Mixing
Does $d x^mu=g^{munu} partial_nu$, or $partial_nu=g_{nu mu}dx^mu$?
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As the title suggests: does $d x^mu=g^{munu} partial_nu$, or $partial_nu=g_{nu mu}dx^mu$?
This is merely my own guess, from the material that I'm reading; so I'm not sure... The partial derivative operator is still an differential operator in my point of view.
This may be too basic for you, but thank you very much.
differential-geometry riemannian-geometry general-relativity
edited Jan 22 at 6:01
Ivo Terek
46.4k954142
asked Jan 22 at 5:09
CollinCollin
1447
$endgroup$
add a comment |
$begingroup$
As the title suggests: does $d x^mu=g^{munu} partial_nu$, or $partial_nu=g_{nu mu}dx^mu$?
This is merely my own guess, from the material that I'm reading; so I'm not sure... The partial derivative operator is still an differential operator in my point of view.
This may be too basic for you, but thank you very much.
differential-geometry riemannian-geometry general-relativity
edited Jan 22 at 6:01
Ivo Terek
46.4k954142
asked Jan 22 at 5:09
CollinCollin
1447
$endgroup$
add a comment |
$begingroup$
As the title suggests: does $d x^mu=g^{munu} partial_nu$, or $partial_nu=g_{nu mu}dx^mu$?
This is merely my own guess, from the material that I'm reading; so I'm not sure... The partial derivative operator is still an differential operator in my point of view.
This may be too basic for you, but thank you very much.
differential-geometry riemannian-geometry general-relativity
edited Jan 22 at 6:01
Ivo Terek
46.4k954142
asked Jan 22 at 5:09
CollinCollin
1447
$endgroup$
As the title suggests: does $d x^mu=g^{munu} partial_nu$, or $partial_nu=g_{nu mu}dx^mu$?
This is merely my own guess, from the material that I'm reading; so I'm not sure... The partial derivative operator is still an differential operator in my point of view.
This may be too basic for you, but thank you very much.
differential-geometry riemannian-geometry general-relativity
differential-geometry riemannian-geometry general-relativity
edited Jan 22 at 6:01
Ivo Terek
46.4k954142
asked Jan 22 at 5:09
CollinCollin
1447
edited Jan 22 at 6:01
Ivo Terek
46.4k954142
asked Jan 22 at 5:09
CollinCollin
1447
edited Jan 22 at 6:01
Ivo Terek
46.4k954142
edited Jan 22 at 6:01
Ivo Terek
46.4k954142
edited Jan 22 at 6:01
Ivo Terek
46.4k954142
46.4k954142
asked Jan 22 at 5:09
CollinCollin
1447
asked Jan 22 at 5:09
CollinCollin
1447
asked Jan 22 at 5:09
CollinCollin
1447
1447
add a comment |
add a comment |
2 Answers
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Not quite. Although the index balance is correct, in ${rm d}x^mu = g^{munu}partial_nu$ you have that the left side is a $1$-form, while the right side is a vector field. What happens here is that if $(M,g)$ is your pseudo-Riemannian manifold (which I assume that in your case should just be a spacetime), the metric induces the so called musical isomorphisms $flatcolon T_x M to T_x^*M$ and $sharp colon T^*_xM to T_xM$ for every point $x in M$. Said isomorphisms are characterized by $v_flat(w) = g_x(v,w)$ for $v, w in T_xM$, and also for $xi in T_x^*M$, the vector $xi^sharp in T_xM$ is characterized by the relation $xi(v) = g_x(v, xi^sharp)$, for all $v in T_xM$. You can put all of these isomorphisms together to get maps $flat : TM to T^*M$ and $sharpcolon T^*Mto TM$ between tangent bundles (which I'm still denoting by the same symbols), and then they can be "upgraded" to the section level, yielding isomorphisms $flatcolon mathfrak{X}(M) to Omega^1(M)$ and $sharpcolon Omega^1(M) to mathfrak{X}(M)$ between vector fields and $1$-forms. What you actually have for a given coordinate system are the formulas $${rm d}x^mu = g^{munu} (partial_nu)_flat quadmbox{and}quad partial_mu = g_{mu nu}({rm d}x^nu)^sharp.$$What happens is that people just sweep these isomorphisms under the rug and write just the formulas you have seen in your textbook. For example, to prove the first one, you just have to check that both sides act the same in an arbitrary coordinate vector field $partial_lambda$, say. Indeed, we have $$g^{munu}(partial_nu)_flat(partial_lambda) = g^{munu}g(partial_nu,partial_lambda) = g^{munu}g_{nulambda} = delta^mu_{;lambda} = {rm d}x^mu(partial_lambda),$$as wanted. The second one follows from the first one (up to index relabeling) by doing $${rm d}x^mu = g^{munu}(partial_nu)_flat implies g_{lambda mu}{rm d}x^mu = g_{lambda mu}g^{munu}(partial_nu)_flat = delta^nu_{lambda}(partial_nu)_flat = (partial_lambda)_flat implies partial_lambda = (g_{lambdamu}{rm d}x^mu)^sharp = g_{lambdamu}({rm d}x^mu)^sharp.$$(Relabel $mu to nu$ and $lambda to mu$ in this order if it makes you more comfortable in the last equality above.)
