Conference rusure::math

    no - the earth is spinning, and the ocean has waves.  furthermore
    the presence of near-by mountain ranges etc. has an effect.
    
    /bevi{

Also, the Earth is somewhat flattened at the poles, as I recall.

On the other hand, the sum of Earth's gravity and centrifugal forces from
rotation ought to be just about constant at sea level.  If it wasn't, the
water would flow until sea level *was* the place where these forces were
equal.

Re .1:

Waves from wind and currents can be negated by taking a average over
time.  Tides are accounted for by defining sea level as the altitude of
Mean High Tide.  (Ocean depths are defined from Mean Low Tide, however.
"Don't drive the boat where the seagulls are walking" and all that).

To be nitty about the precise phrasing of .0 and your answer (which
I also ignored in my answer), Earth's essentially constant rotation doesn't
change it's mass, or it's position w.r.t. an observer on the surface at sea
level, so it doesn't affect gravity (at least using Newtonian physics).
I assume you were talking about centrifugal forces directly affecting
the observer, and not their change on the distribution of mass in the
Earth indirectly changing the effect of gravity.

Note also that if Earth didn't rotate, and you ignore waves and tides, then
gravity would be the same at sea level everywhere, regardless of mass
concentrations underground.  It's just that sea level would not be at
a fixed distance from the center of the Earth.

Re .2:

I remember seeing a crude illustration (in the Time-Life book of Space?)
which showed the Earth as roughly pear-shaped.  The north pole was 25
miles high, the south pole was 25 miles depressed, the northern tropics
were 50 miles depressed and the southern tropics were 50 miles high.
I think this was based on orbital data from Explorer I.  Don't ask me
to explain the asymmetries - I didn't make the planet.  However, it
would seem there is more than just equitorial centrifugal force displacing
the shape from true sphere.
				/AHM

>Note also that if Earth didn't rotate, and you ignore waves and tides, then
>gravity would be the same at sea level everywhere, regardless of mass
>concentrations underground.  It's just that sea level would not be at
>a fixed distance from the center of the Earth.

Yes, this seems obvious.  If 'g' were different in two nearby locations,
or if the force were not perpendicular to the surface, then the water
would flow from the smaller 'g' to the larger 'g'.

    Re .3, .4:
    
    I believe the surface of a waveless, tideless body of idealized water
    not subject to centrifugal or other non-gravitational forces will be at
    points of equal gravitational potential.  I do not think this is
    necessarily the same as points of equal gravitational field strength.
    
    
    				-- edp 

    
    The point about rotation is that it provides an acceleration NOT
    accounted for by "g", hence the force at two nearby points can be
    the same w/o "g" being the same.
    
    Similarly waves and mountain masses provide forces in directions
    other than perpendicular - same again.
    
    /Bevin

re .3:  I believe the irregularity of the shape of the earth in that book is in
Feet, not miles (that is, the south pole is depressed by 25 feet from a sphere,
etc.)  50 miles is quite a bit.

The accelleration due to gravity does vary over the surface of the earth at sea
level, this is believed to be due mostly to irregularities in the density of
matter in the earth's interior.   The acceleration due to gravity isn't
necessarily the same at all points on the surface of the ocean (ignoring tides,
etc) since the ocean will seek the state of the least energy, while the
acceleration due to gravity determines the force on the water.  Since the force
and energy are related, there is a definite effect of local variation of
gravitational acceleration and sea level at a point, however.

-Mike

    I'm sure you chose the phrasing of the question very carefully - And
from it, I can only conclude that you wanted to beg some discussion
around the assumption of a spherical earth or sea level being the same
everywhere, and so forth.

Nevertheless, one can invoke two explicit answers:  Consider g to be
a vector field whose direction and magnitude at every point are determined
by a central force.  In that case, all points equidistant from the center
of mass of the earth, NO MATTER WHAT ITS SHAPE, possess the same magnitude
but with a different direction.  Therefore, at all points equidistant from
the center of mass, g is DIFFERENT...Remember, we're talking vectors here.

The other answer is that the value of 'g' depends upon the frame of
reference. If, as Einstein's general theory postulates, gravity is 
a manifestation of the curvature of space then it's value depends on
the geodesic lying between the point at which 'g' is measured and where
something (a point mass, for example) is motivated to go...

Personally, I'm compelled by toilet physics whose first postulate is
"There is no gravity, the earth sucks..."

/mtp

I disagree with the claim that if gravity were unequal, water would
flow.

