Conference rusure::math

b) a laminar infinite mass body.  The gravitational force is the same no
matter how far away you are.  Just like if it were  an infinite electro
luminescent plane light, the strength of the light is the same no matter how
far away you get from it.  Or if it were electrically charged with a constant
planar charge density, the electric field strength would be the same no
matter how far away you were.  Or if it were chocolate, your attraction to the
infinite chocolate plane would be the same no matter how far away you were.

You do an integral over the plane from the position of the force resulting frmo
infitesimal masses in the plane.

Tom

>  	(a) Sketch a proof that the gravitational force between two point
>     	    masses is inversely proportional to the square of the
>  	    distance between them.  (NO EMPIRICAL EVIDENCE ALLOWED!)

The inverse-square law (which is only known to work to some precision)
is a summary of empirical evidence.  Physics formulas are all like that;
they summarize measurements.  If the measurements are wrong, the equations
are wrong, too.  

Anyway,  what else am I allowed to know to do this problem?

Tom

    I recall seeing a computer-generated film several years ago depicting
    the behavior of two point masses in planetary rotation about each
    other. You could twiddle the value of the exponent in the gravitational
    equations and show that values of the exponent other than two led
    to unstable orbits: in one case (>2) the bodies eventually collided
    and in the other (<2) the bodies drifted apart. Only for the exponent
    exactly two were the orbits stable. My guess is that you can derive
    the exponent 2 from the differential equations of motion and the
    assumption of stability.

    Re .3:
    
    I think you have the cases backwards.  For example, if the exponent is
    one or less, the energy needed by an object to escape another is
    infinite, so drifting apart is not possible.  Also, I do not believe
    all or even most orbits are unstable with other values of exponents,
    just as there are unstable orbits with an exponent of two (viz., two
    objects headed directly for each other or parabolic or hyperbolic
    paths).  In fact, permanent escape is impossible if the kinetic energy
    of the system is less than the gravitational potential energy with the
    bodies infinitely far apart minus the current gravitational potential
    energy, regardless of the exponent, and I think collision may be
    impossible with any exponent as long as the tangential component of
    velocity is non-zero. 
    
    
    				-- edp

2b. A point object of mass m lies distance x from an infinite (planar)
    lamina of finite mass per unit area.  What is the gravitational
    force on the mass?

    Let	G = the gravitational constant,
	m = mass of the object,
	x = distance from the lamina, and
	� = mass per area of the lamina.

    Consider concentric rings on the lamina, centered around the perpendicular
    from the mass to the lamina.  Let an annulus have radius R and 'width' dR.

    The area of this is 2RdR, and its mass is 2�RdR, and all this mass is at
    distance sqrt(R�+x�) from the object.  However, when summing the forces
    between the object and this annulus (these are vector forces), we see that
    the 'effectiveness' of the mass, the component toward the lamina, is only
    x/sqrt(R�+x�) of the whole force.  Putting this together, we have the force:

	 R=inf      2Gxm�RdR             R=inf         RdR
	Integral  ------------- = 2Gxm� Integral  -------------
	  R=0     (R�+x�)^(3/2)           R=0     (R�+x�)^(3/2)

    But,
		       RdR
	Integral  ------------- = -1/sqrt(R�+x�)
		  (R�+x�)^(3/2)

    And so,

	 R=inf	       RdR
	Integral  ------------- = -0 - (-1/x) = 1/x
	  R=0	  (R�+x�)^(3/2)

    So the force is simply 2Gm�, which is independent of x.

re .2
>The inverse-square law (which is only known to work to some precision)
>is a summary of empirical evidence.  Physics formulas are all like that;
>they summarize measurements.  If the measurements are wrong, the equations
>are wrong, too.  

True, the exponent CAN be determined experimentally.  But it happens
that this constant is THE MOST accurately measured physical constant.
And right now I wish I had a reference to this little fact.

re .3
>You could twiddle the value of the exponent in the gravitational
>equations and show that values of the exponent other than two led
>to unstable orbits.

Interesting.  Some years ago I wrote a program with which you could
'steer' a spaceship through a galaxy of stars.  I believe we played
with the exponent, to make it easier to establish a stable orbit.

It should be easy for us to recreate the computational experiment you
describe.

re .3
>You could twiddle the value of the exponent in the gravitational
>equations and show that values of the exponent other than two led
>to unstable orbits.

