Mixing
Generalization of curvature to “mixed units”
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Every discussion of curvature I have ever seen involves an independent variable $x$ and a dependent variable $y(x)$, and both $x$ and $y$ represent some kind of distance. What if this is not the case? What if, for instance $x$ is a time and $y$ is a voltage ... or a stock price ... or whatever? Is there a useful, meaningful generalization of curvature that covers these sorts of cases?
curvature
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bob.sacamento
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Every discussion of curvature I have ever seen involves an independent variable $x$ and a dependent variable $y(x)$, and both $x$ and $y$ represent some kind of distance. What if this is not the case? What if, for instance $x$ is a time and $y$ is a voltage ... or a stock price ... or whatever? Is there a useful, meaningful generalization of curvature that covers these sorts of cases?
curvature
asked yesterday
bob.sacamento
2,3841819
You could still use a formula for curvature in those cases, but I don't think it would have any real use.
– Mark S.
yesterday
You can define distances in terms of all of those things. "Distance" doesn't have to mean "Euclidean distance".
– user3482749
yesterday
add a comment |
up vote
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favorite
up vote
0
down vote
favorite
Every discussion of curvature I have ever seen involves an independent variable $x$ and a dependent variable $y(x)$, and both $x$ and $y$ represent some kind of distance. What if this is not the case? What if, for instance $x$ is a time and $y$ is a voltage ... or a stock price ... or whatever? Is there a useful, meaningful generalization of curvature that covers these sorts of cases?
curvature
asked yesterday
bob.sacamento
2,3841819
Every discussion of curvature I have ever seen involves an independent variable $x$ and a dependent variable $y(x)$, and both $x$ and $y$ represent some kind of distance. What if this is not the case? What if, for instance $x$ is a time and $y$ is a voltage ... or a stock price ... or whatever? Is there a useful, meaningful generalization of curvature that covers these sorts of cases?
curvature
curvature
asked yesterday
bob.sacamento
2,3841819
asked yesterday
bob.sacamento
2,3841819
asked yesterday
bob.sacamento
2,3841819
asked yesterday
bob.sacamento
2,3841819
asked yesterday
bob.sacamento
2,3841819
2,3841819
You could still use a formula for curvature in those cases, but I don't think it would have any real use.
– Mark S.
yesterday
You can define distances in terms of all of those things. "Distance" doesn't have to mean "Euclidean distance".
– user3482749
yesterday
add a comment |
You could still use a formula for curvature in those cases, but I don't think it would have any real use.
– Mark S.
yesterday
You can define distances in terms of all of those things. "Distance" doesn't have to mean "Euclidean distance".
– user3482749
yesterday
You could still use a formula for curvature in those cases, but I don't think it would have any real use.
– Mark S.
yesterday
You could still use a formula for curvature in those cases, but I don't think it would have any real use.
– Mark S.
yesterday
You can define distances in terms of all of those things. "Distance" doesn't have to mean "Euclidean distance".
– user3482749
yesterday
You can define distances in terms of all of those things. "Distance" doesn't have to mean "Euclidean distance".
– user3482749
yesterday
add a comment |
1 Answer
1
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oldest
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up vote
1
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accepted
Curvature of a curve $gamma$ is defined by $$kappa = left|frac{d^2gamma}{ds^2}right|$$ where $s$ is the arclength along $gamma$. There we encounter the difficulty: if the dimensions of $x, y, z$ are not the same, what do we mean by that norm?
Assuming we are using the canonical inner product on $Bbb R^3$ (or the dimension of your choice), the norm is $sqrt{x^2 + y^2 + z^2}$. What units is this going to have?
While multiplying and dividing disparate units of measure is a common thing with meaning in the real world, adding and subtracting disparate units is not. What kind of unit would $sqrt{text{meters}^2 + text{ dollars}^2}$ be? How would you interpret it?
The reason multiplying and dividing are okay is that they behave nicely with rescaling. But addition and subtraction do not. Suppose I am measuring a speed in $frac ms$. If I switch from meters to centimeters, then, regardless of my speed in meters, I know that the speed in cm will be 100 times as large. But if instead I had added the distance and time, units of $m + s$, how will it change in $cm + s$? If my original value was $100 (m+s)$, that could be $50 m + 50 s$ or $1 m + 99 s$ or maybe $100 m + 0 s$. Switching meters to centimeters gives a different value for each. If all I know is how many $m + s$ there are, not how it breaks down into meters and seconds separately, then I have no way of choosing the appropriate value.