answered Jan 22 at 5:59
Ivo TerekIvo Terek
46.4k954142
$endgroup$
$begingroup$
Can you see the example at the end of my answer? I suspect that $(dx^mu)_b=partial_{x^mu}$ requires the metric to be the identity matrix. Thanks!
$endgroup$
– Giuseppe Negro
Jan 22 at 15:40
1
$begingroup$
It definitely requires $g_{munu}=delta_{munu}$. The reason is why coordinate systems can be completely arbitrary while the isomorphisms depend on the given metric. We cannot expect the isomorphisms to behave well with respect to an arbitrary chart without any geometrical relation to the metric.
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– Ivo Terek
Jan 22 at 18:07
add a comment |
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SUMMARY. This is formally wrong, as Ivo already showed. If the metric is $delta_{mu nu}dx^muotimes dx^nu$, then it is morally correct. Otherwise, it is not; see the example at the end of this post.
My understanding of these things is that you can raise or lower indices in the coordinates of a tensor, not in the tensor itself. To explain what I mean, let $mu_0in {1, 2, ldots n}$ be fixed. The 1-form $dx^{mu_0}$ is the tensor whose coordinates in the basis $dx^1, dx^2, ldots, dx^n$ are
$$
(delta^{mu_0}{}_{nu} : nu=1, 2, ldots, n), $$
and we denote them by $delta^{mu_0}{}_{nu}$. (This $delta$ is the Kronecker symbol, which equals $1$ if both indices agree and $0$ otherwise. The spacing in the indices is not important).
Dually, the vector field $frac{partial}{partial x^{mu_0}}$ is the tensor whose coordinates are $delta_{mu_0}{}^nu$. And this is the end of the story unless we introduce a Riemannian metric.
If we introduce a such metric $g_{munu}$ (or $g_{munu}dx^muotimes dx^nu$, if you prefer the extended version), then we can raise or lower indices by contracting with this metric, which is another definition of the "musical isomorphisms" of Ivo's answer. In the case of the tensor we introduced previously, if $g_{munu}=delta_{munu}$, then raising the second index in the first tensor we obtain the second;
$$
delta^{mu_0}{}_nu delta^{nu rho}=delta^{mu_0rho}. $$
WARNING! If the metric is not $delta_{mu nu}dx^muotimes dx^nu$, then the above fails, and (in the language of Ivo's answer) it is not true that
$$
left(frac{partial}{partial x^mu}right)_b =dx^mu, quad frac{partial}{partial x^mu}=(dx^mu)^{#}.$$
Let me make an explicit example: parametrize $mathbb S^2$ (minus a semicircle) as
$$
(cos theta, sin thetacos phi), qquad text{where }thetain(0, pi),phiin(0, 2pi).$$
Then the metric tensor of $mathbb S^2$ is
$$
g = dtheta^2 + sin^2phi dphi^2,$$
hence the matrix $g_{mu nu}$ is
$$
begin{bmatrix} 1 & 0 \ 0 & sin^2phiend{bmatrix}. $$
In particular, $g_{mu nu}$ is not the Kronecker symbol.
Now, let us compute $(dphi)_b$. The components of this tensor, in the basis $dtheta, dphi$, are $(0, 1)$. We should contract then with the metric, which amounts to compute
$$
(0, 1)begin{bmatrix} 1 & 0 \ 0 & sin^2phiend{bmatrix} = (0, sin^2phi).$$
Hence
$$
(dphi)_b= sin^2phi frac{partial}{partial phi}.$$
answered Jan 22 at 11:18
Giuseppe NegroGiuseppe Negro
17.3k331126
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I would say that it is morally correct even if the metric matrix is not the identity, the only thing making it formally wrong being the omission of $flat$ and $sharp$.
