Talk:Geodesics in general relativity

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When explaining geodesics, this looks very obscure. How is a non-relativist to understand the null geodesic for example from this - especially without any diagrams.

Hi User:AugPi, good job on the article. I have several requests, though:

• Can you please not use boxed equations? This is a kind of house-style used in the math articles; see Wikipedia:How to write a Wikipedia article on Mathematics. The discussion about boxes around equations comes up every now and then, and the boxes tend to loose. linas 00:53, 2 August 2005 (UTC)[reply]

I have removed them. Boxes which run right up to the right margin don't look too good, so I had cut down on the length of the boxes to center the equations inside them. I did this in Safari, then when I logged on with Netscape the boxes became shorter than the equations, so boxes can be problematic.

Thanks! linas 00:29, 4 August 2005 (UTC)[reply]

• Can you please consider moving much/most of the long derivation off to a subpage called Proofs for the article Geodesic (general relativity), as per the discussion in Wikipedia:WikiProject Mathematics/Proofs ? See, for example, Cardioid proofs and Laplace_operator/Proofs. The goal here was to provide the "rigor" that is desired, without actually cluttering up the article. linas 00:53, 2 August 2005 (UTC)[reply]

I moved the calculations to Geodesic (general relativity)/Proofs. Incidentally, I didn't know that I could create article titles with slashes: that could come in handy. —AugPi 05:23, 3 August 2005 (UTC).[reply]

They're called "sub-pages", and actually, according to WP policy, one is not supposed to use sub-pages for articles. (its OK to use them on personal pages, talk pages, etc.) Appearently, subpages tended to get misused, which is why there's a policy against them. I've asked for special dispensation, that an exception be made for the case of these types of proof pages. Not clear what the final ruling will be, if any. (the alternative is to have a strict naming convention "XXX article proofs" or something like that.) linas 00:29, 4 August 2005 (UTC)[reply]

There is a sentence: A geodesic between two events could also be described as the curve joining those two events which has the maximum possible length. From other sources I've heard, that the geodesic is actually a minimum possible length.

(The maximum-length curve goes arround the whole universe before getting to the target...)

Could you, please, verify that?

In GR, there is an "invariant interval" which for timelike paths is the proper time experienced as one travels along that path. This proper time acts as a length in relativity and is maximized for inertial motion (which is also geodesic motion in relativity). So it is correct that geodesics in realtivity maximize the length between events, while purely spatial geodesics minimize the length between positions. (For an example of how acceleration decreases the elapsed peoper time between events, see the twin paradox article.) --EMS | Talk 22:50, 24 October 2006 (UTC)[reply]

As noted in the leader, geodesics can be spacelike or timelike or null. Timelike geodesics are particularly interesting because they are the paths that free massive particles follow, but they are not the only geodesics. Also, I think it's good policy to try to incorporate the answers to legitimate questions (like this one) into the text of the article itself. I'm trying to do this, and brush up a few other points. --MOBle 03:44, 4 December 2006 (UTC)[reply]

The section on geodesics seems to have been written by someone with scant knowledge of general relativity or differential geometry. Let me clarify a few points.

1) In a Riemannian manifold (one that has a positive-definite metric) the norm of a non-zero vector is always strictly positive. But in a Lorentzian manifold you can have non-zero vectors that have zero or negative norm. The zero-norm case is that of lightlike vectors. In particular, every lightlike worldline has exactly zero lenght. Thus the "length" of a lightlike worldline has nothing at all to do with whether it's a geodesic or not.

2) In the case of a timelike worldline, the "length" is defined by integrating the square root of the _absolute value_ of the norm (because otherwise one would get imaginary numbers). Any perturbations of a timelike geodesic will be _shorter_ not longer. In particular, you can perturb the timelike worldline of an object that's at rest in Minkowski space (such as one whose worldline is the time axis) to a worldline that moves in one direction at nearly the speed of light and then comes back to the same point again at nearly the speed of light. This wordline will have length that's nearly zero, thus smaller than the lenght of the original worldline. This, by the way, is how the famous paradox of the twins arises.

3) In a general Riemannian manifold, geodesics are local, *not* necessarily global, length minimizers. Similarly, in a general (non-Minkowski) Lorentzian manifold, spacelike geodesics do not necessarily minimize length globally, and timelike geodesics do not necessarily maximize it. For example in a 2-D cylindrical spacetime (with infinite time and finite space) you can have two timelike geodesics between point A and point B that have different lengths. — Preceding unsigned comment added by Csperelli (talk • contribs) 15:43, 6 August 2011 (UTC)[reply]

