Re: Error in Weisstein's redshift

On 28 Jan 2006 18:39:20 -0800, xxein@xxxxxxxxxxxxx wrote:

>
>John C. Polasek wrote:
>> On 28 Jan 2006 16:44:42 -0800, xxein@xxxxxxxxxxxxx wrote:
>>
>> >
>> >John C. Polasek wrote:
>> >> Take a look at this site:
>> >> http://scienceworld.wolfram.com/physics/GravitationalRedshift.html
>> >> I believe they have the wrong formula for change in L (lambda) vs
>> >> gravity. They show
>> >> L/L0 = sqrt(1 - 2MG/r^2c),
>> >> where L0 is the "rest" WL and L is the "shifted" WL. It shows the WL
>> >> would be shifted lower, a blue shift, when gravity makes red shift.
>> >> It would be OK with f/f0, but they derived it using Newton, at which
>> >> point they got it upside down, or so it seems to me.
>> >>
>> >> John Polasek
>> >> http://www.dualspace.net
>> >
>> >xxein: L/L0 = sqrt(1 - 2MG/r^2c) is not the same as sqrt(1 -
>> >2MG/rc^2).
>> >
>> >It should not seem upside down with this correction to your copy.
> > I mis-typed and should have put rc^2. But it still looks wrong. The
>> "shifted" WL L can't be lower than the "rest" L.
>> John Polasek
>
>xxein: Why would you expect L0 to be greater than L?
>
>What is your primary logic here?
The frequency of an oscillator is reduced by gravity. Then its WL in
gravity should be greater; we are talking about the "shifted" L clock.

The clock removed from gravity will run faster; its WL is shorter;
that's the L0 customer with the short WL.

Their equation says L = L0*(1-eps), making L lower, but it should be
higher.
John Polasek
.

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