$endgroup$
– Ivo Terek
Jan 22 at 18:09
add a comment |
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$begingroup$
Not quite. Although the index balance is correct, in ${rm d}x^mu = g^{munu}partial_nu$ you have that the left side is a $1$-form, while the right side is a vector field. What happens here is that if $(M,g)$ is your pseudo-Riemannian manifold (which I assume that in your case should just be a spacetime), the metric induces the so called musical isomorphisms $flatcolon T_x M to T_x^*M$ and $sharp colon T^*_xM to T_xM$ for every point $x in M$. Said isomorphisms are characterized by $v_flat(w) = g_x(v,w)$ for $v, w in T_xM$, and also for $xi in T_x^*M$, the vector $xi^sharp in T_xM$ is characterized by the relation $xi(v) = g_x(v, xi^sharp)$, for all $v in T_xM$. You can put all of these isomorphisms together to get maps $flat : TM to T^*M$ and $sharpcolon T^*Mto TM$ between tangent bundles (which I'm still denoting by the same symbols), and then they can be "upgraded" to the section level, yielding isomorphisms $flatcolon mathfrak{X}(M) to Omega^1(M)$ and $sharpcolon Omega^1(M) to mathfrak{X}(M)$ between vector fields and $1$-forms. What you actually have for a given coordinate system are the formulas $${rm d}x^mu = g^{munu} (partial_nu)_flat quadmbox{and}quad partial_mu = g_{mu nu}({rm d}x^nu)^sharp.$$What happens is that people just sweep these isomorphisms under the rug and write just the formulas you have seen in your textbook. For example, to prove the first one, you just have to check that both sides act the same in an arbitrary coordinate vector field $partial_lambda$, say. Indeed, we have $$g^{munu}(partial_nu)_flat(partial_lambda) = g^{munu}g(partial_nu,partial_lambda) = g^{munu}g_{nulambda} = delta^mu_{;lambda} = {rm d}x^mu(partial_lambda),$$as wanted. The second one follows from the first one (up to index relabeling) by doing $${rm d}x^mu = g^{munu}(partial_nu)_flat implies g_{lambda mu}{rm d}x^mu = g_{lambda mu}g^{munu}(partial_nu)_flat = delta^nu_{lambda}(partial_nu)_flat = (partial_lambda)_flat implies partial_lambda = (g_{lambdamu}{rm d}x^mu)^sharp = g_{lambdamu}({rm d}x^mu)^sharp.$$(Relabel $mu to nu$ and $lambda to mu$ in this order if it makes you more comfortable in the last equality above.)
answered Jan 22 at 5:59
Ivo TerekIvo Terek
46.4k954142
$endgroup$
$begingroup$
Can you see the example at the end of my answer? I suspect that $(dx^mu)_b=partial_{x^mu}$ requires the metric to be the identity matrix. Thanks!
$endgroup$
– Giuseppe Negro
Jan 22 at 15:40
1
$begingroup$
It definitely requires $g_{munu}=delta_{munu}$. The reason is why coordinate systems can be completely arbitrary while the isomorphisms depend on the given metric. We cannot expect the isomorphisms to behave well with respect to an arbitrary chart without any geometrical relation to the metric.
$endgroup$
– Ivo Terek
Jan 22 at 18:07
add a comment |
$begingroup$
Not quite. Although the index balance is correct, in ${rm d}x^mu = g^{munu}partial_nu$ you have that the left side is a $1$-form, while the right side is a vector field. What happens here is that if $(M,g)$ is your pseudo-Riemannian manifold (which I assume that in your case should just be a spacetime), the metric induces the so called musical isomorphisms $flatcolon T_x M to T_x^*M$ and $sharp colon T^*_xM to T_xM$ for every point $x in M$. Said isomorphisms are characterized by $v_flat(w) = g_x(v,w)$ for $v, w in T_xM$, and also for $xi in T_x^*M$, the vector $xi^sharp in T_xM$ is characterized by the relation $xi(v) = g_x(v, xi^sharp)$, for all $v in T_xM$. You can put all of these isomorphisms together to get maps $flat : TM to T^*M$ and $sharpcolon T^*Mto TM$ between tangent bundles (which I'm still denoting by the same symbols), and then they can be "upgraded" to the section level, yielding isomorphisms $flatcolon mathfrak{X}(M) to Omega^1(M)$ and $sharpcolon Omega^1(M) to mathfrak{X}(M)$ between vector fields and $1$-forms. What you actually have for a given coordinate system are the formulas $${rm d}x^mu = g^{munu} (partial_nu)_flat quadmbox{and}quad partial_mu = g_{mu nu}({rm d}x^nu)^sharp.$$What happens is that people just sweep these isomorphisms under the rug and write just the formulas you have seen in your textbook. For example, to prove the first one, you just have to check that both sides act the same in an arbitrary coordinate vector field $partial_lambda$, say. Indeed, we have $$g^{munu}(partial_nu)_flat(partial_lambda) = g^{munu}g(partial_nu,partial_lambda) = g^{munu}g_{nulambda} = delta^mu_{;lambda} = {rm d}x^mu(partial_lambda),$$as wanted. The second one follows from the first one (up to index relabeling) by doing $${rm d}x^mu = g^{munu}(partial_nu)_flat implies g_{lambda mu}{rm d}x^mu = g_{lambda mu}g^{munu}(partial_nu)_flat = delta^nu_{lambda}(partial_nu)_flat = (partial_lambda)_flat implies partial_lambda = (g_{lambdamu}{rm d}x^mu)^sharp = g_{lambdamu}({rm d}x^mu)^sharp.$$(Relabel $mu to nu$ and $lambda to mu$ in this order if it makes you more comfortable in the last equality above.)
