Phase Velocity Definition: A or B?

In summary, the author suggests that the radius of curvature comes from the fact that the tangent vector T is a unit vector tangential to the path.
  • #1
Helios
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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)?
 
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  • #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?
 
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  • #3
Helios said:
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##.
 
  • #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.
 
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  • #5
@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.
 
  • #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.
 
  • #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)
 
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  • #8
Helios said:
V = V(x)

What is V?

Helios said:
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?

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

Where does this equation come from?

Helios said:
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?

Helios said:
γmv2/R is relativistic centripetal force.

What does centripetal force have to do with it?
 
  • #9
Thank you for the reply.
PeterDonis said:
What is V?
V is called the potential energy. It is part of the total energy.
E = γmc2 + V
PeterDonis said:
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.
PeterDonis said:
Where does this equation come from?
It is deduced. It is also validated because it works in the ray equation.
PeterDonis said:
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 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

PeterDonis said:
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
PeterDonis said:
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?
 
  • #10
Helios said:
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.
 
  • #11
PeterDonis said:
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.
 
  • #12
Helios said:
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.
 
  • #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.
 

1. What is the definition of phase velocity?

The phase velocity is a measure of how fast the phase of a wave propagates through space or a medium. It is defined as the rate of change of the phase of a wave with respect to position.

2. What is the difference between phase velocity and group velocity?

Phase velocity and group velocity are both measures of how fast a wave propagates, but they describe different aspects of the wave. While phase velocity measures the speed at which the phase of the wave travels, group velocity measures the speed at which the overall shape or envelope of the wave travels.

3. How is phase velocity calculated?

Phase velocity can be calculated by dividing the frequency of the wave by its wavelength. It can also be calculated using the wave's angular frequency and wave number.

4. What are some applications of phase velocity?

Phase velocity is commonly used in fields such as optics, acoustics, and electromagnetics to study and analyze wave behavior. It is also important in the study of quantum mechanics and in the development of various technologies such as radar and telecommunications.

5. Can the phase velocity of a wave exceed the speed of light?

No, according to the theory of relativity, the phase velocity of a wave cannot exceed the speed of light in a vacuum. However, the group velocity can exceed the speed of light in certain situations, such as in certain types of materials or in the presence of a medium with a negative refractive index.

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