So before we can even talk about the idea of curvature, we need to have our coordinates all measured in the same way. $x$ in meters and $y$ in dollars means that is no general meaningful comparison between the two. For any particular situation, there may be a meaningful way to compare the units (e.g., in relativity, one can consider time as being like space, using the speed of light to convert between them), but there is no such method that holds in general.
However, if you can establish such a conversion between your dimensions in your particular application, you can use it to convert $x, y, z,$ etc to the same unit. Then the definition of curvature works just fine, regardless of what that unit is. The norm will have the same unit, as will arclength. And the unit for the curvature $kappa$ will be the inverse of it.
answered yesterday
Paul Sinclair
19k21440
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Curvature of a curve $gamma$ is defined by $$kappa = left|frac{d^2gamma}{ds^2}right|$$ where $s$ is the arclength along $gamma$. There we encounter the difficulty: if the dimensions of $x, y, z$ are not the same, what do we mean by that norm?
Assuming we are using the canonical inner product on $Bbb R^3$ (or the dimension of your choice), the norm is $sqrt{x^2 + y^2 + z^2}$. What units is this going to have?
While multiplying and dividing disparate units of measure is a common thing with meaning in the real world, adding and subtracting disparate units is not. What kind of unit would $sqrt{text{meters}^2 + text{ dollars}^2}$ be? How would you interpret it?
The reason multiplying and dividing are okay is that they behave nicely with rescaling. But addition and subtraction do not. Suppose I am measuring a speed in $frac ms$. If I switch from meters to centimeters, then, regardless of my speed in meters, I know that the speed in cm will be 100 times as large. But if instead I had added the distance and time, units of $m + s$, how will it change in $cm + s$? If my original value was $100 (m+s)$, that could be $50 m + 50 s$ or $1 m + 99 s$ or maybe $100 m + 0 s$. Switching meters to centimeters gives a different value for each. If all I know is how many $m + s$ there are, not how it breaks down into meters and seconds separately, then I have no way of choosing the appropriate value.
So before we can even talk about the idea of curvature, we need to have our coordinates all measured in the same way. $x$ in meters and $y$ in dollars means that is no general meaningful comparison between the two. For any particular situation, there may be a meaningful way to compare the units (e.g., in relativity, one can consider time as being like space, using the speed of light to convert between them), but there is no such method that holds in general.
However, if you can establish such a conversion between your dimensions in your particular application, you can use it to convert $x, y, z,$ etc to the same unit. Then the definition of curvature works just fine, regardless of what that unit is. The norm will have the same unit, as will arclength. And the unit for the curvature $kappa$ will be the inverse of it.
answered yesterday
Paul Sinclair
19k21440
add a comment |
up vote
1
down vote
accepted
Curvature of a curve $gamma$ is defined by $$kappa = left|frac{d^2gamma}{ds^2}right|$$ where $s$ is the arclength along $gamma$. There we encounter the difficulty: if the dimensions of $x, y, z$ are not the same, what do we mean by that norm?
Assuming we are using the canonical inner product on $Bbb R^3$ (or the dimension of your choice), the norm is $sqrt{x^2 + y^2 + z^2}$. What units is this going to have?
While multiplying and dividing disparate units of measure is a common thing with meaning in the real world, adding and subtracting disparate units is not. What kind of unit would $sqrt{text{meters}^2 + text{ dollars}^2}$ be? How would you interpret it?
The reason multiplying and dividing are okay is that they behave nicely with rescaling. But addition and subtraction do not. Suppose I am measuring a speed in $frac ms$. If I switch from meters to centimeters, then, regardless of my speed in meters, I know that the speed in cm will be 100 times as large. But if instead I had added the distance and time, units of $m + s$, how will it change in $cm + s$? If my original value was $100 (m+s)$, that could be $50 m + 50 s$ or $1 m + 99 s$ or maybe $100 m + 0 s$. Switching meters to centimeters gives a different value for each. If all I know is how many $m + s$ there are, not how it breaks down into meters and seconds separately, then I have no way of choosing the appropriate value.
So before we can even talk about the idea of curvature, we need to have our coordinates all measured in the same way. $x$ in meters and $y$ in dollars means that is no general meaningful comparison between the two. For any particular situation, there may be a meaningful way to compare the units (e.g., in relativity, one can consider time as being like space, using the speed of light to convert between them), but there is no such method that holds in general.
