Enthalpy Posted February 21, 2016 Posted February 21, 2016 Hello dear friends!Whether a charged particle radiates when falling in a gravity field has been debatedhttp://www.scienceforums.net/topic/80770-do-electrons-radiate-from-electostatic-acceleration/page-4#entry861035https://www.scribd.com/doc/100745033/Dewitt-1964Whatever the prediction by plain Relativity is, a disagreeing observation would be a precious hint towards a more general theory, possibly a unification of gravitation and electromagnetism.Here I propose elements for an experiment to check it. "Elements", because the figures I have right now aren't a complete solution - but this can only improve The radiation, if any, is badly small. A chunk of lead attracting an electron would bring no chance to observe something. My proposal instead is to (try to) observe at an accelerator the synchrotron radiation of ultrarelativistic particles falling in Earth's gravity field.----------Synchrotron radiation provides more light to an observer facing the quick particle's path, because it makes a shorter but much taller pulse that totals more energy. The observer sees first the part of the pulse emitted as the particle's deflection begins, but because the particle nears almost as quickly as the electromagnetic field, the part of the pulse emitted as the particle's deflection ends is seen very shortly after. This near compensation of the travel times occurs only in a narrow cone around the particle's instantaneous direction, narrower or broader than the particle's deflection depending on the conditions.At least when a magnetic field deflects the particle, a known and verified formula gives the emitted power, thanks: ----------A particule rushing horizontally through Earth's gravity falls at 1g for the Earth-bound observer. Reasons: Its kinetic energy increases the inertia, hence increases the gravity force the particle creates and experiences. Our Sun, Galaxy, other stars and galaxies create gravity fields where the Earth, the observer and the particle fall freely. If the particle's behaviour differed from the Earth's and observer's one, we would know from local observations if we're in the gravity field of some astronomical object. This is precious to make the particle less sensitive to influences other than gravity, for instance the residual geomagnetic field.----------My figures show three big difficulties up to now - hopefully no worse one will appear: Radiation by the falling particle is faint, even with the synchrotron amplification. Radiation by the particle in magnetic fields is strong. Accelerators' storage rings need dipole magnets, even linear accelerators need quadrupoles. The residual magnetic field at the free fall zone must be tiny, since its radiation contribution occurs at the bad location. The existing accelerators were not designed for this experiment, that's a strong constraint.I'll expose my answers - or elements as I said, because the initial figures look discouraging . But that's usual life at accelerators, isn't it?Marc Schaefer, aka Enthalpy
Enthalpy Posted February 21, 2016 Author Posted February 21, 2016 This is a setup example. Neither the scale not the shape are significant on the drawing. Especially, such racetracks are rare, and storage rings use to have a dozen of short straight sections instead. Of interest is that two magnets, as weak as possible, separate the free fall section from the rest of the noisy track. They offset the beam just enough to extract it from the upstream light. Not all upstream light can be removed because some is parallel to the beam.A linear accelerator would offer more room, added if needed, for a longer free fall section and longer surrounding magnets, weaker hence more silent. Alas, linear accelerators' beam current is small.The appended spreadsheet (for Gnumeric, Excel, or see its screenshot) compares the useful synchrotron signal with the parasitic one and tells also how small the residual induction must be.GravityMagnetRadiation.zip I lack much data, especially the available straight sections, some beam sizes too. Among present machines, the LHC (large hadron collider) gives more signal photons at a more convenient wavelength. It also accepts a stronger residual induction (difficult) because it has the biggest energy per unit charge. Winner. The future CLIC is a potential candidate, providing a stronger signal, but as X-rays. The operation of the colliders and light sources thrives to concentrate the beams at the interaction points, which the present experiment doesn't need. This may enable a stronger beam, especially with heavy ions which would radiate well.The small deflections and offsets suggest that the whole experiment fits in a but wider tube, including the shutters, filters and detectors. A supersoft magnetic tube may dampen the stray induction from the ends of the magnets; over the rest of the free fall section, an active compensation must reduce the geomagnetic field from ~40µT to ~pT. The residual electric field is easier.The signal-to-noise is scary... Winning 120dB twice is a goal. I plan to describe means that tend towards these figures.Marc Schaefer, aka Enthalpy