To the first point it is related a paper W. Belayev “Application of Lagrange Mechanics for Analysis of the Light-Like Particle Motion in Pseudo-Riemann Space”, International Journal of Theoretical and Mathematical Physics , 2012; 2(2): 10-15 . Here the variational method without violation of conformity of the light motion to the null path is proposed. This method is shown to agree with Fermat's principle for the stationary gravity field, which in one’s turn gives for the static metrics the same solution that principle of geodesics yields. — Preceding unsigned comment added by WBBelayev (talk • contribs) 19:17, 14 June 2012 (UTC)[reply]

Timelike curves that are locally maximal are geodesics, but I believe that the claim above that spacelike curves have local extremal length ("shortest") is incorrect. Penrose (in The Road to Reality) describes geodesics as "stationary", not "extremal". Variations to a spacelike geodesic in a timelike direction will shorten it, even locally, and variations in a spacelike direction will lengthen it. Hence it is not locally extremal, but it is stationary in the sense that it is a saddle point in any smoothly parameterized variation. One can also vary the geodesic in a way in which the length remains exactly constant. —Quondum 11:33, 11 March 2014 (UTC)[reply]

Hi, I have added a section showing the derivation of the geodesic equation via a action principle since I believe it is important. — Preceding unsigned comment added by 207.161.187.38 (talk) 02:55, 15 August 2014 (UTC)[reply]

Thanks; you used Euler-Lagrange, you could also use Hamilton; its worth saying that too. 67.198.37.16 (talk) 16:02, 21 February 2016 (UTC)[reply]

This derivation is incorrect. It is not true that

${\displaystyle \delta {\sqrt {A}}={\frac {\delta A}{2{\sqrt {A}}}}}$

because ${\displaystyle \delta }$ is not a differentiation. A simple counterexample is

${\displaystyle \delta {\sqrt {{\dot {x}}^{2}}}=\delta {\dot {x}}=0\neq {\frac {\delta {\dot {x}}^{2}}{2{\sqrt {{\dot {x}}^{2}}}}}=-{\frac {\ddot {x}}{\dot {x}}}}$

You happen to get the correct answer by accident. It is also easy to see that your derivation is incorrect here. You make no assumption about the physical meaning of ${\displaystyle \lambda }$, so it can be any parameter. But the usual form of the geodesic equation is only valid for when ${\displaystyle \lambda }$ is the proper time along the geodesic, and it is not valid for other choices of ${\displaystyle \lambda }$. Since the action you've written is valid for any choice of ${\displaystyle \lambda }$ and the equation of motion isn't, the equation of motion can't follow from the action unless the special parametrization is fixed somewhere in the derivation. The correct derivation is much more involved and involves "fixing" reparametrization symmetry by requiring ${\displaystyle \lambda }$ to be the proper time along the geodesic, not any parameter. --- Anon — Preceding unsigned comment added by 193.176.211.83 (talk) 17:17, 28 April 2020 (UTC)[reply]

"${\displaystyle \delta }$" is a kind of differentiation. Your counter-example is non-sense. I restored the section. There is nothing wrong with it. JRSpriggs (talk) 00:34, 29 April 2020 (UTC)[reply]

${\displaystyle \delta ={\frac {\partial }{\partial x}}-{\frac {d}{dt}}{\frac {\partial }{\partial {\dot {x}}}}}$ is not a differentiation, because it doesn't satisfy the rules for differentiation. Do not be so quick to disregard what you don't understand as "non-sense". If you think my counterexample is really non-sense, please be so kind to explain what is wrong with it. The derivation is clearly incorrect for reasons I described earlier. — Preceding unsigned comment added by 193.176.211.83 (talk) 03:21, 29 April 2020 (UTC)[reply]

You are misinterpreting the definition in Functional derivative; see Functional derivative#Properties.

Replace "f" in "J (f)" by "f + ε g" and differentiate "J (f + ε g)" with respect to the real variable "ε"; then evaluate when "ε" is zero. Where "g" represents the variation of "f".

OK? JRSpriggs (talk) 01:31, 30 April 2020 (UTC)[reply]

The problem with your counter-example is that your formula for "δ" is only true on-shell, but you are using it off-shell.

In other words, your formula for "δ" (the Euler-Lagrange equation) is derived by assuming certain conditions (variation of action is zero for any variation of path; and end-points of path are fixed) while those conditions do not hold in your application of the formula. JRSpriggs (talk) 00:48, 2 May 2020 (UTC)[reply]

I reverted this edit and a previous one as introducing an error. First, let's quote the source (The Road to Reality):

p 319: When g is not positive definite, the argument is basically the same, but now the geodesics do not minimize ∫ds, the integral being what is called ‘stationary’ for a geodesic.

p. 474: More correctly, this may not be actually a ‘minimum’, but the term ‘stationary’ would be appropriate. The situation is basically similar to that which happens in ordinary calculus (see §6.2), where the occurrence of a minimum of a smooth real-valued function f(x) requires df/dx = 0, but where sometimes df/dx = 0 occurs when the function f is not a minimum: it might be a maximum or possibly a point of inflexion or, in higher dimensions, what is called a saddle point (Fig. 20.4b). All places where df(x)/dx = 0 are called stationary. See Figs. 6.4 and 20.4. Recall the basically similar characterization of a geodesic in (pseudo)Riemannian space, given in §14.8, §17.9, and §18.3 as a ‘minimum-length path’ in the positive-definite case (locally), and sometimes as a ‘maximum length timelike path’ in the Lorentzian case, although merely of ‘stationary length’ in the general case.

