Calculating Acceleration of Gravity w/ Geodesic Deviation: Troubleshooting

[Jim Hasty]

I have tried twice now to calculate acceleration of gravity using the general relativistic equation of geodesic deviation and both times my solution is twice the correct answer. What am I doing wrong? As an example here is one problem: Calculate the acceleration of gravity g at the earth’s surface using the equation for geodesic deviation and the Schwarzschild metric. Neglect the earth’s rotation.
Given: the mass of the Earth in geometrized units M=4.434 x 10E(-3) meters, the mean radius of the Earth R= 6.371 x 10E(6) meters.

 

[Jim Hasty]

Is this thread active?

 

[berkeman]

Is this thread active?

It's not in Moderation, if that's what you are asking. It looks like it's had over 300 views since yesterday. Hopefully you will get some replies over the weekend.

 

[Ibix]

It'd be easier to read if you used LaTeX to put your maths in a post. Mucking around with the peculiar behaviour of the PDF viewer on this phone isn't worth the hassle.

How to use LaTeX on PF: https://www.physicsforums.com/help/latexhelp/

 

[strangerep]

[...] my solution is twice the correct answer. What am I doing wrong?

Imho, the mistake is in your step 2, where you base the rest of your computation on taking $dr/d\tau=0$. Rather, you need a formula for the radial acceleration which is valid for any radial velocity. (Beware: the relationship between radial velocity and radial acceleration is rather tricky in spherical coordinates.)

Anyway, I suggest you try doing something like what Schutz does on p285, just before eq(11.15). I.e., differentiate $(dr/d\tau)^2 = \dots$ by $\tau$, then cancel a $dr/d\tau$ from both sides. That should leave you with an extra factor of 2 on the LHS, which is what I think you're missing.

HTH.

 

[pervect]

Can the OP latexify the calculation they did? WIthout reading their attempt, I'll suggest the following.

Generally, I wouldn't use the geodesic deviation to calculate the 'acceleration of gravity'. Instead I'd use the following to calculate the 4-acceleration of worldline of a stationary observer.

$$a^b = u^a \nabla_a u^b$$

Depending on the question, I might also the magnitude of the 4-acceleration $\sqrt{|a^a\,a_a|}$ to get the magnitude of the proper acceleration.

Possibly the OP wants to calculate something different than the 4-acceleration and it's magnitude, the proper acceleration, but that's what I'd calculate.

See <<wiki>>

In the general case, the four-acceleration of a particle is defined as the derivative of the 4-velocity $u^\lambda$ with repsect to proper time $\tau$.

$$A^\lambda := \frac{D u^\lambda }{D\tau} = u^{\mu} \nabla_\mu u^\lambda = u^{\mu}\partial _\mu u^\lambda + \Gamma^\lambda{}_{\mu \nu} u^\mu u^\nu = \frac{d u^\lambda }{d\tau} + \Gamma^\lambda{}_{\mu \nu} u^\mu u^\nu . $$

 

[PeterDonis]

I have tried twice now to calculate acceleration of gravity using the general relativistic equation of geodesic deviation

These two aren't the same thing, so I'm not sure how you would do this calculation. Geodesic deviation is tidal gravity, not "acceleration of gravity". For example, if you drop two rocks that are separated radially, the equation of geodesic deviation tells you how the radial separation between the two rocks will change with time. But it doesn't tell you how either of the rocks' altitudes above the Earth changes with time.

 

[strangerep]

For example, if you drop two rocks that are separated radially, the equation of geodesic deviation tells you how the radial separation between the two rocks will change with time. But it doesn't tell you how either of the rocks' altitudes above the Earth changes with time.

I believe in this case that the 2 bodies in question are a single rock and the Earth itself -- the latter being idealized as having all its gravitating mass at the centre -- cf. the Shell theorem, etc.