Enthalpy Posted March 6, 2016 Author Posted March 6, 2016 A particle beam focused at the experiment helps reduce the stray light in an adapted setup. Here 2*10µrad rms let converge from 2*150µm rms to 2*24µm rms within 30m almost as at the collision points. Focusing (2m, 180mT at r=1*sigma) adds stray light (2*10-22 J/proton) with both polarizations, but a subsequent bending magnet separates the beam from previous light.A first collimator with D~6mm L~5m removes the light beyond +-1.2mrad but diffracts a bit.A 3.1T dipole extracts the beam from previous light by 2mrad in 15m. Its accurately vertical field creates stray light (9*10-18 J/proton) but very little is vertically polarized.A second collimator removes much stray light, including in the beam, because its small openings diffract it between the many stages. The central 4.2m have 2*250µm wide square holes that spread the first lobe to +-1.2mrad. The next stage 520mm away lets -9dB through, 1+8 stages -74dB. The previous 2.7m have 2*400µm holes. First lobe +-1.5mrad, 667mm spacing. 4 stages let -36dB through. The following 2.7m gain 36dB too. The first hole get about -8dB of the bending magnet's stray light, so -154dB emerge from the collimator. To preseve the clean polarization, the collimator steps are split in pairs of successive horizontal and vertical slits. The blades (graphite?) are straight, accurately oriented and their position adjusted dynamically. I won't check the stray light created by residual gas, by rogue particles hitting the blades (put the upstream ones narrower), nor the cosmic rays and natural or induced radioactivity. Collider people know that in detail. Over 30m free fall, the particles emit (or not) a few vertically polarized photons around 600nm wavelength.Something deflects most photons (first lobe +-4mm) away from the beam (+-300µm rms): a flat mirror, a concentrating mirror, the analyzer... It must preserve the clean polarization. A separate analyzer can't stay upstream since it can't have a hole.The filter was a false hope and drops away.The analyzer removes the remaining horizontal photons created by the last bending magnet. More details should come.The detector must be very silent. More hints should come.Marc Schaefer, aka Enthalpy
Enthalpy Posted March 13, 2016 Author Posted March 13, 2016 Even if stray light can be reduced below the signal (hints about the polarization filtering should come later), detecting 4.7*10-6 photons per second looks extraordinarily difficult, but here are some thoughts.With sensors that give one electron per photon, the amplifier introduces a stronger noise. I exclude them as is generally done.---------- Focus the lightThe particle beam emit light over 30m path in a 2*130µrad cone, or rather 2*325µrad to catch most light. Focussing right at the end of this path, as on the drawing, would let the big depth of field blurr the image and impose a too big sensor.The concentrating lens or mirror at (excessive) 100m +-15m instead, after a first deflection, can focus to 0.1m +-16µm. From typical R=13mm, diffraction at 540nm makes an R=2.5µm spot, so R=6µm contains most power, and the depth of field adds R=0.5µm. With a difficult adjustment, the sensor's active area can be R=15µm.Maybe the optics could attenuate stray light produced upwards the particle beam's free-fall section. Especially, several out-of-axis lenses or mirrors would aims at the middle of the free-fall section and attenuate the start and the end. This has other drawbacks and I don't consider it here.The surroundings must be cold to emit no stray light, the optics probably too. Cold detectors may demand a coupling fibre not considered here.An expert would do it better and more accurately.---------- Transition Edge Sensor (Tes) : candidateA thin small superconductor is hold just below its transition temperature, the heat of each absorbed photon creates a short-lived resistive spot.https://en.wikipedia.org/wiki/Transition_edge_sensorMeasures on an optimized Tes are given therehttp://arxiv.org/abs/1509.02064v1 concise http://arxiv.org/abs/1502.07878v1 detailed it detects 95% of the photons and has an intrinsic dark count rate of 1*10-4/s at 80mK. The pulse rises in 200ns and drops in 4µs.Supposedly radioactivity and cosmic muons create at a rate of 1*10-2/s stronger and longer pulses that can be discriminated. For any sensor, the underground Lhc is quieter, and purer materials can be chosen. Also, our photons around 540nm versus 1064nm might help reduce the dark count rate by reducing the sensor's sensitivity: colder or less current. Ideally, the signal detection