[evan_au]

Will the (gravitational) wave be sinusoid, or rectified sine wave?

The sudden change in direction of a rectified sinusoid produces lots of high-frequency harmonics.
- I don't see anything in the path of two orbiting neutron stars that would produce such high frequency components. So a gravitational wave can't be a rectified sine wave.
- The discontinuity of derivative (in fact, the reversal of derivative) in a rectified sine wave implies some unreal forces to reverse the acceleration of a neutron star twice every orbit. Another reason it can't be a rectified sine wave.

I expect that the amplitude of a gravitational wave will be more sin(x)^2, where x is the angle in the orbit.
- But remember that gravitational waves distort more than one dimension of space, and also distorting the passage of time, so a 1-dimensional function can't do it credit!
- And in all of these measures, there is an arbitrary additive constant that can move the entire graph up or down...


SIN_squared.png (15.71 kB . 462x241 - viewed 2225 times)

[alancalverd]

Quantum mottle is part of the basic radiography syllabus. howradiologyworks.com/x-ray-cnr/ has some examples of a everyday phenomenon. It's the reason you can't increase the sensitvity of x-ray film beyond a certain limit, which is less for pediatrics than for adults, because you lose spatial resolution as the quantum noise exceeds the image contrast.

Why can't it be explained using wave mechanism?

Photon detector screen reminds me of popcorn analogy to explain half life of radioactive materials.
In popcorn case, the energy is supplied more or less continuously, but the popping events occur discretely. Analogous with detector screen, we can think that the screen receive electromagnetic wave continuously, but the events of change in screen color/brightness occur discretely.

If you reduce the doserate sufficiently you can detect individual photon events in real time. Easier with a geiger counter or photomultiplier. We know that x and γ radiation are the same stuff, so you can replace the x-ray source with a radionuclide which we know can only emit discrete pulses, not continuous radiation, and get the same effect.

But beware of falling into your own trap of thinking that a photon is a wave or a particle. Remember that, as with physics in general,  these are mathematical models of what it does, not what it is.

[hamdani yusuf (OP)]

The sudden change in direction of a rectified sinusoid produces lots of high-frequency harmonics.

I calculated the gravitational force from two identical objects orbiting each other.

[hamdani yusuf (OP)]

If you reduce the doserate sufficiently you can detect individual photon events in real time. Easier with a geiger counter or photomultiplier. We know that x and γ radiation are the same stuff, so you can replace the x-ray source with a radionuclide which we know can only emit discrete pulses, not continuous radiation, and get the same effect.

But beware of falling into your own trap of thinking that a photon is a wave or a particle. Remember that, as with physics in general,  these are mathematical models of what it does, not what it is.

Waves don't have to be continuous. It can also last for a few cycles only.

[Eternal Student]

Hi.

I calculated the gravitational force from two identical objects orbiting each other.

   I'm really not sure what that graph shows so I can't comment on it.  This is a shame because it looks like you've spent a lot of time on it.  Would you like to describe what it is?

Evan_au also made this comment earlier...
   

But remember that gravitational waves distort more than one dimension of space, and also distorting the passage of time, so a 1-dimensional function can't do it credit!

    I was considering replying to Evan_au earlier about this.   In essence gravitational waves are not like most other waves. 

[hamdani yusuf (OP)]

Imagine a twin star orbiting each other on xy plane, radius 1 light second from barycenter, which is at coordinate 0, 0.
An observer is located at coordinate 2, 0.
The graph is a plot in time.
(x1, y1) is position of first star.
(x2,  y2) is position of second star.
g1 is gravitational force from first star to observer.
g2 is gravitational force from second star to observer.
I missed something here. The y component of the forces are cancelling out. The total force should add x component only. Now The picture is corrected.

[alancalverd]

Waves don't have to be continuous. It can also last for a few cycles only.

Nice image, but if you make a fourier analysis you will see that it contains all sorts of harmonics, and a single pulse has an infinite number of harmonics, whereas a single photon only has one associated frequency 'cos it isn't a wave (or a particle).

