# Why is the right hand side of the Equation of Geodesic Deviation 0?

I'm reading through several primers on GR and I keep seeing the same thing over and over again with no explanation: why is the right hand side of this equation zero: $$\frac {D^2\xi^a}{Du^2}+R^{\alpha}_{\beta \gamma \delta}\dot x^{\beta}{\xi}^{\gamma}\dot x^{\delta}=0$$Is there some assumption of Newton's First Law in this equation?

• Is there a reason you expect the right hand side to not be 0? The geodesic deviation of $\xi^a$ from "nearby geodesics" is given precisely by the Riemann tensor - which encodes all there is to know about the intrinsic curvature of the manifold, what extra term were you expecting? Note that any equation X=Y can be made to have "right hand side equal to 0" by moving the Y over to the left side: X-Y=0 so whether terms appear on the left or the right is pretty arbitrary... – enumaris Sep 28 '18 at 23:37
• @DonaldAirey there is no way to really answer your question without knowing what you expected to be on the right hand side. For example, given Pythagoras' theorem: $a^2+b^2=c^2$, one can write it as: $a^2+b^2-c^2=0$...and when presented with "why is the right hand side zero?", one doesn't know how to answer such a question except with "that's what the Pythagoras' theorem says". Similarly I can not shed more light into your confusion beyond "that's what the geodesic deviation equation is" without some more context from you. – enumaris Sep 28 '18 at 23:44
• @enumaris - Suppose the solar system follows a geodesic path around the center of a spiral galaxy. If r = 15 kpc, at $t_0 = 0$, then at $t_1$ you would expect the radius to be 15 kpc without any additional forces (that is to say, a geodesic created by the galaxies energy/stress). But what if the space between the solar system and the center of the galaxy was expanding during the interval between $t_0$ and $t_1$? – Quarkly Sep 28 '18 at 23:49

Indeed, Newton's first law is implicitly assumed throughout general relativity, which is why you don't get extra "spontaneous acceleration" terms. More precisely:

• We define a geodesic to be a path through spacetime that parallel transports its velocity vector. In flat spacetime, this would correspond to a straight line. In curved spacetime it's a path that's locally straight.
• We postulate that a free particle, with no forces acting on it, follows a geodesic.
• Using this postulate, we derive that free particles obey the geodesic equation, the geodesic deviation equation, and so on.

As you said, the postulate is equivalent to Newton's first law. It's baked right in from the start.

That's not a bad thing; Newton's first law is so fundamental to physics that if we ever saw an apparent violation of it, we would attribute that violation to an unseen new force, rather than throw out the law.

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Sep 29 '18 at 12:09

$$\let\a=\alpha \let\b=\beta \let\g=\gamma \let\d=\delta$$ Would you like $${D^2\xi^a \over Du^2} = {R^\a}_{\!\b\g\d}\,\dot x^\b \dot x^\g\xi^\d \tag{1}$$ better? Actually geodesic deviation is the first member of this equation.

Let's review its meaning. You have a geodesic, whose tangent vector is $$\dot x$$, parametrized by $$u$$. Then a second, infinitesimally near geodesic. The separation vector between these geodesics is $$\xi^\a$$. Eq. (1) says that the second covariant derivative of $$\xi$$ (its acceleration) is given by expression at right, i.e. it can be computed via Riemann tensor.

Eq. (1) is purely geometrical, no dynamics is involved or necessary. Dynamics comes into play only if you state the principle of geodesic (which is not an independent principle in GR) as the generalization of Newton's first principle. Then Eq. (1) informs us that if no other forces are acting, two particles will notwithstanding fall apart or come nearer because of the curvature of spacetime.

This phenomenon is also known in Newtonian physics: two bodies moving in a general external gravitational field (neglecting their mutual gravitational attraction) will change their distance in time because of tide force.

• No, I don't like it any better. It appears to me that all you've done is assumed Newton's First Law, then incorrectly moved part of the left hand side to the right hand side (I would have guessed the equation should be $\frac {D^2\xi^a}{Du^2}=-R^{\alpha}_{\beta \gamma \delta}\dot x^{\beta}{\xi}^{\gamma}\dot x^{\delta}$. You simply baked Newton's First Law into the equation and then rode rough-shod over it because you're assuming that space is static (that is, flat space and curve spaced don't change with time). That's a dangerous and unsupported assumption. – Quarkly Sep 29 '18 at 21:54
• Fixed missing minus sign! :-) – magma Sep 30 '18 at 8:53
• Dear Donald, Elio's answer is perfectly fine and there is no need for your sharp comments. Contrary to what you think, the equation is valid under all conditions (expanding, contracting or somersaulting spacetimes). Any non staticity is encoded in the Riemann tensor, which may be time varying. As a matter of fact this equation is used to measure the passing of a gravitational wave, for example with an idealized GW detector as beautifully described in MTW page 444-445. Oh...sorry if MTW is not in the list of "primers" that you are reading through. – magma Sep 30 '18 at 9:25
• @magma I'm afraid that Donald's primers not even said him that ${R^\alpha}_{\beta\gamma\delta}=-{R^\alpha}_{\beta\delta\gamma}$ – Elio Fabri Sep 30 '18 at 10:15
• @magma regarding physics.stackexchange.com/review/suggested-edits/233386 please do not add blank comments just to bypass the 6 character limit for suggested edits. It is there for a reason. If you can't fix 6 characters, just leave a comment and move on. – user191954 Sep 30 '18 at 11:42