Problem: Cartesian tensor to Spherical Tensor

In summary, the problem is to find the potential equation for a dipole in a dielectric medium with a given dielectric tensor. The equation provided in spherical coordinates is incorrect and should be rewritten using the correct expression for the electric field. This can be done by first finding the D field using the dielectric tensor and then using the relationship between D and E to calculate the potential.
  • #1
gaffareee
6
0
Dear Frnds,

my problem is very simple. please help me.

I have a dielectric tensor like:

epsilon_T=
|1 0 0|
|0 1 0| %% note that it is in cartesian co-ordinate system
|0 0 a|

now the potential equation of dipole should be as in spherical system

V=1/(4*pi*epsilon_T (?) *epsilon_0)*(qdcos(theta)/R)

now should i change the epsilon_T value cartesian to spherical system ?

or how i can write down the code ?

should it be V_xx, V_yy and V_zz?

please hurry.
 
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  • #2
gaffareee said:
now the potential equation of dipole should be as in spherical system

V=1/(4*pi*epsilon_T (?) *epsilon_0)*(qdcos(theta)/R)

This equation doesn't make much sense. On the LHS you have a scalar, while on the RHS you seem to have an uncontracted tensor in the denominator.

I believe that you should have an expression for the potential (electric or energy, it doesn't matter) that is proportional to

[tex]\sum_{ij} (\epsilon_T)_{ij} E_i r_j.[/tex]

As a scalar, this quantity can be evaluated in Cartesian coordinates and then written in terms of spherical coordinates, which would avoid having to convert vectors to spherical coordinates.
 
  • #3
You haven't said what type tensor it is.
 
  • #4
First write down the D field. For example, for a single point charge the D field is
[tex]\vec{D} = q/(4\pi r^2) \hat{r}[/tex]

Then use the relationship [tex]\vec{D} = \epsilon \vec{E}[/tex] to get the E field (you have to invert your dielectric tensor).

Then perform the integral to calculate the potential from the E field. The potential will still be a scalar but I think you will find that instead of 1/r you will have something like 1/sqrt(x^2+y^2+(az)^2). But you have to check.
 

Related to Problem: Cartesian tensor to Spherical Tensor

1. What is a Cartesian tensor?

A Cartesian tensor is a mathematical object that describes the relationship between two vectors or sets of coordinates in a Cartesian coordinate system. It has components that depend on the coordinate system and can be used to represent physical quantities such as force or stress.

2. What is a spherical tensor?

A spherical tensor is a type of tensor that is used to represent physical quantities in spherical coordinates. Unlike Cartesian tensors, its components are independent of the coordinate system and can be more convenient for certain applications, such as in quantum mechanics.

3. How do you convert a Cartesian tensor to a spherical tensor?

The conversion from a Cartesian tensor to a spherical tensor involves applying a transformation matrix to the components of the tensor. This matrix takes into account the differences in the coordinate systems and allows for the translation of the tensor between the two systems.

4. What are the advantages of using a spherical tensor over a Cartesian tensor?

One advantage of using a spherical tensor is that its components are independent of the coordinate system, making it more convenient for certain applications. Additionally, it can simplify calculations in systems with spherical symmetry, such as atoms or molecules.

5. When would you use a Cartesian tensor instead of a spherical tensor?

A Cartesian tensor may be more appropriate in situations where the system does not possess spherical symmetry. It may also be used when working with data that is already represented in a Cartesian coordinate system, as converting to a spherical tensor could introduce unnecessary complexity.

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