MOND's similarity to Le Sage's Model

On 26 Sep 2006 00:31:53 -0700, "George Dishman"
<geo...@xxxxxxxxxxxxxxxxx> wrote:

On Mon, 25 Sep 2006 20:41:28 +0100, "George Dishman"
<geo...@xxxxxxxxxxxxxxxxx> wrote:

"Aetherist" <TheAether...@xxxxxxxx> wrote in message
news:jgkdh25q1lpcma72t9lafsc04lfhrpsqn5@xxxxxxxxxx
In (MO)dified (N)ewtonian (D)ynamics it is proposed that the flat
galactic rotation profiles observed is due to a modification to
Newton's force equation such that,

F = ma

is modified to be,

F = kma

Where,

k = µ(x)

and µ and x are arbitrarily determined constants.

<snip my explanation>

You have a fundamental misunderstanding of my comments
which I cannot help you with...

Paul, you have a fundamental misunderstanding
of MOND if you believe what you wrote. µ(x) is
a function which is intended to make F=ma
non-linear.

Clearly MOND is 'intended' to modify F = ma.

Yes, I said that.

That modification
is also clearly specified as F = ma µ(x).

Yes, no problem so far.

Just as clearly,
x is defined as |a|/a_o where a_o is an 'arbitrary' constant

Right, it is a_0 that is the constant, not mu or x.
If your statement that mu and z were constants was
just a slip of the finger, why not just say so, I
make typos all the time.

picked to match observations. Of course |a|/a_o is a dimensionless
ratio (scalar). Further, JUST AS CLEARLY MOND defines this as,

µ(x) << 1 becomes x
and
µ(x) >> 1 become unity...


Let's quote the Reference shall we,

"The exact form of µ is unspecified, ...
only its behavior when the argument x is small
or large."

Given that µ is also dimensionless, we have a vaguely defined
'function' µ(x) such that when x << 1 µ(x) = x and when x >> 1,
unity .

Again all of that is correct.

Now what continuous mathematical form behaves in such
a manner?

Lots, as you say it is vaguely defined and not too
important.

Given this form, by its very definition, µ(x) MUST!
be non-linear for all x's > 0 & < infinity...

Well let's remind ourselves of what I wrote now
that you have inserted enough text to separate
your comment from it:

µ(x) is
a function which is intended to make F=ma
non-linear.

Sound familiar?

Further, is it a discontinuous function???

Actually, I think it is at x=0. Consider the effect
µ(x)=tanh(x) would have for small positive and
negative accelerations such that |a| << a_0.

[Paul wrote]:
and µ and x are arbitrarily determined constants.

It is _not_ a constant, arbitrary or otherwise.

That's funny, quote...

"The term a_o is a proposed new constant,

Yes, as I pointed out, a_0 is the constant, not
the function mu or its argument x as you said.

...
Milgrom proved in his original paper, the form of µ
does not change most of the consequences of the
theory, such as the flattening of the rotation
curve. ..."

The exists a strong correlation of a characteristic inverse of
a property to the property itself, such as the standard radiation
transport equation,

<snip non-ascii characters>

Where t is the thickness of a shield and h the 'linear attenuation
coefficient' which has dimensional units of inverse length. If
one wanted to know how much of the radiation's momentum was lost
in transiting t it is simply,

-ht
dp = p(1 - e )

Nothing stops me from writing this as,

dp = p f(ht)

And further defining f(ht) as,

-ht
f(ht) = 1 - e

Certainly, when ht << 1 we get,

f(ht) = ht

and when ht >> 1 we get,

f(ht) = 1 (unity)

So George, what IS the difference in the mathematical
characteristics of f(ht) and µ(x)?

There are many exponentials in physics and the
form of the equation does not indicate a physical
connection. For Newtonian gravity we have:

f = GMm/r^2 (1)

and

f = ma (2)

hence

a = GM/r^2 (3)


For a << a_0 in MOND, eqn (2) becomes:

f = m a^2 /a_0 (4)

hence from (1) and (4) we get:

a^2 = a_0 GM/r^2

a = sqrt(a_0 GM) / r (5)


If you want to show a similarity between MOND and Le
Sage, you need to start with your previous derivation
that showed that shadowing of the momentum flux was
equivalent to eqn (3) for |a| >> a_0 but that it
transitions smoothly to eqn (5) when |a| << a_0. The
nature of that transition will then determine the
function mu.

It is NOT clear to me how anyone can misunderstand these obvious
mathematical facts...

It was not clear to me how anyone could misunderstand the
Wiki page and think that "µ and x are arbitrarily determined
constants."

AFAIK the proof Le Sage offerred that his theory was
equaivalent to Newtonian gravity for a >> a_0 holds
for small accelerations, i.e. as r >> infinity so it
does not match MOND.

George

.

Relevant Pages

  • MONDs similarity to Le Sages Model
    ... galactic rotation profiles observed is due to a modification to ... Clearly MOND is 'intended' to modify F = ma. ... "The exact form of µ ... <snip non-ascii characters> ...
    (sci.astro)
  • Re: MONDs similarity to Le Sages Model
    ... galactic rotation profiles observed is due to a modification to ... Newton's force equation such that, ... Clearly MOND is 'intended' to modify F = ma. ... is also dimensionless, ...
    (sci.astro)