# How the last part relative to the azimuthal angle works

1. Dec 5, 2008

### stevebd1

While I understand how coordinates work, I'm still trying to figure out how the last part relative to the azimuthal angle works, for example, in Minkowski metric-

$$c^2 {d \tau}^{2} = c^2 dt^2 - dr^2 - r^2\left(d\theta^2 + \sin^2\theta \, d\phi^2\right)$$

while I get that $\sin^2\theta$ produces a figure that ranges from 0-1, I'd appreciate it if someone could define the range of figures $d\theta^2$ and $d\phi^2$ produce and what units are used (i.e. degrees, radians, degrees expressed as a fraction of 360).

Last edited: Dec 5, 2008
2. Dec 5, 2008

### Ich

Re: coordinates

Theta: 0-pi
Phi: 0-2pi

3. Dec 5, 2008

4. Dec 5, 2008

### DrGreg

Re: coordinates

$\theta$ and $\phi$ are "latitude" and "longitude", both measured in radians. Except that "latitude" (really "colatitude" or "zenith") is measured from the north pole instead of the equator.

$r\,d\theta$ is the distance you move (southwards along a meridian of radius r) when you change your colatitude by a small amount $d\theta$.

$r\,\sin\theta\,d\phi$ is the distance you move (eastwards along a circle of latitude of radius $r\,\sin\theta$) when you change your longitude by a small amount $d\phi$.

5. Dec 5, 2008

### stevebd1

Re: coordinates

Thanks for the replies. So basically zenith plane = 180 degrees = pi and azimuth plane = 360 degrees = 2pi (which I recognise as the denomination of radians).

To summarize, the way I understand it (in the context of approaching a sphere radially) is-

An object drops 10 degrees in the zenith plane-

$$d\theta^2 = (10/180)\pi^2 = 0.031$$

the same object, at the same time, moves 30 degrees in the azimuth plane-

$$d\phi^2 = (30/360)2\pi^2 = 0.274$$

The next part I'm not 100% about but I'm assuming $sin^2\theta$ remains based on degrees and $\theta$ is the change in angle (in this case, 10 degrees)-

$$sin^2\theta = sin^2(10) = 0.030$$

which means-

$$r^2\left(d\theta^2 + \sin^2\theta \ d\phi^2\right)\ =\ r^2\left(0.031 + 0.030 \cdot 0.274\right)\ =\ r^2 0.039$$

This looks like it might be a bit low but I'm guessing the effects on space time (relative to a spherical object) are more related to changes in radial coordinates than changes in spherical coordinates.

Last edited: Dec 5, 2008
6. Dec 5, 2008

### atyy

Re: coordinates

What effect are you trying to understand?

7. Dec 5, 2008

### stevebd1

Re: coordinates

I was referring to $r^2\left(d\theta^2 + \sin^2\theta \, d\phi^2\right)$ in the context of Schwarzschild metric.

Last edited: Dec 5, 2008
8. Dec 5, 2008

### DrGreg

Re: coordinates

The formula is only true (approximately) for small values of $dt$, $dr$, $d\phi$, $d\theta$, that is, in the calculus limit as each of these quantities tends to zero. Both $r$ and $\theta$ can be considered constant -- these are the actual radius and zenith not the changes.

To calculate an accurate (rather than approximate) value over a line or curve in spacetime you would need to integrate along the curve.

9. Dec 5, 2008

### stevebd1

Re: coordinates

Thanks DrGreg, I was almost tempted to leave $\theta$ as a plain angle relative to the z axis.

Would the following then be correct? (putting a max of 5° $d\theta$ and $d\phi$)

constants- r=10,000 m, $\theta$=45° (approach angle relative to z axis)

An object approaching a sphere at $\theta = 45^o$ drops 5° in the zenith plane at r=10,000 m over dr=1 m-

$$d\theta^2 = (10/180)\pi^2 = 0.008$$

the same object, at the same time, has moved 5° in the azimuth plane-

$$d\phi^2 = (5/360)2\pi^2 = 0.008$$

$sin^2\theta$ remains based on degrees and $\theta$ is the (approx?) angle of approach (in this case, 45°) -

$$sin^2\theta = sin^2(45) = 0.5$$

which means-

$$r^2\left(d\theta^2 + \sin^2\theta \ d\phi^2\right)\ =\ r^2\left(0.008 + 0.5 \cdot 0.008\right)\ =\ r^2 0.012$$

$$=10,000^2 \cdot 0.012$$

_______________

I've realised that $r^2\left(d\theta^2 + \sin^2\theta \ d\phi^2\right)$ is simply a geometric solution to finding the true distance travelled on the surface of a sphere (over small increments) using the radius, zenith and azimuth angle. Following on from above-

actual distance travelled based on change in zenith & azimuth angle, angle of approach and radius -

$$=\sqrt{10,000^2 \cdot 0.012}$$

= 1095.445 m

Last edited: Dec 6, 2008
10. Dec 8, 2008

### DrGreg

Re: coordinates

Your calculation and interpretation are both correct (except that you put "10" instead of "5" into the first formula, but nevertheless got the right answer! a typing error I assume). Just remember it's an approximation, which gets more accurate the smaller the increments are.