[MUSIC] So we were solving this equation, and spherically symmetric and static solution of this equation we have found to be like this. C2 over r dt squared- dr squared divided by 1 + c2 over r- r squared d omega squared. Here, c2 is some constant, which is independent of t and r, and the angles. And we want to find the physical meaning and the value of c2 right now for the given situation. So we observe that this metric as r goes to infinity, this term becomes smaller and smaller, and we arrive at flat space time. So to let us use this, let us consider non-relativistic particle very far from the gravitational center. It means that we want to consider that its velocity, so the trajectory of the particle or world line of the particle is t and z(t). This is a world line of this particle, z mu(t). So when it is very far from the gravitational center, we know that its action from classical non-relativistic mechanics. We know that its action is proportional to -m integral 1- z dot squared over 2 of unit mass + V(z modulus) dt. So let me stress again. So we can see the, no, not unit, a particle of the mass m, interacting with the gravitational center very far from it, and for very slow velocity. So we can see that z dot is much less than 1. So this is an approximation of the relativistic action, approximation because like this term contains the trivial contribution. And this guy is nothing but the Newton potential. It means that v, a function of z, which is the same as v as a function of r, is Newton potential minus kappa m over r. Where m is the mass of the gravitating center. So we know that the action that the relativistic action should be equal to this. And this is just its approximation when we are very far and for slow velocities. Then, from this and this, we have approximate relation. The ds is approximately equal to 1- z dot square lower 2 + v of modulus z, dt. So this approximation is valid as r goes to infinity and z dot modulus is much less than 1. Now to linear order at z dot squared and v, to linear order in v, we can square this to obtain that ds squared approximately again = [1 + 2V(r)]dt squared- z dot squared, dt squared. But this is equal to [1 + 2V(r)]dt squared- dz squared. So, we have the following relation for the metric when we are very far. Then we have to have the gtt component of the metric, so this component of this metric has to have the following form 1 + 2 V(r), which is 1- 2 kappa M divided by r. Now to have this, to have this to be true, we have to have that C2 = -2 kappa M. It has also the other name, rg and it is called as a Gravitational Radius for the given mass, M. Thus, their metric under consideration has the following form. ds squared = (1- rg divided by r)dt squared, dr squared divided by 1- rg divided by r- r squared d omega squared. So this is the metric which solves Einstein equations. And describes spherically symmetric gravitational field created by a massive body of the mass m around itself. [MUSIC]