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Is it true to say that:
$omega =(x+y)dx+(x^3+2Ày)dy+(x^5+y)dz=(x+y)hati+(x^3+2Ày)hatj+(x^5+y)hatk$?
multivariable-calculus differential-forms
asked Aug 2 at 18:24
newhere
740310
put on hold as off-topic by John Ma, Mostafa Ayaz, amWhy, Isaac Browne, Taroccoesbrocco Aug 3 at 0:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Mostafa Ayaz, amWhy, Isaac Browne, Taroccoesbrocco
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up vote
2
down vote
favorite
Is it true to say that:
$omega =(x+y)dx+(x^3+2Ày)dy+(x^5+y)dz=(x+y)hati+(x^3+2Ày)hatj+(x^5+y)hatk$?
multivariable-calculus differential-forms
asked Aug 2 at 18:24
newhere
740310
put on hold as off-topic by John Ma, Mostafa Ayaz, amWhy, Isaac Browne, Taroccoesbrocco Aug 3 at 0:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Mostafa Ayaz, amWhy, Isaac Browne, Taroccoesbrocco
No, it is not. The first piece is a differential form whereas the second is a vector field. Vector fields and differential forms are not the same, they are dual concepts.
– PtF
Aug 2 at 18:48
1
The notation ($hat i$, etc.) used makes it obvious that the OP has no modern differential geometric background. Maybe his physics professor has told him that these two things are "essentially the same", without explaining the underlying concepts. Now he comes here, and asks for an explanation. Why does the 5VS (five votes suffice) squad throw the hammer at him?
– Christian Blatter
Aug 3 at 8:10
I am not sure why we closed this question, it is pretty clear to me what OP is asking here...seems like an innocent question.
– Nameless
Aug 3 at 9:47
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up vote
2
down vote
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up vote
2
down vote
favorite
Is it true to say that:
$omega =(x+y)dx+(x^3+2Ày)dy+(x^5+y)dz=(x+y)hati+(x^3+2Ày)hatj+(x^5+y)hatk$?
multivariable-calculus differential-forms
asked Aug 2 at 18:24
newhere
740310
Is it true to say that:
$omega =(x+y)dx+(x^3+2Ày)dy+(x^5+y)dz=(x+y)hati+(x^3+2Ày)hatj+(x^5+y)hatk$?
multivariable-calculus differential-forms
asked Aug 2 at 18:24
newhere
740310
asked Aug 2 at 18:24
newhere
740310
asked Aug 2 at 18:24
newhere
740310
asked Aug 2 at 18:24
newhere
740310
740310
put on hold as off-topic by John Ma, Mostafa Ayaz, amWhy, Isaac Browne, Taroccoesbrocco Aug 3 at 0:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Mostafa Ayaz, amWhy, Isaac Browne, Taroccoesbrocco
put on hold as off-topic by John Ma, Mostafa Ayaz, amWhy, Isaac Browne, Taroccoesbrocco Aug 3 at 0:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Mostafa Ayaz, amWhy, Isaac Browne, Taroccoesbrocco
No, it is not. The first piece is a differential form whereas the second is a vector field. Vector fields and differential forms are not the same, they are dual concepts.
– PtF
Aug 2 at 18:48
1
The notation ($hat i$, etc.) used makes it obvious that the OP has no modern differential geometric background. Maybe his physics professor has told him that these two things are "essentially the same", without explaining the underlying concepts. Now he comes here, and asks for an explanation. Why does the 5VS (five votes suffice) squad throw the hammer at him?
– Christian Blatter
Aug 3 at 8:10
I am not sure why we closed this question, it is pretty clear to me what OP is asking here...seems like an innocent question.
– Nameless
Aug 3 at 9:47
add a comment |Â
No, it is not. The first piece is a differential form whereas the second is a vector field. Vector fields and differential forms are not the same, they are dual concepts.
– PtF
Aug 2 at 18:48
1
The notation ($hat i$, etc.) used makes it obvious that the OP has no modern differential geometric background. Maybe his physics professor has told him that these two things are "essentially the same", without explaining the underlying concepts. Now he comes here, and asks for an explanation. Why does the 5VS (five votes suffice) squad throw the hammer at him?
– Christian Blatter
Aug 3 at 8:10
I am not sure why we closed this question, it is pretty clear to me what OP is asking here...seems like an innocent question.
– Nameless
Aug 3 at 9:47
No, it is not. The first piece is a differential form whereas the second is a vector field. Vector fields and differential forms are not the same, they are dual concepts.
– PtF
Aug 2 at 18:48
No, it is not. The first piece is a differential form whereas the second is a vector field. Vector fields and differential forms are not the same, they are dual concepts.
