Group: sci.physics.particle
From: carlip-nospam@physics.ucdavis.edu
Date: Tuesday, March 04, 2008 11:59 AM
Subject: Re: Why does light bend under gravity?

In sci.physics Koobee Wublee wrote:
> On Mar 3, 4:53 pm, carlip-nos...@physics.ucdavis.edu wrote:

>> > First, derive a set of geodesic equations a massed particle traveling
>> > at high speed near the sun. Then, gradually reducing the mass to zero
>> > and increasing the speed to c, do you see a discontinuity at mass = 0
>> > and speed = c?
>>
>> This is definitely a worthwhile exercise. I recommend that you do it.

> Yes, indeed.

[...]
> 2 O ~= pi + 2 u (1 + 1 / B^2)

This is correct.

> If (B^2 = 1), we have a deflection of (4 u) which is twice the
> Newtonian result.

That's right.

> However, at the speed just below 1, the deflection
> angle is higher than twice the Newtonian result.

No. Now calculate the *Newtonian* prediction. You will find that,
in your notation

2 O ~= pi + 2u/B^2

(If you get stuck on this, look at chapter 3 of Goldstein, _Classical
Mechanics_.) The GR prediction is thus, as I said, (1+B^2) times the
Newtonian one. For speeds just below c, the GR deflection is just
below twice the Newtonian deflection; as v becomes small, the GR
deflection approaches the Newtonian value.

Steve Carlip




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