Planck constant is Lorentz invariant?

[Dale]

See the contour plot I posted above, sorry about editing at the same time you were editing.

 

[keji8341]

See the contour plot I posted above, sorry about editing at the same time you were editing.

Yes, it is very familar. But I don't think your imposed Lorentz transformation creates such a picture. k.x=0 is a plane.

Your Lorentz transformation is not a standard Lorentz transformation. In standard Lorentz transformations, (k,w/c) and (x,ct) are completely independent.

 

[keji8341]

First, you have provided no sound justification for this.

Let me repeat: Einstein’s Doppler formula is not applicable when a moving point light source is close enough to the observer; for example, it may break down or cannot specify a determinate value when the point source and the observer overlap.

The Doppler effect of wave period actually describes the relation between the time interval in which one moving observer emits two δ-light signals and the time interval in which the lab observer receives the two δ-signals at the same place. The lab observer cannot know the period before he receives the second δ-light signal. Based on this physics, no sigularity should exist in the moving point-source Doppler effect.

 

[Dale]

But I don't think your imposed Lorentz transformation creates such a picture.

It does. I simply plotted the formula I posted above. I can post my code.

Your Lorentz transformation is not a standard Lorentz transformation.

It is. I can post the details later today when I return.

 

[keji8341]

It does. I simply plotted the formula I posted above. I can post my code.

I guess you plotted the contours under the condition w't'-|k'||x'|=0 or ct'=|x'|. If so, that's the problem.

 

[keji8341]

It is. I can post the details later today when I return.

I have all the derivations. Just I am unable to write LaTex words, and so I cannot post it to show you.

 

[Dale]

Attached is my code, completely open for inspection.

I guess you plotted the contours under the condition w't'-|k'||x'|=0 or ct'=|x'|. If so, that's the problem.

I did not do that. I told you exactly the conditions I used:
"Here is a contour plot of the lines of constant phase for t=1, z=0, and v=-.6."

There is no problem. The plot is correct and accurately reflects the familiar behavior of Doppler-shifted spherical wavefronts. This familiar behavior emerges naturally from the formalism of four-vectors and how they transform.

But I don't think your imposed Lorentz transformation creates such a picture.

You are calling me a liar? Here is my code, you can check for yourself that it is as I say.

Your Lorentz transformation is not a standard Lorentz transformation.

Yes it is.

\left(<br /> \begin{array}{cccc}<br /> \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} &amp; -\frac{v}{c \sqrt{1-\frac{v^2}{c^2}}} &amp; 0 &amp; 0 \\<br /> -\frac{v}{c \sqrt{1-\frac{v^2}{c^2}}} &amp; \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right)

Compare to http://en.wikipedia.org/wiki/Lorentz_transformation#Matrix_form

It is completely standard.

In standard Lorentz transformations, (k,w/c) and (x,ct) are completely independent.

I don't know what would lead you to believe this. The Lorentz transformations will not decouple two dependent quantities. In the frame where the point source is at rest k' depends on r', so I don't know why you would think that the Lorentz transform would decopule them in the moving frame so that k would be independent of r. Your understanding of the Lorentz transform seems to be incorrect.

 

[Dale]

keji8341, your understanding of the wave four-vector, the Lorentz transform, and relativistic Doppler shift seems to be fundamentally flawed. Unfortunately, I don't know where your basic misunderstanding lies, so I am sorry that I cannot be more helpful.

The derivation that I posted shows how the wave four-vector formalism is applicable to spherical waves, how the wave vector transforms correctly, how the resulting frequency reduces to the standard relativistic Doppler shift formula, and how the resulting phase shows the familiar pattern of non-concentric spheres propagating outward at c. Furthermore, there is expermiental validation of this behavior, in particular the seminal experiment by Stilwell and Ives and subsequent similar experiments.

The wave four-vector is a legitimate four-vector, is therefore mathematically self-consistent and compatible with current physical theory, as well as being experimentally validated. You have no basis to object to its use in justifying the invariance of Planck's constant.

 

[keji8341]

keji8341, your understanding of the wave four-vector, the Lorentz transform, and relativistic Doppler shift seems to be fundamentally flawed. Unfortunately, I don't know where your basic misunderstanding lies, so I cannot be more helpful.

