Planck constant is Lorentz invariant?

[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'

$$k'=\left(1,\frac{x'}{\sqrt{x'^2+y'^2+z'^2}},\frac{y'}{\sqrt{x'^2+y'^2+z'^2}},\frac{z'}{\sqrt{x'^2+y'^2+z'^2}}\right)$$

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.

 

[PeterDonis]

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.

Ah, forgot about that! Thanks for the clarification.

 

[PeterDonis]

Just to expand a bit and make sure I've got this right, if we rewrite DaleSpam's formulas from post #73 in conventional units (i.e., putting back in the factors of $\omega$ and c so that everything is in conventional units of length), we get (in the primed frame):

$$r' = (ct', x', y', z')$$

$$k' = \left( \frac{\omega'}{c}, \frac{\omega' x'}{c \sqrt{x'^{2} + y'^{2} + z'^{2}}}, \frac{\omega' y'}{c \sqrt{x'^{2} + y'^{2} + z'^{2}}}, \frac{\omega' z'}{c \sqrt{x'^{2} + y'^{2} + z'^{2}}} \right)$$

$$\phi = \eta_{\mu \nu} k'^{\mu} r'^{\nu} = \omega' t' - \frac{\omega'}{c} \sqrt{x'^{2} + y'^{2} + z'^{2}}$$

 

[rickrherbert]

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[sciencewatch]

OK keji8341, here we go. Without loss of generality I will use two reference frames in the standard configuration with the point source at rest at the origin in the primed frame, and I will use units of time such that in the primed frame w=1 and units of distance such that c=1. Then, in the primed frame we have:
$$r'=\left( t',x',y',z' \right)$$
$$k'=\left(1,\frac{x'}{\sqrt{x'^2+y'^2+z'^2}},\frac{y'}{\sqrt{x'^2+y'^2+z'^2}},\frac{z'}{\sqrt{x'^2+ y'^2+z'^2}}\right)$$
$$\eta_{\mu\nu}k'^{\mu}k'^{\nu}=0$$
$$\phi=\eta_{\mu\nu}k'^{\mu}r'^{\nu}=t'-\sqrt{x'^2+y'^2+z'^2}$$

Boosting to the unprimed frame we get
$$r^{\mu}=\Lambda^{\mu}_{\nu'}r'^{\nu'}=\left(t,x,y,z\right)$$
$$k^{\mu}=\Lambda^{\mu}_{\nu'}k'^{\nu'}=\left( \begin{array}{c} \frac{v (t v+x)}{\left(v^2-1\right) \sqrt{-\frac{(t v+x)^2}{v^2-1}+y^2+z^2}}+\frac{1}{\sqrt{1-v^2}} \\ \frac{t v+x}{\left(1-v^2\right) \sqrt{-\frac{(t v+x)^2}{v^2-1}+y^2+z^2}}-\frac{v}{\sqrt{1-v^2}} \\ \frac{y}{\sqrt{-\frac{(t v+x)^2}{v^2-1}+y^2+z^2}} \\ \frac{z}{\sqrt{-\frac{(t v+x)^2}{v^2-1}+y^2+z^2}} \end{array} \right)$$
$$\eta_{\mu\nu}k^{\mu}k^{\nu}=0$$
$$\phi=\eta_{\mu\nu}k^{\mu}r^{\nu}=\frac{-\sqrt{1-v^2} \sqrt{-\frac{(t v+x)^2}{v^2-1}+y^2+z^2}+t+v x}{\sqrt{1-v^2}}$$

k behaves as you would expect for the wave four-vector. E.g. for y=0 and z=0 we get
$$k^t=\frac{1-v \; \text{sgn}(t v+x)}{\sqrt{1-v^2}}$$
which is the standard expression for the relativistic Doppler effect including the sign change as the point source passes a given location on the x axis. Off of the x-axis the Doppler shift depends on both position and time, as you would expect from everyday experience. Also, as I stated above, the spacelike part of k is not generally parallel to the spacelike part of r.

phi also behaves as you would expect for the phase of a moving point source. Surfaces of constant phase form light cones centered on the location of the point source at $t=\phi$. The formula is valid as close to the point source as you like.

