Symmetry - Department of Physics | USU

1 SymmetryMarch 31, 20131 The Lie DerivativeWith or without the covariant derivative, which requires a connection on all of spacetime, there is anothersort of derivation called the Lie derivative, which requires only a :R Mbe a curve inMwith tangent vectors, =dd , with components = x =dx x The Lie derivative generalizes the directional derivative of a function,dfd = f x to higher rank tensors. First, consider a vector field,v, defined onM. We define the Lie derivative ofvata pointPalongCto beL v= lim 0v(P+ ) v(P) wherev(P+ )is the Lie transport ofvalong the curve. For simplicity, letP=C( = 0). Lie trans-port involves taking the value of the vector field at a point onC, say,v( ), and performing a coordinatetransformation to bring the pointC( )back toP=C(0).

2 The coordinate transformation we require is, forinfinitesimal = ,y =x (0)The components ofv change as v (0) =v ( ) y x =[v (x (0) + )]( )=[v (0) + v (0)]( )=v (0) + v (0) v (0) The derivative is thenL v= lim 0v(P+ ) v(P) = lim 0v (0) + v (0) v (0) v(0) = v (0) v (0) An easy proof of the covariance of this result is that it equals the commutator of the two vectors,L v= [ ,v]1which has the same form when andvare expanded in components,[ ,v] =[ ,v ]= v v =( v v ) The generalization to higher rank tensors is immediate because derivations must satisfy the Leibnitz , for an outer product of two vectors,T =u v we haveL T =L (u v )= (L u )v +u (L v )= ( u u )v +u ( v v )= (u v ) u v u v = T T T and so on for higher ranks, with one correction term, T.

3 , for each index. For forms, we use thedirectional derivative of a scalar,L = x together with =v , for arbitraryv , (v ) x =L (v ) ( v ) +v = (L v ) +v L ( v ) +v = ( v ) v ( ) +v L v = v ( ) +v L v L =v +v ( ) Since this must hold for allv ,L = + 2 SymmetryThe Lie derivative is just the right tool for finding Symmetry of a metric. The Lie derivative of the metrictensor isL g = g +g +g Now suppose we have a congruence of curves, so that the collected tangent vectors form a vector further we choose coordinates so that is one of the coordinates,x 0, for 0a single fixed the components of are constant, =dx d = 0 For example, if the curve is timelike, then we chooset= and we have = (1,0,0,0).

4 For such a choice, = 0and the Lie derivative of the metric is justL g =dx d g = g 2 This means that if this Lie derivative vanishes, the metric is independent of the coordinate . Since themetric then does not change along the congruence of curves, we have a Symmetry of the spacetime. Anydirection in which the metric is not changing is called an can find a differential equation to describe such Symmetry directions. Setting the Lie derivative ofthe metric to zero, we have0 = g +g +g = g + [ (g ) g ] + [ (g ) g ]= g + g + g = + g g + g = 12( g + g g ) + 12( g + g g )= + = + =D +D resulting in the Killing equation, ; + ; = 0 Given the metric, we can ask for all solutions to this equation.

5 Solutions, if they exist, represent symmetrydirections of the spacetime, , directions in which the metric is Example: Symmetries of Minkowski spacetimeConsider flat spacetime, for which the metric is Minkowski, . In Cartesian coordinates, = 1111 and the Christoffel connection vanishes, = 0. Then we may replace the covariant derivatives by partialderivatives, and the Killing equation is simply , + , = 0 Taking a further derivative, we have , + , = 0 Now, cycle the indices twice, to give , + , = 0 , + , = 0 , + , = 0 Adding the first two and subtracting the third we find0 = , + , + , + , , , = 2 , so that the second derivative of vanishes. This means that must be linear in the coordinates, =a +b x 3 Substituting this into the Killing equation,0 = , + , =b +b so thata is arbitrary whileb bmust be now have 10 independent vector fields, each of the form =a +b x for independent choices of the 10 constantsa andb = b.

