Mixing
Apparent paradox when we use the Kelvin–Stokes theorem and there is a time dependency
I am having trouble to understand what is going on with the Maxwell–Faraday equation:
$$nabla times E = - frac{partial B}{partial t},$$
where $E$ is the electric firld and $B$ the magnetic field. The equation is local, in the sense that any change at point $x$ will not affect what happens at another point $x'$, at least not instantaneously. That is, if there is a change in $B$ only at position $x$, then the change will need time to propagate to $x'$.
But we can use the Kelvin–Stokes theorem and write the equation in integral form:
$$int_{partial Sigma} E.dell = - frac{partial}{partial t}int_Sigma B cdot dS,$$
which is basically telling you that a change in $B$ at the center of the surface will affect instantaneously $E$ at the edge.
What is it wrong with my interpretation of these equations?
stokes-theorem electromagnetism
edited Dec 31 '18 at 16:02
anomaly
17.4k42664
asked Dec 29 '18 at 22:01
Wolphram jonnyWolphram jonny
3291731
add a comment |
I am having trouble to understand what is going on with the Maxwell–Faraday equation:
$$nabla times E = - frac{partial B}{partial t},$$
where $E$ is the electric firld and $B$ the magnetic field. The equation is local, in the sense that any change at point $x$ will not affect what happens at another point $x'$, at least not instantaneously. That is, if there is a change in $B$ only at position $x$, then the change will need time to propagate to $x'$.
But we can use the Kelvin–Stokes theorem and write the equation in integral form:
$$int_{partial Sigma} E.dell = - frac{partial}{partial t}int_Sigma B cdot dS,$$
which is basically telling you that a change in $B$ at the center of the surface will affect instantaneously $E$ at the edge.
What is it wrong with my interpretation of these equations?
stokes-theorem electromagnetism
edited Dec 31 '18 at 16:02
anomaly
17.4k42664
asked Dec 29 '18 at 22:01
Wolphram jonnyWolphram jonny
3291731
I think you should look up the concept of the retarded potential: EM affects travel at the speed of light. So no, information does not travel instantaneously.
– Adrian Keister
Dec 31 '18 at 18:08
@AdrianKeister thanks, I know that it does not, but it is what the second equation is saying, or not? E(t) at the edge depends on B(t) at the center, without any delays
– Wolphram jonny
Jan 1 at 0:21
The integral version is not telling you what you wrote. Instead, think of it this way: an instantaneous change in the magnetic field in the center would affect the rate of change of the electric field on the surface, not the value of the electric field itself.
– Adrian Keister
Jan 1 at 0:35
I am not sure I agree, let us say we have circular simmetry, then you have E=l*d/dt(int Bds), the E is at the edge of the surface, but the change in B can happen anywhere, like just at the center of the surface.
– Wolphram jonny
Jan 1 at 1:40
add a comment |
I am having trouble to understand what is going on with the Maxwell–Faraday equation:
$$nabla times E = - frac{partial B}{partial t},$$
where $E$ is the electric firld and $B$ the magnetic field. The equation is local, in the sense that any change at point $x$ will not affect what happens at another point $x'$, at least not instantaneously. That is, if there is a change in $B$ only at position $x$, then the change will need time to propagate to $x'$.
But we can use the Kelvin–Stokes theorem and write the equation in integral form:
$$int_{partial Sigma} E.dell = - frac{partial}{partial t}int_Sigma B cdot dS,$$
which is basically telling you that a change in $B$ at the center of the surface will affect instantaneously $E$ at the edge.
What is it wrong with my interpretation of these equations?
stokes-theorem electromagnetism
edited Dec 31 '18 at 16:02
anomaly
17.4k42664
asked Dec 29 '18 at 22:01
Wolphram jonnyWolphram jonny
3291731
I am having trouble to understand what is going on with the Maxwell–Faraday equation:
$$nabla times E = - frac{partial B}{partial t},$$
where $E$ is the electric firld and $B$ the magnetic field. The equation is local, in the sense that any change at point $x$ will not affect what happens at another point $x'$, at least not instantaneously. That is, if there is a change in $B$ only at position $x$, then the change will need time to propagate to $x'$.
But we can use the Kelvin–Stokes theorem and write the equation in integral form:
$$int_{partial Sigma} E.dell = - frac{partial}{partial t}int_Sigma B cdot dS,$$
which is basically telling you that a change in $B$ at the center of the surface will affect instantaneously $E$ at the edge.
