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Let us denote by $X^i=(1,\vec 0)$ the Killing vector field and by $u^i(s)$ a tangent vector field of a geodesic, where $s$ is some affine parameter.

What physical significance do the scalar quantity $X_iu^i$ and its conservation hold? If any...? I have seen this in may books and exam questions. I wonder what it means...

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up vote 10 down vote accepted

In general, if $\xi^\mu$ is a Killing vector field on a spacetime, and if $u^\mu$ is a tangent field along a geodesic in that spacetime, then $\xi_\mu u^\mu$ is a conserved quantity along the geodesic. (See for example Wald's GR proposition C.3.1).

To illustrate the physical significance of this, consider a particle moving in $2$-dimensional Minkowski space with metric

$$ds^2 = -dt^2 + dx^2.$$

This metric admits killing vectors $\xi_{(t)} = (1,0)$ and $\xi_{(x)} = (0,1)$. It follows that for a geodesic $(t(\lambda),x(\lambda))$ with tangent $u^\mu(\lambda)=(dt/d\lambda, dx/d\lambda)$, we obtain two conserved quantities $$ \xi_{(t)}^\mu u_\mu = -\frac{dt}{d\lambda}, \qquad \xi_{(x)}^\mu u_\mu = \frac{dx}{d\lambda} $$ If we imagine that our geodesic represents the path of a particle of mass $m$ through spacetime, then if we choose the parameter $\lambda$ to be arc length, which for timelike curves is called proper time $\tau$, namely if we choose $$ -1 = u^\mu u_\mu = \left(\frac{dt}{d\tau}\right)^2 + \left(\frac{dx}{d\tau}\right)^2 $$ then $p^\mu = m u^\mu$ is the four-momentum of the particle, and the conservation equation obtained from $\xi_{(t)}$ gives conservation of $p^t = m u^t$, which is the energy of the particle, and the conservation equation obtained from $\xi_{(x)}$ gives conservation of $p^x = m u^x$ which is the momentum of the particle.

So in this context, Killing vectors of the given metric gave conserved quantities that could be interpreted as the energy and momentum of a particle moving freely in flat spacetime.

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Thanks, Josh! :) – Fernandez Jan 25 '13 at 1:54
You want a subscript $x$ for on the third equation. – kηives Jan 25 '13 at 2:28
@kηives Thanks. – joshphysics Jan 25 '13 at 3:34

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