Phase Invariance: Is It an Invariant?

[bernhard.rothenstein]

phase invariance?

I read papers claiming that the phase of a plane wave (acoustic and electromagnetic) is not an invariant. Taking into account that invoking its invariance, we could derive formulas that account for well tested experimental facts (Doppler shift, aberratiion of light, wave vector-frequency four vector) I consider that it is an invariant. What is your oppinion? Thanks in advance.

 

[Dale]

I read papers claiming that the phase of a plane wave (acoustic and electromagnetic) is not an invariant.

Could you cite that? My understanding is that the phase is an invariant since it is the Minkowski scalar product of two four-vectors.

 

[bernhard.rothenstein]

Could you cite that? My understanding is that the phase is an invariant since it is the Minkowski scalar product of two four-vectors.

The invariance of the phase of waves among inertial frames is questionable

Young-Sea Huang 2007 EPL 79 10006 (4pp) doi: 10.1209/0295-5075/79/10006 Help


PDF (125 KB) | HTML | References | Articles citing this article

 

[Dale]

I don't have access to the full article, but from the abstract it seems that the author's conclusion is based on his objection to negative frequencies. I don't have any problem with negative frequencies, so I doubt that I would agree with the conclusion. But it is hard to know for sure.

 

[Huang XingBin]

Study on the invariance of the phase of plane waves

Abstract. In recent study, the invariance of the phase of plane light waves among inertial frames is challenged by a counter-example where the four-vector (k, ω/c) of plane waves is exceptionally not Lorentz-covariant. A new argument that the conventional invariance of the phase of plane light waves should be replaced by the invariance of the phase difference of plane light waves among all inertial frames is presented in this paper. Based on the physics meaning of negative frequency of plane waves in the case of “superluminal” motion of the medium, the explanation to the counter-example is worked out. The result shows that the four-vector of plane waves is still Lorentz-covariant.

full text:http://www.sciencenet.cn/upload/blog/file/2010/10/201010278923170244.pdf

 

[jason12345]

I don't have access to the full article, but from the abstract it seems that the author's conclusion is based on his objection to negative frequencies. I don't have any problem with negative frequencies, so I doubt that I would agree with the conclusion. But it is hard to know for sure.

The misunderstanding was pointed out in:

Is the phase of plane waves a frame-independent quantity?
Authors: Aleksandar Gjurchinovski

http://arxiv.org/abs/0801.3149

It turns out that the time component of the wave four vector should be k/c dot u rather than k*u/c = w/c, because of relativistically induced optical anisotropy. This doesn't matter in vacuum where k dot u = k*u

 

[Huang XingBin]

I think Gjurchinovski's conclusion is questionable. The reason is that in order to let the frequency changes from negative value to positive value, the direction of the wave vector and the direction of the wave surfaces transport (i.e., the phase speed) point in opposite directions in the case of “superluminal” motion of the medium based on Gjurchinovski's conclusion. This clearly does not meet the definition of a plane wave. Therefore, Gjurchinovski's conclusion is not correct.

 

[jason12345]

I think Gjurchinovski's conclusion is questionable. The reason is that in order to let the frequency changes from negative value to positive value, the direction of the wave vector and the direction of the wave surfaces transport (i.e., the phase speed) point in opposite directions in the case of “superluminal” motion of the medium based on Gjurchinovski's conclusion. This clearly does not meet the definition of a plane wave. Therefore, Gjurchinovski's conclusion is not correct.

Yes, but I don't see how this refutes Gjurchinovski's conclusion that the true invariant phase of the wave is k dot r - k dot u t rather than k dot r - wt

In the case of "superluminal" motion of the medium, k dot u becomes negative, hence preserving the invariance of the phase, whereas Y. S. Huang introduced negative frequencies to get around this problem with his wt term, according to Gjurchinovsk.

Young-Sea Huang's original paper asked:

Is the phase of plane waves an invariant?

Gjurchinovski concluded yes, providing the correct expression for the phase is used.






Y.-S. Huang's

 

[Huang XingBin]

Yes, but I don't see how this refutes Gjurchinovski's conclusion that the true invariant phase of the wave is k dot r - k dot u t rather than k dot r - wt

In the case of "superluminal" motion of the medium, k dot u becomes negative, hence preserving the invariance of the phase, whereas Y. S. Huang introduced negative frequencies to get around this problem with his wt term, according to Gjurchinovsk.

Young-Sea Huang's original paper asked:

Is the phase of plane waves an invariant?

Gjurchinovski concluded yes, providing the correct expression for the phase is used.

Y.-S. Huang's

Thank you for your message, but, I think in the case of "superluminal" motion of the medium, k dot u becomes negative is not correct. If this conclusion is true, then the directions of k and u (In exceptional case of "superluminal" motion of the medium, u is the phase speed) point in opposite directions. According to the definition of a plane wave, the the directions of k and u have to be the same. Therefore, k dot u = w. However, Gjurchinovski's hypothesis, i.e., k dot u becomes negative, is unfounded. In other words, k dot u becomes negative (when the directions of k and u are the same) and negative frequencies are the same.

