Re: A Forgotten Prediction of Einstein

Oops.

On Wed, 25 Jan 2006 22:12:51 -0500, sal wrote:

> On Tue, 17 Jan 2006 21:57:10 +0000, John Kennaugh wrote:
>
>> Harry wrote:
>>>
>>>"John Kennaugh" <JKNG@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
>>>news:Y8zuruDG$qyDFwnz@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>
>>>> >> If the sphere is rotating and the
>>>> >> clock with it then it is subject to constant acceleration. I was
>>>> >> under the impression that the basis of relativity was that constant
>>>> >> acceleration and gravity are equivalent.
>>>> >
>>>> >That is GRT, not SRT.
>>>>
>>>> Of course it is that is what I am asking. I know what SR says. It says
>>>> there is an SR component dependent only on speed and independent of
>>>> the radius of the circle. It is the same whether the clock moves in a
>>>> straight line or whether it goes in a circle.
>>>
>>>GRT was designed to give the same predictions as SRT.
>>
>> Then how can the GR correction cancel out the SR correction
>
> Fascinating! Acceleration "fields" behave the same as gravitational
> "fields" so if we could figure out how to model the Earth's gravity as an
> "acceleration field" we should get the GR prediction in this case. Of
> course, Einstein couldn't see how to fit gravity into SR as a well-behaved
> vector field, so I doubt I could, either, but it's still something to
> think about. In this case the SR "correction" ignored the gravitational
> field completely, which is why the result is different from GR.
>
> In the mean time, while we wonder how to model a spherically symmetric
> central gravitational "field" as a (necessarily flat) uniform acceleration
> "field" (and neither one's really a "field", hence the quotes), it's worth
> recalling that there is a simple, intuitive justification for concluding
> that clocks run at exactly the same rate everywhere on the geoid.
>
> The ocean defines an "equipotential" surface -- going from the pole to the
> equator or back you don't gain or lose energy on this surface. If you
> did, the water would "slosh" in the downhill direction until it was
> equalized.

So far so good.


> Coat the Earth with a friction-free surface (floating on the surface of
> the ocean). Slide a hockey puck from the pole to the equator along this
> smooth surface. By the assumption that the geoid is just that, the puck
> must neither gain nor lose energy on its trip (but you do need to guide it
> with a perpendicular force to overcome the Coriolis effect, of course).

I realized after I posted this that it's bogus, I've proposed
something which apparently violates conservation of momentum, angular
momentum, and energy.

It seems like there should be a plausible argument buried in here
somewhere but the way I've posed it does not work.

Viewed from an observation point hovering over the pole the puck has
clearly gained a boatload of energy as it accelerated up to 1000 mph
(tangential velocity) at the equator. Converting it into light, if all is
as I described it, we lose track of that energy, plus the angular momentum
the puck had gained.

Oops, again.


>
> At the Equator turn the puck into light, and send the beam back to the
> pole. At the pole, turn the light back into a puck.
>
> If energy is to be conserved the final mass must be equal to the
> original mass, which, since the puck didn't gain or lose energy on its
> trip, implies that the light must not have been red or blue shifted on
> _its_ trip.
>
> And that's that.
>
> (Tom Roberts, if he's following this, will no doubt interject that
> energy isn't necessarily conserved in this case because the Earth is
> rotating and the field is non-static.

And perhaps he would be right -- but I think there's a more basic mistake
here.

> By the same token this is no
> "proof" but it's a mighty strong _suggestion_ that time should go at the
> same rate all over the Earth's surface. First-law violations in
> closed-loop situations are highly unappealing.)

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