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How important is SR inorder to calc the precession of Mercury



 
 
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  #1  
Old October 15th 04 posted to sci.physics.relativity
Nicolaas Vroom
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Posts: 230
Default How important is SR inorder to calc the precession of Mercury

Starting point of this posting is chapter 15.3
"Advance of the perihelion of Mercury"
the pages 195 to 198 in the book
"Introducing Einstein's Relativity" by Ray d'Inverno.

IMO the whole purpose of this exercise is to calculate
with a model the future positions of the planets
(i.e. Mercury) solely based on past observations
(positions) as acurate as possible.

In order to do this you need a 3D grid of measuring
rods and clocks. The clocks are located at the cross
sections of the rods and the clocks are all synchronised
with a clock at the origin.
In order to predict you need a model. One model can be
eq. 15.25 which is the relativistic version of Binet's eq.
and differs from Newton's by the presence of the last term.

Using this 3D grid and the clocks you can perform
the past observations of the positions of the planets.
This is important because eq. 15.25 contains constants.
IMO the only correct way to calculate those constants
(for example m) is to use eq. 15.25 based on past
observations.

One parameter discussed is proper time tau.
In the grid there are no moving clocks involved.
On the other hand if you attach a clock onto Mercury
and you synchronise this clock with the nearest clock
from the grid, you will see that this moving clock
constantly runs behind the nearest clock from the grid
(based on its moving position) and that this difference
is increasing (at variable rate).

I expect that in order to calculate proper time tau
you can also use eq. 8.16 i.e. as a function of v and c.
I expect that v is measured with the clocks from the
3D grid.

IN eq 15.22 a constant k is calculated by means
of a factor dtau/dt (multiplied by 1-2m/r).
I expect that k is only a constant because r is variable.

What amases me that in equation 15.25 the factor k
has disappeared. Does this mean that the concept
proper time is of less importance inorder to calculate
the precession of Mercury ?

How do I compare the above with the following
sentence from
http://arxiv.org/PS_cache/gr-qc/pdf/0103/0103044.pdf
The Meaning of Einstein's Equation
Authors: John C. Baez, Emory F. Bunn
at page 3 of 19:
"Thus the concept of inertial frame, so important in SR
is banned from GR"

Nicolaas Vroom
http://users.pandora.be/nicvroom/






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  #2  
Old October 15th 04 posted to sci.physics.relativity
Nicolaas Vroom
external usenet poster
 
Posts: 230
Default How important is SR inorder to calc the precession of Mercury


"Nicolaas Vroom" schreef in bericht
...

In order to do this you need a 3D grid of measuring
rods and clocks. The clocks are located at the cross
sections of the rods and the clocks are all synchronised
with a clock at the origin.


One of the best url's to study such a grid is the following:
http://www.astro.utu.fi/EGal/elg/ELG3D.html
The two main questions a
1. Is such a grid the right tool to study GR
(Finally in order to simulate the planet Mercury)
2. If yes: What is the metric (tensor) involved.

The "centre" of the grid shows the Milkway galaxy
as a large yellow dot and 3 other major galaxies.
Each of those galaxies is surrounded by a cloud
of smaller galaxies in red.

However the same grid can be used as a part of
our Milkyway galaxy.
The Yellow dot in the centre is than the Sun
surrounded by local stars.

At an even smaller scale the centre is still the Sun
surrounded by planets.

What ever the scale at the crossing points of the
grids there are clocks (and a light), all synchronised.
When you look at clocks on the grid, all clocks show
exactly the same time.
However that is not what you see when you are
at the center of the grid.
When you are at the center of the grid and when there
are no masses involved and when you look along the
line x=0 all clocks at different distances show a different
time (as a function of distance and c).
In fact you only see the first clock (light).

When the whole grid only contains one object (one mass)
the object moves in a perfect straight line through the grid.

Suppose this object crosses the line x=0 very close to the
clock which shows 6.00
(Suppose all the clocks show ONE hour difference)
Suppose the clock at the center shows 12.00.
The question is what will be observed by an observer
at the center ?
The observer will not see the clocks at 11.00, 10.00, 9.00
8.00 7.00 and 6.00 but the observer will be able
to see the clocks (light from the clocks) at 5.00 and earlier
because light from those clocks is bended by the (moving)
mass. (This only for a small period of time)

However, and this is important, you do not have to include
this light bending in order to describe the movent of your
moving object. (i.e. all the objects)

Suppose the center of the grid shows the Sun and there
is only one planet (the Earth)
Suppose the Earth crosses the line x=0 twice
at x= x0 and x= -x0. Suppose there are two clocks
fixed at the grid and there is one moving clock.
Suppose you synchronise your moving clock with
the clock at x=x0.
What will happen that your moving clock will run
behind the two fixed clocks after one revolution
and that this discrepancy will increase after each
revolution.
However, again, you do not have to include this behavior
(slow down) of the moving clock to include in order to
describe the movent of your moving object. (and objects)

But you must take it into account when you convert
earth based observations into grid based "observations"
and vice versa.
The same with light bending.
This becomes more complex when the sun itself is moving
in your grid, but the concept is the same.

