[QUOTE=davieddy;267929]BTW I deliberately call them "Lagrange points" for two reasons:

1) The "Lagrangian" is of massive importance in theoretical physics,
as is his work on small oscillations in many particle systems.[/QUOTE]Yes, but I'd argue that it's been a mistake to have unqualified "Lagrangian" refer to only one specific function, when Lagrange is well-known for several other things. "Lagrangian function" wouldn't confuse anyone or lay exclusive claim to a single meaning of the adjective, and its contraction to simply "Lagrangian" within a discussion about that function, once the full term had been used once to establish the context, would be quite in line with practice regarding other items named for their inventors.

[quote]2) I am inclined to think that "Lagrangian Points" have become accidently so-called merely through confusion of the ideas.[/quote]No accident. No confusion. Both arise from Lagrange's mathematics. The L~ points arose from the L~ mechanics calculations using the L~ function.

[quote]I would be extremely surprised if either Euler or Laplace weren't aware of them earlier. Or even Sir Isaac for that matter;)[/quote][quote=http://en.wikipedia.org/wiki/Lagrangian_point]The three collinear Lagrange points were first discovered by [URL="http://en.wikipedia.org/wiki/Leonhard_Euler"]Leonhard Euler[/URL] around 1750.[/quote]

The others became apparent only after
[quote=http://en.wikipedia.org/wiki/Lagrangian_point]Lagrange re-formulated the classical Newtonian mechanics to give rise to Lagrangian mechanics.[/quote]So one couldn't expect them from Newton. (Besides, Isaac had all that alchemy to investigate.)

[QUOTE=Uncwilly;267930]A nice com-sat in orbit around Terra-Luna L2 would make a far side landing mission possible. It could also facilitate a lunar radio telescope (creating a super-long base line array.)[/QUOTE]But should not the ultra-long base line array come before super-long base line array? [size=1]Assuming the HF/VHF/UHF/SHF/EHF/THF model for naming is to be followed.[/size] :evil:

[QUOTE=retina;267942]But should not the ultra-long base line array come before super-long base line array? [SIZE=1]Assuming the HF/VHF/UHF/SHF/EHF/THF model for naming is to be followed.[/SIZE] :evil:[/QUOTE][SIZE=1]Or we could follow the astronomical telescope model and go directly to overwhelmingly-long base line array (OWLBLA).[/SIZE]

[QUOTE=cheesehead;267938]
So one couldn't expect them from Newton. (Besides, Isaac had all that alchemy to investigate.)[/QUOTE]
Not to mention the Royal Mint and rivals like Hooke, Huygens and Leibnitz
to contend with:smile:

[QUOTE=cheesehead;267938]Yes, but I'd argue that it's been a mistake to have unqualified "Lagrangian" refer to only one specific function, when Lagrange is well-known for several other things. "Lagrangian function" wouldn't confuse anyone or lay exclusive claim to a single meaning of the adjective, and its contraction to simply "Lagrangian" within a discussion about that function, once the full term had been used once to establish the context, would be quite in line with practice regarding other items named for their inventors.

[/QUOTE]

Perhaps I've led a sheltered life amid the Dreaming Spires, but I have
to confess I've never heard a "Lagrangian" called a "Lagrangian function"
in my life. The same goes for "Hamiltonian".

And my favourite textbook was Goldstein "Classical Mechanics". (A Yank).

Newton's Laws, Ohm's Law, Hamilton's Principle, Sod's Law,
Maxwell's Equations/Demon and Lagrange's Points.
BTW did you know that the last Fermatian theorem
has been proved?
Searching for Mersennian primes is such fun!

David

BTW Do I recall correctly that Laplace was especially
proud of "Mechaniques Celeste" because it had no diagrams?

PS I will allow "Gaussian Curvature" and "Euclidean Geometry".
Archimedean Solids, but not his Principle, Screw or Tombstone.

[QUOTE=retina;267942]But should not the ultra-long base line array come before super-long base line array? [size=1]Assuming the HF/VHF/UHF/SHF/EHF/THF model for naming is to be followed.[/size] :evil:[/QUOTE]
That would be this: [url]http://spaceflightnow.com/news/n1107/25spektr/index.html[/url]

[QUOTE=cheesehead;267938]
The others became apparent only after
So one couldn't expect them from Newton. (Besides, Isaac had all that alchemy to investigate.)[/QUOTE]

Prove that if we have three different masses (total M) at the
vertices of an equilateral triangle side a, and take the C of M as the origin,
the acceleration of the ith mass at [B][U]r[/U][/B][SUB]i[/SUB] is -GM[U][B]r[/B][/U][SUB]i[/SUB]/a[SUP]3[/SUP]

eg circular orbits with angular velocity (GM/a[SUP]3[/SUP])[SUP]1/2[/SUP]

(So trivial that Isaac, Leonard and Pierre could hardly have missed it)

David

PS if they were released from rest, they would collide simultaneously
at the C of M. When?
I think the same answer would apply to four masses
at the vertices of a regular tetrahedron.