rate is 4.2*10-6/s. The active area, R=15µm versus 25µm*25µm, keeps 1.1*10-4/s dark count rate. The Tes won't discriminate the 230ps bunches nor their 2.5ns spacing period, but let's say that the 2808 bunches can come in groups longer than 200ns and spaced by more than 200ns: then the experiment can be receptive for 7µs over a 89µs storage ring turn. The dark count rate drops to 8.7*10-6/s over the active experiment time. The rest of the time serves to estimate properly the dark count rate. The signal-to-noise achieves 1*sigma after 0.5Ms = 6 days and 5*sigma after 5 months. The Lhc has already run with 3* as many bunches; this would slash the experiment time by 9. More particles per bunch act the same way. A Tes with dark count rate /10 would slash the experiment time by 10.---------- Superconducting Nanowire Single Photon Detector (Snspd) : less goodHere an absorbed photon break a Cooper pair, and the increased current density in the nanowire creates a resistive section.https://en.wikipedia.org/wiki/Superconducting_nanowire_single-photon_detectorMeasures on an optimized Snspd are given therehttp://arxiv.org/abs/1307.0893v1this one detects 5% of the photons and has a wildly extrapolated intrinsic dark count rate of 1*10-4/s at 2.3K and 20.5µA - 540nm may help too. It resolves 230ps but doesn't resolve the energy, so radioactivity and cosmic rays must be minimized. Ideally, the signal detection rate is 2.4*10-7/s. The active area, R=15µm versus 10µm*10µm, raises the dark count rate to 7.1*10-4/s. The experiment is sensitive during 2808 bunches of 230ps, or 1/138 of a turn. This squeezes the dark count rate to 5.1*10-6/s of the active experiment time. The signal-to-noise achieves 1*sigma after 89Ms = 2.8 years active time and 5*sigma after 71 years. The bunches and particles per bunch act the same way, the dark count rate too, and the detection probability acts squared.---------- Single Photon Avalanche Diode (Spad) : noIn the "Geiger" mode, the voltage lowered below the breakdown stops the avalanche and is then set above the breakdown until noise or a converted photon triggers the avalanche - other modes would be more noisy.https://en.wikipedia.org/wiki/Single-photon_avalanche_diodeOnly 5% of the photons triggerhttp://www.perkinelmer.de/CMSResources/Images/44-3477DTS_C30902.pdfwhile the -20°C dark count rate is 450/s, 77K reduce it only /200https://indico.cern.ch/event/51276/session/14/contribution/278/attachments/967044/1373321/278_collazuol.pdfsilicon works badly at deep cold, other semiconductors tend to leak more.The active area can shrink from C30902SH's 0.2mm2 to R=15µm, reducing the dark count rate /283 ideally.This leaves 8*10-3/s dark count rate, a bad figure combined with the 5% detection efficiency - unless a better component exists.---------- Photomultiplier Tube (Pmt): the challenger?https://en.wikipedia.org/wiki/Photomultiplierhttps://www.hamamatsu.com/resources/pdf/etd/PMT_handbook_v3aE.pdfa GaAs photocathode offers 15% quantum efficiency (handbook p33 = pdf p47) and a bold extrapolation of the fig 4-40 (pdf p84) would bring the dark count rate from 1/s @-40°C to 1*10-3/s around -100°C. The pulse can be 1.5ns wide (pdf p62). Energy discrimination is possible. Signal detection rate 7.1*10-7/s. The experiment is sensitive for 1.25ns (half cavity period) of 2808 bunches, or 1/25 of a turn, reducing the dark count rate to 3.9*10-5/s of the active experiment time. The signal-to-noise achieves 1*sigma after 77Ms = 2.4 years. BUT: I see no hard limit to really cold operation. The dark count rate must sink. It has fewer causes than in semiconductors, and the vacuum can improve too. The glass and metals can be synthesized nearly free of radioactivity. Na seals glass on Mo, no K is necessary. Some constructions resolve the 230ps bunches, reducing the experiment time /5.4. The photocathode uses to be huge: 10mm*10mm, or 100,000* our needs. The main dark current cause, the thermoionic current, must shrink with the area. Manufacture it tiny by semiconductor processes, or just deposit a material with a big electron work function (Au, Pt, Ir...) over most photocathode area but the useful D=30µm. If the smaller photocathode works, the photomultiplier tube is also a challenger for fibre optics and more. Marc Schaefer, aka Enthalpy
