Mixing
Confusion about the tensor product and matrix multiplication
$begingroup$
This is a question I came across looking at special relativity and tensor products.
For example, we have the metric tensor and its corresponding matrix representation
$$ g_{munu} = g^{munu} =begin{bmatrix}
1 & 0 & 0 & 0 \
0 & -1 & 0 & 0 \
0 & 0 & -1 & 0 \
0 & 0 & 0 & -1 \
end{bmatrix}$$
which looks the same in co- and contravariant form. Another example would be the Lorentz transformation along the $x$-axis given by:
$$ {Lambda^mu}_{nu} =begin{bmatrix}
gamma & -gammabeta & 0 & 0 \
-gammabeta & gamma & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
end{bmatrix}$$
and the electromagnetic field:
$$ F^{munu} = begin{bmatrix}
0 & E_x & E_y & E_z \
E_x & 0 & -B_z & B_y \
E_y & B_z & 0 & -B_x \
E_z & -B_y & B_x & 0
end{bmatrix} $$
If we want to calculate the form of $F^{munu}$ in a moving coordinate frame we have to transform the elelectromagnetic field tensor:
$$F'^{munu} = {Lambda^mu}_{alpha} {Lambda^nu}_{beta} F^{alphabeta}$$
This is sometimes calculated as a matrix product
$$ F' = Lambda F Lambda $$
My question is now whether there is a simple way to see this product from the above notation using the indices. I think the right matrix $Lambda$ should actually be a transpose (it does not matter for the given example). Is there a generalization for this?
matrices notation tensor-products
asked Jan 12 at 23:00
HerpDerpingtonHerpDerpington
1155
$endgroup$
add a comment |
$begingroup$
This is a question I came across looking at special relativity and tensor products.
For example, we have the metric tensor and its corresponding matrix representation
$$ g_{munu} = g^{munu} =begin{bmatrix}
1 & 0 & 0 & 0 \
0 & -1 & 0 & 0 \
0 & 0 & -1 & 0 \
0 & 0 & 0 & -1 \
end{bmatrix}$$
which looks the same in co- and contravariant form. Another example would be the Lorentz transformation along the $x$-axis given by:
$$ {Lambda^mu}_{nu} =begin{bmatrix}
gamma & -gammabeta & 0 & 0 \
-gammabeta & gamma & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
end{bmatrix}$$
and the electromagnetic field:
$$ F^{munu} = begin{bmatrix}
0 & E_x & E_y & E_z \
E_x & 0 & -B_z & B_y \
E_y & B_z & 0 & -B_x \
E_z & -B_y & B_x & 0
end{bmatrix} $$
If we want to calculate the form of $F^{munu}$ in a moving coordinate frame we have to transform the elelectromagnetic field tensor:
$$F'^{munu} = {Lambda^mu}_{alpha} {Lambda^nu}_{beta} F^{alphabeta}$$
This is sometimes calculated as a matrix product
$$ F' = Lambda F Lambda $$
My question is now whether there is a simple way to see this product from the above notation using the indices. I think the right matrix $Lambda$ should actually be a transpose (it does not matter for the given example). Is there a generalization for this?
matrices notation tensor-products
asked Jan 12 at 23:00
HerpDerpingtonHerpDerpington
1155
$endgroup$
2
$begingroup$
Have you tried writing out the implicit sums in that expression explicitly?
$endgroup$
– amd
Jan 12 at 23:44
add a comment |
$begingroup$
This is a question I came across looking at special relativity and tensor products.
For example, we have the metric tensor and its corresponding matrix representation
$$ g_{munu} = g^{munu} =begin{bmatrix}
1 & 0 & 0 & 0 \
0 & -1 & 0 & 0 \
0 & 0 & -1 & 0 \
0 & 0 & 0 & -1 \
end{bmatrix}$$
which looks the same in co- and contravariant form. Another example would be the Lorentz transformation along the $x$-axis given by:
$$ {Lambda^mu}_{nu} =begin{bmatrix}
gamma & -gammabeta & 0 & 0 \
-gammabeta & gamma & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
end{bmatrix}$$
and the electromagnetic field:
$$ F^{munu} = begin{bmatrix}
0 & E_x & E_y & E_z \
E_x & 0 & -B_z & B_y \
E_y & B_z & 0 & -B_x \
E_z & -B_y & B_x & 0
end{bmatrix} $$
If we want to calculate the form of $F^{munu}$ in a moving coordinate frame we have to transform the elelectromagnetic field tensor:
$$F'^{munu} = {Lambda^mu}_{alpha} {Lambda^nu}_{beta} F^{alphabeta}$$
This is sometimes calculated as a matrix product
$$ F' = Lambda F Lambda $$
My question is now whether there is a simple way to see this product from the above notation using the indices. I think the right matrix $Lambda$ should actually be a transpose (it does not matter for the given example). Is there a generalization for this?
matrices notation tensor-products
asked Jan 12 at 23:00
HerpDerpingtonHerpDerpington
1155
$endgroup$
This is a question I came across looking at special relativity and tensor products.
