| Topic: |
Science > Physics |
| User: |
"Ken S. Tucker" |
| Date: |
05 Feb 2005 07:41:08 PM |
| Object: |
Newtonian two body fun! (kst)... |
Because of the title "two body fun" you
must be 30 years or younger to really
enjoy this post, (70 if you're a girl)
otherwise leave!
I'm studying Einstein's Law G=T, for two
approximately equal bodies "a" and "b" in
free-fall toward each other, but I need
to understand that in Newtonian terms.
Using F = -G*M_a*M_b/r^2 I figure the
accelerations to be, (set pesky G=1),
A_a = -M_b/r^2 and A_b = -M_a/r^2
Suppose I choose to consider body "a"
at rest, then F_a=0 and Force on "b" is,
F_b = M_b*(A_a + A_b), or setting "b"
as the reference,
F_a = M_a*(A_a + A_b).
There are only two possible FoR's, "a"
or "b", and I want to find the invariants
in this inter-action.
So I set,
F = F_a + F_b and
M = M_a + M_b and find
F = M^2/r^2 is invariant.
By invariant I mean it's true for an observer
on either body "a" or "b" who may fix themselves
to be at rest, as I spec'd above.
I'll stop here pending comments.
Ken S. Tucker
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| User: "RP" |
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| Title: Re: Newtonian two body fun! (kst)... |
05 Feb 2005 11:26:15 PM |
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Ken S. Tucker wrote:
Because of the title "two body fun" you
must be 30 years or younger to really
enjoy this post, (70 if you're a girl)
otherwise leave!
I'm studying Einstein's Law G=T, for two
approximately equal bodies "a" and "b" in
free-fall toward each other, but I need
to understand that in Newtonian terms.
Using F = -G*M_a*M_b/r^2 I figure the
accelerations to be, (set pesky G=1),
A_a = -M_b/r^2 and A_b = -M_a/r^2
Suppose I choose to consider body "a"
at rest, then F_a=0 and Force on "b" is,
F_b = M_b*(A_a + A_b), or setting "b"
as the reference,
F_a = M_a*(A_a + A_b).
There are only two possible FoR's, "a"
or "b", and I want to find the invariants
in this inter-action.
So I set,
F = F_a + F_b and
F = M_a*(A_a + A_b) + M_b*(A_a + A_b)
F = (M_a + M_b)(A_a + A_b)
F = (M_a + M_b)(-M_b/r^2 + -M_a/r^2)
F = -((M_a + M_b)^2)/r^2
Which reduces to your conclusion below when M is substituted.
OTOH, on the surface this form seems to be valid for any two masses,
whether similar or not, and as it so happens when M_a >> M_b, this
should reduce to Newton. It does not, thus there is an error in the
premises or in the conclusion that it is general.
The solution appears to be here:
"A_a = -M_b/r^2 and A_b = -M_a/r^2"
Because of the symmetry involved we must have
A_a = A_b
Thus you've assumed here that M_a = M_b.
M = M_a + M_b and find
F = M^2/r^2 is invariant.
Thus
F = (2M_a)^2/r^2 = (2M_b)^2/r^2
By invariant I mean it's true for an observer
on either body "a" or "b" who may fix themselves
to be at rest, as I spec'd above.
I'll stop here pending comments.
Ken S. Tucker
Very nice observation :)
What do you derive for unequal masses?
Richard Perry
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| User: "RP" |
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| Title: Re: Newtonian two body fun! (kst)... |
06 Feb 2005 08:05:28 AM |
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Ken S. Tucker wrote:
Because of the title "two body fun" you
must be 30 years or younger to really
enjoy this post, (70 if you're a girl)
otherwise leave!
I'm studying Einstein's Law G=T, for two
approximately equal bodies "a" and "b" in
free-fall toward each other, but I need
to understand that in Newtonian terms.
Using F = -G*M_a*M_b/r^2 I figure the
accelerations to be, (set pesky G=1),
A_a = -M_b/r^2 and A_b = -M_a/r^2
Suppose I choose to consider body "a"
at rest, then F_a=0 and Force on "b" is,
F_b = M_b*(A_a + A_b), or setting "b"
as the reference,
F_a = M_a*(A_a + A_b).
There are only two possible FoR's, "a"
or "b", and I want to find the invariants
in this inter-action.
So I set,
F = F_a + F_b
Ok, this is incorrect.
The force that each will observe is not constant unless the masses are
equal. And it won't equal F_a + F_b, since these are already the
forces observed in the accelerated frames a and b respectively.
You should have
F = F_a = F_b
Substituting
F = M_a*(A_a + A_b)
F = M_a(M_b/r^2 + M_a/r^2)
F = 2(M_a^2)/r^2
or
F = M^2 / 2r^2
That's more like it.