 

[PeterDonis]

I believe in this case that the 2 bodies in question are a single rock and the Earth itself

If by "the Earth itself", you mean "the geodesic describing the motion of the Earth's center of mass", then we would be dealing with two geodesics, yes; but they're not nearby geodesics, so the geodesic deviation equation doesn't apply. The Riemann tensor's components change significantly from one geodesic to the other, and the geodesic deviation equation does not account for that; it assumes that the Riemann tensor is constant from one geodesic to the other.

 

[PeterDonis]

they're not nearby geodesics

In addition to that, the distance between a rock falling freely near the Earth's surface and the Earth's center of mass cannot be described by the "separation vector" in the geodesic deviation equation, because, heuristically, space on that scale around the Earth is not Euclidean and you can't describe displacements on that scale using a vector space.

 

[strangerep]

To clarify... I'd assumed that the original problem should be regarded in the context of Schutz Sect 7.6, Ex 9 (p183), wherein the student is asked to show that the geodesic deviation eqn implies $$\frac{d^2 \xi^i}{dt^2} ~=~ -\, \Phi,_{ij} \, \xi^j $$to lowest order in $\Phi$ and velocities (where $\Phi$ is the 1st order part of $g_{00}$).

I.e., the context of the original question (iiuc) is for a weak field, slow motion case. Within the limitations of those approximations, one can derive the desired acceleration due to gravity at the Earth's surface.

 

[PeterDonis]

the context of the original question (iiuc) is for a weak field, slow motion case.

I agree, and I wasn't assuming anything different in my previous posts.

Within the limitations of those approximations, one can derive the desired acceleration due to gravity at the Earth's surface.

What is $\xi^i$ in this equation? If it's supposed to be a local displacement vector in a local inertial frame near the Earth's surface, I don't see any problem with that. But if it's supposed to be a displacement vector from the Earth's center to an object on the Earth's surface, I'm not so sure that can be justified. Unfortunately I don't have Schutz's textbook handy to check the section you reference.

 

[PeterDonis]

Within the limitations of those approximations, one can derive the desired acceleration due to gravity at the Earth's surface.

Also, the equation you give in post #11 isn''t for the "acceleration due to gravity", is it? It's for tidal acceleration. If it were for "acceleration due to gravity", we would have the first derivative of $\Phi$ on the RHS, not the second derivative.

 

[strangerep]

Also, the equation you give in post #11 isn''t for the "acceleration due to gravity", is it? It's for tidal acceleration. If it were for "acceleration due to gravity", we would have the first derivative of $\Phi$ on the RHS, not the second derivative.

It might be more productive to wait until you can access the section of Schutz that I mentioned.

But to answer your question briefly,... consider the case when $\xi^i$ corresponds to a purely radial vector. I.e., substitute $r$ for $\xi^i$ in the equation I gave, and assuming $\Phi = \Phi(r)$ ...

[Edit: actually, there seems to be a rogue factor of 2 again. I'll have to go searching for it...]

 

[PeterDonis]

consider the case when ξi\xi^i corresponds to a purely radial vector. I.e., substitute $r$ for $\xi^i$

But $r$ is not a vector, it's a coordinate. It seems like you want to view $r$ as a displacement, but displacements are not vectors, except if we confine ourselves to a local inertial frame. So I don't think the method you seem to be describing, which basically amounts to pretending $r$, the displacement from Earth's center to its surface, is a vector, and then plugging that vector into the geodesic deviation equation, is valid.

 

[PeterDonis]

substitute $r$ for $\xi^i$ in the equation I gave, and assuming $\Phi = \Phi(r)$ ...

Putting aside the concern I raised in my previous post, let's try this. The equation becomes

$$
\frac{d^2 r}{dt^2} = - \Phi_{, rr} r
$$

We have $\Phi(r) = - M / r$, so $\Phi_{, r} = M / r^2$ and $\Phi_{, rr} = - 2M / r^3$, and the equation becomes

$$
\frac{d^2 r}{dt^2} = \frac{2M}{r^3} r = \frac{2M}{r^2}
$$

So not only is there a factor of $2$ that shouldn't be there, the sign is wrong, if we are trying to interpret this as "acceleration due to gravity". But the sign is correct if we are interpreting this as the tidal deviation between radially separated free-falling particles (but on that interpretation we shouldn't be plugging $r$ in as the separation vector to begin with, so the equation would still have $\xi^i$ in it and the RHS would be what we expect for tidal deviation, $\xi^i 2M / r^3$).

