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Proof of euclidean distance

  1. Jan 21, 2010 #1
    How do you prove that for a set of coordinates you are supposed to take
    [tex]\mathrm{d}s^2=\mathrm{d}x_i\mathrm{d}x^i[/tex]
    for the distance? I mean in a very abstract fashion. All I know is that there is some coordinate mesh. Why don't I take other powers for the distance for example?

    Or if that bilinear equation is only a special case, then why is our space obeying it?
     
  2. jcsd
  3. Jan 21, 2010 #2
    The formula only works in Euclidean coordinates but not in others. Area elements can have many forms and can differ from point to point.
     
  4. Jan 21, 2010 #3
    So back to my question: why is the reason that our real world has this property? And why is curved space in general relativity also of this form an no other?
     
  5. Jan 21, 2010 #4

    D H

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    Our everyday world is pretty much Euclidean. The Euclidean 2-norm measures the distance of a line segment (a line in the Euclidean sense, that is.) This question is somewhat tautological. You are essentially asking why is the straight-line distance between two points the Euclidean norm. The answer is that is that is how a straight line is defined.

    It doesn't have this form in general. One key point of general relativity is that spacetime is not Euclidean.
     
  6. Jan 21, 2010 #5
    It does. Please look back at the equation I wrote. I did not write down Euclidean space in particular since there are covariant and contravariant (hope thats the correct word) forms.
     
  7. Jan 21, 2010 #6
    What property? You haven't actually named one... (I know, sometimes the correct question is the hardest part of asking.) By design, ||ds^2|| is invariant under a coordinate transformation. By design, the contravariant coordinates are defined to leave ||ds^2|| unchanged.

    It's sometimes useful, such as considering spacetime, to model it as a manifold having a metric, so that distances are definable. Is this getting at what you're asking?
     
    Last edited: Jan 21, 2010
  8. Jan 21, 2010 #7
    the real world does not have that property, The world metric is curved not Euclidean.

    If you are asking why the distance measure comes from an inner product it is - I think - because the world not only has distances distances but also angles. Without an inner product there is no concept of angle.
     
  9. Jan 22, 2010 #8
    Actually I don't know what property is the key, so I'm looking for someone proposing a sensible axiom (maybe invariance under rotation?) from which the bilinear form derives. I vaguely know the concepts of this notation. What exactly is a *general* coordinate transformation? But in case the invariance under coordinate transformation is the key, then why don't we take other invariant expressions for the distance?

    Or is it possible to relate the distance with changing x and y to distance of constant y?
    So basically I have some squared paper and it is easy to count distances between cells if I stay horizontally or vertically. So I define how to get the distance if only one variable changes. But now I need another axioms to related diagonal distance to these. First I thought about rotations (rotate diagonal onto horizontal), but then I don't know how to define rotations without having a concept of distance.

    First I repeat that the above notation is not restricted to Euclidean! Look up the concepts of tensor notation.
    I'm not sure what you mean by the second reasoning. The world has angle because we defined distances and thus rotations. I don't think you can argue the other way round.
     
  10. Jan 22, 2010 #9

    D H

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    Start with Euclidean space. That definition for distance, the Euclidean norm, is the distance measured by a straight line. That the Euclidean norm is the length of a straight line is the Pythagorean theorem. Your question is essentially tautological. The Euclidean norm is the straight line distance. You are asking why the straight line distance is the distance between two points. The answer is because that is how distance is typically defined.

    The Euclidean norm is not the only definition for distance, even in Euclidean space. Others include the L1-norm or taxicab norm, the L-infinity norm, and in general, the Lp-norm. Yet another is the Mahalanobis distance, widely used in statistics. Note that this latter norm is very similar to the concept of a metric tensor.

    The covariant tensor is related to the contravariant tensor via the metric tensor. The definition of length (or rather, differential length) in a curved metric space is

    [tex]ds^2 \equiv g_{\mu,\nu} dx^{\mu}dx^{\nu}[/tex]

    The metric tensor relates the covariant and contravariant vectors via

    [tex]dx_{\mu} \equiv g_{\mu,\nu} dx^{\nu}[/tex]

    Thus [itex]ds^2 = dx_{\mu}dx^{\nu}[/itex] is tautological.


