Physics 725:2 Special and General Relativity
The L Equations and the Christoffel Symbols
(c) 2014 Andisheh Mahdavi
To calculate orbits in General Relativity, we must find paths of extremal proper time,
1 Z
Z p
Z 1 Z
dx α dx β 2
α
β 2
2
τ=
= Ldσ,
−ds =
−gαβ dx dx
= dσ −gαβ
dσ dσ
where the variational calculus is done with respect to some parameter σ along the world line of the particle.
The path of extremal proper time can be obtained by treating the integrand as the Lagrangian, and using the
Lagrange equations
"
#
d
∂L
∂L
−
+ α =0
α
dx
dσ ∂ dσ
∂x
With the equality d(L2 ) = 2LdL, these equations can be rewritten as
"
#
d 1 ∂L2
1 ∂L2
−
=0
+
α
dσ L ∂ dx
L ∂x α
dσ
And now using L =
dτ
dσ
(1)
we write this as
d
−
dτ
"
#
1 ∂L2
1 ∂L2
+
=0
α
L ∂ dx
L2 ∂x α
dσ
(2)
Now, taking a look at the actual definition of L, we can make use of the fact that the metric gαβ is only a function
of the space variable x α , and not a function of the velocity variables dx α /dσ. Specifically, since dτ = Ldσ,
∂L2
dx β
dx β dτ
dx β
∂L2 (3)
= −2gαβ
= −2Lgαβ
=L
α = −2gαβ
α dσ
dτ dσ
dτ
∂ dx
∂ dx
dσ
dσ
σ→τ
If you don’t understand where the “2” comes from, see the separate handout entitled “Where does the 2 come
from?” For the second term, we have
β
γ
2
∂gβγ dx β dx γ
∂gβγ dx β dτ
∂L2
dx γ dτ
2 ∂gβγ dx dx
2 ∂L =− α
=− α
×
= −L
=L
(4)
∂x α
∂x dσ dσ
∂x dτ dσ
dτ dσ
∂x α dτ dτ
∂x α σ→τ
Why did we switch from αβ to βγ in the metric? Because we need to maintain x α as our free variable, the
variable whose orbit we are calculating. Now that we are differentiating with respect to x α , the other indeces
need to change in order to not violate the Einstein summation convention (note that this wasn’t the case with
the velocities above, because the differentiation with respect to dx α /dσ could be carried out without violating
the summation convention).
Combining (2), (3), and (4) we obtain the L2 equation for the motion of a particle in GR:
(
)
d
∂L2 ∂L2 +
=0
−
(5)
α dτ ∂ dx
∂x α σ→τ
dσ
σ→τ
In my opinion, the above equation is the most straightforward way to derive equations of motion from a line
element. The standard formulation takes this further, towards the calculation of the Christoffel symbols. Inserting
equation (4) and (3) back into equation (5) gives us
∂gβγ dx β dx γ
d
dx β
−2
gαβ
+
=0
(6)
dτ
dτ
∂x α dτ dτ
which upon expanding the leftmost derivative becomes
−2
∂gαβ dx β
∂gβγ dx β dx γ
d 2x β
− 2gαβ
+
=0
dτ dτ
dτ 2
∂x α dτ dτ
Now, it would be most convenient to transform derivatives of the metric with respect to τ into derivatives with
respect to the coordinate variables x, because the metric is an explcit function of them. Therefore the first term
in the above equation can be recast to yield:
−2
∂gβγ dx β dx γ
∂gαβ dx γ dx β
d 2x β
+
−
2g
=0
αβ
dx γ dτ dτ
dτ 2
∂x α dτ dτ
or, rearranging the terms on the other side,
1 dx β dx γ
d 2x β
=
gαβ
dτ 2
2 dτ dτ
∂gβγ
∂gαβ
−2 γ
α
∂x
dx
(7)
Now notice that the term inside the parenthesis in the right hand side of the above isn’t symmetric in β,γ. We
β ∂x γ
would like it to be symmetric, because the multiplied velocities ∂x
∂τ ∂τ are symmetric. We can easily recast it in
a form that is symmetric by noticing first that:
dx β dx γ ∂gαβ
dx γ dx β ∂gαγ
=
dτ dτ ∂x γ
dτ dτ ∂x β
As a result,
2
dx β dx γ ∂gαβ
dx β dx γ ∂gαβ
dx γ dx β ∂gαγ
=
+
dτ dτ ∂x γ
dτ dτ ∂x γ
dτ dτ ∂x β
This gives:
d 2x β
1 dx β dx γ
gαβ
=
dτ 2
2 dτ dτ
∂gβγ
∂gαβ
∂gαγ
−
−
α
β
∂x
dx γ
dx
(8)
Now, if we define the Christoffel symbols such that
β
γ
d 2x δ
δ dx dx
=
−Γ
βγ
dτ 2
dτ dτ
or, multiplying both sides by gαδ ,
β
γ
d 2x δ
δ dx dx
=
−g
Γ
αδ
βγ
dτ 2
dτ dτ
Comparing the above with equation (8), we have
gαδ
gαδ Γδβγ
1
=
2
∂gαβ
∂gβγ
∂gαγ
+
−
γ
β
dx
∂x α
dx
(9)
An easy way to remember the above formula is to realize that the last term in the parenthesis (with the minus
sign) goes with the derivative in α, the free index on the metric in the left hand side.
Another important point: in the above, going from equation (7) to equation (8) we split the 2 ∂gαβ /∂x γ
term into ∂gαβ /∂x γ + ∂gαγ /∂x β . However, it is not in general true that ∂gαβ /∂x γ should equal ∂gαγ /∂x β . They
β ∂x γ
are different! Only the fact that the Christoffel symbols act as a sum over ∂x
∂τ ∂τ , which is the same as the sum
in equation (7), makes the step possible.