The final question to answer is what is the metric of the grid.

What is the proper time in the grid ?

Nicolaas Vroom
http://users.pandora.be/nicvroom/







  #3  
Old October 15th 04 posted to sci.physics.relativity
Nicolaas Vroom
external usenet poster
 
Posts: 230
Default How important is SR inorder to calc the precession of Mercury


"Nicolaas Vroom" schreef in bericht
...

"Nicolaas Vroom" schreef in bericht
...

In order to do this you need a 3D grid of measuring
rods and clocks. The clocks are located at the cross
sections of the rods and the clocks are all synchronised
with a clock at the origin.


One of the best url's to study such a grid is the following:
http://www.astro.utu.fi/EGal/elg/ELG3D.html


In this grid all the objects have grid positions based on
grid coordinates.
We measure (observe) each of those are (Earth based)
positions.
As I explained in my previous posting in order to use the
grid you have to convert your (Earth based) observations
in grid coordinates by taking into account the following
two concepts:
Time dilation and light bending.

Using the positions in grid coordinates and using
a set of rules you can now predict future grid positions.

If you want to test those predicted grid positons with actual
observed (earth based) positions you have to convert
the grid coordinates in Earth based coordinates taking into
account the following two concepts (inverse form):
Time dilation and light bending.

The final question to answer is what is the metric of the grid.


The question is which laws apply to describe the movement
of the objects using the grid coordinates.

IMO there is no time dilation (there are no moving clocks)
and no light bending involved (light bending has to do
with observations but light does not influence the movement)

There is length contraction involved, however I do not know
if that influences the movement of the objects.
(My guess is no)

As a first approximation you can use Newton's Law in order
to describe the movement of the objects in the grid.

One of the most important question to answer is what is the
function of the speed of light within the grid.
(As explained above in order to covert earth based to grid
based coordinates c is important)

IMO the most important parameter to describe the movement
more accurate (beside Newton's law and the calculated mass
parameter m for each object )
is the speed of gravity propagation parameter cg

Nicolaas Vroom
http://users.pandora.be/nicvroom/






  #4  
Old October 27th 04 posted to sci.physics.relativity
Nicolaas Vroom
external usenet poster
 
Posts: 230
Default How important is GR inorder to calc the precession of Mercury (Was SR)

As a result of this discussion I received the following by private e-mail:

"it is not clear what the point is that you are trying to make."

The point I try to make is very simple if you change SR into GR.

What I did in a simple wording is remove "the noise of observations"
in order to transform real observations into positions in a
reference frame consisting of rods and synchronised clocks.
I did this by using the laws of SR.

My question is how do you describe the movements of the
objects in this reference frame ?
By using SR ?
By using GR ?
Do you need everything (all the complexity) described in the book
a) Introducing Einstein's Relativity ?
b) GRAVITATION ?
Do you need the c the speed of light ?
Or should that be c the speed of electro-magnetic radiation ?
What about cg the speed of gravitation ?

(When you are finished you have to "add" the noise of observations,
in order to transform the calculated positions into real observations.)

I have doubts If you need all that is described in the two above
mentioned books and that you can do this in a much simpler way.
IMO the most important parameter is cg,
i.e. you have to mimic the behaviour of the gravitons,
for what ever this is worth.

Anyone responds ?

Nicolaas Vroom
http://users.pandora.be/nicvroom/




  #5  
Old October 28th 04 posted to sci.physics.relativity
Ken S. Tucker
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Posts: 7,674
Default How important is GR inorder to calc the precession of Mercury (Was SR)

"Nicolaas Vroom" wrote in message ...
....
IMO the most important parameter is cg,
i.e. you have to mimic the behaviour of the gravitons,
for what ever this is worth.

Anyone responds ?

Nicolaas Vroom
http://users.pandora.be/nicvroom/


It's always difficult to determine which specific
physical thing causes GR effects.
The easiest way to get to the precession is
to take the derivative of the energy

E = 1/sqrt(1-2GM/rc^2) of a unit mass

dE/dr ~ -(GM/r^2)*(1+3GM/rc^2) = force

That is from E = mc^2/sqrt(g_00).

Regards
Ken S. Tucker
 




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