Is it really possible to solve for how fast an apple will fall on earth with GR

[zeromodz]

I have recently been studying GR to find out if you really can find out how fast an apple will fall on the surface of the Earth in GR. Its simple with Newtonian mechanics, just use

F = GMM / R^2
A = GM / R^2

I started to with the geodesic equation. Which Christoffel symbols should I solve for? Would it be the 02 component ( time y component) or what? Could someone please show me how to actually do the calculation? Thanks

 

[tom.stoer]

You do not need the Christoffel symbols; the Schwarzschild metric is sufficient. Use a radial geodesic (with vanishing angular contribution):

d\tau^2 = g_{00} dt^2 - g_{rr} dr^2

Here t means the coordinate time (which is identical to the proper time of an observer sitting a infinity = i.e. where g becomes the flat Minkowski metric). Using the Schwarzschild metric you can solve for the proper time of the falling object between two different values for r (which again is just a coordinate and not a proper distance).

Keep in mind that strictly speaking you can only compare proper times. For the experiment with falling apple instead of using the the proper time measure at infinity you could use the proper time of an observer sitting closed to the tree, i.e. having fixed r.

 

[Bill_K]

Without the Christoffel symbols you will not be able to say which radial curves are geodesics. You will need all the symbols that have indices r and/or t.

 

[Mentz114]

This paper does the Schwarzschild radial geodesics -

 

[LAHLH]

I thought I would have a go at this after recently calculating the time a radial observer measures assuming he started at rest and fell all the way into the singularity at centre of BH, as I believe the mathematics is similar (since the Earth's EH if it was a Schwarzschild BH is tiny)

If the Earth radius is R and the height I think the equation you get to is

\tau=-\frac{1}{\sqrt{2M}}{\int^{R}_{R+h}\,\mathrm{d}r\,\frac{1}{\sqrt{\frac{1}{r}-\frac{1}{R+h}}}

(this should be for the time the apple measures on its wristwatch, if it indeed had wrists)

I did this integral on Maple and plugged in R=6.371x10^6m, h=10m, and the mass of the Earth as 0.0044356m, and it led me to a proper time of 4.278019266x10^8m which is 1.426 seconds

This sounds reasonable. The Newtonian value calculated from old s=\frac{1}{2}gt^2 for s=10m, g=10m/s^2 yeilds t=sqrt(2)=1.41421seconds.

------------------------------------------------------------------

The function you get is \frac{1}{\sqrt{2M}}\left[\sqrt{hR}-\frac{R+h}{2}\arcsin{\left(\frac{R-h}{R+h}\right)+\frac{(R+h)\pi}{4}\right]\sqrt{R+h}

This is in geometrized natural units, so the input mass M of the Earth needs to be converted from kg to m by multiplication of G/c^2, the answer is then in metres of time, and needs to be converted back to seconds where 1s=3x10^m so division by c is required.

 

[zeromodz]

I thought I would have a go at this after recently calculating the time a radial observer measures assuming he started at rest and fell all the way into the singularity at centre of BH, as I believe the mathematics is similar (since the Earth's EH if it was a Schwarzschild BH is tiny)

If the Earth radius is R and the height I think the equation you get to is

\tau=-\frac{1}{\sqrt{2M}}{\int^{R}_{R+h}\,\mathrm{d}r\,\frac{1}{\sqrt{\frac{1}{r}-\frac{1}{R+h}}}

(this should be for the time the apple measures on its wristwatch, if it indeed had wrists)

I did this integral on Maple and plugged in R=6.371x10^6m, h=10m, and the mass of the Earth as 0.0044356m, and it led me to a proper time of 4.278019266x10^8m which is 1.426 seconds

This sounds reasonable. The Newtonian value calculated from old s=\frac{1}{2}gt^2 for s=10m, g=10m/s^2 yeilds t=sqrt(2)=1.41421seconds.

This is exactly what I wanted to see. Thank you. The only problem is how on Earth did you come to that equation?

 

[LAHLH]

This is exactly what I wanted to see. Thank you. The only problem is how on Earth did you come to that equation?

You can get this equation from looking at the timelike radial geodesics in Schw, see https://www.physicsforums.com/showthread.php?t=473753 as a starting place. From this you have the derivative of r wrt proper time.

How do you convert this to the time of the guy sitting next to the tree as tom.stoer mentioned, would they be a "shell observer" as Wheeler calls them?

 

[tom.stoer]

The observer sitting near the tree measure his tau equals t correced with the effect of the graviational potential

 

[LAHLH]

So what would people say the time measured by the guy sat on the ground by the tree be? for a height of 10m?

I think one must use \dot{t}=\frac{E}{1-\frac{2M}{r}} to first get how much coordinate time lapses as the apple falls? One can also find that E=\sqrt{1-\frac{2M}{R+h}} conserved along trajectory.

Then convert this coordinate time to that of the proper time of the shell observer sat at r=R? or is there a way to obtain this shell observers proper time from that of the falling apple directly?