What is it wrong with my interpretation of these equations?
stokes-theorem electromagnetism
stokes-theorem electromagnetism
edited Dec 31 '18 at 16:02
anomaly
17.4k42664
asked Dec 29 '18 at 22:01
Wolphram jonnyWolphram jonny
3291731
edited Dec 31 '18 at 16:02
anomaly
17.4k42664
asked Dec 29 '18 at 22:01
Wolphram jonnyWolphram jonny
3291731
edited Dec 31 '18 at 16:02
anomaly
17.4k42664
edited Dec 31 '18 at 16:02
anomaly
17.4k42664
edited Dec 31 '18 at 16:02
anomaly
17.4k42664
17.4k42664
asked Dec 29 '18 at 22:01
Wolphram jonnyWolphram jonny
3291731
asked Dec 29 '18 at 22:01
Wolphram jonnyWolphram jonny
3291731
asked Dec 29 '18 at 22:01
Wolphram jonnyWolphram jonny
3291731
3291731
I think you should look up the concept of the retarded potential: EM affects travel at the speed of light. So no, information does not travel instantaneously.
– Adrian Keister
Dec 31 '18 at 18:08
@AdrianKeister thanks, I know that it does not, but it is what the second equation is saying, or not? E(t) at the edge depends on B(t) at the center, without any delays
– Wolphram jonny
Jan 1 at 0:21
The integral version is not telling you what you wrote. Instead, think of it this way: an instantaneous change in the magnetic field in the center would affect the rate of change of the electric field on the surface, not the value of the electric field itself.
– Adrian Keister
Jan 1 at 0:35
I am not sure I agree, let us say we have circular simmetry, then you have E=l*d/dt(int Bds), the E is at the edge of the surface, but the change in B can happen anywhere, like just at the center of the surface.
– Wolphram jonny
Jan 1 at 1:40
add a comment |
I think you should look up the concept of the retarded potential: EM affects travel at the speed of light. So no, information does not travel instantaneously.
– Adrian Keister
Dec 31 '18 at 18:08
@AdrianKeister thanks, I know that it does not, but it is what the second equation is saying, or not? E(t) at the edge depends on B(t) at the center, without any delays
– Wolphram jonny
Jan 1 at 0:21
The integral version is not telling you what you wrote. Instead, think of it this way: an instantaneous change in the magnetic field in the center would affect the rate of change of the electric field on the surface, not the value of the electric field itself.
– Adrian Keister
Jan 1 at 0:35
I am not sure I agree, let us say we have circular simmetry, then you have E=l*d/dt(int Bds), the E is at the edge of the surface, but the change in B can happen anywhere, like just at the center of the surface.
– Wolphram jonny
Jan 1 at 1:40
I think you should look up the concept of the retarded potential: EM affects travel at the speed of light. So no, information does not travel instantaneously.
– Adrian Keister
Dec 31 '18 at 18:08
I think you should look up the concept of the retarded potential: EM affects travel at the speed of light. So no, information does not travel instantaneously.
– Adrian Keister
Dec 31 '18 at 18:08
@AdrianKeister thanks, I know that it does not, but it is what the second equation is saying, or not? E(t) at the edge depends on B(t) at the center, without any delays
– Wolphram jonny
Jan 1 at 0:21
@AdrianKeister thanks, I know that it does not, but it is what the second equation is saying, or not? E(t) at the edge depends on B(t) at the center, without any delays
– Wolphram jonny
Jan 1 at 0:21
The integral version is not telling you what you wrote. Instead, think of it this way: an instantaneous change in the magnetic field in the center would affect the rate of change of the electric field on the surface, not the value of the electric field itself.
– Adrian Keister
Jan 1 at 0:35
The integral version is not telling you what you wrote. Instead, think of it this way: an instantaneous change in the magnetic field in the center would affect the rate of change of the electric field on the surface, not the value of the electric field itself.
– Adrian Keister
Jan 1 at 0:35
I am not sure I agree, let us say we have circular simmetry, then you have E=l*d/dt(int Bds), the E is at the edge of the surface, but the change in B can happen anywhere, like just at the center of the surface.
– Wolphram jonny
Jan 1 at 1:40
I am not sure I agree, let us say we have circular simmetry, then you have E=l*d/dt(int Bds), the E is at the edge of the surface, but the change in B can happen anywhere, like just at the center of the surface.
– Wolphram jonny
Jan 1 at 1:40
add a comment |
2 Answers
2
active
oldest
votes
Since we have also $nablatimes B = 0$, you can only change $B$ by adding an entire loop. In this case, it will either cross the surface $S$ once in each direction, so be 0, or it will actually go around the perimeter current, and induce a current, which will change $E$.
answered Jan 1 at 11:15
wendy.kriegerwendy.krieger
5,75911426
good point, I believe it might answer the question, but I need to think a little more a few issues.
– Wolphram jonny
Jan 2 at 20:46
basically my doubt is how a field line that go outside the loop is formed: instantaneously across space or is slowly built, growing in diameter with time? I guess it is an electromagnetic wave that leaves loops of growing diameter behind the front?
– Wolphram jonny
Jan 2 at 20:54
add a comment |
Good question!
