I have recently been studying the tensors on the left side of the Einstein field equations, but I have been studying and deriving them in 3-D. I would now like to move on to adding time into the mixture. I have some questions regarding the Minkowski metric [itex]\eta[/itex][itex]\mu[/itex][itex]\nu[/itex]. First, I know that the elements of this metric are: x00= -1. The other elements on the diagonal are 1. Every other element is 0. My question about this is: Where does the negative come from? What makes the temporal component opposite to that of the spatial elements? 2nd: I know that ds2= g[itex]\mu[/itex][itex]\nu[/itex]dx[itex]\mu[/itex]dx[itex]\nu[/itex]. In normal Euclidean space, ds2 is the square of the distance between two points in space. It is essentially like the c2 in Pythagorean's theorem a2 + b2 = c2 However, what would ds2 represent if I throw in the temporal component and do something like: [itex]\eta[/itex][itex]\mu[/itex][itex]\nu[/itex]dx[itex]\mu[/itex]dx[itex]\nu[/itex] I heard that it might be the square of proper time, but I am not sure about this. I know that in Minkowski Cartesian coordinates, x0 = ct (where c is usually set to equal 1). x1= x. x2= y. x3 = z. When I used this information to calculate [itex]\eta[/itex][itex]\mu[/itex][itex]\nu[/itex]dx[itex]\mu[/itex]dx[itex]\nu[/itex] , I ended up getting this as my result: ds2 = -c2t2 + x2 + y2+ z2= -t2+ x2 + y2+ z2 (since c = 1). If it were just spatial components then the meaning of ds2 would be apparent to me, but that temporal component makes it not so as apparent because it looks to me like you are either subtracting a measure of time from the square of a distance in space, or you are just subtracting a measure of space from what would be the normal 3D distance in space (because c is a velocity and speed times time equals distance). Can someone please confirm for me what ds2 is in the case of 4 dimensions? Finally, I need to know how I should go about deriving 4 dimensional metric tensors in coordinate systems other than Cartesian. Take spherical coordinates for instance. I can easily derive the 3D metric tensor for spherical coordinates because I know the transformation properties from Cartesian to spherical. Therefore I know the vector that I would differentiate to get the tangential vectors: R= <rsin(θ)cos(ø), rsin(θ)sin(ø), rcos(θ)> and x1=r x2=θ x3=ø However, if I try to switch to 4 dimensions and transform between Cartesian and spherical in 4D, then how would the temporal component of Cartesian coordinates (x0= ct) transform into a temporal component for spherical components? Would the ct of Cartesian coordinates just remain the same ct in spherical? Does this apply to all types of coordinate systems? Is the temporal component just always ct (as opposed to some spatial components changing to stuff like r or θ or ø)? Also, would this mean that to derive the tangential vectors for a 4D metric tensor that I would have to differentiate the vector with respect to t in addition to all of the other components? That is all. Thank you.