– PtF
Aug 2 at 18:48
1
1
The notation ($hat i$, etc.) used makes it obvious that the OP has no modern differential geometric background. Maybe his physics professor has told him that these two things are "essentially the same", without explaining the underlying concepts. Now he comes here, and asks for an explanation. Why does the 5VS (five votes suffice) squad throw the hammer at him?
– Christian Blatter
Aug 3 at 8:10
The notation ($hat i$, etc.) used makes it obvious that the OP has no modern differential geometric background. Maybe his physics professor has told him that these two things are "essentially the same", without explaining the underlying concepts. Now he comes here, and asks for an explanation. Why does the 5VS (five votes suffice) squad throw the hammer at him?
– Christian Blatter
Aug 3 at 8:10
I am not sure why we closed this question, it is pretty clear to me what OP is asking here...seems like an innocent question.
– Nameless
Aug 3 at 9:47
I am not sure why we closed this question, it is pretty clear to me what OP is asking here...seems like an innocent question.
– Nameless
Aug 3 at 9:47
add a comment |Â
1 Answer
1
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up vote
2
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accepted
Physically they amount to the same thing, but you should not put an equality sign between them.
In "Advanced Calculus" we use vector fields for two completely different things: To describe (i) "force fields", which are integrated over lines, and (ii) "flow fields", which are integrated over (resp., across) surfaces. In the case at hand it seems that
$$bf F(x,y,z):=(x+y, x^3+2pi y, x^5+y)$$
is viewed as a "force field". The work needed to push a cart against this field from $bf r$ to $bf r+dbf r$ is $$dW=bf F(bf r)cdot dbf r .$$ Note that we have used a scalar product to compute this work. The scalar product has allowed us to view the "force" as a single vector, even though this "force" acts in different directions with different intensities.
Now there is a different view of the same physical situation. Here the idea of "force" felt when we push a cart from $bf r$ to $bf r+bf X$ is encoded in a linear form $omega(bf r)in T_bf r^*$, and one has $$dW=omega(bf r).bf X ,$$ whereby the . denotes application of the linear form (functional) $ omega(bf r)$ to the tangent vector $bf Xin T_bf r$. The coordinate differentials $dx$, $dy$, $dz$ form a basis of the space $T_bf r^*$; hence $omega(bf r)$ can be written as a linear combination of these. And lo and behold: It turns out that one has
$$omega(x,y,z)=(x+y)dx+(x^3+2pi y)dy+ (x^5+y) dz$$
when $omega$ is the linear form corresponding to the above $bf F$.
answered Aug 2 at 19:35
Christian Blatter
163k7106305
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Physically they amount to the same thing, but you should not put an equality sign between them.
In "Advanced Calculus" we use vector fields for two completely different things: To describe (i) "force fields", which are integrated over lines, and (ii) "flow fields", which are integrated over (resp., across) surfaces. In the case at hand it seems that
$$bf F(x,y,z):=(x+y, x^3+2pi y, x^5+y)$$
is viewed as a "force field". The work needed to push a cart against this field from $bf r$ to $bf r+dbf r$ is $$dW=bf F(bf r)cdot dbf r .$$ Note that we have used a scalar product to compute this work. The scalar product has allowed us to view the "force" as a single vector, even though this "force" acts in different directions with different intensities.
Now there is a different view of the same physical situation. Here the idea of "force" felt when we push a cart from $bf r$ to $bf r+bf X$ is encoded in a linear form $omega(bf r)in T_bf r^*$, and one has $$dW=omega(bf r).bf X ,$$ whereby the . denotes application of the linear form (functional) $ omega(bf r)$ to the tangent vector $bf Xin T_bf r$. The coordinate differentials $dx$, $dy$, $dz$ form a basis of the space $T_bf r^*$; hence $omega(bf r)$ can be written as a linear combination of these. And lo and behold: It turns out that one has
$$omega(x,y,z)=(x+y)dx+(x^3+2pi y)dy+ (x^5+y) dz$$
when $omega$ is the linear form corresponding to the above $bf F$.
answered Aug 2 at 19:35
Christian Blatter
163k7106305
add a comment |Â
up vote
2
down vote
accepted
Physically they amount to the same thing, but you should not put an equality sign between them.
In "Advanced Calculus" we use vector fields for two completely different things: To describe (i) "force fields", which are integrated over lines, and (ii) "flow fields", which are integrated over (resp., across) surfaces. In the case at hand it seems that
$$bf F(x,y,z):=(x+y, x^3+2pi y, x^5+y)$$
is viewed as a "force field". The work needed to push a cart against this field from $bf r$ to $bf r+dbf r$ is $$dW=bf F(bf r)cdot dbf r .$$ Note that we have used a scalar product to compute this work. The scalar product has allowed us to view the "force" as a single vector, even though this "force" acts in different directions with different intensities.
