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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?

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  • $\begingroup$ 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. $\endgroup$ – Physics_Et_Al Aug 30 '18 at 15:35
  • $\begingroup$ 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?" $\endgroup$ – Physics_Et_Al Aug 30 '18 at 15:47
  • $\begingroup$ $\theta$ is not a "longitude" in that equation and $\phi$ is not a latitude. In fact $\phi$ is a longitude and $\theta$ is a colatitude. $\endgroup$ – Rob Jeffries Aug 30 '18 at 17:09
  • $\begingroup$ 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 $\pi/2\le\theta\le\pi\,.$ $\endgroup$ – Physics_Et_Al Aug 30 '18 at 19:39
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The Schwarzschild metric is for a spherically symmetric, non-rotating black hole (or in fact any spherically symmetric potential outside the object of interest). There is no preferred axis so you can set up the "zeropoints" of the spherical polar coordinate system in any way that makes a problem easier to solve.

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 $\phi=0$.

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  • $\begingroup$ So does this mean that we would generally define the latitude $\varphi$ to be the same as the longitude $\theta$? $\endgroup$ – Anders Gustafson Aug 29 '18 at 16:04
  • $\begingroup$ @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. $\endgroup$ – Rob Jeffries Aug 29 '18 at 16:12
  • $\begingroup$ 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. $\endgroup$ – Michele Grosso Aug 30 '18 at 11:39
  • $\begingroup$ @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. $\endgroup$ – Rob Jeffries Aug 30 '18 at 14:20
  • $\begingroup$ @Rob Jeffries. Sorry that I posted the comment about the signature under your answer. I should have posted under the question. $\endgroup$ – Michele Grosso Aug 30 '18 at 16:23

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