[MUSIC] So we have postulated on general grounds the following action for the relativistic particle. T1, T2, dtau, square root of gmunu [z(tau)], z mu of dot znu dot. Postulated in arbitrary curvilinear space-time with the metric g mu nu. This action is invariant under the general coordinate transformations, obviously because what is standing here is just a line element along the world line of the particle. A line element, sorry ds squared / dtau squared with this standing under the square root. And it is invariant under the reparameterizations tau to f(tau). Now we want to find equations of motion following from the least or minimal action principle for this action. So let us vary this action. It means that we have world line of a particle, zmu(tau) and we change it a bit. So at every moment tau, we shift zmu(tau) to zmu(tau) + delta zmu(tau). So we vary. And we want to see how this action changes with respect to this variation namely. So delta S is just S[z + delta z]- S(z) taken to the linear order, linear in delta z. So let us do this variation. Delta S = -m integral of tau1, tau2, dtau. Let me use the notation, that g mu nu z mu dot. z mu dot, I just, for simplicity sake, to simplify the formulas, I just denote as z dot squared. So during this variation, we just vary, differentiate this guy, which is just 1 2 over square root of z dot squared, and the variation of g mu nu o f(z) z mu dot, z nu dot, variation of this guy. Now we want to make one more thing, important thing. We want to multiply by square root of z dot squared and divide by square root of z dot squared. We combine this together with this into ds. In fact dtau square root of z dot squared = ds. Due to the fact that this stands here. So now we vary this thing and then put the square roots inside, inside the brackets. What we get? We get the following thing. Integral of ds /2, -m of course is here. And here stands delta g mu nu z dot mu z dot mu + g mu nu delta z dot mu z dot nu + g mu nu z mu dot delta z nu dot. Where dot here is now differential with respect to s, rather that tau. Because in this differential, every dtau is combined with square root of z dot squared from here. Now minimal action principle demands us to put this to be equal to zero for any delta z. But we notice here that there is a delta z dot and here delta z dot, which is just, it means that we have here delta z dot is nothing. Let me write delta z dot is nothing, just d over ds of delta z. So to get rid of d over ds of delta z here and here, we have to integrate by parts. By integrating by parts, we gain here boundary terms at s1 and s2 at initial moment of time and final moment of time. But the variation delta z here and here equal are to zero. So these terms are just vanishing, we do not write them explicitly. So let me write now. One more thing, this is just obviously equal to the following thing. So let me continue. This is equal to this. - m, s1, s2, ds / 2. So I write this term first. This term is d alpha g mu nu delta z alpha. Which is just a variation of the metric to the linear order in delta z, is just this guy. Because we just Taylor expand g mu nu of z. * z mu dot z mu dot- after integrating by parts this and this term we obtain the following thing. d over ds of g mu nu of z, which is a function of s, in its own right. z mu dot, delta, z mu,- d/ds [g mu nu, z mu dot] delta z nu. Sorry. This is curly brackets corresponding to these curly brackets. So we have arrived at the following expression for the first variation of the action. It looks as follows. s1 to s2 integral over ds/2{ d alpha g mu nu delta z alpha z mu dot z mu dot- d/ds (g mu nu(z), z nu dot), delta z mu- d/ds[ g mu nu (z) z mu dot] delta z nu. And I would like to stress one thing, that in obtaining this expression, I have used the fact I dropped off the following terms. g mu nu of z delta z mu z nu dot at s1 s2 + g mu nu of z, z mu dot delta z nu at s1 s2. I drop them off because they're 0. They are 0, because this variations are 0 at the initial and final points of the trajectory. Not trajectory, world line. We've arrived at the world line at all intermediate points, keeping fixed the initial and final points. And one more thing, here we have used that delta g mu nu of z is nothing but g mu nu of z + delta z- [COUGH] g mu nu of z. To linear order in delta z which is approximately valid to be d alpha g mu nu, Taylor expanding this in delta z. And subtracting this, we obtain d alpha g mu nu(z) times delta z, alpha. That's how we arrived at this expression. Now we differentiate, take this differential. Open their brackets, this bracket and this bracket, and take the differential of g mu / ds as a complicated function. In the sense that g mu nu depends on s while the fact that z depends on s. So g mu nu depends on z, and in its own right z depends on s. So taking this differential, we arrive from this expression, we arrive at the following expression for the variation. -m s1 s2 ds / 2 d alpha g mu nu delta z alpha, z mu dot z nu dot. This is just the first term, I rewrite it. Then I get d alpha g mu nu z alpha dot z dot nu, delta z mu- d alpha g mu nu z alpha dot z mu dot delta z nu- twice g mu nu z mu dot dot delta z nu. To obtain the last term, the last term appears both from here and here, two expressions where I renamed the indices mu and nu. And use the fact that the metric is symmetric under the exchange of its two indices. That's how I obtained the last term. Both from here and from here. Now the last thing, I have to rename indices here and subtract, Extract delta z. So this is equivalent to this. ds[1/2 (d alpha g mu nu- d mu g alpha nu- d nu g mu alpha) z mu dot z nu dot- g mu alpha z mu dot dot ] delta z alpha. So that what we obtained. Now this variation of the action has to be 0 for any delta z alpha. It means that this should be 0. As a result we obtain the following equation. z mu dot dot + Gamma mu nu alpha z mu dot z alpha dot = 0. Where Gamma is so-called Christoffel symbol. Christoffel symbols and they have the following form as follows from this expression. Gamma mu nu alpha = 1/2 g mu beta ( d mu g alpha beta + d alpha g beta nu- d beta g mu alpha). Here this guy is just inverse metric tensor. It means that g mu beta times g beta nu is just delta mu nu. This is a Kronecker symbol. It has unit matrix. So this guy is just inverse metric tensor with respect to g beta alpha which is original metric tensor. So this guy stands here. That's the end of this lecture. [MUSIC]