Enthalpy Posted June 19, 2021 Author Posted June 19, 2021 I like ever more a photoemissive detector, but with a semiconductor anode. Hamamatsu add optional semiconductor avalanche at the anode and call it "hybrid photo-detector". I prefer a PIN diode at the anode with no or little polarisation. scienceforums Hamamatsu indicate dark count rates like 100/s for a GaAs photocathode at -40°C, but 13930K=1.20eV activation energy would reduce this cause to 10-6/s @-78°C and 10-53/s @77K, wow. The emissive area shrunk from 10mm*10mm to 30µm*30µm would also gain 105. In a usual photomultiplier, other sources limit the improvement below some -50°C. The pulse count mode can't remove them all, for instance the stray emission by the first dynode, because the gain per stage is small and fluctuates. The single semiconductor anode will improve that at once. The very constant pulse area by a photoelectron hitting the semiconductor anode lets two thresholds remove most stray pulses by cosmic rays and radioactivity. As the experiment shall produce 5*10-6 photons per second, bunches of pulses within a µs can be all suppressed too. ========== I initially wanted for maximum power a proton free fall as long as possible, 50m, but photocathodes aren't very sensitive to the resulting 900nm peak photon wavelength. GaAs would respond only below 850nm, and with 15% quantum efficiency. Better 30m free fall to emit 0.36* as many photons, but around 540nm, where a transmissive GaAsP photocathode offers 40% quantum efficiency. This gives twice as many counts as GaAs that detects every second photon with 15%. If more length is available, install several mirrors and detectors. Marc Schaefer, aka Enthalpy
Enthalpy Posted June 20, 2021 Author Posted June 20, 2021 The polarization analyser shall remove much of the stray light created at the bending magnet, as on the diagram of 06 March 2016. After the last deflection and focussing magnets, which produce much stray light of badly defined polarisation, a first set of diaphragms (ex collimator) shall remove most light, before a long and weak bend magnet turns the particle beam away from this stray light. This bend magnet has a very stable orientation and, somehow, a field of uniform orientation. Field uniformity is less critical if the wavefunction of each particle is delocalized over the beam width. If this is achieved, the stray light is produced essentially by the bend magnet, with accurate polarization, so an analyser can contribute remove it. On 21 February 2016, I had suggested 120dB cleaning by the analyser, but meanwhile the diaphragms provide 154dB in my estimate, so the analyser can provide about 100dB. Since semiconductor patterning achieves 5nm over 20mm, and a milling machine few µm over 1m, an accurate analyser is feasible. It must also keep its orientation to 10-5rad, or 10µm over 1m, which is less difficult than a laser resonant cavity. The adjustments are not done with 5*10-6 photons/second. To adjust the analyser, the diaphragms are widened, so up to 2*1012 photons/second from the bending magnet emerge from the analyser to an auxiliary detector. To observe the efficiency of the diaphragms, light is measured before the analyser, where 80 photons/second from the bending magnet are still present. ========== Maybe more diaphragms, packed more closely, or a continuous black part, perform better. ========== The geomagnetic field must be shielded, say to 0.25pT so it creates 0.1* as many photons as gravity does. The shielded section includes the last diaphragm section and the free falling section up to the optical deflector. The Lorenz factor eases that because the Earth attracts the kinetic energy too, or as seen by the photon, because the speeding Earth is heavier. Surrounding cylinders of soft magnetic material can't achieve 0.25pT alone. Hysteresis in mumetal leaves a field about 106* too strong inside. Imperfections in the demagnetization would limit what concentric cylinders achieve, and I believe gaining one million is impossible. Active coils can't compensate the 50µT down to 0.25pT, because the 0.58A beam creates 0.1µT at R=1m, so I believe measuring 0.25pT is too difficult. But just a cylinder of type I superconductor (or a type II used in region I) can expel the 50µT geomagnetic field. Niobium's critical field Hc1 nears 0.2T at 4.2K, for MgB2 it's 0.15T, so to my understanding, the superconductor needs no help by soft magnetic material nor by coils, except maybe near the bending magnet. The cylinder must be longer than the protected section for the field to drop enough. A small diameter helps. Assembled parts can constitute the cylinder if they overlap closely and long enough. Marc Schaefer, aka Enthalpy
Enthalpy Posted June 20, 2021 Author Posted June 20, 2021 At least one paper claims to "prove" that a falling charged particle doesn't radiate for an equally falling observer. It's a hypothesis by Relativity that we can't notice locally a gravitational field uniform enough, so said "proof" is a tautology. The LHC sits on Earth, it's not free falling.
swansont Posted June 22, 2021 Posted June 22, 2021 On 2/21/2016 at 7:26 AM, Enthalpy said: Its kinetic energy increases the inertia, hence increases the gravity force the particle creates and experiences. Are you claiming that a particle moving at high speed will deflect at an acceleration greater than 1g? If not, how does this matter? What is the frequency of the photons that would be emitted by the particles?
