00:01
Okay, so in this question we're asked to prove that for spherical coordinates, spherical coordinates that i would write down afterwards, that the jacobian transformation or the volume element, however you want to call it.
00:14
So i'm going to note by d x, y, z over d, and the order here will be row phi theta, that this is given by row squared sine of phi.
00:28
I guess that's what we want to prove.
00:30
So let's just go straight ahead for the solution.
00:34
I'm just going to briefly recall what circular coordinates are.
00:39
So this will be x is equal to something, y is equal to something, z is equal to something.
00:45
Okay, and i'm just going to write them down and then i'll make a picture.
00:48
So here, x and y, so these are notes, sort of the standard row cost theta, row sine theta for polar coordinates.
00:56
But then we have an extra angle to identify.
00:58
So this shows up in here as row cause phi, and for the extra angle we get sine phi, sine phi on both sides, right? and the picture here that you should have in mind is that you have a point given here, let's call it upon p in space, and the way to identify it, so you project it onto the x -axis, and here you kind of use, so this distance here will be row, this distance here will also be row, and then we have our theta angle here at the basis.
01:30
Okay, so this would be our theta angle between the x and y axis, and we have a phi angle in here.
01:39
All right, this is how you should think of spherical coordinates.
01:44
So, and then the jacobian of this matrix will be, so the matrix here for the transformation will be the matrix of derivatives, so we're going to write it down, and then we want the jacobian of it.
01:55
Okay, so let's just write what we would get.
01:59
The first row would be the derivatives of x with respect to row, phi, and theta.
02:03
So with respect to row, we get cost theta, sine phi.
02:08
With respect to theta, we would get just a cost theta becomes a minus sine theta.
02:13
Okay, so sorry, it's phi first.
02:14
So we respect to phi, sine phi becomes cost phi.
02:17
So this is this, row, cos theta, cause phi.
02:22
And in respect to theta, the cost theta becomes a minus sine theta.
02:25
So minus r, sine theta, sign.
02:28
Fine 5 okay now for why with respect to row we get sign theta sign phi with respect to phi sign 5 becomes cause 5 so we get a row sine of theta cause of phi and with respect to theta we get row cause theta sign then last the last one would be respect to row we get the cause phi with direct to phi we get minus row sine phi with respect to theta, we get zero.
03:04
Okay.
03:05
And the amount that we want to calculate, so the d, x, y, z over d, rho, phi, theta is the determinant of this matrix.
03:16
And of course, we have a zero in here, so let's take advantage of it.
03:19
Let's make an expansion with respect the last row.
03:24
And you know how this goes, right? so first element, i'm going to try to use colors here to make it clear, is a cost phi.
03:30
So i have cos phi and then we delete this row in here to obtain the sub matrix in here on the top corner.
03:40
So cost phi will multiply by row, cause, theta, cause, phi minus row sine theta, sine phi.
03:51
Rho, sine theta, cause phi.
03:55
Then row, cos theta, sine, fine...