The answer, at least to me, lies in that the integral form holds for arbitrary surfaces $Sigma$. This can be interpreted in two ways:
- For a given closed loop $ellinmathbb{R}^3$, there are infinitely many smooth surfaces $Sigma$ such that $partialSigma=ell$;
- The closed loop $ell$ itself could also be arbitrarily specified.
Therefore, while the integral form appears non-local, it is actually local, as you may take a "small" closed loop $ell$ (e.g., a circle with an infinitesimal radius).
Further, even if you take a "large" closed loop $ell$, you may still choose different surface $Sigma$, such that a local change of $mathbf{B}$ in $Sigma$ would not effect the value of $mathbb{E}$ on $ell=partialSigma$.
With these arguments, your question could be interpreted as follows. Suppose you have chosen some $ell$ and $Sigma$ with $ell=partialSigma$. Suppose $mathbf{B}$ observes a tiny change in the interior of $Sigma$. Then according to
$$
oint_{ell}mathbf{E}cdot{rm d}mathbf{l}=-frac{partial}{partial t}int_{Sigma}mathbf{B}cdot{rm d}mathbf{S},
$$
it seems as if $mathbf{E}$ also yields some changes along $ell$. But wait! Since the change in $mathbf{B}$ is tiny, you may want to find some $Sigma'$, such that (1) $partialSigma'=ell$, and that (2) $mathbf{B}$ does not have any change on $Sigma'$. In this sense, you will obtain, at least for the moment,
$$
oint_{ell}mathbf{E}cdot{rm d}mathbf{l}=-frac{partial}{partial t}int_{Sigma'}mathbf{B}cdot{rm d}mathbf{S}=0,
$$
with which you would have no idea whether or not $mathbf{E}$ changes along $ell$. For tiny changes in $mathbf{B}$, you may apply the integral form around each point on $ell$ with small closed loops $ell'$ and surfaces $Sigma''$ with $partialSigma''=ell'$ on which $mathbf{B}$ does not find any change, and the arbitrariness of the choice of $ell'$ and $Sigma''$ would imply the free of change in $mathbf{E}$. This trick fails only if the change in $mathbf{B}$ hits $ell$, which exactly indicates the locality of its physics.
Hope this could be helpful for you.
answered Jan 1 at 8:50
hypernovahypernova
3,899313
Great point, but I disagree with the conclusion, is you get diferent E's depending on how do you chose the surface, would not that mean that E is not defined, rather than unknown?
– Wolphram jonny
Jan 2 at 20:42
@Wolphramjonny $mathbf{E}$ has always been pre-determined before you choose any $ell$ or $Sigma$. While you may take different $ell$ and $Sigma$ to figure out the value of $mathbf{E}$, your choice does not change its value.
– hypernova
Jan 3 at 13:38
But is not your argument, which seems correct, that a good choice of $Sigma$ will result in an area integral =0? in which case the computed fiels prediction is also zero? but with another surface choice the result would be different than zero? In such a case, regardless of physical reality(I agree E is well defined), the equations make inconsistent predictions, so the question is: what is it wrong with the equations?
– Wolphram jonny
Jan 3 at 14:14
@Wolphramjonny: Not really. If you can find some $Sigma_1$ such that $oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$, then for any other $Sigma_2$ with $partialSigma_2=partialSigma_1$, it is a must that $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}=0$. My argument is thus as follows. If $mathbf{B}$ observes a tiny change on $Sigma_2$, it is hard to figure out $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}$. Nevertheless, if you can find some $Sigma_1$ (with $partialSigma_1=partialSigma_2$) on which $mathbf{B}$ does not change for the moment,
– hypernova
Jan 3 at 15:09
@Wolphramjonny: (con't) then it is straightforward that $oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$. This result, in turn, implies $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}=oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$. This is a trick to help figure out $mathbf{E}$. After all, your doubt to me lies in that, from the integral form, it is hard to tell if a tiny change in $mathbf{B}$ influences the value of $mathbf{E}$ at a remote location. The above trick is to clarify that there is no such influence.
– hypernova
Jan 3 at 15:10
|
show 4 more comments
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2 Answers
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2 Answers
2
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oldest
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Since we have also $nablatimes B = 0$, you can only change $B$ by adding an entire loop. In this case, it will either cross the surface $S$ once in each direction, so be 0, or it will actually go around the perimeter current, and induce a current, which will change $E$.
answered Jan 1 at 11:15
wendy.kriegerwendy.krieger
5,75911426
good point, I believe it might answer the question, but I need to think a little more a few issues.
– Wolphram jonny
Jan 2 at 20:46
basically my doubt is how a field line that go outside the loop is formed: instantaneously across space or is slowly built, growing in diameter with time? I guess it is an electromagnetic wave that leaves loops of growing diameter behind the front?