[hamdani yusuf (OP)]

Waves don't have to be continuous. It can also last for a few cycles only.

Nice image, but if you make a fourier analysis you will see that it contains all sorts of harmonics, and a single pulse has an infinite number of harmonics, whereas a single photon only has one associated frequency 'cos it isn't a wave (or a particle).

Equation [9] states that the Fourier Transform of the Gaussian is the Gaussian! The Fourier Transform operation returns exactly what it started with. This is a very special result in Fourier Transform theory.

"TheFourierTransform.com - Fourier Transform of the Gaussian" https://www.thefouriertransform.com/pairs/gaussian.php

[Eternal Student]

The graph is a plot in time......

....  and something to do with force on an observer at a fixed poisiton (1.5, 0).

    O.K.  Thanks for explaining that Hamdani.   Then it's a nice graph.  I thought the previous discussions were about gravitational waves and that graph might have been something to represent this.   Sorry for my confusion but thanks for taking the time to explain it.

[Bored chemist]

Waves don't have to be continuous. It can also last for a few cycles only.

Nice image, but if you make a fourier analysis you will see that it contains all sorts of harmonics, and a single pulse has an infinite number of harmonics, whereas a single photon only has one associated frequency 'cos it isn't a wave (or a particle).

That's why people use these
https://en.wikipedia.org/wiki/Wavelet

[alancalverd]

...and if you read the section on fourier transforms of wavelets you can see why they don't reflect the spectral characteristics of a single photon.

The simplest proof that a photon isn't a particle or a wave is to look at the diffraction pattern of a monochromatic beam. It's the same color as the original beam.

Now we know that we can generate a diffraction pattern when only one photon is present at a time (the classic double slit experiment), which is why we can model the photon as a wave..

If the single photon split and reappeared simultaneously in several positions, conservation of energy demands that the color of the diffraction pattern would be different. If not, we could focus the diffracted light back into a single beam with more energy than the incoming photon! So whilst wavelet analysis predicts the shape of the diffraction pattern, it can't predict the color.   

Therefore modelling a photon as a wave proves that it is an inadequate model. But modelling it as a particle doesn't predict a diffraction pattern at all, so that is also an inadequate model..     

[hamdani yusuf (OP)]

If the photon split and reappeared simultaneously in several positions, conservation of energy demands that the color of the diffraction pattern would be different. If not, we could focus the diffracted light back into a single beam with more energy than the incoming photon! So whilst wavelet analysis predicts the shape of the diffraction pattern, it can't predict the color.   

How do you split a single photon?

[alancalverd]

Precisely the point! If it's a particle, it can't split so it must go through one slit or the other, without losing energy. If it's a wave, it can split and pass though both slits, but the energy at any point on the receiver will be less than in the primary beam. What actually happens? It appears to go though both slits and arrives with the same energy as it set out!

[hamdani yusuf (OP)]

Precisely the point! If it's a particle, it can't split so it must go through one slit or the other, without losing energy. If it's a wave, it can split and pass though both slits, but the energy at any point on the receiver will be less than in the primary beam. What actually happens? It appears to go though both slits and arrives with the same energy as it set out!

Have you watched the video in my reply #56? Using a photon as a particle analogy, we are forced to conclude that the photon must interact with its own duplicate which has already arrived at the screen a few moments ago. Moreover, we must assume that a macroscopic beam splitter made of large number of atoms can consistently split each photon to intended directions without much loss due to scattering randomly. It seems like the results become significantly randomized in the screen only. This is in line with the popcorn analogy which I mentioned earlier.

[alancalverd]

Which illustrates the dangers of analogy - you end up with a menagerie of quasiparticles that travel faster than light and get randomised by a photographic film but not by a scintillation counter that does exactly the same thing, but faster.

My approach to teaching physics is to avoid analogies which I find confusing at best and positively misleading at worst. Start with what you observe and construct a purely abstract mathematical model that produces the observed result. If you need to use two models, just accept that they are both approximations. All that matters is that they should be consistent with other models (such as conservation of energy, momentum, mass....) and adequately predictive.