The derivation that I posted shows how the wave four-vector formalism is applicable to spherical waves, how the wave vector transforms correctly, how the resulting frequency reduces to the standard relativistic Doppler shift formula, and how the resulting phase shows the familiar pattern of non-concentric sphers propagating outward at c. Furthermore, there is expermiental validation of this behavior, in particular the seminal experiment by Stilwell and Ives and subsequent similar experiments.

The wave four-vector is a legitimate four-vector, is therefore mathematically self-consistent and compatible with current physical theory, as well as being experimentally validated. You have no basis to object to its use in justifying the invariance of Planck's constant.

DaleSpam, Thank you very much for all your calculations.

1. My main question is: your Doppler formula has a sigularity at the overlap-point (you realize that), which is not physical at all.

2. Your Doppler formula is different from Einstein's formula.

3. From your Doppler formula, the observed frequency in the lab frame changes with locations, as you indicated in your post #73:

"Off of the x-axis the Doppler shift depends on both position and time..."

That means a photon's frequency changes during propagation, clearly challenging the energy conservation of Einstein's light-quantum hypothesis if the Planck constant is a "universal constant".

Therefore, actually it is you yourself who is chanlleging the invariance of Planck constant.

You go further than I do. Am I right?

 

[keji8341]

It does. I simply plotted the formula I posted above. I can post my code.

It is. I can post the details later today when I return.

Of course, I believe your calculations, but I realized that your Lorentz transformation is not a "standard" Lorentz transformation. Let me explain.

The wave 4-vector (k',w'/c) in the source-rest frame and (x',ct') are completely independent originally, belonging to two different spaces. You first redefine the wave 4-vector by setting k'=|k'|(x'/|x'|), this is a kind of "transfromation". Then you set (|k'|(x'/|x'|),w'/c) to follow Lorentz transformation. So from the original (k',w'/c) to (k,w/c), something more than Lorentz transformation is imposed.

Nothing wrong with itself; just the final result for Doppler formula has a sigularity, which is not acceptable physically.

 

[keji8341]

Attached is my code, completely open for inspection.I did not do that. I told you exactly the conditions I used:
"Here is a contour plot of the lines of constant phase for t=1, z=0, and v=-.6."

There is no problem. The plot is correct and accurately reflects the familiar behavior of Doppler-shifted spherical wavefronts. This familiar behavior emerges naturally from the formalism of four-vectors and how they transform.

You are calling me a liar? Here is my code, you can check for yourself that it is as I say.

Yes it is.

\left(<br /> \begin{array}{cccc}<br /> \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} &amp; -\frac{v}{c \sqrt{1-\frac{v^2}{c^2}}} &amp; 0 &amp; 0 \\<br /> -\frac{v}{c \sqrt{1-\frac{v^2}{c^2}}} &amp; \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right)

Compare to http://en.wikipedia.org/wiki/Lorentz_transformation#Matrix_form

It is completely standard.

I don't know what would lead you to believe this. The Lorentz transformations will not decouple two dependent quantities. In the frame where the point source is at rest k' depends on r', so I don't know why you would think that the Lorentz transform would decopule them in the moving frame so that k would be independent of r. Your understanding of the Lorentz transform seems to be incorrect.

I believe your calculations, but I realized that your Lorentz transformation is not a "standard" Lorentz transformation. Let me explain.

The wave 4-vector (k',w'/c) in the source-rest frame and (x',ct') are completely independent originally, belonging to two different spaces. You first redefine the wave 4-vector by setting k'=|k'|(x'/|x'|), this is a kind of "transformation". Then you set (|k'|(x'/|x'|),w'/c) to follow Lorentz transformation. So from the original (k',w'/c) to (k,w/c), something more than Lorentz transformation is imposed.

What difference between "standard transformation" and "not standard transformation"?

1. Standard: (k,w/c) and (x,ct) are completely independent. Since (x,ct) must follows Lorentz transformation, the invariance of phase phi=wt-k.x and the covariance of (k,w/c) are equivalent; that is, a sufficient and necessary condition for the invariance of phase is the covariance of (k,w/c).

