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Problem about taking measurements in flat metric spaces

  1. Apr 28, 2014 #1
    Hello,

    I am having a problem about the nature of the measurements of the intervals ds's forming out of infinitesimal displacements dx's of the coordinates and the actual meaning of the measurements of the same dx's, in flat metric spaces.

    I am certain that this must be a trivial problem, but I, somehow, can't see how to get around it.

    Well, to start with, I do not have this problem with non-flat metric spaces because there the dx's and ds's are clearly distinguished from one another as they clearly belong to different dimensions.

    So, I take an orthogonal triangle on a plane and measure the hypotenuse there to be, say: ds=[itex]\sqrt{dX^{2}+dY^{2}+2adXdY}[/itex] for some a, 0<a<1.
    How do I know that my infinitesimal triangle is orthogonal in this metric space?
    By Euclid's definition and construction: I take a line segment and draw the vertical at its middle point. How?
    By drawing two circles radius the line segment and center one of its end points. Thereby joining the two points where the circles intersect by a straight line, I am left with two vertical lines.

    Of course, to a creature living in a space with ds=[itex]\sqrt{dX^{2}+dY^{2}}[/itex], this would look like I have used a transformation of coordinates:

    x=X +Ycosθ
    y=Ysinθ,

    but my metric space is different altogether than his or hers. I have no problem understanding the nature of such a metric space, either.

    But, what is wrong with ds's like: ds=[itex]\sqrt{adX^{2}+bdY^{2}}[/itex], a,b≠1 ?

    What does my infinitesimal triangle look like there? What are the lengths of its sides? Are the vertical sides of length dX and dY with hypotenuse ds=[itex]\sqrt{adX^{2}+bdY^{2}}[/itex], or are they [itex]\sqrt{a}[/itex]dX and [itex]\sqrt{b}[/itex]dY with the same hypotenuse ds=[itex]\sqrt{adX^{2}+bdY^{2}}[/itex]?

    How come for example, since both dX and ds belong on the same plane, to measure differently for dY=0? Shouldn't ds measure dX for dY=0, too?

    I would like someone to please explain this to me, since in my mind it is an impredicativity I don't seem to be able to go around...
     
  2. jcsd
  3. Apr 29, 2014 #2

    Ben Niehoff

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    Science Advisor
    Gold Member

    Your use of terminology is very confusing, so I'm not entirely sure what you're asking. However, if you think about the following, it may clear things up:

    Suppose you measure distances along the X axis in meters, and distances along the Y axis in feet. Then what does Pythagoras' theorem look like?
     
  4. Apr 29, 2014 #3
    I am sorry in advance, if I lack the expertise and terminology to explain rigorously what I mean, but I am trying my best.

    To give you an idea, consider a metric with ds=[itex]\sqrt{dx^{2}-dy^{2}}[/itex]. A creature applies the Pythagoras' theorem there, with the manner I described above.

    If a line segment is on the x-axis with end-points 0 and 1, and the creature wants to find the vertical at point (1/2,0):

    It draws two circles radius 1 centered at points (0,0) and (1,0). In semi-Euclidean space (semi or pseudo-euclidean I think is the term ) these circles are the hyperbolas: y= [itex]\pm[/itex][itex]\sqrt{x^{2}+1}[/itex] and y=[itex]\pm[/itex] [itex]\sqrt{(x-1)^{2}+1}[/itex]. They both cross on the x=1/2 line, which is the requested vertical.
    The creature then verifies the Pythagoras theorem for the hypotenuse: ds=[itex]\sqrt{dx^{2}-dy^{2}}[/itex]
     
    Last edited: Apr 29, 2014
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