Newton's universal law of gravitation

Click For Summary

Discussion Overview

The discussion revolves around the derivation of Newton's law of universal gravitation, exploring the mathematical framework and metrics involved in non-relativistic contexts. Participants analyze the relationships between gravitational acceleration, metric identities, and the Einstein tensor, while questioning the correctness of the derived equations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants present derivations involving the non-relativistic Schwarzschild metric and the Einstein tensor, leading to expressions for gravitational acceleration and force.
  • Participants question the validity of the equation g = - \frac{d^2 s}{dt^2} = - g_{00} \nu', seeking clarification on the derivation.
  • One participant challenges the assumption that leads to the expression for gravitational acceleration, suggesting an alternative formulation that accounts for radial motion.

Areas of Agreement / Disagreement

There is no consensus on the correctness of the equations presented, as participants express differing views on the derivations and assumptions involved.

Contextual Notes

Participants highlight the dependence on specific assumptions regarding motion and the definitions of the metrics used, which may affect the validity of the derived equations.

Orion1
Messages
961
Reaction score
3

Derivation of Newton's law of universal gravitation...

Non-relativistic Schwarzschild metric:
c^2 {d \tau}^{2} = e^{\nu} c^2 dt^2 - e^{\lambda} dr^2 - r^2 d\theta^2 - r^2 \sin^2 \theta d\phi^2

metric identity:
g_{00} = e^{\nu} c^2 = \frac{ds^2}{dt^2}

Non-relativistic Einstein tensor:
G_{11} = \frac{- r \nu' + e^{\lambda} - 1}{r^2} = 0

Einstein tensor metric differential:
\nu' = \frac{e^{\lambda} - 1}{r} = \frac{d}{ds}

Non-relativistic metric identity:
e^{\nu} = \frac{1}{2} \left(1 - \frac{r_s}{r} \right)

metric identity:
e^{\lambda} - 1 = \frac{r_s}{r - r_s}

Gravitational acceleration:
g = - \frac{d s^2}{dt^2} \cdot \frac{d}{ds} = - \frac{d^2 s}{dt^2} = - g_{00} \nu' = - \frac{c^2 e^{\nu} (e^{\lambda} - 1)}{r} = - \frac{c^2}{2r} \left(1 - \frac{r_s}{r} \right) \left(\frac{r_s}{r - r_s} \right) = - \frac{c^2 r_s}{2 r^2} = - \frac{G M_1}{r^2}

Newton's second law and universal law of gravitation:
F_g = M_2 g = - M_2 g_{00} \nu' = - \frac{G M_1 M_2}{r^2}

Are these equations correct?
[/Color]
Reference:
http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation"
 
Last edited by a moderator:
Physics news on Phys.org
Orion1 said:
Derivation of Newton's law of universal gravitation...

Non-relativistic Newton-Schwarzschild metric:
c^2 {d \tau}^{2} = \frac{1}{2} \left(1 - \frac{r_s}{r} \right) c^2 dt^2 - \frac{dr^2}{1 - \frac{r_s}{r}} - r^2 d\theta^2 - r^2 \sin^2 \theta d\phi^2

metric identity:
g_{00} = e^{\nu} c^2

Non-relativistic Einstein tensor:
G_{11} = \frac{- r \nu' + e^{\lambda} - 1}{r^2} = 0

Einstein tensor metric differential:
\nu' = \frac{e^{\lambda} - 1}{r}

metric identity:
e^{\nu} = \frac{1}{2} \left(1 - \frac{r_s}{r} \right)

metric identity:
e^{\lambda} - 1 = \frac{r_s}{r - r_s}

Gravitational acceleration:
g = - \frac{d^2 s}{dt^2} = - g_{00} \nu' = - \frac{c^2 e^{\nu} (e^{\lambda} - 1)}{2r} = - \frac{c^2}{2r} \left(1 - \frac{r_s}{r} \right) \left(\frac{r_s}{r - r_s} \right) = - \frac{c^2 r_s}{2 r^2} = - \frac{G M_1}{r^2}

Newton's second law and universal law of gravitation:
F_g = M_2 g = - M_2 g_{00} \nu' = - \frac{G M_1 M_2}{r^2}

Are these equations correct?
[/Color]
Reference:
http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation"

How did you get g = - \frac{d^2 s}{dt^2} = - g_{00} {\nu}'?

AB
 
Last edited by a moderator:

Altabeh said:
How did you get g = - \frac{d^2s}{dt^2} = - g_{00} \nu'?

metric identity:
g_{00} = e^{\nu} c^2 = \frac{ds^2}{dt^2}

Einstein tensor metric differential:
\nu' = \frac{e^{\lambda} - 1}{r} = \frac{d}{ds}

Gravitational acceleration:
g = - \frac{d s^2}{dt^2} \cdot \frac{d}{ds} = - \frac{d^2 s}{dt^2}
[/Color]
 
Last edited:
Orion1 said:
metric identity:
g_{00} = e^{\nu} c^2 = \frac{ds^2}{dt^2}

Einstein tensor metric differential:
\nu' = \frac{e^{\lambda} - 1}{r} = \frac{d}{ds}

Gravitational acceleration:
g = - \frac{d s^2}{dt^2} \cdot \frac{d}{ds} = - \frac{d^2 s}{dt^2}
[/Color]

This is not true! If we assume that the motion is radial, then

g=\frac{-e^\lambda \dot{r}\dot{{\dot{r}}}}{(e^{\nu}c^2-e^{\lambda}{\dot{r}}^2)^{1/2}},

where a dot over r refers to the derivative wrt time.

AB
 
Last edited:

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 43 ·
2
Replies
43
Views
5K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
947
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K