Christoffel Symbols PHYS 471: Introduction to Relativity and Cosmology 1 The Metric and Coordinate Basis Let's recap some properties of the metric, before deriving the Christoffel symbols. First, we know the metric is a tensor gµν, which can be written as a 4 × 4 diago- nal matrix. The line element ds2 is the length-squared of an infinitesimal vector in spacetime: 2 µ ν ds = gµνdx dx (1) The components of the metric will depend on the coordinate system, however. For flat spacetime in Cartesian coordinates, this is the familiar Minkowski metric 2 2 2 2 2 ds = dt − dx − dy − dz ; gµν = ηµν = diag(+1; −1; −1; −1) But if we were to change coordinates to, say, spherical ones, the line element would look different: ds2 = dt2 − dr2 − r2dθ2 − r2 sin2 θdφ2 These metric coefficients can be \encoded" in what's called a \basis vector"e ^µ (more technically a vierbein!). Formally, we define the metric as the product gµν =e ^µ · e^ν (2) Great, but what is this basis vector thing?? You've actually seen them before. When you first start talking about vectors, we define the unit vectorsx; ^ y;^ z^ that point in those orthogonal directions. A general vector is then written ~s = Axx^ + Ayy^ + Azz^ When we change coordinates to spherical ones, we want to re-express these coefficients as some other set ~s = Arr^ + Aθθ^ + Aφφ^ So, consider a vector that has infinitesimal length for all its components: Ax = dx; Ay = dy; Az = dz. Then, ds~ = dx x^ + dy y^ + dz z^ If we switch to spherical coordinates, the appropriate transformation (using the Ja- cobian) gives ds~ = dr r^ + rdθ θ^ + r sin θ^dφφ But what we really want to do is express any such vector in a way that \looks like" the Cartesian version: ~ ds = dx^1 + dx^2 + dx^3 In order to do so, we define the coordinate basis vector component e^i to be the thing that allows us to write ~ ds = dx^1 + dx^2 + dx^3 = dx1e^1 + dx2e^2 + dx3e^3 Note that the \basis" quality has been taken away from dxi, which is now only a number. So, for the Cartesian system, dx^1 = dx e^x, dx^2 = dy e^y, dx^3 = dz e^z, which tells us that the components of the coordinate basis vector aree ^x =e ^y =e ^z = 1. Using this formalism, the line element is 2 ~ ~ 2 2 2 2 2 2 ds = ds · ds = (^e1) dx1 + (^e2) dx2 + (^e3) dx3 2 Comparing the above to Equations ?? and ??, we see that g11 =e ^1 · e^1 = (^e1) , and so forth. If we did the same thing in spherical coordinates, we'd find ~ ^ ^ ds = dr^ + dθ + dφ = dr e^r + dθ e^θ + dφ e^φ =) e^r = 1 ; e^θ = r ; e^φ = r sin θ 2 2 2 2 2 2 =) grr = (^er) = 1 ; gθθ = (^eθ) = r ; gφφ = (^eφ) = r sin θ Note that we're used to seeing these as negative, which they still are in spacetime! The above discussion was only for spatial components. 2 Properties of the Christoffel Symbols The coordinate basis vectors have an obvious coordinate dependence in some systems, so we must account for this when we compute things like derivatives. The Christoffel symbols encapsulate this dependence when you take the derivative of a coordinate basis vector. We define this as λ 0 1 2 3 @µe^ν = Γ µνe^λ = Γ µνe^0 + Γ µνe^1 + Γ µνe^2 + Γ µνe^3 That is, the derivative of the basis vector is a linear combination of the four com- ponents of the basis vector! This accounts for any motion the basis vector might undergo as it moves around a non-trivial geometry. Since we know gµν =e ^µe^ν, it's straightforward to show that the derivative of the metric is λ λ @αgµν = (@αe^µ)^eν +e ^µ(@αe^ν) = Γ αµe^λe^ν + Γ ανe^λe^µ λ λ =) @αgµν = Γ αµgλν + Γ ανgλµ (3) From this, we can derive an expression for the Chirstoffel symbols in terms of first derivatives of the metric. When we cycle the order of the indices α; µ, ν, we get two more equations: λ λ =) @µgνα = Γ µνgλα + Γ µαgλν (4) λ λ =) @νgαµ = Γ ναgλµ + Γ νµgλα (5) As luck would have it (or for geometric reasons, if you prefer), the Christoffel symbols are symmetric in their lower indices, which reveals that many of these terms are λ λ the same. For example, Γ µνgλα in Equation ?? is the same as Γ νµgλα in Equation ??, λ λ because we can write Γ µν = Γ νµ. Re-writing the above equations to make this obvious, we get λ λ (3) becomes −! @νgαµ = Γ µνgλα + Γ ανgλµ (6) λ λ (1) becomes −! @αgµν = Γ αµgλν + Γ ανgλµ (7) λ λ (2) becomes −! @µgνα = Γ αµgλν + Γ µνgλα (8) Note each term in Equation ?? (formerly Eq 1) appears in Equation ?? and ??, re- spectively. So, we combine these three equations and solve for the Christoffel symbols: Equation 4 + Equation 6 − Equation 4 = @νgαµ + @µgνα − @αgµν λ λ = Γ µνgλα + Γ µνgλα λ = 2Γ µνgλα (9) Success! All we need to do is isolate the Γ, and we do so by multiplying both sides by the inverse metric gλα: λ 2Γ µνgλα = @νgαµ + @µgνα − @αgµν 1 Γα = gαλ (@ g + @ g − @ g ) (10) µν 2 ν µλ µ λν λ µν OK, so now that we have a way to calculate these coefficients, what good are they? As mentioned before, these tell us something about how basis vectors change their direction as a vector moves around some space(time) with curvature. They also tell us something about the nature of lines in the curved spacetime, specifically straight ones (which are kinda sorta important to General Relativity). 3 Uses of the Christoffel Symbols There are varied applications of the Christoffel symbols in General Relativity, and more broadly in differential geometry. We'll encounter a few key ones in our discus- sions, however, so let's review three fundamental ones right now. 3.1 The Covariant Derivative Christoffel symbols are defined in terms of the derivative of the coordinate basis vector components. Their first and most important application is in the computation µ of derivatives in non-flat spacetimes. If we take the derivative of a vector A = A e^µ, we get ν ν ν @µA = @µ(A e^ν) = (@µA )^eν + A (@µe^ν) The second term in the derivative is the Christoffel symbol, so ν ν ν α @µ(A e^ν) = @µA e^ν + A Γ µνe^α Since the ν index in the first term is paired, we can arbitrarily replace it with α to find α ν α α ν α @µA e^α + A Γ µνe^α = @µA + A Γ µν e^α The term in parenthesis shows how the derivative of the vector A also depends on the coordinate basis terms. We call this the covariant derivative of Aν: ν ν ν α =) DµA = @µA + Γ µαA Note the lower in the Christoffel symbol combine the derivative index with the vector index (hence, \mixing" the coordinates due to derivatives). In General Relativity, we also deal with objects with lower indices, though, so some extra care must be taken. It turns out that the derivative of a covariant vector Aµ looks almost the same, except the sign in front of the Christoffel symbol is negative: α =) DµAν = @µAν − Γ µνAα So the rule is: upstairs indices get a positive Christoffel symbol, downstairs indices get a negative one. You can remember this easily, since up is positive, down is negative. Based on these rules, we can take the covariant derivative of a mixed tensor of any type, as long as we have a Christoffel symbol for each index. For example, the µν µ covariant derivative of R , Rµν, and R ν are µν µν α βν α µβ Positive for upper indices =) DαR = @αR + Γ µβR + Γ βνR β β Negative for lower indices =) DαRµν = @αRµν − Γ αµRβν − Γ ανRµβ µ µ µ β β µ + and − for mixed indices =) DαR ν = @αR ν + Γ αβR ν − Γ ανR β 3.2 The Christoffel Symbols and the Metric Taking a derivative is \moving" something in a space, or spacetime. If I move in the @ x^ direction, I think \ @x "! This is great for flat spaces, but as the above discussion suggests, moving in curved spaces (or spacetimes) means we have to think covariant derivative. So { what happens if we take the covariant derivative of the metric itself? (Whoa!) The answer is: NOTHING! Since the metric defines the intrinsic geometry, it shouldn't change as we move from place to place. A curved spacetime has the same curvature properties everywhere. So, that means Dαgµν = 0 =) Metric is the same everywhere! Writing this out in terms of Christoffel symbols, we get β β @αgµν − Γ αµgβν − Γ ανgµβ = 0 because we need a negative sign when we take the covariant derivative of things with covariant indices. We can rewrite this as β β @αgµν = Γ αµgβν + Γ ανgµβ A If we cycle through the indices as we did before (i.e. cyclically swap α; µ, and ν) we get β β @νgαµ = Γ ναgβµ + Γ νµgαβ B β β @µgνα = Γ µνgβα + Γ µαgνβ C I've called these A; B; C. As before, we note that because of the symmetry of the metric under exchanges of indices, terms with gαµ are the same as those with gµα, and so forth.