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?






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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















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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







share|cite|improve this question






















  • 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













up vote
0
down vote

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







share|cite|improve this question













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






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asked yesterday









bob.sacamento

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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


















  • 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










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1
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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.






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    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.






    share|cite|improve this answer

























      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.






      share|cite|improve this answer























        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Paul Sinclair

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