Now there is a different view of the same physical situation. Here the idea of "force" felt when we push a cart from $bf r$ to $bf r+bf X$ is encoded in a linear form $omega(bf r)in T_bf r^*$, and one has $$dW=omega(bf r).bf X ,$$ whereby the . denotes application of the linear form (functional) $ omega(bf r)$ to the tangent vector $bf Xin T_bf r$. The coordinate differentials $dx$, $dy$, $dz$ form a basis of the space $T_bf r^*$; hence $omega(bf r)$ can be written as a linear combination of these. And lo and behold: It turns out that one has
$$omega(x,y,z)=(x+y)dx+(x^3+2pi y)dy+ (x^5+y) dz$$
when $omega$ is the linear form corresponding to the above $bf F$.
answered Aug 2 at 19:35
Christian Blatter
163k7106305
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Physically they amount to the same thing, but you should not put an equality sign between them.
In "Advanced Calculus" we use vector fields for two completely different things: To describe (i) "force fields", which are integrated over lines, and (ii) "flow fields", which are integrated over (resp., across) surfaces. In the case at hand it seems that
$$bf F(x,y,z):=(x+y, x^3+2pi y, x^5+y)$$
is viewed as a "force field". The work needed to push a cart against this field from $bf r$ to $bf r+dbf r$ is $$dW=bf F(bf r)cdot dbf r .$$ Note that we have used a scalar product to compute this work. The scalar product has allowed us to view the "force" as a single vector, even though this "force" acts in different directions with different intensities.
Now there is a different view of the same physical situation. Here the idea of "force" felt when we push a cart from $bf r$ to $bf r+bf X$ is encoded in a linear form $omega(bf r)in T_bf r^*$, and one has $$dW=omega(bf r).bf X ,$$ whereby the . denotes application of the linear form (functional) $ omega(bf r)$ to the tangent vector $bf Xin T_bf r$. The coordinate differentials $dx$, $dy$, $dz$ form a basis of the space $T_bf r^*$; hence $omega(bf r)$ can be written as a linear combination of these. And lo and behold: It turns out that one has
$$omega(x,y,z)=(x+y)dx+(x^3+2pi y)dy+ (x^5+y) dz$$
when $omega$ is the linear form corresponding to the above $bf F$.
answered Aug 2 at 19:35
Christian Blatter
163k7106305
Physically they amount to the same thing, but you should not put an equality sign between them.
In "Advanced Calculus" we use vector fields for two completely different things: To describe (i) "force fields", which are integrated over lines, and (ii) "flow fields", which are integrated over (resp., across) surfaces. In the case at hand it seems that
$$bf F(x,y,z):=(x+y, x^3+2pi y, x^5+y)$$
is viewed as a "force field". The work needed to push a cart against this field from $bf r$ to $bf r+dbf r$ is $$dW=bf F(bf r)cdot dbf r .$$ Note that we have used a scalar product to compute this work. The scalar product has allowed us to view the "force" as a single vector, even though this "force" acts in different directions with different intensities.
Now there is a different view of the same physical situation. Here the idea of "force" felt when we push a cart from $bf r$ to $bf r+bf X$ is encoded in a linear form $omega(bf r)in T_bf r^*$, and one has $$dW=omega(bf r).bf X ,$$ whereby the . denotes application of the linear form (functional) $ omega(bf r)$ to the tangent vector $bf Xin T_bf r$. The coordinate differentials $dx$, $dy$, $dz$ form a basis of the space $T_bf r^*$; hence $omega(bf r)$ can be written as a linear combination of these. And lo and behold: It turns out that one has
$$omega(x,y,z)=(x+y)dx+(x^3+2pi y)dy+ (x^5+y) dz$$
when $omega$ is the linear form corresponding to the above $bf F$.
answered Aug 2 at 19:35
Christian Blatter
163k7106305
edited Aug 3 at 8:53
answered Aug 2 at 19:35
Christian Blatter
163k7106305
answered Aug 2 at 19:35
Christian Blatter
163k7106305
answered Aug 2 at 19:35
Christian Blatter
163k7106305
163k7106305
add a comment |Â
add a comment |Â
No, it is not. The first piece is a differential form whereas the second is a vector field. Vector fields and differential forms are not the same, they are dual concepts.
– PtF
Aug 2 at 18:48
1
The notation ($hat i$, etc.) used makes it obvious that the OP has no modern differential geometric background. Maybe his physics professor has told him that these two things are "essentially the same", without explaining the underlying concepts. Now he comes here, and asks for an explanation. Why does the 5VS (five votes suffice) squad throw the hammer at him?
– Christian Blatter
Aug 3 at 8:10
I am not sure why we closed this question, it is pretty clear to me what OP is asking here...seems like an innocent question.
– Nameless
Aug 3 at 9:47