Ex. 10 time dependence and w_V

[SaraPJensen]

In exercise 10, w_V is treated as a constant, whereas V, and thereby w_z, w_plus and w_min are time-dependent. However, we are asked to find the relationship between w_V and w_z, w_plus and w_min. In order to find this, I assume it is necessary to print out the values of w_z, w_plus and w_min, but at what time-step are we meant to do this? And is the relationship between them and w_V expected to be time-independent?

[anderkve]

Hi @SaraPJensen!

I probably should have clarified this in the text (I'll add a note about it), but the definition of w_z in Problem 10 remains the same as before, i.e. in terms of the constant V_0. (But in Problem 10 V_0 has a new value.) So the value of w_z, and thus w_plus and w_minus, remains time-independent also in problem 10.

Hope that helps!

[SaraPJensen]

Thank you. However, assuming we use eqs. 18-20 to calculate the trajectories of the particles, and w_z is only dependent on the initial potential V_0, where does the time-dependence even enter into our calculations?

[anderkve]

Eqs. 18-20 are only eqs. 13-15 extended to the case of many particles. That is, just as eqs. 13-15, they are for the case of a constant external electric field. Since we don't encourage you to design your code such that you explicitly write out these long equations in the code (see note below eq. 20 and the comment below the code snippet), we haven't presented how eqs 18-20 look in the case of Problem 10.

However, if you want to write down the full equations for Problem 10, all you need to do is to take eqs 18-20, replace w_z^2 with it's definition (2 q V_0) / (m d^2) and then make the replacement V_0 --> V_0 (1 + f cos(w_V t)). (To convince yourself that this replacement is OK, you can always go back and redo Problem 1 using V(x,y,z,t) = (V_0 / (2 d^2)) (1 + f cos(w_V t)) (2 z^2 - x^2 - y^2) to see how the term arising from the external E field changes.)