Throwing Balls on a Large Planet: GR Approximation Problem

In summary, the conversation discusses a problem involving two balls being dropped simultaneously from a train station and the comparison between Newtonian mechanics and Relativity in predicting when the balls will hit the ground. The approach proposed involves using Schwarzschild coordinates and taking the limit of a large launch coordinate to approximate the motion of the balls. However, it is pointed out that this method may not be accurate and a different approach, similar to the Rindler wedge in Minkowski spacetime, may be needed. The possibility of using the orbit formula for Mercury is also mentioned.
  • #1
Ibix
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In a recent thread, @sweet springs asked:
sweet springs said:
In IFR of the train station let two balls start falling at the same time and the same place with initial vertical speed 0 and with horizontal speed each 0 and say 0.99c. Newtonian mechanics says they touch the station ground simultaneously. I would like to know how Relativity says.
I pointed out that the claim about Newton is only strictly true in a uniform gravitational field, and proposed an approximation in GR. The idea was to take a Schwarzschild spacetime, restrict the "acceleration due to gravity" to some specified value, and take the limit as the launch ##r## coordinate grows large. I've had a go at this now, and have come out with a silly answer. I can't see what I'm doing wrong - any comments gratefully received.

Initial conditions
The idea is that we are throwing the ball horizontally. We can use Schwarzschild coordinates, and set ##\theta=\pi/2## and work in the equatorial plane without loss of generality. That means that the initial velocity of the ball is in the ##t##-##\phi## plane. By insisting that the initial four-velocity of the ball have inner product with the four-velocity of a hovering observer of ##\gamma## and requiring normalisation, I can write $$\begin{eqnarray*}v^t&=&{{1}\over{\sqrt{1-{{{R_s}}\over{{r_0}}}}\sqrt{1-v^2}}}\\
v^\phi&=&{{v}\over{{r_0}\sqrt{1-v^2}}}\end{eqnarray*}$$Then I can plug those values into the geodesic equations (see Carroll's GR notes, equations 7.43 and 7.44) to get the constants of motion, ##E## and ##L##:
$$\begin{eqnarray*}E&=&{{\sqrt{1-{{{R_s}}\over{{r_0}}}}}\over{\sqrt{1-v^2}}}\\
L&=&{{{r_0}v}\over{\sqrt{1-v^2}}}\end{eqnarray*}$$

Equations of motion
We are interested in comparing the radial motions of balls launched at different velocities on different trajectories. So we don't want to work in terms of their proper times - rather we want to look at the coordinate velocities in some coordinate system that treats hovering as not moving. Schwarzschild fits the bill - so we want to find ##dr/dt=(dr/d\tau)\div(dt/d\tau)##. Carroll's 7.43, 7.47 and 7.48 are what we're looking for, and we get$$\left(\frac{dr}{dt}\right)^2=\frac{R_sr^2(r_0-r)(r-R_s)^2-v^2r_0(r_0^2-r^2)(r-R_s)^3}{r^5(r_0-R_s)}$$

Other constraints
We want the "acceleration due to gravity" (the proper acceleration of a hovering observer) to be a specified value, ##a##. Since this is given by $$a=\frac{R_s}{2r_0^2\sqrt{1-R_s/r_0}}$$we can solve for ##R_s## and pick the positive solution:$$R_s=2ar_0^2\left(\sqrt{a^2r_0^2+1}-ar_0\right)$$Since we're also going to take the limit as ##r_0## gets very large while considering a relatively small drop, it's helpful to write ##r=r_0-\delta##, where ##\delta## is the coordinate distance fallen.

Putting it all together
Finally, we can substitute in the expression for ##R_s## and replace ##r##. This leads to:$$\left(\frac{dr}{dt}\right)^2=\frac{P_1+P_2\sqrt{a^2r_0^2+1}}{P_3+P_4\sqrt{a^2r_0^2+1}}$$where the ##P_n## are polynomials in ##r_0##. Since they are rather lengthy, I won't write them out here - there is Maxima code below if you want to see.

Approximation
The final step is to take the limit of large ##r_0##. We can write ##\sqrt{a^2r_0^2+1}\simeq ar_0+1/2##, which reduces top and bottom to polynomials in ##r_0##, for which we can take the leading term. Doing that yields$$\left(\frac{dr}{dt}\right)^2=-16a^4r_0^3\delta\left(2v^2+1\right)$$

My problem
The problem is that all of those terms are positive - so I have a negative ##(dr/dt)^2##, which is obviously problematic. I must be doing something illegitimate, but I can't see what. I did all the algebra in Maxima, so I don't think I've made a slip there. And I think I've managed to use +--- consistently throughout. And I don't see anything wrong with the basic idea - but the answer is absurd.

