Re: Challenge for Paul; Probe that with Mercury ds^2>0 and the solution is spacelike

Liste des GroupesRevenir à sp relativity 
Sujet : Re: Challenge for Paul; Probe that with Mercury ds^2>0 and the solution is spacelike
De : hertz778 (at) *nospam* gmail.com (rhertz)
Groupes : sci.physics.relativity
Date : 24. Feb 2025, 01:37:54
Autres entêtes
Organisation : novaBBS
Message-ID : <9562356f4c43590086f583b7906cd3a5@www.novabbs.com>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Rocksolid Light
Schwarzschild metric
ds² = -(1-2GM/c²r) c²dt² + 1/(1-2GM/c²r) dr² + r²(dθ² + sin² θ dϕ²)
A general solution is spacetime S = ∫ds
S = ∫√[-(1-2GM/c²r) c²dt² + 1/(1-2GM/c²r) dr² + r²(dθ² + sin² θ dϕ²)]
Now, genius, explain this stupid simplification that you wrote, enabling
to wrote the general solution S, or its derivate ds, when you can't
integrate a function with several differentials that is under a square
root operation.
Even if you work only with ds², who ILLUMINATED YOU to write that
s² = −c²T² + D²?
This is another proof that you are a self-entitled pompous retarded.
This was part of your post (quite common my ass):
*************************************************************************
It is quite common to use s² as the interval, but it is more 'natural'
to call the interval s, so that's what I will do.
's' consists of two components, a temporal and a spatial.
If we call the temporal component cT and the spatial component D,
we have: s² = −c²T² + D²
If D > cT then S is spacelike  (s² > 0)  D/T > c
If D = cT then S is lightlike  (s² = 0)  D/T = c
If D < cT then S is timelike   (s² < 0)  D/T < c
Two events on the worldline of a massive object will always be
separated by a timelike interval, because the object's speed D/T
is always less than c, and D < cT.
*************************************************************************
Dig this example of spacelike events in the surface of the Sun. No FTL
here.
/////////////////////////////////////////////////////////////////////
Numerical example using the Schwarzschild metric to determine
spacelike-separated events around a single star. To calculate whether
two events that
on the surface of the star (at different points) are
spacelike-separated,
it's assumed that the Schwarzschild metric applies to the spacetime
around
the star.
The Schwarzschild metric is given by:
ds² = -(1-2GM/c²r) c²dt² + 1/(1-2GM/c²r) dr² + r²(dθ² + sin² θ dϕ²)
Focusing on two events that occur at different positions on the surface
of the star, where both are located at the same radius r, but the
spatial separation between them is along the ϕ-direction (longitude) on
the surface.
For two events to be spacelike-separated, the spacetime interval ds²
between them must be positive. This means ds²>0
Consider two events that happen at the same time t1=t2 but at different
points on the surface of the star (θ=π/2 and different ϕ -coordinates).
For events on the surface of the star, r is constant, and the only
nonzero spatial separation is along the ϕ-direction.
Therefore, the differential dr=0, dθ=0, and the only spatial
component left is dϕ.
The spacetime interval between these events reduces to:
ds² = r²dϕ²
For these two events to be spacelike-separated, ds² must be greater than
zero, which is trivially true since the spatial distance between them is
nonzero.
Numerical Example
Assumptions:
• The mass of the Sun M ≈ 1.989×10^30 kg.
• The radius of the Sun r = 7×10^8m (about the radius of the Sun).
• Gravitational constant G = 6.67430×10^−11 m^3 kg^−1 s^−2
• The two events occur at angular separation Δϕ=0.1 radians (which
corresponds to 5.7°).
Using the formula ds² = r²dϕ², and substituting the values:
ds² = (7×10^8m)² × (0.1 radians)² = 4.9×10^15m²
ds = (7×10^8m) × (0.1 radians) = +70km
Since ds²>0, these events are spacelike-separated. The spatial
separation
between the two events is enough that no signal (including light) can
travel between them in the time frame defined by their identical time
coordinates.
The distance between the events on the Sun’s surface is: +70km.
This calculation demonstrates how two events on the surface of the Sun
can be spacelike-separated, as the spacetime interval ds² is positive.
The events are +70 kilometers apart in space, and since they occur
at the same time in the chosen coordinate system, there’s no way
for information (traveling at or below the speed of light) to
travel between them. They are causally disconnected, which is
the hallmark of spacelike separation.
/////////////////////////////////////////////////////
You can't generalize your stupid s² = −c²T² + D² with Schwarzschild'd
metric.
Why don't you try with the Kerr metric instead?

Date Sujet#  Auteur
12 Feb 25 * Challenge for Paul; Probe that with Mercury ds^2>0 and the solution is spacelike12rhertz
12 Feb 25 `* Re: Challenge for Paul; Probe that with Mercury ds^2>0 and the solution is spacelike11Richard Hachel
13 Feb 25  `* Re: Challenge for Paul; Probe that with Mercury ds^2>0 and the solution is spacelike10rhertz
13 Feb 25   `* Re: Challenge for Paul; Probe that with Mercury ds^2>0 and the solution is spacelike9rhertz
15 Feb 25    +* Re: Challenge for Paul; Probe that with Mercury ds^2>0 and the solution is spacelike6rhertz
17 Feb 25    i`* Re: Challenge for Paul; Probe that with Mercury ds^2>0 and the solution is spacelike5rhertz
18 Feb 25    i `* Re: Challenge for Paul; Probe that with Mercury ds^2>0 and the solution is spacelike4Maciej Wozniak
21 Feb 25    i  `* Re: Challenge for Paul; Probe that with Mercury ds^2>0 and the solution is spacelike3rhertz
22 Feb 25    i   `* Re: Challenge for Paul; Probe that with Mercury ds^2>0 and the solution is spacelike2rhertz
24 Feb 25    i    `- Re: Challenge for Paul; Probe that with Mercury ds^2>0 and the solution is spacelike1rhertz
19 Feb 25    `* Re: Challenge for Paul; Probe that with Mercury ds^2>0 and the solution is spacelike2J. J. Lodder
19 Feb 25     `- Re: Challenge for Paul; Probe that with Mercury ds^2>0 and the solution is spacelike1Maciej Wozniak

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