How Do I Calculate Time as a Function of Distance for Two Electrons in a Vacuum?

[rebeka]

Five years ago I posted this question. At the time it was difficult for me to understand almost all of what I was trying to learn and was very frustrated. I put it down and studied other stuff including math. I just looked at this question again and am finding myself still unable to derive any answers.

I'm having some trouble with a pretty basic question and I'm not sure what I'm missing that will correct my thinking. If I have two electrons in a vacuum and they are set at arbitrary origins at a correspondingly arbitrary fixed distance between them with initial velocity 0 how do I find time as a function of distance?

I'm looking at this like so:
a(r) = k_{e} \frac{q^{2}}{m \cdot r^{2}}

where a is acceleration, r is the distance between the two electrons, k_{e} is the Coulomb Constant, m is two times the electron mass and q is the charge on one electron. Both electrons are allowed to move freely!

I feel that plotting acceleration as a function of distance would be useful but I'm not seeing how to integrate in time? What am I missing about the mathematics which is also probably rather elementary and is preventing me from logically thinking this through?

 

[weejee]

For these kind of problems, you use the energy conservation.
E=\frac{1}{2}m\left(\frac{dx}{dt}\right)^{2} + V(x) \ \ \rightarrow \ \ \frac{dx}{dt} = \pm \sqrt{2[E-V(x)]/m}
\rightarrow \ \ dt=\pm \frac{dx}{\sqrt{2[E-V(x)]/m}}

\pm in the expression is determined by the initial condition of the problem. (for example, whether they approaching or moving away from each other?)

 

[rebeka]

Thank you for redirecting my approach!

 

[qbert]

While weejee is absolutely correct in everything he did, We can solve this problem
from first principles, just the way you asked. [and it includes a common physics-trick,
which is probably good to see used.] The energy is sometimes called a FIRST INTEGRAL
of the motion. meaning we get it when we integrate the equations of motion 1 time.

start from a = (kq^2)/(mr^2)
recognize a = dv/dt = d^2r/dt^2
[the trick]
\frac{dv}{dt} = \frac{dv}{dr}\frac{dr}{dt} = v \frac{dv}{dr}

So that
v\frac{dv}{dr} = \frac{k q^2}{m r^2}
or
\int_{v_0}^v v dv = \frac{k q^2}{m} \int_{r_0}^r r^{-2} dr
or
\frac{v^2}{2} - \frac{v_0^2}{2} = \frac{k q^2}{m} <br /> \left( \frac{1}{r_0} - \frac{1}{r} \right)

Now this can be rearranged a-la weejee.

---
of course, we really wouldn't want to do this integral from scratch everytime
for each different force F that we run into. If the force is conservative (i.e. the
gradient of some potential) we can do this one time to arive at the law of
conservation of energy -- and from then on just use conservation of energy.

 

[rebeka]

That's awesome ... I wish I had come up with this myself with all the time I spent trying to figure it out ... but I'm glad to get it out of my head, it's been stuck there for a while. Thanks again!