 

[jason12345]

Thank you for your message, but, I think in the case of "superluminal" motion of the medium, k dot u becomes negative is not correct. If this conclusion is true, then the directions of k and u (In exceptional case of "superluminal" motion of the medium, u is the phase speed) point in opposite directions.

Yes, they point in opposite direction.

According to the definition of a plane wave, the the directions of k and u have to be the same.

I see an EM plane wave as one where the EM field at any point is the same for a set of points lying within some plane, these planes being parallel to one another, and the common EM value for each plane vaying sinusoidally from plane to plane. I think the direction of k and u relative to one another is irrelevant for the definition of what a plane wave is.

Therefore, k dot u = w

Yes. It's the component of u along k, rather than just u, that determines w.

However, Gjurchinovski's hypothesis, i.e., k dot u becomes negative, is unfounded. In other words, k dot u becomes negative (when the directions of k and u are the same) and negative frequencies are the same.

Gjurchinovski considers a plane wave traveling with velocity u'_x' along k'_x'. Transforming to S and for the case of u'_x' < V < c, he gets k_x > 0, u < 0. They point in opposite directions and so k dot u becomes negative.

 

[Huang XingBin]

Yes, they point in opposite direction.

I think the direction of k and u relative to one another is irrelevant for the definition of what a plane wave is.

From the definition of a plane wave, direction of k is the direction of the wave surface (i.e., plane) moving, and direction of u is also the direction of the wave surface moving, it should be stressed here that the u is the phase velocity. Hence, the directions of k and u must be the same. How do we visualize a wave surface propagating in two directions?

Gjurchinovski considers a plane wave traveling with velocity u'_x' along k'_x'. Transforming to S and for the case of u'_x' < V < c, he gets k_x > 0, u < 0. They point in opposite directions and so k dot u becomes negative.

Gjurchinovski's mistake is that he gets k_x > 0, u < 0. Frome his Fig.5, correct conclusion should be k_x < 0, u < 0.

Therefore, Gjurchinovski does not answer Y.-S. Huang's question.

 

[Huang XingBin]

I have submitted my paper to Europhysics Letters and pointed out their mistake, but Europhysics Letters have refused to correct the error. Therefore we cannot easily believe the paper published in Europhysics Letters. Europhysics Letters may not be a true science journal.

 

[jason12345]

From the definition of a plane wave, direction of k is the direction of the wave surface (i.e., plane) moving, and direction of u is also the direction of the wave surface moving, it should be stressed here that the u is the phase velocity. Hence, the directions of k and u must be the same. How do we visualize a wave surface propagating in two directions?

Principles of Optics - Born Wolf, page 15:

$$\nabla^2V - 1/v^2 \partial V/\partial t^2 = 0 \ (1)$$

Let $$\textbf{r}(x,y,z)$$ be a position vector of a point P in space and $$\textbf{s}(s_x,s_y,s_z)$$ a unit vector in a fixed direction. Any solution of (1) of the form:

$$V = V(\textbf{r}\cdot \textbf{s}, t)$$

is said to represent a plane wave, since at each instant of time V is constant over each of the planes

$$\textbf{r} \cdot \textbf{s} = constant$$

which are perpendicular to the unit vector s.

There is no mention of the wave velocity playing a part in this definition of a plane wave.

You decompose the velocity into components co-planar and orthogonal to the wave front. Hence you can visualise the plane wave moving along k and the wave front at the same time, but only for u < c.

Gjurchinovski's mistake is that he gets k_x > 0, u < 0. Frome his Fig.5, correct conclusion should be k_x < 0, u < 0.

Therefore, Gjurchinovski does not answer Y.-S. Huang's question.

The wave four vector is $$(\textbf{k}, \textbf{k} \cdot \textbf{u}/c)$$

In s', k' = (k',0,0)

Transforming to s which moves with velocity +v relative to s' gives

$$k = \gamma (k'- vk'u/c^2)$$

so k > 0 for u < v < c

 

[Huang XingBin]

Principles of Optics - Born Wolf, page 17, after Eq.(19):

"It is also convenient to define vectors k_0 and k in the direction s of propagation,...
......
The vector k= ks is called the wave vector or the propagation vector in the medium,..."

From above definition of the k, we know that the direction of k is the direction of the wave surface (i.e., plane) moving or propagating.

From Gjurchinovski's Fig. 5, he gets "so k > 0 for u < v < c". It is obvious that his wave vector points in opposite direction of the wave surface moving or propagating.

Therefore, Gjurchinovski's method is invalid.

Thank you very much for your discussion!

 

[jason12345]

Principles of Optics - Born Wolf, page 17, after Eq.(19):

"It is also convenient to define vectors k_0 and k in the direction s of propagation,...
......
The vector k= ks is called the wave vector or the propagation vector in the medium,..."

From above definition of the k, we know that the direction of k is the direction of the wave surface (i.e., plane) moving or propagating.

