Why is the first term on the right-hand side of this equation transposed?

[Living_Dog]

From Box 3.3, p. 85:

Since $$S^{\alpha}_{\phantom{\alpha}\beta\gamma} = S(\omega^\alpha, e_\beta, e_\gamma)$$

and

since S$$=S^{\alpha}_{\phantom{\alpha}\beta\gamma}e_\alpha\otimes\omega^\beta\otimes\omega^\gamma$$

is it then true that

S$$=S(\omega^\alpha, e_\beta, e_\gamma)e_\alpha\otimes\omega^\beta\otimes\omega^\gamma\ ?$$

Also, to get a new tensor from an old tensor, one of the techniques is to contract two of the indexes with each other. Is this another form of contraction, namely:

$$T_\gamma = S^{\alpha}_{\phantom{\alpha}\alpha\gamma} = S^{\alpha}_{\phantom{\alpha}\beta\gamma}\eta^{\beta}_{\phantom{\beta}\alpha} = S^{\alpha}_{\phantom{\alpha}\beta\gamma}\eta^{\beta\lambda}\eta_{\lambda\alpha}\ ?$$

Finally, why is the 1^st term on the rhs of this equation transposed??

$$\nabla($$R$$\otimes$$M$$) = (\nabla$$R$$\otimes$$M$$)^T\ +\$$R$$\otimes\nabla$$M

 

[qbert]

is it then true that
S$$=S(\omega^\alpha, e_\beta, e_\gamma)e_\alpha\otimes\omega^\beta\otimes\omega^\gamma\ ?$$

yes

Also, to get a new tensor from an old tensor, one of the techniques is to contract two of the indexes with each other. Is this another form of contraction, namely:

$$T_\gamma = S^{\alpha}_{\phantom{\alpha}\alpha\gamma} = S^{\alpha}_{\phantom{\alpha}\beta\gamma}\eta^{\beta}_{\phantom{\beta}\alpha} = S^{\alpha}_{\phantom{\alpha}\beta\gamma}\eta^{\beta\lambda}\eta_{\lambda\alpha}\ ?$$

yes. but the reason is easy if you remember $\eta^\alpha_\beta = \delta^\alpha_\beta$. where $\eta^\alpha_\beta$ is defined
by your second equality.

Finally, why is the 1^st term on the rhs of this equation transposed??

$$\nabla($$R$$\otimes$$M$$) = (\nabla$$R$$\otimes$$M$$)^T\ +\$$R$$\otimes\nabla$$M

Just a reminder of the way the indices line up. look at the lhs
(R_a M_b),_c = R_a,_c M_b + R_aM_b,_c.

The order of indices is abc. So how do you make the 1st term on the right have
the same order? Transpose the last two entries.

 

[Living_Dog]

...

Just a reminder of the way the indices line up. look at the lhs
(R_a M_b),_c = R_a,_c M_b + R_aM_b,_c.

The order of indices is abc. So how do you make the 1st term on the right have
the same order? Transpose the last two entries.

Thanks for the explanation. So ok, my transpositioning skillz are weak... is it:

(R_a,_c M_b)^T = M_b^T(R_a,_c) ^T = M_b R_c,_a?

But then the order is bca != abc.

 

[qbert]

no. they're using transpose differently here.

(and not in an especially great way - in my oppinion
/see earlier in the chapter where they only transpose
the last two indices of a rank 3 tensor/ )

let's go slowly and see how this all works.

start with two rank 1 tensors, R and M. They act on vectors to give numbers.
We can form a rank two tensor R $\otimes$ M which "eats" two vectors and spits out a number. from this we can form a rank 3 tensor by using the "gradient".

ok. say we had a rank two tensor S. the definition for the gradient
says given 3 vectors "u,v,w" we have
$$\nabla S (u, v, w) = \frac{\partial S_{ab}}{\partial x^c} u^a v^b w^c$$
or in the case of the S = R $\otimes$ M
we have
$$\nabla (R\otimes M) (u, v, w) = \frac{\partial (R_a M_b)}{\partial x^c} u^a v^b w^c = \frac{\partial R_a}{\partial x^c}M_b u^a v^b w^c + R_a \frac{\partial M_b}{\partial x^c} u^a v^b w^c$$

Now we want to make sense of these coordinate independently
The second term is: $(R \otimes \nabla M )(u, v, w)$.
But the first term is: $(\nabla R \otimes M )(u, w, v)$.

Notice we've switched the order only of the last two slots. so we define a new
tensor Transpose $(\nabla R \otimes M)$ such that
for any three vectors u, v, w
Transpose $(\nabla R \otimes M)$ (u,v,w) = $(\nabla R \otimes M)$ (u, w, v).

That's it.