[Eternal Student]

Hi.

the end-result is curved trajectory accelerations in K'1, and K'2 frames.

    That is an interesting situation but I'm not sure what differences you are trying to imply would be seen.   

Take some care to distinguish between a curved trajectory and an acceleration which is altered or "curved" as you phrased it.

    The acceleration experienced by the charged particle is the same in all of those inertial reference frames.
The acceleration vector is  a   in   frame  K'1    and also the same vector in  K'2 and the original frame which I'll call K0.   For convenience let's say the acceleration is in in the z-axis direction, it will be in the z-axis direction for every frame and assuming the gap between plates and the potential across them remains constant,  this acceleration is just a constant vector a  independant of time for every experiment.
     The acceleration is not curved or altered in any way.

    Now we can imagine the experiment is run theree times,   where the plates are at rest in the three different frames (and the charged particle is initially at rest on the blue plate)  and you also maintain observers at rest in the frames K'1, k'2 and K0  for every experiment.    Then the observers will each see two experiments where the charged particle takes a curved trajectory and only one experiment where it was straight.

The Larmor formula   ( https://en.wikipedia.org/wiki/Larmor_formula ) depends only on the acceleration and not on whether the path was straight or curved.   Using the simple Larmor formula which ignores relativistic effects (so the particle never reaches a high speed.   Late Editing:  and also we make no relativistic correction between frames, all relative speeds << c ) we see that every observer records exactly the same total amount of power radiated in the e-m field in every experiment.

There will be some differences in the e-m radiation.    We have a Doppler effect,  so if the apparatus and particle are moving toward an observer for half the experiment and then away for the other half,   the radiation incident on them changes frequency and energy content.    However, if they just collect all the radiation coming off the particle   (for example construct a sphere around the apparatus that moves with the apparatus, and just measure the total power incident on that sphere at all times)  the Larmor formula applies  and everyone sees the same total radiated power.

Is a stationary charged particle release photons when seen by an inertially moving observer?

   No.  If the charged particle is at rest in some inertial frame,   then an observer at rest in any other inertial frame records the same acceleration for the charged particle   (that will be 0 since it was at rest in some inertial frame).   So the Larmor formula shows 0  e-m radiation from it.
    If you wish to include General Relativity where you can examine non-inertial or accelerated reference frames properly then you could get some more interesting effects.   There is the Unruh effect for example   ( https://en.wikipedia.org/wiki/Unruh_effect ) together with associated Unruh radiation (which is a bit more controversial again).  An accelerated particle, let's say a detector,  may detect photons that were not there in an inertial frame.

Best Wishes.

[Jaaanosik]

Hi.

the end-result is curved trajectory accelerations in K'1, and K'2 frames.

    That is an interesting situation but I'm not sure what differences you are trying to imply would be seen.   

Take some care to distinguish between a curved trajectory and an acceleration which is altered or "curved" as you phrased it.

    The acceleration experienced by the charged particle is the same in all of those inertial reference frames.
The acceleration vector is  a   in   frame  K'1    and also the same vector in  K'2 and the original frame which I'll call K0.   For convenience let's say the acceleration is in in the z-axis direction, it will be in the z-axis direction for every frame and assuming the gap between plates and the potential across them remains constant,  this acceleration is just a constant vector a  independant of time for every experiment.
     The acceleration is not curved or altered in any way.

    Now we can imagine the experiment is run theree times,   where the plates are at rest in the three different frames (and the charged particle is initially at rest on the blue plate)  and you also maintain observers at rest in the frames K'1, k'2 and K0  for every experiment.    Then the observers will each see two experiments where the charged particle takes a curved trajectory and only one experiment where it was straight.

The Larmor formula   ( https://en.wikipedia.org/wiki/Larmor_formula ) depends only on the acceleration and not on whether the path was straight or curved.   Using the simple Larmor formula which ignores relativistic effects (so the particle never reaches a high speed ) we see that every observer records exactly the same total amount of power radiated in the e-m field in every experiment.

