Show Isotropic Tensor: \epsilon _{{{\it ijm}}}\epsilon _{{{\it mkl}}}

[latentcorpse]

I've been asked to show that $\epsilon _{{{\it ijm}}}\epsilon _{{{\it mkl}}}$ is an isotropic tensor using $\epsilon _{{{\it ijk}}}\det \left( M \right) =\epsilon _{{\alpha <br /> \beta \gamma }}m_{{i\alpha }}m_{{j\beta }}m_{{k\gamma }}$.

Then to take the most general form for a fourth rank tensor and show $\epsilon _{{{\it ijm}}}\epsilon _{{{\it mkl}}}=\delta_{{{\it ik}}}<br /> \delta_{{{\it jl}}}-\delta_{{{\it il}}}\delta_{{{\it jk}}}$

The first part I tried and got completely lost on.

As for the second part all I've managed so far is to ascertain that the most general fourth rank tensor is $c_{{{\it ijkl}}}=\lambda \delta _{{{\it ij}}}\delta_{{{\it kl}}}+\mu <br /> \delta _{{{\it ik}}}\delta_{{{\it jl}}}+\upsilon \delta _{{{\it il}}}<br /> \delta_{{{\it jk}}}$

 

[olgranpappy]

I've been asked to show that $\epsilon _{{{\it ijm}}}\epsilon _{{{\it mkl}}}$ is an isotropic tensor using $\epsilon _{{{\it ijk}}}\det \left( M \right) =\epsilon _{{\alpha <br /> \beta \gamma }}m_{{i\alpha }}m_{{j\beta }}m_{{k\gamma }}$.

Then to take the most general form for a fourth rank tensor and show $\epsilon _{{{\it ijm}}}\epsilon _{{{\it mkl}}}=\delta_{{{\it ik}}}<br /> \delta_{{{\it jl}}}-\delta_{{{\it il}}}\delta_{{{\it jk}}}$

The first part I tried and got completely lost on.

As for the second part all I've managed so far is to ascertain that the most general fourth rank tensor is $c_{{{\it ijkl}}}=\lambda \delta _{{{\it ij}}}\delta_{{{\it kl}}}+\mu <br /> \delta _{{{\it ik}}}\delta_{{{\it jl}}}+\upsilon \delta _{{{\it il}}}<br /> \delta_{{{\it jk}}}$

How does the tensor
$$T_{ijk}$$
transform in general (under rotations)?

Apply this transformation to the tensor
$$\epsilon_{ijk}$$
and you will see that you get something that looks like the rhs of the first line you wrote... but what is it?

 

[latentcorpse]

$\epsilon^{'} _{{{\it ijk}}}= det(M)m_{{i\alpha }}m_{{j\beta }}m_{{k\gamma }}<br /> \epsilon _{{\alpha \beta \gamma }}$

 

[latentcorpse]

ok so using the fact that $\epsilon _{{{\it ijk}}}$ is invariant and that
$(det(M))^{-1}=det(M)$. I can rearrange and get that $\epsilon _{{{\it ijk}}} det(M) = \epsilon _{{\alpha \beta \gamma }} m_{{i\alpha }}m_{{j\beta }}m_{{k\gamma }}$

 

[latentcorpse]

However, that was the given identity. I need to use that to show that $(\epsilon _{{{\it ijm}}}\epsilon _{{{\it mkl}}})^{'}=\epsilon _{{{\it ijm}}}\epsilon _{{{\it mkl}}}$ Can you offer any advice? Cheers.

 

[olgranpappy]

$\epsilon^{'} _{{{\it ijk}}}= det(M)m_{{i\alpha }}m_{{j\beta }}m_{{k\gamma }}<br /> \epsilon _{{\alpha \beta \gamma }}$

What is M?

 

[latentcorpse]

M is the transformation matrix and since we're considering a rotation which is a passive transformation, we know that det(M)=1 as it's a proper transformation. How does that help?

 

[olgranpappy]

M is the transformation matrix and since we're considering a rotation which is a passive transformation, we know that det(M)=1 as it's a proper transformation. How does that help?

Ok... the transformation for a tensor T_{ijk} under rotation M is
$$T_{ijk}^{'} = m_{ia}m_{jb}m_{kc}T_{abc}\;.$$

So, the transformation of the tensor \epsilon_{ijk} is
$$\epsilon_{ijk}^{'}=m_{ia}m_{jb}m_{kc}\epsilon_{abc}\;,$$
which is not what you wrote down exactly.

