Time dilation explanation for gravity?

yuiop
Messages
3,962
Reaction score
20
Hi,

I seem to recall reading a post a long time ago (that I cannot find) that gravity in the Newtonian limit (eg the Solar system) can be completely explained in terms of time dilation alone. Is that true and if so, how does that work?
 
Physics news on Phys.org
In the weak field limit, when GR goes to Newtonian gravity, a function of g_00 acts like the potential, and g_00 is the time dilation factor ( or it's square root).

In fact,

-g_{00} = 1 + \frac{2\phi}{c^2}

Is this what you mean ?
 
Last edited:
Mentz114 said:
In the weak field limit, when GR goes to Newtonian gravity, a function of g_00 acts like the potential, and g_00 is the time dilation factor ( or it's square root).

In fact,

-g_{00} = 1 + \frac{2\phi}{c^2}

Is this what you mean ?

Not sure. Assuming \phi = GM/R

and -g_{00} = 1 + \frac{2GM}{Rc^2}

is obtained from the binomial aproximation of \frac{1}{\sqrt{1-2GM/Rc^2}}

How is the Newtonian of gravitational potential of -2GM/R obtained from that?
Where do the +1 and c^2 go and how is the inverse square law of gravity recovered?

Also, as I understand it. GR does not have gravitational forces acting on a free falling body, but there still are gravitational forces acting on a stationary body. Does GR alter the force felt by a stationary body or is it the usual GM/R^2?
 
Kev,
it takes four pages to demonstrate properly, but if you start with the EFE with a simple source, and let most of the gradients involved disappear, then solve the equations of motion you get an inverse square force law, with a function of g_00 playing the part of the potential. Newtonian gravity is recovered in full. Quite a triumph, in fact.

In my earlier post, phi is not the Newtonian potential, but a function of it.

Here's a good explanatory article.

http://www.mth.uct.ac.za/omei/gr/chap7/node3.html
 
Last edited by a moderator:
If you want to use GR without a curved spacetime you have to to alter the constants that GR uses like C and the Planck length and make them rubbery. You can then use euclidian flat space still using G as a constant.
 
mojocujo said:
If you want to use GR without a curved spacetime you have to to alter the constants that GR uses like C and the Planck length and make them rubbery. You can then use euclidian flat space still using G as a constant.

Sounds dodgy to me. GR does not have a length scale. Care to elucidate ?
 
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Thread 'Dirac's integral for the energy-momentum of the gravitational field'
See Dirac's brief treatment of the energy-momentum pseudo-tensor in the attached picture. Dirac is presumably integrating eq. (31.2) over the 4D "hypercylinder" defined by ##T_1 \le x^0 \le T_2## and ##\mathbf{|x|} \le R##, where ##R## is sufficiently large to include all the matter-energy fields in the system. Then \begin{align} 0 &= \int_V \left[ ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g}\, \right]_{,\nu} d^4 x = \int_{\partial V} ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g} \, dS_\nu \nonumber\\ &= \left(...
Abstract The gravitational-wave signal GW250114 was observed by the two LIGO detectors with a network matched-filter signal-to-noise ratio of 80. The signal was emitted by the coalescence of two black holes with near-equal masses ## m_1=33.6_{-0.8}^{+1.2} M_{⊙} ## and ## m_2=32.2_{-1. 3}^{+0.8} M_{⊙}##, and small spins ##\chi_{1,2}\leq 0.26 ## (90% credibility) and negligible eccentricity ##e⁢\leq 0.03.## Postmerger data excluding the peak region are consistent with the dominant quadrupolar...

Similar threads

Replies
46
Views
4K
Replies
16
Views
2K
Replies
18
Views
1K
Replies
103
Views
5K
Replies
40
Views
6K
Back
Top