Mixing
“Parameterizationâ€, “Change of Variablesâ€, “Transformationâ€, “Change of Coordinatesâ€: What's the Difference?
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What is the difference between parameterization and change of variables? How do I know when I'm doing one or the other? Also, under which one of these does "transformations" and "change of coordinates" fall?
For instance, I've recently been working on the following problem:
The function $u(x,y)$ satisfies $u_y + u_x = 0$ in $x > 0$, $y > 0$ together with the initial condition $u(x, 0) = sin(x)$, $x > 0$ and the boundary condition $u(0, y) = sin(y)$, $y > 0$. Determine the values of $u$ in the whole quarter plane $x > 0$, $y > 0$.
Which requires us to change from $(x(s), y(s))$ to $(x(s, tau), y(s, tau))$, as follows:
$$x(s) = s + C_1(tau)$$
$$y(s) = s + C_2(tau)$$
$$x(0) = tau$$
$$therefore C_1(tau) = tau$$
$$y(0) = 0$$
$$therefore C_2(tau) = 0$$
Therefore, we have
$$x(s, tau) = s + tau$$
$$y(s, tau) = s$$
Thank you for any help you can provide in making this clear.
vector-analysis transformation parametric parametrization change-of-variable
asked Aug 2 at 13:44
handler's handle
1277
add a comment |Â
up vote
0
down vote
favorite
What is the difference between parameterization and change of variables? How do I know when I'm doing one or the other? Also, under which one of these does "transformations" and "change of coordinates" fall?
For instance, I've recently been working on the following problem:
The function $u(x,y)$ satisfies $u_y + u_x = 0$ in $x > 0$, $y > 0$ together with the initial condition $u(x, 0) = sin(x)$, $x > 0$ and the boundary condition $u(0, y) = sin(y)$, $y > 0$. Determine the values of $u$ in the whole quarter plane $x > 0$, $y > 0$.
Which requires us to change from $(x(s), y(s))$ to $(x(s, tau), y(s, tau))$, as follows:
$$x(s) = s + C_1(tau)$$
$$y(s) = s + C_2(tau)$$
$$x(0) = tau$$
$$therefore C_1(tau) = tau$$
$$y(0) = 0$$
$$therefore C_2(tau) = 0$$
Therefore, we have
$$x(s, tau) = s + tau$$
$$y(s, tau) = s$$
Thank you for any help you can provide in making this clear.
vector-analysis transformation parametric parametrization change-of-variable
asked Aug 2 at 13:44
handler's handle
1277
IMO are more or less the same... different terminology may depends on context.
– Mauro ALLEGRANZA
Aug 2 at 13:57
@MauroALLEGRANZA You mean they're all about the same? So we would be correct to call the example I gave any of the 4?
– handler's handle
Aug 2 at 14:00
1
You can see Change of variables.
– Mauro ALLEGRANZA
Aug 2 at 14:13
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What is the difference between parameterization and change of variables? How do I know when I'm doing one or the other? Also, under which one of these does "transformations" and "change of coordinates" fall?
For instance, I've recently been working on the following problem:
The function $u(x,y)$ satisfies $u_y + u_x = 0$ in $x > 0$, $y > 0$ together with the initial condition $u(x, 0) = sin(x)$, $x > 0$ and the boundary condition $u(0, y) = sin(y)$, $y > 0$. Determine the values of $u$ in the whole quarter plane $x > 0$, $y > 0$.
Which requires us to change from $(x(s), y(s))$ to $(x(s, tau), y(s, tau))$, as follows:
$$x(s) = s + C_1(tau)$$
$$y(s) = s + C_2(tau)$$
$$x(0) = tau$$
$$therefore C_1(tau) = tau$$
$$y(0) = 0$$
$$therefore C_2(tau) = 0$$
Therefore, we have
$$x(s, tau) = s + tau$$
$$y(s, tau) = s$$
Thank you for any help you can provide in making this clear.
vector-analysis transformation parametric parametrization change-of-variable
asked Aug 2 at 13:44
handler's handle
1277
What is the difference between parameterization and change of variables? How do I know when I'm doing one or the other? Also, under which one of these does "transformations" and "change of coordinates" fall?
For instance, I've recently been working on the following problem:
The function $u(x,y)$ satisfies $u_y + u_x = 0$ in $x > 0$, $y > 0$ together with the initial condition $u(x, 0) = sin(x)$, $x > 0$ and the boundary condition $u(0, y) = sin(y)$, $y > 0$. Determine the values of $u$ in the whole quarter plane $x > 0$, $y > 0$.
