Inertial Gravity in Relativistic Geodesics - Forum Discussion

[Thytanium]

TL;DR Summary

The inertial acceleration in free fall is zero but the ordinary acceleration is not zero in free fall due to the relative movement and for this reason there are differences between the two that I want to know and above all to be able to obtain Newton's kinematic equations using the parametric geodesic equations of relativity.

Hello friends of the Forum. I want to ask you why the inertial acceleration in free fall in the relativistic geodesic equations is assumed equal to zero in free fall and equal to 9.8 m/s at rest on the earth's surface. On the other hand, assuming that zero acceleration in free fall, what would be the metric tensor g(ij) to use in Christoffel Symbols and obtain x¨β(τ)=Γijβx˙i(τ)x˙j(τ)=9.8m/s in the vicinity of the earth's surface in free fall so that when doing the double integration of x¨β(τ) obtain the parametric equation x(t)=v.t+9.8.t22 assuming τ=t and v a constant.

 

[Ibix]

What you are calling the "ordinary acceleration" is usually called "coordinate acceleration". That's just $d^2r/dt^2$, which (for the case you want) you can obtain directly from the geodesic equations in Schwarzschild coordinates.

Actually, $d^2r/d\tau^2$ will do just as well, since you only expect this to match the Newtonian approximation when $r\gg R_S$ and $v\ll c$, in which case $dt/d\tau\approx 1$ anyway.

 

[Dale]

The weak field metric is $$ds^2 = -(1+2\Phi)dt^2+(1-2\Phi)(dx^2+dy^2+dz^2)$$ where $\Phi$ is the Newtonian gravitational field. So if we have near the surface of the earth $\Phi = g z$ then the Christoffel symbols are $$\Gamma^{t}{}_{tz} = \Gamma^{t}{}_{zt}=\frac{g}{1+2 \Phi}$$$$\Gamma^{x}{}_{xz}=\Gamma^{x}{}_{zx}=\Gamma^{y}{}_{yz}=\Gamma^{y}{}_{zy}=\Gamma^{z}{}_{zz}=\frac{g}{-1+2 \Phi}$$$$\Gamma^{z}{}_{tt}=\Gamma^{z}{}_{xx}=\Gamma^{z}{}_{yy}=\frac{g}{1-2\Phi}$$

 

[Thytanium]

Hello friend Dale. Very good answer. Where you locate the origin of the coordinate system txyz to use $\Phi = gz$ so that in the vicinity of the earth's surface $\Phi = g$. Forgive me friend but I don't understand that equation. Thanks my friend.

 

[Thytanium]

Friend Dale if you can explain the equation $\Phi = gz$ to me that is very simple I appreciate it.

 

[Thytanium]

What you are calling the "ordinary acceleration" is usually called "coordinate acceleration". That's just $d^2r/dt^2$, which (for the case you want) you can obtain directly from the geodesic equations in Schwarzschild coordinates.

Actually, $d^2r/d\tau^2$ will do just as well, since you only expect this to match the Newtonian approximation when $r\gg R_S$ and $v\ll c$, in which case $dt/d\tau\approx 1$ anyway.

Hello friend Ibix. I really appreciate your answer. The "t" coordinate is really "ct" right? so it must be $c.dt/d\tau = c$. Thanks friend Ibix.

 

[Ibix]

Hello friend Ibix. I really appreciate your answer. The "t" coordinate is really "ct" right? so it must be $c.dt/d\tau = c$. Thanks friend Ibix.

Usually in relativity we work in units where $c=1$ to save keeping track of all the factors of $c$ that pop up everywhere.

If you want to work in other units you need to insert factors of $c$ for dimensional consistency. Both the $dt^2$ and $d\tau^2$ terms need to be multiplied by $c^2$ and you need to make sure that all the $GM/r$ have a $c^2$ in the denominator.

Friend Dale if you can explain the equation $\Phi = gz$ to me that is very simple I appreciate it.

It's just the Newtonian gravitational potential for a uniform gravitational field. In other contexts you usually write $gh$ so the potential energy is $mgh$, but here @Dale is using the coordinate he's assigned as vertical, $z$, instead of $h$.

 

[Dale]

Where you locate the origin of the coordinate system txyz to use $\Phi = gz$ so that in the vicinity of the earth's surface $\Phi = g$.

Usually you would put the origin at ground level and the $z$ axis pointing vertically upwards so that $z$ is the height above the ground. Then $\Phi$ is the gravitational potential with respect to the ground. Since the gravitational potential energy is $U=mgh$ and since $\Phi=U/m$ and $z=h$ we get $\Phi= gz$

 

[pervect]

TL;DR Summary: The inertial acceleration in free fall is zero but the ordinary acceleration is not zero in free fall due to the relative movement and for this reason there are differences between the two that I want to know and above all to be able to obtain Newton's kinematic equations using the parametric geodesic equations of relativity.

Hello friends of the Forum. I want to ask you why the inertial acceleration in free fall in the relativistic geodesic equations is assumed equal to zero in free fall and equal to 9.8 m/s at rest on the earth's surface. On the other hand, assuming that zero acceleration in free fall, what would be the metric tensor g(ij) to use in Christoffel Symbols and obtain x¨β(τ)=Γijβx˙i(τ)x˙j(τ)=9.8m/s in the vicinity of the earth's surface in free fall so that when doing the double integration of x¨β(τ) obtain the parametric equation x(t)=v.t+9.8.t22 assuming τ=t and v a constant.

