Mixing
Metric Tensor inner product and summation notation
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I have a really simple question:
Let $$g = beginbmatrix 1&0&0&0\ 0&-1&0&0\0&0&-1&0\0&0&0&-1 endbmatrix$$
be the metric tensor, then it has elements $g_kappa nu$. In einstein summation notation, what is the inner product between the two tensors $g_kappa nu g^mu nu$? Obviously here, we are summing over $nu$ in both terms, but $kappa$ and $mu$ are different, do we just say that $kappa = mu$ otherwise the terms are zero, giving us the inner product $g_kappa nu g^mu nu = g_mu nu g^mu nu = 1 +(-1)^2 + (-1)^2 + (-1)^2$. Or am I missing something fundamental about the tensor relationships / summation notation?
Cheers.
linear-algebra tensors
asked Jul 25 at 1:43
jamesmartini
319210
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up vote
1
down vote
favorite
I have a really simple question:
Let $$g = beginbmatrix 1&0&0&0\ 0&-1&0&0\0&0&-1&0\0&0&0&-1 endbmatrix$$
be the metric tensor, then it has elements $g_kappa nu$. In einstein summation notation, what is the inner product between the two tensors $g_kappa nu g^mu nu$? Obviously here, we are summing over $nu$ in both terms, but $kappa$ and $mu$ are different, do we just say that $kappa = mu$ otherwise the terms are zero, giving us the inner product $g_kappa nu g^mu nu = g_mu nu g^mu nu = 1 +(-1)^2 + (-1)^2 + (-1)^2$. Or am I missing something fundamental about the tensor relationships / summation notation?
Cheers.
linear-algebra tensors
asked Jul 25 at 1:43
jamesmartini
319210
1
The indices $kappa$ and $mu$ are free in the expression $g_kappa nug^mu nu$. Free indices as a usual rule appear on both sides of an equation, the missing free indices in $1+(-1)^2+(-1)^2+(-1)^2$ is a dead give-away that your idea is off. The answer explains further...
– James S. Cook
Jul 25 at 2:25
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a really simple question:
Let $$g = beginbmatrix 1&0&0&0\ 0&-1&0&0\0&0&-1&0\0&0&0&-1 endbmatrix$$
be the metric tensor, then it has elements $g_kappa nu$. In einstein summation notation, what is the inner product between the two tensors $g_kappa nu g^mu nu$? Obviously here, we are summing over $nu$ in both terms, but $kappa$ and $mu$ are different, do we just say that $kappa = mu$ otherwise the terms are zero, giving us the inner product $g_kappa nu g^mu nu = g_mu nu g^mu nu = 1 +(-1)^2 + (-1)^2 + (-1)^2$. Or am I missing something fundamental about the tensor relationships / summation notation?
Cheers.
linear-algebra tensors
asked Jul 25 at 1:43
jamesmartini
319210
I have a really simple question:
Let $$g = beginbmatrix 1&0&0&0\ 0&-1&0&0\0&0&-1&0\0&0&0&-1 endbmatrix$$
be the metric tensor, then it has elements $g_kappa nu$. In einstein summation notation, what is the inner product between the two tensors $g_kappa nu g^mu nu$? Obviously here, we are summing over $nu$ in both terms, but $kappa$ and $mu$ are different, do we just say that $kappa = mu$ otherwise the terms are zero, giving us the inner product $g_kappa nu g^mu nu = g_mu nu g^mu nu = 1 +(-1)^2 + (-1)^2 + (-1)^2$. Or am I missing something fundamental about the tensor relationships / summation notation?
Cheers.
linear-algebra tensors
asked Jul 25 at 1:43
jamesmartini
319210
asked Jul 25 at 1:43
jamesmartini
319210
asked Jul 25 at 1:43
jamesmartini
319210
asked Jul 25 at 1:43
jamesmartini
319210
319210
1
The indices $kappa$ and $mu$ are free in the expression $g_kappa nug^mu nu$. Free indices as a usual rule appear on both sides of an equation, the missing free indices in $1+(-1)^2+(-1)^2+(-1)^2$ is a dead give-away that your idea is off. The answer explains further...
– James S. Cook
Jul 25 at 2:25
add a comment |Â
1
The indices $kappa$ and $mu$ are free in the expression $g_kappa nug^mu nu$. Free indices as a usual rule appear on both sides of an equation, the missing free indices in $1+(-1)^2+(-1)^2+(-1)^2$ is a dead give-away that your idea is off. The answer explains further...
– James S. Cook
Jul 25 at 2:25
1
1
The indices $kappa$ and $mu$ are free in the expression $g_kappa nug^mu nu$. Free indices as a usual rule appear on both sides of an equation, the missing free indices in $1+(-1)^2+(-1)^2+(-1)^2$ is a dead give-away that your idea is off. The answer explains further...
