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When does Del x B =0?
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In my lab, I use electromagnets to apply a magnetic gradient force to lots of very small (superparamagnetic) nanoparticles embedded in an elastic medium. I believe that these can be treated as magnetic dipoles, with a dipole moment $mathbfm$.
It is well known that the force on a magnetic dipole moment in a magnetic field is given by
$$mathbfF = mathbfnabla(mathbfm cdot mathbfB).$$
I need to prove to myself that this can be reduced to
$$mathbfF = (mathbfm cdot mathbfnabla)mathbfB.$$
I know that we can rewrite the first equation using one of those vector calculus identities that appears, e.g. on the inside covers of Jackson:
$$mathbfF = mathbfnabla(mathbfm cdot mathbfB) = (mathbfm cdot mathbfnabla)mathbfB + (mathbfB cdot mathbfnabla)mathbfm + mathbfm times (mathbfnabla times mathbfB) + mathbfB times (mathbfnabla times mathbfm).$$
- The first term is good -- it can stay!
- For the second term, can I use the commutative property of the dot product to say that $mathbfB cdot mathbfnabla = mathbfnabla cdot mathbfB = 0$ because magnetic monopoles don't exist?
- On page 374 of Andrew Zangwill's Modern Electrodynamics (2013), he writes, "When the sources of $mathbfB$ are far away so $mathbfnabla times mathbfB = 0$, blah blah blah." This is the thing I'm most confused about. How do we know that $mathbfnabla times mathbfB = 0$? Can someone show me a proof and/or help me understand what the "far away" criterion means in real life? (Far away relative to what?)
- I think the fourth term is simple -- since a dipole moment is just 1 vector, the curl is always zero.
electromagnetism magnetic-fields dipole-moment
asked 3 hours ago
Bunji
538
add a comment |Â
up vote
4
down vote
favorite
In my lab, I use electromagnets to apply a magnetic gradient force to lots of very small (superparamagnetic) nanoparticles embedded in an elastic medium. I believe that these can be treated as magnetic dipoles, with a dipole moment $mathbfm$.
It is well known that the force on a magnetic dipole moment in a magnetic field is given by
$$mathbfF = mathbfnabla(mathbfm cdot mathbfB).$$
I need to prove to myself that this can be reduced to
$$mathbfF = (mathbfm cdot mathbfnabla)mathbfB.$$
I know that we can rewrite the first equation using one of those vector calculus identities that appears, e.g. on the inside covers of Jackson:
$$mathbfF = mathbfnabla(mathbfm cdot mathbfB) = (mathbfm cdot mathbfnabla)mathbfB + (mathbfB cdot mathbfnabla)mathbfm + mathbfm times (mathbfnabla times mathbfB) + mathbfB times (mathbfnabla times mathbfm).$$
- The first term is good -- it can stay!
- For the second term, can I use the commutative property of the dot product to say that $mathbfB cdot mathbfnabla = mathbfnabla cdot mathbfB = 0$ because magnetic monopoles don't exist?
- On page 374 of Andrew Zangwill's Modern Electrodynamics (2013), he writes, "When the sources of $mathbfB$ are far away so $mathbfnabla times mathbfB = 0$, blah blah blah." This is the thing I'm most confused about. How do we know that $mathbfnabla times mathbfB = 0$? Can someone show me a proof and/or help me understand what the "far away" criterion means in real life? (Far away relative to what?)
- I think the fourth term is simple -- since a dipole moment is just 1 vector, the curl is always zero.
electromagnetism magnetic-fields dipole-moment
asked 3 hours ago
Bunji
538
To add to the clarification in the accepted answer. $mathbfBcdotnabla$ is an operator itself, $B_ipartial_i$, that can be used on other quantites. This is why it is not the same as $nablacdotmathbfB=partial_iB_i=0$
– Aaron Stevens
11 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
In my lab, I use electromagnets to apply a magnetic gradient force to lots of very small (superparamagnetic) nanoparticles embedded in an elastic medium. I believe that these can be treated as magnetic dipoles, with a dipole moment $mathbfm$.