There will be some differences in the e-m radiation.    We have a Doppler effect,  so if the apparatus and particle are moving toward an observer for half the experiment and then away for the other half,   the radiation incident on them changes frequency and energy content.    However, if they just collect all the radiation coming off the particle   (for example construct a sphere around the apparatus that moves with the apparatus, and just measure the total power incident on that sphere at all times)  the Larmor formula applies  and everyone sees the same total radiated power.
...
Best Wishes.

Do you know that magnetic field generated by moving charged particle is velocity dependent?
The magnetic field generated by an accelerated charged particle is changing in magnitude and direction because it is velocity/trajectory dependent.
K0 does not predict this directional change but K'1 and K'2 do predict the change.

[Jaaanosik]

From this book:





The B field forms an ellipsoid around the moving charged particle.
When the velocity grows the ellipsoid flattens and the B grows.
The growing B field causes spin like effect around the charged particle.
When the velocity changes direction the ellipsoid changes direction.
The moving frames predict a 'spin-flip' due to the 'curved' trajectory of the acceleration but the straight line acceleration frame does not.
There is a 4-tourque evolution disagreement between the inertial frames.
This is all textbook... somehow missed in our understanding of physics.

That is an interesting situation but I'm not sure what differences you are trying to imply would be seen.   

This is a huge difference...

[Eternal Student]

Hi.

Do you know that magnetic field generated by moving charged particle is velocity dependent?

   Yes  (but thanks for the information).

The magnetic field generated by an accelerated charged particle is changing in magnitude and direction because it is velocity/trajectory dependent.
K0 does not predict this directional change but K'1 and K'2 do predict the change.

Yes,  the magnetic and electric fields that the observers see will be different.   How or why does that prevent a linearly accelerated charged particle from emitting radiation that you may detect as a photon of some frequency?

1.   The base line of the B field at position x and time t₀   does not prevent an oscillation above and below that being obtained as time evolves.   We can still identify oscillations in the E and B fields which may be asscoiated with photons (conventional oscillatons of the E and B field at right angles and propogating away at c).   Any oscillation in the fields can be decomposed into a sum of conventional sinusoidal oscillations of different frequencies.
For each observer, the frequencies and corresponding amplitudes involved in this Fourier decomposition of the oscillations at any given event (place and time allocated by observer according to their rest frame) could be different, that is not in dispute.    As already mentioned there is a Doppler effect and we would fully expect the frequencies of radiation that an observer would see being thrown out in front of a moving charge will be different to the frequencies thrown out to the back of the moving particle.

2.   Although the E and B fields may have different values for each observer, the transfer of some value from the E field to the B field changes the power being transferred in the e-m fields in a complicated way,  we have to examine the product  of magnitudes   |E| |B| and the orientation of the fields.
    The total power radiated in the e-m field is obtained by examining the Poynting vector  S   ~   E x B.

3.   The Poynting vector can still be different at various events  (an event = place and time) for observers at rest in different frames, that is also not in dispute.    However, the  integral of the Poynting vector over a spherical surface around the apparatus remains constant for all observers who are at rest in any one of the frames.
     This is the information being encapsulated by the Larmor formula.   It's not attempting to describe anything about the intensity or frequency of some photons that may be produced and emitted in various places or directions, only to determine the total radiated power carried in the E and B fields.

    Just to be clear,  there are no "photons" to be found in the classical theory of elctrodynamics, nothing is discretised like this.  You can attempt to identify a stream of photons with a classical sinusoidal e-m wave of a particular frequency.  Then the frequency of each photon is assumed to be the frequency of the e-m wave and the number intensity of photons (number of photons impacting on a surface per second) is considered to be proportional to the intensity (the square of the amplitude) of the classical e-m wave.   That identification is what is usualy done.
    For all observers, classical electrodynamics suggests the same total power is radiated.  You need some quantum theory to identify any photons that may be emitted. 

    (Although I had started to discuss the conventional way to link classical electrodynamics to a model involving photons, it's not essential and the post is long, so it'll go under a spoiler).

