Showing scalars are unchanged by rotation

[MathematicalPhysics]

Hello, just hoping someone can give me a hand here.

I have a second-order tensor P, which has components p_{ij} and I want to show that the following scalar quantities are unchanged by rotation:

p_{ii}
p_{ij}p_{ji}
p_{ij}p_{jk}p_{ki}

Now, I know scalars are zero'th order tensors, I know I am going to have to use the tensor transformation law, I know I must keep in mind the orthogonality of the rotation matrix and I must use the substitution property.

This is what I've done but I am not happy that its valid as a solution to my problem.

The transformation law tells us that {p^'}_{ii} = \alpha_{ia} \alpha_{ib} p_{ab}

If it is isotropic then l.h.s = p_{ii} & r.h.s = \alpha_{ia} \alpha_{ia} by the substitution property. This is equal to p_{ii} by the orthogonality of the rotation matrix.

Im not happy with this, any help is much appreciated! Thanks, Matt.

p.s. this is only the first quantity!

 

[MathematicalPhysics]

I worked out how to do it now, if anyone wants to know..

p'(ii) = alpha (ia) alpha (ib) p (ab)

p'(ii) = delta (ab) p (ab)

p'(ii) = p (aa)

which in this case can be rewritten to look like

p'(ii) = p(ii)

This implies p(ii) is invariant, or unchanged by rotation.

 

[M98Ranger]

Hi Matt,

Your approach is correct so far. To show that a scalar is unchanged by rotation, we need to show that its components in the rotated coordinate system are equal to its components in the original coordinate system. Let's look at each of the three scalar quantities separately:

1. p_{ii}

Using the transformation law, we have:

{p^'}_{ii} = \alpha_{ia} \alpha_{ib} p_{ab}

Since the rotation matrix is orthogonal, we have:

\alpha_{ia} \alpha_{ib} = \delta_{ab}

Substituting this into the above equation, we get:

{p^'}_{ii} = \delta_{ab} p_{ab}

Since the Kronecker delta is equal to 1 when the indices are the same (i.e. a=b), and 0 when the indices are different, we can rewrite the above equation as:

{p^'}_{ii} = p_{ii}

Thus, the scalar p_{ii} is unchanged by rotation.

2. p_{ij}p_{ji}

Using the transformation law, we have:

{p^'}_{ij} {p^'}_{ji} = (\alpha_{ia} \alpha_{jb} p_{ab})(\alpha_{ja} \alpha_{ib} p_{ab})

Using the orthogonality of the rotation matrix, we can simplify this to:

{p^'}_{ij} {p^'}_{ji} = \delta_{ab} \delta_{ab} p_{ab} p_{ab}

Again, using the properties of the Kronecker delta, we can rewrite this as:

{p^'}_{ij} {p^'}_{ji} = p_{ab} p_{ab}

Thus, the scalar p_{ij}p_{ji} is also unchanged by rotation.

3. p_{ij}p_{jk}p_{ki}

Using the transformation law, we have:

{p^'}_{ij} {p^'}_{jk} {p^'}_{ki} = (\alpha_{ia} \alpha_{jb} p_{ab})(\alpha_{jb} \alpha_{kc} p_{bc})(\alpha_{ka} \alpha_{ic} p_{ac})

Using the orthogonality of the rotation matrix, we can simplify this to:

{p^'}_{ij} {p^'}_{jk