On Tue, Apr 26, 2022 at 11:07 AM John Clark <johnkcl...@gmail.com> wrote:

> On Mon, Apr 25, 2022 at 6:42 PM Bruce Kellett <bhkellet...@gmail.com>
> wrote:
>>> *> Sure, a spring balance needs to be calibrated against some standard
>> mass. But we do not calibrate every day. Once the scale is set, we assume
>> that the spring constant or whatever remains the same, so that
>> recalibration is not necessary.*
> You're right, it's not necessary because as long as the test mass and the
> mass standard decrease by an equal percentage you're always gonna get the
> same result and you'll never notice that anything has changed.

No need to compare with a standard mass.

> *> So if all energies (including mass) drop by 90%, we will be able to
>> detect this as long as the spring constant does not also change by this
>> amount. Springs tend to rely on the electromagnetic properties of metals,
>> and these will not change just because we measure a spin component in the
>> next room.*
> If Many Worlds is correct then of course the spring constant will change
> because the world will split due to ANY measurement, and the absolute
> non-relative amount of energy of EVERY type will decrease.

The Schrodinger equation does not change the spring constant. The spring
constant itself is not an energy, so it does not change. It merely reflects
the change in any mass placed on the scales.

* > I used a spring balance to compare a mass against the gravitational
>> field, where I assumed that Newton's constant does not change on a spin
>> measurement. If all energies (and masses) drop by 50% in each branch of the
>> spin measurement, then the mass of the earth decreases by 50%, and the
>> local acceleration due to gravity, g, also drops by 50%. Now consider a
>> simple pendulum: the period of swing is T = 2*pi*sqr(L/g), where L is the
>> length of the pendulum. If g drops by 50%*,[...]
> But g does NOT drop by 50% and I never said it did,

If you actually calculate g, it is g = GM/r^2, where G is Newton's
constant, M is the mass of the earth, and r is the radius of the earth. So
of course g change in exactly the same proportion as every energy and mass.

I said the gravitational potential energy drops by 50%, and that will
> happen if the mass/energy of a gravitationally bound system drops by 50%
> even if g remains constant.

g does not remain constant. The gravitational PE is PE = mgh, so if both m
and g decreased by 50%, the PE decreases by 75%. Showing, once again, that
the idea that all energies decrease by the same proportion is contrary to
the physics of the situation

> If yesterday I measured the mass/energy of a pendulum and of the entire
> earth against an energy standard and I measure those things again today
> against today's energy standard, and if the mass/energy of the pendulum and
> the earth and today's energy standard have all decreased by 50%, then I
> will get the same measured value that I got yesterday even if g really is
> the same as it was yesterday.

We are comparing the pendulum period before and after the split, not masses
against some standard every day.

And yes the force that the earth is pulling down on that pendulum would
> only be half as strong as it was yesterday, HOWEVER  the inertia (which is
> proportional to the mass/energy) of the pendulum would only be half as much
> as it was yesterday, so the two changes with cancel out and the pendulum
> would fall with the same acceleration that it did yesterday, and the
> period of its swing would be the same too.

The period of the pendulum is T = 2pisqrt(L/g). In this equation, only g
changes given your theory of the split, so the period certainly changes.
The period does not depend on the pendulum's mass, so inertial effects are
included in the calculation of the above formula for the period..


You received this message because you are subscribed to the Google Groups 
"Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send an email 
to everything-list+unsubscr...@googlegroups.com.
To view this discussion on the web visit 

Reply via email to