- #1
EsmeeDijk
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Homework Statement
A tensor t has the following components in a given orthonormal basis of R3
tij(x) = a(x2)xixj + b(x2) \deltaij x2 + c(x2) \epsilonijk xk (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 := xixi, i.e. a = a(x2) depends on xi through the rotationally-symmetric combination xixi only.
The trace of the tensor t is defined as:
Tr(t) = tii (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