Mixing
Interpretation of stokes theorem
$begingroup$
I recently solved the following task:
Let $A = [0,1]^3$ and $omega = dfrac{x_1^2 x_2^3}{1+x_3^2} dx_1 wedge dx_3$ Show that this fulfills stokes theorem by showing that $displaystyle int_A domega = displaystyle int_{partial A} omega $.
That worked out really well. As the solution I got for both sides $-dfrac{pi}{12}$. The question is how to interpret this (maybe physically). I just learned about diffrential forms a few days ago and Im not sure how to interpret them. I think that integrating a 1-form over a curve gives the work that the vectorfield applies on a particle that walks along this curve. Is $-dfrac{pi}{12}$ something like the work that the vectorfield applies on the cube or something like that? Please keep in mind that Im not a physicist at all.
calculus differential-forms stokes-theorem
asked Jan 27 at 17:28
ArjihadArjihad
390112
$endgroup$
|
show 4 more comments
$begingroup$
I recently solved the following task:
Let $A = [0,1]^3$ and $omega = dfrac{x_1^2 x_2^3}{1+x_3^2} dx_1 wedge dx_3$ Show that this fulfills stokes theorem by showing that $displaystyle int_A domega = displaystyle int_{partial A} omega $.
That worked out really well. As the solution I got for both sides $-dfrac{pi}{12}$. The question is how to interpret this (maybe physically). I just learned about diffrential forms a few days ago and Im not sure how to interpret them. I think that integrating a 1-form over a curve gives the work that the vectorfield applies on a particle that walks along this curve. Is $-dfrac{pi}{12}$ something like the work that the vectorfield applies on the cube or something like that? Please keep in mind that Im not a physicist at all.
calculus differential-forms stokes-theorem
asked Jan 27 at 17:28
ArjihadArjihad
390112
$endgroup$
$begingroup$
I assume that $[0,1^3]$ should be $[0,1]^3$.
$endgroup$
– md2perpe
Jan 27 at 20:35
$begingroup$
Oh yes. Im going to change that immediately.
$endgroup$
– Arjihad
Jan 27 at 20:36
$begingroup$
Why do you feel a need to interpret this physically?
$endgroup$
– md2perpe
Jan 27 at 20:42
$begingroup$
I just wonder what this result tells me. What does it mean to get a negative value? Using the standard integrals the value would stand for the area under a curve or a volume but what does this value stand for? If this would just be a random number I guess the whole chapter would not be interesting.
$endgroup$
– Arjihad
Jan 27 at 20:47
$begingroup$
This exercise was just an example to have you test the theorem. It doesn't need an interpretation just like $int_0^1 3x^2 + 7x + 5 , dx$ doesn't need an interpretation. It could represent work, but it could as well just represent the area under a function.
$endgroup$
– md2perpe
Jan 27 at 20:55
|
show 4 more comments
$begingroup$
I recently solved the following task:
Let $A = [0,1]^3$ and $omega = dfrac{x_1^2 x_2^3}{1+x_3^2} dx_1 wedge dx_3$ Show that this fulfills stokes theorem by showing that $displaystyle int_A domega = displaystyle int_{partial A} omega $.
That worked out really well. As the solution I got for both sides $-dfrac{pi}{12}$. The question is how to interpret this (maybe physically). I just learned about diffrential forms a few days ago and Im not sure how to interpret them. I think that integrating a 1-form over a curve gives the work that the vectorfield applies on a particle that walks along this curve. Is $-dfrac{pi}{12}$ something like the work that the vectorfield applies on the cube or something like that? Please keep in mind that Im not a physicist at all.
calculus differential-forms stokes-theorem
asked Jan 27 at 17:28
ArjihadArjihad
390112
$endgroup$
I recently solved the following task:
Let $A = [0,1]^3$ and $omega = dfrac{x_1^2 x_2^3}{1+x_3^2} dx_1 wedge dx_3$ Show that this fulfills stokes theorem by showing that $displaystyle int_A domega = displaystyle int_{partial A} omega $.
