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Relationship between between “calculus differentials” and exterior differentials
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I'm having a hard time understanding differentials. I know this topic has been covered here in StackExchange, but I really can't find an answer to my question here.
So, I've been taught two different notions of differential and I don't know how to reconcile them. The first one was in my first course on differential equations, in which $text{d}y$ was just the linear part of $Delta{y}$, which can be defined iff $f text{ such, that } y=f(x)$ is differentiable (makes sense) as
$$text{d}y=frac{partial f}{partial x^i}Delta{x^i}$$
and, as $Delta{x^i}$ depends linearly on $Delta{x^i}$, we can say it equals $text{d}x^i$ and get the usual expression for the differential of a scalar field. This sort of operator $text{d}$ is linear and satisfies $text{d}(gh)=htext{d}g+gtext{d}h$, Leibniz's rule.
The second one was in the context of differential forms. Here, $text{d}y$ was a tensor field of type $(1,0)$ such, that
$$text{d}y=frac{partial{f}}{partial{x}^i}text{d}x^i,spacespacespacespacespacetext{d}x^i(partial_j)=delta^i{}_j.$$
where we're using the exterior differential operator $text{d}$, which is also linear and also satisfies Leibniz's rule.
Up to this point, both notions seem to be very compatible. Troubles come when we try to multiply or divide them. As "calculus differentials" represent functions from $(Delta{x^1},...,Delta{x^n})$ to $mathbb{R}$ (if we're doing real-variable calculus), we can multiply and divide them pointwise just fine. In fact, we can prove that
$$left(frac{text{d}y}{text{d}u}right)_{u=varphi(a)}left(frac{text{d}w}{text{d}x}right)_{x=a}=left(frac{text{d}y}{text{d}x}right)_{x=a}left(frac{text{d}w}{text{d}u}right)_{u=varphi(a)}$$
for one-variable differentiable functions, which seem to imply that is totally legit to manipulate first-order total differentials algebraically as long as we work only with differentiable functions.
But, if we want to multiply differential forms, we can only do that by means of the exterior product. This means that we can't divide differential forms and it also implies that $text{d}x^1text{d}x^2$ as "calculus differentials" is not the same as $text{d}x^1wedgetext{d}x^2$, as exterior differentials.
We can also talk about $text{d}^2y$ for "calculus differentials", but it doesn't make much sense to do it with exterior differentials because there $text{d}^2$ is the $0$ operator.
Clearly, calculus differentials and exterior differentials are different things; but if they are both called differentials, there must be a connection between them, and that is what I'm asking for: I would like a brief explanation of their connection or references to notes or books which discus it.
differential-forms
asked 2 days ago
TeicDaun
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I'm having a hard time understanding differentials. I know this topic has been covered here in StackExchange, but I really can't find an answer to my question here.
So, I've been taught two different notions of differential and I don't know how to reconcile them. The first one was in my first course on differential equations, in which $text{d}y$ was just the linear part of $Delta{y}$, which can be defined iff $f text{ such, that } y=f(x)$ is differentiable (makes sense) as
$$text{d}y=frac{partial f}{partial x^i}Delta{x^i}$$
and, as $Delta{x^i}$ depends linearly on $Delta{x^i}$, we can say it equals $text{d}x^i$ and get the usual expression for the differential of a scalar field. This sort of operator $text{d}$ is linear and satisfies $text{d}(gh)=htext{d}g+gtext{d}h$, Leibniz's rule.
The second one was in the context of differential forms. Here, $text{d}y$ was a tensor field of type $(1,0)$ such, that
$$text{d}y=frac{partial{f}}{partial{x}^i}text{d}x^i,spacespacespacespacespacetext{d}x^i(partial_j)=delta^i{}_j.$$
where we're using the exterior differential operator $text{d}$, which is also linear and also satisfies Leibniz's rule.
Up to this point, both notions seem to be very compatible. Troubles come when we try to multiply or divide them. As "calculus differentials" represent functions from $(Delta{x^1},...,Delta{x^n})$ to $mathbb{R}$ (if we're doing real-variable calculus), we can multiply and divide them pointwise just fine. In fact, we can prove that
$$left(frac{text{d}y}{text{d}u}right)_{u=varphi(a)}left(frac{text{d}w}{text{d}x}right)_{x=a}=left(frac{text{d}y}{text{d}x}right)_{x=a}left(frac{text{d}w}{text{d}u}right)_{u=varphi(a)}$$
for one-variable differentiable functions, which seem to imply that is totally legit to manipulate first-order total differentials algebraically as long as we work only with differentiable functions.
But, if we want to multiply differential forms, we can only do that by means of the exterior product. This means that we can't divide differential forms and it also implies that $text{d}x^1text{d}x^2$ as "calculus differentials" is not the same as $text{d}x^1wedgetext{d}x^2$, as exterior differentials.
We can also talk about $text{d}^2y$ for "calculus differentials", but it doesn't make much sense to do it with exterior differentials because there $text{d}^2$ is the $0$ operator.
Clearly, calculus differentials and exterior differentials are different things; but if they are both called differentials, there must be a connection between them, and that is what I'm asking for: I would like a brief explanation of their connection or references to notes or books which discus it.
differential-forms
asked 2 days ago
TeicDaun
7116
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm having a hard time understanding differentials. I know this topic has been covered here in StackExchange, but I really can't find an answer to my question here.