Enthalpy Posted June 22, 2021 Author Posted June 22, 2021 (edited) As seen from the lab, the particle falls at 9.8m/s2 despite being heavier. As seen from the particle, it accelerates much more than 9.8m/s2 to drop by the same height in a shorter time, and this happens because the Earth moving quickly is heavier. Frequency: the wavelength for a free fall over 30m horizontal distance is around 540nm. It's a benefit from the synchrotron effect, and the reason to make the experiment at an accelerator, as I propose. Without this effect, the radiation would be impossible to measure practically, as the other people noted before my proposal. Edited June 22, 2021 by Enthalpy
swansont Posted June 22, 2021 Posted June 22, 2021 37 minutes ago, Enthalpy said: As seen from the lab, the particle falls at 9.8m/s2 despite being heavier. As seen from the particle, it accelerates much more than 9.8m/s2 to drop by the same height in a shorter time, and this happens because the Earth moving quickly is heavier. This is relativistic mass, which is not an actual change in mass. If your proposed effect were true, you could form a black hole by moving fast enough. This doesn't happen. 37 minutes ago, Enthalpy said: Frequency: the wavelength for a free fall over 30m horizontal distance is around 540nm. It's a benefit from the synchrotron effect, and the reason to make the experiment at an accelerator, as I propose. Without this effect, the radiation would be impossible to measure practically, as the other people noted before my proposal. 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.
Enthalpy Posted June 23, 2021 Author Posted June 23, 2021 18 hours ago, swansont said: This is relativistic mass, which is not an actual change in mass. Kinetic energy creates gravity. You can measure on scales its contribution to the mass. The kinetic energy of baryons in an atom is easily measured. Relativistic particles too fall with 9.8m/s2 on Earth. This results directly from the relativity principle, as explained in the first message. 18 hours ago, 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. It's 12 orders of magnitude less than the proton's kinetic energy. Where would you suppose the energy of the synchrotron light comes from, when a magnetic field deflects the relativistic particle?
swansont Posted June 23, 2021 Posted June 23, 2021 1 hour ago, Enthalpy said: Kinetic energy creates gravity. Transverse KE does not. Quote It's 12 orders of magnitude less than the proton's kinetic energy. Your OP mentions electrons, not protons, and the acceleration is vertical. Compare the momentum of a vertically-directed photon with the momentum of the proton or electron Quote Where would you suppose the energy of the synchrotron light comes from, when a magnetic field deflects the relativistic particle? 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
John Cuthber Posted June 24, 2021 Posted June 24, 2021 I know that the model of an atom as "electrons orbiting a nucleus" is ... unhelpful, but the fundamental point is still there. The electrons of an atom must be doing something; we know that there are interesting effects like the colour of gold and the melting point of mercury- which arise from the relativistic effects of electrons. So there is some significant similarity between the electrons in an atom, and those in an accelerator. If we expect 540 nm radiation from the electrons in an accelerator interacting with gravity, why don't we expect some similar outcome from the electrons in a dropped sandwich? 23 hours ago, Enthalpy said: Where would you suppose the energy of the synchrotron light comes from, when a magnetic field deflects the relativistic particle? Ask the guy who pays the electricity bill for the synchrotron.
swansont Posted June 24, 2021 Posted June 24, 2021 3 hours ago, John Cuthber said: If we expect 540 nm radiation from the electrons in an accelerator interacting with gravity We need to separate the effect that Enthalpy is incorrectly including from the nonexistent increase in mass from the established physics prediction of a dropped or deflected particle. Which is why we need Enthalpy's calculation edit: and an observer that's co-moving with the particle will see it at rest, so the question is where is that 1 eV photon coming from. Even a proton has much less than an eV of kinetic energy after falling 30m, and the recoil from emitting a 540 nm photon would be quite noticeable
swansont Posted June 24, 2021 Posted June 24, 2021 On 3/6/2016 at 12:30 PM, Enthalpy said: Over 30m free fall, the particles emit (or not) a few vertically polarized photons around 600nm wavelength. If this is a beam in a particle accelerator, it will have traveled upwards of 750 million km in the roughly 2.5 seconds it takes to drop 30m Not happening in any accelerator around here. Also, the Larmor formula says that the radiated energy in a 1g field is around 10^-50 joules in this period of time (just a rough calculation)
John Cuthber Posted June 24, 2021 Posted June 24, 2021 6 hours ago, swansont said: We need to separate the effect that Enthalpy is incorrectly including from the nonexistent increase in mass from the established physics prediction of a dropped or deflected particle. Which is why we need Enthalpy's calculation edit: and an observer that's co-moving with the particle will see it at rest, so the question is where is that 1 eV photon coming from. Even a proton has much less than an eV of kinetic energy after falling 30m, and the recoil from emitting a 540 nm photon would be quite noticeable Let's face it, when I said "If we expect 540 nm radiation", I was being ironic. We don't expect that. Partly because, if we did,we should expect it from any electron in a strong em field, and falling under gravity- for example, in a dropped sandwich. His claim is as plausible as if he said we ought to see a sandwich light up before it hits the floor.
