# How to formalize "does not move appreciably" in a mechanics problem?

I'm working on an exercise from Keith Symon's "Mechanics", ch.2 no.4 (self-study, not homework). The problem is about analyzing the motion of an electron, initially at rest, when a proton flies nearby with constant speed.

A high-speed proton of electric charge $$e$$ moves with constant speed $$v_0$$ in a straight line past an electron of mass $$m$$ and charge $$-e$$, initially at rest. The electron is at a distance $$a$$ from the path of the proton.
a) Assume that the proton passes so quickly that the electron does not have time to move appreciably from its initial position until the proton is far away.

This assumption allows me to consider electron's location to be fixed, and calculate the force in the perpendicular/parallel directions to the proton's movement, the impulse delivered to the electron over all time from $$t=-\infty$$ to $$t=\infty$$, and the final momentum and kinetic energy of the electron. I've done these parts of the exercise myself. What I'm having trouble with is the final part,

e) Show that the condition for the original assumption in part (a) to be valid is $$(e^2/4\pi\epsilon_0) \ll \frac{1}{2}mv_0^2$$

I'm not sure how to approach this. What should "appreciably" and "far away" mean here? And isn't it strange, from physical considerations, that the suggested answer doesn't depend on the distance $$a$$ between the electron and the proton's path?

One thing I tried was to arbitrarily say "consider the range $$[-t_0,t_0]$$ over which the proton delivers 99% of its overall impulse to the electron; then the final velocity $$v$$ I calculated based on the assumption of the electron not moving allows it to move during the time $$2t_0$$ a distance that's $$\ll a$$". I calculated the condition for this and it comes out more complicated than the suggested answer, and depends on $$a$$. I'd appreciate some ideas or a complete working out (but hints are fine too) for a better? more correct? way to approach this.

P.S. I'll record for the sake of the context the missing parts of the exercise:

a) Assume that the proton passes so quickly that the electron does not have time to move appreciably from its initial position until the proton is far away. Show that the component of force in a direction perpendicular to the line along which the proton moves is $$F = \frac{e^2a}{4\pi\epsilon_0(a^2+v_0^2t^2)^{3/2}}$$ (mks units).
b) Calculate the impulse delivered by this force.
c) Write the component of the force in a direction parallel to the proton velocity and show that the net impulse in that direction is zero.
d) Using these results, calculate the (approximate) final momentum and final kinetic energy of the electron.
e) Show that the condition for the original assumption in part (a) to be valid is $$(e^2/4\pi\epsilon_0) \ll \frac{1}{2}mv_0^2$$

• Hm, the condition $e^2/(4\pi \epsilon_0)\ll \frac{1}{2} m v_0^2$ is not dimensionally consistent, the LHS has units of energy times length, while the RHS has units of energy. I suspect the LHS is supposed to be divided by $a$. May 16, 2021 at 23:26

This kind of problem is all about limits, and in particular trying to understand what you can from various limiting cases without doing an exact calcultion.

The first limit that's interesting to think about is if $$a\rightarrow 0$$. Now of course when $$a=0$$ exactly there will be some collision, but let's just imagine $$a$$ is extremely small but there is some tiny gap so the proton doesn't bump into the electron. In this situation, will there be any net force imparted to the electron? No! Any work the proton does while approaching the electron, will be undone by work of the same magnitude, in the opposite direction, as the proton recedes from the electron.

Therefore the net momentum transfer to the electron must only occur at finite $$a$$. This gives us the idea that the proton will transfer momentum to the electron while the distance "along the direction of motion" is comparable to the distance "perpendicular to the direction of motion." Note that this intuition we get from taking limits is backed up by your exact calculation of the force in the perpendicular direction -- this expression is peaked at $$t=0$$, and the time interval over which the force is large is $$t \sim a/v$$, which is the time it takes to proton to travel a distance $$a$$.

The proton is in this interval for a time of approximately $$\delta t \sim v / a$$. During this time, the acceleration of the electron is approximately (but less than) the acceleration at the moment when the proton is exactly a distance $$a$$ away: $$\ddot{x} = F/m \lesssim e^2/(4\pi \epsilon_0 a^2 m_e)$$. Assuming that the electron doesn't move much (we'll revisit this in a second), the amount the electron moves $$\delta x$$ in the time interval $$\delta t$$ via which the proton traverses the distance $$a$$ is $$\delta x \sim \ddot{x} (\delta t)^2 \lesssim e^2/(4\pi \epsilon_0 a^2 m_e) (a/v_0)^2 = e^2/(4\pi \epsilon_0 m_e v_0^2)$$. Multiplying the numerator and denominator of the last expression by $$a$$, and rearranging, we can rewrite this as $$$$\frac{\delta x}{a} \sim \frac{e^2}{4\pi \epsilon_0 a} \frac{1}{m v_0^2}$$$$

What does it mean for the electron not to move much? It means that we can treat the distance between the electron and the proton as approximately constant, during the time the proton is exerting a force, or in other words that $$|\delta x| \ll a$$. This means that we need $$$$\frac{e^2}{4\pi \epsilon_0 a} \ll m v_0^2$$$$ At this level of precision we haven't been too careful about factors of 2. However, given the similarity of the last expression to the kinetic energy, it's natural to rephrase this as, and equivalent to the rough level of precision we are working, to say $$$$\frac{e^2}{4\pi \epsilon_0 a} \ll \frac{1}{2} m v_0^2$$$$ (You could also say this is a more conservative bound than the one where we don't include the 1/2).

I suspect this is at the level of precision your book is asking the question. But if you wanted to give a more precise answer, you could work out the first order correction to the impulse accounting for the motion of the electron; you would find there is some correction to the displacement $$\delta x$$ of the electron that has the form $$\delta x = \delta x_0 + C e^2/(4\pi \epsilon_0 m v_0^2)$$, where $$\delta x_0$$ is the answer you get for the displacement ignoring the motion of the electron, and $$C$$ is some order 1 dimensionless constant you could compute. By demanding the size of the first order correction is below a certain amount relative to the leading order term, you would get a more precise answer.

• What a wonderful answer! Hugely appreciate the time and effort you took to give it. May 18, 2021 at 15:44
• @AnatolyVorobey Glad to help! May 18, 2021 at 15:46