How do I find the metric tensor for an arbitrary coordinate system?

  • Thread starter MalleusScientiarum
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    2d Laplacian
In summary, the Laplacian is a differential operator that is used to measure the distribution of a function's values from its average value. It is given by g^{ij}\nabla_{i}\nabla_{j} in two dimensions and reduces to \frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2} in Cartesian coordinates. It is important in various mathematical concepts such as heat flow, harmonic and holomorphic functions, and cohomology classes. The metric tensor for an arbitrary coordinate system can be found by using the chain rule and plugging in the appropriate variables.
  • #1
MalleusScientiarum
Does anybody out there know what the Laplacian is for two dimensions?
 
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  • #2
The Laplacian is the trace of the Hessian operator and is undoubtedly given by

[tex] \Delta =g^{ij}\nabla_{i}\nabla_{j} [/tex]

,where [itex] i,j=1,2 [/itex]. Incidentally, for a manifold on which Christoffel's symbols are 0, it reduces to the standard form

[tex] \Delta=g^{ij}\partial_{i}\partial_{j} [/tex]

Daniel.
 
  • #3
That is probably one of the most complicated answers to a simple question I've ever seen.
 
  • #4
It depends upon the coordinate system. Dextercioby gave the general formula for any coordinate system, thereby making it, as MalleusScientarum said, "one of the most complicated answers to a simple question I've ever seen"!

In Cartesian coordinates it is [tex]\frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}[/tex].

By the way, the Laplacian is the simplest differential operator that is "invarient under rigid motions".
 
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  • #5
lets see now, isn't the laplacian the differential operator that vanishes on harmonic functions? if so, wouldn't that say it is the one that governs steady state heat flow? and characterizes the real and imaginary parts of holomorphic functions?


harmonic forms on the other hand are very sueful in discussing cohomology. the de rham theorem says that every cohomology class on a compact oriented? diff manifold can be represented by a smooth differential form, but there is no uniqueness.

by imposing a metric and hence defining a laplacian, one can define harmonic foirms and thewn there is a unique harminic representative for each cohomology class.

e.g. on an elliptic curve, formed as a quotient of the complex numbers by a lattice, one has the natural harmonic basis dz and "dzbar".
 
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  • #6
yes that stuff about harmonic & holomorphic functions sounds right. i don't know anything about heat flow though.
 
  • #7
yes, the heat-diffusion equation is:

[tex]T_p = k\Delta T[/tex]

where in cartesian coordinates [itex]T_p[/itex] is the change in temperature at the point [itex]p[/itex]

[tex]T(t,x,y,z)[/tex] and [tex]p: (x,y,z)[/tex]


[tex]T_p = k(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial ^2 T}{\partial z^2})[/tex]
 
  • #8
Apologies, I guess I should specify for polar coordinates.
 
  • #10
That's ultimately what I decided to do, but it always makes me nervous when I do something like that without knowing what mathematical reasoning I could use to justify it. As for the original answer, how would one find the metric tensor for an arbitrary coordinate system? Examples would be greatly appreciated.
 

1. How do I determine the metric tensor for a given coordinate system?

The metric tensor for a coordinate system can be determined by calculating the inner product of the basis vectors for that system. This inner product is represented by the metric tensor, which describes the geometry of the coordinate system.

2. What is the purpose of the metric tensor in a coordinate system?

The metric tensor is used to define the distance and angles between points in a given coordinate system. It also determines the volume element and curvature of the space, and is essential for performing calculations in general relativity and differential geometry.

3. Can the metric tensor be written in terms of coordinates?

Yes, the metric tensor can be expressed in terms of the coordinates of the system. This allows for a more intuitive understanding of the tensor, as well as making it easier to calculate and work with.

4. How does the metric tensor change under a coordinate transformation?

The metric tensor transforms according to the rules of tensor transformation. This means that under a change of coordinates, the metric tensor will transform in a specific way, depending on the type of transformation.

5. Are there any special cases where the metric tensor is easy to calculate?

Yes, there are certain coordinate systems where the metric tensor can be easily calculated. These include Cartesian coordinates, spherical coordinates, and cylindrical coordinates. In these cases, the metric tensor has a diagonal form, making it easier to determine its components.

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