Planck constant is Lorentz invariant?

[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]

I found it really fascinating to browse I might love to find out you create far more on this topic.

 

[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&#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+<br /> y&#039;^2+z&#039;^2}}\right)
\eta_{\mu\nu}k&#039;^{\mu}k&#039;^{\nu}=0
\phi=\eta_{\mu\nu}k&#039;^{\mu}r&#039;^{\nu}=t&#039;-\sqrt{x&#039;^2+y&#039;^2+z&#039;^2}

Boosting to the unprimed frame we get
r^{\mu}=\Lambda^{\mu}_{\nu&#039;}r&#039;^{\nu&#039;}=\left(t,x,y,z\right)
k^{\mu}=\Lambda^{\mu}_{\nu&#039;}k&#039;^{\nu&#039;}=\left(<br /> \begin{array}{c}<br /> \frac{v (t v+x)}{\left(v^2-1\right) \sqrt{-\frac{(t<br /> v+x)^2}{v^2-1}+y^2+z^2}}+\frac{1}{\sqrt{1-v^2}} \\<br /> \frac{t v+x}{\left(1-v^2\right) \sqrt{-\frac{(t<br /> v+x)^2}{v^2-1}+y^2+z^2}}-\frac{v}{\sqrt{1-v^2}} \\<br /> \frac{y}{\sqrt{-\frac{(t v+x)^2}{v^2-1}+y^2+z^2}} \\<br /> \frac{z}{\sqrt{-\frac{(t v+x)^2}{v^2-1}+y^2+z^2}}<br /> \end{array}<br /> \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 definition 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.

 

[keji8341]

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

Any references? Please.

 

[Dale]

I would start here, particularly the introductory paragraph.
http://en.wikipedia.org/wiki/Fourier_optics

 

[keji8341]

I would start here, particularly the introductory paragraph.
http://en.wikipedia.org/wiki/Fourier_optics

I did not find the theorem as you mentioned: "A spherically symmetric wave can indeed be expanded as an infinite sum of plane waves. "

Be specific, Please.

 

[Dale]

Did you not even read the first paragraph? "In Fourier optics, by contrast, the wave is regarded as a superposition of plane waves which are not related to any identifiable sources; instead they are the natural modes of the propagation medium itself."

If you want something more in depth you can get a textbook:
http://www.wiley.com/WileyCDA/WileyTitle/productCd-0780334116.html

Or Google "Fourier optics" or "plane wave spectrum".

 

[keji8341]

Did you not even read the first paragraph? "In Fourier optics, by contrast, the wave is regarded as a superposition of plane waves which are not related to any identifiable sources; instead they are the natural modes of the propagation medium itself." ...

Thanks. Yes, I read these words:

“In Fourier optics, by contrast, the wave is regarded as a superposition of plane waves which are not related to any identifiable sources; instead they are the natural modes of the propagation medium itself.”

Did you get your conclusion from the above words? It seems to me that, the above ambiguous words cannot result in your conclusion:

"A spherically symmetric wave can indeed be expanded as an infinite sum of plane waves."

Probably that is your own work, related to your intellectual property. If so, I won’t ask.

I don’t have much math. But to my knowledge, whether a function f(x) can be represented as a Fourier integral depends on the property which the function has. Here is a theorem I got from
http://mathworld.wolfram.com/FourierTransform.html

A function f(x) has a Fourier transform if
1. The integral of absolute f(x) exists.
2. There are a finite number of discontinuities.
3. The function has bounded variation. A sufficient weaker condition is fulfillment of the Lipschitz condition.

I think the field produced by a point source has a singularity. I am not sure if the first and the third math conditions can be satisfied.

 

[Dale]

“In Fourier optics, by contrast, the wave is regarded as a superposition of plane waves which are not related to any identifiable sources; instead they are the natural modes of the propagation medium itself.”

Did you get your conclusion from the above words? It seems to me that, the above ambiguous words cannot result in your conclusion:

"A spherically symmetric wave can indeed be expanded as an infinite sum of plane waves."

Why not? They are saying the same thing.

 

[PhilDSP]

A function f(x) has a Fourier transform if
1. The integral of absolute f(x) exists.
2. There are a finite number of discontinuities.
3. The function has bounded variation. A sufficient weaker condition is fulfillment of the Lipschitz condition.

I think the field produced by a point source has a singularity. I am not sure if the first and the third math conditions can be satisfied.

Probably even more basic considerations are what domain you are interested into evaluate or define the Fourier Transform. For the field at a single point you only need a one dimensional Fourier Transform and at that point the transform (and wave equation) will be the same whether the wave is a plane wave or whether it is spherical.

It's only when you wish to evaluate a spatially extended region that you will see a difference between values for a plane wave or spherical wave originating from a particular point. In that case you'll need to define and evaluate a Fourier Transform in at least 2 dimensions. You will see a phase variation across the extended region which will be visible in the FT.

 

[keji8341]

Probably even more basic considerations are what domain you are interested into evaluate or define the Fourier Transform. For the field at a single point you only need a one dimensional Fourier Transform and at that point the transform (and wave equation) will be the same whether the wave is a plane wave or whether it is spherical.

It's only when you wish to evaluate a spatially extended region that you will see a difference between values for a plane wave or spherical wave originating from a particular point. In that case you'll need to define and evaluate a Fourier Transform in at least 2 dimensions. You will see a phase variation across the extended region which will be visible in the FT.

Thanks. I think you are talking about numerical computations about FT. Actually my question is a pure theoretical problem. The point source has a sigularity at r=0; the function to be transformed is ~expi[wt-(w/c)*r]/r with r=sqrt(x**2+y**2+z**2), the frequency w is given, monochromatic wave. I am not sure if it can be represented by uniform plane waves, although some scientists often firmly claim “The plane wave decomposition is mathematically universal.”

The point-source solution actually is closely related so-called "invariant Green's function" in classical electrodynamic.

 

[Dale]

You may be interested in this page:
http://lmb.informatik.uni-freiburg.de/papers/download/wa_report01_08.pdf

 

[PhilDSP]

Thanks. I think you are talking about numerical computations about FT. Actually my question is a pure theoretical problem. The point source has a sigularity at r=0; the function to be transformed is ~expi[wt-(w/c)*r]/r with r=sqrt(x**2+y**2+z**2), the frequency w is given, monochromatic wave. I am not sure if it can be represented by uniform plane waves, although some scientists often firmly claim “The plane wave decomposition is mathematically universal.”

The point-source solution actually is closely related so-called "invariant Green's function" in classical electrodynamic.

I'm not sure what you really mean by "mathematically universal". Sure, very often EM problems or SR problems are expressed and evaluated in terms of plane waves for simplicity. It seems that a spherical wave decomposition of a plane wave is much more commonly described than the converse. But even that is rather complicated and potentially fraught with technical problems such as in this description:

http://farside.ph.utexas.edu/teaching/jk1/lectures/node102.html

In short, it seems far more complicated to force a spherical wave solution to a problem described in terms of a plane wave than to re-describe the problem in terms of spherical waves.

By the term "evaluate" I mean evaluate analytically (not numerically). Usually to evaluate a FT numerically you need to use a Discrete Fourier Transform which at best only approximates a FT (Continuous Fourier Transform)

If you could rewrite your function so that it doesn't already have terms for frequency then it might be simple to determine the FT analytically. Can you somehow replace the frequency term with a partial derivative (involving dr/dt with a constant c) ?