Spoiler: show

 There is no reason to assume all the radiation can be identified as a collection of photons that are each just in a single state.   Indeed, if we make the usual identification suggested above (freq of classical sinusoidal wave --->  freq. of the photon ;    classical intensity ----> number intensity of photons) and recall the decomposition of the classical E and B fields into elementary sinusoidal components (a Fourier expansion of the oscillations in the E and B fields, if you like) discussed in (1) above,  then we have good reason to believe the photons must be in a superposition of states.  Over a small time interval, we may assume only one photon is released and if it is to match the sum of sinusoidal terms exhibited in the classical E and B field representation,  then it must be in a superposition of states.   Each single frequency component that appears in the Fourier sum corresponds to a single frequency state of a photon and the coeffeicients for the combination of states must correspond to the amplitude coefficient in that Fourier expansion for the classical E and B fields.   
    If you did place detectors around the charged particle and force the collapse of a quantum particle, a photon, to a definite state,  then you would expect to detect photons of various frequencies with probabilities corresponding to the coeffeicient of that state in the superposition.   For example, it is very likely that you'll find more blue photons being thrown out to the front of a moving charge and more red photons coming out at the back.

    The magnetic and electric fields for each observer will be different and that is not a problem.   It doesn't prevent photons from being detected in some frame (e.g. the one where the particle was linearly accelerated and took a straight path).   It merely changes the prevalence (frequency of detection) of photons of a given colour (oscillation frequency) that can be detected in various directions.
   
    I hope that makes some sense.

Best Wishes.

[hamdani yusuf]

The only minor issue that will arise is whether all of this oscillation in E and B fields would always be recognisable as "a photon", a discrete packet of energy.   Classically, it's an oscillation in the E and B fields but it should be a continous emission rather than something that occurrs in bursts at discrete places and times.

Someone may argue that the issue is fundamental rather than minor.

[hamdani yusuf]

The unit for h is actually Joule second per cycle.

Er, no.

The SI units are defined in such a way that, when the Planck constant is expressed in SI units, it has the exact value h = 6.62607015?10⁻³⁴ J⋅Hz⁻¹

Since 1 Hz = 1/second, the term Hz^-1 has the dimension of time, so h is simply 6.62607015?10⁻³⁴ joule.sec.

h/2π = ħ is still joule.sec but 2π turns up in so many calculations that it is simply a more compact way of writing equations.

What do you think about this Wikipedia entry?

In many applications, the Planck constant
ℎ naturally appears in combination with 2π as ℎ/(2π), which can be traced to the fact that in these applications it is natural to use the angular frequency (in radians per second) rather than plain frequency (in cycles per second or hertz). For this reason, it is often useful to absorb that factor of 2π into the Planck constant by introducing the reduced Planck constant[38][39]: 482  (or reduced Planck's constant[40]: 5  [41]: 788 ), equal to the Planck constant divided by
2π[38] and denoted by ħ (pronounced h-bar[42]: 336 ).
https://en.wikipedia.org/wiki/Planck_constant#Reduced_Planck_constant_%E2%84%8F

[hamdani yusuf]

Quote from: hamdani yusuf on Yesterday at 13:09:12
Is a stationary charged particle release photons when seen by an inertially moving observer?

   No.  If the charged particle is at rest in some inertial frame,   then an observer at rest in any other inertial frame records the same acceleration for the charged particle   (that will be 0 since it was at rest in some inertial frame).   So the Larmor formula shows 0  e-m radiation from it.

Let a positively charged particle staying still at Cartesian coordinate (0,1).
An observer moving along x axis from left to the right at 1 m/s.

We can think of the implications of this situation:
Before passing the point of origin, he feels increasing electric field.
After passing the point of origin, he feels decreasing electric field.
Changing of electric field is often said to create magnetic field.
The rate of change of the electric field in this case is not constant.
Which means the magnetic field is also changing.
Changing of electromagnetic field is often called radiation.

Among the statements above, which one(s) do you think is wrong?

[alancalverd]

What do you think about this Wikipedia entry?

Here's what Wikipedia thinks about it (panel at the top right of the article)

Reduced Planck constant
Common symbols   
ℏ
SI unit   joule-seconds

exactly the same units as h

[hamdani yusuf]

What do you think about this Wikipedia entry?

