You are right at 4:48 . A should remain, B should have been dropped. This mistake is not fixed in even 3rd and 4th edition. I am trying to understand where we can be wrong now... Edit: Nope, griffiths is not wrong. He just skipped a few steps there. I have explained in the comment below.
I found it! If the TE wave is propagating through -z Direction, then it obeys B_z (x,y) = X(x) Y(y). Which can be put into B_x = ( i / [(w/c)^2-k^2] ) * [ k * (Del B_z / Del x) - w/c^2 * (Del E_z / Del y) ] . E_z=0 taking derivative of B_z =X(x)Y(y) wrt x diminishes A, rather than B.
@@enestepe For other people who might be struggling ^^: Answering the question why B_x = 0 (page 428). You know from the boundary condition that B_perpendicular = 0 (with respect to the sides of the rectangular walls). So this means that B_x is perpendicular at x = 0 and x = a (both the (y,z)-planes, the upper and lower plane of the conductor). B_x vanishes, and then, using 9.180iii, one can see that the Del E_z/Del_y term vanishes as well due to the fact that Ez equals zero along all four planes (as the wave propagates in z-hat direction, so Ez is parallel everywhere with respect to the walls, applying boundary condition: E_parallel = 0). To conclude, you end up with only the Del B_z/Del x, which has to equal zero. When applying separation of variables, you can just plug it in, which results in dX(x)/dx * Y(y), which means dX(x)/dx = 0.
Since the perpendicular components of B vanishes at the interface, Bx and By are zeros. Applying this in the equation of Bx and By, we get dX and dY as zeros
The boundary conditions are really confusing. Perhaps you could explain why the boundary conditions require A=0 etc., as well as why ky and kz are what they are...The book is no help in this regard hence why i am on youtube. Thanks!
The equation that says By is proportional to κ∂Bx/∂y -(ω/c^2)∂Ex/∂z says that both ∂Bx/∂y and ∂Ex/∂z are 0 at y = 0 and y = a where By is 0. (Ex and hence ∂Ex/∂z is always 0 anyway.) (We could arrive at the same conclusion for ∂Bx/∂y from the equation of Ez since Ez also has to be 0 at y = 0 and y = a by the boundary condition that E parallel = 0.). ∂Bx/∂y = ZdY/dy, but Z is not a function of y, so it can't depend on y, hence dY/dy = 0 at y = 0 and a. Now dY/dy = Aky cos(ky y) -Bky sin(ky y). In order for dY/dy to be 0 at y = 0, A has to be 0 (since cos 0 = 1, but sin 0 = 0). So we're left with -Bky sin(ky y) = 0 at y = 0 and a. At y = 0 it's obvious, but in order for it to be 0 at y = a, ky has to be = mπ/a.
Is the general solution to this incorrect due to the boundary condition? I thought the general solutions would be two sin functions? I don't want to question the book, especially since my pde is rusty. but..
(a) I was banging these videos out with minimum effort, long before I realized people would actually use them. (b) I plan on revisiting these when I have upgraded my capacity to produce quality content. This last month I've played with lighting, video capture, and audio capture. Over the coming months I'll be experimenting with style and presentation too. Check out my new series to see the evolution.
Will do. Perhaps my response was a bit curt. I appreciate the videos anyway, and if you've the time, we're headed into thermo next quarter and would love a rundown of that book. Thanks either way
Which book are you using for thermo? When I did it back in the 90s we used Hallidary, Resnick and Krane's book. I would love to do a series on thermodynamics, it's one of my favorite topics of all time.
"An introduction to thermal physics" Daniel V Schroeder. The average for that class at this uni is consistently the lowest of all the upper div courses. It would be a great help.