– Wolphram jonny
Jan 2 at 20:54
add a comment |
Since we have also $nablatimes B = 0$, you can only change $B$ by adding an entire loop. In this case, it will either cross the surface $S$ once in each direction, so be 0, or it will actually go around the perimeter current, and induce a current, which will change $E$.
answered Jan 1 at 11:15
wendy.kriegerwendy.krieger
5,75911426
good point, I believe it might answer the question, but I need to think a little more a few issues.
– Wolphram jonny
Jan 2 at 20:46
basically my doubt is how a field line that go outside the loop is formed: instantaneously across space or is slowly built, growing in diameter with time? I guess it is an electromagnetic wave that leaves loops of growing diameter behind the front?
– Wolphram jonny
Jan 2 at 20:54
add a comment |
Since we have also $nablatimes B = 0$, you can only change $B$ by adding an entire loop. In this case, it will either cross the surface $S$ once in each direction, so be 0, or it will actually go around the perimeter current, and induce a current, which will change $E$.
answered Jan 1 at 11:15
wendy.kriegerwendy.krieger
5,75911426
Since we have also $nablatimes B = 0$, you can only change $B$ by adding an entire loop. In this case, it will either cross the surface $S$ once in each direction, so be 0, or it will actually go around the perimeter current, and induce a current, which will change $E$.
answered Jan 1 at 11:15
wendy.kriegerwendy.krieger
5,75911426
answered Jan 1 at 11:15
wendy.kriegerwendy.krieger
5,75911426
answered Jan 1 at 11:15
wendy.kriegerwendy.krieger
5,75911426
answered Jan 1 at 11:15
wendy.kriegerwendy.krieger
5,75911426
5,75911426
good point, I believe it might answer the question, but I need to think a little more a few issues.
– Wolphram jonny
Jan 2 at 20:46
basically my doubt is how a field line that go outside the loop is formed: instantaneously across space or is slowly built, growing in diameter with time? I guess it is an electromagnetic wave that leaves loops of growing diameter behind the front?
– Wolphram jonny
Jan 2 at 20:54
add a comment |
good point, I believe it might answer the question, but I need to think a little more a few issues.
– Wolphram jonny
Jan 2 at 20:46
basically my doubt is how a field line that go outside the loop is formed: instantaneously across space or is slowly built, growing in diameter with time? I guess it is an electromagnetic wave that leaves loops of growing diameter behind the front?
– Wolphram jonny
Jan 2 at 20:54
good point, I believe it might answer the question, but I need to think a little more a few issues.
– Wolphram jonny
Jan 2 at 20:46
good point, I believe it might answer the question, but I need to think a little more a few issues.
– Wolphram jonny
Jan 2 at 20:46
basically my doubt is how a field line that go outside the loop is formed: instantaneously across space or is slowly built, growing in diameter with time? I guess it is an electromagnetic wave that leaves loops of growing diameter behind the front?
– Wolphram jonny
Jan 2 at 20:54
basically my doubt is how a field line that go outside the loop is formed: instantaneously across space or is slowly built, growing in diameter with time? I guess it is an electromagnetic wave that leaves loops of growing diameter behind the front?
– Wolphram jonny
Jan 2 at 20:54
add a comment |
Good question!
The answer, at least to me, lies in that the integral form holds for arbitrary surfaces $Sigma$. This can be interpreted in two ways:
- For a given closed loop $ellinmathbb{R}^3$, there are infinitely many smooth surfaces $Sigma$ such that $partialSigma=ell$;
- The closed loop $ell$ itself could also be arbitrarily specified.
Therefore, while the integral form appears non-local, it is actually local, as you may take a "small" closed loop $ell$ (e.g., a circle with an infinitesimal radius).
Further, even if you take a "large" closed loop $ell$, you may still choose different surface $Sigma$, such that a local change of $mathbf{B}$ in $Sigma$ would not effect the value of $mathbb{E}$ on $ell=partialSigma$.
With these arguments, your question could be interpreted as follows. Suppose you have chosen some $ell$ and $Sigma$ with $ell=partialSigma$. Suppose $mathbf{B}$ observes a tiny change in the interior of $Sigma$. Then according to
$$
oint_{ell}mathbf{E}cdot{rm d}mathbf{l}=-frac{partial}{partial t}int_{Sigma}mathbf{B}cdot{rm d}mathbf{S},
$$
it seems as if $mathbf{E}$ also yields some changes along $ell$. But wait! Since the change in $mathbf{B}$ is tiny, you may want to find some $Sigma'$, such that (1) $partialSigma'=ell$, and that (2) $mathbf{B}$ does not have any change on $Sigma'$. In this sense, you will obtain, at least for the moment,
$$
oint_{ell}mathbf{E}cdot{rm d}mathbf{l}=-frac{partial}{partial t}int_{Sigma'}mathbf{B}cdot{rm d}mathbf{S}=0,
$$
with which you would have no idea whether or not $mathbf{E}$ changes along $ell$. For tiny changes in $mathbf{B}$, you may apply the integral form around each point on $ell$ with small closed loops $ell'$ and surfaces $Sigma''$ with $partialSigma''=ell'$ on which $mathbf{B}$ does not find any change, and the arbitrariness of the choice of $ell'$ and $Sigma''$ would imply the free of change in $mathbf{E}$. This trick fails only if the change in $mathbf{B}$ hits $ell$, which exactly indicates the locality of its physics.
