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Observer Dependant vs Invariant

  1. Oct 12, 2003 #1

    pmb

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    "Observer Dependant" vs "Invariant"

    The point I want to make in this thread is on something with is rather subtle.

    Define the quantity, m, as the quantity such that, in an inertial frame, mv is conserved. Call this quantity "mass" (some call this "relativistic mass")

    For tardyon's (particles which move at speeds less than c) the mass is a function of speed, i.e. m = m(v).

    Definition of symbols:
    m = mass of particle (tardyon) = m_o/sqrt[1-(v/c)^2]
    m_o = rest mass of particle (tardyon) = m(0)
    p = mv = momentum
    P = (mc, p) = 4-momentum of particle
    U_obs = 4-velocity of observer


    Certain quantities are observer dependant. The mass of a particle, m, as defined above, is one of them. Each observer will meausure the mass of a particle and obtain a quantitity which will depend of the relative velocity of observer and particle. That is what is called an "observer dependant quantity."

    There are quantities in relativity which are called 'invariants' (aka
    'scalar' aka 'tensor of rank zero'). They are quantities which remain
    unchanged when the coordinates are changed. That's the very meaning of the term 'invariant. '

    The norm of a 4-vector remains unchanged by a change in coordinates and is thus is an invariant. The scalar product of two 4-vectors also remains unchanged by a change in coordinates and thus is also an invariant.

    Now consider the quantity P*U_obs/c which is the scalar product of the 4-momentum of a partricle and the 4-velocity of the observer divided by c, the speed of light. Since this is a scalar product of two 4-vectors it is an invariant. It can be easily shown that

    m = P*U_obs/c = m_o/sqrt[1-(v/c)^2]

    This is called the "mass measured by the observer"

    And thus the magntiude of this scalar product, m, is an invariant. This is overly obvious since U_obs represents the observer and thus its not surprising that U_obs dotted with a quantity will be observer dependant.

    However the components of a 4-vector change upon a change in coordinate system so in that context "mc" and thus "m" is not an invariant.

    Example from Euclidian geometry - The dot product the position vector, R = (x,y,z), and the unit vector in the x-direction, "i", is an invariant. i.e.

    R*i = invariant

    This invariant has the value R*i = x.

    If I change coordinates to primed coordinates, x', then the unit vector in the x' direction is i'.

    In the new system R = (x',y',z'). The dot product of the vector with the new unit vector will now be R*i' = x' which will in general be different than R*i = x. However i and i' are quite different vectors and therefore i != i'. The quantity R*i is a Euclidean scalar and has the value of x.

    Thus x can be interpreted in two ways, one way as the component of a vector (and as such not an invariant) and another way as the component of R along a given direction - the quantity then being invariant.

    Pete
     
  2. jcsd
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