However, if you can establish such a conversion between your dimensions in your particular application, you can use it to convert $x, y, z,$ etc to the same unit. Then the definition of curvature works just fine, regardless of what that unit is. The norm will have the same unit, as will arclength. And the unit for the curvature $kappa$ will be the inverse of it.
answered yesterday
Paul Sinclair
19k21440
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Curvature of a curve $gamma$ is defined by $$kappa = left|frac{d^2gamma}{ds^2}right|$$ where $s$ is the arclength along $gamma$. There we encounter the difficulty: if the dimensions of $x, y, z$ are not the same, what do we mean by that norm?
Assuming we are using the canonical inner product on $Bbb R^3$ (or the dimension of your choice), the norm is $sqrt{x^2 + y^2 + z^2}$. What units is this going to have?
While multiplying and dividing disparate units of measure is a common thing with meaning in the real world, adding and subtracting disparate units is not. What kind of unit would $sqrt{text{meters}^2 + text{ dollars}^2}$ be? How would you interpret it?
The reason multiplying and dividing are okay is that they behave nicely with rescaling. But addition and subtraction do not. Suppose I am measuring a speed in $frac ms$. If I switch from meters to centimeters, then, regardless of my speed in meters, I know that the speed in cm will be 100 times as large. But if instead I had added the distance and time, units of $m + s$, how will it change in $cm + s$? If my original value was $100 (m+s)$, that could be $50 m + 50 s$ or $1 m + 99 s$ or maybe $100 m + 0 s$. Switching meters to centimeters gives a different value for each. If all I know is how many $m + s$ there are, not how it breaks down into meters and seconds separately, then I have no way of choosing the appropriate value.
So before we can even talk about the idea of curvature, we need to have our coordinates all measured in the same way. $x$ in meters and $y$ in dollars means that is no general meaningful comparison between the two. For any particular situation, there may be a meaningful way to compare the units (e.g., in relativity, one can consider time as being like space, using the speed of light to convert between them), but there is no such method that holds in general.
However, if you can establish such a conversion between your dimensions in your particular application, you can use it to convert $x, y, z,$ etc to the same unit. Then the definition of curvature works just fine, regardless of what that unit is. The norm will have the same unit, as will arclength. And the unit for the curvature $kappa$ will be the inverse of it.
answered yesterday
Paul Sinclair
19k21440
Curvature of a curve $gamma$ is defined by $$kappa = left|frac{d^2gamma}{ds^2}right|$$ where $s$ is the arclength along $gamma$. There we encounter the difficulty: if the dimensions of $x, y, z$ are not the same, what do we mean by that norm?
Assuming we are using the canonical inner product on $Bbb R^3$ (or the dimension of your choice), the norm is $sqrt{x^2 + y^2 + z^2}$. What units is this going to have?
While multiplying and dividing disparate units of measure is a common thing with meaning in the real world, adding and subtracting disparate units is not. What kind of unit would $sqrt{text{meters}^2 + text{ dollars}^2}$ be? How would you interpret it?
The reason multiplying and dividing are okay is that they behave nicely with rescaling. But addition and subtraction do not. Suppose I am measuring a speed in $frac ms$. If I switch from meters to centimeters, then, regardless of my speed in meters, I know that the speed in cm will be 100 times as large. But if instead I had added the distance and time, units of $m + s$, how will it change in $cm + s$? If my original value was $100 (m+s)$, that could be $50 m + 50 s$ or $1 m + 99 s$ or maybe $100 m + 0 s$. Switching meters to centimeters gives a different value for each. If all I know is how many $m + s$ there are, not how it breaks down into meters and seconds separately, then I have no way of choosing the appropriate value.
So before we can even talk about the idea of curvature, we need to have our coordinates all measured in the same way. $x$ in meters and $y$ in dollars means that is no general meaningful comparison between the two. For any particular situation, there may be a meaningful way to compare the units (e.g., in relativity, one can consider time as being like space, using the speed of light to convert between them), but there is no such method that holds in general.
However, if you can establish such a conversion between your dimensions in your particular application, you can use it to convert $x, y, z,$ etc to the same unit. Then the definition of curvature works just fine, regardless of what that unit is. The norm will have the same unit, as will arclength. And the unit for the curvature $kappa$ will be the inverse of it.
answered yesterday
Paul Sinclair
19k21440
answered yesterday
Paul Sinclair
19k21440
answered yesterday
Paul Sinclair
19k21440
answered yesterday
Paul Sinclair
19k21440
19k21440
add a comment |
add a comment |
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You could still use a formula for curvature in those cases, but I don't think it would have any real use.
– Mark S.
yesterday
You can define distances in terms of all of those things. "Distance" doesn't have to mean "Euclidean distance".
– user3482749
yesterday