r=2M to r=0 which is 3/2 c.

It works like this

dr = drrain [notice the free falling rain observer measures the interval drrain to be equal to dr the Schwarzchild r-coordinate.]

and

dt = dtrain

also

dr/dtrain = - (2M/r)1/2

so

dtdtrain = r1/2dr/(2M)1/2

and this integral becomes

train = (1/3)(2/M)1/2r3/2

so

(r/train)average = r/[(1/3)(2/M)1/2r3/2]

Since M is the mass in meters and the event horizon = 2M substitute this for r and solve. Remember this is an average velocity over the distance 2M (r=2M to r=0).

for r = 2M to r = 0 the average velocity is 3/2 c.
for r = M to r = 0 the average velocity is ~ 2.1 c.
for r = (1/10)M tp r = 0 the average velocity is ~ 6.7 c.

The proper velocity approaches infinity as r -> 0.

During the entire journey the free falling rain observer would always measure the local coordinate velocity of light to be c as relativity requires.

Hope I didn't flub some HTML.

• I did flub.. - bruce - November 29, 2001 - 21:29 UTC
  • Flubber! - luis_hamburgh - November 30, 2001 - 13:58 UTC
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      • Sometimes light... - bruce - December 1, 2001 - 06:01 UTC
    • I wished I had gone into science as a young man... - bruce - December 1, 2001 - 06:04 UTC
      • Young Australian LOVES astronomy...! - natwozhere - April 17, 2002 - 04:25 UTC