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Phase Velocity Definition

  1. Dec 22, 2015 #1
    As I'm seeing things, there are these choices for a definition of relativistic phase velocity u;
    a) u = (mc^2)/ p
    b) u = E / p
    Now I like choice a) because it leads to a correct-looking index of refraction via n = c / u. It leads to a correct-looking solution to the ray equation. However, references give choice b). Who says b)? Why not a)?
     
  2. jcsd
  3. Dec 22, 2015 #2
    Phase velocity is E/p. It reduces to c^2/v.
    Not sure what "a" is, but it's not phase velocity.

    By definition phase velocity is v = w/k, which is E/p.

    What are you doing with refraction exactly?
     
  4. Dec 22, 2015 #3

    bcrowell

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    Version a is clearly wrong because for a light wave, which has m=0, it gives zero.

    Version b is a universally valid identity for any object in special relativity that can be described by a world-line. It doesn't have anything specifically to do with waves or phase velocity. It simply says that for such an object, the energy-momentum four-vector is tangent to the world line. That's always true, because there's no other choice for the direction that wouldn't violate Lorentz invariance.

    As DuckAmuck says, phase velocity is ##\omega/k##.
     
  5. Dec 22, 2015 #4
    The phase velocity I refer to is for a material wave, not light or a massless wave. Definition a) fails for light, obviously! I agree that references give the definition of phase velocity as v = w/k, which is E/p, which reduces to c^2/v. But it leads to an index of refraction n = v/c. Is this so?

    We begin with the integral for optical path length (OPL).
    https://en.wikipedia.org/wiki/Optical_path_length
    We employ the calculus of variations to get the ray equation.
    We plug in ( trial ) index of refraction n to see if the optico-mechanical analogy is apparent. I try a) and it looks right, very nice in fact. I get the correct expression for relativistic centripetal force, normal to path. Trying b) doesn't lead anywhere, so that's what's fishy.
     
    Last edited: Dec 22, 2015
  6. Dec 22, 2015 #5

    bcrowell

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    @Helios - We're not mind readers. If you have calculations that are turning out wrong, and you want to know why they're turning out wrong, you need to post the calculations.
     
  7. Dec 22, 2015 #6
    Let's consider the following as axiomatic, with V = V(x), ( please forgive exponents not raised )

    v / c = SQR[ 1 - m2c4 / ( E - V )2 ]
    γmc2 = ( E - V )

    γmcv = SQR[ ( E - V )2 - m2c4 ]

    γmv2 = [ ( E - V )2 - m2c4 ] / ( E - V )
    --------------------------------------------------------------
    Posit that
    n = SQR{ [ ( E - V )/mc2 ]2 - 1 } = γv / c
    Plug into ray equation with T=tangent vector, N=normal = unit normal/R, R = radius of curvature.
    GRAD( n ) - [ GRAD( n ) . T ]T - n N = 0
    It looks like a) is right to me.
     
  8. Dec 23, 2015 #7
    The minimum optical path lengthn ds leads to the ray equation. With n = √{ [ ( E - V )/mc2 ]2 - 1 },
    the result when the math is carried out is
    (-V) - [ (-V) ⋅ τ ]τ - ( γmv2/R )η = 0
    γmv2/R is relativistic centripetal force.
    If we accept n and n = c/u , then u = (mc2)/ p, which I gave in case a)
     
    Last edited: Dec 23, 2015
  9. Dec 23, 2015 #8

    PeterDonis

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    What is V?

    Where do these equations come from?

    Where does this equation come from?

    Where does the radius of curvature come from? What physical scenario are you trying to describe?

    What does centripetal force have to do with it?
     
  10. Dec 23, 2015 #9
    Thank you for the reply.
    V is called the potential energy. It is part of the total energy.
    E = γmc2 + V
    These are basic relativistic relationships that include potential energy. There's no invention here. These are all just algebraic variants of what is basic to special relativistic kinematics.
    It is deduced. It is also validated because it works in the ray equation.
    I quote from Mechanics ( Landau, Lifshitz ) page 143
    t = dr/dl is a unit vector tangential to the path
    The difference F - ( F.t)t is the component Fn of the force normal to the path
    the derivative d2r/dl2 = dt/dl is known from differential geometry to be n/R where R is the radius of curvature of the path and n is the principal normal

    a relativistic particle moving through space subjected to a force derivable from a potential which moves through space such that the optical path length is a minimum
    It comes about as a consequence of the math. It is recognized. There it is, that's the relativistic centripetal force. Because it emerges where it is supposed to, it confirms the index of refraction.
    My opinion, as of now, is that I have show that the phase velocity of a particle is not u = E/p but E0/p = mc2/p. Does anyone disagree?
     
  11. Dec 23, 2015 #10

    PeterDonis

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    I don't think this is the correct way to model potential energy in SR. The equations you write don't look covariant to me.
     
  12. Dec 23, 2015 #11
    I agree, not absolutely perfect because the potential energy would only work for one reference frame.
     
  13. Dec 23, 2015 #12

    PeterDonis

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    But the whole point of defining "relativistic phase velocity" is to make it covariant. So I don't understand what you're trying to do.
     
  14. Dec 23, 2015 #13
    I confess that I can't reformulate the whole scheme into a perfect covariant form. A 4-dimensional covariant potential energy is out of my depths. I'll listen if someone else can do this.
     
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