Jump to content

Observe a Falling Charge Radiate


Enthalpy
 Share

Recommended Posts

I'm confident now that the path length has no effect on the power emitted by a deflected charge, because light emitted from different positions doesn't add.

ConstantPowerCorrect.png.a0cbaf2e3cc8babd36dcfcd3d15489c7.png

So the working mental representation is that a detector receives a Dirac pulse from each position of the emitting particle, and the spread of path lengths defines the pulse duration (the pulse shape isn't flat since emission to the sides is weaker) hence the spectrum.

==========

My previous message mentions very thin transparent elements around the proton beam. To shorten the pulse to 540nm/2 at the detector, they must delay by 270nm the light emitted by previous section(s), so a material with index N=1+0.5 must be 540nm thin. Maybe semiconductor processes can support such a membrane with a truss of narrow thick fins, possibly with several thickness levels and truss periods. The material would be a bulk transparent one or an oxidized semiconductor, among which nanoporous silica has a low index hence can be thicker.

Full-energy protons at the LHC (9*10-9 below c) can fall over 30m to create a 270nm pulse before reaching a retarding membrane, while lead ions (57*10-9) need a membrane every 4.7m.

My previous message's sketch lengthens the light path for every section, so only the path difference is so small. I believe membranes here add E rather than P to give a stronger signal.

==========

My previous argument about sending information back in time is doubtful, because observing the effect on a proton perturbs it, and all sections may no more add the amplitudes efficiently.

Marc Schaefer, aka Enthalpy

(I come back)

Link to comment
Share on other sites

On 8/23/2021 at 12:09 AM, John Cuthber said:

The electrons near the nuclei of atoms are in very large fields and experience huge accelerations- much bigger than those in an accelerator.

So, if you think that electrons in an accelerator should emit light due to the effects of gravity, why don't those accelerated in, for example, a sandwich?

The synchrotron radiation needs a speed very close to c to produce more light than the same acceleration does with little speed. While relativistic effects are easily observed in atoms (electrons deep in heavy atoms have 0.1MeV mean kinetic energy versus 0.5MeV rest mass), the "speed" is still far from c, while protons at the LHC move with (1-9*10-9)c. This makes already a numerical difference.

A qualitative difference: "movement" is subtle in QM. Electrons in atoms have a kinetic energy, possibly an orbital momentum, and things you may or not want to call "speed" and "acceleration" because the phase or the amplitude of the wave function changes over the position. But radiation needs in the wavefunction a bulge that moves and accelerates over time, and this is not the case with orbitals, which are "stationary". You can translate this with "immobile" for the sake of light emission. That's how QM explains that atoms in their ground state don't radiate.

In a non-stationary state, for instance (2p+1s)/sqrt(2) during a transition, an electron does have a bulged wavefunction that moves at the frequency of the emitted light. The speed isn't favourable, but the number of electrons in a sandwich is (or the number of colour centres in  a garnet).

My expectation, not fully thought-through, is that the not fantastic speed of the electron in a atom distinguishes it too little from the proton, and that any net emission on a time scale corresponding to the drop duration would be extraordinarily faint. At the LHC, the speed shifts the photon half-period from seconds to femtoseconds and it increases the power too.

An operating laser, where electron transitions are synchronous, would be a first improvement. Maybe the fall modulates the emitted light with a faint off-frequency or off-polarisation component. But 240dB had to be found at the LHC to see something despite the huge speed (and not so few protons) so electrons in atoms must be badly discouraging.

Link to comment
Share on other sites

57 minutes ago, Enthalpy said:

The synchrotron radiation needs a speed very close to c to produce more light than the same acceleration does with little speed

The emission has to happen in the rest frame of the particle, too. Synchrotron radiation is associated with large speeds because that correlates to a large acceleration.

Are you going to provide your calculation of the wavelength of the emitted radiation? 

Link to comment
Share on other sites

The sleeve mirror suggested on 22 Aug 2021
scienceforums
helps to focus the light emitted by the charged particle, as this light exits from a smaller area with a divergence as good. Without sleeve nor other help, light emitted over a huge field depth gives a blurred spot.

SleeveMirror.png.35b1b329df2abb5d0234841872592f69.png

The mirror must reflect light many times without wasting power. Visible wavelength and grazing incidence help. I imagine only metals, since dichroic layers and metamaterials would spread the half-wave pulse.

The mirror lives dangerously around a particle beam, but the diaphragms are narrower and upstream. Near slowly the mirror parts to the stabilized beam according to sensor indications, maybe.

The mirror is maybe 1mm away from the beam, but as extreme darkness is needed, checking the Cherenkov emission at distance would be cautious.

Marc Schaefer, aka Enthalpy

Link to comment
Share on other sites

On 6/22/2021 at 9:16 AM, swansont said:

540 nm? Where did you get that? It's 9 or 10 orders of magnitude more than the KE of an electron dropping that distance.

On 6/23/2021 at 4:45 AM, swansont said:

The KE of the particle, in the plane of the acceleration. So you’re viewing this as a synchrotron with a bend downward.

Synchrotron radiation depends on the bend radius, which gets larger as you give the particle transverse KE 

Show your calculation

On 7/10/2021 at 3:29 PM, swansont said:

And you’ve been asked to support this claim, which you have yet to do.

citation, or calculation, please.

On 7/12/2021 at 7:09 AM, swansont said:

You've been asked for your calculation of the wavelength of the emitted light, and your response is to insult me, which strongly suggests you don't have one to share.

On 8/23/2021 at 4:36 AM, swansont said:

Can you link to where you showed the calculation for the synchrotron emission wavelength or frequency?

On 8/28/2021 at 8:49 AM, swansont said:

Are you going to provide your calculation of the wavelength of the emitted radiation? 

31 minutes ago, swansont said:

Where is your calculation that shows the wavelength?

!

Moderator Note

Enthalpy, you don't get to soapbox your ideas here like a blog. You've ignored repeated requests for clarity, and that's not discussion, it's preaching. Your rigor is much better than a lot of folks but you have a tendency to shrug off tough questions and plow on past them. You know the rules, you've been around enough. This is unacceptable, thread closed.

 
Link to comment
Share on other sites

Guest
This topic is now closed to further replies.
 Share

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.