Hope this could be helpful for you.
answered Jan 1 at 8:50
hypernovahypernova
3,899313
Great point, but I disagree with the conclusion, is you get diferent E's depending on how do you chose the surface, would not that mean that E is not defined, rather than unknown?
– Wolphram jonny
Jan 2 at 20:42
@Wolphramjonny $mathbf{E}$ has always been pre-determined before you choose any $ell$ or $Sigma$. While you may take different $ell$ and $Sigma$ to figure out the value of $mathbf{E}$, your choice does not change its value.
– hypernova
Jan 3 at 13:38
But is not your argument, which seems correct, that a good choice of $Sigma$ will result in an area integral =0? in which case the computed fiels prediction is also zero? but with another surface choice the result would be different than zero? In such a case, regardless of physical reality(I agree E is well defined), the equations make inconsistent predictions, so the question is: what is it wrong with the equations?
– Wolphram jonny
Jan 3 at 14:14
@Wolphramjonny: Not really. If you can find some $Sigma_1$ such that $oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$, then for any other $Sigma_2$ with $partialSigma_2=partialSigma_1$, it is a must that $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}=0$. My argument is thus as follows. If $mathbf{B}$ observes a tiny change on $Sigma_2$, it is hard to figure out $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}$. Nevertheless, if you can find some $Sigma_1$ (with $partialSigma_1=partialSigma_2$) on which $mathbf{B}$ does not change for the moment,
– hypernova
Jan 3 at 15:09
@Wolphramjonny: (con't) then it is straightforward that $oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$. This result, in turn, implies $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}=oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$. This is a trick to help figure out $mathbf{E}$. After all, your doubt to me lies in that, from the integral form, it is hard to tell if a tiny change in $mathbf{B}$ influences the value of $mathbf{E}$ at a remote location. The above trick is to clarify that there is no such influence.
– hypernova
Jan 3 at 15:10
|
show 4 more comments
Good question!
The answer, at least to me, lies in that the integral form holds for arbitrary surfaces $Sigma$. This can be interpreted in two ways:
- For a given closed loop $ellinmathbb{R}^3$, there are infinitely many smooth surfaces $Sigma$ such that $partialSigma=ell$;
- The closed loop $ell$ itself could also be arbitrarily specified.
Therefore, while the integral form appears non-local, it is actually local, as you may take a "small" closed loop $ell$ (e.g., a circle with an infinitesimal radius).
Further, even if you take a "large" closed loop $ell$, you may still choose different surface $Sigma$, such that a local change of $mathbf{B}$ in $Sigma$ would not effect the value of $mathbb{E}$ on $ell=partialSigma$.
With these arguments, your question could be interpreted as follows. Suppose you have chosen some $ell$ and $Sigma$ with $ell=partialSigma$. Suppose $mathbf{B}$ observes a tiny change in the interior of $Sigma$. Then according to
$$
oint_{ell}mathbf{E}cdot{rm d}mathbf{l}=-frac{partial}{partial t}int_{Sigma}mathbf{B}cdot{rm d}mathbf{S},
$$
it seems as if $mathbf{E}$ also yields some changes along $ell$. But wait! Since the change in $mathbf{B}$ is tiny, you may want to find some $Sigma'$, such that (1) $partialSigma'=ell$, and that (2) $mathbf{B}$ does not have any change on $Sigma'$. In this sense, you will obtain, at least for the moment,
$$
oint_{ell}mathbf{E}cdot{rm d}mathbf{l}=-frac{partial}{partial t}int_{Sigma'}mathbf{B}cdot{rm d}mathbf{S}=0,
$$
with which you would have no idea whether or not $mathbf{E}$ changes along $ell$. For tiny changes in $mathbf{B}$, you may apply the integral form around each point on $ell$ with small closed loops $ell'$ and surfaces $Sigma''$ with $partialSigma''=ell'$ on which $mathbf{B}$ does not find any change, and the arbitrariness of the choice of $ell'$ and $Sigma''$ would imply the free of change in $mathbf{E}$. This trick fails only if the change in $mathbf{B}$ hits $ell$, which exactly indicates the locality of its physics.
Hope this could be helpful for you.
answered Jan 1 at 8:50
hypernovahypernova
3,899313
Great point, but I disagree with the conclusion, is you get diferent E's depending on how do you chose the surface, would not that mean that E is not defined, rather than unknown?