answered Jan 22 at 5:59
Ivo TerekIvo Terek
46.4k954142
$endgroup$
$begingroup$
Can you see the example at the end of my answer? I suspect that $(dx^mu)_b=partial_{x^mu}$ requires the metric to be the identity matrix. Thanks!
$endgroup$
– Giuseppe Negro
Jan 22 at 15:40
1
$begingroup$
It definitely requires $g_{munu}=delta_{munu}$. The reason is why coordinate systems can be completely arbitrary while the isomorphisms depend on the given metric. We cannot expect the isomorphisms to behave well with respect to an arbitrary chart without any geometrical relation to the metric.
$endgroup$
– Ivo Terek
Jan 22 at 18:07
add a comment |
$begingroup$
Not quite. Although the index balance is correct, in ${rm d}x^mu = g^{munu}partial_nu$ you have that the left side is a $1$-form, while the right side is a vector field. What happens here is that if $(M,g)$ is your pseudo-Riemannian manifold (which I assume that in your case should just be a spacetime), the metric induces the so called musical isomorphisms $flatcolon T_x M to T_x^*M$ and $sharp colon T^*_xM to T_xM$ for every point $x in M$. Said isomorphisms are characterized by $v_flat(w) = g_x(v,w)$ for $v, w in T_xM$, and also for $xi in T_x^*M$, the vector $xi^sharp in T_xM$ is characterized by the relation $xi(v) = g_x(v, xi^sharp)$, for all $v in T_xM$. You can put all of these isomorphisms together to get maps $flat : TM to T^*M$ and $sharpcolon T^*Mto TM$ between tangent bundles (which I'm still denoting by the same symbols), and then they can be "upgraded" to the section level, yielding isomorphisms $flatcolon mathfrak{X}(M) to Omega^1(M)$ and $sharpcolon Omega^1(M) to mathfrak{X}(M)$ between vector fields and $1$-forms. What you actually have for a given coordinate system are the formulas $${rm d}x^mu = g^{munu} (partial_nu)_flat quadmbox{and}quad partial_mu = g_{mu nu}({rm d}x^nu)^sharp.$$What happens is that people just sweep these isomorphisms under the rug and write just the formulas you have seen in your textbook. For example, to prove the first one, you just have to check that both sides act the same in an arbitrary coordinate vector field $partial_lambda$, say. Indeed, we have $$g^{munu}(partial_nu)_flat(partial_lambda) = g^{munu}g(partial_nu,partial_lambda) = g^{munu}g_{nulambda} = delta^mu_{;lambda} = {rm d}x^mu(partial_lambda),$$as wanted. The second one follows from the first one (up to index relabeling) by doing $${rm d}x^mu = g^{munu}(partial_nu)_flat implies g_{lambda mu}{rm d}x^mu = g_{lambda mu}g^{munu}(partial_nu)_flat = delta^nu_{lambda}(partial_nu)_flat = (partial_lambda)_flat implies partial_lambda = (g_{lambdamu}{rm d}x^mu)^sharp = g_{lambdamu}({rm d}x^mu)^sharp.$$(Relabel $mu to nu$ and $lambda to mu$ in this order if it makes you more comfortable in the last equality above.)