An object "at rest on the Earth's surface" is not following a geodesic. The object must be sitting on some surface (usually called the floor) to remain at rest. The floor exerts a force on the object causing it to have a non-zero proper acceleration.

[add]Rephrasing this slightly, which may be clearer, the force the floor exerts on the object causes it to follow a path that is not a geodesic. If the floor did not exert a force on the object, it'd fall through the floor towards the center of the Earthy.

The relative acceleration between an object "at rest" on the floor and an object following a free-fall geodesic is a property of the Earth, and is approximately 9.8 m/s^2, though it varies slightly with location. Another poster has given the details of the calculation of this value from a specific coordinate system, called the "Earth Centered Inertial Coordinate System", according to the mechanics of GR. The name "inertial coordinate system" may unfortunately be a bit misleading, as there is no such thing as an inertial frame in any curved space-time, of which the space-time near and around the Earth is an example. However, for historical reasons, that's what this coordinate system is called. See for instance https://arxiv.org/abs/gr-qc/9508043, "Precis of General Relativity" for details on the ECI coordinate system and it's associated metric. (Note that the above link is to the abstract of the paper, clicking on PDF will give the PDF version, which resolves to https://arxiv.org/pdf/gr-qc/9508043.pdf).

In Newtonian theory, gravity is regarded as a "real force" which acts on an object. GR uses a different model and different terminology - the object "at rest" is only at rest in some specific coordiantes (the ECI coordinates), which, however, are in common use. It's not following a geodesic, that's why the geodesic equation does not evaluate to zero. It's not following a geodesic because the floor is exerting a force on the object.

 

[A.T.]

I want to ask you why the inertial acceleration in free fall in the relativistic geodesic equations is assumed equal to zero in free fall and equal to 9.8 m/s at rest on the earth's surface.

If by "inertial acceleration" you mean "proper acceleration", as measured by an accelerometer, then it is not just "assumed", but observed to be zero in free fall, and 9.8 m/s² upwards at rest on the Earth's surface.

The key point of GR is to accept this locally measured, frame invariant acceleration as the deviation from inertial motion.

 

[Thytanium]

An object "at rest on the Earth's surface" is not following a geodesic. The object must be sitting on some surface (usually called the floor) to remain at rest. The floor exerts a force on the object causing it to have a non-zero proper acceleration.

[add]Rephrasing this slightly, which may be clearer, the force the floor exerts on the object causes it to follow a path that is not a geodesic. If the floor did not exert a force on the object, it'd fall through the floor towards the center of the Earthy.

The relative acceleration between an object "at rest" on the floor and an object following a free-fall geodesic is a property of the Earth, and is approximately 9.8 m/s^2, though it varies slightly with location. Another poster has given the details of the calculation of this value from a specific coordinate system, called the "Earth Centered Inertial Coordinate System", according to the mechanics of GR. The name "inertial coordinate system" may unfortunately be a bit misleading, as there is no such thing as an inertial frame in any curved space-time, of which the space-time near and around the Earth is an example. However, for historical reasons, that's what this coordinate system is called. See for instance https://arxiv.org/abs/gr-qc/9508043, "Precis of General Relativity" for details on the ECI coordinate system and it's associated metric. (Note that the above link is to the abstract of the paper, clicking on PDF will give the PDF version, which resolves to https://arxiv.org/pdf/gr-qc/9508043.pdf).

In Newtonian theory, gravity is regarded as a "real force" which acts on an object. GR uses a different model and different terminology - the object "at rest" is only at rest in some specific coordiantes (the ECI coordinates), which, however, are in common use. It's not following a geodesic, that's why the geodesic equation does not evaluate to zero. It's not following a geodesic because the floor is exerting a force on the object.

Thanks friend Pervect. Very good explanation. Thanks for your answer. I am studying these answers to try to understand them. Good day.

 

[Thytanium]

The weak field metric is $$ds^2 = -(1+2\Phi)dt^2+(1-2\Phi)(dx^2+dy^2+dz^2)$$ where $\Phi$ is the Newtonian gravitational field. So if we have near the surface of the earth $\Phi = g z$ then the Christoffel symbols are $$\Gamma^{t}{}_{tz} = \Gamma^{t}{}_{zt}=\frac{g}{1+2 \Phi}$$$$\Gamma^{x}{}_{xz}=\Gamma^{x}{}_{zx}=\Gamma^{y}{}_{yz}=\Gamma^{y}{}_{zy}=\Gamma^{z}{}_{zz}=\frac{g}{-1+2 \Phi}$$$$\Gamma^{z}{}_{tt}=\Gamma^{z}{}_{xx}=\Gamma^{z}{}_{yy}=\frac{g}{1-2\Phi}$$

Hello PF friends. Hello friend Dale. Studying the metric that you propose friend Dale based on the gravitational potential in the coordinate system "t,x,y,z" and trying to solve them for "z", and for movement only in the coordinate "z" that is to say $\ddot t_\tau = dx =dy = 0$ and considering that for this reason I only have the equation: $\ddot z = \Gamma^z_{ij}\dot x_{i}\dot x_{j}$ and with $dx/dt = dy/dt = 0$ because there is no movement in those coordinates we would get $\ddot z = \Gamma^z_{ij}\dot x_{i}\dot x_{j}$ for i = t, z and j = t,z and making the double summation for equal indices it would be, taking into account that $\Gamma^z_{ij} = 0$ for "i" different from "j" we get finally: $\ddot z= \Gamma^z_{tt}.1.1 + \Gamma^z_{zz}\dot z \dot z$ and using the Gamma values that you have given Dale, would you solve the equation?, am I right or am I wrong?