– Wolphram jonny
Jan 2 at 20:42
@Wolphramjonny $mathbf{E}$ has always been pre-determined before you choose any $ell$ or $Sigma$. While you may take different $ell$ and $Sigma$ to figure out the value of $mathbf{E}$, your choice does not change its value.
– hypernova
Jan 3 at 13:38
But is not your argument, which seems correct, that a good choice of $Sigma$ will result in an area integral =0? in which case the computed fiels prediction is also zero? but with another surface choice the result would be different than zero? In such a case, regardless of physical reality(I agree E is well defined), the equations make inconsistent predictions, so the question is: what is it wrong with the equations?
– Wolphram jonny
Jan 3 at 14:14
@Wolphramjonny: Not really. If you can find some $Sigma_1$ such that $oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$, then for any other $Sigma_2$ with $partialSigma_2=partialSigma_1$, it is a must that $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}=0$. My argument is thus as follows. If $mathbf{B}$ observes a tiny change on $Sigma_2$, it is hard to figure out $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}$. Nevertheless, if you can find some $Sigma_1$ (with $partialSigma_1=partialSigma_2$) on which $mathbf{B}$ does not change for the moment,
– hypernova
Jan 3 at 15:09
@Wolphramjonny: (con't) then it is straightforward that $oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$. This result, in turn, implies $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}=oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$. This is a trick to help figure out $mathbf{E}$. After all, your doubt to me lies in that, from the integral form, it is hard to tell if a tiny change in $mathbf{B}$ influences the value of $mathbf{E}$ at a remote location. The above trick is to clarify that there is no such influence.
– hypernova
Jan 3 at 15:10
|
show 4 more comments
Good question!
The answer, at least to me, lies in that the integral form holds for arbitrary surfaces $Sigma$. This can be interpreted in two ways:
- For a given closed loop $ellinmathbb{R}^3$, there are infinitely many smooth surfaces $Sigma$ such that $partialSigma=ell$;
- The closed loop $ell$ itself could also be arbitrarily specified.
Therefore, while the integral form appears non-local, it is actually local, as you may take a "small" closed loop $ell$ (e.g., a circle with an infinitesimal radius).
Further, even if you take a "large" closed loop $ell$, you may still choose different surface $Sigma$, such that a local change of $mathbf{B}$ in $Sigma$ would not effect the value of $mathbb{E}$ on $ell=partialSigma$.
With these arguments, your question could be interpreted as follows. Suppose you have chosen some $ell$ and $Sigma$ with $ell=partialSigma$. Suppose $mathbf{B}$ observes a tiny change in the interior of $Sigma$. Then according to
$$
oint_{ell}mathbf{E}cdot{rm d}mathbf{l}=-frac{partial}{partial t}int_{Sigma}mathbf{B}cdot{rm d}mathbf{S},
$$
it seems as if $mathbf{E}$ also yields some changes along $ell$. But wait! Since the change in $mathbf{B}$ is tiny, you may want to find some $Sigma'$, such that (1) $partialSigma'=ell$, and that (2) $mathbf{B}$ does not have any change on $Sigma'$. In this sense, you will obtain, at least for the moment,
$$
oint_{ell}mathbf{E}cdot{rm d}mathbf{l}=-frac{partial}{partial t}int_{Sigma'}mathbf{B}cdot{rm d}mathbf{S}=0,
$$
with which you would have no idea whether or not $mathbf{E}$ changes along $ell$. For tiny changes in $mathbf{B}$, you may apply the integral form around each point on $ell$ with small closed loops $ell'$ and surfaces $Sigma''$ with $partialSigma''=ell'$ on which $mathbf{B}$ does not find any change, and the arbitrariness of the choice of $ell'$ and $Sigma''$ would imply the free of change in $mathbf{E}$. This trick fails only if the change in $mathbf{B}$ hits $ell$, which exactly indicates the locality of its physics.
Hope this could be helpful for you.
answered Jan 1 at 8:50
hypernovahypernova
3,899313
Good question!
The answer, at least to me, lies in that the integral form holds for arbitrary surfaces $Sigma$. This can be interpreted in two ways:
- For a given closed loop $ellinmathbb{R}^3$, there are infinitely many smooth surfaces $Sigma$ such that $partialSigma=ell$;
- The closed loop $ell$ itself could also be arbitrarily specified.
Therefore, while the integral form appears non-local, it is actually local, as you may take a "small" closed loop $ell$ (e.g., a circle with an infinitesimal radius).
Further, even if you take a "large" closed loop $ell$, you may still choose different surface $Sigma$, such that a local change of $mathbf{B}$ in $Sigma$ would not effect the value of $mathbb{E}$ on $ell=partialSigma$.
