Introduction
We’ve all heard the modern physics mantra and it’s true: symmetries lead to conservation laws. So one thing you might be wondering is how can we find symmetries in relativity when the theory is so geometric? Geometrically speak- ing, a symmetry is hiding somewhere when we find that the metric is the same from point to point. Move from over here to over there, and the metric remains the same. That’s a symmetry.
It turns out that there is a systematic way to tease out symmetries by finding a special type of vector called a Killing vector. A Killing vector X satisfies
Killing’s equation, which is given in terms of covariant derivatives as
∇bXa + ∇aXb =0 (8.1)
Note that this equation also holds for contravariant components; i.e., ∇bXa+
∇aXb =0. Killing vectors are related to symmetries in the following way: if X is a vector field and a set of points is displaced by Xadxa and all distance
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relationships remain the same, thenX is a Killing vector. This kind of distance preserving mapping is called anisometry. In a nutshell, if you move along the direction of a Killing vector, then the metric does not change. This is important because as we’ll see in later chapters, this will lead us to conserved quantities. A free particle moving in a direction where the metric does not change will not feel any forces. This leads to momentum conservation. Specifically, if X is a Killing vector, then
X ·u=const
X · p=const
along a geodesic, whereuis the particle four velocity and pis the particle four momentum.
Killing’s equation can be expressed in terms of the Lie derivative of the metric tensor, as we show in this example.
EXAMPLE 8-1
Show that if the Lie derivative of the metric tensor vanishes, then
LXgab= 0
This implies Killing’s equation for X, given in (8.1).
SOLUTION 8-1
The Lie derivative of the metric is
LXgab = Xc∂cgab+gcb∂aXc+gac∂bXc
Let’s recall the form of the covariant derivative. It’s given by
∇cXa =∂cXa+abcXb (8.2)
Now, the covariant derivative of the metric tensor vanishes, ∇cgab =0. The
covariant derivative of the metric tensor is given by∇cgab=∂cgab−dacgdb− d
bcgad. Since this vanishes, we can write
Let’s use this to rewrite the Lie derivative of the metric tensor. We have LXgab= Xc∂cgab+gcb∂aXc+gac∂bXc = Xcd acgdb+dbcgad +gcb∂aXc+gac∂bXc =gdbdacXc+gaddbcXc+gcb∂aXc+gac∂bXc
Some manipulation can get this in the form we need to write down covariant derivatives of X. Note that
∇bXc =∂bXc+cbdXd
In our expression we just derived for the Lie derivative the last term we have isgac∂bXc. Can we find the other term needed to write down a covariant
derivative? Yes, we can. Remember, a repeated index is a dummy index, and we are free to call it whatever we want. Looking at the last line we got for the Lie derivative, consider the second term
gaddbcXc
The indices candd are repeated, so they are dummy indices. Let’s switch themc↔dand rewrite this term as
gaccbdXd
First, let’s write down the result for the Lie derivative and rearrange the terms so that they are in the order we want for a covariant derivative.
LXgab= gdbdacXc+gaddbcXc+gcb∂aXc+gac∂bXc
= gac∂bXc +gdbdacXc+gaddbcXc+gcb∂aXc
= gac∂bXc +gaddbcXc +gcb∂aXc+gdbdacXc
At this point, we put in the change of indices we used on what is now the second term in this expression.
LXgab=gac∂bXc+gaddbcXc+gcb∂aXc+gdbdacXc =gac∂bXc+gaccbdXd+gcb∂aXc +gdbdacXc =gac ∂bXc+cbdXd +gcb∂aXc+gdbdacXc =gac∇bXc+gcb∂aXc+gdbdacXc
Now we switch indices on the last term and get LXgab= gac∇bXc+gcb∂aXc+gdbdacXc =gac∇bXc+gcb∂aXc+gcbcadXd =gac∇bXc+gcb ∂aXc+cadXd =gac∇bXc+gcb∇aXc
Since the covariant derivative of the metric vanishes, we move the metric tensor inside the derivative and lower indices. For the first term we find
gac∇bXc = ∇b(gacXc)= ∇bXa
and for the second term we obtain
gcb∇aXc = ∇a(gcbXc)= ∇aXb
Therefore, we have
LXgab= gac∇bXc+gcb∇aXc = ∇bXa+ ∇aXb
Since we are given thatLXgab =0, this implies (8.1).
Often, we need to find the Killing vectors for a specific metric. We consider an explicit example by finding the Killing vectors for the 2-sphere.
