Geodesics with arbitrary parametrization

In summary: And if you start with the acceleration being zero, you can find a parametrization such that the proportionality holds.In summary, the conversation discusses the definition of a geodesic curve in a manifold, and the conditions under which a curve can be considered geodesic. It is mentioned that there are two equivalent definitions, one involving a specific system of equations and the other involving the proportionality of acceleration and velocity. The speaker concludes that the formula in question does not contradict their intuition and must be correct.
  • #1
wrobel
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Let ##x=(x^1,\ldots,x^m)## be local coordinates in a manifold ##M##; and let ##\{\Gamma^i_{jk}(x)\}## be a connection. Assume that we have a curve ##x=x(t),\quad \dot x\ne 0##. Is this curve geodesic or not?
My guess is that the answer is "yes" iff for all ##k,n## the function ##x(t)## satisfies the following system
$$\dot x^k(\ddot x^n+\Gamma^n_{rj}\dot x^r\dot x^j)=\dot x^n(\ddot x^k+\Gamma^k_{rj}\dot x^r\dot x^j).$$
In my taste this system looks strange. Or not?
 
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  • #2
wrobel said:
Let ##x=(x^1,\ldots,x^m)## be local coordinates in a manifold ##M##; and let ##\{\Gamma^i_{jk}(x)\}## be a connection. Assume that we have a curve ##x=x(t),\quad \dot x\ne 0##. Is this curve geodesic or not?
My guess is that the answer is "yes" iff for all ##k,n## the function ##x(t)## satisfies the following system
$$\dot x^k(\ddot x^n+\Gamma^n_{rj}\dot x^r\dot x^j)=\dot x^n(\ddot x^k+\Gamma^k_{rj}\dot x^r\dot x^j).$$
In my taste this system looks strange. Or not?
Well, when it comes to your taste only you can say if it is strange or not. It is not strange of you notice the following: a curve is geodesic if the acceleration is proportional to the velocity i.e. ##a^i=\lambda v^i##. Equivalently (eliminating lambda) you have ##v^ja^i=v^ia^j##, which is your "strange" formula.
 
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  • #3
Ok, thanks. I employed the following definition of geodesic: the geodesic is a curve that can be parametrized such that ##x=x(s),##
$$x^i_{ss}+\Gamma^i_{kj} x^k_s x^j_s=0.$$ Here ##\Gamma## is not obliged to be generated by a Riemann metric.
I suspect that your remark about proportionality of the acceleration and the velocity is not an independent fact but follows from this definition by the same argument as I used. Anyway that is good that the formula from #1 does not contradict to your intuition. Thus it must be correct.
 
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  • #4
Yes, they are equivalent definitions. If you have the proportionality, you can reparametrize to get that the acceleration is zero.
 

Related to Geodesics with arbitrary parametrization

1. What is a geodesic with arbitrary parametrization?

A geodesic with arbitrary parametrization is a curve on a surface that follows the shortest path between two points. It is defined by a set of parametric equations that describe the path of the curve.

2. How is a geodesic with arbitrary parametrization different from a regular geodesic?

A regular geodesic is defined by a specific parametrization, such as arc length or time. An arbitrary parametrization allows for more flexibility in defining the curve, as it is not limited to a specific parameter.

3. What is the importance of studying geodesics with arbitrary parametrization?

Studying geodesics with arbitrary parametrization allows for a better understanding of the behavior of curves on surfaces. It also has applications in fields such as differential geometry, physics, and computer graphics.

4. How are geodesics with arbitrary parametrization calculated?

Geodesics with arbitrary parametrization can be calculated using various methods, such as the variational principle or the geodesic equation. These methods involve finding the shortest path between two points on a surface.

5. Can geodesics with arbitrary parametrization exist on any surface?

Yes, geodesics with arbitrary parametrization can exist on any surface, as long as the surface has a well-defined metric. This means that the surface must have a way to measure distance between points, which is necessary for calculating geodesics.

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