Planck constant is Lorentz invariant?

[PeterDonis]

Thanks. I know Lorentz transformation is more than the constancy of light speed: (1) isotropy of time and space, (2) homogeneous, (3) constancy of light speed, (4) physical laws are covariant.

That's all true, but it's still not what I was saying about null cones. Do you understand my description of what a null cone is? It is *not* just a description of a Lorentz transformation or the hypothesis of constant light speed.

 

[keji8341]

That's all true, but it's still not what I was saying about null cones. Do you understand my description of what a null cone is? It is *not* just a description of a Lorentz transformation or the hypothesis of constant light speed.

Not really. I just roughly know (1) within the cone (|ct|>|x|) absolute causalty events, or time-like events, with t>0 as future and t<0 as past; (2) outside the cone, space-like events. If the interval is space-like, there is an inertial frame where the two events are simultaneous, but there is no frame in which the two events take place at the same place. If the interval is time-like, there is no frame in which the two evets occurs at the same time but there is an frame where the two events take place at the same place.

Space-like events: delta_t=0 but delta_x not=0;
Time-like evets: delta_t not=0 but delta_x=0.

 

[Dale]

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

 

[PeterDonis]

Not really. I just roughly know (1) within the cone (|ct|>|x|) absolute causalty events, or time-like events, with t>0 as future and t<0 as past; (2) outside the cone, space-like events. If the interval is space-like, there is an inertial frame where the two events are simultaneous, but there is no frame in which the two events take place at the same place. If the interval is time-like, there is no frame in which the two evets occurs at the same time but there is an frame where the two events take place at the same place.

Space-like events: delta_t=0 but delta_x not=0;
Time-like evets: delta_t not=0 but delta_x=0.

All this is fine. But what about *on* the cone itself? There you have |ct| = |x|. Do you see that a Lorentz transformation must map the null cone into itself? I.e., if we have |ct| = |x| in the unprimed frame, we must have |ct'| = |x'| in the primed frame?

Do you also see that, given any event, we can set up coordinates (t, x, y, z) such that that event is at the origin (0, 0, 0, 0)? And that, if we do this, the null cone |ct| = |x| is simply the set of all light rays that pass through our chosen event? Which means that the *future* null cone (where t > 0) is the set of all light rays *emitted* from that event? So if our chosen event is on the worldline of our point light source, the future null cone of that event is just the "world-sheet" of the spherical wavefront emitted from that event by the source?

And given all the above, do you see how, for each event on the point source's worldline, the wavefront emitted from that event is Lorentz covariant?

(Btw, this is all consistent with what DaleSpam posted as well. I'm just taking a different route to seeing how, when you represent the light emitted by a point source properly, you find that it *is* Lorentz covariant.)

 

[keji8341]

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

Thanks a lot. Let me iterate your idea to see if I understand what you said. (Sorry, I am not able to use LaTex words.)

1. In the source-rest (X'Y'Z') frame, the wave number vector is given by k'=(w'/c)(x'/|x'|), and the corresponding 4-vector is (k',w'/c), so that (k',w'/c).(k',w'/c)=0.

2. 1. In the XYZ frame, the wave number vector is given by k=(w/c)(x/|x|), and the corresponding 4-vector is (k,w/c), so that (k,w/c).(k,w/c)=0.

3. Then let (k',w'/c) and (k,w/c) follow the Lorentz transformation.

At first thought, it's ok. On second thoughts, problems come.
When we set (k',w'/c) and (k,w/c) to follow Lorentz transformations, we can obtain another x'-->x transformation, which is not compatible with the original x'-->x Lorentz transformation. That is why I said that there is a strong constraint between x and k, which destroyed the covariance.

 

[Dale]

2 is incorrect, see above. Your constraint is false.

 

[keji8341]

2 is incorrect, see above. Your constraint is false.

Sorry. Let me reiterate it to see if I got your idea.

1. In the source-rest (X'Y'Z') frame, the wave number vector is given by k'=(w'/c)(x'/|x'|), with k'//x', and the corresponding 4-vector is (k',w'/c), so that (k',w'/c).(k',w'/c)=0. phi'=w't'-|k'||x'|, phi'=constant is a spherical surface.

2. From Lorentz transformation, get (k,w/c), with |k|=w/c. phi=wt-k.x, but with k not //x.

3. The Doppler formula is w'=w*gamma*(1-beta.x/|x|).

No math problems, but two physical problems come:
(1) The frequency is indeterminate at overlap-point, which is the same as Einstein's formula.
(2) Because k is not //x, phi=wt-k.x=constat is not a spherical surface, which is not physical (physical law not covariant).

That is why I said that there is a strong constraint between x' and k', which destroyed the covariance.

 

[keji8341]

In the frame where the source is moving the wavefront fired at t=t'=0 and x=x'=0 w varies over the wavefront. I will work out the problem in detail for you later, probably tomorrow.