It seems to me that gravity can be unequal at two points, and that as
long as the two different forces are still DOWNWARD, water won't flow.
In order to have flow, there needs to be a force in a non-downward
direction.

/Eric

    Re .9:
    
    Consider this situation.  I have very carefully placed each molecule
    of water so that I have two columns next to each other in a container:
    
    	|	      |		     | -- container wall
    	|	wwwwww|              - -- bottom of container
    	|	wwwwww|              w -- water
        |wwwwwwwwwwwww|
    	|wwwwwwwwwwwww|
    	---------------
    
    Surely you cannot tell me the water will not flow from the higher
    gravitational potential to the lower one simply because it cannot go
    directly downward!
    
    In reality, the water molecules vibrate slightly, move away from the
    edge of the column, find there is nothing but air under them, and fall.
    There is also more pressure from the water on one side of a water
    molecule than from air on the other side, which pushes the molecule
    toward the air.  These things very quickly bring the water down.
    
    
    				-- edp 

We have two constraints on the shape -- the gravitational force must
be perpendicular to the surface, and the potential energy must be
minimized.

    Re .11:
    
    
    Conjecture:  If the potential energy is at a local minimum, the
    gravitational force is perpendicular to the surface.
    
    
    				-- edp

    FALSE - I have already pointed out that gravity is NOT the only
    force acting on the surface.
    
    /Bevin

    Of course, there's also the issue that not all the seas are contiguous,
    there seems little reason to believe the gravitional force on the
    Dead Sea is as low as that in the mid-Atlantic...
    
    /Bevin
    

    Oh yes, one more point - the existence of tides proves that the
    force on the water "at sea level" is not constant, and since tides
    are strongly associated with the relative positions of the
    sun/moon/earth the is reason to believe that the dominant force
    is gravity...
    
    /Bevin

    Re .13:
    
    To what response are you referring?  Recall this:
    
    .5> I believe the surface of a waveless, tideless body of idealized water
    .5> not subject to centrifugal or other non-gravitational forces will be at
    .5> points of equal gravitational potential.
    
    Now I think that is true for a contiguous body of water.  For separated
    bodies, see below. 
    
    
    Re .14:
    
    Even if all seas are not contiguous, the water (even considered as a
    whole) will still be at a local minimum in the gravitational potential
    energy even if the gravitational forces vary among the difference seas,
    subject to the above conditions.  And the gravitational force will
    be perpendicular to the surface, pointing into the water.
    
    Conjecture:  If the gravitational force at the surface of a body of
    water is pointed into the water and perpendicular to the surface at
    every point, the water is at a local minimum in the gravitational
    potential energy.  (The surface is the set of points where the water
    and the air are next to each other, and does not include points where
    water is stopped by land or other items.  So the two columns of water I
    had in a previous example have a surface along the side of one of the
    columns, where gravity is parallel to the surface.) 
                                                       
    
    				-- edp 

    The local gravitational attraction (and thus the acceleration
    due to gravity) at an arbitrary point on the earth's surface
    depends not only on the total mass of the earth and the fact
    that the earth is not perfectly spherical, but also on the
    density of the material in that area (and thus on the local
    thickness of the earth's crust).  These local gravitational 
    fluctuations are measurable.
    
	-Neil

Let's first deal with the idealized situation -- the only gravitational
body is the one with the water on its surface, the mass of the fluid is
negligable, the fluid completely covers the surface of this lonely orb,
there are no winds, nor tides, the system has reached steady state, and
thermal, coreolis, and magnetic effects can be ignored.  The only relevant
features in this simplified problem are shape and density of the planet,
and the amount of water.  Our interest is in the surface of the water. 

Now the potential energy of the system must be at a minimum (just a local
minimum if there are 'holes' in the planet).  Around and in the planet are
surfaces of gravitational equipotential.  These I maintain are smooth.  All
points on the water's surface are on the same surface of gravitational
equipotential -- suppose they were not: then there is some water at a
greater potential that could flow to a lower potential, which contradicts
the premise that the potential energy has reached a local minimum.

Now can someone explain why the surface of gravitational equipotential must
everywhere be perpendicular to the gravitational force? 

    Re .18:
    
    > Now can someone explain why the surface of gravitational
    > equipotential must everywhere be perpendicular to the gravitational
    > force? 
    
    If it were not, an iota of water could move along the surface of
    supposedly equal potential but in the process also moving along the
    component of gravity parallel to the surface.  Moving along a non-zero
    component of the gravity vector means reducing the gravitational
    potential energy, which is a contradiction. 
    