From Gjurchinovski's Fig. 5, he gets "so k > 0 for u < v < c". It is obvious that his wave vector points in opposite direction of the wave surface moving or propagating.

Therefore, Gjurchinovski's method is invalid.

Thank you very much for your discussion!

Yes, the book defines k to be in the same direction as the direction of propagation of the wave. With this strict definition then Gjurchinovski does have k pointing in the wrong direction, as you say.

However, Gjurchinovski defines k as normal to the surface of the plane wave, independent of u, and so he has two choices, each in opposite directions to one another. Since it's clear in his paper that this is how he is defining k, then his conclusions are still valid.

 

[Huang XingBin]

However, Gjurchinovski defines k as normal to the surface of the plane wave, independent of u, and so he has two choices, each in opposite directions to one another. Since it's clear in his paper that this is how he is defining k, then his conclusions are still valid.

Yes, but I don't see how Gjurchinovski redefine the wave vector in his paper. I see that he give a new velocity, the new velocity u of plane waves is only described as the velocity of the profile of plane waves. From his new velocity, he has gotten the conclusion contradicted with the definition of wave vector. Finally, it should be stressed here that the wave vector k is a very very important concept; even God cannot readily redefine it.
To sum up: In my opinion Gjurchinovski’s method is not correct.

 

[jason12345]

Yes, but I don't see how Gjurchinovski redefine the wave vector in his paper. I see that he give a new velocity, the new velocity u of plane waves is only described as the velocity of the profile of plane waves. From his new velocity, he has gotten the conclusion contradicted with the definition of wave vector. Finally, it should be stressed here that the wave vector k is a very very important concept; even God cannot readily redefine it.
To sum up: In my opinion Gjurchinovski’s method is not correct.

If you have a plane wave traveling normal to, but also parallel to its plane, how do you define k?

You could stick with Born and define it to be in the same direction as u in which case it's no longer normal to the plane. Or you could define it as being normal to the plane, independent of u as Gjurchinovski did.

u is the velocity of a fixed point on the plane wave which travels normal and along the plane generally. Of course, most textbooks don't consider this general case, but instead have u normal to the plane.

 

[Huang XingBin]

If you have a plane wave traveling normal to, but also parallel to its plane, how do you define k?

You could stick with Born and define it to be in the same direction as u in which case it's no longer normal to the plane. Or you could define it as being normal to the plane, independent of u as Gjurchinovski did.

u is the velocity of a fixed point on the plane wave which travels normal and along the plane generally. Of course, most textbooks don't consider this general case, but instead have u normal to the plane.

Principles of Optics - Born Wolf, page 794, after Eq.(6):
We must distinguish between the phase velocity and the velocity of energy transport. The former, the phase velocity, is in the direction of the unit vector s……. (i.e., the direction of wave vector k)


Gjurchinovski’s velocity u is the velocity of energy transport (or light ray or a fixed point on the plane wave as you say). If the directions of the two velocities are not the same, the direction of k and the direction of phase velocity are the same (according to the admitted definition of k).

However, in Gjurchinovski’s Fig 5, the directions of the two velocities are the same. Therefore the directions of phase velocity, a fixed point on the plane wave as you say and wave vector k must be the same.

We could not redefine k, such as Gjurchinovski’s conclusion. In other words, the direction of k is only the direction of phase velocity. Otherwise, it will bring physics about great confusion.

 

[jason12345]

Principles of Optics - Born Wolf, page 794, after Eq.(6):
We must distinguish between the phase velocity and the velocity of energy transport. The former, the phase velocity, is in the direction of the unit vector s……. (i.e., the direction of wave vector k)


Gjurchinovski’s velocity u is the velocity of energy transport (or light ray or a fixed point on the plane wave as you say). If the directions of the two velocities are not the same, the direction of k and the direction of phase velocity are the same (according to the admitted definition of k).

However, in Gjurchinovski’s Fig 5, the directions of the two velocities are the same. Therefore the directions of phase velocity, a fixed point on the plane wave as you say and wave vector k must be the same.

We could not redefine k, such as Gjurchinovski’s conclusion. In other words, the direction of k is only the direction of phase velocity. Otherwise, it will bring physics about great confusion.

As defined by Gjurchinovski, k only has to be normal to the plane of the wave and hence parallel to the phase velocity, which means it can point in the same or opposite direction to the phase velocity.

If k1 = -k2 then any point r on the wave satisfies either expression

$$(\textbf{k1} \cdot \textbf{u}/c + \textbf{k1} \cdot \textbf{r})$$ = constant

$$(\textbf{k2} \cdot \textbf{u}/c - \textbf{k2} \cdot \textbf{r})$$ = constant

k being part of Gjurchinovski's wave 4 vector means the LT transformation takes care of the direction of k.

I think you're only objecting to him using the symbol k to define a vector which only has to be normal to the plane, whereas it's conventionally used to define that vector which also points in the same direction as the phase velocity.