- #1

- 396

- 88

- 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)##.

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!

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!