Number of independent components of a vector satisfying a differential constraint?
Edited question Consider a vector field $vec{A}(vec{x})$ such that in one case $nablacdotvec{A}=0$. It looks like that this condition gives rise to a differential equation constraint $$partial_xA_x+partial_yA_y+partial_zA_z=0.$$
Consider the next case where for a known vector $vec{k}=k_xhat{x}+k_yhat{y}+k_zhat{z}$, one finds that $vec{k}cdotvec{A}=0$. Therefore, $$k_xA_x+k_yA_y+k_zA_z=0.$$
The latter case definitely reduces the number of independent components of $vec{A}$ from 3 to 2. What about the first case? Does that also reduce the number of independent components of $vec{A}$? If not, how to see that simply?
differential-equations vectors mathematical-physics vector-fields constraints
asked Dec 1 '18 at 13:14
mithusengupta123
1069
add a comment |
Edited question Consider a vector field $vec{A}(vec{x})$ such that in one case $nablacdotvec{A}=0$. It looks like that this condition gives rise to a differential equation constraint $$partial_xA_x+partial_yA_y+partial_zA_z=0.$$
Consider the next case where for a known vector $vec{k}=k_xhat{x}+k_yhat{y}+k_zhat{z}$, one finds that $vec{k}cdotvec{A}=0$. Therefore, $$k_xA_x+k_yA_y+k_zA_z=0.$$
The latter case definitely reduces the number of independent components of $vec{A}$ from 3 to 2. What about the first case? Does that also reduce the number of independent components of $vec{A}$? If not, how to see that simply?
differential-equations vectors mathematical-physics vector-fields constraints
asked Dec 1 '18 at 13:14
mithusengupta123
1069
add a comment |
Edited question Consider a vector field $vec{A}(vec{x})$ such that in one case $nablacdotvec{A}=0$. It looks like that this condition gives rise to a differential equation constraint $$partial_xA_x+partial_yA_y+partial_zA_z=0.$$
Consider the next case where for a known vector $vec{k}=k_xhat{x}+k_yhat{y}+k_zhat{z}$, one finds that $vec{k}cdotvec{A}=0$. Therefore, $$k_xA_x+k_yA_y+k_zA_z=0.$$
The latter case definitely reduces the number of independent components of $vec{A}$ from 3 to 2. What about the first case? Does that also reduce the number of independent components of $vec{A}$? If not, how to see that simply?
differential-equations vectors mathematical-physics vector-fields constraints
asked Dec 1 '18 at 13:14
mithusengupta123
1069
Edited question Consider a vector field $vec{A}(vec{x})$ such that in one case $nablacdotvec{A}=0$. It looks like that this condition gives rise to a differential equation constraint $$partial_xA_x+partial_yA_y+partial_zA_z=0.$$
Consider the next case where for a known vector $vec{k}=k_xhat{x}+k_yhat{y}+k_zhat{z}$, one finds that $vec{k}cdotvec{A}=0$. Therefore, $$k_xA_x+k_yA_y+k_zA_z=0.$$
The latter case definitely reduces the number of independent components of $vec{A}$ from 3 to 2. What about the first case? Does that also reduce the number of independent components of $vec{A}$? If not, how to see that simply?
differential-equations vectors mathematical-physics vector-fields constraints
differential-equations vectors mathematical-physics vector-fields constraints
asked Dec 1 '18 at 13:14
mithusengupta123
1069
asked Dec 1 '18 at 13:14
mithusengupta123
1069
edited Dec 1 '18 at 13:43
asked Dec 1 '18 at 13:14
mithusengupta123
1069
asked Dec 1 '18 at 13:14
mithusengupta123
1069
asked Dec 1 '18 at 13:14
mithusengupta123
1069
1069
add a comment |
add a comment |
1 Answer
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That the rotation is zero $∇times vec A=0$ means that locally (on small balls) you can find a potential function $phi$ so that the vector field is its gradient, $vec A = ∇phi$. It may not be possible to extend the potential function to the full domain, especially in case it is not simply connected.
The second case $vec k×vec A=0$ can be stated even more restrictively, $vec A$ has to be a multiple of $vec k$, there is only one independent quantity.
More deeply, $∇×vec A=0$ means that the one-form $alpha=A_xdx+A_ydy+A_zdz$ is exact, $dα=0$, so that it is also locally closed, $α=dϕ$.
For the divergence something similar holds, $∇⋅vec A=0$ is equivalent to the differential 2-form $α=A_xdywedge dz+A_ydz∧dx+A_zdx∧dy$ being exact, $dα=0$, so again locally closed, which means that $α=domega$ for some one-form $ω$ which is only determined up to gage transformations, as also $α=d(ω+dϕ)$. Or $vec A=∇×(vec V+∇ϕ)$. $ϕ$ can be chosen to fix one of the components of $vec V$.
answered Dec 1 '18 at 13:38
LutzL
56k42054
I wanted to ask something different. I am editing the question now. Thanks for your answer though
– mithusengupta123
Dec 1 '18 at 13:41
add a comment |
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1 Answer
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That the rotation is zero $∇times vec A=0$ means that locally (on small balls) you can find a potential function $phi$ so that the vector field is its gradient, $vec A = ∇phi$. It may not be possible to extend the potential function to the full domain, especially in case it is not simply connected.