I've already made the point above under § Really maximum length? above, but I'd like to stress again that a spacelike geodesic is not a local minimum, and that this is true in special relativity (not only general relativity). It is possible to locally vary a spacelike geodesic so that its proper length either increases or decreases. The statement "the space-like curve with the shortest proper length between two events that are space-like separated is a space-like geodesic" is completely false for a Lorentz metric. A simple illustration: in flat Minkowski space with two spacelike-separated events (simultaneous events in our frame of reference), vary the straight line by moving the centre point in the direction of our time axis, keeping each of the two path segments connecting this point to the end events straight. With this variation the proper path length decreases. Hence it was not a minimum. —Quondum 06:11, 13 September 2014 (UTC)[reply]

Yes, I see your point, my mistake. From a geodesic in Minkowski space that's purely spatial in some lab frame, a variation of the geodesic that's purely spatial in the lab frame will increase the proper length, but a variation of the geodesic that's purely temporal in the lab frame will decrease it. Red Act (talk) 16:47, 13 September 2014 (UTC)[reply]

It seems that with one spatial dimension there is symmetry between space and time (with c = 1): they can be interchanged, replacing proper time by proper length, and replacing timelike by spacelike. So it seems proper length is locally always a maximum on a spacelike geodesic.

Also, the distinction between "curve that differs from the geodesic purely spatially" and "curve that differs from the geodesic purely temporally" is not clear. It is like shifting an oblique line, doing that horizontally or vertically is the same. - Patrick (talk) 12:30, 10 March 2015 (UTC)[reply]

The two-dimensional case (one space and one time dimension) is, as you say, symmetric, and both spacelike and timelike geodesics are extremal. (edit: inserting accidentally omitted words, underlined) In any case of two or one dimensions, though, spacelike geodesics are not extremal, and these are the primary cases of interest, mainly the 3+1 case. In considering the matter of whether a geodesic is extremal, "oblique" lines need not be considered. Consider an orthogonal set of axes in Minkowski space, with a geodesic along each of the axes. Consider the effect of perturbing each of these geodesics as described by adding a small triangular function to one of the coordinates not in the direction of the geodesic, and compute the length. A change that does not change the time coordinates along the curve is "purely spatial". A change that does not change any space coordinates along the curve is "purely temporal". I'm not too sure what is not clear here. —Quondum 15:01, 10 March 2015 (UTC)[reply]

I understand now, thanks!. - Patrick (talk) 18:38, 10 March 2015 (UTC)[reply]

—Quondum 19:29, 10 March 2015 (UTC)[reply]

The article says

${\displaystyle {{\rm {{\ddot {x}}^{i}=-\sum _{j=1}^{4}\sum _{k=1}^{4}\ \Gamma _{jk}^{i}\ {{\dot {x}}^{j}}\ {{\dot {x}}^{k}}+q\ {F^{ik}}\ {{\dot {x}}^{j}}}}\ {g_{\rm {jk}}}}}$

but shouldn't it be

${\displaystyle {{\rm {{\ddot {x}}^{i}=-\sum _{j=1}^{4}\sum _{k=1}^{4}\ \Gamma _{jk}^{i}\ {{\dot {x}}^{j}}\ {{\dot {x}}^{k}}-q\ {F_{ik}}\ {{\dot {x}}^{j}}}}\ {g^{\rm {jk}}}}}$ ?

The version as it is given in the article right now gives some nasty Csc[2θ] terms in the equation for φ" in the Kerr-Newman-metric, which lead to infinite acceleration at the equator, while the shifted indicies (version below) reduces to the Coulomb-law at far distances. So it seems to me we need the covariant electromagnetic tensor multiplied with the contravariant metric tensor, not the other way round. If no objections come, I'll change that in the next few days.--Yukterez (talk) 03:47, 24 September 2017 (UTC)[reply]

What happened to Wikipedia's requirement of wp:verifiability and wp:reliable sources? If you can find a reliable source for your version (or for the original version) then feel free to make the change, referring to the source (or leave it as it was, and add the source). Thanks! - DVdm (talk) 19:41, 24 September 2017 (UTC)[reply]

Note also that the version that you seem to prefer violates basic conservation of indices.TR 21:56, 24 September 2017 (UTC)[reply]