That worked out really well. As the solution I got for both sides $-dfrac{pi}{12}$. The question is how to interpret this (maybe physically). I just learned about diffrential forms a few days ago and Im not sure how to interpret them. I think that integrating a 1-form over a curve gives the work that the vectorfield applies on a particle that walks along this curve. Is $-dfrac{pi}{12}$ something like the work that the vectorfield applies on the cube or something like that? Please keep in mind that Im not a physicist at all.
calculus differential-forms stokes-theorem
calculus differential-forms stokes-theorem
asked Jan 27 at 17:28
ArjihadArjihad
390112
asked Jan 27 at 17:28
ArjihadArjihad
390112
edited Jan 28 at 9:45
Arjihad
asked Jan 27 at 17:28
ArjihadArjihad
390112
asked Jan 27 at 17:28
ArjihadArjihad
390112
asked Jan 27 at 17:28
ArjihadArjihad
390112
390112
$begingroup$
I assume that $[0,1^3]$ should be $[0,1]^3$.
$endgroup$
– md2perpe
Jan 27 at 20:35
$begingroup$
Oh yes. Im going to change that immediately.
$endgroup$
– Arjihad
Jan 27 at 20:36
$begingroup$
Why do you feel a need to interpret this physically?
$endgroup$
– md2perpe
Jan 27 at 20:42
$begingroup$
I just wonder what this result tells me. What does it mean to get a negative value? Using the standard integrals the value would stand for the area under a curve or a volume but what does this value stand for? If this would just be a random number I guess the whole chapter would not be interesting.
$endgroup$
– Arjihad
Jan 27 at 20:47
$begingroup$
This exercise was just an example to have you test the theorem. It doesn't need an interpretation just like $int_0^1 3x^2 + 7x + 5 , dx$ doesn't need an interpretation. It could represent work, but it could as well just represent the area under a function.
$endgroup$
– md2perpe
Jan 27 at 20:55
|
show 4 more comments
$begingroup$
I assume that $[0,1^3]$ should be $[0,1]^3$.
$endgroup$
– md2perpe
Jan 27 at 20:35
$begingroup$
Oh yes. Im going to change that immediately.
$endgroup$
– Arjihad
Jan 27 at 20:36
$begingroup$
Why do you feel a need to interpret this physically?
$endgroup$
– md2perpe
Jan 27 at 20:42
$begingroup$
I just wonder what this result tells me. What does it mean to get a negative value? Using the standard integrals the value would stand for the area under a curve or a volume but what does this value stand for? If this would just be a random number I guess the whole chapter would not be interesting.
$endgroup$
– Arjihad
Jan 27 at 20:47
$begingroup$
This exercise was just an example to have you test the theorem. It doesn't need an interpretation just like $int_0^1 3x^2 + 7x + 5 , dx$ doesn't need an interpretation. It could represent work, but it could as well just represent the area under a function.
$endgroup$
– md2perpe
Jan 27 at 20:55
$begingroup$
I assume that $[0,1^3]$ should be $[0,1]^3$.
$endgroup$
– md2perpe
Jan 27 at 20:35
$begingroup$
I assume that $[0,1^3]$ should be $[0,1]^3$.
$endgroup$
– md2perpe
Jan 27 at 20:35
$begingroup$
Oh yes. Im going to change that immediately.
$endgroup$
– Arjihad
Jan 27 at 20:36
$begingroup$
Oh yes. Im going to change that immediately.
$endgroup$
– Arjihad
Jan 27 at 20:36
$begingroup$
Why do you feel a need to interpret this physically?
$endgroup$
– md2perpe
Jan 27 at 20:42
$begingroup$
Why do you feel a need to interpret this physically?
$endgroup$
– md2perpe
Jan 27 at 20:42
$begingroup$
I just wonder what this result tells me. What does it mean to get a negative value? Using the standard integrals the value would stand for the area under a curve or a volume but what does this value stand for? If this would just be a random number I guess the whole chapter would not be interesting.
$endgroup$
– Arjihad
Jan 27 at 20:47
$begingroup$
I just wonder what this result tells me. What does it mean to get a negative value? Using the standard integrals the value would stand for the area under a curve or a volume but what does this value stand for? If this would just be a random number I guess the whole chapter would not be interesting.
$endgroup$
– Arjihad
Jan 27 at 20:47
$begingroup$
This exercise was just an example to have you test the theorem. It doesn't need an interpretation just like $int_0^1 3x^2 + 7x + 5 , dx$ doesn't need an interpretation. It could represent work, but it could as well just represent the area under a function.
$endgroup$
– md2perpe
Jan 27 at 20:55
$begingroup$
This exercise was just an example to have you test the theorem. It doesn't need an interpretation just like $int_0^1 3x^2 + 7x + 5 , dx$ doesn't need an interpretation. It could represent work, but it could as well just represent the area under a function.