Any ideas? Foot at the ready to kick myself if it's something obvious...

Maxima code that I used is hidden in the spoiler tag below.
Code:
/* sweet springs asked if a ball dropped next to a ball thrown */
/* horizontally hit the ground simultaneously in GR as they do */
/* in Newton.                                                  */

/* Initial conditions: U is 4-velocity of a hovering observer and */
/* V is the 4-velocity of the ball. c=1, work in equatorial plane */
/* so theta=pi/2 and dTheta=0, r=r0, and use g_ab U^a V^b = gamma */
/* and g_ab V^a V^b=1                                             */

assume(v<1,v^2<1);
gamma:1/sqrt(1-v^2);
gtt:1-Rs/r0;
gpp:-r0^2;
Ut:1/sqrt(gtt);
Vt:gamma/(gtt*Ut);
Vp:rhs(solve(1=gtt*Vt^2+gpp*Vp^2,Vp)[2]);

/* Plug into Carroll's 7.43 and 7.44 to get conserved quantities, L and E */
E:Vt*(1-Rs/r0);
L:substitute(-gamma*v,sqrt(gamma^2-1),r0^2*Vp); /* (gamma v)^2 = 1 - gamma^2 */

/* Plug back into 7.43 to get dt/dtau at arbitrary r, and into  */
/* 7.47 and 7.48 to get (dr/dtau)^2 at arbitrary r. epsilon=1   */
/* and lambda=tau for a massive particle. Use Phi for effective */
/* potential, so as not to confuse with four-velocity V.        */
dtdtau:E/(1-Rs/r);
Phi:(1-Rs/r)*(1+L^2/r^2)/2;
drdtau2:ratsimp(E^2-2*Phi);

/* We want to compare dr/dt for various gamma. Calculate */
/* (dr/dtau)^2 / (dt/dtau)^2                             */
drdt2:ratsimp(drdtau2/dtdtau^2);

/* We want to choose our "acceleration due to gravity" rather  */
/* than Rs so write down the proper acceleration of a hovering */
/* observer, solve for Rs, and substitute.                     */
a=(Rs/(2*r0^2*sqrt(gtt)));
Rs:rhs(solve(lhs(%)^2=rhs(%)^2,Rs)[2]);

/* Put it all together */
substitute(r0-delta,r,drdt2);
substitute(Rs,'Rs,%);
drdt2final:ratsimp(%);

/* Approximate and find leading order terms */
drdt2approx:substitute(a*r0+1/2,sqrt(a^2*r0^2+1),drdt2final);

drdt2num:collectterms(expand(num(drdt2approx)),r0);
drdt2numHi:hipow(drdt2num,r0);
drdt2num:r0^drdt2numHi*coeff(drdt2num,r0,drdt2numHi);

drdt2denom:collectterms(expand(denom(drdt2approx)),r0);
drdt2denomHi:hipow(drdt2denom,r0);
drdt2denom:r0^drdt2denomHi*coeff(drdt2denom,r0,drdt2denomHi);

drdt2LeadingOnly:ratsimp(drdt2num/drdt2denom);
[/spolier]
 
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  • #2
Ibix said:
The idea was to take a Schwarzschild spacetime, restrict the "acceleration due to gravity" to some specified value, and take the limit as the launch ##r## coordinate grows large.

I don't think this is right. What you want is to find a patch of Schwarzschild spacetime that has the following properties: there is a "hovering" observer at radial coordinate ##r## whose proper acceleration is ##a## (since that will determine the "acceleration due to gravity" of an object dropped radially by the observer), and curves of constant proper acceleration are "horizontal" to a good enough approximation over a large enough horizontal distance to allow the ball moving sideways to fall by a sufficient amount. Or, more briefly, you want a patch of Schwarzschild spacetime that looks, to a good enough approximation, like a patch of the "Rindler wedge" of Minkowski spacetime, as seen by a Rindler observer with proper acceleration ##a##.

The way to get that is not to make ##r## large. It's to make ##r## very close to ##2M##, the horizon radius, of a very, very large black hole (the smaller you want ##a## to be, the larger the hole has to be). This issue arose in the course of discussing the solution to a puzzle I posed a couple of months ago; the posts towards the end of that thread might be helpful:

https://www.physicsforums.com/threads/general-relativity-rocket-puzzle.965384/
The element that is added in the scenario under discussion here is sideways (tangential) motion (the motion in the puzzle considered in the thread I just linked to is purely radial). But the general idea of what approximation you want to use should be the same.
 
