Differentiation of a tensor equation with respect to an undefined variable
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I'm just wrestling with some fundamentals of tensor calculus. And here is a confusion corresponding to a differentiation case. In this book, pages 45 and 46, one reads:
Consider two alternative coordinate systems $Z^i$ and $Z^{i^{'}}$ in
an $N$-dimensional space. Notice that we placed the prime next to the
index rather than the letter $Z$. Let us call the coordinates $Z^i$
unprimed and the coordinates $Z^{i^{'}}$ primed. We also use the
symbols $Z^{i^{'}}$ and $Z^i$ to denote the functions that express the
relationships between the coordinates:
$Z^{i^{'}} = Z^{i^{'}}(Z)$
$Z^i = Z^i(Z^{'})$
Then, the following two identities are introduced:
$Z^i(Z^{'}(Z)) equiv Z^i$
$Z^{i^{'}}(Z(Z^{'})) equiv Z^{i^{'}}$
Finally, a $Z^{j}$ comes from the middle of nowhere with respect to which the book differentiates the identities above. In particular the book says:
We differentiate the identity $Z^i(Z^{'}(Z)) equiv Z^i$ with respect
to $Z^j$ . It is essential that the differentiation is to take place
with respect to $Z^j$ rather than $Z^i$, because our intention is to
differentiate each of the identities in $Z^i(Z^{'}(Z)) equiv Z^i$
with respect to each of the variables.
Can anyone explain what it does mean?
tensors
asked Dec 21 '18 at 9:58
RoboticistRoboticist
244315
$endgroup$
add a comment |
$begingroup$
I'm just wrestling with some fundamentals of tensor calculus. And here is a confusion corresponding to a differentiation case. In this book, pages 45 and 46, one reads:
Consider two alternative coordinate systems $Z^i$ and $Z^{i^{'}}$ in
an $N$-dimensional space. Notice that we placed the prime next to the
index rather than the letter $Z$. Let us call the coordinates $Z^i$
unprimed and the coordinates $Z^{i^{'}}$ primed. We also use the
symbols $Z^{i^{'}}$ and $Z^i$ to denote the functions that express the
relationships between the coordinates:
$Z^{i^{'}} = Z^{i^{'}}(Z)$
$Z^i = Z^i(Z^{'})$
Then, the following two identities are introduced:
$Z^i(Z^{'}(Z)) equiv Z^i$
$Z^{i^{'}}(Z(Z^{'})) equiv Z^{i^{'}}$
Finally, a $Z^{j}$ comes from the middle of nowhere with respect to which the book differentiates the identities above. In particular the book says:
We differentiate the identity $Z^i(Z^{'}(Z)) equiv Z^i$ with respect
to $Z^j$ . It is essential that the differentiation is to take place
with respect to $Z^j$ rather than $Z^i$, because our intention is to
differentiate each of the identities in $Z^i(Z^{'}(Z)) equiv Z^i$
with respect to each of the variables.
Can anyone explain what it does mean?
tensors
asked Dec 21 '18 at 9:58
RoboticistRoboticist
244315
$endgroup$
add a comment |
$begingroup$
I'm just wrestling with some fundamentals of tensor calculus. And here is a confusion corresponding to a differentiation case. In this book, pages 45 and 46, one reads:
Consider two alternative coordinate systems $Z^i$ and $Z^{i^{'}}$ in
an $N$-dimensional space. Notice that we placed the prime next to the
index rather than the letter $Z$. Let us call the coordinates $Z^i$
unprimed and the coordinates $Z^{i^{'}}$ primed. We also use the
symbols $Z^{i^{'}}$ and $Z^i$ to denote the functions that express the
relationships between the coordinates:
$Z^{i^{'}} = Z^{i^{'}}(Z)$
$Z^i = Z^i(Z^{'})$
Then, the following two identities are introduced:
$Z^i(Z^{'}(Z)) equiv Z^i$
$Z^{i^{'}}(Z(Z^{'})) equiv Z^{i^{'}}$
Finally, a $Z^{j}$ comes from the middle of nowhere with respect to which the book differentiates the identities above. In particular the book says:
We differentiate the identity $Z^i(Z^{'}(Z)) equiv Z^i$ with respect
to $Z^j$ . It is essential that the differentiation is to take place
with respect to $Z^j$ rather than $Z^i$, because our intention is to
differentiate each of the identities in $Z^i(Z^{'}(Z)) equiv Z^i$
with respect to each of the variables.
Can anyone explain what it does mean?
tensors
asked Dec 21 '18 at 9:58
RoboticistRoboticist
244315
$endgroup$
I'm just wrestling with some fundamentals of tensor calculus. And here is a confusion corresponding to a differentiation case. In this book, pages 45 and 46, one reads:
Consider two alternative coordinate systems $Z^i$ and $Z^{i^{'}}$ in
an $N$-dimensional space. Notice that we placed the prime next to the
index rather than the letter $Z$. Let us call the coordinates $Z^i$
unprimed and the coordinates $Z^{i^{'}}$ primed. We also use the
symbols $Z^{i^{'}}$ and $Z^i$ to denote the functions that express the
relationships between the coordinates:
$Z^{i^{'}} = Z^{i^{'}}(Z)$
$Z^i = Z^i(Z^{'})$
Then, the following two identities are introduced:
$Z^i(Z^{'}(Z)) equiv Z^i$
$Z^{i^{'}}(Z(Z^{'})) equiv Z^{i^{'}}$
Finally, a $Z^{j}$ comes from the middle of nowhere with respect to which the book differentiates the identities above. In particular the book says:
We differentiate the identity $Z^i(Z^{'}(Z)) equiv Z^i$ with respect
to $Z^j$ . It is essential that the differentiation is to take place
with respect to $Z^j$ rather than $Z^i$, because our intention is to
differentiate each of the identities in $Z^i(Z^{'}(Z)) equiv Z^i$
with respect to each of the variables.
Can anyone explain what it does mean?
tensors
tensors
asked Dec 21 '18 at 9:58
RoboticistRoboticist
244315
asked Dec 21 '18 at 9:58
RoboticistRoboticist
244315
asked Dec 21 '18 at 9:58
RoboticistRoboticist
244315
asked Dec 21 '18 at 9:58
RoboticistRoboticist
244315
asked Dec 21 '18 at 9:58
RoboticistRoboticist
244315
244315
add a comment |
add a comment |
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