Planck constant is Lorentz invariant?

[Dale]

k and x must be parallel

This is not true in general. It is only true for a spherical wave centered at the origin. In general k and x can have any arbitrary relationship. Perhaps that is the source of your misunderstanding?

 

[keji8341]

Can you demonstrate that there are errors which become large at smaller distances? I certainly see no indication of that.

The relative error is proportional to (atom's radiation wavelength)/(distance between the source and the observer). Usually, radiation wavelength is < microns, and the distance >cm ----> the error is smaller than 10**(-4), all past experiments cannot identify the error. As you know the experimentally obtained Planck constant also has errors of ~10**(-4).

 

[keji8341]

This is not true in general. It is only true for a spherical wave centered at the origin. In general k and x can have any arbitrary relationship. Perhaps that is the source of your misunderstanding?

As you kow, I am talking about the Doppler effect from a moving point light source. For a moving point light source, the phase factor is exp[i(wt-|k||x|)] required by wave equation, the wave vector k is always parallel to the radial vector (the point source fixed at x'=0; at t=t'=0, x=x'=0).

 

[Dale]

As you kow, I am talking about the Doppler effect from a moving point light source. For a moving point light source, the phase factor is exp[i(wt-|k||x|)] required by wave equation, the wave vector k is always parallel to the radial vector (the point source fixed at x'=0; at t=t'=0, x=x'=0).

A point source cannot be both fixed at the origin and moving. In any frame where the source is moving k is not parallel to x in general.

 

[keji8341]

A point source cannot be both fixed at the origin and moving. In any frame where the source is moving k is not parallel to x in general.

In the source-rest frame X'Y'Z', phi'=w't'-|k'||x'|, where x' denotes the radial vector from the source point to the observation point. Under Lorentz transformation, phi=phi'=wt-|k||x|, because the wavefront fired at t=t'=0 and x=x'=0 is always a spherical surface observed in both frames at any of the same times.

 

[Dale]

In the source-rest frame X'Y'Z', phi'=w't'-|k'||x'|. Under Lorentz transformation, phi=phi'=wt-|k||x|, because the wavefront fired at t=t'=0 and x=x'=0 is always a spherical surface observed in both frames at any of the same times.

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.

 

[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.

I think that, 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. Do you agree? This is easy to judge.

 

[PeterDonis]

Thanks. That is a math description of the hypothesis of constancy of light speed, which is used in the derivation of Lorentz transformation.

You're not reading very carefully. The Lorentz transformation is defined as keeping the interval constant; the interval is defined as the quantity t^{2} - x^{2} - y^{2} - z^{2}. (Strictly speaking, that's the interval of a given point (t, x, y, z) from the origin (0, 0, 0, 0).)

The equation I wrote down says more than that; it says that this quantity, the interval, is not just constant, but *equal to zero*. In other words, it defines a set of points in spacetime that are separated from the origin (0, 0, 0, 0) by a zero interval. This set of points is called a null cone. The "future" portion of the null cone is the subset of these points for which t > 0; in other words, it's the portion of the null cone that lies to the future of the origin (0, 0, 0, 0).

Since a Lorentz transformation keeps the interval constant, it must map the null cone into itself; in other words, it maps null rays into other null rays. The mapping is conformal, so it preserves the inner product; thus, the null cone is Lorentz covariant.

 

[keji8341]

You're not reading very carefully. The Lorentz transformation is defined as keeping the interval constant; the interval is defined as the quantity t^{2} - x^{2} - y^{2} - z^{2}. (Strictly speaking, that's the interval of a given point (t, x, y, z) from the origin (0, 0, 0, 0).)

The equation I wrote down says more than that; it says that this quantity, the interval, is not just constant, but *equal to zero*. In other words, it defines a set of points in spacetime that are separated from the origin (0, 0, 0, 0) by a zero interval. This set of points is called a null cone. The "future" portion of the null cone is the subset of these points for which t > 0; in other words, it's the portion of the null cone that lies to the future of the origin (0, 0, 0, 0).

Since a Lorentz transformation keeps the interval constant, it must map the null cone into itself; in other words, it maps null rays into other null rays. The mapping is conformal, so it preserves the inner product; thus, the null cone is Lorentz covariant.

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.

 

[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.

x and x' should be taken the radial vectors from the source point to the observation point. The point source is fixed in the X'Y'Z' frame, and the source point is constant observed in X'Y'Z' frame, but it is moving observed in the XYZ frame.

 

[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&#039;=\left( t&#039;,x&#039;,y&#039;,z&#039; \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

 

[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&#039;=\left( t&#039;,x&#039;,y&#039;,z&#039; \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

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.