    
    				-- edp 

re .18,.19:

    That done, what can we say about little g at the surface?

re: .18

> Now the potential energy of the system must be at a minimum (just a local
> minimum if there are 'holes' in the planet).  

This is false.  

In point of fact, there is a famous physics principle called Hamilton's 
principle which states that the difference between the kinetic energy and the 
potential energy of a system are minimized.  This is sometimes called "the 
principle of least action."  Mathematically, one can derive Newton's laws using 
Hamilton's principle. The derivations are usually found in discussions of the 
Calculus of Variations in math texts, or in junior level physics texts.

Your statement above fails to take kinetic energy into account.  And it
does count!

> All points on the water's surface are on the same surface of gravitational 
> equipotential.

If you mean that all points on the water's surface are at the same gravitational
potential energy, you are patently wrong.  Please clarify what you mean, maybe
I'm missing something?

> -- suppose they were not: then there is some water at a greater potential that 
> could flow to a lower potential, which contradicts the premise that the 
> potential energy has reached a local minimum.

Your premise is wrong (see Hamilton's principle, above).

One of the problems with this whole discussion is that we are not being
very explicit as to just what gravitational force we're talking about!  
I'm assuming we're talking about the NET GRAVITATIONAL FORCE exerted on a 
particle of water at the surface. In this case, as is the case for 
all fields derived from central forces, the force field vector points to the 
CENTER OF MASS (CoM) of the earth-moon-etc. system with a magnitude inversely 
proportional to the square of the distance between the point (R^2) and the CoM.  

The gravitational field is constantly shifting because the CoM constantly 
changes.  The surface of the earth is moving, its oceans are in motion, etc.
Now, let's define some surface for which all points possess the same value
of gravitational potential energy - An equipotential energy surface.  If
one examines the gravitational force field at every point on the equipotential
surface, you will find that the force vectors will vary in direction and
magnitude.  In special cases the magnitudes will all be the same (e.g.,
a spherical equipotential energy surface), but in all cases I can think
of the directions will always be different!

The confusion in this whole discussion seems to stem from the confusion between
the gravitational force field (a vector field) and the gravitational potential
energy field (a scalar field).  The original question dealt explicitly with
'g' - The vector associated with the gravitational force field.  This value,
as argued in the previous paragraph, varies from point to point along the
gravitational equipotential energy surface.

/mtp

    Since I see no easy way to account for kinetic energy, I'll assume
    that it is negligible, and *doesn't* count (aren't simplifying
    assumptions fun?).

    I think you're mistaken about the force vector pointing toward the
    center of mass.  Were this so, rocks on the visible side of the
    moon might fall toward the earth; actually everything would be pulled
    toward the center of mass of the universe.

>                                                                       If
> one examines the gravitational force field at every point on the equipotential
> surface, you will find that the force vectors will vary in direction and
> magnitude.

    That the force varies in direction is obvious.  The variation in magnitude
    at the surface is the crux of the problem -- that it varies is not obvious.
    Further explanation would help (though I think you're right).

    Re .21:
    
    > In point of fact, there is a famous physics principle called
    > Hamilton's principle which states that the difference between the
    > kinetic energy and the potential energy of a system are minimized.
    > This is sometimes called "the principle of least action." 
    
    Is this a principle, a theorem, or a physical law?  My guess is you are
    referring to dynamic systems of some sort.  I can guarantee the
    principle as you have stated it does not hold in many situations.  For
    example, a spring with friction and with a weight on the end will come
    to rest in a gravitational field with no kinetic energy but with
    potential energy.  In fact, one just needs to consider the potential
    energy represented by mass to see that kinetic energy is often
    overwhelmed.  Also, kinetic energy and potential energy vary depending
    on what frame of reference you use.  Your statement of Hamilton's
    principle is not invariant in inertial frames of reference!
    
    > Mathematically, one can derive Newton's laws using Hamilton's
    > principle. 
    
    I know that has got to be an overly simplistic statement.
    
    
    				-- edp

>Is this [Hamilton's Principle] a principle, a theorem, or a physical law?  

It is usually discussed as a physical law.  It's basis in physical theory
comes from the notion that a particle, in motion, will always move along
a geodesic.  The geodesic is determined by the geometry of the surface
over which the particle is constrained to move, and the energy
(potential and kinetic) of the particle.  For example, in a cartesian
coordinate system a particle will move in a straight line (the geodesic)
unless one adds energy to the system.  As a second example, particles
on the surface of a sphere will move in a great circle (the geodesic) unless one 
does something (i.e., add energy) to displace it.