If the exponent is other than two, the orbits are not always conic sections. 
They are stable in the two body case (I dont know about 3 or more body cases)
in the sense that the bodys will not always collide (or escape).

Phil

re: .4
>and I think collision may be
>    impossible with any exponent as long as the tangential component of
>    velocity is non-zero. 
>
Is that why that Russian satellite hit Canada's tundra a few years ago?
8-)
Tom

    2(a)
    	The key to this, I think, is geometry, not physics.
    In addition to the fact that a given line of force does
    not decay, I also assume that it does not increase.
    	Say we represent the gravitational field of a point
    mass as a bunch of lines of force radiating out from the
    object in 3 dimensions (euclidean space).  The number of
    lines is proportional to the gravitional strength of the
    point mass (who knows? the constant of proportionality might
    have something to do with the mass of the object, but it
    doesn't matter).  Now look at what happens as you move
    away from the point mass:  The surface area of a spherical
    shell at a distance R from the object is (4/3)*pi*(R**2).
    The number of lines of force per unit area is proportional
    to 1/(surface area) since the number of lines of force
    remains constant.  Therefore, the number of lines of force
    per unit area is proportional to 1/(R**2).  All someone
    has to prove is that the force another object would feel
    is proportional to the number of lines of force per unit
    area.
    

re .4
>   For example, if the exponent is one or less, the energy needed by an
>   object to escape another is infinite, so drifting apart is not possible.

    Yes, but what about exponents in the range of (say) 1.5-2.5?

    Re .10:
    
    Then it depends on the particular situation.
    
    
    				-- edp

    Re .9:
    
    The number of lines emanating from a point and intersecting an object
    with non-zero cross-section is infinite (the same as the number
    of points in a plane).  Or, if two points are being considered,
    the number of lines emanating from one and intersecting the other
    is always one, regardless of distance.  I think the problem needs
    to be better defined.
    
    
    				-- edp

    re .-1
    OK, ok.  Obviously "lines of force" is just a concept that makes
    it easy to draw a force field without resorting to some
    half-tone technique to show the "amount" of force at a given point.
    When you see a picture of a magnetic field surrounding a bar magnet
    in a physics book, you certainly don't assume that the force is
    zero between the discrete "lines of force" illustrated in the book.
    I guess the problem is one of trying to represent a continuous
    phenomenon with a discrete illustration.  Sorry about that.  Is
    there a way to invoke limits which will turn the discrete notion
    of "lines of force" into something more continuous like "force field"?
    I still think there's a geometric approach lurking in this problem
    somewhere.  I seem to remember that Einstein's general theory of
    relativity had a lot of geometry in it (albeit non-Euclidean).
    
    Carl
    

    Re .13:
    
    Flux.  The total gravitational/electric/magnetic flux "passing through"
    a closed surface is equal to the mass/charge/"magnetic charge" (the last
    of these has never been observed to be anything other than zero) inside
    the volume enclosed by the surface, multiplied by some constant.
    
    The flux passing through a given surface is the integral over the
    surface over the dot product of field strength and d(area).  For a
    sphere around a single mass, the field strength is constant and pointed
    perpendicular to the surface, because of symmetry, so the integral is
    equal to the field strength multiplied by the surface area.  If the
    field strength at a radius of r is G(r), then kM = G(r)*4*pi*r^2, so
    G(r) = kM/(4*pi*r^2), and we see the field strength is inversely
    proportional to the square of distance. 

    An intuitive interpretation of this is that the mass is emitting
    "flux", and any closed surface around that mass will catch all of the
    "flux", no matter what shape the surface has.  Any "flux" being emitted
    by mass outside of the surface goes both into and out of the surface,
    so if there were such a mass complicating the problem, it would be
    accounted for and eliminated.  The field strength is the "density of
    flux", so multiplying the "density of flux" over a certain area by the
    amount of area gives the "flux passing through" the area, which
    completes the relationship between the mass emitting "flux" and the
    field strength. 

    Once understood, this relationship is actually quite simple and is very
    useful in a variety of situations in which a surface can be arbitrarily
    shaped to make integration simple. 