With these arguments, your question could be interpreted as follows. Suppose you have chosen some $ell$ and $Sigma$ with $ell=partialSigma$. Suppose $mathbf{B}$ observes a tiny change in the interior of $Sigma$. Then according to
$$
oint_{ell}mathbf{E}cdot{rm d}mathbf{l}=-frac{partial}{partial t}int_{Sigma}mathbf{B}cdot{rm d}mathbf{S},
$$
it seems as if $mathbf{E}$ also yields some changes along $ell$. But wait! Since the change in $mathbf{B}$ is tiny, you may want to find some $Sigma'$, such that (1) $partialSigma'=ell$, and that (2) $mathbf{B}$ does not have any change on $Sigma'$. In this sense, you will obtain, at least for the moment,
$$
oint_{ell}mathbf{E}cdot{rm d}mathbf{l}=-frac{partial}{partial t}int_{Sigma'}mathbf{B}cdot{rm d}mathbf{S}=0,
$$
with which you would have no idea whether or not $mathbf{E}$ changes along $ell$. For tiny changes in $mathbf{B}$, you may apply the integral form around each point on $ell$ with small closed loops $ell'$ and surfaces $Sigma''$ with $partialSigma''=ell'$ on which $mathbf{B}$ does not find any change, and the arbitrariness of the choice of $ell'$ and $Sigma''$ would imply the free of change in $mathbf{E}$. This trick fails only if the change in $mathbf{B}$ hits $ell$, which exactly indicates the locality of its physics.
Hope this could be helpful for you.
answered Jan 1 at 8:50
hypernovahypernova
3,899313
answered Jan 1 at 8:50
hypernovahypernova
3,899313
answered Jan 1 at 8:50
hypernovahypernova
3,899313
answered Jan 1 at 8:50
hypernovahypernova
3,899313
3,899313
Great point, but I disagree with the conclusion, is you get diferent E's depending on how do you chose the surface, would not that mean that E is not defined, rather than unknown?
– Wolphram jonny
Jan 2 at 20:42
@Wolphramjonny $mathbf{E}$ has always been pre-determined before you choose any $ell$ or $Sigma$. While you may take different $ell$ and $Sigma$ to figure out the value of $mathbf{E}$, your choice does not change its value.
– hypernova
Jan 3 at 13:38
But is not your argument, which seems correct, that a good choice of $Sigma$ will result in an area integral =0? in which case the computed fiels prediction is also zero? but with another surface choice the result would be different than zero? In such a case, regardless of physical reality(I agree E is well defined), the equations make inconsistent predictions, so the question is: what is it wrong with the equations?
– Wolphram jonny
Jan 3 at 14:14
@Wolphramjonny: Not really. If you can find some $Sigma_1$ such that $oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$, then for any other $Sigma_2$ with $partialSigma_2=partialSigma_1$, it is a must that $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}=0$. My argument is thus as follows. If $mathbf{B}$ observes a tiny change on $Sigma_2$, it is hard to figure out $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}$. Nevertheless, if you can find some $Sigma_1$ (with $partialSigma_1=partialSigma_2$) on which $mathbf{B}$ does not change for the moment,
– hypernova
Jan 3 at 15:09
@Wolphramjonny: (con't) then it is straightforward that $oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$. This result, in turn, implies $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}=oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$. This is a trick to help figure out $mathbf{E}$. After all, your doubt to me lies in that, from the integral form, it is hard to tell if a tiny change in $mathbf{B}$ influences the value of $mathbf{E}$ at a remote location. The above trick is to clarify that there is no such influence.
– hypernova
Jan 3 at 15:10
|
show 4 more comments
Great point, but I disagree with the conclusion, is you get diferent E's depending on how do you chose the surface, would not that mean that E is not defined, rather than unknown?
– Wolphram jonny
Jan 2 at 20:42
@Wolphramjonny $mathbf{E}$ has always been pre-determined before you choose any $ell$ or $Sigma$. While you may take different $ell$ and $Sigma$ to figure out the value of $mathbf{E}$, your choice does not change its value.
– hypernova
Jan 3 at 13:38
But is not your argument, which seems correct, that a good choice of $Sigma$ will result in an area integral =0? in which case the computed fiels prediction is also zero? but with another surface choice the result would be different than zero? In such a case, regardless of physical reality(I agree E is well defined), the equations make inconsistent predictions, so the question is: what is it wrong with the equations?
– Wolphram jonny
Jan 3 at 14:14
@Wolphramjonny: Not really. If you can find some $Sigma_1$ such that $oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$, then for any other $Sigma_2$ with $partialSigma_2=partialSigma_1$, it is a must that $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}=0$. My argument is thus as follows. If $mathbf{B}$ observes a tiny change on $Sigma_2$, it is hard to figure out $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}$. Nevertheless, if you can find some $Sigma_1$ (with $partialSigma_1=partialSigma_2$) on which $mathbf{B}$ does not change for the moment,
– hypernova
Jan 3 at 15:09
@Wolphramjonny: (con't) then it is straightforward that $oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$. This result, in turn, implies $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}=oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$. This is a trick to help figure out $mathbf{E}$. After all, your doubt to me lies in that, from the integral form, it is hard to tell if a tiny change in $mathbf{B}$ influences the value of $mathbf{E}$ at a remote location. The above trick is to clarify that there is no such influence.