answered Jan 22 at 5:59
Ivo TerekIvo Terek
46.4k954142
$endgroup$
Not quite. Although the index balance is correct, in ${rm d}x^mu = g^{munu}partial_nu$ you have that the left side is a $1$-form, while the right side is a vector field. What happens here is that if $(M,g)$ is your pseudo-Riemannian manifold (which I assume that in your case should just be a spacetime), the metric induces the so called musical isomorphisms $flatcolon T_x M to T_x^*M$ and $sharp colon T^*_xM to T_xM$ for every point $x in M$. Said isomorphisms are characterized by $v_flat(w) = g_x(v,w)$ for $v, w in T_xM$, and also for $xi in T_x^*M$, the vector $xi^sharp in T_xM$ is characterized by the relation $xi(v) = g_x(v, xi^sharp)$, for all $v in T_xM$. You can put all of these isomorphisms together to get maps $flat : TM to T^*M$ and $sharpcolon T^*Mto TM$ between tangent bundles (which I'm still denoting by the same symbols), and then they can be "upgraded" to the section level, yielding isomorphisms $flatcolon mathfrak{X}(M) to Omega^1(M)$ and $sharpcolon Omega^1(M) to mathfrak{X}(M)$ between vector fields and $1$-forms. What you actually have for a given coordinate system are the formulas $${rm d}x^mu = g^{munu} (partial_nu)_flat quadmbox{and}quad partial_mu = g_{mu nu}({rm d}x^nu)^sharp.$$What happens is that people just sweep these isomorphisms under the rug and write just the formulas you have seen in your textbook. For example, to prove the first one, you just have to check that both sides act the same in an arbitrary coordinate vector field $partial_lambda$, say. Indeed, we have $$g^{munu}(partial_nu)_flat(partial_lambda) = g^{munu}g(partial_nu,partial_lambda) = g^{munu}g_{nulambda} = delta^mu_{;lambda} = {rm d}x^mu(partial_lambda),$$as wanted. The second one follows from the first one (up to index relabeling) by doing $${rm d}x^mu = g^{munu}(partial_nu)_flat implies g_{lambda mu}{rm d}x^mu = g_{lambda mu}g^{munu}(partial_nu)_flat = delta^nu_{lambda}(partial_nu)_flat = (partial_lambda)_flat implies partial_lambda = (g_{lambdamu}{rm d}x^mu)^sharp = g_{lambdamu}({rm d}x^mu)^sharp.$$(Relabel $mu to nu$ and $lambda to mu$ in this order if it makes you more comfortable in the last equality above.)
answered Jan 22 at 5:59
Ivo TerekIvo Terek
46.4k954142
answered Jan 22 at 5:59
Ivo TerekIvo Terek
46.4k954142
answered Jan 22 at 5:59
Ivo TerekIvo Terek
46.4k954142
answered Jan 22 at 5:59
Ivo TerekIvo Terek
46.4k954142
46.4k954142
$begingroup$
Can you see the example at the end of my answer? I suspect that $(dx^mu)_b=partial_{x^mu}$ requires the metric to be the identity matrix. Thanks!
$endgroup$
– Giuseppe Negro
Jan 22 at 15:40
1
$begingroup$
It definitely requires $g_{munu}=delta_{munu}$. The reason is why coordinate systems can be completely arbitrary while the isomorphisms depend on the given metric. We cannot expect the isomorphisms to behave well with respect to an arbitrary chart without any geometrical relation to the metric.
$endgroup$
– Ivo Terek
Jan 22 at 18:07
add a comment |
$begingroup$
Can you see the example at the end of my answer? I suspect that $(dx^mu)_b=partial_{x^mu}$ requires the metric to be the identity matrix. Thanks!
$endgroup$
– Giuseppe Negro
Jan 22 at 15:40
1
$begingroup$
It definitely requires $g_{munu}=delta_{munu}$. The reason is why coordinate systems can be completely arbitrary while the isomorphisms depend on the given metric. We cannot expect the isomorphisms to behave well with respect to an arbitrary chart without any geometrical relation to the metric.
$endgroup$
– Ivo Terek
Jan 22 at 18:07
$begingroup$
Can you see the example at the end of my answer? I suspect that $(dx^mu)_b=partial_{x^mu}$ requires the metric to be the identity matrix. Thanks!
$endgroup$
– Giuseppe Negro
Jan 22 at 15:40
$begingroup$
Can you see the example at the end of my answer? I suspect that $(dx^mu)_b=partial_{x^mu}$ requires the metric to be the identity matrix. Thanks!
$endgroup$
– Giuseppe Negro
Jan 22 at 15:40
1
1
$begingroup$
It definitely requires $g_{munu}=delta_{munu}$. The reason is why coordinate systems can be completely arbitrary while the isomorphisms depend on the given metric. We cannot expect the isomorphisms to behave well with respect to an arbitrary chart without any geometrical relation to the metric.
