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Four vector component. how to recognize it?

  1. Feb 9, 2009 #1
    What is the simplest way to find out if a physical quantity is the component of a four vector? There are opinions that the Lorentz contracted length is not, the Lorentz dilate length being.
     
  2. jcsd
  3. Feb 10, 2009 #2
    I add some details.
    Consider that observers from I detect simultaneously the space coordinates of the ends of a moving rod, at rest in the I', located along the x' axis. The result is that the proper length of the rod L0 measured in I' and its contracted length L are related by the formula that accounts for the Lorentz contraction. The same formula could be derived without imposing the simultaneous detection of the space coordinates of the two ends.
    The space coordinates of the ends measured in the stationary frame do not depend on the time when they are measured. Consider that observers from I' detect simultaneously the space coordinates of the stationary rod, the measurements being associated with the space-time coordinates (x'1,t') and (x'2,t'). Observers from I perform the Lorentz transformations for the space coordinates of the two ends obtaining
    L=[tex]\gamma[/tex]L0 (1)
    which accounts for the Lorentz expansion (dilation?).
    Equation (1) is derived by radar detection of the space coordinates of the moving rod.
    Virtues of (1) are mentioned.
    Which is the simplest way to show that (1) is the component of a four vector the length contracted length being not?
     
  4. Feb 10, 2009 #3

    DrGreg

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    A set of quantites form a 4-vector if they tranform via the Lorentz tranform. Thus if [itex]\textbf{X} = (ct, x, y, z)^T[/itex] and [itex]\textbf{X'} = (ct', x', y', z')^T[/itex] are coordinates for the same event in two different inertial frames (in "standard configuration", i.e. with aligned spatial axes and in relative motion along their common x axis), we can say this is a 4-vector because

    [tex]\textbf{X'} = \Lambda \, \textbf{X}[/tex]​

    where

    [tex]\Lambda = \left[ \begin{array}{cccc}
    \gamma & -\gamma v & 0 & 0 \\
    -\gamma v & \gamma & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \\
    \end{array} \right][/tex]​

    "Lorentz-contracted length" doesn't satisfy this because the length is measured between two events that are simultaneous in the frame where the measurement is made. When you change frames, you also change events. The Lorentz transform applies only when you measure the same pair of events in two different frames.

    Other examples of 4-vectors are

    Energy-momentum............ [tex]\textbf{P} = (\frac{E}{c}, p_1, p_2, p_3)^T = m\frac{d\textbf{X}}{d\tau}[/tex]
    4-force........................... [tex]\textbf{F} = \frac{d\textbf{P}}{d\tau}[/tex]
    4-current......................... [tex]\textbf{J} = (\rho c, j_1, j_2, j_3)^T[/tex]
    electromagnetic 4-potential [tex]\Psi = (\phi, a_1 c, a_2 c, a_3 c)^T[/tex] ​

    (See four-vector on Wikipedia.)
     
  5. Feb 11, 2009 #4
    Thank you. Please have a look at the following lines.
    Consider that in a 2D approach A is the scalar component of a "two vector" and let A0 be its proper (invariant) magnitude related to its relativitic magnitude as
    A=[tex]\gamma[/tex](u)A0. (1)
    Let A be the vector component of the same "two vector",related to A by
    A=[tex]\gamma[/tex](u)A0u (2)
    (1) and (2) ensuring the invariance of the relativistic interval.
    As we see the Lorentz contracted length is not the component of a four vector
    whereas the dilated length L=L0[tex]\gamma[/tex] is.
    The dilated length could be obtained simultaneously detecting the ends of the rod in its rest frame. Did Einstein take into account that measurement variant?
    Would you accept such an approach?
     
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