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Gravitational acceleration in GR 
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#1
Apr2809, 12:51 PM

P: 83

I tried to search everywhere but couldn't find an answer, so here it goes.
In Newtonian mechanics, the gravitational acceleration g at a distance r from the gravitating object is given by Does this equation apply in general relativity aswel? If not, what is the equivalent in GR, ie. how do you calculate the acceleration due to gravity in GR? 


#2
Apr2809, 02:53 PM

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PF Gold
P: 39,345

No, not precisely. There is no simple version of that equation in General Relativity. In relativity, there is an equation involving the "Gravitation tensor" which depends upon the metric tensor of the space. From that, you can (theoretically!) find the metric tensor and use that to find the Geodesics in 4space. All free falling objects move along geodesics with constant "4 speed" which, if you try to force it to a flat 3space, appears as an acceleration.



#3
Apr2809, 07:00 PM

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PF Gold
P: 1,843

There is an equation in GR
[tex]g =  \frac{Gm}{r^2 \sqrt{1  2Gm/rc^2}} [/tex]However, the r in that equation is not "radius" in the sense of something you could measure with a stationary ruler next to a black hole. In fact you can't measure such a radius, because any ruler that approached the hole would fall to pieces. r is the circumference of an orbiting circle divided by [itex]2 \pi[/itex], which in GR is not the same thing a rulermeasured radius. Actually proving the formula is no easy thing. Reference: Woodhouse, N M J (2007), General Relativity, Springer, London, ISBN 9781846284861, page 99 


#4
May1009, 02:44 PM

P: 3,914

Gravitational acceleration in GR



#5
May1009, 03:26 PM

Mentor
P: 6,231

Note that in the limit Gm/rc^2 is small compared to 1, this acceleration is the same as the Newtonian value. 


#6
May1009, 04:41 PM

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Thanks!



#7
Sep2711, 04:19 PM

P: 1,555

The graph below shows the proper distance to the singularity for both a free falling (from infinity) observer and a stationary observer (but note that passed the event horizon there cannot be any stationary observers, but we can still calculate the proper distance passed the event horizon): 


#8
Sep2711, 05:20 PM

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P: 7,598

The apparent reason for the disagreement is that the notion of simultaneity for the Fermi normal distance is different than Passionflowers. On the plus side, the numbers are a lot easier to crunch using Passionflower's definition, it's not a particularly bad approach to measuring distance, it's just not unique. The other big drawback is that it's not computed via a Bornrigid set of observers. 


#9
Sep2711, 10:27 PM

P: 1,555

My integrals are for 'stationary': [tex]\int _{{\it ri}}^{{\it ro}}\!{\frac {1}{\sqrt { \left 1{\frac {{\it rs}}{r}} \right }}}{dr}[/tex] And free falling (from infinity) [tex]\int _{{\it ri}}^{{\it ro}}\!\sqrt { \left 1{\frac {{\it rs}}{r}} \right }{\frac {1}{\sqrt { \left 1{\frac {{\it rs}}{{\it r}}} \right }}}{dr} [/tex] which obviously becomes: rori 


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