# How are longitude and latitude defined in the Schwarzchild Metric?

The Schwarzchild metric is for the gravitational field of an object of mass $M$ with no electric charge and no angular momentum. The metric is

$${ds}^{2} = \frac{dr^2}{1 - \frac{r_\mathrm{s}}{r}} - c^2dt^2\left(1-\frac{r_\mathrm{s}}{r}\right) + r^2 \left(d\theta^2 + \sin^2\theta \ d\varphi^2\right)$$

with $r_s = \frac{2GM}{c^2},$ $\theta$ being longitude, and $\varphi$ being the latitude, and $ds^2$ being the spacetime interval between two points in spacetime.

In this equation are the longitude $\theta$, $\varphi$ defined based on the two points in space, or are they defined by a coordinate system that was defined independently of the points in spacetime? For instance if after finding a planet that had no rotation and no electric charge we were to assign a coordinate system for the longitude and latitude of this planet, for the Schwarzchild metric for two points in spacetime near this planet would we use the same coordinate system we had defined previously for longitude and latitude or would we define a new coordinate system for longitude and latitude based on these two points in spacetime?

• As Michele Grosso pointed out in a comment on Rob Jeffries' answer, which is not relevant to Rob's answer, the term $-r^2(d\theta^2+\sin^2\theta\,d\varphi^2)$ should be $r^2(d\theta^2+\sin^2\theta\,d\varphi^2)\,.$ This is because the sign in front of $c^2dt^2$ is a minus sign. Aug 30, 2018 at 15:35
• I think that basically you are asking the following. If you take a sphere of radius $R$ and mass $M\,,$ a person living on the surface of the sphere can choose a co-ordinate system that can be used as latitude and longitude. Let these co-ordinates be called $(\alpha, \beta)\,.$ Now the line element, i.e. the space-time interval between two infinitesimally separated events, for the space-time on this sphere and outside it is the Schwarzschild line-element. Is your question: how are the coordinates $(\alpha, \beta)$ related to $(\theta, \varphi)$ on the surface of the sphere and outside?" Aug 30, 2018 at 15:47
• $\theta$ is not a "longitude" in that equation and $\phi$ is not a latitude. In fact $\phi$ is a longitude and $\theta$ is a colatitude. Aug 30, 2018 at 17:09
• So since $\theta$ is the colatitude and $\varphi$ is the longitude, the relationship between $(\alpha, \beta)$ and $(\theta, \varphi)$ is $\varphi=\beta$ and $\alpha = \pi/2 - \theta$ for $0\le\theta\le\pi/2$ and $\alpha=\theta-\pi/2$ for $\pi/2\le\theta\le\pi\,.$ Aug 30, 2018 at 19:39

e.g. If we are considering something on a circular orbit around a spherically symmetric potential, we would probably use the plane of the orbit to define $\theta = \pi/2$ and the azimuthal position at some fiducial time ($t=0$?) to define $\phi=0$.
• So does this mean that we would generally define the latitude $\varphi$ to be the same as the longitude $\theta$? Aug 29, 2018 at 16:04
• @AndersGustafson I am not sure what you mean. They are orthogonal spherical polar coordinates. Note that $\theta$ is the "polar angle", not the "latitude" - it is an angle that increase from the "north pole" of the spherical coordinate system. Aug 29, 2018 at 16:12
• Here the reference in wikipedia for Spherical coordinates: en.wikipedia.org/wiki/Spherical_coordinate_system. However in the question the - sign in front of $r^2$ is wrong, should be a + sign. Aug 30, 2018 at 11:39
• @MicheleGrosso I don't understand your comment. I have not used $r^2$ in my answer. If it is a comment on the question, attach it to the question. Aug 30, 2018 at 14:20