Parallel transport on a cardioid

In summary, the conversation discusses the process of calculating an explicit example of parallel transport of a vector along a cardioid curve. The necessary components for this calculation include a vector field on the manifold (the cardioid itself) and a curve on the manifold. The parametrization of the cardioid in Cartesian coordinates is provided and the process of finding a vector field is discussed. The general formula for parallel transport is mentioned, as well as the need to calculate the Levi-Civita connection with the metric. Further clarification and simplification is requested for the process.
  • #1
JonnyMaddox
74
1
Hi guys,
I want to calculate an explicit example of a vector parallel transported along a cardioid to see what happens. Maybe someone could help me with that since no author of any book or pdf on the topic is capable of showing a single numerical example.

So we need a vector field on a manifold (which is the cardioid itself) [itex]X=\frac{dx^{i}}{dt}\frac{\partial}{\partial x^{i}}[/itex] and a curve [itex]x^{i}=x^{i}(t)[/itex]. My problem is, I'm not sure how to make up a curve + vector field on a manifold. Let's take the parametrization of the cardioid in Cartesian coordinates as

[itex]x(t)=a(1+2\cos t + \cos 2t)[/itex]

[itex]y(t)=a(2\sin t + \sin 2t)[/itex]

(I think this could be written in polar coordinates which would make more sense, but I'm not sure what happens there)

So I think this should be the curve on which the vector is transported. Now I'm not sure how to make up the vector field. For the vector field I also need a function [itex]f[/itex], but what function? A vector function? For example could I just take [itex]f=r(\phi, \rho)= (\rho \cos \phi, \rho \sin \phi)[/itex] (polar coordinates) and then [itex]X=\frac{dx^{i}}{dt}\frac{\partial}{\partial x^{i}}= \frac{dx(t)}{dt}\frac{\partial r(\phi, \rho)}{\partial \phi}+\frac{dy(t)}{dt}\frac{\partial r(\phi,\rho)}{\partial \rho}[/itex] ? I think this looks right since the [itex]\frac{\partial}{\partial x^{i}}[/itex] span the tangent space. Now how exactly does the condition for parallel transport in coordinates for this looks like?
The general formula is

[itex]\frac{\partial X^{\mu}}{dt}+ \Gamma^{\mu}_{v\lambda} \frac{\partial x^{v}(c(t))}{dt}X^{\lambda}=0[/itex]

(I know how to calculate the Levi-Civita connection with the metric,but I'm not sure about the rest)
 
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  • #2
JonnyMaddox said:
Hi guys,
I want to calculate an explicit example of a vector parallel transported along a cardioid to see what happens. Maybe someone could help me with that since no author of any book or pdf on the topic is capable of showing a single numerical example.

So we need a vector field on a manifold (which is the cardioid itself) [itex]X=\frac{dx^{i}}{dt}\frac{\partial}{\partial x^{i}}[/itex] and a curve [itex]x^{i}=x^{i}(t)[/itex]. My problem is, I'm not sure how to make up a curve + vector field on a manifold. Let's take the parametrization of the cardioid in Cartesian coordinates as

[itex]x(t)=a(1+2\cos t + \cos 2t)[/itex]

[itex]y(t)=a(2\sin t + \sin 2t)[/itex]

(I think this could be written in polar coordinates which would make more sense, but I'm not sure what happens there)

So I think this should be the curve on which the vector is transported. Now I'm not sure how to make up the vector field. For the vector field I also need a function [itex]f[/itex], but what function? A vector function? For example could I just take [itex]f=r(\phi, \rho)= (\rho \cos \phi, \rho \sin \phi)[/itex] (polar coordinates) and then [itex]X=\frac{dx^{i}}{dt}\frac{\partial}{\partial x^{i}}= \frac{dx(t)}{dt}\frac{\partial r(\phi, \rho)}{\partial \phi}+\frac{dy(t)}{dt}\frac{\partial r(\phi,\rho)}{\partial \rho}[/itex] ? I think this looks right since the [itex]\frac{\partial}{\partial x^{i}}[/itex] span the tangent space. Now how exactly does the condition for parallel transport in coordinates for this looks like?
The general formula is

[itex]\frac{\partial X^{\mu}}{dt}+ \Gamma^{\mu}_{v\lambda} \frac{\partial x^{v}(c(t))}{dt}X^{\lambda}=0[/itex]

(I know how to calculate the Levi-Civita connection with the metric,but I'm not sure about the rest)
Sorry Johhny, is there anything else you can add or simplify if you still need an answer?
 
  • #3
The cardiod is a curve in the plane and I guess you want to parallel translate a vector along it using the standard inner product on the plane.

- what do you get for the covariant derivative in the plane?

Once you have that all else follows quickly.
 
Last edited:

1. What is parallel transport on a cardioid?

Parallel transport on a cardioid is a mathematical concept that involves moving a vector along a path on a cardioid without changing its direction. It is similar to parallel transport on a sphere, but with a different geometric shape.

2. How is parallel transport on a cardioid different from parallel transport on a circle?

The main difference between parallel transport on a cardioid and a circle is the curvature of the path. The circle has a constant curvature, while the curvature of a cardioid changes along the path. This makes parallel transport on a cardioid more complex and challenging to visualize.

3. What is the importance of parallel transport on a cardioid?

Parallel transport on a cardioid has applications in various fields, such as physics, engineering, and computer graphics. It is used to understand the behavior of vectors in curved spaces and can help in the design and analysis of curved objects.

4. How is parallel transport on a cardioid calculated?

The calculation of parallel transport on a cardioid involves using the curvature of the path and the direction of the vector at each point. This can be done using mathematical formulas or through graphical methods. Software programs, such as MATLAB, can also be used for numerical calculations.

5. Can parallel transport on a cardioid be applied to other geometric shapes?

Yes, parallel transport can be applied to any curved surface or shape, including ellipses, hyperbolas, and more complex surfaces. The principles of parallel transport remain the same, but the calculations may vary depending on the specific shape and its curvature.

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