# Killing Vectors in Schwarzschild Metric

Given the Schwarzschild metric with $(-,+,+,+)$ signature,

$$\text ds^2=-\left(1-\frac{2M}{r}\right)dt^2+\left(1-\frac{2M}{r}\right)^{-1}dr^2+r^2(d\theta^2+\sin^2\theta\,d\phi^2)$$

the lack of dependence of the metric on $t$ and $\phi$ allow us to read off the Killing vectors $K_1=\partial_t$ and $K_2=\partial_{\phi}$. These vectors, in their coordinate representations, are given by

$$K_1=\left(-\left(1-\frac{2M}{r}\right),0,0,0\right)$$

$$K_2=\left(0,0,0,r^2\sin^2\theta\right)$$

How does one immediately read off those vector components for $K_1$ and $K_2$? What is the logic behind reading them off? How would I "read off the Killing vectors" if I, while maintaining no explicit dependence on $t$ or $\phi$, added some off-diagonal terms to the metric? Please help me intuitively understand what's going on here.

• your question as it stands is not very clear. What you really want to know are the covariant components of the killing vectors, given the simple contravariant ones. You just go from the contravariant to the covariant components simply by multiplying with the metric as described by Chris White in his answer. Commented Oct 29, 2015 at 13:43

If all components of the metric are independent of some particular $x^\nu$, then you have the killing vector $\vec{K}$ with components $K^\mu = \delta^\mu_\nu$. That is, the contravariant form just has a constant in the appropriate slot and zeros elsewhere. In Schwarzschild, you have $K^\mu = (1, 0, 0, 0)$ and $R^\mu = (0, 0, 0, 1)$ ($\vec{K}$ and $\vec{R}$ being your $K_1$ and $K_2$, respectively).

To find the covariant forms, simply lower with the metric. In Schwarzschild we have \begin{align} K_\mu & = g_{\mu\nu} K^\nu = g_{\mu t} = \big({-}(1-2M/r), 0, 0, 0\big) \\ R_\mu & = g_{\mu\nu} R^\nu = g_{\mu\phi} = \big(0, 0, 0, r^2 \sin^2\!\theta\big). \end{align} This is where off-diagonal terms would come in. For example, in Boyer-Lindquist we also have no $t$-dependence, so we have $K^\mu = (1, 0, 0, 0)$ and $$K_\mu = g_{\mu t} = \big({-}(1-2Mr/\Sigma), 0, 0, -(2Mar/\Sigma)\sin^2\!\theta\big),$$ where the fourth component is precisely $g_{t\phi}$.

• Oh my gosh thank you so much. This makes so much sense. Commented Oct 25, 2015 at 4:12
• @user10851 is this related to the fact, that if some $\frac{dP_{x_0}}{d\tau}=0$ and $K^{\mu}$ is a Killing vector, then $K^{\mu}P_{\mu}=P_{x_0}$? Commented Jun 25, 2022 at 20:38

Conceptually:

If we leave the mathematical definition aside for a while, we can define the killing vector:

Killing vector $K^\mu(x)$ leaves metric unchanged under infinitesimal coordinate changes

Time coordinate

Change in $t$ does nothing to the metric:

Change in $t\rightarrow t+dt$:

$g_{\mu \nu}=g_{\mu \nu}(t)=g_{\mu \nu}(t+dt)=g_{\mu \nu}$

So one of the killing vectors should be along $t$:

$K^\mu=(1,0,0,0)$

That's it!

Note: The form you have has the indice lowered: $K_\mu=g_{\mu \nu}K^\nu=(g_{\mu0}K^0,0,0,0)=(-(1-2M/r),0,0,0)$

• Note: Similar logic can be used for any metric. Example: If $g_{\mu \nu} \rightarrow g_{\mu \nu}(x,y,z)$, then one of the killing vectors should be $K^\mu=(1,0,0,0)$
– OTH
Commented Oct 25, 2015 at 5:17