$endgroup$
– Ivo Terek
Jan 22 at 18:07
$begingroup$
It definitely requires $g_{munu}=delta_{munu}$. The reason is why coordinate systems can be completely arbitrary while the isomorphisms depend on the given metric. We cannot expect the isomorphisms to behave well with respect to an arbitrary chart without any geometrical relation to the metric.
$endgroup$
– Ivo Terek
Jan 22 at 18:07
add a comment |
$begingroup$
SUMMARY. This is formally wrong, as Ivo already showed. If the metric is $delta_{mu nu}dx^muotimes dx^nu$, then it is morally correct. Otherwise, it is not; see the example at the end of this post.
My understanding of these things is that you can raise or lower indices in the coordinates of a tensor, not in the tensor itself. To explain what I mean, let $mu_0in {1, 2, ldots n}$ be fixed. The 1-form $dx^{mu_0}$ is the tensor whose coordinates in the basis $dx^1, dx^2, ldots, dx^n$ are
$$
(delta^{mu_0}{}_{nu} : nu=1, 2, ldots, n), $$
and we denote them by $delta^{mu_0}{}_{nu}$. (This $delta$ is the Kronecker symbol, which equals $1$ if both indices agree and $0$ otherwise. The spacing in the indices is not important).
Dually, the vector field $frac{partial}{partial x^{mu_0}}$ is the tensor whose coordinates are $delta_{mu_0}{}^nu$. And this is the end of the story unless we introduce a Riemannian metric.
If we introduce a such metric $g_{munu}$ (or $g_{munu}dx^muotimes dx^nu$, if you prefer the extended version), then we can raise or lower indices by contracting with this metric, which is another definition of the "musical isomorphisms" of Ivo's answer. In the case of the tensor we introduced previously, if $g_{munu}=delta_{munu}$, then raising the second index in the first tensor we obtain the second;
$$
delta^{mu_0}{}_nu delta^{nu rho}=delta^{mu_0rho}. $$
WARNING! If the metric is not $delta_{mu nu}dx^muotimes dx^nu$, then the above fails, and (in the language of Ivo's answer) it is not true that
$$
left(frac{partial}{partial x^mu}right)_b =dx^mu, quad frac{partial}{partial x^mu}=(dx^mu)^{#}.$$
Let me make an explicit example: parametrize $mathbb S^2$ (minus a semicircle) as
$$
(cos theta, sin thetacos phi), qquad text{where }thetain(0, pi),phiin(0, 2pi).$$
Then the metric tensor of $mathbb S^2$ is
$$
g = dtheta^2 + sin^2phi dphi^2,$$
hence the matrix $g_{mu nu}$ is
$$
begin{bmatrix} 1 & 0 \ 0 & sin^2phiend{bmatrix}. $$
In particular, $g_{mu nu}$ is not the Kronecker symbol.
Now, let us compute $(dphi)_b$. The components of this tensor, in the basis $dtheta, dphi$, are $(0, 1)$. We should contract then with the metric, which amounts to compute
$$
(0, 1)begin{bmatrix} 1 & 0 \ 0 & sin^2phiend{bmatrix} = (0, sin^2phi).$$
Hence
$$
(dphi)_b= sin^2phi frac{partial}{partial phi}.$$
answered Jan 22 at 11:18
Giuseppe NegroGiuseppe Negro
17.3k331126
$endgroup$
$begingroup$
I would say that it is morally correct even if the metric matrix is not the identity, the only thing making it formally wrong being the omission of $flat$ and $sharp$.
$endgroup$
– Ivo Terek
Jan 22 at 18:09
add a comment |
$begingroup$
SUMMARY. This is formally wrong, as Ivo already showed. If the metric is $delta_{mu nu}dx^muotimes dx^nu$, then it is morally correct. Otherwise, it is not; see the example at the end of this post.
My understanding of these things is that you can raise or lower indices in the coordinates of a tensor, not in the tensor itself. To explain what I mean, let $mu_0in {1, 2, ldots n}$ be fixed. The 1-form $dx^{mu_0}$ is the tensor whose coordinates in the basis $dx^1, dx^2, ldots, dx^n$ are
$$
(delta^{mu_0}{}_{nu} : nu=1, 2, ldots, n), $$
and we denote them by $delta^{mu_0}{}_{nu}$. (This $delta$ is the Kronecker symbol, which equals $1$ if both indices agree and $0$ otherwise. The spacing in the indices is not important).
Dually, the vector field $frac{partial}{partial x^{mu_0}}$ is the tensor whose coordinates are $delta_{mu_0}{}^nu$. And this is the end of the story unless we introduce a Riemannian metric.