The "a few moments ago" is confusing because students know (or should know, if they stayed awake in the previous lecture) that a single photon is unlikely to produce a spot on a photographic film. You need two visible photons to strike a halide grain within a fairly short time in order to produce blackening. This phenomenon is known as reciprocity failure - the apparent sensitivity of film decreases at very low (and very high) photon intensities. Astronomers used to "pre-fog" their photographic plates to lift the sensitivity above the RF level, and radiographers exploit it to mask noise and produce clean white bones. 

[Bored chemist]

If the single photon split and reappeared simultaneously in several positions,

It doesn't.
It forms a probability density that is higher than zero, but less than one simultaneously in several positions.
It's sort of more like a quarter of the photon is in each of 4 place, but that's not the whole story either.

[hamdani yusuf (OP)]

The "a few moments ago" is confusing because students know (or should know, if they stayed awake in the previous lecture) that a single photon is unlikely to produce a spot on a photographic film. You need two visible photons to strike a halide grain within a fairly short time in order to produce blackening.

What's your reference to two photons requirement?

This breakdown in the usual tradeoff between aperture and shutter speed is known as reciprocity failure. Each different film type has a different response at low light levels. Some films are very susceptible to reciprocity failure, and others much less so. Some films that are very light sensitive at normal illumination levels and normal exposure times lose much of their sensitivity at low light levels, becoming effectively "slow" films for long exposures. Conversely some films that are "slow" under normal exposure duration retain their light sensitivity better at low light levels.

https://en.wikipedia.org/wiki/Reciprocity_(photography)#Reciprocity_failure

[alancalverd]

Check the paragraph before the one you just quoted!

Unlike camera film, which may stay in the camera for hours or years before being developed, "x-ray" film is processed within minutes of exposure so doesn't suffer from latent image fade but still needs at least two photons to switch on a halide grain.  "X-ray" in inverted commas because photographic film (a) is actually not very sensitive to x-rays and (b) doesn't display reciprocity failure!  In practice we use double-emulsion film sandwiched between fluorescent screens that generate several visible (or low energy x-ray) photons for each incoming high energy photon. 

[hamdani yusuf (OP)]

Which illustrates the dangers of analogy - you end up with a menagerie of quasiparticles that travel faster than light and get randomised by a photographic film but not by a scintillation counter that does exactly the same thing, but faster.

That's when you use particle analogy. But if you use wave analogy as suggested by the video, the result can be understood less surprisingly.

I have an idea to test the validity of the wave explanation. Let's make the long path of the interfering light beam much longer than the short path, so there's a significant time difference in their arrival at the screen. To balance the intensity, the short path can be reduced further using a filter.

First, the light source is turned on continuously. The single photon production is done using attenuating filter alone. As demonstrated in the video, the screen shows interference pattern.

Second the light source is turned on in short pulses. The pulse width is shorter than time difference between the paths of the light beams. The silence period between the pulses should be randomized to minimize the effect from resonance or echo. If the wave explanation is correct, then the interference pattern should not be observed significantly.

[hamdani yusuf (OP)]

Check the paragraph before the one you just quoted!

You mean this one?

At very low light levels, film is less responsive. Light can be considered to be a stream of discrete photons, and a light-sensitive emulsion is composed of discrete light-sensitive grains, usually silver halide crystals. Each grain must absorb a certain number of photons in order for the light-driven reaction to occur and the latent image to form. In particular, if the surface of the silver halide crystal has a cluster of approximately four or more reduced silver atoms, resulting from absorption of a sufficient number of photons (usually a few dozen photons are required), it is rendered developable. At low light levels, i.e. few photons per unit time, photons impinge upon each grain relatively infrequently; if the four photons required arrive over a long enough interval, the partial change due to the first one or two is not stable enough to survive before enough photons arrive to make a permanent latent image center.

It doesn't specifically say that two photons is needed. It says a few dozen instead.
More modern detectors might be able to detect absorbtion of a single photon.