6 The simplest choice of the 10 vector fieldsis to take only one of the constants nonzero. If we takeb = 0and one of the components (say,mform=,0,1,2,3) ofa nonzero, we get four constant vector fields, (m)= mThis represents a unit vector in each of the coordinate directions. Since they are constant, the integral curvesare just the Cartesian coordinate axes, and the metric is indeed independent of each of settinga = 0and choosing one of the six antisymmetric matricesb , we have either rotations orboosts. For example, withb21= b12= 1, with all the rest zero, the vector field is = =( b x ) =x y y xThis is the generator of a rotation around thezaxis. Similarly,b23= b32andb31= b13lead to thegenerators of rotations around thexandyaxes.

7 If one of the nonzero indices is time, then we have a boostbecause of the sign change. Forb10= b01= 1, we find = =( b x ) =x t+t xThis is a generator for a Lorentz transformation. To see this, exponentiate the generator with a parameter, = exp[ (x t+t x)]= n=01n! n(x t+t x)nConsider the effect on the coordinates(t,x,y,z). Clearly, y= z= 0. Fortwe need(x t+t x)t=x(x t+t x)2t=(x t+t x)x=tand so on, alternating betweenxandt. The even and odd parts of the series therefore sum separately, t= n=01n! n(x t+t x)nt4= m=01(2m+ 1)! 2m+1x+ m=01(2m)! 2mt=xsinh +tcosh Similarly, acting onxwe get x=tsinh +xcosh We recognize as the rapidity, and the full transformation, t=xsinh +tcosh x=tsinh +xcosh y=y z=zas a boost in therefore find exactly 10 isometries in Minkowski space.

8 This is the maximum number of independentsolutions to the Killing equation. The static, spherically symmetric Schwarzschild solution had one timelikeKilling field and three spatial rotational Killing fields for a total of three. A generic spacetime has Example: Static, Spherically Symmetric SpacetimesWe may now say what we mean by a static, spherically symmetric spacetime. To be static, there must bea timelike Killing vector field; to be spherically symmetric, we require a full set of three rotational (hencespacelike) Killing use the Lie derivative to say restrict the form of the metric for a static, spherically we want a static spacetime, it means that we want there to exist atimelikeKilling vector field. Choosingthe time coordinate to be the parametert= , the Symmetry condition becomes0 =L g = g + g + g However, withx0=t= , the components of are constant, so that = 0 Therefore,0 = g = t(g )and we have a coordinates system in which the metric is independent of the time the spherical Symmetry , we know that we have three rotational Killing vector fields which togethergenerateSO(3).

9 We can pick two of these for coordinates, but they will not commute with one another, sothe metric will not be independent of both coordinates. Starting with the familiar form 1=y z z y 2=z x x z 3=x y y x5it is natural to choose one coordinate, , such that =x y y xis a Killing vector. To describe a second direction, we want a linear combination of the remaining tworotations, ( ) 1+ ( ) 2and we want this to remain orthogonal to 3,0 = 3, 1+ 2 = x y y x, (y z z y)+ (z x x z) =x y, (y z z y)+ (z x x z) y x, (y z z y)+ (z x x z) =x y, z y y x, z x = zx y, y zy x, x = z( x+ y)= rsin z( cos + sin )To get zero, we can take = sin = cos Then we have 4= 1sin + 2cos = sin (y z z y) cos (z x x z)= sin (rsin sin z rcos y) cos (rcos x rsin cos z)=rsin sin sin z rcos sin y rcos cos x+rsin cos cos z= cos (rcos x+rsin y)+rsin z= cos (x x+y y)

10 Rsin zCompare x=xr r+1 x2+y2xzr2 yx2+y2 = sin cos r+1rcos cos 1rsin sin y=yr r+1 x2+y2yzr2 +xx2+y2 6= sin sin r+1rcos sin +1rcos sin z=zr r x2+y2r2 = cos r sin r so we have 4= cos (x x+y y)+rsin z= cos (rsin cos2 r+ cos cos2 +rsin sin2 r+ cos sin2 ) rsin (cos r sin r )=rsin cos r+ cos2 rsin cos r+ sin2 = cos2 + sin2 = We may therefore take two of the Killing vectors to be 4= 3= giving two coordinates, , , corresponding to Symmetry directions. Since these do not commute, the metriccannot be independent of