It is well known that the force on a magnetic dipole moment in a magnetic field is given by
$$mathbfF = mathbfnabla(mathbfm cdot mathbfB).$$
I need to prove to myself that this can be reduced to
$$mathbfF = (mathbfm cdot mathbfnabla)mathbfB.$$
I know that we can rewrite the first equation using one of those vector calculus identities that appears, e.g. on the inside covers of Jackson:
$$mathbfF = mathbfnabla(mathbfm cdot mathbfB) = (mathbfm cdot mathbfnabla)mathbfB + (mathbfB cdot mathbfnabla)mathbfm + mathbfm times (mathbfnabla times mathbfB) + mathbfB times (mathbfnabla times mathbfm).$$
- The first term is good -- it can stay!
- For the second term, can I use the commutative property of the dot product to say that $mathbfB cdot mathbfnabla = mathbfnabla cdot mathbfB = 0$ because magnetic monopoles don't exist?
- On page 374 of Andrew Zangwill's Modern Electrodynamics (2013), he writes, "When the sources of $mathbfB$ are far away so $mathbfnabla times mathbfB = 0$, blah blah blah." This is the thing I'm most confused about. How do we know that $mathbfnabla times mathbfB = 0$? Can someone show me a proof and/or help me understand what the "far away" criterion means in real life? (Far away relative to what?)
- I think the fourth term is simple -- since a dipole moment is just 1 vector, the curl is always zero.
electromagnetism magnetic-fields dipole-moment
asked 3 hours ago
Bunji
538
In my lab, I use electromagnets to apply a magnetic gradient force to lots of very small (superparamagnetic) nanoparticles embedded in an elastic medium. I believe that these can be treated as magnetic dipoles, with a dipole moment $mathbfm$.
It is well known that the force on a magnetic dipole moment in a magnetic field is given by
$$mathbfF = mathbfnabla(mathbfm cdot mathbfB).$$
I need to prove to myself that this can be reduced to
$$mathbfF = (mathbfm cdot mathbfnabla)mathbfB.$$
I know that we can rewrite the first equation using one of those vector calculus identities that appears, e.g. on the inside covers of Jackson:
$$mathbfF = mathbfnabla(mathbfm cdot mathbfB) = (mathbfm cdot mathbfnabla)mathbfB + (mathbfB cdot mathbfnabla)mathbfm + mathbfm times (mathbfnabla times mathbfB) + mathbfB times (mathbfnabla times mathbfm).$$
- The first term is good -- it can stay!
- For the second term, can I use the commutative property of the dot product to say that $mathbfB cdot mathbfnabla = mathbfnabla cdot mathbfB = 0$ because magnetic monopoles don't exist?
- On page 374 of Andrew Zangwill's Modern Electrodynamics (2013), he writes, "When the sources of $mathbfB$ are far away so $mathbfnabla times mathbfB = 0$, blah blah blah." This is the thing I'm most confused about. How do we know that $mathbfnabla times mathbfB = 0$? Can someone show me a proof and/or help me understand what the "far away" criterion means in real life? (Far away relative to what?)
- I think the fourth term is simple -- since a dipole moment is just 1 vector, the curl is always zero.
electromagnetism magnetic-fields dipole-moment
electromagnetism magnetic-fields dipole-moment
asked 3 hours ago
Bunji
538
asked 3 hours ago
Bunji
538
asked 3 hours ago
Bunji
538
asked 3 hours ago
Bunji
538
asked 3 hours ago
Bunji
538
538
To add to the clarification in the accepted answer. $mathbfBcdotnabla$ is an operator itself, $B_ipartial_i$, that can be used on other quantites. This is why it is not the same as $nablacdotmathbfB=partial_iB_i=0$
– Aaron Stevens
11 mins ago
add a comment |Â
To add to the clarification in the accepted answer. $mathbfBcdotnabla$ is an operator itself, $B_ipartial_i$, that can be used on other quantites. This is why it is not the same as $nablacdotmathbfB=partial_iB_i=0$
– Aaron Stevens
11 mins ago
To add to the clarification in the accepted answer. $mathbfBcdotnabla$ is an operator itself, $B_ipartial_i$, that can be used on other quantites. This is why it is not the same as $nablacdotmathbfB=partial_iB_i=0$
– Aaron Stevens
11 mins ago
To add to the clarification in the accepted answer. $mathbfBcdotnabla$ is an operator itself, $B_ipartial_i$, that can be used on other quantites. This is why it is not the same as $nablacdotmathbfB=partial_iB_i=0$
– Aaron Stevens
11 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Ampere's law says that
$$nabla times boldsymbolB = mu_0 boldsymbolJ + epsilon_0mu_0 fracpartialpartial t boldsymbolE$$
so "far away from sources" means that the current density $boldsymbolJ$ can be taken to be zero, and that there are no time-varying electric fields. The latter is actually a general approximation that can often be made for relatively low frequency (including steady-state) phenomena.
