I was pondering this thread https://www.physicsforums.com/showthread.php?t=401713" when it occurred to me that we should be be able to answer simple questions like "what does General Relativity predict that the weight of a 1kg mass on the surface of a very massive gravitational body to be?" or "what is the initial acceleration of 1kg mass when released from a short distance above a very massive gravitational body?".

Not being able to find a clear, simple definitive answer to that question on the internet, I decided to do some "back of the envelope" calculations and this is what I came up with (using a combination of hunches, intuition and guesses):

Assuming Schwarzschild geometry (uncharged non-rotating body) and defining the gravitational gamma factor as:

[tex] \gamma = \frac{1}{\sqrt{1-\frac{2GM}{rc^2}}} [/tex]

the initial coordinate acceleration of a test mass released at r is (according to the observer at infinity) is:

[tex]a= \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)[/tex]

The initial acceleration of the test mass when released at r, as measured by a local observer at r is:

[tex]a_0 = \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}} = a \gamma^3[/tex]

This is equal to the proper acceleration of the test mass when it is stationary in the gravitational field at r, as measured by an accelerometer.

The proper force acting on the stationary test mass on the surface of the gravitational body (i.e its weight as measured by a set of weighing scales on the surface) is:

[tex] F_0 = \frac{GMm_0}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}} = m_0a_0[/tex]

The Schwarzschild coordinate force as measured by an observer at infinity is:

[tex] F = \frac{GMm}{r^2} [/tex]

Note that:

[tex]a_0 = a\gamma ^3[/tex]

and:

[tex]F_0 = F = m_0 a_0 [/tex]

just as in the linear analogue in Special Relativity for longitudinal acceleration and force.

Does that seem about right? Can anyone find any solutions in "the literature"?

P.S. I have not shown my working or referenced any sources that I based my calculations on, but I can if anyone is interested.

Not being able to find a clear, simple definitive answer to that question on the internet, I decided to do some "back of the envelope" calculations and this is what I came up with (using a combination of hunches, intuition and guesses):

Assuming Schwarzschild geometry (uncharged non-rotating body) and defining the gravitational gamma factor as:

[tex] \gamma = \frac{1}{\sqrt{1-\frac{2GM}{rc^2}}} [/tex]

the initial coordinate acceleration of a test mass released at r is (according to the observer at infinity) is:

[tex]a= \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)[/tex]

The initial acceleration of the test mass when released at r, as measured by a local observer at r is:

[tex]a_0 = \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}} = a \gamma^3[/tex]

This is equal to the proper acceleration of the test mass when it is stationary in the gravitational field at r, as measured by an accelerometer.

The proper force acting on the stationary test mass on the surface of the gravitational body (i.e its weight as measured by a set of weighing scales on the surface) is:

[tex] F_0 = \frac{GMm_0}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}} = m_0a_0[/tex]

The Schwarzschild coordinate force as measured by an observer at infinity is:

[tex] F = \frac{GMm}{r^2} [/tex]

Note that:

[tex]a_0 = a\gamma ^3[/tex]

and:

[tex]F_0 = F = m_0 a_0 [/tex]

just as in the linear analogue in Special Relativity for longitudinal acceleration and force.

Does that seem about right? Can anyone find any solutions in "the literature"?

P.S. I have not shown my working or referenced any sources that I based my calculations on, but I can if anyone is interested.

Last edited by a moderator: