Re: Infinite 3d lattice

On Jul 27, 4:43 am, "I.N. Galidakis" <morph...@xxxxxxxxxxxx> wrote:
cbr...@xxxxxxxxxxxxxxxxx wrote:

<snip>

If we call the force at x caused by ABCD f(x), then the force at x
caused by EFGH is f(x-1), and the force caused by E'F'G'H' is f(x+1);
and what we then will want to find is

F(x) = f(x) + sum(n=1 to oo)(f(x+n) + f(x-n))

I get that

f(x) = (4*g*M*m)*x/((x^2 + 1/2)^(3/2))

(derivation below at [*]) and since we are really only interested in
whether the result is less than, greater than, or equal to 0, we might
as well choose M and m so that 4*g*m*M = 1, and then

f(x) = x/((x^2+1/2)^(3/2))

.... with the later correction since the force is attractive ...

f(x) = -x/((x^2 + 1/2)^(3/2))



<snip>

I think I understand the problem a little better now. Using your diagram above,
which I rotate by Pi/4 radians (so |Ao| = sqrt(2)/2), so we are looking at one
diagonal plane to the central cube:

... E' A E ...
| |\ |
----o------o-x----o----
| |/ |
... F' C F ...

Let us concentrate on the upper half only. As you correctly say, the force will
be AT LEAST:

F(x) = f(x) + sum(n=1 to oo)(f(x+n) + f(x-n))

and I get:

f(x) = m*M*G_k*cos(arctan((sqrt(2)/2/x)))/(1/2 + x^2)

(whether this actually agrees with your magnitude is relatively unimportant at
this point)


To simplify a bit:

Given a right triangle with sides a, b and hypotenuse h = sqrt(a^2 +
b^2),

cos(arctan(a/b)) = b/h
= b/sqrt(a^2 + b^2)

allows us to reduce

cos(arctan( (sqrt(2)/2)/x ))

to

x/sqrt( (sqrt(2)/2)^2 + x^2 )

and then to

x / sqrt( 1/2 + x^2)

and then, letting K = m*M*G_k, we find that

f(x) = K*cos(arctan((sqrt(2)/2/x)))/(1/2 + x^2)
= K*x / sqrt( 1/2 + x^2) )/(1/2 + x^2)
= K * (x /( 1/2 + x^2)^(3/2))

which is the function I gave previously (normalized for ease of
readability):

f(x) = -x/(x^2 + 1/2)^(3/2)

Note the sign of f(x): this is an /attractive/ force; so actually your
G_k must be negative.

The problem is that F(x) approximates the forces ONLY close to o. In other words
this is valid for x SMALL.


I disagree.

Firstly, the force f(x) is valid for /any/ x no matter how small or
large; it's not an approximation, it's an exact value for the force
exerted by a /single/ face at the (signed) distance x along the x-
axis. It has a graph somewhat reminiscent of a sigmoid function.

Secondly, letting n be the x coordinate of any square face, the
distance from the nth face (measured along the x axis) to a point at x
is going to be x - n. So the force from the nth face on x is going to
be exactly f(x-n).

And then the total force at x is going to be exactly the limit of the
sum of forces from each face at each integer n (and in this case the
limit exists); which is exactly F(x).

Now, because of symmetries in the problem, it suffices to only look at
F(x) in the range |x| <= 1/2; but this is just another way of saying
that F(x) = F(x+a) for all integer a.

This follows because

F(x) = f(x) + sum(n=1 to oo)(f(x-n) + f(x+n))
= ... + f(x-2) + f(x-1) + f(x) + f(x+1) + f(x+2) +...

F(x+1) = f(x+1) + sum(n=1 to oo)(f(x+1+n) + f(x+1-n))
= ... f(x+1-2) + f(x+1-1) + f(x+1) + f(x+1+1) + f(x+1+2) +...
= ... f(x-1) + f(x) + f(x+1) + f(x+2) + f(x+3) + ...
= F(x)

and then the claim follows by induction.

So F is a cyclic function with period 1.

In fact, since f(-x) = -f(x) we can deduce also that

F(-x) = f(-x) + sum(n=1 to oo)(f(-x-n) + f(-x+n))
= - f(x) + sum(n=1 to oo)(-f(x+n) - f(x-n))
= - (f(x) + sum(n=1 to oo)(f(x+n) + f(x-n)))
= -F(x)

so F is an /even/ cyclic function with period 1, looking a bit like -
sin(x*2*pi) if you graph it.

To see what happens near x = 1/2, we need another analysis, because close to
1/2, the force from A is balanced out by the force from E (1/2 is inertial). Let
us therefore choose x small to THE LEFT of 1/2:

A 1/2 E...
| /| |
o----x-o-----o
| \| |
C 1/2 F...


There really is no difference between this analysis and the previous
one; except for a change in the origin and orientatation of the x-
axis, which doesn't change any of the distances involved, and thus
doesn't change any of the forces involved.

This time, the forces at x will be from A and from E, plus the rest of the chain
as you've noticed:

G(x) = g(x) - g(-x) + sum(g(n-x)-g(n+x),n=1..infinity)

where:

g(x) = m*M*G_k*cos(arctan((sqrt(2)/2/(1/2-x))))/((1/2-x)^2+1/2)


Sure; and then after eliminating the trig functions, it pretty easy to
see that

g(x) = K*f(1/2-x)

and from this and the fact that f(-x) = -f(x), I think you'll find
that G(x) = K*F(1/2-x) and thus G(0) = F(1/2) = 0.

All you've done is change where the origin from which x is to be
measured, as well as the sense of which direction is the positive
direction for x. But we originally chose x=0 to be the center of a
face purely for convenience; it has no essential role.

I think now, G(x) correctly approximates what happens close to 1/2, for x
approaching 0 from the left.

Maple seems to verify that these points are indeed inertial points:

We must be careful not to let our new Computer Overlords do /
everything/ for us!

F(1/2) = ... + f(-5/2) + f(-3/2) +
f(-1/2) + f(1/2) +
f(3/2) + f(5/2) + ...

which, since f(-x) = -f(x), sums neatly to 0. And since F(x) = F(x+a)
for all integers a, it also follows that F(a/2) = 0 for all odd
integers a.

Similarly,

F(0) = ... + f(-2) + f(-1) +
f(0) +
f(1) + f(2) + ...

and since f(0) = 0, F(0) = 0.


For the case you analyzed:

f:=x->m*M*G_k*cos(arctan((sqrt(2)/2/x)))/(x^2+1/2);
F:=x->f(x)+sum(f(n-x)-f(n+x),n=1..infinity);

In my formulation, that would be

f:=x->m*M*G_k*x/(x^2+1/2)^(3/2);
F:=x->f(x)+sum(f(x-n)+f(x+n),n=1..infinity);

which gives an F which is slightly different from what you write above
for non-integral x; but since -f(x) = f(-x) and f(0) = 0, it comes to
the same thing in this case, namely...

limit(F(x),x=0,right);

0

For the case close to 1/2:

g:=x->m*M*G_k*cos(arctan((sqrt(2)/2/(1/2-x))))/((1/2-x)^2+1/2);
G:=x->g(x)-g(-x)+sum(g(n-x)-g(n+x),n=1..infinity);
limit(G(x),x=0,left);

0

Of course, the forces F and G are STILL lower bounds (in terms of magnitude),
because the grid extends upwards and downwards to infinity and there are
contributions from there, too (although I suspect that because the forces vary
as 1/r^2, the corresponding series might converge).

I agree; but I haven't explored this yet.


By Jove, I think we've made some progress...

Inch by inch, row by row, ...

Cheers - Chas
.

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