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I Observation of distances w.r.t. metric

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  1. May 11, 2016 #1
    Hello. I don't know exactly if my question can be treated physically but so...
    Let us have a 3D space with non-constant metric. We are in the first region with a euclidian metric.
    [itex]ds^2=dx^2+dy^2+dz^2[/itex]
    So the distance between two points is got through pythagorean theorem
    Then near us we have the second region that has another metric such as
    [itex]ds^2=a^2dx^2+b^2dy^2+c^2dz^2[/itex]
    where [itex]a,b,c[/itex] - some coefficients
    In our region we have line with [itex]A(x_1,y_1,z_1)[/itex] and [itex]B(x_2,y_2,z_2)[/itex] at the ends. So the length of the line is [itex]L_1=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}[/itex]
    Then we take our line and put in the second region, so the line ends have coordiantes [itex]C(x_3,y_3,z_3)[/itex] and [itex]D(x_4,y_4,z_4)[/itex]. Its length is
    [itex]L_2=\sqrt{a^2(x_3-x_4)^2+b^2(y_3-y_4)^2+c^2(z_3-z_4)^2}[/itex]
    The question is:
    Will we really observe the line smaller or bigger after putting it into the second region?
     
  2. jcsd
  3. May 11, 2016 #2

    Orodruin

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    What do you mean by this? The two are different lines, they are not the same lines at different locations.
     
  4. May 11, 2016 #3
    By that I mean continuous translation
     
  5. May 11, 2016 #4

    fresh_42

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    Without to know how you define continuity at the border between the two different metric spaces or even the presumably common border itself, and what you mean by "observe", because to live in one metric means to me you only have this metric for comparison, the closest picture I can come up with is someone standing in front of a convex or concave mirror as they can be found in, e.g. amusement parks.
     
  6. May 11, 2016 #5

    Orodruin

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    This still is not well defined. You need to specify what this continuous translation is and how it affects the points on the lines. These things are not unique on curved manifolds. In particular, the way I suspect you will want to define it is going to be coordinate dependent.
     
  7. May 11, 2016 #6
    For example, that region has a border as a circle. But the circle has some width. So the metric is changing smoothly passing through border

    So how can one derive the accurate picture that the observer sees?
     
  8. May 11, 2016 #7

    Orodruin

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    This answers nothing about the issues that were raised, or what you mean by an "observer". There is no such thing as "translation" of an extended object in a general manifold. You can define continuous transforms based on parallel transport, but this has issues to. What you are asking is simply not well defined.
     
  9. May 23, 2016 #8
    Let it be 3D differential manifold. Let we have metrics [itex]g[/itex] that is euclidian and [itex]\eta[/itex] that is
    [itex]\eta = \begin{pmatrix}
    a & 0 & 0 \\
    0 & b & 0 \\
    0 & 0 & c
    \end{pmatrix}[/itex]
    The circle in image represents sphere as a border of region with metric [itex]\eta[/itex](omitted here z coordinate)
    Untitled-1.png
    Let metric moving in arrow direction changes in a way that when you moving closer to the circle interior it looks like
    [itex]\mu = \begin{pmatrix}
    \alpha & 0 & 0 \\
    0 & \beta & 0 \\
    0 & 0 & \gamma
    \end{pmatrix}[/itex]
    where [itex]\alpha \rightarrow a, \beta \rightarrow b, \gamma \rightarrow c[/itex]
    And vice versa, if you moving in opposite direction of the arrow the metric is the same but
    [itex]\alpha \rightarrow 1, \beta \rightarrow 1, \gamma \rightarrow 1[/itex]

    So do the question in the first post have answer in such conditions?
    And again in such conditions can one get a picture how distorted an observer being outside the sphere will see the line?

    When I say "observer" I mean the one who just stays there if such manifold was a part of our physical space.
     
  10. May 23, 2016 #9

    Orodruin

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    Your statements are still not mathematically sound, you are just repeating the same thing you have already said.
     
  11. May 23, 2016 #10
    Could you explain what things I need to define mathematically or maybe guide me that?
    Thanks.
     
  12. May 23, 2016 #11

    Orodruin

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    You have defined the manifold and the metric, that has never been in question. Your problem is that you have not defined what you mean by "observer" or what the process of this "observation" is supposed to be. You have also not defined what you think it means to perform a "continuous translation". If by this you mean to just make a coordinate translation ##x^\mu \to x^\mu + a^\mu## where ##a^\mu## is a set of constants, what you are trying to do is coordinate dependent and not something which is inherent in the manifold. Any geometrical property you wish to define should not depend on the coordinates you use on the manifold.

    From what I understand from your attempts, what you are trying to ask is simply undefined.
     
  13. May 23, 2016 #12
    I see "observating the line" as perceiving light rays that were reflected from the line.
    So let's regard continuous translation as translation along geodesic curve. AFAIK we can derive a geodesic from the connection and the connection from the metric.
    What else do I need?
     
  14. May 23, 2016 #13

    Orodruin

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    The way you have defined it, you are using a fully Riemannian manifold. What do you mean by light ray? Light rays are things in space-times that are 4-dimensional and have a pseudo metric.

    It is also not clear whether or not you are "observing" from the outside, i.e., whether or not you assume some sort of embedding or similar which is not really a property of the manifold itself.

    This is not enough. The line is an extended object so you will need one geodesic per point on the line and so you will have to properly define this. Once done, whether the line length is changed or not is going to depend on the metric.
     
  15. May 23, 2016 #14
    Ok, let it be 4d pseudo riemannian manifold and for [itex]\eta[/itex] I'll add the time coordinate that is invatiant for tensor, so
    [itex]\eta = \begin{pmatrix}
    -1 & 0 & 0 & 0 \\
    0 & a & 0 & 0 \\
    0 & 0 & b & 0\\
    0 & 0 & 0 & c
    \end{pmatrix} [/itex]

    Why isn't it enough if connection is defined everywhere so do geodesics?
     
  16. May 23, 2016 #15

    Orodruin

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    Because of the exact reason I just mentioned.
     
  17. May 23, 2016 #16
    So did you mean I just needed to find more than one geodesic?
     
  18. May 23, 2016 #17

    Orodruin

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    You need to assign a method of assigning a geodesic to each point in the line, i.e., you need to define the 4-velocity of each point.

    You also need to note that if you have an actual physical object, it is going to be held together by forces such as EM forces. This will mean the object's parts are not moving along geodesics.
     
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