Which requires us to change from $(x(s), y(s))$ to $(x(s, tau), y(s, tau))$, as follows:
$$x(s) = s + C_1(tau)$$
$$y(s) = s + C_2(tau)$$
$$x(0) = tau$$
$$therefore C_1(tau) = tau$$
$$y(0) = 0$$
$$therefore C_2(tau) = 0$$
Therefore, we have
$$x(s, tau) = s + tau$$
$$y(s, tau) = s$$
Thank you for any help you can provide in making this clear.
vector-analysis transformation parametric parametrization change-of-variable
asked Aug 2 at 13:44
handler's handle
1277
asked Aug 2 at 13:44
handler's handle
1277
asked Aug 2 at 13:44
handler's handle
1277
asked Aug 2 at 13:44
handler's handle
1277
1277
IMO are more or less the same... different terminology may depends on context.
– Mauro ALLEGRANZA
Aug 2 at 13:57
@MauroALLEGRANZA You mean they're all about the same? So we would be correct to call the example I gave any of the 4?
– handler's handle
Aug 2 at 14:00
1
You can see Change of variables.
– Mauro ALLEGRANZA
Aug 2 at 14:13
add a comment |Â
IMO are more or less the same... different terminology may depends on context.
– Mauro ALLEGRANZA
Aug 2 at 13:57
@MauroALLEGRANZA You mean they're all about the same? So we would be correct to call the example I gave any of the 4?
– handler's handle
Aug 2 at 14:00
1
You can see Change of variables.
– Mauro ALLEGRANZA
Aug 2 at 14:13
IMO are more or less the same... different terminology may depends on context.
– Mauro ALLEGRANZA
Aug 2 at 13:57
IMO are more or less the same... different terminology may depends on context.
– Mauro ALLEGRANZA
Aug 2 at 13:57
@MauroALLEGRANZA You mean they're all about the same? So we would be correct to call the example I gave any of the 4?
– handler's handle
Aug 2 at 14:00
@MauroALLEGRANZA You mean they're all about the same? So we would be correct to call the example I gave any of the 4?
– handler's handle
Aug 2 at 14:00
1
1
You can see Change of variables.
– Mauro ALLEGRANZA
Aug 2 at 14:13
You can see Change of variables.
– Mauro ALLEGRANZA
Aug 2 at 14:13
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
To parametrize a set $S$ is to write it as $$S=f(s)mid sin D$$ where $D$ is some domain where the parameter $s$ lives (it can be of any kind, but more often than not it is a $k$-tuple of real numbers).
Now if you start from a set endowed with a parametrization, you can change the parametrization by applying a bijection $varphi: D'to D$, resulting in
$$S=f(varphi(t))mid tin D'$$
This would be called a change of variables in general, but in the particular case where $D$ is a subset of $Bbb R^n$, you think of a parameter as an $n$-tuple of coordinates, and such a change of variables would then be called a change of coordinates.
For the last one, transformation, this can mean many things in many contexts, so you would have to make the question more precise.
Note that replacing $f(s), sin D$ with $f(varphi(t)), tin D'$ is a change of variable in many contexts, not only when there is an underlying parametrization. $f$ can be an unknown function occurring in a differential equation for instance. In most contexts, there would be additional requirements on $varphi$, not only being a bijection. It would need to respect a structure of some sort: be a diffeomorphism, linear, etc.
answered Aug 2 at 14:04
Arnaud Mortier
17.8k21857
Ahh, thanks for that. So what would my example be?
– handler's handle
Aug 2 at 14:06
@handler'shandle Your example looks like a change of variables, but it is hard to tell given we don't know what you mean by $x(s)$ in the first place.
– Arnaud Mortier
Aug 2 at 14:11
add a comment |Â
up vote
1
down vote
I love questions about notation and terminology. Let's begin with parametrization vs. change of variable.
If we have a curve $Gamma$ (which we can think of as a $1$-dimensional subset of $mathbbR^n$ with an orientation defined), we call any function $alphaspacecolon IsubseteqmathbbRlongrightarrowmathbbR^n$ which satisfies $alpha(I)=Gamma$ a parametrization of the curve. Usually, you will want that parametrization to be one-to-one (injective).
