-trivial.
 
69911e said:
Just for my knowledge, can the Acc =1/x^2 be entered into Wolframalpha or similar program to get a graph of distance/time? If so, what would I type in?
Yes, you type in

x''= -1/x^2, x(0)=1,x'(0)=0​
1607641299051.png


You can also leave out e.g. the x(0)=0
Wolfie then 'samples' x(0) and scales the axes

1607641559763.png


If you leave out both, a 'solution' is shown

1607641729470.png

But I don't get it... (more o:) :cry: )
 
Nugatory said:
I seem to remember someone (maybe @Redbelly98?) working out an analytical solution here a while back. The integral was not trivial.
The full solution for ##r(t)## is nontrivial but the time until collision is just ##\frac T 2 ## where ##T## is the period for a highly eccentric Keplerian (coulomb force) orbit. $$Gm^2 \to q^2/4\pi\epsilon_0$$ where m is probably the reduced mass

That was the OP question.
Am I missing something here ?

correction: Thanks to @TSny who pointed out (twice!) that it should be T/2 and not T/4.
 
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hutchphd said:
The full solution for ##r(t)## is nontrivial but the time until collision is just ##\frac T 4## where ##T## is the period for a highly eccentric Keplerian (coulomb force) orbit. $$Gm^2 \to q^2/4\pi\epsilon_0$$ where m is probably the reduced mass

That was the OP question.
Am I missing something here ?
No, you're right. Grabbing the Keplerian solution is a good end-run around various unpleasant integrals.
 
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