As for the other questions, that identity actually does not apply here, because $mathbfmcdot B$ is not the dot product of two vector fields. In particular, the spatial derivatives of $mathbfm$ are not defined.
Instead, we can use index notation to get the actual identity we're looking for. Note that
$$[nabla(mathbfmcdot B)]_i = partial_i m_j B_j = m_jpartial_i B_j$$
This doesn't have an immediately obvious vector form, but we can do the following sorcery (which, full disclosure, I did backwards):
$$m_jpartial_iB_j = (m_jpartial_iB_j - m_jpartial_jB_i) + m_jpartial_jB_i$$
$$ = (delta_ildelta_jm - delta_imdelta_jl) m_j partial _l B_m + m_jpartial_j B_i$$
$$ =epsilon_ijk epsilon_klm m_jpartial_lB_m + m_jpartial_j B_i$$
$$ =epsilon_ijk m_j (epsilon_klmpartial_l B_m) + m_j partial_j B_i$$
$$ = [ mathbfmtimes (nabla times mathbfB) + (mathbfmcdot nabla)mathbfB]_i$$
and so if $mathbfm$ and $mathbfB$ are a constant vector and a vector field respectively, the applicable vector identity is
$$nabla(mathbfmcdot B) = mathbfm times (nabla times mathbfB) + (mathbfm cdot nabla)mathbfB$$
This is, of course, what we would get if we treated $mathbfm$ as a spatially constant vector field, so you could wave your hands and say that $nabla mathbfm$ and $nabla times mathbfm$ are equal to zero. However, you should remember that those expressions are formally not defined.
Lastly, I want to clarify that there is no "commutative property of the dot product" when it comes to the divergence operator, because divergence is not a dot product. It only looks like one (and only in Cartesian coordinates), so $div(mathbfB) = nabla cdot mathbf B$ is nothing more than a useful mnemonic device.
answered 1 hour ago
J. Murray
5,9302519
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Ampere's law says that
$$nabla times boldsymbolB = mu_0 boldsymbolJ + epsilon_0mu_0 fracpartialpartial t boldsymbolE$$
so "far away from sources" means that the current density $boldsymbolJ$ can be taken to be zero, and that there are no time-varying electric fields. The latter is actually a general approximation that can often be made for relatively low frequency (including steady-state) phenomena.
As for the other questions, that identity actually does not apply here, because $mathbfmcdot B$ is not the dot product of two vector fields. In particular, the spatial derivatives of $mathbfm$ are not defined.
Instead, we can use index notation to get the actual identity we're looking for. Note that
$$[nabla(mathbfmcdot B)]_i = partial_i m_j B_j = m_jpartial_i B_j$$
This doesn't have an immediately obvious vector form, but we can do the following sorcery (which, full disclosure, I did backwards):
$$m_jpartial_iB_j = (m_jpartial_iB_j - m_jpartial_jB_i) + m_jpartial_jB_i$$
$$ = (delta_ildelta_jm - delta_imdelta_jl) m_j partial _l B_m + m_jpartial_j B_i$$
$$ =epsilon_ijk epsilon_klm m_jpartial_lB_m + m_jpartial_j B_i$$
$$ =epsilon_ijk m_j (epsilon_klmpartial_l B_m) + m_j partial_j B_i$$
$$ = [ mathbfmtimes (nabla times mathbfB) + (mathbfmcdot nabla)mathbfB]_i$$
and so if $mathbfm$ and $mathbfB$ are a constant vector and a vector field respectively, the applicable vector identity is
$$nabla(mathbfmcdot B) = mathbfm times (nabla times mathbfB) + (mathbfm cdot nabla)mathbfB$$
This is, of course, what we would get if we treated $mathbfm$ as a spatially constant vector field, so you could wave your hands and say that $nabla mathbfm$ and $nabla times mathbfm$ are equal to zero. However, you should remember that those expressions are formally not defined.
