Levi-Civita: very small problem, need two steps explained

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In summary, the conversation is about a math exercise involving the Levi-Civita symbol and a tensor. The last two steps involve a term with only two indices and then something distributive with three indices. The question is whether the term with two indices is correct and why it equals zero. The last step involves swapping an index and pulling the symbol out. The solution to the exercise should include a factor of one half in both the second to last and last step.
  • #1
Waxterzz
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Hi all,

kLW43Jw.png


Can someone explain me the last two steps?

I don't know why suddenly there is a term with only two indices, and then in the last step you do something distributive and again three indices.

Thanks in advance
 
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  • #2
Obviously indices are missing in the middle step as well as a definition of what D is and what properties it has. The latter is probably in the text you are quoting but we have no way of knowing this and it is important in order to be able to help you.
 
  • #3
Orodruin said:
Obviously indices are missing in the middle step as well as a definition of what D is and what properties it has. The latter is probably in the text you are quoting but we have no way of knowing this and it is important in order to be able to help you.

Hi,

Here is the complete exercise and solution from Zangwhil Modern Electrodynamics. Since my question is about the levi-civita symbol, I thought it would be better to post it here since it is a tensor.

So in the step e(ijk) D(jk) = e(ijk)D(jk) + e(kj)D(kj)

So The e(kj)D(kj) part is wrong and it should be e(ikj)D(kj) or e(kji)D(kj) ? And since the other terms ( e(ijk) D(jk) )cancel since they are equal, this e(ikj)D(kj) or e(kji)D(kj) term should be zero, why is this?The very last step is just swapping again an index so it turns out to be negative and then you can pull the symbol out, so I think the last step I understand (if it's indeed the case if an index was missing in the former step)

i9vnbM2.png
 
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  • #4
The second to last step should be
$$\frac 12 (\epsilon_{ijk} D_{jk} + \epsilon_{ikj} D_{kj}).$$
Note the factor of one half. There should also be a factor of a half in front of the last expression.
 

Related to Levi-Civita: very small problem, need two steps explained

1. What is the Levi-Civita symbol?

The Levi-Civita symbol, also known as the permutation symbol, is a mathematical symbol used in vector calculus and differential geometry to represent the orientation of a coordinate system. It is usually denoted by the Greek letter epsilon (ε) and is defined as 1 if the indices are in an even permutation, -1 if they are in an odd permutation, and 0 if any index is repeated.

2. How is the Levi-Civita symbol used to solve problems?

The Levi-Civita symbol is used to simplify vector and tensor operations, particularly in cross and dot products. It helps to determine the direction of a vector or the orientation of a surface in a given coordinate system. In differential geometry, it is used to define the curvature of a surface and to express differential operators.

3. What is the "very small problem" referred to in the question?

The "very small problem" refers to the limit of taking infinitesimal steps in a mathematical problem. In the context of the Levi-Civita symbol, it is used to represent the infinitesimal rotation of a coordinate system and is essential in defining the concept of curvature in differential geometry.

4. Can you explain the two steps involved in solving a Levi-Civita problem?

Sure, the first step is to identify the indices in the expression and determine if they are in an even or odd permutation. This will give a value of either 1 or -1. The second step is to multiply the value obtained in the first step with the corresponding components of the vector or tensor. The final result will be a simplified expression that represents the direction or orientation in the given coordinate system.

5. Are there any applications of the Levi-Civita symbol outside of mathematics?

Yes, the Levi-Civita symbol has applications in various fields such as physics, engineering, and computer graphics. It is used to represent the direction of magnetic fields, the orientation of molecules, and to calculate the angular momentum of a system. In computer graphics, it is used to determine the orientation of 3D objects and to simulate fluid dynamics.

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