For example, we have the metric tensor and its corresponding matrix representation
$$ g_{munu} = g^{munu} =begin{bmatrix}
1 & 0 & 0 & 0 \
0 & -1 & 0 & 0 \
0 & 0 & -1 & 0 \
0 & 0 & 0 & -1 \
end{bmatrix}$$
which looks the same in co- and contravariant form. Another example would be the Lorentz transformation along the $x$-axis given by:
$$ {Lambda^mu}_{nu} =begin{bmatrix}
gamma & -gammabeta & 0 & 0 \
-gammabeta & gamma & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
end{bmatrix}$$
and the electromagnetic field:
$$ F^{munu} = begin{bmatrix}
0 & E_x & E_y & E_z \
E_x & 0 & -B_z & B_y \
E_y & B_z & 0 & -B_x \
E_z & -B_y & B_x & 0
end{bmatrix} $$
If we want to calculate the form of $F^{munu}$ in a moving coordinate frame we have to transform the elelectromagnetic field tensor:
$$F'^{munu} = {Lambda^mu}_{alpha} {Lambda^nu}_{beta} F^{alphabeta}$$
This is sometimes calculated as a matrix product
$$ F' = Lambda F Lambda $$
My question is now whether there is a simple way to see this product from the above notation using the indices. I think the right matrix $Lambda$ should actually be a transpose (it does not matter for the given example). Is there a generalization for this?
matrices notation tensor-products
matrices notation tensor-products
asked Jan 12 at 23:00
HerpDerpingtonHerpDerpington
1155
asked Jan 12 at 23:00
HerpDerpingtonHerpDerpington
1155
asked Jan 12 at 23:00
HerpDerpingtonHerpDerpington
1155
asked Jan 12 at 23:00
HerpDerpingtonHerpDerpington
1155
asked Jan 12 at 23:00
HerpDerpingtonHerpDerpington
1155
1155
2
$begingroup$
Have you tried writing out the implicit sums in that expression explicitly?
$endgroup$
– amd
Jan 12 at 23:44
add a comment |
2
$begingroup$
Have you tried writing out the implicit sums in that expression explicitly?
$endgroup$
– amd
Jan 12 at 23:44
2
2
$begingroup$
Have you tried writing out the implicit sums in that expression explicitly?
$endgroup$
– amd
Jan 12 at 23:44
$begingroup$
Have you tried writing out the implicit sums in that expression explicitly?
$endgroup$
– amd
Jan 12 at 23:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In index notation, matrix multiplication takes the following form:
$$
(AB)_{i}^j=A_{i}^kB_{k}^j,
$$
which for three matrices (relevant to the case at hand) looks like this:
$$
(ABC)_{i}^j=A_{i}^kB_{k}^lC_l^j.
$$
Thus, we may write
$$
(Lambda FLambda)_{mu}^{nu}=Lambda_mu^alpha F_alpha^betaLambda_beta^nu.
$$
We can rewrite this as
$$
Lambda_mu^alpha Lambda_beta^nu F_alpha^beta.
$$
Finally, note that $Lambda_mu^alpha F_alpha^beta = Lambda^mu_alpha F^{alphabeta}$ (since $Lambda$ is a symmetric matrix), giving the expression
$$
Lambda_alpha^mu Lambda_beta^nu F^{alphabeta}.
$$
answered Jan 13 at 3:14
pre-kidneypre-kidney
12.9k1748
$endgroup$
1
$begingroup$
This takes nicely care of the numeric values, but I can't help thinking there ought to be some notation to handle how the left index of $F$ is apparently being raised and lowered using $I=Delta(1,1,1,1)$ instead of $g=Delta(1,-1,-1,-1)$. Perhaps it would be clearer to stagger the indices?
$endgroup$
– Henning Makholm
Jan 13 at 3:35
add a comment |
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$begingroup$
In index notation, matrix multiplication takes the following form:
$$
(AB)_{i}^j=A_{i}^kB_{k}^j,
$$
which for three matrices (relevant to the case at hand) looks like this:
$$
(ABC)_{i}^j=A_{i}^kB_{k}^lC_l^j.
$$
Thus, we may write
$$
(Lambda FLambda)_{mu}^{nu}=Lambda_mu^alpha F_alpha^betaLambda_beta^nu.
$$
We can rewrite this as
$$
Lambda_mu^alpha Lambda_beta^nu F_alpha^beta.
$$
Finally, note that $Lambda_mu^alpha F_alpha^beta = Lambda^mu_alpha F^{alphabeta}$ (since $Lambda$ is a symmetric matrix), giving the expression
$$
Lambda_alpha^mu Lambda_beta^nu F^{alphabeta}.
$$
answered Jan 13 at 3:14
pre-kidneypre-kidney
12.9k1748
$endgroup$
1
$begingroup$
This takes nicely care of the numeric values, but I can't help thinking there ought to be some notation to handle how the left index of $F$ is apparently being raised and lowered using $I=Delta(1,1,1,1)$ instead of $g=Delta(1,-1,-1,-1)$. Perhaps it would be clearer to stagger the indices?
