Phase Velocity Definition: A or B?

[Helios]

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)?

 

[DuckAmuck]

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?

 

[bcrowell]

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)?

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 $\omega/k$.

 

[Helios]

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.

 

[bcrowell]

@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.

 

[Helios]

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.

 

[Helios]

The minimum optical path length ∫n 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)

 

[PeterDonis]

V = V(x)

What is V?

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

γmcv = SQR[ ( E - V )2 - m2c4 ]

γmv2 = [ ( E - V )2 - m2c4 ] / ( E - V )

Where do these equations come from?

Posit that
n = SQR{ [ ( E - V )/mc2 ]2 - 1 } = γv / c

Where does this equation come from?

Plug into ray equation with T=tangent vector, N=normal = unit normal/R, R = radius of curvature.

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

γmv2/R is relativistic centripetal force.

What does centripetal force have to do with it?

 

[Helios]

Thank you for the reply.

What is V?

V is called the potential energy. It is part of the total energy.
E = γmc² + V

Where do these equations come from?

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.

Where does this equation come from?

It is deduced. It is also validated because it works in the ray equation.

Where does the radius of curvature come from?

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 F_n 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

What physical scenario are you trying to describe?

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

What does centripetal force have to do with it?

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?

 

[PeterDonis]

V is called the potential energy. It is part of the total energy.
E = γmc2 + V

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.

 

[Helios]

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.

I agree, not absolutely perfect because the potential energy would only work for one reference frame.

 

[PeterDonis]

I agree, not absolutely perfect because the potential energy would only work for one reference frame.

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.

 

[Helios]

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.