Setting the trace of a tensor equal to zero

[EsmeeDijk]

Homework Statement

A tensor t has the following components in a given orthonormal basis of R³

t_ij(x) = a(x²)x_ix_j + b(x²) \delta_ij x² + c(x²) \epsilon_ijk x_k (1)​

where the indices i,j,k = 1, 2, 3.
We use the Einstein summation convention. We will only consider orthogonal transformations on R³; therefore we will not make a distinction between upper and lower indices.
a,b,c are real functions of x² := x_ix_i, i.e. a = a(x²) depends on x_i through the rotationally-symmetric combination x_ix_i only.
The trace of the tensor t is defined as:

Tr(t) = t_ii (2)​

1. Calculate the trace (2) and set it to zero, i.e. find a constraint among the functions a,b,c such that Tr(t) = 0 for all x.

Homework Equations

We only need to use equation (1) and (2)

The Attempt at a Solution

Since the indices go from 1 to 3 we have to do t₁₁ + t₂₂ + t₃₃ = 0
So the index i has to be equal to j. We get:

a(x²)x₁x₁ + b(x²)x² + a(x²)x₂x₂ + b(x²)x² + a(x²)x₃x₃ + b(x²)x²​

The kronecker delta after the b is going to be one in every case so we can leave it out. The epsilon tensor in the part with c will be zero because we won't be able to get 3 different numbers, making the entire part with c 0.
We can now simplify to :

x²(a(x₁)² + a(x₂)² + a(x₃)² + 3b(x²)) = 0​

So either x² = 0
or (a(x₁)² + a(x₂)² + a(x₃)² + 3b(x²) = 0
So x = 0 or (a(x₁)² + a(x₂)² + a(x₃)² = -3b(x²)

Now my question is, is it the case that x₁, x₂ and x₃ will they become just a normal x together?
such that we will have ax² = -3bx²

a = -3b

​

 

[Orodruin]

x2 := xixi

This should already contain the answer to your question.

 

[samalkhaiat]

Contract your tensor equation with \delta_{ij} and set it equal to zero.
\mbox{Tr}(T) = \delta_{ij}T_{ij} = 0, \ \Rightarrow \ a = - 3b.
So
T_{ij} = -3b X_{i}X_{j} + b X^{2} \delta_{ij} + c \ \epsilon_{ijk} X_{k},
with b and c are arbitrary functions of the invariant X^{2}.

 

[Samy_A]

Never mind, others already reacted while I was composing my post.