Now, what is the LHS of the final equation I wrote equal to according to the first equation you gave in the OP, and also using det(M)=1?

 

[latentcorpse]

hi $\epsilon _{{{\it ijk}}}^{'}=det(M)m_{{i\alpha }}m_{{j\beta }}m_{{k\gamma }}<br /> \epsilon _{{\alpha \beta \gamma }}$
we need that det(M) there because it's a rank three pseudotensor so surely in the equation you wrote in your last post you're missing a det(M), no?

 

[latentcorpse]

however if we're just assumin det(M)=1 everywhere then,
$\epsilon _{{{\it ijk}}}=\epsilon _{{\alpha \beta \gamma }}m_{{i\alpha }}m_{{j\beta }}m_{{k\gamma }}$ from my original post and setting det(M) =1
But
$\epsilon _{{\alpha \beta \gamma }}m_{{i\alpha }}m_{{j\beta }}m_{{k\gamma }}=\epsilon _{ijk}det(M)$
However det(M)=1 can be used again here to show
$\epsilon _{ijk}^{'}=\epsilon _{ijk}det(M)=\epsilon _{ijk}$
Is this a sufficient proof?
If so, any ideas for the next part?

 

[olgranpappy]

however if we're just assumin det(M)=1 everywhere then,
$\epsilon _{{{\it ijk}}}=\epsilon _{{\alpha \beta \gamma }}m_{{i\alpha }}m_{{j\beta }}m_{{k\gamma }}$ from my original post and setting det(M) =1
But
$\epsilon _{{\alpha \beta \gamma }}m_{{i\alpha }}m_{{j\beta }}m_{{k\gamma }}=\epsilon _{ijk}det(M)$
However det(M)=1 can be used again here to show
$\epsilon _{ijk}^{'}=\epsilon _{ijk}det(M)=\epsilon _{ijk}$
Is this a sufficient proof?
If so, any ideas for the next part?

So, then, since you just showed that \epsilon_{ijk} is an isotropic tensor it follows immediately that the multiplication of two \epsilon tensors is also isotropic (this is the first part you asked about).

As for the next part, you have:
$$\epsilon_{ijm}\epsilon_{mkl}<br /> =<br /> \lambda \delta_{ij}\delta_{kl}<br /> +<br /> \mu \delta_{ik}\delta_{jl}<br /> +<br /> \nu \delta_{il}\delta_{jk}\;.$$

What happens if you set i=j=1 and k=l=2?

 

[latentcorpse]

Hi again. How does that follow immediately? Am I just being really stupid? I guess I'm just not really satisfied with the rigour of this proof-I can't really follow it myself without constantly referring back to your posts.

and for the next part i get
$\epsilon _{{11\,m}}\epsilon _{{{\it m22}}}=\lambda=0$
which is good because we don't want a $\delta_{{{\it ij}}}\delta_{{{\it kl}}}$ term. How do I go about getting the values of $\mu$ and $\nu$?

 

[olgranpappy]

Hi again. How does that follow immediately?

\epsilon is isotropic

i.e., the components \epsilon_{ijm} (and \epsilon_{mkl}) are the same in different rotated frames.

The components are the same in diffferent frames, so multiplication of two componets gives the same result in different frames, and summing multiplied components gives the same result in different frames.

Am I just being really stupid?

no.

I guess I'm just not really satisfied with the rigour of this proof-I can't really follow it myself without constantly referring back to your posts.

and for the next part i get
$\epsilon _{{11\,m}}\epsilon _{{{\it m22}}}=\lambda=0$
which is good because we don't want a $\delta_{{{\it ij}}}\delta_{{{\it kl}}}$ term. How do I go about getting the values of $\mu$ and $\nu$?

Try plugging in some other values for i, j, k, and l.

 

[latentcorpse]

yeah that's what i tried but i didn't have any paper so did it in my head and messed it up the first time but now that you recommended it and I've been over it i get it. thanks for all your help. how did you know to start plugging in numbers on that second part or is that just a typical method when your looking at removing terms from the most general isotropic tensor of a particular rank?

 

[olgranpappy]

yeah that's what i tried but i didn't have any paper

Invest in some paper. In my experience it is a very useful tool. Pencils too.

so did it in my head and messed it up the first time but now that you recommended it and I've been over it i get it. thanks for all your help. how did you know to start plugging in numbers on that second part or is that just a typical method when your looking at removing terms from the most general isotropic tensor of a particular rank?

It's a pretty standard method. Cheers and good luck with your studies.