# How to calculate a trace of a tensor?

• Haorong Wu
In summary, the textbook gives the following examples for cosine: cosine of zero, cosine of pi/2, and cosine of 3 pi. cosine of zero is just P1(cos0). cosine of pi/2 is P2(cos(pi/2)). cosine of 3 pi is P3(cos(3 pi)).
Haorong Wu
Homework Statement
Use the tensor approach to work out the legendre polynomials ##P_4 \left ( cos \theta \right)=P_4 ^0 \left ( cos \theta \right)##.
Relevant Equations
None
The textbook gives some examples for ##P_1 \left ( cos \theta \right)##, ##P_2 \left ( cos \theta \right)##, and ##P_3 \left ( cos \theta \right)##.

Consider the unit vector ##V^i##. In spherical coordinates, ##V^3=\hat z=cos \theta##, which is just ##P_1 \left ( cos \theta \right)##.

Consider the symmetric traceless tensor ##T^{ij}=V^i V^j -\frac 1 3 \delta ^{ij} \left |V \right |^2 = V^i V^j - \frac 1 3 \delta ^{ij}##. Then ##T^{33}=cos^2 \theta - 1/3##, which is ##P_2 \left ( cos \theta \right)##.

Consider the symmetric traceless 3-indexed tensor ##T^{ijk}=V^i V^j V^k-\frac 1 5 \left ( \delta ^{ij} V^k+ \delta ^{jk} V^i+ \delta ^{ki} V^j \right ) ##. Then ##T^{333}=cos^3 \theta -\frac 3 5 cos\theta##, which is ##P_3 \left ( cos \theta \right)##.

The procedure is clear. The key to the problem is to find the symmetric traceless tensor for the proper rank.

I first construct a symmetric tensor ##U^{ijkl}=V^i V^j V^k V^l##.

Then I am not sure how to determine its trace. There is no clear definition of the trace of a high rank tensor. I infered that a trace of a high rank tensor is given by contraction of indices from the example for ##P_3 \left ( cos \theta \right)## given above. Then I write the trace of ##U^{ijkl}## as ##U^{kl}=\delta ^{ij} U^{ijkl}=V^i V^i V^k V^l =V^k V^l ## where the repeated index summation is implied. Next I would subtract the trace out of ##U^{ijkl}## to get a symmetric traceless tensor ##T^{ijkl}##.

But I am wrong because in the answer, the tensor is given by ##T^{ijkl}=V^i V^j V^k V^l - \frac 1 7 \left ( \delta ^{ij} V^k V^l +\delta ^{jk} V^l V^i +\delta ^{ki} V^j V^l +\delta ^{il} V^j V^k +\delta ^{jl} V^k V^i +\delta ^{kl} V^i V^j \right ) +\frac 1 {35} \left ( \delta ^{ij}\delta ^{kl}+\delta ^{jk}\delta ^{il}+\delta ^{ki}\delta ^{jl} \right )##.

I can get the terms following ##- \frac 1 7##, but not the terms following ##\frac 1 {35}##.

I believe the problem is at the calculation of the trace. Should I contract ##U^{kl}## further and subtract it from ##U^{kl}##?

Thanks!

Haorong Wu and jim mcnamara
Fred Wright said:
Thanks, Fred Wright. It is just what I need!

## 1. What is a tensor?

A tensor is a mathematical object that describes the relationship between different coordinate systems in a multi-dimensional space. It is represented as an array of numbers that follow specific transformation rules.

## 2. How do you calculate the trace of a tensor?

The trace of a tensor is calculated by summing the elements along the main diagonal of the tensor. This can be done by multiplying the tensor with the identity matrix and then taking the sum of the resulting diagonal elements.

## 3. What is the significance of the trace of a tensor?

The trace of a tensor gives the sum of the eigenvalues of the tensor, which can provide important information about the behavior of the tensor in different coordinate systems. It is also used in various mathematical and physical applications, such as calculating the curvature of a space.

## 4. Can the trace of a tensor be negative?

Yes, the trace of a tensor can be negative if the tensor has at least one negative eigenvalue. This means that the tensor has a negative curvature in at least one direction in the multi-dimensional space.

## 5. How is the trace of a tensor related to its determinant?

The trace of a tensor is equal to the sum of the eigenvalues, while the determinant is equal to the product of the eigenvalues. Therefore, the trace and determinant are related through the characteristic equation of the tensor. In some cases, the determinant can be calculated using the trace and other properties of the tensor.

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