The mass of your cup increases by the addition of the energy done in raising it
against the force of gravity. This is easily seen if one converts the results
of measurement so that they are made with the same units of measurment (choose
one or the other elevations as a reference).

Don't get hung up with the conclusions of GR, their derivation contains a basic
mathematical error and is faulty. For more information see
http://www.members.aol.com/einsteinhoax/gravity.htm . For response, E-mail

(Esuaceb14) wrote in message ...
The mass of your cup increases by the addition of the energy done in raising it
against the force of gravity. This is easily seen if one converts the results
of measurement so that they are made with the same units of measurment (choose
one or the other elevations as a reference).

Don't get hung up with the conclusions of GR, their derivation contains a basic
mathematical error and is faulty. For more information see
http://www.members.aol.com/einsteinhoax/gravity.htm . For response, E-mail


Mr. Retic,

It seems that you and Dr. Carlip are in agreement on this point,
although he has not specifically answered this case. In the case of
the massive shell, he claims that the "quasilocal energy", newtonian
like potential energy, will be the same as the kenetic energy
released.
But this creates a paradox because he claims that the total mass
will not increase. So that if the shell colapses and the sphere loses
radient energy it will have less mass than it did before it was
raised, for the same number of atoms.
My own view is that gravitational potential energy does not have
mass but kenetic energy does, a raised cup has less mass. A shell
with the same number of atoms will have the same mass as a sphere but
when it colapses it will have that mass plus the mass of the radient
energy.
BTW, do you think space has a euclidean geometry? You have
published quite a tome and you accept an aether, what is its density?

Regards,
Stephen Kearney

(Old Physics)
Message-id:

In the case of the massive shell, he [Dr. Carlip] claims that the "quasilocal
energy", newtonian
like potential energy, will be the same as the kenetic energy
released.
But this creates a paradox because he claims that the total mass
will not increase. So that if the shell colapses and the sphere loses
radient energy it will have less mass than it did before it was
raised, for the same number of atoms.

How it is a paradox? So far the total energy (together with
the radiated out) is the same. If nothing radiates out then
both masses (of the shell and of the sphere) are equal.

My own view is that gravitational potential energy does not have
mass but kenetic energy does, a raised cup has less mass. A shell
with the same number of atoms will have the same mass as a sphere but
when it colapses it will have that mass plus the mass of the radient
energy.

It seems to me as exact equivalent of what Dr. Carlip says.
How is it different in your opinion?

When you say "gravitational potential energy does not have
mass but kenetic energy does" it is obviously true but only
because there is no such thing as "gravitational potential
energy" (if it existed it had mass of course since "mass"
and "energy" is the same thing, E=mc^2) in Einsteinian
world (a.k.a. real world).

When you lift your cup you move it into place where time
runs faster, and so speed of light is a little bit greater (from
your point of view, not form the point of view of the cup
of course), and so its internal energy (E=mc^2, from your
point of view) is a little bit greater. So you have to do work
that you can recover lowering the cup. No change in mass
needed (or happening).

Of course when you just let the cup go (not lowering it
slowly, recovering the energy) then its mass goes up
until the cup stops, and this will be the mass of the
kinetic energy. When you calculate everything carefully
you'll see that during the whole trip down of the cup its
mass increases as much as its kinetic energy (from
your point of view) increases but also c drops keeping
the total energy constant -- a very informative exercise,
with a few tricks however, so you might not get the right
result the first time, but just think harder and everything
will fall into place).

That's the whole mechanism corresponding to Newtonian
"potential energy". So as you can see there is no special
"gravitational energy" involved, only the properties of
spacetime, too subtle to notice, that simulate the
existence of "gravitational potential energy".

Similarly as other properties of spacetime simulate the
accelerating expansion of the universe. It turns out that
Einstein discovered more things that he is credited so
far by the contemporary scientists for.

-- Jim

But this creates a paradox because he claims that the total mass
will not increase. So that if the shell colapses and the sphere loses
radient energy it will have less mass than it did before it was
raised, for the same number of atoms.

How it is a paradox? So far the total energy (together with
the radiated out) is the same. If nothing radiates out then
both masses (of the shell and of the sphere) are equal.

My own view is that gravitational potential energy does not have
mass but kenetic energy does, a raised cup has less mass. A shell
with the same number of atoms will have the same mass as a sphere but
when it colapses it will have that mass plus the mass of the radient
energy.

It seems to me as exact equivalent of what Dr. Carlip says.
How is it different in your opinion?

When you say "gravitational potential energy does not have
mass but kenetic energy does" it is obviously true but only
because there is no such thing as "gravitational potential
energy" (if it existed it had mass of course since "mass"
and "energy" is the same thing, E=mc^2) in Einsteinian
world (a.k.a. real world).

When you lift your cup you move it into place where time
runs faster, and so speed of light is a little bit greater (from
your point of view, not form the point of view of the cup
of course), and so its internal energy (E=mc^2, from your
point of view) is a little bit greater. So you have to do work
that you can recover lowering the cup. No change in mass
needed (or happening).

