On Tue, 26 Apr 2005 07:25:18 GMT, "Dirk Van de moortel"
<dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote:

>
>"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message news:i88r615m648jtquicfctpsm8el6abfncma@4ax.com...
>> On Mon, 25 Apr 2005 21:49:22 GMT, "Dirk Van de moortel"
>> <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote:
>
>[snip]
>
>> >It's about a year ago since I read this and made the
>> >exercises of this project, so it's a bit rusty.
>> >I'll just copy the relevant equations from project D
>> >of Taylor and Wheeler's "Exploring Black Holes",
>> >    http://www.eftaylor.com/general.html
>> >
>> >Beware the typos.
>> >
>> >Schwarzschild metric
>> >Coordinates t, r,f  (f=phi)    - theta is taken constant
>> >b = impact parameter
>> >M = solar mass
>> >R = sun radius
>> >space = multiplication
>> >
>> >From the radial and angular components of motion of light:
>> >    dr/dt = +/- (1-2 M/r) sqrt( 1-(1-2 M/r) b^2/r^2 )

VELOCITY = DIMENSIONLESS

>> >    r df/dt = +/- b/r  (1-2 M/r)

VELOCITY = DIMENSIONLESS
Both the above equations are inhomogeneous.

>> >on page 5-8 of chapter 5
>> >you get
>> >[1]    (df/dt)^2 = b^2/r^4 (1-2 M/r)^2

INVERSE SECONDS^2 = DIMENSIONLESS

>> >[2]    (dr/dt)^2 = (1-2 M/r)^2 - (1-2 M/r)^3 b^2/r^2

VELOCITY^2 = DIMENSIONLESS
If you are using c = 1 I'm out of here. It won't wash. There's no clue
where to put the c back in, (which must be the problem).

>> >from [1] and [2]:
>> >[4]    df = dr / ( r^2 sqrt( 1/b^2 - 1/r^2 (1-2 M/r) ) )
>> >
>> >with u = R/r this gives
>> >[5]    df = -du / sqrt( R^2/b2 - u^2 + 2 M/R u^3 )
>> >and since r goes from R to infinity, u goes from 1 to 0.
>> >
>> >Eliminating the impact parameter (such that the light just
>> >grazes the sun) with
>> >[7]    R^2/b^2 = 1-2 M/R
>> >gives
>> >
>> >[8]    df = -(1-u^2)^(-1/2) du / sqrt( 1 - 2 M/R (1-u^3)/(1-u^2) )
>> >
>> >Approximate for small M/R:
>> >[10]    df ~= -du / sqrt(1-u^2)
>> integral 0 to 1 is pi/2
>>
>> >                  - M/R du / (1-u^2)^(3/2)
>>
>> integral is infinity
>
>Of course, this one seperately gives infinity.
>
>>
>> >                  + M/R u^3 du / (1-u^2)^(3/2)
>>
>> integral = infinity
>> I get infinities.
>
>If you do it properly and take the intergrals together,
>you don't get infinities:
>Try Int{ - 1/(1-u^2)^(3/2) + u^3 / (1-u^2)^(3/2) }
>
>
>> >Integrate [10] with u between 1 and 0
>> >[12]    f_total = Int{ u=1 to 0 ; df }
>> >                  = pi + Delta(f)
>> >where a tricky integration gives
>> >[13]    Delta(f) = ... = 4 M/R
>> >
>> >Using the solar values:
>> >    M = 1477 m
>> >    R = 6.9598 10^8 m
>> >gives
>> >    Df = 8.489 10^(-6)     (radians)
>> >or
>> >    Df_arcseconds = 1.75
>> >
>> >Dirk Vdm
>> >
>>
>> Thank you for your effort but it's still murky, and check out those
>> integrals.
>
>I have checked them. They are 100% okay.
>If you have a problem with one of them, ask away.
>
>> And I guess you mean M is the mass radius of the Sun.
>
>Of course.
>
>> Haven't got time to decode it all.
>
>Take your time.
>
>Dirk Vdm
>

Thanks for your effort, but it's not easy doing math in ASCII-type,
and letting c = 1 is not all that efficient. 
John Polasek
http://www.dualspace.net