    I think the question you are really asking is "Why is this definition useful?"
     
  11. Jan 22, 2010 #10
    You need to be more clear. I am beginning to think though that you are not thinking carefully enough about the replies that people are sending you.

    To repeat.

    - The only tensor that gives you a metric is a quadratic tensor.

    - tensors are invariant under any coordinate transformation

    - the idea of a tensor is a quantity which when measured in any set of coordinates gives the same answer.

    - the metric ds^2 = sum dxi^2 only works for an orthonormal frame and is not coordinate invariant unless you only change coordinates only by orthogonal transformations. If you do not have an orthonormal frame the formula fails.

    - A metric that does not have an idea of angle is not a tensor. The only metrics with an idea of angle are quadratic metrics. Without an idea of angle there would be no physics as we know it. Everything we know about the world is based on the idea of angle. Distance falls out from the idea of angle.
     
  12. Jan 22, 2010 #11
    Please have a look at the "squared paper" thought I wrote about. It's not clear what a straight line is when only grid points are defined. A straight line can only be defined once you admit for an operation that rotates a diagonal line onto a horizontal line.
    In any case the idea of straight line is not trivial. Isn't even a straight line defined by minimization of distance? So obviously straight line and distance metric are the same and so they cannot be the answer to my question why the distance isn't [itex]\mathrm{d}s^2=\mathrm{d}x^2+\mathrm{d}y^3[/itex]

    Thanks for this idea. I'm interested in reading up about them. My question is related to that: Why is our space not one of these extraordinary metrics or even more general than Mahalanobis?
    As I said, maybe rotational invariance can be used as an answer to my question I just don't know how.

    I'm not sure what you mean by tautological, but that's basically my point. My metric is general. Btw, I did some "google science" and found that my question is probably related to asking why people use
    http://mathworld.wolfram.com/RiemannianGeometry.html
    for all of our space instead of
    http://mathworld.wolfram.com/FinslerSpace.html

    I think it's the other way round. Some people only read my (slightly misleading) headline and do not care what I write further on. Instead they quote some random textbook knowledge. D.H. also explained that my metric isn't restricted to Euclidean.

    Apparently this is incorrect, if I understood that correctly:
    http://mathworld.wolfram.com/FinslerSpace.html

    I wrote asked before: So is the neccessity of coordinate transformations the answer to my question? If so, what exactly are general coordinate transformations and how do you define them without having a concept of distance and so on?

    That is true, but not what my equation states.
     
  13. Jan 22, 2010 #12
    The things I have told you about metrics and tensors are true. Why don't you try proving it for yourself?

    The Finsler metric is not a tensor. But it determines a quadratic tensor i.e. a metric tensor.

    If you do not want to use tensors there are all kinds of ways to define distance. Like I said before, the key to quadratic tensors is that they give you an idea of angle.

    A bilinear tensor is the tensor way to get a metric. It is not necessary to use a tensor in order to have a metric - but in Physics is has been the predominately useful metic type because it gives an idea of an inner product - or equivalently an idea of angle.
     
    Last edited: Jan 22, 2010
  14. Jan 22, 2010 #13
    That's the point. The thing are correct textbook knowledge, but not related to my question. For example the line element I wrote down is not Euclidean only.

    Saying you want to measure angles would be one step of an explanation, but now one has to explain why angles are important. And that has to be done in a mathematical abstract way best with my squared paper problem I mentioned.
    Because all "intuitive" concept implicitily assume a quadric metric already.
     
  15. Jan 22, 2010 #14
    You are still not being clear.
    A mesh by itself has no metric.
    Are you saying that you are assuming that the mesh does partially imply a metric because you can count the number of mesh points in any direction? By the way I think that methodology already gives you a metric on the mesh. But you have no idea of angle. Further you have no way to find distance for points not on the mesh.

    You were confusing because you wrote originally down a metric tensor in orthogonal coordinates and then asked why the world has to be that way. So we assumed that's what you were talking about. But you are not talking about tensors. What are you talking about?