Any other injective function $IsubseteqmathbbRlongrightarrowmathbbR^n$ will be another parametrization of $Gamma$, and as they are both injective, there will always exist a bijection – a change of variable of a function that represents a parametrization, which we call reparametrization – that takes you from one to the other and vice versa. So, for example: if you're computing line integrals, reparametrizing the curve in which you are integrating will be the same as performing a change of variable on the Riemann integral you have to compute to do the line integral.
The words variable and parameter are similae. We would say that the parameter $t$ of your parametric curve $vecalpha(t)$ is the independent variable of the function $vecr=vecalpha(t)$.
You usually talk about parameters when parametrizing manifolds, and you talk about variables when you have any sort of function, specially in summations, integrals and stuff like that. So, if you have a function $fcolonspacemathbbR^nlongrightarrowmathbbR^m$, you would would say that you have $n$ independent variables $(x^1,...,x^n)$ and $m$ dependent variables $(r^1,...,r^m)$. A reparametrization is just a change of variables that represent parameters.
Let's talk about coordinates, now. When you're dealing with points in space; you call those $(x^1,...,x^n)$ the coordinates of a point.
A coordinate system allows you to identify every point in your space with a unique ordered set of numbers (well, you can allow some points to be labelled with more than one ordered set of numbers as long as the set of all of these numbers has $0$ $n$-dimensional volume). The change of one coordinate system to another is called a change of coordinates.
A change of coordinates tells you how to go from one way of labelling all points of space to another way of labelling all points in space; you can think of this as parametrizing all space instead of just a curve, surface or manifold in it. You can use a change of coordinates as a change of variables in a multiple integral, as you may already know:
$$intcdotsintf(q^1,...,q^n)space textdq^1cdotstextdq^n=intfracpartial(q^1,...,q^n)partial(x^1,...,x^n)right$$
where $vecPhi(q^1,...,q^n)=(x^1,...,x^n)$ is the change of variables.
A transformation isn't the same as a change of variable, though. The word 'transformation' usually implies leaving things different than they were before, while change of variables implies leaving things as they are, but with a different name.
For example: a matrix can represent both a linear endomorphism (a linear transformation of a vector space) and a change of basis matrix (a matrix that takes in a vector expressed with respect to one basis and changes its coordinates to the ones with respect to another basis). Those are two different operations at a conceptual level, as one changes the areas, volumes... of your space while the other keeps everything the way it was.
In fact, a transformation need not be bijective at all, while a change of variables HAS to be bijective (again, except for a $0$ $n$-dimentional volume set of points in the most general case, definitely not in the linear case).
answered Aug 2 at 14:32
Pablo T.
242113
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
To parametrize a set $S$ is to write it as $$S=f(s)mid sin D$$ where $D$ is some domain where the parameter $s$ lives (it can be of any kind, but more often than not it is a $k$-tuple of real numbers).
Now if you start from a set endowed with a parametrization, you can change the parametrization by applying a bijection $varphi: D'to D$, resulting in
$$S=f(varphi(t))mid tin D'$$
This would be called a change of variables in general, but in the particular case where $D$ is a subset of $Bbb R^n$, you think of a parameter as an $n$-tuple of coordinates, and such a change of variables would then be called a change of coordinates.
For the last one, transformation, this can mean many things in many contexts, so you would have to make the question more precise.
Note that replacing $f(s), sin D$ with $f(varphi(t)), tin D'$ is a change of variable in many contexts, not only when there is an underlying parametrization. $f$ can be an unknown function occurring in a differential equation for instance. In most contexts, there would be additional requirements on $varphi$, not only being a bijection. It would need to respect a structure of some sort: be a diffeomorphism, linear, etc.
answered Aug 2 at 14:04
Arnaud Mortier
17.8k21857
Ahh, thanks for that. So what would my example be?
– handler's handle
Aug 2 at 14:06
@handler'shandle Your example looks like a change of variables, but it is hard to tell given we don't know what you mean by $x(s)$ in the first place.
– Arnaud Mortier
Aug 2 at 14:11
add a comment |Â
up vote
1
down vote
accepted
To parametrize a set $S$ is to write it as $$S=f(s)mid sin D$$ where $D$ is some domain where the parameter $s$ lives (it can be of any kind, but more often than not it is a $k$-tuple of real numbers).
