00:01
In this question, we're asked about some geometrical stuff to do with the two sphere, or spherical polar coordinates, to be precise.
00:12
So in spherical polar coordinates, we have ds squared equals d r squared, plus r squared, d theta, d theta, d phi squared.
00:30
And this means that we can read off from this equation, that the metric tensor has the following components.
00:40
1, r squared, r squared, sine squared, theta, and then the rest of zeros.
00:49
Well, this would be the rr component, this is the theta -theta -theta component.
00:52
This is the phi phi component.
00:57
Now we need to find the expression for the divergence of a scalar field, and we're going to use this equation for that.
01:05
And for that, we're going to need the determinant of the metric.
01:08
So the determinant of the metric is, the product of these, r to the 4, sine squared theta, which means that root g is r squared sine theta.
01:21
So now we can plug that into here, and then we're going to work this out.
01:35
So it's d by d, d by d, r of root g, which r squared sine theta, vr, plus d by d by d theta, of, same thing v theta plus d by d phi v phi so now we can work this out so d by d r we're going to get 2 r here so that's going to become a 2 over r when we divide by this so we get 2 over r vr plus so that's when it hits the r when it hits the sign theta we don't get anything and then when it hits vr, we're going to get a d by dr of vr.
02:42
Now here, we get a term when it hits the theta, and that's going to give us cos theta.
02:50
So when we divide out by r squared sine theta, we're going to get a cotan theta, and then the standard bit.
03:06
And then here, there's no phi term in here, so we just get a d by d phi, v5.
03:13
And this is the expression.
03:18
Now the last thing we have to do is calculate the christophil symbols.
03:25
And we're going to use the formula that i wrote up there.
03:31
I'm going to write it again.
03:47
So first of all, let's put an r at the top.
03:51
So gamma r, j, k is...
03:54
And now because g is diagonal, the l is going to have to be an r as well.
03:59
And then we get an inverse of gr, which is 1.
04:03
D -j, g -k -r.
04:07
D -k -g -j -r minus d -r g -j -k.
04:17
So now if we have, if we let, so let's do gamma r -r -r, so two r downstairs, we're going to get a dr -g -r, g -r is just one, so that goes away, d -r -g -r, goes away, this goes away as well, so that's just zero.
04:48
R -r -theta.
04:52
So now dr -g -theta -r, that goes away.
04:57
D -theta -g -r, well, that's 1, so that goes away.
05:02
And then this is mixed as well, so that's 0.
05:12
And then by similar argument, that's going to be 0 -2...