Kartazion Posted July 10, 2021 Posted July 10, 2021 @Enthalpy Very interesting this thread. I think the collapse of the charged particle, by its own mass, occurs effortlessly with the natural gravitational force. The charge is however not necessary. On the other hand, the emergence of the particle would then be the energy supplied, contrary to this gravitational force, but now with a charge. It would then take a vertical accelerator/collider to verify this.
Enthalpy Posted July 10, 2021 Author Posted July 10, 2021 Wow, so much interest for this topic! Here a few answers - I probably forget some interrogations. Much is also in the early messages. My preferred particles are (since the beginning) protons at the LHC, as they give an easier wavelength. Other particles are interesting too: Pb ions at the LHC, electrons at the future CLIC. I had also thought at alphas at the LHC. My central idea is that horizontal near-light speed lets the falling particle emit more light than if dropped from zero speed, just as the synchrotron radiation increases sharply with the particle energy. This gives a chance to observe the effect. The protons fall freely over 30m horizontal path. In the 100ns, they lose 50fm altitude, not 30m. The energy a proton may radiate comes from its kinetic energy, just like when magnets deflect the proton. A speeding particle falls in a gravity field with the same vertical acceleration as an initially immobile particle. This is not a far hence uncertain consequence of Relativity but its fundamental hypothesis, that we can't notice locally a gravity field uniform enough. That is, if some unknown matter somewhere in our galaxy attracts our Sun, the Earth with the LHC, an initially immobile particle and the speeding particle, all get an identical contribution to their path from that unknown matter. If anyone believes the speeding particle is less accelerated by gravity, he's kindly invited to propose his theory to replace Relativity. The kinetic energy of the electrons makes 2ppm of the mass of a lead atom but 0.1ppm of a carbon atom. I hope everyone agrees that kinetic energy increases the inert mass of objects: that's why particle accelerators must be big. But the identity of inertial mass, active mass (which creates gravity) and passive mass (which experiences gravity) has been proven identical to 10-13 experimentally. So kinetic energy creates and experiences gravitation. That's why lead and carbon atoms fall to Earth with the same acceleration. Swansont, this is a persistent misconception from you. Miscalling it "established physics" won't help you admit it and then debug it. Every single experiment up to now tells that kinetic energy experiences gravitation, and to a very high accuracy: that's the "equivalence principle". I know that, for some reason, some people here believe a thesis more easily when other people than I explain it, so here are few external sources: arxiv.org Wikipedia or just google for "inertial mass" "passive mass" As for the massive star not collapsing to a black hole for a speedy observer: I've heard of models for immobile black holes and spinning black holes, not for speeding black holes. Many subtleties can happen, gammas apply or not and to what, that make the argument uncertain. Even if the increased mass of the star didn't create a stronger gravitation, the star is flatter in the speed direction hence denser, which would increase the gravitation field and should let some stars collapse. But this doesn't happen, so clearly the models known for immobile black holes don't apply as is to speeding ones.
swansont Posted July 10, 2021 Posted July 10, 2021 1 hour ago, Enthalpy said: Wow, so much interest for this topic! Here a few answers - I probably forget some interrogations. Much is also in the early messages. My preferred particles are (since the beginning) protons at the LHC, as they give an easier wavelength. Other particles are interesting too: Pb ions at the LHC, electrons at the future CLIC. I had also thought at alphas at the LHC. My central idea is that horizontal near-light speed lets the falling particle emit more light than if dropped from zero speed, just as the synchrotron radiation increases sharply with the particle energy. This gives a chance to observe the effect. And you’ve been asked to support this claim, which you have yet to do. 1 hour ago, Enthalpy said: The protons fall freely over 30m horizontal path. In the 100ns, they lose 50fm altitude, not 30m. The energy a proton may radiate comes from its kinetic energy, just like when magnets deflect the proton. So this is what 30m freefall meant. OK, analyze this in the rest frame of the proton. The physics has to work in all frames. 1 hour ago, Enthalpy said: The kinetic energy of the electrons makes 2ppm of the mass of a lead atom but 0.1ppm of a carbon atom. citation, or calculation, please. 1 hour ago, Enthalpy said: I hope everyone agrees that kinetic energy increases the inert mass of objects: that's why particle accelerators must be big. Rest mass is unaffected. One reason accelerators are big because the radiation charged particles emit depends on the acceleration, and a smaller radius means larger acceleration at a given speed (a=v^2/r; multiply by gamma for relativistic systems). Once the radiated energy gets large enough you can’t get them moving faster. 