Now if you start from a set endowed with a parametrization, you can change the parametrization by applying a bijection $varphi: D'to D$, resulting in
$$S=f(varphi(t))mid tin D'$$
This would be called a change of variables in general, but in the particular case where $D$ is a subset of $Bbb R^n$, you think of a parameter as an $n$-tuple of coordinates, and such a change of variables would then be called a change of coordinates.
For the last one, transformation, this can mean many things in many contexts, so you would have to make the question more precise.
Note that replacing $f(s), sin D$ with $f(varphi(t)), tin D'$ is a change of variable in many contexts, not only when there is an underlying parametrization. $f$ can be an unknown function occurring in a differential equation for instance. In most contexts, there would be additional requirements on $varphi$, not only being a bijection. It would need to respect a structure of some sort: be a diffeomorphism, linear, etc.
answered Aug 2 at 14:04
Arnaud Mortier
17.8k21857
Ahh, thanks for that. So what would my example be?
– handler's handle
Aug 2 at 14:06
@handler'shandle Your example looks like a change of variables, but it is hard to tell given we don't know what you mean by $x(s)$ in the first place.
– Arnaud Mortier
Aug 2 at 14:11
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
To parametrize a set $S$ is to write it as $$S=f(s)mid sin D$$ where $D$ is some domain where the parameter $s$ lives (it can be of any kind, but more often than not it is a $k$-tuple of real numbers).
Now if you start from a set endowed with a parametrization, you can change the parametrization by applying a bijection $varphi: D'to D$, resulting in
$$S=f(varphi(t))mid tin D'$$
This would be called a change of variables in general, but in the particular case where $D$ is a subset of $Bbb R^n$, you think of a parameter as an $n$-tuple of coordinates, and such a change of variables would then be called a change of coordinates.
For the last one, transformation, this can mean many things in many contexts, so you would have to make the question more precise.
Note that replacing $f(s), sin D$ with $f(varphi(t)), tin D'$ is a change of variable in many contexts, not only when there is an underlying parametrization. $f$ can be an unknown function occurring in a differential equation for instance. In most contexts, there would be additional requirements on $varphi$, not only being a bijection. It would need to respect a structure of some sort: be a diffeomorphism, linear, etc.
answered Aug 2 at 14:04
Arnaud Mortier
17.8k21857
To parametrize a set $S$ is to write it as $$S=f(s)mid sin D$$ where $D$ is some domain where the parameter $s$ lives (it can be of any kind, but more often than not it is a $k$-tuple of real numbers).
Now if you start from a set endowed with a parametrization, you can change the parametrization by applying a bijection $varphi: D'to D$, resulting in
$$S=f(varphi(t))mid tin D'$$
This would be called a change of variables in general, but in the particular case where $D$ is a subset of $Bbb R^n$, you think of a parameter as an $n$-tuple of coordinates, and such a change of variables would then be called a change of coordinates.
For the last one, transformation, this can mean many things in many contexts, so you would have to make the question more precise.
Note that replacing $f(s), sin D$ with $f(varphi(t)), tin D'$ is a change of variable in many contexts, not only when there is an underlying parametrization. $f$ can be an unknown function occurring in a differential equation for instance. In most contexts, there would be additional requirements on $varphi$, not only being a bijection. It would need to respect a structure of some sort: be a diffeomorphism, linear, etc.
answered Aug 2 at 14:04
Arnaud Mortier
17.8k21857
edited Aug 2 at 14:14
answered Aug 2 at 14:04
Arnaud Mortier
17.8k21857
answered Aug 2 at 14:04
Arnaud Mortier
17.8k21857
answered Aug 2 at 14:04
Arnaud Mortier
17.8k21857
17.8k21857
Ahh, thanks for that. So what would my example be?
– handler's handle
Aug 2 at 14:06
@handler'shandle Your example looks like a change of variables, but it is hard to tell given we don't know what you mean by $x(s)$ in the first place.
– Arnaud Mortier
Aug 2 at 14:11
add a comment |Â
Ahh, thanks for that. So what would my example be?
– handler's handle
Aug 2 at 14:06
@handler'shandle Your example looks like a change of variables, but it is hard to tell given we don't know what you mean by $x(s)$ in the first place.
– Arnaud Mortier
Aug 2 at 14:11
Ahh, thanks for that. So what would my example be?
– handler's handle
Aug 2 at 14:06
Ahh, thanks for that. So what would my example be?
– handler's handle
Aug 2 at 14:06
@handler'shandle Your example looks like a change of variables, but it is hard to tell given we don't know what you mean by $x(s)$ in the first place.