If we introduce a such metric $g_{munu}$ (or $g_{munu}dx^muotimes dx^nu$, if you prefer the extended version), then we can raise or lower indices by contracting with this metric, which is another definition of the "musical isomorphisms" of Ivo's answer. In the case of the tensor we introduced previously, if $g_{munu}=delta_{munu}$, then raising the second index in the first tensor we obtain the second;
$$
delta^{mu_0}{}_nu delta^{nu rho}=delta^{mu_0rho}. $$
WARNING! If the metric is not $delta_{mu nu}dx^muotimes dx^nu$, then the above fails, and (in the language of Ivo's answer) it is not true that
$$
left(frac{partial}{partial x^mu}right)_b =dx^mu, quad frac{partial}{partial x^mu}=(dx^mu)^{#}.$$
Let me make an explicit example: parametrize $mathbb S^2$ (minus a semicircle) as
$$
(cos theta, sin thetacos phi), qquad text{where }thetain(0, pi),phiin(0, 2pi).$$
Then the metric tensor of $mathbb S^2$ is
$$
g = dtheta^2 + sin^2phi dphi^2,$$
hence the matrix $g_{mu nu}$ is
$$
begin{bmatrix} 1 & 0 \ 0 & sin^2phiend{bmatrix}. $$
In particular, $g_{mu nu}$ is not the Kronecker symbol.
Now, let us compute $(dphi)_b$. The components of this tensor, in the basis $dtheta, dphi$, are $(0, 1)$. We should contract then with the metric, which amounts to compute
$$
(0, 1)begin{bmatrix} 1 & 0 \ 0 & sin^2phiend{bmatrix} = (0, sin^2phi).$$
Hence
$$
(dphi)_b= sin^2phi frac{partial}{partial phi}.$$
answered Jan 22 at 11:18
Giuseppe NegroGiuseppe Negro
17.3k331126
$endgroup$
$begingroup$
I would say that it is morally correct even if the metric matrix is not the identity, the only thing making it formally wrong being the omission of $flat$ and $sharp$.
$endgroup$
– Ivo Terek
Jan 22 at 18:09
add a comment |
$begingroup$
SUMMARY. This is formally wrong, as Ivo already showed. If the metric is $delta_{mu nu}dx^muotimes dx^nu$, then it is morally correct. Otherwise, it is not; see the example at the end of this post.
My understanding of these things is that you can raise or lower indices in the coordinates of a tensor, not in the tensor itself. To explain what I mean, let $mu_0in {1, 2, ldots n}$ be fixed. The 1-form $dx^{mu_0}$ is the tensor whose coordinates in the basis $dx^1, dx^2, ldots, dx^n$ are
$$
(delta^{mu_0}{}_{nu} : nu=1, 2, ldots, n), $$
and we denote them by $delta^{mu_0}{}_{nu}$. (This $delta$ is the Kronecker symbol, which equals $1$ if both indices agree and $0$ otherwise. The spacing in the indices is not important).
Dually, the vector field $frac{partial}{partial x^{mu_0}}$ is the tensor whose coordinates are $delta_{mu_0}{}^nu$. And this is the end of the story unless we introduce a Riemannian metric.
If we introduce a such metric $g_{munu}$ (or $g_{munu}dx^muotimes dx^nu$, if you prefer the extended version), then we can raise or lower indices by contracting with this metric, which is another definition of the "musical isomorphisms" of Ivo's answer. In the case of the tensor we introduced previously, if $g_{munu}=delta_{munu}$, then raising the second index in the first tensor we obtain the second;
$$
delta^{mu_0}{}_nu delta^{nu rho}=delta^{mu_0rho}. $$
WARNING! If the metric is not $delta_{mu nu}dx^muotimes dx^nu$, then the above fails, and (in the language of Ivo's answer) it is not true that
$$
left(frac{partial}{partial x^mu}right)_b =dx^mu, quad frac{partial}{partial x^mu}=(dx^mu)^{#}.$$
Let me make an explicit example: parametrize $mathbb S^2$ (minus a semicircle) as
$$
(cos theta, sin thetacos phi), qquad text{where }thetain(0, pi),phiin(0, 2pi).$$
Then the metric tensor of $mathbb S^2$ is
$$
g = dtheta^2 + sin^2phi dphi^2,$$
hence the matrix $g_{mu nu}$ is
$$
begin{bmatrix} 1 & 0 \ 0 & sin^2phiend{bmatrix}. $$
In particular, $g_{mu nu}$ is not the Kronecker symbol.
