Prove the transformation is scalar

[Whitehole]

Homework Statement

1.) Prove that the infinitesimal volume element d³x is a scalar
2.) Let A^ijk be a totally antisymmetric tensor. Prove that it transforms as a scalar.

Homework Equations

The Attempt at a Solution

[/B]
1.) R^kh = ∂x'^h/∂x^k

By coordinate transformation, x'^h = R^kh x^k

dx'^h = (∂x'^h/∂x^k) (∂x^k/∂x'^j) (∂x'ⁱ/∂x^j) dx^j

dx'^h = δ^ih R^ji dx^j

dx'^h = R^jh dx^j

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 )
= R^imR^jnR^klT^mnl - R^klR^jnR^imT^lnm
= R^imR^jnR^kl(T^mnl - T^lnm)

What should I do next here?

 

[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?

 

[Whitehole]

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?

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

 

[ChrisVer]

I'm trying to prove that it is a scalar under that transformation.

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?

 

[Whitehole]

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?

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?

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?

 

[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)

 

[ChrisVer]

d3x′≡(dx′1)i(dx′2)j(dx′3)k=RimRjsRkt(dx1)m(dx2)s(dx3)td3x′≡(dx′1)i(dx′2)j(dx′3)k=RimRjsRkt(dx1)m(dx2)s(dx3)td^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

No indices over dx^1,2,3 and dx'^1,2,3 ...
in other words just replace the dx'^h= R^hidxⁱ 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:
$\epsilon_{ijk}R^{1i}R^{2j} R^{3k}=?$

 

[Whitehole]

No indices over dx^1,2,3 and dx'^1,2,3 ...
in other words just replace the dx'^h= R^hidxⁱ 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:
$\epsilon_{ijk}R^{1i}R^{2j} R^{3k}=?$

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

 

[ChrisVer]

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

nothing, it concerns question 1.

 

[Whitehole]

No indices over dx^1,2,3 and dx'^1,2,3 ...
in other words just replace the dx'^h= R^hidxⁱ 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:
$\epsilon_{ijk}R^{1i}R^{2j} R^{3k}=?$

No indices over dx^1,2,3 and dx'^1,2,3 ...
in other words just replace the dx'^h= R^hidxⁱ 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:
$\epsilon_{ijk}R^{1i}R^{2j} R^{3k}=?$

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?

 

[TerryD]

No indices over dx^1,2,3 and dx'^1,2,3 ...
in other words just replace the dx'^h= R^hidxⁱ 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:
$\epsilon_{ijk}R^{1i}R^{2j} R^{3k}=?$

I know this is an old thread, but I'm hoping someone might help out. I'm able to get:
$d^3 x' = dx'^1 dx'^2 dx'^3 = R^{1i}R^{2j} R^{3k} dx^i dx^j dx^k$

How do we get from this to:
$R^{1i}R^{2j} R^{3k} \epsilon^{ijk} dx^1 dx^2 dx^3$

I know this step basically completes the proof but I'm not able to get to it.

 

[ergospherical]

Recall that, if $\sigma$ is some permutation of $(123)$, then $dx^{\sigma(1)} \wedge dx^{\sigma(2)} \wedge dx^{\sigma(3)} = \mathrm{sgn}(\sigma) dx^1 \wedge dx^2 \wedge dx^3$. Writing this in terms of the antisymmetric symbol $\epsilon$,
\begin{align*}
d^3 \tilde{x} = d\tilde{x}^1 \wedge d\tilde{x}^2 \wedge d\tilde{x}^3 &= {R^1}_i {R^2}_j {R^3}_k dx^i \wedge dx^j \wedge dx^k \\
&= {R^1}_i {R^2}_j {R^3}_k \epsilon^{ijk} dx^1 \wedge dx^2 \wedge dx^3 \\
&= |R| dx^1 \wedge dx^2 \wedge dx^3 \\
&= |R| d^3 x
\end{align*}

 

[TerryD]

Recall that, if $\sigma$ is some permutation of $(123)$, then $dx^{\sigma(1)} \wedge dx^{\sigma(2)} \wedge dx^{\sigma(3)} = \mathrm{sgn}(\sigma) dx^1 \wedge dx^2 \wedge dx^3$. Writing this in terms of the antisymmetric symbol $\epsilon$,
\begin{align*}
d^3 \tilde{x} = d\tilde{x}^1 \wedge d\tilde{x}^2 \wedge d\tilde{x}^3 &= {R^1}_i {R^2}_j {R^3}_k dx^i \wedge dx^j \wedge dx^k \\
&= {R^1}_i {R^2}_j {R^3}_k \epsilon^{ijk} dx^1 \wedge dx^2 \wedge dx^3 \\
&= |R| dx^1 \wedge dx^2 \wedge dx^3 \\
&= |R| d^3 x
\end{align*}

Thanks! I'd gotten an incorrect idea about a volume element which was the source of confusion.