– Arnaud Mortier
Aug 2 at 14:11
@handler'shandle Your example looks like a change of variables, but it is hard to tell given we don't know what you mean by $x(s)$ in the first place.
– Arnaud Mortier
Aug 2 at 14:11
add a comment |Â
up vote
1
down vote
I love questions about notation and terminology. Let's begin with parametrization vs. change of variable.
If we have a curve $Gamma$ (which we can think of as a $1$-dimensional subset of $mathbbR^n$ with an orientation defined), we call any function $alphaspacecolon IsubseteqmathbbRlongrightarrowmathbbR^n$ which satisfies $alpha(I)=Gamma$ a parametrization of the curve. Usually, you will want that parametrization to be one-to-one (injective).
Any other injective function $IsubseteqmathbbRlongrightarrowmathbbR^n$ will be another parametrization of $Gamma$, and as they are both injective, there will always exist a bijection – a change of variable of a function that represents a parametrization, which we call reparametrization – that takes you from one to the other and vice versa. So, for example: if you're computing line integrals, reparametrizing the curve in which you are integrating will be the same as performing a change of variable on the Riemann integral you have to compute to do the line integral.
The words variable and parameter are similae. We would say that the parameter $t$ of your parametric curve $vecalpha(t)$ is the independent variable of the function $vecr=vecalpha(t)$.
You usually talk about parameters when parametrizing manifolds, and you talk about variables when you have any sort of function, specially in summations, integrals and stuff like that. So, if you have a function $fcolonspacemathbbR^nlongrightarrowmathbbR^m$, you would would say that you have $n$ independent variables $(x^1,...,x^n)$ and $m$ dependent variables $(r^1,...,r^m)$. A reparametrization is just a change of variables that represent parameters.
Let's talk about coordinates, now. When you're dealing with points in space; you call those $(x^1,...,x^n)$ the coordinates of a point.
A coordinate system allows you to identify every point in your space with a unique ordered set of numbers (well, you can allow some points to be labelled with more than one ordered set of numbers as long as the set of all of these numbers has $0$ $n$-dimensional volume). The change of one coordinate system to another is called a change of coordinates.
A change of coordinates tells you how to go from one way of labelling all points of space to another way of labelling all points in space; you can think of this as parametrizing all space instead of just a curve, surface or manifold in it. You can use a change of coordinates as a change of variables in a multiple integral, as you may already know:
$$intcdotsintf(q^1,...,q^n)space textdq^1cdotstextdq^n=intfracpartial(q^1,...,q^n)partial(x^1,...,x^n)right$$
where $vecPhi(q^1,...,q^n)=(x^1,...,x^n)$ is the change of variables.
A transformation isn't the same as a change of variable, though. The word 'transformation' usually implies leaving things different than they were before, while change of variables implies leaving things as they are, but with a different name.
For example: a matrix can represent both a linear endomorphism (a linear transformation of a vector space) and a change of basis matrix (a matrix that takes in a vector expressed with respect to one basis and changes its coordinates to the ones with respect to another basis). Those are two different operations at a conceptual level, as one changes the areas, volumes... of your space while the other keeps everything the way it was.
In fact, a transformation need not be bijective at all, while a change of variables HAS to be bijective (again, except for a $0$ $n$-dimentional volume set of points in the most general case, definitely not in the linear case).
answered Aug 2 at 14:32
Pablo T.
242113
add a comment |Â
up vote
1
down vote
I love questions about notation and terminology. Let's begin with parametrization vs. change of variable.
If we have a curve $Gamma$ (which we can think of as a $1$-dimensional subset of $mathbbR^n$ with an orientation defined), we call any function $alphaspacecolon IsubseteqmathbbRlongrightarrowmathbbR^n$ which satisfies $alpha(I)=Gamma$ a parametrization of the curve. Usually, you will want that parametrization to be one-to-one (injective).
Any other injective function $IsubseteqmathbbRlongrightarrowmathbbR^n$ will be another parametrization of $Gamma$, and as they are both injective, there will always exist a bijection – a change of variable of a function that represents a parametrization, which we call reparametrization – that takes you from one to the other and vice versa. So, for example: if you're computing line integrals, reparametrizing the curve in which you are integrating will be the same as performing a change of variable on the Riemann integral you have to compute to do the line integral.
The words variable and parameter are similae. We would say that the parameter $t$ of your parametric curve $vecalpha(t)$ is the independent variable of the function $vecr=vecalpha(t)$.