				-- edp

    
    I was intrigued to learn it was the decay of the force carrying
    particle that leads to short range forces.  In fact therefore the
    closer form of the force equation is

                 kx 2
    		e  x
                 
                                kx
    The fact that for gravity  e   approx= 1  shows that either (1)
    it is fundamentally different, or (2) is carried by a particle with
    an extremely long half-life.
    
    
    /Bevin

  Problem 2a appears to have resulted in more argument than I thought.
  It is true of course that the physicists have missed out on this
  one.  The idea of proving this one occurred to me when considering
  a topic in celestial mechanics concerning eg. inverse cubed law.
  The argument was that the rule was inversed squared because orbits
  would not otherwise be stable,  although it occurred to me that
  this was a most unsatisfactory approach when I was convinced
  that a proof existed and yes geometry and calculus are indispensable
  here.
  
  The idea of the invariant line of force does not mean that you
  have to stop there.  The approach I prefer out of the many I
  can envisage here is as follows:
  
  Start with one point mass and a circular section of a spherical
  lamina whose centre (of the sphere,  not the circle)  is at the
  point mass.  The measure of force per unit area will follow from
  the invariancy of any one line of force,  but you must start with
  finite lines of force with scalar value say delta-F which then tend
  in number to infinity and delta-F to zero.  It can now be shown
  that if you increase the radius of the spherical shell but keep
  constant the area of the circular section inscribed in its surface
  (This gets flatter as the radius of the sphere increases,  and
  of course the radius of the circle measured along the sphere
  does not vary whereas the linear radius of the section increases)
  that the force is proportional to the area of the circular section
  divided by the area of the complete sphere.  Since the area of the
  circular section does not vary,  the result is proportional to the
  reciprocal of the area of the complete sphere alone,  which of
  course is itself proportional to the square of the radius of the
  sphere.  To finish,  simply show that the relation holds if the area
  of the circular section tends to zero,  thus providing the
  conditions for a point mass.

This is a good example that although Gauss's lines of force did not catch
on big in the face of vector theory, they are an OK model.

Trouble is, you needed empirical evidence to justify using lines of force.
Fortunately, you didn't have to make the astronomical measurements yourself,
but becuase of this, you forgot it was the motion of planets that led to
lines of force.

So the question is false, and your answer violates the conditions of the
problem. 

Problems must be more carefully designed than solved.
Questons must be more carefully asked than answered.
Tom

    re .-1
    Right on, right on, right on.  The problem with doing mathematical
    physics is in correctly translating empirical reality into math.
    After that is done, the math itself is relatively easy.
    
    By the way, does anyone know how Einstein (or whoever) showed that
    the mass in the grav equation: F = GMm/(r**2) was the same as the
    mass in the acceleration equation: F = ma ?
    Was it all based on experimental observation?  Is there a physics
    notes file where this question might better be addressed?
    

  re .17, .18
  I don't understand.  I admit that I have included the assumption
  "that any one line of force does not decay [OK, sorry 'vary']"
  But I am not suggesting that the problem is valid in the absence
  of such an assumption.  The case for the prosecution appears
  to deny this fact which is on view in the opening phrase of the
  problem as stated.  So thus far:  "I deny the charges,  your honour!"

    re .-1
    Actually, I didn't think there was anything wrong with the original
    statement of the problem.  Maybe if it were restated, some objections
    would be overcome.  How about:

    "Let's imagine a force which can be represented by lines of force
    which do not decay (actually, the continuous analog, which involves
    flux, is more precise).  Let's also assume that this force between
    two point masses acts in a direction parallel to a vector connecting
    the two points (I believe that was a tacit assumption of the original).
    Given those two assumptions (and probably a few more I can't think
    of), show that the force follows an inverse square law."
    Is that a fair restatement of the problem?  I still think it's
    interesting.

    <flame-on>
      I am afraid that it's no fun laying the groundwork
    for a mathematical proof, but it's also necessary.  I saw the
    difference when learning vector calculus.  I learned all the vector
    calculus I needed to do physics in one semester of freshman physics
    while learning other physics things too.  It took a whole year of
    multivariate calculus from the math department to teach the same
    stuff, but it sure was more rigorous.  I think both approaches have
    their place.  It would have taken years to get through thermodynamics
    without a very cavalier attitude towards flinging partial derivatives
    around.
    <flame-off>