$endgroup$
– Henning Makholm
Jan 13 at 3:35
add a comment |
$begingroup$
In index notation, matrix multiplication takes the following form:
$$
(AB)_{i}^j=A_{i}^kB_{k}^j,
$$
which for three matrices (relevant to the case at hand) looks like this:
$$
(ABC)_{i}^j=A_{i}^kB_{k}^lC_l^j.
$$
Thus, we may write
$$
(Lambda FLambda)_{mu}^{nu}=Lambda_mu^alpha F_alpha^betaLambda_beta^nu.
$$
We can rewrite this as
$$
Lambda_mu^alpha Lambda_beta^nu F_alpha^beta.
$$
Finally, note that $Lambda_mu^alpha F_alpha^beta = Lambda^mu_alpha F^{alphabeta}$ (since $Lambda$ is a symmetric matrix), giving the expression
$$
Lambda_alpha^mu Lambda_beta^nu F^{alphabeta}.
$$
answered Jan 13 at 3:14
pre-kidneypre-kidney
12.9k1748
$endgroup$
1
$begingroup$
This takes nicely care of the numeric values, but I can't help thinking there ought to be some notation to handle how the left index of $F$ is apparently being raised and lowered using $I=Delta(1,1,1,1)$ instead of $g=Delta(1,-1,-1,-1)$. Perhaps it would be clearer to stagger the indices?
$endgroup$
– Henning Makholm
Jan 13 at 3:35
add a comment |
$begingroup$
In index notation, matrix multiplication takes the following form:
$$
(AB)_{i}^j=A_{i}^kB_{k}^j,
$$
which for three matrices (relevant to the case at hand) looks like this:
$$
(ABC)_{i}^j=A_{i}^kB_{k}^lC_l^j.
$$
Thus, we may write
$$
(Lambda FLambda)_{mu}^{nu}=Lambda_mu^alpha F_alpha^betaLambda_beta^nu.
$$
We can rewrite this as
$$
Lambda_mu^alpha Lambda_beta^nu F_alpha^beta.
$$
Finally, note that $Lambda_mu^alpha F_alpha^beta = Lambda^mu_alpha F^{alphabeta}$ (since $Lambda$ is a symmetric matrix), giving the expression
$$
Lambda_alpha^mu Lambda_beta^nu F^{alphabeta}.
$$
answered Jan 13 at 3:14
pre-kidneypre-kidney
12.9k1748
$endgroup$
In index notation, matrix multiplication takes the following form:
$$
(AB)_{i}^j=A_{i}^kB_{k}^j,
$$
which for three matrices (relevant to the case at hand) looks like this:
$$
(ABC)_{i}^j=A_{i}^kB_{k}^lC_l^j.
$$
Thus, we may write
$$
(Lambda FLambda)_{mu}^{nu}=Lambda_mu^alpha F_alpha^betaLambda_beta^nu.
$$
We can rewrite this as
$$
Lambda_mu^alpha Lambda_beta^nu F_alpha^beta.
$$
Finally, note that $Lambda_mu^alpha F_alpha^beta = Lambda^mu_alpha F^{alphabeta}$ (since $Lambda$ is a symmetric matrix), giving the expression
$$
Lambda_alpha^mu Lambda_beta^nu F^{alphabeta}.
$$
answered Jan 13 at 3:14
pre-kidneypre-kidney
12.9k1748
answered Jan 13 at 3:14
pre-kidneypre-kidney
12.9k1748
answered Jan 13 at 3:14
pre-kidneypre-kidney
12.9k1748
answered Jan 13 at 3:14
pre-kidneypre-kidney
12.9k1748
12.9k1748
1
$begingroup$
This takes nicely care of the numeric values, but I can't help thinking there ought to be some notation to handle how the left index of $F$ is apparently being raised and lowered using $I=Delta(1,1,1,1)$ instead of $g=Delta(1,-1,-1,-1)$. Perhaps it would be clearer to stagger the indices?
$endgroup$
– Henning Makholm
Jan 13 at 3:35
add a comment |
1
$begingroup$
This takes nicely care of the numeric values, but I can't help thinking there ought to be some notation to handle how the left index of $F$ is apparently being raised and lowered using $I=Delta(1,1,1,1)$ instead of $g=Delta(1,-1,-1,-1)$. Perhaps it would be clearer to stagger the indices?
$endgroup$
– Henning Makholm
Jan 13 at 3:35
1
1
$begingroup$
This takes nicely care of the numeric values, but I can't help thinking there ought to be some notation to handle how the left index of $F$ is apparently being raised and lowered using $I=Delta(1,1,1,1)$ instead of $g=Delta(1,-1,-1,-1)$. Perhaps it would be clearer to stagger the indices?
$endgroup$
– Henning Makholm
Jan 13 at 3:35
$begingroup$
This takes nicely care of the numeric values, but I can't help thinking there ought to be some notation to handle how the left index of $F$ is apparently being raised and lowered using $I=Delta(1,1,1,1)$ instead of $g=Delta(1,-1,-1,-1)$. Perhaps it would be clearer to stagger the indices?
$endgroup$
– Henning Makholm
Jan 13 at 3:35
add a comment |
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$begingroup$
Have you tried writing out the implicit sums in that expression explicitly?
$endgroup$
– amd
Jan 12 at 23:44