Lastly, I want to clarify that there is no "commutative property of the dot product" when it comes to the divergence operator, because divergence is not a dot product. It only looks like one (and only in Cartesian coordinates), so $div(mathbfB) = nabla cdot mathbf B$ is nothing more than a useful mnemonic device.
answered 1 hour ago
J. Murray
5,9302519
add a comment |Â
up vote
4
down vote
accepted
Ampere's law says that
$$nabla times boldsymbolB = mu_0 boldsymbolJ + epsilon_0mu_0 fracpartialpartial t boldsymbolE$$
so "far away from sources" means that the current density $boldsymbolJ$ can be taken to be zero, and that there are no time-varying electric fields. The latter is actually a general approximation that can often be made for relatively low frequency (including steady-state) phenomena.
As for the other questions, that identity actually does not apply here, because $mathbfmcdot B$ is not the dot product of two vector fields. In particular, the spatial derivatives of $mathbfm$ are not defined.
Instead, we can use index notation to get the actual identity we're looking for. Note that
$$[nabla(mathbfmcdot B)]_i = partial_i m_j B_j = m_jpartial_i B_j$$
This doesn't have an immediately obvious vector form, but we can do the following sorcery (which, full disclosure, I did backwards):
$$m_jpartial_iB_j = (m_jpartial_iB_j - m_jpartial_jB_i) + m_jpartial_jB_i$$
$$ = (delta_ildelta_jm - delta_imdelta_jl) m_j partial _l B_m + m_jpartial_j B_i$$
$$ =epsilon_ijk epsilon_klm m_jpartial_lB_m + m_jpartial_j B_i$$
$$ =epsilon_ijk m_j (epsilon_klmpartial_l B_m) + m_j partial_j B_i$$
$$ = [ mathbfmtimes (nabla times mathbfB) + (mathbfmcdot nabla)mathbfB]_i$$
and so if $mathbfm$ and $mathbfB$ are a constant vector and a vector field respectively, the applicable vector identity is
$$nabla(mathbfmcdot B) = mathbfm times (nabla times mathbfB) + (mathbfm cdot nabla)mathbfB$$
This is, of course, what we would get if we treated $mathbfm$ as a spatially constant vector field, so you could wave your hands and say that $nabla mathbfm$ and $nabla times mathbfm$ are equal to zero. However, you should remember that those expressions are formally not defined.
Lastly, I want to clarify that there is no "commutative property of the dot product" when it comes to the divergence operator, because divergence is not a dot product. It only looks like one (and only in Cartesian coordinates), so $div(mathbfB) = nabla cdot mathbf B$ is nothing more than a useful mnemonic device.
answered 1 hour ago
J. Murray
5,9302519
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Ampere's law says that
$$nabla times boldsymbolB = mu_0 boldsymbolJ + epsilon_0mu_0 fracpartialpartial t boldsymbolE$$
so "far away from sources" means that the current density $boldsymbolJ$ can be taken to be zero, and that there are no time-varying electric fields. The latter is actually a general approximation that can often be made for relatively low frequency (including steady-state) phenomena.
As for the other questions, that identity actually does not apply here, because $mathbfmcdot B$ is not the dot product of two vector fields. In particular, the spatial derivatives of $mathbfm$ are not defined.
