Understanding proper distance in Schwarzschild solution.

  • #1
peter46464
37
0
I'm trying to understand the Schwarzschild solution concept of proper distance. Given the proper distance equation
[tex]
d\sigma=\frac{dr}{\left(1-\frac{R_{S}}{r}\right)^{1/2}}
[/tex]
how would I calculate the coordinate distance. For example - assuming the distance from the Earth to the Sun is 150,000,000km, is it a valid question to ask what the coordinate distance is, and how would I calculate it?

I know [itex]R_{S}[/itex] is about 3km.

Many thanks
 
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  • #2
peter46464 said:
I'm trying to understand the Schwarzschild solution concept of proper distance. Given the proper distance equation
[tex]
d\sigma=\frac{dr}{\left(1-\frac{R_{S}}{r}\right)^{1/2}}
[/tex]
how would I calculate the coordinate distance. For example - assuming the distance from the Earth to the Sun is 150,000,000km, is it a valid question to ask what the coordinate distance is, and how would I calculate it?

I know [itex]R_{S}[/itex] is about 3km.

Many thanks

If you use Schwarzschild coordinates, then the radial coordinate distance is simply the difference in the radial coordinate "r" between two points: |rA-rB|

To get the radial proper distance you have to integrate your formula between rA and rB.
 
  • #3
Thanks. For me, integrating that looks hard. Can I make any simplifying approximations?
 
  • #4
peter46464 said:
Thanks. For me, integrating that looks hard. Can I make any simplifying approximations?

That general form of integral actually appears in standard tables of integrals, so you can find an exact antiderivative for it (you may have to modify the form of the integrand somewhat to fit it into a standard form). Try here for one such table:

http://integral-table.com/

Or, particularly if r is very large compared to R_s (which it certainly is in your case), you can expand the binomial in the denominator in a power series, as here:

http://hyperphysics.phy-astr.gsu.edu/hbase/alg3.html

You should only need the first couple of terms to see how things will go for the case of R_s / r very small.
 
  • #5
It is very easy to solve it with a math program such as Mable, Mathematica or Matlab. But effectively R = rho as the Sun's mass is not large enough to be considered a strong field. You have to go many numbers behind the decimal point to find a discrepancy.
 
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