ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Thu, 28 Nov 2019 22:55:38 +0100- Check if a curve is a geodesichttps://ask.sagemath.org/question/48915/check-if-a-curve-is-a-geodesic/Suppose you have a curve `gamma` on a riemannian manifold `(M,g)`. Can you check if `gamma` is a geodesic?
I tried
nabla =g.connection()
nabla(gamma.nabla.tangent_vector_fields())
But it seems to be impossible to compute covariant derivatives of vectors field along curves. Is there any way to do this?Thu, 28 Nov 2019 16:37:01 +0100https://ask.sagemath.org/question/48915/check-if-a-curve-is-a-geodesic/
- Answer by eric_g for <p>Suppose you have a curve <code>gamma</code> on a riemannian manifold <code>(M,g)</code>. Can you check if <code>gamma</code> is a geodesic?</p>
<p>I tried</p>
<pre><code>nabla =g.connection()
nabla(gamma.nabla.tangent_vector_fields())
</code></pre>
<p>But it seems to be impossible to compute covariant derivatives of vectors field along curves. Is there any way to do this?</p>
https://ask.sagemath.org/question/48915/check-if-a-curve-is-a-geodesic/?answer=48916#post-id-48916Indeed, at the moment, one cannot take covariant derivatives along curves. This shall be improved in the future. Meanwhile, one can check whether a curve $\gamma(t)$ is a geodesic by evaluating the acceleration vector $a = \nabla_v v$ , where $v = \gamma'(t)$, via the formula
$$a^i = \frac{\mathrm{d}v^i}{\mathrm{d}t} + \Gamma^i_{\ jk} v^j v^k, \qquad (1)$$
where $(a^i)$ and $(v^i)$ are the comments of $a$ and $v$ with respect to a coordinate system $(x^i)$ and $\Gamma^i_{\ jk}$ are the Christoffel symbols of the metric with respect to $(x^i)$. One can get $\Gamma^i_{\ jk}$ via `g.christoffel_symbols()`. The curve $\gamma(t)$ is a geodesic iff $a=0$.
Here is a full example, involving a curve in the hyperbolic plane, described by the PoincarĂ© half-plane. We first introduce the manifold $M$, the PoincarĂ© half-plane coordinates $(x,y)$ and the metric $g$:
sage: M = Manifold(2, 'M', structure='Riemannian')
sage: X.<x,y> = M.chart(r'x y:(0,+oo)')
sage: g = M.metric()
sage: g[0, 0], g[1, 1] = 1/y^2, 1/y^2
sage: g.display()
g = y^(-2) dx*dx + y^(-2) dy*dy
We then consider a curve describing a half circle orthogonal to the boundary $y=0$ of PoincarĂ© half-plane:
sage: R.<t> = RealLine()
sage: gamma = M.curve([cos(t), sin(t)], (t, 0, pi), name='gamma')
sage: v = gamma.tangent_vector_field()
sage: v.display()
gamma' = -sin(t) d/dx + cos(t) d/dy
The formula (1) is implemented as follows:
sage: cc = g.christoffel_symbols()
sage: a = gamma.domain().vector_field(dest_map=gamma)
sage: for i in M.irange():
....: a[i] = diff(v[i].expr(), t) \
....: + sum(sum(cc[i, j, k](*X(gamma(t)))*v[j].expr()*v[k].expr()
....: for j in M.irange()) for k in M.irange())
....:
sage: print(a)
Vector field along the Real interval (0, pi) with values on the 2-dimensional Riemannian manifold M
sage: a.display()
cos(t) d/dx - cos(t)^2/sin(t) d/dy
We notice that $a \not=0$, which means that $\gamma$ is not geodesic. However $a$ is collinear to $v$:
sage: a == - cos(t)/sin(t) * v
True
which implies that $\gamma$ is *pregeodesic*: a reparametrization would turn it into a geodesic curve.Thu, 28 Nov 2019 22:55:38 +0100https://ask.sagemath.org/question/48915/check-if-a-curve-is-a-geodesic/?answer=48916#post-id-48916