Null curves vs. straight curves on Minkowski space

[quasar_4]

I understand that we could think of a null curve in Minkowski space as being the curve c(s) such that the tangent vector dc(s)/ds = 0 at all s.

So suppose that we have a curve c(s) = (t(s), x(s), y(s), z(s)) and we want to ask ourselves what conditions would make c a straight line. I guess I'm having trouble understanding how c(s) as a straight line relates to tangency, if at all. Certainly one can think of a tangent vector at s as an equivalence class of curves passing through s, but I am not sure that's helpful.

Can anyone clarify this a bit?

 

[bcrowell]

I understand that we could think of a null curve in Minkowski space as being the curve c(s) such that the tangent vector dc(s)/ds = 0 at all s.

I don't think you want the tangent vector to *be* zero, I think you just want it to have a norm of zero.

So suppose that we have a curve c(s) = (t(s), x(s), y(s), z(s)) and we want to ask ourselves what conditions would make c a straight line. I guess I'm having trouble understanding how c(s) as a straight line relates to tangency, if at all. Certainly one can think of a tangent vector at s as an equivalence class of curves passing through s, but I am not sure that's helpful.

There are a bunch of different ways of doing this. One way is to say that a geodesic is a curve that parallel-transports its own tangent vector. Another is by using the geodesic equation written in a particular coordinate system. Another is to say that a geodesic is a curve that locally extremizes its own length.

 

[quasar_4]

Ok, I think I see what I was missing.

I guess another question is, if I have some curve and I imagine taking the tangent vector at each point along the curve, then what I've really got is a vector field. If I could show that this vector field is constant (i.e., that the vector field has constant components), is that equivalent to showing that the curve parallel transports its own tangent vector?

 

[Fredrik]

I recently studied these things in Lee's "Riemannian manifolds: an introduction to curvature", so it might help me as well as you if I write down a summary of it here. (I'm writing indices as i,j,k rather than with greek letters because i,j,k is easier to type).

We need to distinguish between "a vector field" and "a vector field along a curve". Let $\gamma:[a,b]\rightarrow M$ be a smooth curve in a manifold M. A vector field on an open subset U of M takes each $p\in U$ to a tangent vector at p. A vector field along $\gamma$ is a function that takes each $t\in[a,b]$ to a tangent vector at $\gamma(t)$. The most obvious example of vector field along $\gamma$ is its velocity, defined by

$$\dot\gamma(t)=\gamma_*D_t$$

where $\gamma_*$ is the pushforward of $\gamma$ and $D_t$ is the operator that takes $f:[a,b]\rightarrow R$ to f'(t). Note that this is just the definition of the tangent vector of $\gamma$ at the point $\gamma(t)$.

$$\dot\gamma(t)f=D_t(f\circ\gamma)=(f\circ\gamma)'(t)$$

If the manifold is equipped with a connection, we can use it to define a covariant derivative of vector fields along $\gamma$. The book explains it well.

$\gamma$ is said to be a geodesic if the covariant derivative along $\gamma$ of $\dot\gamma$ is 0, i.e. if

$$0=D\dot\gamma(t)=(\nabla_V V)_{\gamma(t)}$$

for all t, where D is the covariant dervative operator associated with $\gamma$, V is any vector field along $\gamma$ such that $V_{\gamma(t)}=\dot\gamma(t)$ for all t. Such a V is said to be an extension of $\dot\gamma$. The second equality is explained in the book as well.

The definition of a connection tells us that

$$\nabla_XY=X^i\nabla_{\partial_i}(Y^j\partial_j)=X^i(\partial_i Y^j\partial_j+Y^j\Gamma^k_{ij}\partial_k)=(XY^k+\Gamma^k_{ij}X^iY^j)\partial_k$$

so

$$0=(\nabla_V V)_{\gamma(t)}^k=V_{\gamma(t)}V^k+\Gamma^k_{ij}(\gamma(t))V^i(\gamma(t))V^j(\gamma(t))$$

But we have

$$V_{\gamma(t)}V^k=\dot\gamma(t)V^k=(V^k\circ\gamma)'(t)=\frac{d}{dt}V^k(\gamma(t))$$

This is what needs to be zero for the components of the tangent vector to be constant along gamma. So the definition of a geodesic is telling us that this happens if and only if the Christoffel symbols [itex]\Gamma^k_{ij}(\gamma(t))[/tex] all vanish in the coordinate system we're using.<br /> <br /> If we restrict ourselves to metric compatible symmetric connections (that includes any metric of GR, and of course the Euclidean metric as well), the Christoffel symbols can be expressed as<br /> <br /> $$\Gamma^k_{ij}=\frac 1 2 g^{kl}(\partial_i g_{jl}+\partial_j g_{il} -\partial_l g_{ij})$$<br /> <br /> So they all vanish if the components of the metric are constant in the coordinate system we're using. The components of the Minkowski metric in any global inertial coordinate system are of course just $g_{ij}(p)=\eta_{ij}$ for all p, so the Christoffel symbols vanish, and the components of the tangent vector of a geodesic are constant along the geodesic.[/itex]