Now, let us compute $(dphi)_b$. The components of this tensor, in the basis $dtheta, dphi$, are $(0, 1)$. We should contract then with the metric, which amounts to compute
$$
(0, 1)begin{bmatrix} 1 & 0 \ 0 & sin^2phiend{bmatrix} = (0, sin^2phi).$$
Hence
$$
(dphi)_b= sin^2phi frac{partial}{partial phi}.$$
answered Jan 22 at 11:18
Giuseppe NegroGiuseppe Negro
17.3k331126
$endgroup$
SUMMARY. This is formally wrong, as Ivo already showed. If the metric is $delta_{mu nu}dx^muotimes dx^nu$, then it is morally correct. Otherwise, it is not; see the example at the end of this post.
My understanding of these things is that you can raise or lower indices in the coordinates of a tensor, not in the tensor itself. To explain what I mean, let $mu_0in {1, 2, ldots n}$ be fixed. The 1-form $dx^{mu_0}$ is the tensor whose coordinates in the basis $dx^1, dx^2, ldots, dx^n$ are
$$
(delta^{mu_0}{}_{nu} : nu=1, 2, ldots, n), $$
and we denote them by $delta^{mu_0}{}_{nu}$. (This $delta$ is the Kronecker symbol, which equals $1$ if both indices agree and $0$ otherwise. The spacing in the indices is not important).
Dually, the vector field $frac{partial}{partial x^{mu_0}}$ is the tensor whose coordinates are $delta_{mu_0}{}^nu$. And this is the end of the story unless we introduce a Riemannian metric.
If we introduce a such metric $g_{munu}$ (or $g_{munu}dx^muotimes dx^nu$, if you prefer the extended version), then we can raise or lower indices by contracting with this metric, which is another definition of the "musical isomorphisms" of Ivo's answer. In the case of the tensor we introduced previously, if $g_{munu}=delta_{munu}$, then raising the second index in the first tensor we obtain the second;
$$
delta^{mu_0}{}_nu delta^{nu rho}=delta^{mu_0rho}. $$
WARNING! If the metric is not $delta_{mu nu}dx^muotimes dx^nu$, then the above fails, and (in the language of Ivo's answer) it is not true that
$$
left(frac{partial}{partial x^mu}right)_b =dx^mu, quad frac{partial}{partial x^mu}=(dx^mu)^{#}.$$
Let me make an explicit example: parametrize $mathbb S^2$ (minus a semicircle) as
$$
(cos theta, sin thetacos phi), qquad text{where }thetain(0, pi),phiin(0, 2pi).$$
Then the metric tensor of $mathbb S^2$ is
$$
g = dtheta^2 + sin^2phi dphi^2,$$
hence the matrix $g_{mu nu}$ is
$$
begin{bmatrix} 1 & 0 \ 0 & sin^2phiend{bmatrix}. $$
In particular, $g_{mu nu}$ is not the Kronecker symbol.
Now, let us compute $(dphi)_b$. The components of this tensor, in the basis $dtheta, dphi$, are $(0, 1)$. We should contract then with the metric, which amounts to compute
$$
(0, 1)begin{bmatrix} 1 & 0 \ 0 & sin^2phiend{bmatrix} = (0, sin^2phi).$$
Hence
$$
(dphi)_b= sin^2phi frac{partial}{partial phi}.$$
answered Jan 22 at 11:18
Giuseppe NegroGiuseppe Negro
17.3k331126
edited Jan 22 at 15:39
answered Jan 22 at 11:18
Giuseppe NegroGiuseppe Negro
17.3k331126
answered Jan 22 at 11:18
Giuseppe NegroGiuseppe Negro
17.3k331126
answered Jan 22 at 11:18
Giuseppe NegroGiuseppe Negro
17.3k331126
17.3k331126
$begingroup$
I would say that it is morally correct even if the metric matrix is not the identity, the only thing making it formally wrong being the omission of $flat$ and $sharp$.
$endgroup$
– Ivo Terek
Jan 22 at 18:09
add a comment |
$begingroup$
I would say that it is morally correct even if the metric matrix is not the identity, the only thing making it formally wrong being the omission of $flat$ and $sharp$.
$endgroup$
– Ivo Terek
Jan 22 at 18:09
$begingroup$
I would say that it is morally correct even if the metric matrix is not the identity, the only thing making it formally wrong being the omission of $flat$ and $sharp$.
$endgroup$
– Ivo Terek
Jan 22 at 18:09
$begingroup$
I would say that it is morally correct even if the metric matrix is not the identity, the only thing making it formally wrong being the omission of $flat$ and $sharp$.
$endgroup$
– Ivo Terek
Jan 22 at 18:09
add a comment |
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