# Prove the transformation is scalar

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1. Jan 27, 2016

### Whitehole

1. The problem statement, all variables and given/known data
1.) Prove that the infinitesimal volume element d3x is a scalar
2.) Let Aijk be a totally antisymmetric tensor. Prove that it transforms as a scalar.

2. Relevant equations

3. The attempt at a solution
1.) Rkh = ∂x'h/∂xk

By coordinate transformation, x'h = Rkh xk

dx'h = (∂x'h/∂xk) (∂xk/∂x'j) (∂x'i/∂xj) dxj

dx'h = δih Rji dxj

dx'h = Rjh dxj

This shows that the differential doesn't affect the transformation hence by performing the differential three times it would not affect the transformation, that is, it is a scalar. Can anyone verify if this is correct?

2.)
A'mnl = ( T'mnl - T'lnm )
= RimRjnRklTmnl - RklRjnRimTlnm
= RimRjnRkl(Tmnl - Tlnm)

What should I do next here?

2. Jan 27, 2016

### ChrisVer

1. I am not sure what you are trying to prove with that? In fact I am not even sure that the infinitesimal volume is indeed a scalar...? It should be transformed with the Jacobian determinant of the transformation (take for example the $dV_{spherical}$ and the $dV_{cartesian}$. Obviously going from (x,y,z) to (x',y',z')=(r,θ,φ) doesn't keep dV the same. In cartesian it's dx dy dz (fine), in spherical it's not dr dθ dφ.

2. What is T?

3. Jan 27, 2016

### Whitehole

It is an orthogonal transformation so I'm trying to prove that it is a scalar under that transformation. T is an antisymmetric tensor.

4. Jan 28, 2016

### ChrisVer

You ended up in:
$dx'^h = R^{jh}dx^j$
right?
Now you should check what is:
$d^3x' \equiv dx'^1 dx'^2 dx'^3$
why? because by the form you ended at, I cannot see how you deduce that it's a scalar, but instead that dxi are the components of a vector.

2. Well if you have:
$\epsilon'^{ijk} = R^{im} R^{js} R^{kt} \epsilon^{mst}$
can you show that if $i=j$ is $\epsilon^{ijk}=0$? Similar for the rest cases?

5. Jan 28, 2016

### Whitehole

For the first part I think the argument will be the same as part two because,

$d^3x' \equiv (dx'^1)^i (dx'^2)^j (dx'^3)^k = R^{im}R^{js}R^{kt}(dx^1)^m (dx^2)^s (dx^3)^t$

So I need to do something to loose those R's (which is the case in part two)

For part two, by setting i=j then take the sum,

$Σ\epsilon'^{iik} = ΣR^{im} R^{is} R^{kt} \epsilon^{mst}$

$Σ\epsilon'^{iik} = δ^{ms} R^{kt} \epsilon^{mst}$

$Σ\epsilon'^{iik} = R^{kt} \epsilon^{mmt}$

By definition, $\epsilon^{mmt}=0$ if m=s so, $Σ\epsilon'^{iik} = 0$

Is this what you want me to show? Is it correct that by definition $\epsilon^{mmt}=0$ if m=s?

6. Jan 28, 2016

### ChrisVer

I don't think that "summing" can help... you are trying to prove that for example $\epsilon'^{11a}=\epsilon'^{22a}=\epsilon'^{33a} =0$ and not $\epsilon'^{11a}+\epsilon'^{22a}+\epsilon'^{33a} =0$...
just set the index $\epsilon'^{ijk}$ i=j (don't sum)... and see what you get...

and yes, by definition that's how it is (that's why it's a totally-antisymmetric tensor)...
if $\epsilon^{mnr}$ is a totally antisymmetric tensor then the following is true:
$\epsilon^{mnr}=-\epsilon^{nmr}$ (along with other permutations).
so if $m=n$ let's say it's m=n=1 then:
$\epsilon^{11r}=-\epsilon^{11r}$ (you ended up with a=-a, so a must be 0)

7. Jan 28, 2016

### ChrisVer

No indices over dx^1,2,3 and dx'^1,2,3 ...
in other words just replace the dx'h= Rhidxi you obtained but for h=1,2 and 3:
$d^3 x' = dx'^1 dx'^2 dx'^3 = R^{1i}R^{2j} R^{3k} \epsilon^{ijk} dx^1 dx^2 dx^3$
from linear algebra, what is:
$\epsilon_{ijk}R^{1i}R^{2j} R^{3k}=?$

Last edited: Jan 28, 2016
8. Jan 28, 2016

### Whitehole

It should be $\epsilon_{123}$, but by your post above what does $\epsilon^{ijk}=0$ have to do with this?

9. Jan 28, 2016

### ChrisVer

nothing, it concerns question 1.

10. Jan 28, 2016

### Whitehole

Oh wait, how come $\epsilon^{ijk}$ appeared from the transformation

$d^3 x' = dx'^1 dx'^2 dx'^3 = R^{1i}R^{2j} R^{3k} \epsilon^{ijk} dx^1 dx^2 dx^3$

and I thought the transformation should involve indices? Why use numbers and introduce epsilon?