PPS Are those circular orbits necessarily coplanar?

L4 and L5
 

[QUOTE=davieddy;268015]Prove that if we have three different masses (total M) at the
vertices of an equilateral triangle side a, and take the C of M as the origin,
the acceleration of the ith mass at [B][U]r[/U][/B][SUB]i[/SUB] is -GM[U][B]r[/B][/U][SUB]i[/SUB]/a[SUP]3[/SUP]
(So trivial that Isaac, Leonard and Pierre could hardly have missed it)

I think the same answer would apply to four masses
at the vertices of a regular tetrahedron.
[/QUOTE]

Greeted with incredulous silence.

m[SUB]1[/SUB][B]r[/B][SUB]1[/SUB] + m[SUB]2[/SUB][B]r[/B][SUB]2[/SUB] + m[SUB]3[/SUB][B]r[/B][SUB]3[/SUB] + m[SUB]4[/SUB][B]r[/B][SUB]4[/SUB] = M[B]0 [/B]
[C of M is at the origin]

m[SUB]1[/SUB][B]a[/B][SUB]1[/SUB] = Gm[SUB]1[/SUB](m[SUB]1[/SUB]([B]r[/B][SUB]1[/SUB] - [B]r[/B][SUB]1[/SUB]) + m[SUB]2[/SUB]([B]r[/B][SUB]2[/SUB] - [B]r[/B][SUB]1[/SUB]) + m[SUB]3[/SUB]([B]r[/B][SUB]3[/SUB] - [B]r[/B][SUB]1[/SUB]) + m[SUB]4[/SUB]([B]r[/B][SUB]4[/SUB] - [B]r[/B][SUB]1[/SUB]))/a[SUP]3[/SUP]
[[B]Newton's[/B] 1st and 2nd laws of motion, and [B]Newton's[/B] law of gravitation]

So [B]a[/B][SUB]1[/SUB] = -GM[B]r[/B][SUB]1[/SUB]/a[SUP]3[/SUP]

David

Elegant answers to a beautiful problem
 

[QUOTE=davieddy;268015]Prove that if we have three different masses
(total M) at the vertices of an equilateral triangle side a,
and take the C of M as the origin, the acceleration of the ith mass at
[B][U]r[/U][/B][SUB]i[/SUB] is -GM[U][B]r[/B][/U][SUB]i[/SUB]/a[SUP]3[/SUP][/QUOTE]

Answered (with possibly deceptive concision) in the preceding post
for the tetrahedron case.

[B]Points to note: [/B]

1) The centroid of the equilateral triangle does [B]not [/B]lie at the
centre of mass (barycentre).

2) The position vectors [B][U]r[/U][/B][SUB]1[/SUB] [B][U]r[/U][/B][SUB]2[/SUB] [B][U]r[/U][/B][SUB]3[/SUB]
of the 3 masses relative to the C of M have different magnitudes.
Excercise: what are they in terms of a? (I don't need to know the answer).
The angles between them are not exactly 120 degrees.
However, if the masses remain equidistant from each other, these angles
remain constant as the equilateral rotates about the C of M and
changes size.

3) As the equilateral rotates and the side a changes, the speed and acceleration of m[SUB]i[/SUB] are proportional to r[SUB]i[/SUB] (pure kinematics).
The beauty is that Newton's Laws show this to be the case!

[quote]
eg circular orbits with angular velocity (GM/a[SUP]3[/SUP])[SUP]1/2[/SUP]
[/quote]
Note v = [TEX]\omega[/TEX]r and acceleration = [TEX]\omega[/TEX][SUP]2[/SUP]r.

Note that the acceleration towards the C of M obeys the inverse
square law as the size of the triangle varies.
So if the initial angular velocity differed from (GM/a[SUP]3[/SUP])[SUP]1/2[/SUP], the three masses would trace a conic, with focus at the C of M, just as in a two particle system.
Excercises:
1) Find the escape angular velocity (easy).
2) Find the period of elliptical orbits (harder).
3) If they were released from rest, they would collide simultaneously
at the C of M. When?
[quote] I think the same answer would apply to four masses
at the vertices of a regular tetrahedron. [/quote]
I'm now damn sure it would

[quote]Are the orbits necessarily coplanar?[/quote]
Yes!
The 3 momentum vectors sum to zero; a triangle is planar.
Note that this does not apply to a quadrilateral, so the
tetrahedral case might hold further fascination.

Enough for now.
Hope you get as much fun out of this problem as I have! .

David

Are the orbits necessarily coplanar?

[QUOTE=davieddy;268169]Yes!
The 3 momentum vectors sum to zero; a triangle is planar.
[/QUOTE]

I still think the answer is yes, but this glib argument is
completely broken!

David

Evidence(?) for other universes found in CMB
 

[url]http://www.bbc.co.uk/news/science-environment-14372387[/url]

Far too early to say how good the evidence may be but at least there are some testable predictions and a small amount of supporting evidence. When the Planck telescope comes on-line it should give a clearer picture (ambiguity intentional).

Paul