– hypernova
Jan 3 at 15:10
Great point, but I disagree with the conclusion, is you get diferent E's depending on how do you chose the surface, would not that mean that E is not defined, rather than unknown?
– Wolphram jonny
Jan 2 at 20:42
Great point, but I disagree with the conclusion, is you get diferent E's depending on how do you chose the surface, would not that mean that E is not defined, rather than unknown?
– Wolphram jonny
Jan 2 at 20:42
@Wolphramjonny $mathbf{E}$ has always been pre-determined before you choose any $ell$ or $Sigma$. While you may take different $ell$ and $Sigma$ to figure out the value of $mathbf{E}$, your choice does not change its value.
– hypernova
Jan 3 at 13:38
@Wolphramjonny $mathbf{E}$ has always been pre-determined before you choose any $ell$ or $Sigma$. While you may take different $ell$ and $Sigma$ to figure out the value of $mathbf{E}$, your choice does not change its value.
– hypernova
Jan 3 at 13:38
But is not your argument, which seems correct, that a good choice of $Sigma$ will result in an area integral =0? in which case the computed fiels prediction is also zero? but with another surface choice the result would be different than zero? In such a case, regardless of physical reality(I agree E is well defined), the equations make inconsistent predictions, so the question is: what is it wrong with the equations?
– Wolphram jonny
Jan 3 at 14:14
But is not your argument, which seems correct, that a good choice of $Sigma$ will result in an area integral =0? in which case the computed fiels prediction is also zero? but with another surface choice the result would be different than zero? In such a case, regardless of physical reality(I agree E is well defined), the equations make inconsistent predictions, so the question is: what is it wrong with the equations?
– Wolphram jonny
Jan 3 at 14:14
@Wolphramjonny: Not really. If you can find some $Sigma_1$ such that $oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$, then for any other $Sigma_2$ with $partialSigma_2=partialSigma_1$, it is a must that $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}=0$. My argument is thus as follows. If $mathbf{B}$ observes a tiny change on $Sigma_2$, it is hard to figure out $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}$. Nevertheless, if you can find some $Sigma_1$ (with $partialSigma_1=partialSigma_2$) on which $mathbf{B}$ does not change for the moment,
– hypernova
Jan 3 at 15:09
@Wolphramjonny: Not really. If you can find some $Sigma_1$ such that $oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$, then for any other $Sigma_2$ with $partialSigma_2=partialSigma_1$, it is a must that $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}=0$. My argument is thus as follows. If $mathbf{B}$ observes a tiny change on $Sigma_2$, it is hard to figure out $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}$. Nevertheless, if you can find some $Sigma_1$ (with $partialSigma_1=partialSigma_2$) on which $mathbf{B}$ does not change for the moment,
– hypernova
Jan 3 at 15:09
@Wolphramjonny: (con't) then it is straightforward that $oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$. This result, in turn, implies $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}=oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$. This is a trick to help figure out $mathbf{E}$. After all, your doubt to me lies in that, from the integral form, it is hard to tell if a tiny change in $mathbf{B}$ influences the value of $mathbf{E}$ at a remote location. The above trick is to clarify that there is no such influence.
– hypernova
Jan 3 at 15:10
@Wolphramjonny: (con't) then it is straightforward that $oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$. This result, in turn, implies $oint_{partialSigma_2}mathbf{E}cdot{rm d}mathbf{l}=oint_{partialSigma_1}mathbf{E}cdot{rm d}mathbf{l}=0$. This is a trick to help figure out $mathbf{E}$. After all, your doubt to me lies in that, from the integral form, it is hard to tell if a tiny change in $mathbf{B}$ influences the value of $mathbf{E}$ at a remote location. The above trick is to clarify that there is no such influence.
– hypernova
Jan 3 at 15:10
|
show 4 more comments
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I think you should look up the concept of the retarded potential: EM affects travel at the speed of light. So no, information does not travel instantaneously.
– Adrian Keister
Dec 31 '18 at 18:08
@AdrianKeister thanks, I know that it does not, but it is what the second equation is saying, or not? E(t) at the edge depends on B(t) at the center, without any delays
– Wolphram jonny
Jan 1 at 0:21
The integral version is not telling you what you wrote. Instead, think of it this way: an instantaneous change in the magnetic field in the center would affect the rate of change of the electric field on the surface, not the value of the electric field itself.
– Adrian Keister
Jan 1 at 0:35
I am not sure I agree, let us say we have circular simmetry, then you have E=l*d/dt(int Bds), the E is at the edge of the surface, but the change in B can happen anywhere, like just at the center of the surface.
– Wolphram jonny
Jan 1 at 1:40