Equation of Motion and GeodesicsSo far we’ve talked about how to represent curved spacetime using a metric, and whatquantities are conserved. Now let’s see how test particles move in such a spacetime. Todo this, we need to take a brief diversion to see how derivatives are changed in curvedspacetime.First, let’s simplify and go back to thinking about a flat two-dimensional plane.Consider a scalar quantity. Ask class: pick an example of a scalar quantity. Could betemperature, Newtonian gravitational potential, whatever. Define that quantity to be Φ.Let’s think in terms of a Cartesian coordinate basis. The derivative along the x directionis ∂Φ/∂x, and along the y direction is ∂Φ/∂y. We will adopt the notation that a commadenotes a partial derivative, so these become Φ,xand Φ,y, respectively. If we want thedirectional derivative along some vector v with components vxand vy, that derivative is∂vΦ = Φ,xvx+ Φ,yvy= Φ,αvα. Similarly, if we have a vector field E, its gradient is therank 2 tensor ∇E = Eα,βand its directional derivative along v is the rank-1 tensor Eα,βvβ.As always, the geometric objects formed (e.g., gradient or directional derivative) havecoordinate-free reality, but we express them here in terms of components for concreteness.A particle moving in free space (no forces on it) simply continues moving along in a straightline. That means that the directional derivative of the velocity, along the velocity, is zero(this comes from, e.g., conservation of momentum and energy). Therefore, uα,βuβ= 0 is theequation of motion for a free particle. All these results are also true for free motion in flatspacetime.But what happens when the spacetime is curved? Then the coordinates themselvestwist and turn. One must take this into account by adding another term to the equation fora derivative (you can think of this as a chain rule, if you like; the derivative of a quantitycontains one term for the partial derivative of that quantity, and another for the partialderivative of the coordinates themselves). A result that we’ll simply have to state is thatthe correction terms involve the connection coefficients. For a coordinate basis, these areΓαβγ=12gαµ(gµβ,γ+ gµγ,β− gβγ,µ) . (1)The actual derivative itself is represented by a semicolon. The connection coefficients comein with a + sign when the index to be “corrected” is up, and a − sign when the index isdown. Some examples:Φ;α= Φ,αvα;β= vα,β+ Γαµβvµvα;β= vα,β− ΓµαβvµTβα;γ= Tβα,γ+ ΓβµγTµα− ΓµαγTβµ.(2)Note that for scalar fields there is no correction.What about “straight lines” in curved spacetime? These are called geodesics, and aresimply extensions of what we found in flat spacetime. That is, the directional derivativeof the velocity along itself is zero, or ∇uu = 0, so uα;βuβ= 0. If one defines some affineparameter λ along the geodesic (the proper time τ is a good choice for particles withnonzero rest mass), then this comes out tod2xαdλ2+ Γαµγdxµdλdxγdλ= 0 . (3)From the definition of the four-velocity, uα= dxα/dλ, we can also write thisduαdλ+ Γαµγuµuγ= 0 . (4)We’ll explore some consequences of this in a second, but first a quick check thatthis is reasonable. This is the equation of motion. So, Ask class: what kind of pathshould it give in flat space? It should give a straight line, meaning that the accelerationduα/dλ = 0. Is it so? Let’s look at the connection coefficients again. Ask class: whatis the Minkowski metric? ηαβ= (−1, 1, 1, 1). So, Ask class: what are the derivatives ofthe components of ηαβ? The derivatives are all zero. Ask class: what does that imply forthe connection coefficients? They’re related to derivatives of the metric coefficients, so theconnection coefficients are zero. Therefore, Ask class: what is the equation of motion inflat spacetime? It becomes just duα/dλ = 0, a straight line, as required.For a Schwarzschild spacetime, let’s consider motion in the rφ plane (i.e., no θ motion,which one can always arrange in a spherically symmetric spacetime just by redefinition ofcoords). Then the radial equation of geodesic motion is (here we use the proper time τ asour affine parameter)d2rdτ2+Mr2− (1 − 3M/r)u2φ/r3= 0 . (5)In deriving this we’ve used the fact that the specific angular momentumuφ= gφαuα= gφφuφ=1r2(dφ/dτ). Let’s think about what all this means. First,let’s check this in the Newtonian limit. In that limit, M/r ¿ 1 and can be ignored, and atlow velocities dτ2≈ dt2so we get the usual expressiond2rdt2= −Mr2+ u2φ/r3. (6)In particular, that means that for a circular orbit, d2r/dt2= 0, the specific angularmomentum is given by u2φ= Mr.What about in strong gravity? First consider radial motion, uφ= 0. Thend2r/dτ2= −M/r2. This has the same form as the Newtonian expression, but rememberthat the coordinates mean different things, so you have to be careful. Now consider circularorbits. Again we set d2r/dτ2= 0, to find u2φ= Mr2/(r − 3M). But wait! Something’sstrange here. That r − 3M in the denominator means that the specific angular momentumgoes to infinity at r = 3M, but the horizon is at r = 2M. Have we reached a contradictionof some sort?No, but we have happened upon one of the most important features and predictions ofgeneral relativity. As you will demonstrate in the problem set, all of this implies that thespecific angular momentum has a minimum at a radius well outside the horizon. This isalso the minimum radius of a circular orbit that is stable to small perturbations, hence iscalled the innermost stable circular orbit (ISCO).Ask class: what does this mean for gas spiraling close to a black hole or neutron star?It means that even if the gas was moving in almost circular orbits at larger distances, then(neglecting other forces) when it reaches this critical radius it’ll go right in without havingto lose more angular momentum. This radius is called plays a fundamental role in thephysics of accretion disks around very compact objects.Qualitatively, one can think of it like this. A fundamental feature of the Schwarzschildgeometry is the so-called “pit in the potential”. That is, near a compact object gravityis “stronger” than you would have expected based on an extrapolation of the Newtonianlaw. To compensate for this, the angular velocity has to be higher than it would have beenotherwise, so the angular