Here's what Wikipedia thinks about it (panel at the top right of the article)

Reduced Planck constant
Common symbols   
ℏ
SI unit   joule-seconds

exactly the same units as h

Not exactly. See the panel on top of what you quoted.

Planck constant
Common symbols   
ℎ
SI unit   joule per hertz (joule seconds)

[alancalverd]

Exactly exactly! The unit is  joule.second whether you are talking about h or ħ.

[hamdani yusuf]

Does the unit of the radiation frequency have no effect on the numeric value of their quantum of energy?

[Bored chemist]

Does the unit of the radiation frequency have no effect on the numeric value of their quantum of energy?

The number of joules of energy which a photon carries  doesn't know or care if you measure frequency  in 1/seconds or 1/weeks
The value of Planck's constant (h) would change.

[alancalverd]

Does the unit of the radiation frequency have no effect on the numeric value of their quantum of energy?

No. The unit of energy is joules or electron volts and the number is (by definition) fixed for any quantum.

[alancalverd]

doesn't know or care if you measure frequency  in 1/seconds or 1/weeks

Schlumberger used to boast that they could supply customised oscilloscopes "calibrated in millifurlongs per microfortnight", if you wanted.

[Eternal Student]

Hi.

I'm sorry if this isn't what you ( @hamdani yusuf ) wanted.   It takes hours to write replies like this, so I've got to try and keep it short even if there's a risk of sounding curt.

Among the statements above, which one(s) do you think is wrong?

    The following one:

Changing of electromagnetic field is often called radiation.

    No it isn't.  At best it's an indication there may be some propagating radiation.    If you can identify a propagating wave in the E and B field with appropriate properties (sinusoidal, E orthogonal to B etc.) then you're on to something, you may have some light and hence some photons there.  A photon can be called a form of radiation (along with alpha emission and other things).

    More generally,  radiation through the e-m field means there is some energy being radiated through the electric and magnetic fields.  Radiated just means travelling away from the particle.
    Quantum models suggest that photons are the only way you're going to get energy being carried away.   So if you can find some radiation of energy through the e-m field then you can be fairly sure there will be some photons to be found travelling away from the particle.   The converse doesn't have to hold.   If there's no radiation of energy from the particle then (i) there may not be any photons  BUT it could also be (ii) that there are some phtons, an equal number travelling in and being absorbed as those being thrown out (for example).

    Power is radiated through the Electric and Magnetic fields     if and only if     The Poynting vector illustrates a net flow of energy out of an enclosed surface you can draw around the object.
    It's apparent that in the frame where your positive charge was at rest, nothing too interesing happens.
    In the frame where the observer was at rest,   the particle will be moving left at 1m/s.   Take a simple enclosed surface around the particle, a flat sided box around the particle will do nicely.

    Draw the diagram of the E and B fields that exist around the particle.   An earlier post from @Jaaanosik  has made that easy for you where some pages from Griffith's appear.
    You'll have a diagram a bit like this:

⊕  ⊕  ⊕
      ↑
      ◊
      ↓
?    ?   ?    
    where   ◊  is the particle,  Blue arrows ↑  ↓   are the electric field,    red ⊕ and ? are the magnetic field going into or out of the page respectively.   (The ? were dots but the forum won't support that character).

     Determine  the Poynting vector  which is (proportional to)  E x B   as usual with your preferred rule   (I use a right hand with fingers).   Put that on your diagram.

You should end up with something a bit like this:
              ←   ←
                  ◊
              ←   ←
Draw the field lines carefully and you'll have a slight downward slope on the top left arrow, slight upward slope on the top left arrow   etc.   They'll be symmetric in their up and down slope-ness.

Now put the box around it.   The flux on the left wall is outward,  the flux on the right wall is inward.   These match in magnitude.   There's no flux over the top and bottom  or the "near to you" and "far from you" wall.   Overall, there is no net flux of the energy into or out of the box.