DaleSpam, even for a plane wave in free space, you cannot directly show the Planck constant hbar is Lorentz invariant. Let me show you why.

For a plane wave in free space, from the invariance of phase, we obtain (k,w/c) is Lorentz covariant. Under the condition of the covariance of (k,w/c), the invariance of hbar and the covariance of hbar*(k,w/c) are equivalent. Therefore, to get the hbar-invariance, we have to assume the covariance of hbar*(k,w/c), namely "common sense": momentum and energy constitute a 4-vector; the "common sense" is just an particle-nature assumption.

 

[Dale]

1. In the source-rest (X'Y'Z') frame, the wave number vector is given by k'=(w'/c)(x'/|x'|), with k'//x', and the corresponding 4-vector is (k',w'/c), so that (k',w'/c).(k',w'/c)=0. phi'=w't'-|k'||x'|, phi'=constant is a spherical surface.

Yes.

2. From Lorentz transformation, get (k,w/c), with |k|=w/c. phi=wt-k.x, but with k not //x.

Correct.

3. The Doppler formula is w'=w*gamma*(1-beta.x/|x|).

No, see above. It is more complicated than that because the position of the source is changing wrt time. It reduces to the form that you are familiar with in the appropriate limit, but in general it is not given by that.

No math problems, but two physical problems come:
(1) The frequency is indeterminate at overlap-point, which is the same as Einstein's formula.

Again, so what? The list of formulas that break down at the overlap-point of a point particle is extensive: Coulomb's law, Newton's law of gravity, Schwarzschild metric, EM energy, Lienard Wiechert potential, gravitational potential, electrical potential, basically any formula with an r in the denominator. Are you going to throw out all of those as well? If not, then what justification do you have for rejecting the Doppler formula and not the rest?

IMO, this demonstrates not that any of these equations are wrong, but simply that the very idea of an classical point particle is incorrect. That is clearly borne out by QM.

(2) Because k is not //x, phi=wt-k.x=constat is not a spherical surface, which is not physical (physical law not covariant).

This is incorrect. See above, last paragraph before the PS.

 

[keji8341]

Yes.

Correct.

No, see above. It is more complicated than that because the position of the source is changing wrt time. It reduces to the form that you are familiar with in the appropriate limit, but in general it is not given by that.

It is correct. Here x is the radial vector from the point source to the observation point. Since the point source is moving, x changes with t. But the formula itself is not physical.

Please note: Not all equations or expressions which follow Lorentz transformations give correct physics. For example, (k,w/c) in a uniform dielectric is Lorentz covariant but the phase velocity, parallel to k, cannot not follow Lorentz transformation like (k,w/c).

 

[Dale]

It is correct. Here x is the radial vector from the point source to the observation point. Since the point source is moving, x changes with t.

OK, our notation is different. I am not going to check your notation at this point, I know that mine is correct.

But the formula itself is not physical.

The math checks out, it is consistent with modern physics theory, there is no experimental evidence against it, and there is experimental evidence for it. A formula cannot be more physical than that.

 

[keji8341]

Again, so what? The list of formulas that break down at the overlap-point of a point particle is extensive: Coulomb's law, Newton's law of gravity, Schwarzschild metric, EM energy, Lienard Wiechert potential, gravitational potential, electrical potential, basically any formula with an r in the denominator. Are you going to throw out all of those as well? If not, then what justification do you have for rejecting the Doppler formula and not the rest?

IMO, this demonstrates not that any of these equations are wrong, but simply that the very idea of an classical point particle is incorrect. That is clearly borne out by QM.

This is incorrect. See above, last paragraph before the PS.

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 period should be a measurable physical quantity. The lab observer cannot know the period before he receives the second δ-light signal. Based on this physics, no sigularity should be exist for the Doppler effect of moving point light source.

Coulomb's law is correct as r-->0, no matter whether from math or Maxwell equations. For example, Div (x/|x|**3) = 4*pi*delta(x), we often use such math.

 

[keji8341]

The math checks out, it is consistent with modern physics theory, there is no experimental evidence against it, and there is experimental evidence for it. A formula cannot be more physical than that.

The wavefront or equi-phase surface, observed in the lab frame at the same time, should be always spherical. But wt-k.x=constant is not spherical, unless k.x=|k||x|; however k not //x.

 

[Dale]

Coulomb's law is correct as r-->0, no matter whether from math or Maxwell equations. For example, Div (x/|x|**3) = 4*pi*delta(x), we often use such math.

Exactly. Same here. It is valid as r->0 and only has problems at r=0. If you disagree with this then please use the above formula to derive exactly what r>0 causes problems.

The wavefront or equi-phase surface, observed in the lab frame at the same time, should be always spherical. But wt-k.x=constant is not spherical, unless k.x=|k||x|; however k not //x.

