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-

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

while I get that [itex]\sin^2\theta[/itex] produces a figure that ranges from 0-1, I'd appreciate it if someone could define the range of figures [itex]d\theta^2[/itex] and [itex]d\phi^2[/itex] 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]

[itex]\theta[/itex] and [itex]\phi[/itex] 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.

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

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

 

[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-

[tex]d\theta^2 = (10/180)\pi^2 = 0.031[/tex]

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

[tex]d\phi^2 = (30/360)2\pi^2 = 0.274[/tex]

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

[tex]sin^2\theta = sin^2(10) = 0.030 [/tex]

which means-

[tex] 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[/tex]

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 [itex]r^2\left(d\theta^2 + \sin^2\theta \, d\phi^2\right)[/itex] in the context of Schwarzschild metric.

 

[DrGreg]

The formula is only true (approximately) for small values of [itex]dt[/itex], [itex]dr[/itex], [itex]d\phi[/itex], [itex]d\theta[/itex], that is, in the calculus limit as each of these quantities tends to zero. Both [itex]r[/itex] and [itex]\theta[/itex] 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 [itex]\theta[/itex] as a plain angle relative to the z axis.

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

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

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

[tex]d\theta^2 = (10/180)\pi^2 = 0.008[/tex]

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

[tex]d\phi^2 = (5/360)2\pi^2 = 0.008[/tex]

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

[tex]sin^2\theta = sin^2(45) = 0.5[/tex]

which means-

[tex] 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[/tex]

[tex]=10,000^2 \cdot 0.012[/tex]

_______________

I've realized that [itex]r^2\left(d\theta^2 + \sin^2\theta \ d\phi^2\right)[/itex] 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 -

[tex]=\sqrt{10,000^2 \cdot 0.012}[/tex]

= 1095.445 m

 

[DrGreg]

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

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

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

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

[tex]d\theta^2 = (10/180)\pi^2 = 0.008[/tex]

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

[tex]d\phi^2 = (5/360)2\pi^2 = 0.008[/tex]

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

[tex]sin^2\theta = sin^2(45) = 0.5[/tex]

which means-

[tex] 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[/tex]

[tex]=10,000^2 \cdot 0.012[/tex]

_______________

I've realized that [itex]r^2\left(d\theta^2 + \sin^2\theta \ d\phi^2\right)[/itex] 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 -

[tex]=\sqrt{10,000^2 \cdot 0.012}[/tex]

= 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.