    Incidentally, inner products are important not only for the idea of angle per se but for the idea of breaking a vector down into components. For instance the component of a force in the direction of motion of a mass tells you the amount of work that it does. This kind of thing happens all over in Physics. Calculating work, flux, circulation and other physical quantities always requires an inner product.

    Perhaps you could tell me how to compute the work done by a gravitation field on a mass that is constrained to move in a specific path without using an inner product. That would be interesting.
     
    Last edited: Jan 22, 2010
  16. Jan 22, 2010 #15
    Here is a thought for you.

    One way to geometrize Physics is to replace forces - which are somewhat mystically defined concepts - by geometry. For instance, a particle moving in a potential can be thought of as travelling along a geodesic in a suitably defined metric - and the force in some sense goes away as a force and becomes a parameter of the metric. In this case the metric is a quadratic tensor. The idea applies in general realtivity for gravity and I imagine one might be able to geometrize other fields as well.
     
  17. Jan 22, 2010 #16
    Another thought

    Given your square mesh you could define a set of transformations that compare any line segment to a standard ruler where you already know how to measure distance. These transformations would have to obey certain regularities in order to make sense. You could then declare for instance that these transformation preserve distance. In classical geometries these tranformations usually determine a quadratic metric tensor that makes the transformations into isometries. But I see no reason why this always has to be true.

    But be that as it may - one idea of geometry is those properties of a space that are invariant under some set of transformations. The Euclidean metric - not written as a tensor thanks - makes Euclidean transformations into isometries. Moebius transformations on a disc determine non-Euclidean geometry - Gauss/Lobachevsky geometry.

    From this point of view I do not necessarily think that starting with a mesh is the best way to go. What you really want is to preserve some set of geometric properties.

    Be careful though. There are geometric spaces where there are not enough isometries to compare any two lengths. In fact there are spaces that have no isometries at all.
     
    Last edited: Jan 22, 2010
  18. Jan 22, 2010 #17
    Correct. So basically I'm looking for a mathematical "low-level" ingredient to add from our real world, as to make only Riemannian metrics permissible.

    At least for motion along the mesh connections. It does not specify the diagonal distance.

    There are no points off the mesh. Or maybe one can define the mesh to be arbitrarily fine.

    Actually I believe I didn't put any restrictions on the metric tensor, so I did not restrict myself to orthogonal coordinates?!

    Could there be other artifical ways to break thing down into components? I mean I can break down any integer into primes and say only 4k+1 primes are along the x-axis and the rest is along the y-axis? No inner product necessary.

    I could make up a crazy law. And that's the question here. Why everyone discards these crazy laws from the very start. They use crazy metrics, but all of them are Riemannian.

    I don't think that's an answer I'm looking for, but it's interesting anyway. Can you recommend an introductory reading to that method? This method should work for all particles at once (don't introduce metric for every particle) and not be restricted to light only.
     
  19. Jan 22, 2010 #18
    I think this would be a way to answer what I look for. Just the question is how to define these isometries (with some relation to our real world)? One cannot use distance though, since it hasn't been defined yet or could be defined arbitrarily.

    At the moment I believe the mesh idea might be more artificial but somehow it should be possible to transfer any argument to it.
     
  20. Jan 22, 2010 #19
    Here is a thought of mine:

    What would be different if we lived in a world with taxicab metric (non-Riemannian I believe?)

    One difference is of course that there would be many shortest paths from A to B.

    Another way to find differences is to measure the shortest distances between a set of points A, B, C, D,...

    I believe if you take only three points, then in taxicab and Riemannian the restrictions are the same? Namely the triangle inequality.

    Probably if you take 4 points and 3D space, then Riemannian geometry posed additional contraints on the distances AB, AC, AD, BC, BD, CD?
     
  21. Jan 22, 2010 #20

    There are many ways to define transformations that will give you ageometry -e.g. the examples I mentioned above. I think it would be helpful if you defined what properties of the space that you want to be preserved under transformation.

    BTW: In the case of Euclidean motions or Moebius transformation the distance measure is uniquely determined (up to a constant)- so if done right you do not need to start with an idea of distance.
     
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