You usually talk about parameters when parametrizing manifolds, and you talk about variables when you have any sort of function, specially in summations, integrals and stuff like that. So, if you have a function $fcolonspacemathbbR^nlongrightarrowmathbbR^m$, you would would say that you have $n$ independent variables $(x^1,...,x^n)$ and $m$ dependent variables $(r^1,...,r^m)$. A reparametrization is just a change of variables that represent parameters.
Let's talk about coordinates, now. When you're dealing with points in space; you call those $(x^1,...,x^n)$ the coordinates of a point.
A coordinate system allows you to identify every point in your space with a unique ordered set of numbers (well, you can allow some points to be labelled with more than one ordered set of numbers as long as the set of all of these numbers has $0$ $n$-dimensional volume). The change of one coordinate system to another is called a change of coordinates.
A change of coordinates tells you how to go from one way of labelling all points of space to another way of labelling all points in space; you can think of this as parametrizing all space instead of just a curve, surface or manifold in it. You can use a change of coordinates as a change of variables in a multiple integral, as you may already know:
$$intcdotsintf(q^1,...,q^n)space textdq^1cdotstextdq^n=intfracpartial(q^1,...,q^n)partial(x^1,...,x^n)right$$
where $vecPhi(q^1,...,q^n)=(x^1,...,x^n)$ is the change of variables.
A transformation isn't the same as a change of variable, though. The word 'transformation' usually implies leaving things different than they were before, while change of variables implies leaving things as they are, but with a different name.
For example: a matrix can represent both a linear endomorphism (a linear transformation of a vector space) and a change of basis matrix (a matrix that takes in a vector expressed with respect to one basis and changes its coordinates to the ones with respect to another basis). Those are two different operations at a conceptual level, as one changes the areas, volumes... of your space while the other keeps everything the way it was.
In fact, a transformation need not be bijective at all, while a change of variables HAS to be bijective (again, except for a $0$ $n$-dimentional volume set of points in the most general case, definitely not in the linear case).
answered Aug 2 at 14:32
Pablo T.
242113
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I love questions about notation and terminology. Let's begin with parametrization vs. change of variable.
If we have a curve $Gamma$ (which we can think of as a $1$-dimensional subset of $mathbbR^n$ with an orientation defined), we call any function $alphaspacecolon IsubseteqmathbbRlongrightarrowmathbbR^n$ which satisfies $alpha(I)=Gamma$ a parametrization of the curve. Usually, you will want that parametrization to be one-to-one (injective).
Any other injective function $IsubseteqmathbbRlongrightarrowmathbbR^n$ will be another parametrization of $Gamma$, and as they are both injective, there will always exist a bijection – a change of variable of a function that represents a parametrization, which we call reparametrization – that takes you from one to the other and vice versa. So, for example: if you're computing line integrals, reparametrizing the curve in which you are integrating will be the same as performing a change of variable on the Riemann integral you have to compute to do the line integral.
The words variable and parameter are similae. We would say that the parameter $t$ of your parametric curve $vecalpha(t)$ is the independent variable of the function $vecr=vecalpha(t)$.
You usually talk about parameters when parametrizing manifolds, and you talk about variables when you have any sort of function, specially in summations, integrals and stuff like that. So, if you have a function $fcolonspacemathbbR^nlongrightarrowmathbbR^m$, you would would say that you have $n$ independent variables $(x^1,...,x^n)$ and $m$ dependent variables $(r^1,...,r^m)$. A reparametrization is just a change of variables that represent parameters.
Let's talk about coordinates, now. When you're dealing with points in space; you call those $(x^1,...,x^n)$ the coordinates of a point.
A coordinate system allows you to identify every point in your space with a unique ordered set of numbers (well, you can allow some points to be labelled with more than one ordered set of numbers as long as the set of all of these numbers has $0$ $n$-dimensional volume). The change of one coordinate system to another is called a change of coordinates.
A change of coordinates tells you how to go from one way of labelling all points of space to another way of labelling all points in space; you can think of this as parametrizing all space instead of just a curve, surface or manifold in it. You can use a change of coordinates as a change of variables in a multiple integral, as you may already know:
$$intcdotsintf(q^1,...,q^n)space textdq^1cdotstextdq^n=intfracpartial(q^1,...,q^n)partial(x^1,...,x^n)right$$
where $vecPhi(q^1,...,q^n)=(x^1,...,x^n)$ is the change of variables.