– James S. Cook
Jul 25 at 2:25
The indices $kappa$ and $mu$ are free in the expression $g_kappa nug^mu nu$. Free indices as a usual rule appear on both sides of an equation, the missing free indices in $1+(-1)^2+(-1)^2+(-1)^2$ is a dead give-away that your idea is off. The answer explains further...
– James S. Cook
Jul 25 at 2:25
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The product of $g_kappanug^munu$ will not return a scalar, it will return a mixed tensor, specifically $delta_kappa^mu$, the Kronecker delta. In fact, $g_kappanug^munu=delta_kappa^mu$ is often taken as the definition of the inverse metric $g^munu$. The delta has components
$$
beginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
endbmatrix
.$$
That is, you only perform contraction on the repeated index $nu$, while the other two stay free. This is true for contracting tensors in general. So $g_kappanug^munu$ is not the same as $g_munug^munu$.
*EDIT
in fact this question has been answered at https://physics.stackexchange.com/questions/66394/kronecker-delta-confusion
answered Jul 25 at 2:04
Merk Zockerborg
1429
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The product of $g_kappanug^munu$ will not return a scalar, it will return a mixed tensor, specifically $delta_kappa^mu$, the Kronecker delta. In fact, $g_kappanug^munu=delta_kappa^mu$ is often taken as the definition of the inverse metric $g^munu$. The delta has components
$$
beginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
endbmatrix
.$$
That is, you only perform contraction on the repeated index $nu$, while the other two stay free. This is true for contracting tensors in general. So $g_kappanug^munu$ is not the same as $g_munug^munu$.
*EDIT
in fact this question has been answered at https://physics.stackexchange.com/questions/66394/kronecker-delta-confusion
answered Jul 25 at 2:04
Merk Zockerborg
1429
add a comment |Â
up vote
2
down vote
accepted
The product of $g_kappanug^munu$ will not return a scalar, it will return a mixed tensor, specifically $delta_kappa^mu$, the Kronecker delta. In fact, $g_kappanug^munu=delta_kappa^mu$ is often taken as the definition of the inverse metric $g^munu$. The delta has components
$$
beginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
endbmatrix
.$$
That is, you only perform contraction on the repeated index $nu$, while the other two stay free. This is true for contracting tensors in general. So $g_kappanug^munu$ is not the same as $g_munug^munu$.
*EDIT
in fact this question has been answered at https://physics.stackexchange.com/questions/66394/kronecker-delta-confusion
answered Jul 25 at 2:04
Merk Zockerborg
1429
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The product of $g_kappanug^munu$ will not return a scalar, it will return a mixed tensor, specifically $delta_kappa^mu$, the Kronecker delta. In fact, $g_kappanug^munu=delta_kappa^mu$ is often taken as the definition of the inverse metric $g^munu$. The delta has components
$$
beginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
endbmatrix
.$$
That is, you only perform contraction on the repeated index $nu$, while the other two stay free. This is true for contracting tensors in general. So $g_kappanug^munu$ is not the same as $g_munug^munu$.
*EDIT
in fact this question has been answered at https://physics.stackexchange.com/questions/66394/kronecker-delta-confusion
answered Jul 25 at 2:04
Merk Zockerborg
1429
The product of $g_kappanug^munu$ will not return a scalar, it will return a mixed tensor, specifically $delta_kappa^mu$, the Kronecker delta. In fact, $g_kappanug^munu=delta_kappa^mu$ is often taken as the definition of the inverse metric $g^munu$. The delta has components
$$
beginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
endbmatrix
.$$
That is, you only perform contraction on the repeated index $nu$, while the other two stay free. This is true for contracting tensors in general. So $g_kappanug^munu$ is not the same as $g_munug^munu$.
*EDIT
in fact this question has been answered at https://physics.stackexchange.com/questions/66394/kronecker-delta-confusion
answered Jul 25 at 2:04
Merk Zockerborg
1429
edited Jul 25 at 2:10
answered Jul 25 at 2:04
Merk Zockerborg
1429
answered Jul 25 at 2:04
Merk Zockerborg
1429
answered Jul 25 at 2:04
Merk Zockerborg
1429
1429
add a comment |Â
add a comment |Â
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1
The indices $kappa$ and $mu$ are free in the expression $g_kappa nug^mu nu$. Free indices as a usual rule appear on both sides of an equation, the missing free indices in $1+(-1)^2+(-1)^2+(-1)^2$ is a dead give-away that your idea is off. The answer explains further...
– James S. Cook
Jul 25 at 2:25