EXAMPLE 8-2
Use Killing’s equation to find the Killing vectors for the 2-sphere: ds2 =a2dθ2+a2sin2θdφ2
SOLUTION 8-2
Killing’s equation involves covariant derivatives. Therefore we need to recall the affine connection for this metric. In an earlier chapter we found
θ θθ =φθθ =θφθ =φφφ =0 φ φθ =φθφ =cotθ θ φφ = −sinθ cosθ
Now, we recall that the covariant derivative is given by ∇bVa =∂bVa−cabVc
Starting witha=b=θ in Killing’s equation, we find ∇θXθ + ∇θXθ =0
⇒ ∇θXθ =0
Using the equation for the covariant derivative, and recalling the Einstein sum- mation convention, we obtain
∇θXθ = ∂θXθ −cθθVc =∂θXθ −θθθVθ −φθθVφ
Sinceθθθ =φθθ =0,this reduces to the simple equation ∂θXθ =0
More explicitly, we have
∂Xθ
∂θ =0
Integrating, we find that theXθcomponent of our Killing vector is some function of theφvariable:
Xθ = f (φ) (8.3)
Next, we considera=b=φ. Using Killing’s equation, we obtain ∇φXφ =0
Working out the left-hand side by writing out the covariant derivative, we find ∇φXφ =∂φXφ−cφφ =∂φXφ−θφφXθ −φφφXφ
Now, φφφ =0 and θφφ = −sinθcosθ, and so using (8.3), this equation becomes
∂Xφ
Integrating, we obtain
Xφ = −sinθcosθ )
f(φ) dφ+g(θ) (8.4)
Now, returning to Killing’s equation, by settinga= θandb=φ,we have the last equation for this geometry, namely
∇θXφ+ ∇φXθ =0
Let’s write down each term separately. The first term is
∇θXφ =∂θXφ−cφθXc =∂θXφ−θφθXθ −φφθXφ
Looking at the Christoffel symbols, we see that this leads to ∇θXφ =∂θXφ −cotθXφ
Now we consider the second term in∇θXφ+ ∇φXθ = 0. We obtain
∇φXθ =∂φXθ −cθφXc =∂φXθ −θθφXθ −φθφXφ =∂φXθ −cotθXφ
and so, the equation∇θXφ+ ∇φXθ =0 becomes ∂θXφ +∂φXθ −2 cotθXφ =0
⇒∂θXφ+∂φXθ =2 cotθ Xφ (8.5)
We can refine this equation further using our previous results. Using (8.3) to- gether with (8.4), we find ∂θXφ =∂θ $ sinθcosθ ) f(φ) dφ+g(θ) % =(sin2θ−cos2θ) ) f φ dφ+∂θg(θ) We also have ∂φXθ =∂φf (φ)
Now, we can put all this together and obtain a solution. Adding these terms together, we get
∂θXφ+∂φXθ =(sin2θ−cos2θ)
)
f(φ) dφ+∂θg(θ)+∂φf (φ) Next we want to set this equal to the right-hand side of (8.5). But let’s work on that a little bit. We get
2 cotθXφ =2 cotθ $ −sinθcosθ ) f φdφ+g(θ) %
Now we know that
cotθ(sinθcosθ)= cosθ
sinθ (sinθcosθ)=cos 2θ
Therefore, we can write
2 cotθXφ = −2 cos2θ
)
f(φ) dφ+2 cotθg(θ) Finally, we equate both sides of (8.5) and we have
(sin2θ−cos2θ) ) f(φ) dφ+∂θg(θ)+∂φf (φ) = −2 cos2θ ) f φdφ+2 cotθg(θ)
Our goal is to get allθterms on one side and allφterms on the other. We can do this by adding 2 cos2θ*f (φ) dφ to both sides and then move the ∂θg(θ) on the left-hand side over to the right. When we do this, we get this equation
)
f(φ) dφ+∂φf (φ)=2 cotθg(θ)−∂θg(θ)
If you think back to your studies of partial differential equations, the kind where you used separation of variables, you will recall that when you have an equation in one variable equal to an equation in another variable, they must both be constant. So we will do that here. Let’s call that constantk. Looking at theθ
equation, we have
∂θg(θ)−2 cotθg(θ)= −k
We multiplied through by−1 so that the derivative term would be positive. We can solve this kind of equation using the integrating factor method. Let’s quickly review what that is. Consider your basic differential equation of the form
dy
dt + p(t)y=r(t)
First, we integrate the multiplying term p(t):
p(t)= )
p(s) ds
Then we can solve the ordinary differential equation by writing
y(t)=e−p(t)
)
ep(s)r(s) ds+Ce−p(t)
Here C is our constant of integration. Looking at our equation ∂θg(θ)− 2 cotθg(θ)= −k, we make the following identifications. We set p(θ)= −2 cotθandr(θ)= −k. First, we integrate p:
p(θ)= ) −2 cotθdθ = −2 ) cosθ sinθ dθ = −2 ln (sinθ) Now let’s plug this into the exponential and we get
e−p(θ)=e2 ln(sinθ) =eln(sin2θ) =sin2θ From this we deduce thatep(θ) = 1
sin2θ. Now we can use the integrating factor
formula to write down a solution for the functiong. The formula together with what we’ve just found gives us
g(θ)= sin2θ
) (−
k)
sin2t dt+Csin
2θ
In this case,t is just a dummy variable of integration. When we integrate we will write the functions in terms of theθ variable. It’s the easiest to look up the
integral in a table or use a program like Mathematica, and if you can’t remember you’ll find out that
) 1
sin2t dt = −cotθ
and so, we get the following:
g(θ)=sin2θ
) (−
k)
sin2tdt +Csin
2θ =
sin2θkcotθ+Csin2θ =sin2θ(kcotθ+C)
The final piece is to get a solution for the φ term. Earlier, we had
*
f (φ) dφ+∂φf (φ)=2 cotθg(θ)−∂θg(θ) and we decided to set this equal to some constant that we calledk. So looking at theφ piece, we obtain
)
f(φ)dφ+∂φf (φ)=k
Getting a solution to this one is easy. Let’s differentiate it. That will get us rid of the integral and turn the constant to zero, giving us the familiar equation
d2f
dφ2 + f (φ)=0
This is a familiar equation to most of us, and we know that the solution is given in terms of trignometric functions. More specifically, we have f (φ)=
Acosφ+Bsinφ. We quickly see that∂φf = −Asinφ+ Bcosφ. Even better, we can explicitly calculate the integral that has been hanging around since we started this example. We get the following:
)
f φdφ = )
(Acosφ+ Bsinφ) dφ= Asinφ− Bcosφ
This is a really nice result. Remember, we had defined the constant k in the following way:
)