   Use a small box and you can see there's no net energy lost by the particle.   No energy is radiated by the particle through the Electric and magnetic fields.   This is in total agreement with the Larmor formula discussed earlier, there was no acceleration of the particle.

    The diagram with the Poynting vectors may look like something is being sent out in one direction and coming in from the other direction but it's just a flow of energy in the E-M field at these points in space and is not always a propagating wave in the E and B fields that you could identify as some light.   The situation is similar to having a magnet just sit still with the North pole pointing to a static positively charged particle.   There will be some regions of space where E and B fields cross at some oblique angle and you have a non-zero Poynting vector, so some energy is flowing at that point in space.   However, travelling e-m waves are not found at that point in space, we get no flashes of light from a static magnet pointing to a static charge.   Take a moment to think about this,  we can have a flow of energy indicated by the Poynting vector but there is no light to be found.   In a D.C. electrical circuit, we can identify a flow of energy out of the battery and into the load it is connected to,  there's definitely a flow of energy but we don't have rays of light (any frequency of it) shining out of the battery and being sucked up by the load.   A flow of energy in the e-m field at a point in space can be quiet and invisble rather than forcing the production of photons.

    We can go a bit further, if you wish.   Since there's no net energy transfer in your example we don't expect to identify any photons being thrown out and propagating away from the particle.   However, we can't formally exclude the possibility that there may be some, we may be in the situation where there are also some incoming photons.

   So there may be some photons,  you'd have to examine the E and B fields around the particle carefully to get more information.   You're looking for the usual sinusoidal propagating waves as previously mentioned.   You're free to do that if you wish.   If you found any, then there would be an interesting situation.  For example, light may be thrown out by the particle to the front in the direction of its travel,  but some light must also appear behind it.   The light behind it is interesting because it won't travel away, it would appear in space after the particle has passed and then chase after the particle and be absorbed.  If you find a solution for the E and B fields that exhibit that behaviour, that'll be great and probably a publishable result.

Best Wishes.

[hamdani yusuf]

Does the unit of the radiation frequency have no effect on the numeric value of their quantum of energy?

No. The unit of energy is joules or electron volts and the number is (by definition) fixed for any quantum.

The value of h is 6.62607015x10⁻³⁴
energy of a single photon is h.f.
If the photon frequency is 1 Hz, then its energy is 6.62607015x10⁻³⁴ Joule.
If the photon frequency is 1 GHz, then its energy is not 6.62607015x10⁻³⁴ Joule.
If the photon frequency is 1 rpm, then its energy is not 6.62607015x10⁻³⁴ Joule.
If the photon frequency is 1 rad/s, then its energy is not 6.62607015x10⁻³⁴ Joule.

[Bored chemist]

Does the unit of the radiation frequency have no effect on the numeric value of their quantum of energy?

No. The unit of energy is joules or electron volts and the number is (by definition) fixed for any quantum.

The value of h is 6.62607015x10⁻³⁴
energy of a single photon is h.f.
If the photon frequency is 1 Hz, then its energy is 6.62607015x10⁻³⁴ Joule.
If the photon frequency is 1 GHz, then its energy is not 6.62607015x10⁻³⁴ Joule.
If the photon frequency is 1 rpm, then its energy is not 6.62607015x10⁻³⁴ Joule.
If the photon frequency is 1 rad/s, then its energy is not 6.62607015x10⁻³⁴ Joule.

because

The value of Planck's constant (h) would change.

[hamdani yusuf]

Correct.
The familiar numerical value of h is in the unit of Joule/Hz, or Joule second/cycle. In different unit, the value would be different.

[Bored chemist]

Correct.
The familiar numerical value of h is in the unit of Joule/Hz, or Joule second/cycle. In different unit, the value would be different.

And a grand total of zero people said otherwise.

[hamdani yusuf]

Correct.
The familiar numerical value of h is in the unit of Joule/Hz, or Joule second/cycle. In different unit, the value would be different.

And a grand total of zero people said otherwise.

Are you sure?
The only reason I resurrected this thread was because I saw someone said that the unit cycle in Planck's constant can be ignored. Subsequent discussion shows that he's not the only one.