# Acceleration in gravity

• A
Summary:
accel on a quasar
I want to derive an acceleration in the case for a stationary mass in the gravity field.

I found the total energy in the GR is provided by a simple equation:
https://en.wikipedia.org/wiki/Schwarzschild_geodesics

## E = mc^2\sqrt{1 - rs/r} * \gamma ##

So, this is easy to provide acceleration for that energy definition, using standard math alone: the gradient.

## g(r) = -GM/r^2 \frac{1}{1 - rs/r} ##

And this should be correct (for a stationary body: v = 0).

I have seen many complicated solutions for this problem, and inconsistent with this result.
What is a problem with this?

Last edited:

PeterDonis
Mentor
2020 Award
I want to derive the acceleration in the case for stationary mass in the gravity.
Such a body will not be traveling on a geodesic. But your derivation uses an equation which is only valid for a body that is traveling on a geodesic.

What is a problem with this?
See above.

Do You suggest: the standard math is wrong?

Ibix
2020 Award
Do You suggest: the standard math is wrong?
No - you are misapplying it. ##dt/d\tau## isn't ##\gamma## in general, so your expression is wrong to start with. And an object sat on a solid surface isn't following a geodesic, either, so looking at geodesics won't be an awful lot of help.

The correct way to derive it is to calculate the four-acceleration, ##A^i=U^j\nabla_j U^i##, of the worldline of a hovering observer (one at constant Schwarzschild spatial coordinates) and take the modulus, ##\sqrt{g_{ij}A^iA^j}##.

Last edited:
PeterDonis
Mentor
2020 Award
By which you mean "take the derivative with respect to ##r##", which, in a curved spacetime, is not the same as a gradient (you would need to take a covariant derivative, not a partial derivative).

Do You suggest: the standard math is wrong?
No, just that (a) the "standard math" you claim to be using is irrelevant to the question you are trying to answer, and (b) you are doing the "standard math" that you claim to be using incorrectly.

Irrelevant: I can take derivative wrt time t.
dE/dt = 0.

The result is still the same.

• weirdoguy and Dale
Ibix
2020 Award
Irrelevant: I can take derivative wrt time t.
dE/dt = 0.

The result is still the same.
You already know you have the wrong answer. Yet when we tell you what you're doing wrong, you describe it as irrelevant and come up with another incorrect idea. Do you think you're going to get anywhere with this strategy?

• Dale
PeterDonis
Mentor
2020 Award
I can take derivative wrt time t.
dE/dt = 0.
This says that the energy at infinity of an object following a geodesic worldline is a constant of the motion. Which is (a) a well-known fact which you don't have to explain to anyone here, and (b) irrelevant to the question you are trying to answer, since, as has already been pointed out to you, a stationary object in this spacetime is not following a geodesic worldline.

Your OP question has been answered, both explaining why your claimed answer is wrong and indicating how to get the correct answer. Whether you want to accept those answers or not is (a) your call, and (b) not something it is worth keeping this thread open to find out.