The second case $vec k×vec A=0$ can be stated even more restrictively, $vec A$ has to be a multiple of $vec k$, there is only one independent quantity.
More deeply, $∇×vec A=0$ means that the one-form $alpha=A_xdx+A_ydy+A_zdz$ is exact, $dα=0$, so that it is also locally closed, $α=dϕ$.
For the divergence something similar holds, $∇⋅vec A=0$ is equivalent to the differential 2-form $α=A_xdywedge dz+A_ydz∧dx+A_zdx∧dy$ being exact, $dα=0$, so again locally closed, which means that $α=domega$ for some one-form $ω$ which is only determined up to gage transformations, as also $α=d(ω+dϕ)$. Or $vec A=∇×(vec V+∇ϕ)$. $ϕ$ can be chosen to fix one of the components of $vec V$.
answered Dec 1 '18 at 13:38
LutzL
56k42054
I wanted to ask something different. I am editing the question now. Thanks for your answer though
– mithusengupta123
Dec 1 '18 at 13:41
add a comment |
That the rotation is zero $∇times vec A=0$ means that locally (on small balls) you can find a potential function $phi$ so that the vector field is its gradient, $vec A = ∇phi$. It may not be possible to extend the potential function to the full domain, especially in case it is not simply connected.
The second case $vec k×vec A=0$ can be stated even more restrictively, $vec A$ has to be a multiple of $vec k$, there is only one independent quantity.
More deeply, $∇×vec A=0$ means that the one-form $alpha=A_xdx+A_ydy+A_zdz$ is exact, $dα=0$, so that it is also locally closed, $α=dϕ$.
For the divergence something similar holds, $∇⋅vec A=0$ is equivalent to the differential 2-form $α=A_xdywedge dz+A_ydz∧dx+A_zdx∧dy$ being exact, $dα=0$, so again locally closed, which means that $α=domega$ for some one-form $ω$ which is only determined up to gage transformations, as also $α=d(ω+dϕ)$. Or $vec A=∇×(vec V+∇ϕ)$. $ϕ$ can be chosen to fix one of the components of $vec V$.
answered Dec 1 '18 at 13:38
LutzL
56k42054
I wanted to ask something different. I am editing the question now. Thanks for your answer though
– mithusengupta123
Dec 1 '18 at 13:41
add a comment |
That the rotation is zero $∇times vec A=0$ means that locally (on small balls) you can find a potential function $phi$ so that the vector field is its gradient, $vec A = ∇phi$. It may not be possible to extend the potential function to the full domain, especially in case it is not simply connected.
The second case $vec k×vec A=0$ can be stated even more restrictively, $vec A$ has to be a multiple of $vec k$, there is only one independent quantity.
More deeply, $∇×vec A=0$ means that the one-form $alpha=A_xdx+A_ydy+A_zdz$ is exact, $dα=0$, so that it is also locally closed, $α=dϕ$.
For the divergence something similar holds, $∇⋅vec A=0$ is equivalent to the differential 2-form $α=A_xdywedge dz+A_ydz∧dx+A_zdx∧dy$ being exact, $dα=0$, so again locally closed, which means that $α=domega$ for some one-form $ω$ which is only determined up to gage transformations, as also $α=d(ω+dϕ)$. Or $vec A=∇×(vec V+∇ϕ)$. $ϕ$ can be chosen to fix one of the components of $vec V$.
answered Dec 1 '18 at 13:38
LutzL
56k42054
That the rotation is zero $∇times vec A=0$ means that locally (on small balls) you can find a potential function $phi$ so that the vector field is its gradient, $vec A = ∇phi$. It may not be possible to extend the potential function to the full domain, especially in case it is not simply connected.
The second case $vec k×vec A=0$ can be stated even more restrictively, $vec A$ has to be a multiple of $vec k$, there is only one independent quantity.
More deeply, $∇×vec A=0$ means that the one-form $alpha=A_xdx+A_ydy+A_zdz$ is exact, $dα=0$, so that it is also locally closed, $α=dϕ$.
For the divergence something similar holds, $∇⋅vec A=0$ is equivalent to the differential 2-form $α=A_xdywedge dz+A_ydz∧dx+A_zdx∧dy$ being exact, $dα=0$, so again locally closed, which means that $α=domega$ for some one-form $ω$ which is only determined up to gage transformations, as also $α=d(ω+dϕ)$. Or $vec A=∇×(vec V+∇ϕ)$. $ϕ$ can be chosen to fix one of the components of $vec V$.
answered Dec 1 '18 at 13:38
LutzL
56k42054
edited Dec 1 '18 at 13:56
answered Dec 1 '18 at 13:38
LutzL
56k42054
answered Dec 1 '18 at 13:38
LutzL
56k42054
answered Dec 1 '18 at 13:38
LutzL
56k42054
56k42054
I wanted to ask something different. I am editing the question now. Thanks for your answer though
– mithusengupta123
Dec 1 '18 at 13:41
add a comment |
I wanted to ask something different. I am editing the question now. Thanks for your answer though
– mithusengupta123
Dec 1 '18 at 13:41
I wanted to ask something different. I am editing the question now. Thanks for your answer though
– mithusengupta123
Dec 1 '18 at 13:41
I wanted to ask something different. I am editing the question now. Thanks for your answer though
– mithusengupta123
Dec 1 '18 at 13:41
add a comment |
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