 

[PAllen]

I wonder if this would be valid? Take the world line for the free falling observer and the world line for an r=<const> observer. Call the 4 velocity of the former u2, and latter u1. A t=const r unit vector is 4 orthogonal to u1. Compute the expression for u2 dot <r unit vector> / (u2 dot u1). This gives instantaneous speed measured by static observer. Then take the derivative of this expression by proper time along r=const world line. That would seem to be acceleration as measured by the r = const observer.

 

[PAllen]

Another thought: In the paper cited by Mentz114, eq. 29 gives acceleration of static observer relative to inertial frame. Would the acceleration of a free fall object relative to a static observer be the same? This formula certainly 'looks right', reducing to Newtonian acceleration for realistic cases.

 

[Neandethal00]

I thought I would have a go at this after recently calculating the time a radial observer measures assuming he started at rest and fell all the way into the singularity at centre of BH, as I believe the mathematics is similar (since the Earth's EH if it was a Schwarzschild BH is tiny)

If the Earth radius is R and the height I think the equation you get to is

\tau=-\frac{1}{\sqrt{2M}}{\int^{R}_{R+h}\,\mathrm{d}r\,\frac{1}{\sqrt{\frac{1}{r}-\frac{1}{R+h}}}

(this should be for the time the apple measures on its wristwatch, if it indeed had wrists)

I did this integral on Maple and plugged in R=6.371x10^6m, h=10m, and the mass of the Earth as 0.0044356m, and it led me to a proper time of 4.278019266x10^8m which is 1.426 seconds

This sounds reasonable. The Newtonian value calculated from old s=\frac{1}{2}gt^2 for s=10m, g=10m/s^2 yeilds t=sqrt(2)=1.41421seconds.

------------------------------------------------------------------

The function you get is \frac{1}{\sqrt{2M}}\left[\sqrt{hR}-\frac{R+h}{2}\arcsin{\left(\frac{R-h}{R+h}\right)+\frac{(R+h)\pi}{4}\right]\sqrt{R+h}

This is in geometrized natural units, so the input mass M of the Earth needs to be converted from kg to m by multiplication of G/c^2, the answer is then in metres of time, and needs to be converted back to seconds where 1s=3x10^m so division by c is required.

I'm not good with Tex, so I'm copying your equations. According to your equations, following equality must hold.
\tau=-\frac{1}{\sqrt{2M}}{\int^{R}_{R+h}\,\mathrm{d}r\,\frac{1}{\sqrt{\frac{1}{r}-\frac{1}{R+h}}} = \sqrt\frac{2sc^2}{g}

Does the equation when solved for 'g' produce 9.81 m/sec² and can be used for 'g' of any planet? Just wondering.
Note: note good at Tex, the right side should be c*sqrt(2s/g)

 

[LAHLH]

I'm not good with Tex, so I'm copying your equations. According to your equations, following equality must hold.
\tau=-\frac{1}{\sqrt{2M}}{\int^{R}_{R+h}\,\mathrm{d}r\,\frac{1}{\sqrt{\frac{1}{r}-\frac{1}{R+h}}} = \sqrt\frac{2sc^2}{g}

Does the equation when solved for 'g' produce 9.81 m/sec² and can be used for 'g' of any planet? Just wondering.
Note: note good at Tex, the right side should be c*sqrt(2s/g)

Yes you can solve for g and it will come out somewhere around that, but not exactly. The equating of the LHS with the RHS is just a comparison between Newton and Einstein. The other problem is this is the proper time of the apple, what I would really like to calculate is the time an observer sat by the tree measures given that the apple's watch says such and such.

\frac{\tau}{c} \approx \sqrt{\frac{2s}{g}}

Thus,

g \approx \frac{2sc^2}{\tau^2}

Take the values I found for s=10m, c^2=9x10^16m^2/s^2, tau=4.278019266x10^8m, you find g=9.82168337 m / s^2

 

[LAHLH]

If an observer is at rest on the surface of the Earth then he is not on a geodesic, he is accelerating, he is what Wheeler calls a shell observer I believe. So if he is just sat there then dr=0, d\theta=0, d\phi=0 and thus the time in the shell frame is the proper time dt^2_{s}=d\tau^2=\left(1-\frac{2M}{r}\right)dt^2

This allows one to relate shell time to Schwarzschild time as dt_{s}=\sqrt{1-\frac{2M}{r}}dt, and note that this only makes sense for r&gt;2M, and that as r goes to infinity the times agree. Also this shows that the shell time runs slow compared to the Schwarzschild time (Grav time dilation).

-------------------------------------------------

So all of that helps me relate the proper time of a shell clock to that of a Schwarzschild time and so forth, but now the shell observer is sat there measuring an apple falling and we know how much proper time elapses for the apple, what we want to know is how much proper time the shell observer measures during the apples fall on his watch.

One can also find that in the shell frame the distance between two simultaneous radially separated events is dr_{s}= \frac{dr}{\sqrt{1-\frac{2M}{r}}}

In these shell coordinates the metric can be re-written as ds^2=-dt^2_{s}+dr^2_{s}+angles

Looks a lot like Minkowski metric (but I think it different because the angles contain r^2 not r^2_{s} ?)


Is this the way to go?