PS k is written as a colum vector just so that it would fit on the screen width easily

I found a paper, saying the wave vector and frequency for a moving point light source CANNOT form a Lorentz covariant 4-vector. I am not sure if it is correct.

 

[Dale]

Hi sciencewatch, welcome to PF!

Does the paper come from a mainstream scientific source? If so, can you link to it?

 

[sciencewatch]

Hi sciencewatch, welcome to PF!

Does the paper come from a mainstream scientific source? If so, can you link to it?

seems in arxiv.org. I will check later.

 

[sciencewatch]

Hi sciencewatch, welcome to PF!

Does the paper come from a mainstream scientific source? If so, can you link to it?

http://arxiv.org/ftp/arxiv/papers/1007/1007.0980.pdf

seems not published; everybody can post their own stuffs there.

 

[Dale]

Hi sciencewatch, I looked at the reference. You are correct, it is not published in a mainstream science journal. The arguments presented in the paper are exactly the same as those debunked here in this thread. In fact, the arguments and language are so similar that this paper must be the source of keji8341's misunderstanding.

 

[Phrak]

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

Sorry, that was a typo. I meant to write x/r². But without your defintion of r, it's difficult to know what your equations mean. But that's ok.

 

[Phrak]

Is Plank’s constant Lorentz invariant?

To solve, I use planar waves.

$$k \cdot \lambda = 1$$
First, for k to be a vector, it must be invaiant over the group of continuous rotational transformations SO[3].
A plane including the origin has phase $\phi$. A second plane at distance $r_0$ has phase $\phi + 2\pi$.
The equation of a plan at a normal distance $\lambda = r_0$ is given by
$$\hat{n} \cdot (r-r_0)=0$$
Also
$$r_0 \cdot (r-r_0)=0$$
$r_0$ is the distance to the plane from the origin and r is the vector (x,y,z).
(The r term could be upgraded to its 2-form directed area to produce a well behaved tensor equation good in any unaccelerated coordinate system, i.e.: $*(*r (r-r_0))=0$.)
$$(x_{0}^2+y_{0}^2+z_{0}^2) = x x_0+ y y_0+z z_0$$
where $r_{0}^2 = x_{0}^2 + y_{0}^2 + z_{0}^2$.

The plane intersects the x, y and z axis at:
$$x = \lambda^2 / x_0$$
$$y = \lambda^2 / y_0$$
$$z = \lambda^2 / z_0$$
$(x,y,z) = (\lambda_x, \lambda_y, \lambda_z)$ are the directed wavelengths, and together, do *not* transform as a vector. $(\lambda_x, \lambda_y, \lambda_z)$ is not a vector.

The normal wave number k, however, is proportional to the reciprical of the normal wavelength.
$x^i$ is a vector, $\lambda$ is a constant; $$k^i = x_0^{i} / \lambda^2$$
k is a vector.

The same strategy can be followed for a Lorentz boost in the x direction.
$$r_0 \cdot (r-r_0)=0$$
$$r_0 = (ct_0, x_0)$$
$$r = (ct, x)$$
This obtains
$$k^t = ct_0 / \lambda^2$$
$$k^x = x_0 / \lambda^2$$
$k^t$ is in units of inverse length.
$$\nu /c = ct_0 / \lambda^2$$
$(\nu /c, k^x, 0, 0)$is a 4-vector.
One boost and SO(3) uniquely define the Lorentz group; k is a Lorentz invariant vector. If (E,c/p) is a Lorentz invariant vector then Plank’s constant, h is a scalar constant under the Lorentz group.

The full Poincare group would be a different matter.

 

[sciencewatch]

Is Plank’s constant Lorentz invariant?

...$(\nu /c, k^x, 0, 0)$is a 4-vector.
... k is a Lorentz invariant vector. If (E,c/p) is a Lorentz invariant vector then Plank’s constant, h is a scalar constant under the Lorentz group.

The full Poincare group would be a different matter.

Do you mean:

$(\nu /c, k^x, 0, 0)$ is a 4-vector + h$(\nu /c, k^x, 0, 0)$ is a 4-vector ---->h =a scalar constant under the Lorentz group ?