Instead, we can use index notation to get the actual identity we're looking for. Note that
$$[nabla(mathbfmcdot B)]_i = partial_i m_j B_j = m_jpartial_i B_j$$
This doesn't have an immediately obvious vector form, but we can do the following sorcery (which, full disclosure, I did backwards):
$$m_jpartial_iB_j = (m_jpartial_iB_j - m_jpartial_jB_i) + m_jpartial_jB_i$$
$$ = (delta_ildelta_jm - delta_imdelta_jl) m_j partial _l B_m + m_jpartial_j B_i$$
$$ =epsilon_ijk epsilon_klm m_jpartial_lB_m + m_jpartial_j B_i$$
$$ =epsilon_ijk m_j (epsilon_klmpartial_l B_m) + m_j partial_j B_i$$
$$ = [ mathbfmtimes (nabla times mathbfB) + (mathbfmcdot nabla)mathbfB]_i$$
and so if $mathbfm$ and $mathbfB$ are a constant vector and a vector field respectively, the applicable vector identity is
$$nabla(mathbfmcdot B) = mathbfm times (nabla times mathbfB) + (mathbfm cdot nabla)mathbfB$$
This is, of course, what we would get if we treated $mathbfm$ as a spatially constant vector field, so you could wave your hands and say that $nabla mathbfm$ and $nabla times mathbfm$ are equal to zero. However, you should remember that those expressions are formally not defined.
Lastly, I want to clarify that there is no "commutative property of the dot product" when it comes to the divergence operator, because divergence is not a dot product. It only looks like one (and only in Cartesian coordinates), so $div(mathbfB) = nabla cdot mathbf B$ is nothing more than a useful mnemonic device.
answered 1 hour ago
J. Murray
5,9302519
Ampere's law says that
$$nabla times boldsymbolB = mu_0 boldsymbolJ + epsilon_0mu_0 fracpartialpartial t boldsymbolE$$
so "far away from sources" means that the current density $boldsymbolJ$ can be taken to be zero, and that there are no time-varying electric fields. The latter is actually a general approximation that can often be made for relatively low frequency (including steady-state) phenomena.
As for the other questions, that identity actually does not apply here, because $mathbfmcdot B$ is not the dot product of two vector fields. In particular, the spatial derivatives of $mathbfm$ are not defined.
Instead, we can use index notation to get the actual identity we're looking for. Note that
$$[nabla(mathbfmcdot B)]_i = partial_i m_j B_j = m_jpartial_i B_j$$
This doesn't have an immediately obvious vector form, but we can do the following sorcery (which, full disclosure, I did backwards):
$$m_jpartial_iB_j = (m_jpartial_iB_j - m_jpartial_jB_i) + m_jpartial_jB_i$$
$$ = (delta_ildelta_jm - delta_imdelta_jl) m_j partial _l B_m + m_jpartial_j B_i$$
$$ =epsilon_ijk epsilon_klm m_jpartial_lB_m + m_jpartial_j B_i$$
$$ =epsilon_ijk m_j (epsilon_klmpartial_l B_m) + m_j partial_j B_i$$
$$ = [ mathbfmtimes (nabla times mathbfB) + (mathbfmcdot nabla)mathbfB]_i$$
and so if $mathbfm$ and $mathbfB$ are a constant vector and a vector field respectively, the applicable vector identity is
$$nabla(mathbfmcdot B) = mathbfm times (nabla times mathbfB) + (mathbfm cdot nabla)mathbfB$$
This is, of course, what we would get if we treated $mathbfm$ as a spatially constant vector field, so you could wave your hands and say that $nabla mathbfm$ and $nabla times mathbfm$ are equal to zero. However, you should remember that those expressions are formally not defined.
Lastly, I want to clarify that there is no "commutative property of the dot product" when it comes to the divergence operator, because divergence is not a dot product. It only looks like one (and only in Cartesian coordinates), so $div(mathbfB) = nabla cdot mathbf B$ is nothing more than a useful mnemonic device.
answered 1 hour ago
J. Murray
5,9302519
edited 1 hour ago
answered 1 hour ago
J. Murray
5,9302519
answered 1 hour ago
J. Murray
5,9302519
answered 1 hour ago
J. Murray
5,9302519
5,9302519
add a comment |Â
add a comment |Â
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To add to the clarification in the accepted answer. $mathbfBcdotnabla$ is an operator itself, $B_ipartial_i$, that can be used on other quantites. This is why it is not the same as $nablacdotmathbfB=partial_iB_i=0$
– Aaron Stevens
11 mins ago