I thought so too, but I have doubt since he prefaces that formula by saying:
or, in an older version of his book:
The wording sounds like he's doing more than approximating ##\theta##. But I could be wrong..
For the record, the question with solution for 13.1 is found here...
Hello, I am reading a section in Jackson discussing classical Rutherford scattering, and he mentions incorporating a cutoff in the Coulomb interaction in order to model electric screening. I am trying to understand how he applies this, as described below.
The set up is basically the classic...
Indeed. I guess what I was looking for with respect to my original question was a mathematical demonstration of what Jackson is trying to encapsulate in that paragraph using the regular "vanilla" calculus that he's exclusively worked with thus far in the book. I was hoping that was possible...
Thanks for the reply. Per your last paragraph, I guess it would be nice to see what you're saying explicitly written out mathematically. That would probably help me grasp the fundamental logic of what's going on...but since I haven't studied differential geometry yet, I imagine that's too...
Hello all,
I am reading through the Jackson text as a hobby and have reached a question regarding the Hamiltonian transformation properties. I will paste the relevant section from the text below:
I don't understand what he's getting at in the sentence I highlighted.
To attempt to see what...
Hello all,
I'm reading through Jackson's Classical Electrodynamics book and am working through the derivation of the Legendre polynomials. He uses this ##\alpha## term that seems to complicate the derivation more and is throwing me for a bit of a loop. Jackson assumes the solution is of the...
Hah, I could imagine! I'm actually interested to learn more about this stuff. I work in a completely unrelated field, but I want to increase my knowledge of how circuits are developed, so I'm trying to learn these things on my own.
Ok, I agree with you in that the two sides are independent...
That's true, technically. However, my contention with this problem is that it seems to assume that no current ever flows between the left hand loop and right hand loop and so they can be treated independently. For instance, maybe some of the current ##I_o## from the left side loop flows over the...
I would agree with that. Perhaps, then, my contention rises from the assumption that there is no potential gradient between one side vs the other. On the left side, the potential difference between the voltage source and the common ground (the red circled node) is ##V_s##. On the right side, if...
Hmm, I see your point. I guess my only question would be: is it possible that current coming IN though that connection builds up at the current controlled current source (say, via a capacitor or some sort) and therefore wouldn't need a return path out of the righthand loop? Admittedly, I might...
Homework Statement
This problem is Problem 2.24 from Fundamentals of Electric Circuits (5th edition).
The problem is as follows:
Homework Equations
KVL, KCL
The Attempt at a Solution
In the official solution, they assume that the total current in the lefthand loop is ##I_o##, whereas the...
Hmm, that's interesting, I'd be curious to see that example if you don't mind sending it to me.
As for the "Blv" law, as I was mentioning in my previous reply to tsny, it seems like the calculation of emf as the work done per unit charge perhaps needs some further clarification, whereas the...
I see your point. So, if we recall, there are two ways to calculate emf -- 1) the "snapshot" view (as Griffiths calls it) which only looks at one instance of time and 2) the work per unit charge. I think combined with rude man's response (below yours), it seems to be the case that option 2) is a...
True. So if we ignore the resistance, then, we have ##\int \vec f_{pull} \cdot \vec {dl} ## = ##\int u(t)B \hat x \cdot \vec {dl} ##. Now ##\vec {dl} = u(t)dt \hat z + vdt \hat x## and so we get ##\int u(t)B \hat x \cdot vdt\hat x ## = ##vB\int u(t)dt ## = ##vBh## = ##ξ##, since ##u(t)## is the...