# measuring curvature of space around a star

by lavinia
Tags: curvature, measuring, space, star
 Sci Advisor P: 1,716 I am wondering how space geographers would measure curvature of space around a large isolated star. i am thinking of the set up where there are two nearby spheres surrounding the star whose circumferences are already known. The remaining step is to measure the length of a radial geodesic segment connecting the two spheres. This it seems would give measurements in geodesic polar coordinates and would allow the computation of curvature using the usual formulas. How then does one find a geodesic ray and the measure its length?
 Sci Advisor PF Gold P: 4,862 Aren't radial coordinate lines (t=theta=phi=0, SC coordinates) outside the horizon spacelike geodesics? It looks like this should be so from the geodesic equations, and it seems this is regularly assumed. Then you just integrated the line element along r, with all other coords held to zero. Am I missing what you are asking?
Sci Advisor
P: 1,716
 Quote by PAllen Aren't radial coordinate lines (t=theta=phi=0, SC coordinates) outside the horizon spacelike geodesics? It looks like this should be so from the geodesic equations, and it seems this is regularly assumed. Then you just integrated the line element along r, with all other coords held to zero. Am I missing what you are asking?
yes you are right. I was asking an empirical question, How does the space geographer find the radial geodesic physically? And how does he measure the distance between the spheres with instruments? Suppose he is standing on one of the spheres and the other one is some large structure. Does he use light and mirrors? Does he drop a plumb line? does he drop a stone and measure how long it takes for the stone to land?

Sci Advisor
PF Gold
P: 4,862

## measuring curvature of space around a star

Note, there are other more complex spacelike geodesics, but I assume those are not relevant.

Also, note that a free faller using GP coordinated can foliate a region of spacetime such that the spatial slices are exactly Euclidean flat for the induced metric. Then, all curvature would only be seen by involving time.
Sci Advisor
PF Gold
P: 4,862
 Quote by lavinia yes you are right. I was asking an empirical question, How does the space geographer find the radial geodesic physically? And how does he measure the distance between the spheres with instruments? Suppose he is standing on one of the spheres and the other one is some large structure. Does he you light and mirrors. Does he drop a plumb line? does he drop a stone and measure how long it takes for the stone to land?
For a radial, spacelike geodesic, for static foliation, a plumb line would be the physical analog.
P: 1,555
 Quote by PAllen Note, there are other more complex spacelike geodesics, but I assume those are not relevant. Also, note that a free faller using GP coordinated can foliate a region of spacetime such that the spatial slices are exactly Euclidean flat for the induced metric. Then, all curvature would only be seen by involving time.
Provided the star does not rotate.
Closest to GP coordinates for a rotating star is the Doran metric.
Sci Advisor
PF Gold
P: 4,862
 Quote by Passionflower Provided the star does not rotate. Closest to GP coordinates for a rotating star is the Doran metric.
Yes, I assumed the star was not rotating (which is obviously absurd in the real world). If it were rotating, then a radial line (in typical coordinates) would not be (exactly) a spacelike geodesic, and there wouldn't be a unique static foliation (because the spacetime is not static).
Sci Advisor
P: 1,716
 Quote by PAllen For a radial, spacelike geodesic, for static foliation, a plumb line would be the physical analog.
I can see why the plumb line would find the direction of the radial geodesic. But wouldn't it stretch and give an answer that is too small? Why wouldn't one use the plumb line to first find the radial direction but use reflected light beamed in the radial direction to measure the distance?
Sci Advisor
PF Gold
P: 4,862
 Quote by lavinia I can see why the plumb line would find the direction of the radial geodesic. But wouldn't it stretch and give an answer that is too small? Why wouldn't one use the plumb line to first find the radial direction but use reflected light beamed in the radial direction to measure the distance?
Because the speed of light is not constant (assuming strong gravity). The closest physical analog to radial proper distance would be a plumb line of extremely high tensile strength.

Of course, if you know the geometry, you could mathematically convert round trip light time to geodesic distance.

You could use roundtrip light time * c as a radial distance coordinate directly. You just can't assume it measures proper distance.

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