 

[keji8341]

...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.

Actually the point-source effect only takes place at the microscale. The breaking of the invariance of Planck constant in the microscale might be used for explaining why the Planck's blackbody radiation law breaks down in a nanoscale (http://web.mit.edu/newsoffice/2009/heat-0729.html ).

 

[Dale]

The breaking of the invariance of Planck constant in the microscale might be used for explaining why the Planck's blackbody radiation law breaks down in a nanoscale (http://web.mit.edu/newsoffice/2009/heat-0729.html ).

That is a speculation piled on top of a speculation. Do you have any mainstream scientific references to support the suggestions that:
1) the Planck constant is not frame invariant at small scales
2) that has anything to do with the increased thermal transfer at small distances

If not, then your unsupported speculations have no place on this forum and are a violation of the rules you agreed to when you signed up.

 

[Phrak]

Do you mean:

$(\nu /c, k^x, 0, 0)$ is a 4-vector + h$(\nu /c, k^x, 0, 0)$ is a 4-vector ---->h =a scalar constant under the Lorentz group ?

No. From the original post, it is implicitely assume (E/c, p) is a 4-vector. It is also assumed, by implication, that

$E = h \nu$
$p_x = h k_x$
$p_y = h k_y$
$p_z = h k_z$

in some inertial frame of reference.

I think I have demonstrated, roughly, withing the confines of special relativity, if (E/c,p) is a 4-vector of some wave phenomena, there is an associated 4-vector wave number in invariant proportions in its elements. It's a rough demonstration because it's all rubbish, anyway. It just happens to work because it assumes coordinate axes are orthonormal where it is meaningful to use the dot product between vectors. I'm fairly confident it still works in curvilinear coordinates in Minkowski spacetime, but it would be a lot more work to show it, and no one would understand the archane notation anyway.

 

[keji8341]

...k behaves as you would expect for the wave four-vector. E.g. for y=0 and z=0 we get
$$k^t=\frac{1-v \; \text{sgn}(t v+x)}{\sqrt{1-v^2}}$$
which is the standard expression for the relativistic Doppler effect including the sign change as the point source passes a given location on the x axis. Off of the x-axis the Doppler shift depends on both position and time, as you would expect from everyday experience. ...

ZT: "Most scientists insist that Einstein’s plane wave Doppler formula be applicable to any cases, no matter whether the observer is close to a moving source or not. A strong argument is that the spherical wave produced by a moving point source can be decomposed into plane waves. "

“The plane wave decomposition is mathematically universal.”

Do you think so?

 

[Dale]

No, I have never seen any survey of scientists which would indicate that most insist on that point. I personally don't hold an opinion on what most scientists insist. Do you have a reference supporting that claim?

 

[keji8341]

No, I have never seen any survey of scientists which would indicate that most insist on that point. I personally don't hold an opinion on what most scientists insist. Do you have a reference supporting that claim?

By a lot of private communications. They are surprised when I mentioned a paper taking about moving point-source Doppler effect.

 

[Dale]

Private communications with several scientists does not constitute a mainstream scientific reference concerning the opinion of a majority of scientists. Would you care to rephrase your question without the attempt to give it an artificial level of credibility by pretending that it is a position known to be held by a majority of scientists?

 

[keji8341]

Private communications with several scientists does not constitute a mainstream scientific reference concerning the opinion of a majority of scientists. Would you care to rephrase your question without the attempt to give it an artificial level of credibility by pretending that it is a position known to be held by a majority of scientists?

To tell the truth, you are the first one I met, who realized there is something different.

Some one said to me, “The plane wave decomposition is mathematically universal.” The point-source spherical wave can be decomposed into plane waves. So should be the same.

I don't know much about the math principle: “The plane wave decomposition is mathematically universal.” That is a famous principle?

 

[Dale]

I don't know if it is "universal", but this much is correct.

The point-source spherical wave can be decomposed into plane waves.

A spherically symmetric wave can indeed be expanded as an infinite sum of plane waves. I don't know what the result of an infinite sum of Doppler shifts would be.