1 hour ago, Enthalpy said: But the identity of inertial mass, active mass (which creates gravity) and passive mass (which experiences gravity) has been proven identical to 10-13 experimentally. So kinetic energy creates and experiences gravitation. That's why lead and carbon atoms fall to Earth with the same acceleration. Repeating this doesn’t make it true. There are mass and momentum terms in the GR tensor equation, and a sign difference in the metric. Basically the KE contribution is subtracted out. Relative motion will not turn a star into a black hole. 1 hour ago, Enthalpy said: Swansont, this is a persistent misconception from you. Miscalling it "established physics" won't help you admit it and then debug it. Every single experiment up to now tells that kinetic energy experiences gravitation, and to a very high accuracy: that's the "equivalence principle". I’ve never seen a formulation of the EP that states this. 1 hour ago, Enthalpy said: I know that, for some reason, some people here believe a thesis more easily when other people than I explain it, so here are few external sources: arxiv.org Wikipedia or just google for "inertial mass" "passive mass" These don’t, so far as I see, mention linear motion increasing gravitation. I’m not going to slog through citations; you need to do better providing support. 1 hour ago, Enthalpy said: As for the massive star not collapsing to a black hole for a speedy observer: I've heard of models for immobile black holes and spinning black holes, not for speeding black holes. And for good reason. 1 hour ago, Enthalpy said: Many subtleties can happen, gammas apply or not and to what, that make the argument uncertain. Even if the increased mass of the star didn't create a stronger gravitation, the star is flatter in the speed direction hence denser, which would increase the gravitation field and should let some stars collapse. But this doesn't happen, so clearly the models known for immobile black holes don't apply as is to speeding ones. The changing shape from length contraction is a wholly different argument, and “should let them collapse” is an unsupported assertion.
Enthalpy Posted July 12, 2021 Author Posted July 12, 2021 On 7/10/2021 at 11:29 PM, swansont said: Quote I hope everyone agrees that kinetic energy increases the inert mass of objects: that's why particle accelerators must be big. Rest mass is unaffected. Inertia depends on rest mass + kinetic energy. This is the basic and historically first reason to build accelerators bigger, as the bending magnets can't keep a small radius as kinetic energy increases despite the speed remains nearly constant. If you didn't grasp this, why do you troll threads about Relativity? And for your information: every single form of energy contributes to the mass, without exception. This is why you were asked during your studies to compute reaction energies by comparing the masses of reactants and products. A uranium atom contains kinetic energy, electromagnetic energy, the energy of the strong force, rest mass of some constituents. If any component didn't contribute to the mass, the computation would be wrong. This is established physics. Not you extraordinarily heterodox claim, that kinetic energy wouldn't contribute to the mass and its gravitational effects.
swansont Posted July 12, 2021 Posted July 12, 2021 2 hours ago, Enthalpy said: Inertia depends on rest mass + kinetic energy. And saying rest mass is unaffected does not contradict this. The point behind this is not why accelerators are big, it's your mistaken notion that gravity is affected by linear motion, which, of course, is relative. You are arguing that gravity will increase because some other object moves with respect to it. edit: The bottom line here is that you can do a solution in the rest frame, which must be valid, and the you can do a coordinate transform into the moving frame. Whatever happens has to be able to happen in the rest frame of the particle. 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. Quote And for your information: every single form of energy contributes to the mass, without exception. No, this is not true. Not for rest mass. Quote This is why you were asked during your studies to compute reaction energies by comparing the masses of reactants and products. A uranium atom contains kinetic energy, electromagnetic energy, the energy of the strong force, rest mass of some constituents. If any component didn't contribute to the mass, the computation would be wrong. Which is beside the point, since we are talking about the KE of the CoM. The total energy of something is E2 = m2c4 + p2c2 Any CoM motion, which would result in linear momentum, is accounted for separately from the mass. None of the items you listed result in momentum of the uranium nucleus.
Enthalpy Posted July 16, 2021 Author Posted July 16, 2021 As usual, you revert to rhetorics when your misconceptions are disproven. Rhetorics is not a part of physics. And please read the topic before re-asking information already given.