So, I've been taught two different notions of differential and I don't know how to reconcile them. The first one was in my first course on differential equations, in which $text{d}y$ was just the linear part of $Delta{y}$, which can be defined iff $f text{ such, that } y=f(x)$ is differentiable (makes sense) as
$$text{d}y=frac{partial f}{partial x^i}Delta{x^i}$$
and, as $Delta{x^i}$ depends linearly on $Delta{x^i}$, we can say it equals $text{d}x^i$ and get the usual expression for the differential of a scalar field. This sort of operator $text{d}$ is linear and satisfies $text{d}(gh)=htext{d}g+gtext{d}h$, Leibniz's rule.
The second one was in the context of differential forms. Here, $text{d}y$ was a tensor field of type $(1,0)$ such, that
$$text{d}y=frac{partial{f}}{partial{x}^i}text{d}x^i,spacespacespacespacespacetext{d}x^i(partial_j)=delta^i{}_j.$$
where we're using the exterior differential operator $text{d}$, which is also linear and also satisfies Leibniz's rule.
Up to this point, both notions seem to be very compatible. Troubles come when we try to multiply or divide them. As "calculus differentials" represent functions from $(Delta{x^1},...,Delta{x^n})$ to $mathbb{R}$ (if we're doing real-variable calculus), we can multiply and divide them pointwise just fine. In fact, we can prove that
$$left(frac{text{d}y}{text{d}u}right)_{u=varphi(a)}left(frac{text{d}w}{text{d}x}right)_{x=a}=left(frac{text{d}y}{text{d}x}right)_{x=a}left(frac{text{d}w}{text{d}u}right)_{u=varphi(a)}$$
for one-variable differentiable functions, which seem to imply that is totally legit to manipulate first-order total differentials algebraically as long as we work only with differentiable functions.
But, if we want to multiply differential forms, we can only do that by means of the exterior product. This means that we can't divide differential forms and it also implies that $text{d}x^1text{d}x^2$ as "calculus differentials" is not the same as $text{d}x^1wedgetext{d}x^2$, as exterior differentials.
We can also talk about $text{d}^2y$ for "calculus differentials", but it doesn't make much sense to do it with exterior differentials because there $text{d}^2$ is the $0$ operator.
Clearly, calculus differentials and exterior differentials are different things; but if they are both called differentials, there must be a connection between them, and that is what I'm asking for: I would like a brief explanation of their connection or references to notes or books which discus it.
differential-forms
asked 2 days ago
TeicDaun
7116
I'm having a hard time understanding differentials. I know this topic has been covered here in StackExchange, but I really can't find an answer to my question here.
So, I've been taught two different notions of differential and I don't know how to reconcile them. The first one was in my first course on differential equations, in which $text{d}y$ was just the linear part of $Delta{y}$, which can be defined iff $f text{ such, that } y=f(x)$ is differentiable (makes sense) as
$$text{d}y=frac{partial f}{partial x^i}Delta{x^i}$$
and, as $Delta{x^i}$ depends linearly on $Delta{x^i}$, we can say it equals $text{d}x^i$ and get the usual expression for the differential of a scalar field. This sort of operator $text{d}$ is linear and satisfies $text{d}(gh)=htext{d}g+gtext{d}h$, Leibniz's rule.
The second one was in the context of differential forms. Here, $text{d}y$ was a tensor field of type $(1,0)$ such, that
$$text{d}y=frac{partial{f}}{partial{x}^i}text{d}x^i,spacespacespacespacespacetext{d}x^i(partial_j)=delta^i{}_j.$$
where we're using the exterior differential operator $text{d}$, which is also linear and also satisfies Leibniz's rule.
Up to this point, both notions seem to be very compatible. Troubles come when we try to multiply or divide them. As "calculus differentials" represent functions from $(Delta{x^1},...,Delta{x^n})$ to $mathbb{R}$ (if we're doing real-variable calculus), we can multiply and divide them pointwise just fine. In fact, we can prove that
$$left(frac{text{d}y}{text{d}u}right)_{u=varphi(a)}left(frac{text{d}w}{text{d}x}right)_{x=a}=left(frac{text{d}y}{text{d}x}right)_{x=a}left(frac{text{d}w}{text{d}u}right)_{u=varphi(a)}$$
for one-variable differentiable functions, which seem to imply that is totally legit to manipulate first-order total differentials algebraically as long as we work only with differentiable functions.
But, if we want to multiply differential forms, we can only do that by means of the exterior product. This means that we can't divide differential forms and it also implies that $text{d}x^1text{d}x^2$ as "calculus differentials" is not the same as $text{d}x^1wedgetext{d}x^2$, as exterior differentials.
We can also talk about $text{d}^2y$ for "calculus differentials", but it doesn't make much sense to do it with exterior differentials because there $text{d}^2$ is the $0$ operator.
Clearly, calculus differentials and exterior differentials are different things; but if they are both called differentials, there must be a connection between them, and that is what I'm asking for: I would like a brief explanation of their connection or references to notes or books which discus it.
differential-forms
differential-forms
asked 2 days ago
TeicDaun
7116
asked 2 days ago
TeicDaun
7116
asked 2 days ago
TeicDaun
7116
asked 2 days ago
TeicDaun
7116
asked 2 days ago
TeicDaun
7116
7116
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