Hamilton's principle allows a much more general application of mechanics
than does Newton's laws.  In point of fact, most of the mathematical treatment
of mechanics beyond first year physics, is based on Hamilton's principle.
The other advantage to using Hamilton's Principle is that one can derive
    Newton's equations of motion from the expressions of potential and
    kinetic energy alone.  One does not have to use Force equations.
    
> My guess is you are referring to dynamic systems of some sort.

Nope!  It applies generally.  Moreover, using Hamilton's principle it is
much easier to derive equations of motion, for example, for coordinate
systems other than cartesian.

>I can guarantee the principle as you have stated it does not hold in many 
>situations.  For example, a spring with friction and with a weight on the end 
>will come to rest in a gravitational field with no kinetic energy but with 
>potential energy.  In fact, one just needs to consider the potential energy 
>represented by mass to see that kinetic energy is often overwhelmed.  Also, 
>kinetic energy and potential energy vary depending on what frame of reference 
>you use.  

Sorry, it applies explicitly to the situation you describe.

>Your statement of Hamilton's principle is not invariant in inertial 
>frames of reference!

Not sure what this means.  Can you explain?
    
>> Mathematically, one can derive Newton's laws using Hamilton's
>> principle. 
    
>I know that has got to be an overly simplistic statement.

Right!  It's a wonderfully simple (viz elegant) derivation of the laws
of motion and I'd be pleased to put the derivation in a subsequent
response.  But given time, etc., probably not until this evening.
Look for it tomorrow.
                                                           
Regards,

    Re .24:
    
    >> I can guarantee the principle as you have stated it does not hold in
    >> many situations.  For example, a spring with friction and with a weight
    >> on the end will come to rest in a gravitational field with no kinetic
    >> energy but with potential energy.  In fact, one just needs to consider
    >> the potential energy represented by mass to see that kinetic energy is
    >> often overwhelmed.  Also, kinetic energy and potential energy vary
    >> depending on what frame of reference you use. 
    >
    > Sorry, it applies explicitly to the situation you describe.
    
    How so?  A spring at rest with no kinetic energy but with potential
    energy does not demonstrate kinetic energy tending to equal potential
    energy.  There are two examples above, not just one.  Mass is potential
    energy, and it is immensely larger than kinetic energy in common
    situations, again demonstrating kinetic energy does not tend to equal
    potential energy.
    
    >> Your statement of Hamilton's principle is not invariant in inertial
    >> frames of reference! 
    > 
    > Not sure what this means.  Can you explain?
    
    Physical laws are invariant in inertial frames of reference.  This
    means the law does not change when the observer moves at different
    speeds relative to the objects being observed.  For example, if I see a
    rocket expel some material, the momentum of the system (the rocket and
    the material) will be the same before and after the expulsion.  This
    law is true whether I am not moving relative to the rocket initially or
    whether I am moving at half the speed of light relative to the rocket.
    
    Your statement of Hamilton's principle does not have this property.
    Therefore, it cannot be a valid physical law.  The kinetic energy of
    the system will be much greater if I am moving at half the speed of
    light relative to the rocket than if I were not moving relative to it.
    In these two frames of reference, the kinetic energies and potential
    energies would vary, and so they could not tend to be equal in both
    frames of reference.  In other words, Hamilton's principle would
    necessarily say two different things, depending on whether I am moving
    or not.  But that cannot be, because my moving does not affect the
    rocket.
    
    >> I know that has got to be an overly simplistic statement.
    > 
    > Right!  It's a wonderfully simple (viz elegant) derivation of the laws
    > of motion and I'd be pleased to put the derivation in a subsequent
    > response.
    
    I said "overly simplistic", not "very simple".  I would like to see the
    derivation. 
    
    Please give a citation for Hamilton's principle.
    
    
    				-- edp

    Could Hamilton's principle be described in another note, and can
    we let this one continue with the burning question of whether the
    magnitude of 'g' is constant at sea level?  Thanks.

    Some notes from the CRC Handbook, 49th edition , page F-144.
    
    	Latitude	g cm/s*s
    	0		978.039
    	30		979.329
    	60		981.918
    	90		983.217
    
    The geologists commonly measure Bouger gravity, a metric showing
    the difference between the angle of a plumb bob and true vertical.
    In fact "gravity" tends to point toward the ocean floor and away
    from those heavy looking mountains. The massive basalt base of the
    oceans wins over the lighter, weaker mountains. In reference to
    the problems of oceans flowing see references on "isostasy" for
    example in "Physical Geology" by Leet and Judson.