$endgroup$
– md2perpe
Jan 27 at 20:55
|
show 4 more comments
1 Answer
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votes
$begingroup$
OK, so the physical interpretation is that you're finding the flux of the vector field $vec F = begin{bmatrix} 0 \ -x_1^2x_2^3/(1+x_3^2) \ 0end{bmatrix}$ outwards across $partial A$. In this setting, $domega = (text{div}, vec F) dx_1wedge dx_2wedge dx_3$, and $text{div},vec F$ tells you infinitesimally whether you have a source or a sink (or neither) at each point; adding these up all over $A$ gives the net flux across the boundary.
answered Jan 28 at 19:28
Ted ShifrinTed Shifrin
64.5k44692
$endgroup$
add a comment |
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1 Answer
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$begingroup$
OK, so the physical interpretation is that you're finding the flux of the vector field $vec F = begin{bmatrix} 0 \ -x_1^2x_2^3/(1+x_3^2) \ 0end{bmatrix}$ outwards across $partial A$. In this setting, $domega = (text{div}, vec F) dx_1wedge dx_2wedge dx_3$, and $text{div},vec F$ tells you infinitesimally whether you have a source or a sink (or neither) at each point; adding these up all over $A$ gives the net flux across the boundary.
answered Jan 28 at 19:28
Ted ShifrinTed Shifrin
64.5k44692
$endgroup$
add a comment |
$begingroup$
OK, so the physical interpretation is that you're finding the flux of the vector field $vec F = begin{bmatrix} 0 \ -x_1^2x_2^3/(1+x_3^2) \ 0end{bmatrix}$ outwards across $partial A$. In this setting, $domega = (text{div}, vec F) dx_1wedge dx_2wedge dx_3$, and $text{div},vec F$ tells you infinitesimally whether you have a source or a sink (or neither) at each point; adding these up all over $A$ gives the net flux across the boundary.
answered Jan 28 at 19:28
Ted ShifrinTed Shifrin
64.5k44692
$endgroup$
add a comment |
$begingroup$
OK, so the physical interpretation is that you're finding the flux of the vector field $vec F = begin{bmatrix} 0 \ -x_1^2x_2^3/(1+x_3^2) \ 0end{bmatrix}$ outwards across $partial A$. In this setting, $domega = (text{div}, vec F) dx_1wedge dx_2wedge dx_3$, and $text{div},vec F$ tells you infinitesimally whether you have a source or a sink (or neither) at each point; adding these up all over $A$ gives the net flux across the boundary.
answered Jan 28 at 19:28
Ted ShifrinTed Shifrin
64.5k44692
$endgroup$
OK, so the physical interpretation is that you're finding the flux of the vector field $vec F = begin{bmatrix} 0 \ -x_1^2x_2^3/(1+x_3^2) \ 0end{bmatrix}$ outwards across $partial A$. In this setting, $domega = (text{div}, vec F) dx_1wedge dx_2wedge dx_3$, and $text{div},vec F$ tells you infinitesimally whether you have a source or a sink (or neither) at each point; adding these up all over $A$ gives the net flux across the boundary.
answered Jan 28 at 19:28
Ted ShifrinTed Shifrin
64.5k44692
answered Jan 28 at 19:28
Ted ShifrinTed Shifrin
64.5k44692
answered Jan 28 at 19:28
Ted ShifrinTed Shifrin
64.5k44692
answered Jan 28 at 19:28
Ted ShifrinTed Shifrin
64.5k44692
64.5k44692
add a comment |
add a comment |
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$begingroup$
I assume that $[0,1^3]$ should be $[0,1]^3$.
$endgroup$
– md2perpe
Jan 27 at 20:35
$begingroup$
Oh yes. Im going to change that immediately.
$endgroup$
– Arjihad
Jan 27 at 20:36
$begingroup$
Why do you feel a need to interpret this physically?
$endgroup$
– md2perpe
Jan 27 at 20:42
$begingroup$
I just wonder what this result tells me. What does it mean to get a negative value? Using the standard integrals the value would stand for the area under a curve or a volume but what does this value stand for? If this would just be a random number I guess the whole chapter would not be interesting.
$endgroup$
– Arjihad
Jan 27 at 20:47
$begingroup$
This exercise was just an example to have you test the theorem. It doesn't need an interpretation just like $int_0^1 3x^2 + 7x + 5 , dx$ doesn't need an interpretation. It could represent work, but it could as well just represent the area under a function.
$endgroup$
– md2perpe
Jan 27 at 20:55