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Likes mfb and Ibix
  • #3
You may be able to make use of the familiar orbit formula of Mercury for perihelion movement discussion in investing the case.
 
  • #4
Thanks, Peter. I've been distracted by other stuff for a few days, but will take a look at that example.

Your interpretation of my imaginary ##dr/dt## seems that there's no region of Schwarzschild spacetime that has the properties I'm requiring. I'm going to have to think on why that should be so.
 
  • #5
You put your system inside a black hole.

In Newtonian gravity a constant acceleration at increasing radius means M~r2. If you want to keep your object outside a black hole in GR then you need 2M<r. While GR modifies the first relation a bit the conclusion stays the same: There is a maximal r where this can work.
 
  • #6
mfb said:
You put your system inside a black hole.

This won't work because there is no such thing as an observer "hovering" at constant ##r## inside a black hole.

mfb said:
a constant acceleration at increasing radius

As I pointed out in post #2, I don't think this is actually the right limit to be taking.
 
  • #7
Ibix said:
Your interpretation of my imaginary ##dr/dt## seems that there's no region of Schwarzschild spacetime that has the properties I'm requiring.

I wasn't trying to interpret your imaginary ##dr / dt## (negative ##( dr / dt )^2##). I was saying that I think you derived that result using the wrong limit, the limit of large ##r##. I don't think you will get an imaginary ##dr / dt## if you work in the limit I suggested (##r## close to ##2M## for a black hole with very large ##M##).
 
  • #8
PeterDonis said:
This won't work because there is no such thing as an observer "hovering" at constant ##r## inside a black hole.

As I pointed out in post #2, I don't think this is actually the right limit to be taking.
Yes, these were the points of my post.
 
  • #9
mfb said:
Yes, these were the points of my post.
Actually, I think we're in "Ibix can't do basic maths correctly" territory.

If I were falling into the black hole, that would require ##R_s>r##, and rearranging my exact expression for ##R_s## shows that this is always satisfied. The approximate expression I gave, however, does not satisfy the inequality for large ##r_0##. That struck me as odd because it should be valid for large ##r_0##, which led me to investigate - and I see that my Taylor expansion of the square root is wrong. ##(a^2r_0^2+1)^{1/2}\simeq ar_0(1+1/2a^2r_0^2)##, not ##ar_0(1+1/2ar_0)##.

I will see if that fixes it.
 

1. What is the GR approximation problem?

The GR approximation problem refers to the challenge of accurately predicting the trajectory of a thrown ball on a large planet using Einstein's theory of general relativity (GR). This problem arises because GR predicts that the path of a moving object is affected by the curvature of spacetime, which is influenced by the mass and energy of the planet. Therefore, the traditional equations used to calculate the trajectory of a thrown ball on a flat surface are not sufficient.

2. How does the curvature of spacetime affect the trajectory of a thrown ball?

The curvature of spacetime, as predicted by GR, causes the path of a moving object to bend towards the planet's center of mass. This means that a thrown ball will not travel in a straight line, but rather in a curved path. The amount of curvature is dependent on the mass and energy of the planet, as well as the velocity and trajectory of the ball.

3. Can the GR approximation problem be solved exactly?

No, the GR approximation problem cannot be solved exactly. This is because the equations used to describe the curvature of spacetime are highly complex and non-linear, making it difficult to calculate the exact trajectory of a thrown ball. However, scientists have developed approximations and numerical methods to estimate the trajectory with a high degree of accuracy.

4. How does the size of the planet affect the trajectory of a thrown ball?

The size of the planet does not directly affect the trajectory of a thrown ball, but rather its mass and energy do. However, on a larger planet, the curvature of spacetime will be more significant, resulting in a more pronounced bending of the ball's path. This means that the trajectory of a thrown ball on a large planet will be more affected by GR than on a smaller planet.

5. Are there any real-world applications for the GR approximation problem?

Yes, the GR approximation problem has practical applications in fields such as space exploration and satellite navigation. By accurately predicting the trajectory of a spacecraft or satellite, scientists and engineers can plan and execute missions more effectively. Additionally, understanding the effects of GR on the trajectory of objects can also help improve GPS systems and other technologies that rely on precise positioning.

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