Enthalpy Posted August 22, 2021 Author Posted August 22, 2021 The deflection of protons by gravitation at the LHC is tiny, much smaller than the light cone angle, which doesn't set here the usual pulse duration. So do the usual formulas for emitted power apply, especially in the first message? I believe they do, because among the diverse computations, this one en.wiki obtains a Poynting vector and an emitted power independently of a pulse duration or spectrum. Also: protons at the LHC emit pulses about 1fs long but 0.5A beam intensity puts a proton every <1as, even closer in the bunches. Though, the emitted light cumulates E, not E*H. So why should usual models work, as they cumulate the power from allegedly independent protons? Because only the AC part of the emission propagates, carries power away and is detected as light. For R random protons whose pulses overlap, this fluctuation of E is proportional to sqrt(R) and the power is proportional to R. Signal theory also wants the random sum of independent signals to have the same spectrum as each signal, here light, while the slow deterministic modulation by bunches passes in the light. ========== I had hoped to reduce the synchrotron losses of storage rings by absorbing, diverting, retarding... at many places in a bending magnet the light emitted by one particle by so it doesn't add constructively. This one attempt looks vain presently. What might perhaps work is to reorient the light emitted near the entrance of a bending magnet to inject it near the exit so it adds destructively. An odd number of mirrors should do that. They must need curvature so the injected light mimics the divergence of the fresh one. The path must be inwards over some length so the light catches the particle up. As the particle is delocalized laterally, the reinjection point must follow the delocalisation. Is that feasible, I don't know. At least, synchrotron light is around the visible spectrum at the LHC. Proper interference may need to mirror very short pulses, easier for Pb ions. It needs also subwavelength accuracy over a decametre, as difficult as the VLT. The light of pairs of successive magnets can also cancel out. Light cancelling out means each particle would reaccelerate itself. The width of the particle beam, of the vacuum tube, other dimensions... constrain the design. I didn't even try figures and drawings. A new design may be needed if a retrofit is impossible. Maybe the opposite if feasible too where wanted, increase the emitted light by reinjecting earlier light in phase near the particle. Something reinforces the emitted light at free-electron lasers, I don't know what, maybe that. Or is all that already abandoned? ========== Protons falling at the LHC emit waves a bit too long for quiet detectors. Helium and lead ions are worse. I suggested already to reduce the drop length and accept fewer photons easier to detect, optionally to split the drop length and have several detectors here on 19 Jun 2021 Splitting the drop length in N sections and detectors keeps the light power in /N photons with *N individual energy. Light from individual drop sections could also be sent to a common detector of same area to get more signal with the same noise, as for instance photocathodes are little sensitive to the light's direction: A fascinating option would superpose the pulses of all drop sections at the same time and place to add E rather than P, hence obtain N2 as many photons as from one of these sections. Since light from all sections is coherent, E can add. It looks "only" like a technological challenge similar to the VLT. That would be an other case where hardware downstream influences earlier photon emission, like in EPR experiments. Or is it? At least in principle, the increased photon emission can be noticed from the emitting proton's remaining energy, before the light reaches the influencing optical elements. Acting on these elements would transmit information backwards in time, so my present opinion is that coherent light addition will fail. I carefully neglected to display how light from earlier drop sections passes at later optical elements. The paths can be stacked vertically for instance. But as individual beams arrive at different angles, they may produce an interference pattern that cumulates P over the image, not E. The impossibility might manifest itself as a stubborn technological difficulty to concentrate light within the additive fringe and still transport one light path around the other optical elements, or a difficulty to concentrate the light without spreading the pulse in time. Very thin transparent elements around the proton beam seem impractical. Synchrotron light seems difficult to focus without spreading the pulse over time, since light emitted earlier is broader but arrives earlier. Focussing to a ring could be better than to a spot. A sleeve mirror? Photocathodes can advantageously be big and keep a small dark current. An ironless electromagnet, or the geomagnetic field with removed shield, let emit more photons in the drop region to adjust the optics. ========== More drop segments allow to reduce the energy per nucleon and keep a good wavelength. A smaller gamma reduces identically the gravitation and the magnetic light. Insofar possible, more beam intensity and less energy have advantages: A smaller acceleration at the bending magnets emits less magnetic light, while g stays the same. The same acceleration power makes more gravitation light. The last bending magnet emits light at a longer wavelength, which the diaphragms suppress better. The protons per nucleus may address other limits. Marc Schaefer, aka Enthalpy
John Cuthber Posted August 22, 2021 Posted August 22, 2021 The way in which I expressed my point was light-hearted. However, that is no reason to ignore it. Th 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?
swansont Posted August 23, 2021 Posted August 23, 2021 On 7/16/2021 at 8:47 AM, Enthalpy said: And please read the topic before re-asking information already given. Can you link to where you showed the calculation for the synchrotron emission wavelength or frequency?
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