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Think about SR as different observers using different coordinate

  1. Dec 11, 2005 #1

    daniel_i_l

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    Is it correct to think about SR as different observers using different coordinate systems depending on their speeds? This would meen that their measurements would differ since space, time, mass,etc.. measurements depend on the system used. I find it very easy to understand SR when I think about it this way. Is it correct?
     
  2. jcsd
  3. Dec 11, 2005 #2
    You can definitely think about it that way, just keep in mind that the coordinate systems being used are a little funny, i.e. the invarient in Minkowsky space is [tex]\sqrt{t^2-x^2-y^2-z^2}[/tex] or [tex]\sqrt{-t^2+x^2+y^2+z^2}[/tex], depending on whether the interval is timelike or spacelike, instead of [tex]\sqrt{t^2+x^2+y^2+z^2}[/tex], as one might expect from Cartesian space.

    You can even represent the Lorentz transformation between refrence frames as a matrix, and then demonstrate that the matrix preserves lengths and angles (again, in the sense of Minkowsky coordinates) so that you can view it as being simply a rotation between different coordinate systems.
     
    Last edited: Dec 11, 2005
  4. Dec 11, 2005 #3

    pervect

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    It is correct as far as it goes, but unless you go further and give the Lorentz transforms, or some equivalent statement such as the fact that the Lorentz interval is invariant for all observers, you have not uniquely identified SR.

    You also have to note that the differing coordinate systems reflect the same underlying reality (covariance).
     
  5. Dec 11, 2005 #4

    daniel_i_l

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    Thanks,
    Well, obviously the coordunate system would have to rotate is such a way that the speed of light would be constant for all observers.
     
  6. Dec 11, 2005 #5

    jtbell

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    Not only the speed of light, but also other invariant quantities. For example, the spacetime interval between two events [itex]\Delta s = \sqrt{(c \Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2}[/itex] is the same in any inertial reference frame. Also the quantity [itex]m_0 c^2 = \sqrt{E^2 - (p_x c)^2 - (p_y c)^2 - (p_z c)^2}[/itex] is the same in any inertial reference frame, for any object.

    More generally, the magnitude of any four-vector is the same in all inertial reference frames, just as the magnitude of any ordinary "three-vector" stays the same when you rotate your (3-dimensional) coordinate system. An examples of a four-vector is the energy-momentum four-vector [itex](p_0, p_1, p_2, p_3) = (E/c, p_x, p_y, p_z)[/itex]. Also, the "dot product" of two four-vectors, [itex]a_0 b_0 - a_1 b_1 - a_2 b_2 - a_3 b_3[/itex] is the same in any inertial reference frame.
     
    Last edited: Dec 11, 2005
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