2. Not standard: If (k',w'/c) and (x',ct') are not independent, like (|k'|(x'/|x'|), w'/c) which you used, the invariance of phase and the covariance of (k',w'/c) are not equivalent; in other words, the covariance of (k',w'/c) is a sufficient condition for the invariance of phase, but not a necessary one!

 

[Dale]

1. My main question is: your Doppler formula has a sigularity at the overlap-point (you realize that), which is not physical at all.

Classical point particles themselves are also not physical at all, and many other valid formulas suffer the same problem. So I see no issue here.

2. Your Doppler formula is different from Einstein's formula.

It reduces to his in the appropriate situation.

3. From your Doppler formula, the observed frequency in the lab frame changes with locations, as you indicated in your post #73:

"Off of the x-axis the Doppler shift depends on both position and time..."

That means a photon's frequency changes during propagation

No, it doesn't. Different photons go to different locations. No photon changes frequency.

You are really good at making incorrect conclusions from correct premises. Here is another example:

The wave 4-vector (k',w'/c) in the source-rest frame and (x',ct') are completely independent originally, belonging to two different spaces.

Many physics equations express a dependence between vectors in two different spaces. The fact that they are in different spaces has nothing to do with whether or not two vectors are independent. In this case the dependency between x and k is given by the equation \phi=g_{\mu\nu}x^{\mu}k^{\nu}

 

[Dale]

I realized that your Lorentz transformation is not a "standard" Lorentz transformation. Let me explain.

Then kindly point out in my code exactly where my code deviates from a standard Lorentz transform.

I assert that it is standard and as evidence I have posted the formula I used, a link to the standard formula for comparison, and my code. You are either saying I am a liar and deliberately misrepresenting my code or you are saying I am stupid and don't realize I am misrepresenting it.

So either point out the exact line in my code where I deviate from the standard Lorentz transform or stop impugning my integrity.

 

[PeterDonis]

No, it doesn't. Different photons go to different locations. No photon changes frequency.

I had been trying to figure out how to describe that particular error keji8341 was making, but this sums it up perfectly.

 

[keji8341]

Then kindly point out in my code exactly where my code deviates from a standard Lorentz transform.

I assert that it is standard and as evidence I have posted the formula I used, a link to the standard formula for comparison, and my code. You are either saying I am a liar and deliberately misrepresenting my code or you are saying I am stupid and don't realize I am misrepresenting it.

So either point out the exact line in my code where I deviate from the standard Lorentz transform or stop impugning my integrity.

Dear DaleSpam, you misunderstood my comments. You are a very kind and serious scientist; I am very happy that I had a chance to discuss the question I raised. I completely believe your calculations based on your own math model. Although we have different academic viewpoints, I find that you have a very deep understanding in relativity, you have quick reflections to new physical problems, and you are a very strong discussion rival who I have never met in this field. All my comments are just pointing to academic issues, nothing else. If any of my words make you unhappy, that is never my original intention, and I apologize to you.

 

[Dale]

I completely believe your calculations based on your own math model. ... If any of my words make you unhappy, that is never my original intention, and I apologize to you.

Again, it is not my model, it is the standard Lorentz transform, but I appreciate and accept the apology, provided you don't return to accusing me of falsifying the Lorentz transform.

 

[Phrak]

The transformation matrix to obtain (k_x',w') from (k_x,w) should directly derive from the Lorentz matrix transforming (x,t) to (x',t'), where k_x = 1/x and w = 1/t. What are the matrix elements?

This matrix seems to be something that appears to be a "reciprical" matrix 1/L, from 1/x' = 1/[L(x)], in very schematic notation.

This problem of showing that h is lorentz invariant, given E=hv, amounts to showing that 1/L is also a lorentz transform.

 

[Dale]

The transformation matrix to obtain (k_x',w') from (k_x,w) should directly derive from the Lorentz matrix transforming (x,t) to (x',t'), where k_x = 1/x and w = 1/t.

No, k is a four-vector, so the transformation matrix is the Lorentz matrix. That is essentially the definition of a vector, that it transforms according to the Lorentz matrix.

 

[Phrak]

No, k is a four-vector, so the transformation matrix is the Lorentz matrix. That is essentially the definition of a vector, that it transforms according to the Lorentz matrix.

"No, k is a four vector", or "yes, k is a four vector?" There's nothing in error in what I said that I can see.