A transformation isn't the same as a change of variable, though. The word 'transformation' usually implies leaving things different than they were before, while change of variables implies leaving things as they are, but with a different name.
For example: a matrix can represent both a linear endomorphism (a linear transformation of a vector space) and a change of basis matrix (a matrix that takes in a vector expressed with respect to one basis and changes its coordinates to the ones with respect to another basis). Those are two different operations at a conceptual level, as one changes the areas, volumes... of your space while the other keeps everything the way it was.
In fact, a transformation need not be bijective at all, while a change of variables HAS to be bijective (again, except for a $0$ $n$-dimentional volume set of points in the most general case, definitely not in the linear case).
answered Aug 2 at 14:32
Pablo T.
242113
I love questions about notation and terminology. Let's begin with parametrization vs. change of variable.
If we have a curve $Gamma$ (which we can think of as a $1$-dimensional subset of $mathbbR^n$ with an orientation defined), we call any function $alphaspacecolon IsubseteqmathbbRlongrightarrowmathbbR^n$ which satisfies $alpha(I)=Gamma$ a parametrization of the curve. Usually, you will want that parametrization to be one-to-one (injective).
Any other injective function $IsubseteqmathbbRlongrightarrowmathbbR^n$ will be another parametrization of $Gamma$, and as they are both injective, there will always exist a bijection – a change of variable of a function that represents a parametrization, which we call reparametrization – that takes you from one to the other and vice versa. So, for example: if you're computing line integrals, reparametrizing the curve in which you are integrating will be the same as performing a change of variable on the Riemann integral you have to compute to do the line integral.
The words variable and parameter are similae. We would say that the parameter $t$ of your parametric curve $vecalpha(t)$ is the independent variable of the function $vecr=vecalpha(t)$.
You usually talk about parameters when parametrizing manifolds, and you talk about variables when you have any sort of function, specially in summations, integrals and stuff like that. So, if you have a function $fcolonspacemathbbR^nlongrightarrowmathbbR^m$, you would would say that you have $n$ independent variables $(x^1,...,x^n)$ and $m$ dependent variables $(r^1,...,r^m)$. A reparametrization is just a change of variables that represent parameters.
Let's talk about coordinates, now. When you're dealing with points in space; you call those $(x^1,...,x^n)$ the coordinates of a point.
A coordinate system allows you to identify every point in your space with a unique ordered set of numbers (well, you can allow some points to be labelled with more than one ordered set of numbers as long as the set of all of these numbers has $0$ $n$-dimensional volume). The change of one coordinate system to another is called a change of coordinates.
A change of coordinates tells you how to go from one way of labelling all points of space to another way of labelling all points in space; you can think of this as parametrizing all space instead of just a curve, surface or manifold in it. You can use a change of coordinates as a change of variables in a multiple integral, as you may already know:
$$intcdotsintf(q^1,...,q^n)space textdq^1cdotstextdq^n=intfracpartial(q^1,...,q^n)partial(x^1,...,x^n)right$$
where $vecPhi(q^1,...,q^n)=(x^1,...,x^n)$ is the change of variables.
A transformation isn't the same as a change of variable, though. The word 'transformation' usually implies leaving things different than they were before, while change of variables implies leaving things as they are, but with a different name.
For example: a matrix can represent both a linear endomorphism (a linear transformation of a vector space) and a change of basis matrix (a matrix that takes in a vector expressed with respect to one basis and changes its coordinates to the ones with respect to another basis). Those are two different operations at a conceptual level, as one changes the areas, volumes... of your space while the other keeps everything the way it was.
In fact, a transformation need not be bijective at all, while a change of variables HAS to be bijective (again, except for a $0$ $n$-dimentional volume set of points in the most general case, definitely not in the linear case).
answered Aug 2 at 14:32
Pablo T.
242113
edited yesterday
answered Aug 2 at 14:32
Pablo T.
242113
answered Aug 2 at 14:32
Pablo T.
242113
answered Aug 2 at 14:32
Pablo T.
242113
242113
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IMO are more or less the same... different terminology may depends on context.
– Mauro ALLEGRANZA
Aug 2 at 13:57
@MauroALLEGRANZA You mean they're all about the same? So we would be correct to call the example I gave any of the 4?
– handler's handle
Aug 2 at 14:00
1
You can see Change of variables.
– Mauro ALLEGRANZA
Aug 2 at 14:13