 

[PeterDonis]

It might be more productive to wait until you can access the section of Schutz that I mentioned.

Ok, I'm looking at Schutz section 7.6 exercise 9, and I find the following as item (c):

"Interpret this equation when the geodesics are world lines of freely falling particles which begin from rest at nearby points in a Newtonian gravitational field."

So I don't see how Schutz intends this exercise as a calculation of "acceleration due to gravity". I think he intends it as a calculation of tidal acceleration, just as I've been saying. Interpreted as tidal acceleration, the equation is fine and I have no problem with it. (The interesting case as I see it in the exercise is not the radial one but the tangential one, at least for people used to standard Schwarzschild coordinates; the metric in his equation 7.8, which is referenced in the exercise, is in isotropic coordinates.)

 

[strangerep]

[...] So I don't think the method you seem to be describing, [...] is valid.

I retract my earlier statements. Thank you for the correction.

[I suppose, since the equation for tidal acceleration in Newtonian gravity is derived from the ordinary Newtonian equation of motion, one could work backwards from the tidal equation to recover the ordinary EoM. But that seems a bit silly.]

 

[Jim Hasty]

I very much appreciate all of the responses. I need some time to digest them to reply. Thanks.

 

[Jim Hasty]

The problem for calculating gravity's acceleration at the Earth's surface was one that I "made up" after studying black holes. I was wondering if the tidal forces equation would predict, in a straight forward way. Since mass tells spacetime how to curve, and the Riemann tensor in the geodesic equation is derived from the spacetime metric. In my mind's eye I am looking at two geodesics: one thru the center of the earth, and the other passing thru a tangent point on the Earth's surface. I am interpreting the connecting vector as the radial distance. (See sketch I uploaded.) Is the deviation equation giving me the dr/dt between the tangent point on the other side of the earth? Does that resolve the factor of 2 question?

 

[pervect]

The diagram you draw doesn't represent geodesic deviation. The geodesic deviation equation would apply to the relative acceleration between two nearby geodesics. The geodesic equation, if applied correctly, could give you the relative acceleration between an object instantaneously at rest on the surface of the Earth, and another object instantaneously at rest 1 meter above the surface of the Earth. This would be the tidal acceleration between the two objects. The $\xi^r$ vector would be the 1 meter radial separation between the two objects in this example.

Your $\xi^r$ is too large in the diagram - it doesn't show two nearby geodesics, it shows two geodesics that are quite far apart.

I'd suggest looking at my previous post , #6, for how to calculate the acceleration rather than the tidal acceleration.

 

[PeterDonis]

Does that resolve the factor of 2 question?

No, because, as I pointed out in post #16, the sign is wrong. Nearby geodesics that are separated radially diverge. Think of two geodesics at the top of your diagram instead of one, separated radially by a small amount. Both will curve towards the Earth, but the upper one will curve a little less than the lower one, so the two will diverge. That's what the radial component of the Riemann tensor, which is basically what you're calculating, tells you.

The fact that the geodesics at the top and bottom of your diagram converge just reflects the fact that, as @pervect points out, those two geodesics are not "nearby".

 

[Jim Hasty]

In response to Pervect and PeterDonis, I understand your comments now thank you. I agree that the use of "geodesic deviation" was not justified in my problem statement since the equation is derived by analyzing variations between "nearby geodesics", such as at the Earth's surface and 1 meter above the surface. I like Pervect's reply #6. And thanks to Strangerep for referring me back to Schutz 7.6 ex. 9 (p183). This Physics Forums thread has been helpful for me also by shedding light on other issues. Einstein began with Newtonian gravity (the weak field approximations, etc.) and expanded to a general spacetime. I could have calculated the acceleration (g) directly from the Schwarzschild metric, without of involving Riemann components. Something I will not overlook in the future.