Have you demonstrated that it transforms as a four vector or given a reference? This is a very long thread.

 

[Dale]

"No, k is a four vector", or "yes, k is a four vector?" There's nothing in error in what I said that I can see.

The error was that you needed to derive a different transformation matrix. You simply use the standard Lorentz transformation matrix.

Have you demonstrated that it transforms as a four vector or given a reference? This is a very long thread.

Yes, I demonstrated it for the case of a spherical source.

 

[Phrak]

The error was that you needed to derive a different transformation matrix. You simply use the standard Lorentz transformation matrix.

Yes, I demonstrated it for the case of a spherical source.

I'll look into these two points. In which post did you demonstrate a spherical source?

This is interesting in its own right, that 1/x^mu should transform as x^mu to necessitate h as a scalar.

 

[Dale]

Post 74, with the code posted in 97. If you have Mathematica then I would start with 97 since it is more complete.

 

[Phrak]

Post 74, with the code posted in 97. If you have Mathematica then I would start with 97 since it is more complete.

I no longer have mathematica. It would help if variables were identified.

 

[Phrak]

by the way, in your definition of k'

k&#039;=\left(1,\frac{x&#039;}{\sqrt{x&#039;^2+y&#039;^2+z&#039;^2}},\frac{y&#039;}{\sqrt{x&#039;^2+y&#039;^2+z&#039;^2}},\frac{z&#039;}{\sqrt{x&#039;^2+y&#039;^2+z&#039;^2}}\right)

you will want to remove the square root operators. Instead,

k_x = x/r²

for spatial dimensions.

 

[Dale]

No, that is not correct. That would mean that the wavelength increases as the wave gets further from the source. Why would you think that?

 

[PeterDonis]

Phrak, I think you may be assuming a different convention for the units of k. In the convention DaleSpam is using, the phase \phi has the same units as the "position vector" r; that means k is dimensionless (see the formulas in post #73). You may be thinking of a different convention where k is a "wavenumber vector" and has dimensions of inverse length, so that the phase is dimensionless. The latter is the convention I learned when I took wave mechanics in school; the reason for adopting it was that if you describe a wave as a complex exponential (or sines and cosines), the argument of the exponential (or the sines and cosines), which is the dot product k . r, has to be dimensionless.

Even in that convention, though, I don't think k = 1/r would be correct, since, as DaleSpam points out, that would mean the wavelength increases with distance from the source. I'm not sure how you would modify the formulas in post #73 for the unit convention where k has dimensions of inverse length.

 

[Phrak]

No, that is not correct. That would mean that the wavelength increases as the wave gets further from the source. Why would you think that?

A wave front at r, centered at the origin, has wave numbers k_xⁱ = xⁱ/r.

 

[DrGreg]

Please bear in mind that DaleSpam stated in post #73 that

...I will use units of time such that in the primed frame w=1 and units of distance such that c=1.

This means you can't use dimensional analysis on any equations derived from this assumption. To avoid confusion you would have to reinsert symbols for \omega and c into the formulae.

 

[Dale]

A wave front at r, centered at the origin, has wave numbers k_xⁱ = xⁱ/r.

Yes, as you suggest here x/r is correct, not x/r² as you suggested above. My formula for k is correct.

 

[Dale]

Phrak, I think you may be assuming a different convention for the units of k. In the convention DaleSpam is using, the phase \phi has the same units as the "position vector" r; that means k is dimensionless (see the formulas in post #73). You may be thinking of a different convention where k is a "wavenumber vector" and has dimensions of inverse length, so that the phase is dimensionless. The latter is the convention I learned when I took wave mechanics in school; the reason for adopting it was that if you describe a wave as a complex exponential (or sines and cosines), the argument of the exponential (or the sines and cosines), which is the dot product k . r, has to be dimensionless.

I am using the standard convention. Phase is dimensionless, x has units of distance, and k therefore has units of inverse distance. As DrGreg points out I am using units of time such that w=1 and units of distance such that c=1. So you need to plug the appropriate factors back in wherever the units don't make sense. I apologize for the confusion that has caused. It is a pretty common thing to do, but it is somewhat sloppy and definitely confusing if you aren't looking out for it.