Of course when you just let the cup go (not lowering it
slowly, recovering the energy) then its mass goes up
until the cup stops, and this will be the mass of the
kinetic energy. When you calculate everything carefully
you'll see that during the whole trip down of the cup its
mass increases as much as its kinetic energy (from
your point of view) increases but also c drops keeping
the total energy constant -- a very informative exercise,
with a few tricks however, so you might not get the right
result the first time, but just think harder and everything
will fall into place).

That's the whole mechanism corresponding to Newtonian
"potential energy". So as you can see there is no special
"gravitational energy" involved, only the properties of
spacetime, too subtle to notice, that simulate the
existence of "gravitational potential energy".

Similarly as other properties of spacetime simulate the
accelerating expansion of the universe. It turns out that
Einstein discovered more things that he is credited so
far by the contemporary scientists for.

-- Jim

Jim

Thank you for your informed and well written answer. I have a lot
of thinking to do.
There is something else I have a tough time with. Imagine a shell
about a LY in radius with a thickness of about 4000 miles and the
earth's density. Its total mass would be about three trillion times
the mass of the sun and it would constitute a black hole with the same
radius. The surface gravity if it were of uniform density within,
about a billionth the density of earth, would give it a surface
acceleration of 1g.
Because the mass is concentrated in a thin shell, would the
surface gravity be greater? What would the time dilation be at its
center, relative to a distant observer?
If the BH aspect complicates things, imagine it 2000 miles in
thickness.

With highest regards
Stephen Kearney

I would still like an answer. sk

(Old Physics) wrote in message


There is something else I have a tough time with.
Imagine a shell about a LY in radius with a thickness
of about 4000 miles and the earth's density. Its
total mass would be about three trillion times the
mass of the sun and it would constitute a black hole
with the same radius. The surface gravity if it were
of uniform density within, about a billionth the
density of earth, would give it a surface acceleration
of 1g.
Because the mass is concentrated in a thin shell,
would the surface gravity be greater?

No. If the masses of a solid ball and a shell are
equal then the gravitational field outside them is the
same and so the "surface gravity" (acceleration) is
the same.

What would the time dilation be at its
center, relative to a distant observer?

The time dilation is changing with the distance to the
center as the "Newtonian potential" since the difference
in Newtonian potential between two points is just the
time dilation between those points multiplied by c^2
(with opposite sign, to be exact: when potential drops,
time dilation increases). So when the observer approaches
the surface from outside the time rate with respect to
the observer at infinity drops (time dilation increases)
as integral of "gravitational acceleration" (which is the
same as "gravitational potential"). You will have
the greatest acceleration at the surface of the shell,
and then it drops throughout the shell until it reaches
zero on the other side of the shell, inside it. So inside
the shell the "gravitational potential" is the same
everywhere and so the time runs inside the shell
everywhere at the same rate (the same dilation with
respect to the distant observer). I assume you know how
to calculate Newtonian gravitational potential.

If the BH aspect complicates things, imagine it
2000 miles in thickness.

BH aspect complicates thing in the way that with
stronger fields you have to consider that not only
time dilation as relevant (as it is always where any
"gravitational acceleration" is felt) but also changes
in geometry of space become important since amount of
space increases at the same rate as time gets dilated
(the reason for it is that the nature can't make energy
from nothing and it would have to if it were otherwise
for certain rather complex reasons).

While we immediately detect time dilation feeling
"gravitational acceleration", the changes in amount of
space might be easily overlooked. E.g. at acceleration
1g the change of "gravitational potential" along distance
of 1 meter corresponds to about 9 m^2/s^2. Therefore it
corresponds to time dilation (divide by c^2) of about
10^(-16). And so it corresponds to the same change in
the amount of space (which simulates an effect of
rulers getting relatively "shorter" by this amount).
Such small change as 10^(-16) is hardly possible to
notice, while the same time dilation, by causing 1g
"gravitational acceleration" may be felt very easily
by an average human (as "acceleration" and not as
change in rate of time of course). For an average
photon, because of its speed, and so covering large
distances during its travel, both time dilation and
"length contraction" (in fact the increase in amount
of space which just looks like "contraction of the
rulers") is the same relevant and so a ray of light
bends twice as much while passing near the sun as our
imperfect Newtonian gravity could predicted. It is
because Newtonian gravity predicts only phenomena
caused by time dilation (and so corresponding to the
"gravitational acceleration"). And that's why your
questions might be answered easily just considering
the behavior of Newtonian potential.

It won't be that easy when geometry of space deviates
so much from flat space that it gets important which
might happen when you start messing with "strong
gravitational fields" (a.k.a. "BH"). But at 1g and human
distances it makes not much difference. When you
get to distances like 1 ly it might mess your
calculations because the relative error caused by the
space curvature is greater. So 2000 miles thick shell
is safer if you are interested just in the time dilation
for the time being.

-- Jim

Jim Jastrzebski wrote:
: (Old Physics) wrote in message
: Because the mass is concentrated in a thin shell,
: would the surface gravity be greater?
:
: No. If the masses of a solid ball and a shell are
: equal then the gravitational field outside them is the
: same and so the "surface gravity" (acceleration) is
: the same.

You don't know that. It is possible the
method of using a Cavendish Balance will make it
appear that is true, but a Cavendish Balance
does not measure surface gravity.

And you won't find a hollow shell the size
of the Earth.

Joe Fischer

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