How to calculate a trace of a tensor?

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SUMMARY

This discussion focuses on calculating the trace of a tensor, specifically symmetric traceless tensors in the context of spherical coordinates. The key tensors discussed include the symmetric traceless tensor ##T^{ij}## and the 4-indexed tensor ##T^{ijkl}##. The user outlines the construction of these tensors using the unit vector ##V^i## and expresses confusion regarding the correct method to determine the trace of higher-rank tensors. The correct formulation for ##T^{ijkl}## is provided, highlighting the importance of proper contraction and subtraction methods to achieve a symmetric traceless tensor.

PREREQUISITES
  • Understanding of tensor algebra and contraction methods
  • Familiarity with spherical coordinates and unit vectors
  • Knowledge of symmetric traceless tensors
  • Basic principles of linear algebra
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  • Study the properties of symmetric traceless tensors in detail
  • Learn about tensor contraction techniques and their applications
  • Explore examples of calculating traces for higher-rank tensors
  • Review the derivation of tensor equations from physics textbooks, particularly in electromagnetism
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Students and professionals in physics, particularly those studying electromagnetism, as well as mathematicians and engineers working with tensor calculus and applications in theoretical physics.

Haorong Wu
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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
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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!
 
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