So the integral is just over the (infinitesimal) region where $\delta(r'-r)$ is non-zero, and over this region we can take J constant since the region is infinitesimal. That means we can take J outside the integral to get:
But sir my problem is that $\mathbf J$ is the volume current density and it is a function of $\vec r$’ so what does it mean when we write $\mathbf J (\mathbf r)$
The point is that the integrand is zero everywhere except at $\mathbf r' = \mathbf r$.
We can split our integral into parts:
1. at $\mathbf r' \ne \mathbf r$: in this region $\mathbf J(\mathbf r') \ne \mathbf J(\mathbf r)$ but we don't care because the integrand is zero anyway due to the $\delta$.
2. at $\mathbf r' = \mathbf r$: everywhere in this region $\mathbf J = \mathbf J(\mathbf r)$
This method is used frequently in derivations. I think you have a deeper issue than just this one derivation. You need to go back and consider the mathematical methods used.
e.g. the Fourier transform of a functon $f(x)$ is $\int f(x)e^{ixp}dx$
@Nobodyrecognizeable ah, OK, sorry I misunderstood what you were asking. You're asking why when unpolarised light passes through a polariser its intensity is halved?
You could also think of it in terms of breaking things up into orthogonal components. If the light is randomly polarized, half of the electric field is oriented orthogonal to the polarizer, and half of the electric field is oriented parallel to it. So half goes through, half gets absorbed.
OK, hand-wavey but more precise, you can think of it intuitively like this: without a preference for polarization, perfectly depolarized light must dump half its energy into a polarizer: you can take this as a kind of "definition" of depolarized light if you like.
@JohnRennie Something else. In the process of gas expansion to vacuum, according to the first law of thermodynamics, dU= 0. However, if dU=−PdV+TdS holds, because the expansion process doesn't do work and dS > 0, dU must be greater than 0. This seems to cause a contradiction? dV for the system is zero if we treat the box and not the gas as our system, yet the change in entropy is positive and the change in internal energy is zero, which doesn't sit quite right with me.
@JohnRennie I'm not sure either. My prof said so long as pressure and temperature are well defined at the starting point and endpoint of the process (i.e. there is equilibrium at the start and end), the identity should hold.
@JohnRennie I guess as long as it's quasistatic it holds. So maybe we could consider free expansion but quasistatically, and then sum up all of those processes to get the overall free expansion. That still doesn't explain my confusion, but yeah.
An electron of an Hydrogen atom is in its ground state and having energy equal to that of an excited electron of He+. Find the ratio of magnetic field created by electron in Hydrogen atom's ground state upon nucleus to that of electron in He+ in excited state upon nucleus.