How the last part relative to the azimuthal angle works

[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).

 

[Ich]

Theta: 0-pi
Phi: 0-2pi

 

[neu]

it's in spherical polar coordinates http://en.wikipedia.org/wiki/Spherical_coordinates

 

[DrGreg]

\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.

 

[stevebd1]

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.

 

[atyy]

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.

What effect are you trying to understand?

 

[stevebd1]

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

 

[DrGreg]

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.

 

[stevebd1]

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

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

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

= 1095.445 m

 

[DrGreg]

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

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

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

= 1095.445 m

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.