This is not correct. See above. The math directly contradicts your assertion. As I said in the paragraph before the PS, $\phi$ behaves exactly as you would expect, emitting spherical wavefronts of constant phase centered on the position of the emitter at the time of emission.

You need to spend a little less time making unfounded assertions and a little more time studying. The math is all laid out there for you.

 

[keji8341]

This is not correct. See above. The math directly contradicts your assertion. As I said in the paragraph before the PS, $\phi$ behaves exactly as you would expect, emitting spherical wavefronts of constant phase centered on the position of the emitter at the time of emission.

You need to spend a little less time making unfounded assertions and a little more time studying. The math is all laid out there for you.

I did not check your math in details, just the idea, which is correct mathematically, but the result is not physical. Actually no one gives the EM field solution of a moving point light source. There is no difficult in getting the field-amlitude transformation but no one have given the frequency-shift effect. Almost all people think that Einstein's formula is applicable, but I don't think so. Plane wave and spherical wave from moving point source must be different. Experimentally, it is well known that Gaussian light beam has curve effect.

Do you remember, Einstein used spherical waves to derive Lorentz transformation? The spherical wavefront fired at t=t'=0 and x=x'=0 is always spherical observed in both frames at any of the same times. But the math result in your phi=wt-k.x does not reflect this, and so I think it is not physical.

 

[Dale]

I did not check your math in details, just the idea, which is correct mathematically, but the result is not physical.

You keep saying that, but with no justification. Again, the math is correct, it follows from physical theory, there is experimental evidence for it, and there is no experimental evidence against it. It is not possible to have a more physical result.

Almost all people think that Einstein's formula is applicable, but I don't think so.

First, you have provided no sound justification for this. Second, this site is not for promoting personal theories, it is for learning mainstream physics and Einstein's formula is mainstream physics since it has been experimentally confirmed.

The spherical wavefront fired at t=t'=0 and x=x'=0 is always spherical observed in both frames at any of the same times. But the math result in your phi=wt-k.x does not reflect this, and so I think it is not physical.

Prove it.

 

[keji8341]

You keep saying that, but with no justification. Again, the math is correct, it follows from physical theory, there is experimental evidence for it, and there is no experimental evidence against it. It is not possible to have a more physical result.

1. That math calculation is correct does not necessarily mean that the idea is correct. For example, you assume ((w'/c)x'/|x'|, w'/c) follows Lorentz transformation, but I am questioning such a 4-vector. Actually such a construction cannot be found in any textbooks, and I have never heard.

2. Your Doppler formula is different from Einstein's formula, What does it mean for your words "there is experimental evidence for it, and there is no experimental evidence against it"?

 

[keji8341]

...confirmed.

Prove it.

From analytical geometry, |x|=constant is a spherical surface, and |x|=ct+constant is a spherical surface with the radius increasing with t.

wt-k.x=const is not a spherical surface unless k.x=|k||x|.
For example, for t=0 and const=0, wt-k.x=const ===> k.x=0 is a plane including the x=0-point.

 

[Dale]

From analytical geometry, |x|=constant is a spherical surface, and |x|=ct+constant is a spherical surface with the radius increasing with t.

True.

wt-k.x=const is not a spherical surface unless k.x=|k||x|.

This does not follow. |x|=r is indeed a spherical surface with radius r centered about the origin, but it is not the only spherical surface. For example, |x-x0|=r is also a spherical surface with radius r centered about x0. Remember, the point is moving, so not all of the spheres of constant phase will be centered on the origin. Only the sphere of 0 phase will be centered on the origin.

Here is a contour plot of the lines of constant phase for t=1, z=0, and v=-.6. Note that the contours of constant phase are circles, note that only the line of phase = 0 is centered at the origin, and that it has a radius of 1 as you would expect for a wave front moving at c.

 

[keji8341]

This does not follow. |x|=r is indeed a spherical surface with radius r centered about the origin, but it is not the only spherical surface. For example, |x-x0|=r is also a spherical surface with radius r centered about x0. Remember, the point is moving, so not all of the spheres of constant phase will be centered on the origin. Only the sphere of 0 phase will be centered on the origin.

I realize that the spherical center is moving. but k.x=0 is indeed a plane.

 

[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.

$$\left( \begin{array}{cccc} \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} & -\frac{v}{c \sqrt{1-\frac{v^2}{c^2}}} & 0 & 0 \\ -\frac{v}{c \sqrt{1-\frac{v^2}{c^2}}} & \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

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.

$$\left( \begin{array}{cccc} \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} & -\frac{v}{c \sqrt{1-\frac{v^2}{c^2}}} & 0 & 0 \\ -\frac{v}{c \sqrt{1-\frac{v^2}{c^2}}} & \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

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.