M = M_a + M_b and find
F = M^2/r^2 is invariant.
Except for the miss by a factor of 1/2, and the fact that it isn't
invariant for unequal masses, it is just "equal" for equal masses.
:)
Richard Perry
By invariant I mean it's true for an observer
on either body "a" or "b" who may fix themselves
to be at rest, as I spec'd above.
I'll stop here pending comments.
Ken S. Tucker
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| User: "Ken S. Tucker" |
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| Title: Re: Newtonian two body fun! (kst)... |
07 Feb 2005 08:46:10 AM |
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RP wrote:
Ken S. Tucker wrote:
Because of the title "two body fun" you
must be 30 years or younger to really
enjoy this post, (70 if you're a girl)
otherwise leave!
I'm studying Einstein's Law G=T, for two
approximately equal bodies "a" and "b" in
free-fall toward each other, but I need
to understand that in Newtonian terms.
Using F = -G*M_a*M_b/r^2 I figure the
accelerations to be, (set pesky G=1),
A_a = -M_b/r^2 and A_b = -M_a/r^2
Suppose I choose to consider body "a"
at rest, then F_a=0 and Force on "b" is,
F_b = M_b*(A_a + A_b), or setting "b"
as the reference,
F_a = M_a*(A_a + A_b).
There are only two possible FoR's, "a"
or "b", and I want to find the invariants
in this inter-action.
So I set,
F = F_a + F_b
Ok, this is incorrect.
The force that each will observe is not constant unless the masses
are
equal. And it won't equal F_a + F_b, since these are already the
forces observed in the accelerated frames a and b respectively.
You should have
F = F_a = F_b
Substituting
F = M_a*(A_a + A_b)
F = M_a(M_b/r^2 + M_a/r^2)
F = 2(M_a^2)/r^2
CRASH, no need to set M_a = M_b.
or
F = M^2 / 2r^2
That's more like it.
M = M_a + M_b and find
F = M^2/r^2 is invariant.
Except for the miss by a factor of 1/2, and the fact that it isn't
invariant for unequal masses, it is just "equal" for equal masses.
:)
Richard Perry
By invariant I mean it's true for an observer
on either body "a" or "b" who may fix themselves
to be at rest, as I spec'd above.
I'll stop here pending comments.
Ken S. Tucker
By defining the magnitudes, A = A_a + A_b
and the system mass M = M_a + M_b I obtain,
F = M*A = M^2/r^2
(sorry that's not the same as the F = -G*M_a*M_b/r^2
I used in my first post, pardon the notation
mal-function, people keep stopping by and visiting).
What I want to do here is get to gravitational energy,
e = F*r = M^2/r
where M is the systems mass and now e is the
systems gravitational energy, that could be
+/-, in Newtonian terms.
Adding SR, E=Mc^2, set pesky c=1, then
e = E^2/r and then e*r = E^2 conceptually.
Wait, when we used SR, I think we should
put Newton's "G" back in i.e.
e*r = G*E^2 .
Since e and E have the same units n=e/E
is an invariant scalar and produces,
n*r = G*E.
Subject that to a SR transformation, i.e.
what does a speedy CS measure passing
the system (M_a , M_b).
n*r' = G'*E'
Using SR, r' = r/gamma, E'=E*gamma.
It follows G' = r'/E' = r/E*gamma^2 = G/gamma^2,
and Newton's Gravitational Constant is not an
invariant.
How does that figure into Einstein's Law,
G_uv = K*T_uv, (where K == G) ?
Well given that length "r" is represented
by the contravariant x^u, and that G transforms
like r^2, we should expect Newton's constant
to tranform as a 2nd rank contravariant tensor
K^ab.
So we insert that into Einstein's Law, to get,
G_uv = K^ab T_uv , well that's funny...
I see two options to balance the L and RHS,
1) G = K^uv T_uv
2) G_uv = K^ab T_uvab
If we were to find the contravariant metric
"g^uv" invariantly proportional to K^uv, then
the above would become,
1a) G=T (invariant relation) ,
2a) G_uv = g^ab T_uvab .
The system of Equations(a) looks promising,
especially the tensor T_uvab, as it provides
an opportunity to merge better with the
Riemann Christoffel curvature R_uvab.
Also, it provides more degrees of freedom
to define the meaning of *energy density*.
Remarkably, I haven't replaced Einstein's
Law, nor has it been modified, I merely
suggest Newton's Big "G" is invariantly
proportional to g^uv.
What has been ascertained is that Big "G"
doesn't need to be an invariant, and so
setting G=1, be done as an approximation.
Regards
Ken S. Tucker
(KxSxT)
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