Since I don't have a source for the version which gives the Coulomb force at large distances, nor one for the present version which seems to give the wrong output I won't change it, but as I said, the present version gives an infinite acceleration at the equator: csc(2θ)-term, while the shifted indicies give a meaningful output: φ"-solution. Since the present version follows from a cartesian Minkowski-solution, where the indicies don't matter, maybe the contravariant Maxwell tensor times the covariant metric tensor only works in flat Minkowski, while the general solution needs to be the other way around? Maybe I made some error here, but since there is no source for any of the wo possibilieties, maybe some expert can shine a light on that. I don't even see how one should invert F_μν in the case of the Reissner-Nordström metric where the inversion to F^μν would give a singular matrix... Maybe I missed he point that explains this, but whatever the point is, it doesn't seem to be in the article so other readers might miss it too. Maybe a "citation needed"-tag would fit the present version, since it isn't really clear who derived that solution. --Yukterez (talk) 02:20, 25 September 2017 (UTC)[reply]

Speaking of the Kerr–Newman metric#Equations of motion, there seems to be an unnecessary factor of ${\displaystyle \csc ^{2}\theta }$ in the formula for the derivative of time. It cancels out with factors of the sine which are scattered throughout the numerator. I do not know if that is related to this issue or not. JRSpriggs (talk) 02:56, 25 September 2017 (UTC)[reply]

~~

I added a source for the formula. Moreover, you do realize that indices are not raised and lowered through matrix inversion, do you? If not, it is probably time to pick up a textbook on GR. TR 07:27, 25 September 2017 (UTC)[reply]

No need for a textbook, the rule ${\displaystyle F^{\mu \nu }=g^{\mu \sigma }\ g^{\nu \kappa }\ F_{\sigma \kappa }}$ can also be found online. With that everything becomes clear now. Nevertheless, thanks for the hint; although you comment seems rather snootily, I appreciate it.  Done --Yukterez (talk) 09:02, 25 September 2017 (UTC)[reply]

Please see section Geodesics in general relativity#Deriving the geodesic equation via an action. Although the derivation there gives the correct result, it appears to me that it fails to be mathematically correct. Specifically, one is taking square roots of things which might be negative and dividing by things which might be zero.

The variation of a time-like path might include portions which are space-like or light-like even when shrunk down to an infinitesimal amount. Just imagine adding a multiple of the cube root of t minus t_midpoint to x in a neighborhood of the midpoint of the path. Similarly, a space-like path might gain time-like or light-like parts when varied. A light-like path is even worse since, even without variation, one cannot divide by the time derivative of the proper-time or proper-distance since they are zero.

It appears to me that there is no way to get around these problems without avoiding using the square-root altogether. But how can one talk about the interval (distance or duration) without using the integral of the square-root? JRSpriggs (talk) 20:26, 28 June 2018 (UTC)[reply]

OK, by restricting variations to C² functions of time, we should be able to handle the time-like geodesics and the space-like geodesics since I think that the sets of C² time-like paths and C² space-like paths are open sets. But, the similar derivation for light-like geodesics still appears to be invalid. JRSpriggs (talk) 04:26, 2 July 2018 (UTC)[reply]

Here is meant "becomes one (= 1). A rather strange use of the word "vanish", in my opinion. — Preceding unsigned comment added by Koitus~nlwiki (talk • contribs) 22:04, 22 September 2019 (UTC)[reply]

Regarding the equation in Geodesics in general relativity#Equivalent mathematical expression using coordinate time as parameter. If

${\displaystyle t=x^{0}}$,

then

${\displaystyle {\frac {dx^{0}}{dt}}=1}$

and

${\displaystyle {d^{2}x^{0} \over dt^{2}}=0}$.

In other words, the acceleration in the time-direction does indeed vanish. JRSpriggs (talk) 08:05, 23 September 2019 (UTC)[reply]

At the end of the section Geodesics as curves of stationary interval the text says "If the parameter s is chosen to be affine...", but doesn't define "affine" in this context, nor does it provide a link where you can find out. -- Dr Greg  talk  16:47, 4 July 2021 (UTC)[reply]

The full sentence is "If the parameter s is chosen to be affine, then the right side of the above equation vanishes (because ${\displaystyle U_{\nu }U^{\nu }}$ is constant).". I think that the parenthetical expression defines "affine". In other words, the time parameter is affine if speed (magnitude of velocity) is constant. JRSpriggs (talk) 06:07, 5 July 2021 (UTC)[reply]

Thanks. If what you say is correct, the sentence as it stands doesn't really make that clear, as it suggests the constant length is a consequence of having an affine parameter (a one-way implication) rather than a definition (a two-way implication). -- Dr Greg  talk  10:48, 5 July 2021 (UTC)[reply]