Electric universe theories here.

[GeeMack]

I'm curious how many of your fellow EU antagonists share that viewpoint?

I'm curious how many of your fellow science and math antagonists share the viewpoint that you can see stuff over 4000 kilometers below the Sun's photosphere by looking at a graphical representation of a series of mathematical computations that was created using data acquired several thousand kilometers above the photosphere. My bet would be that, even though being in your league they'd misunderstand science and be scared of math, there aren't any!

When do you figure to take down that LMSAL running difference graph from your web site, Michael? We've already determined, those of us who are sane, intelligent, and not delusional have anyway, that it doesn't show the nonsense that you once claimed it showed.

You're a riot, kid. I'm still waiting for the day you admit you're just a sick troll, and you've been keeping up this ridiculous stupid scientist-wannabe act for so many years just to have a good joke on everyone.

 

[Michael Mozina]

When do you figure to take down that LMSAL running difference graph from your web site, Michael?

I don't have any running difference "graphs" on my website, only running difference "images". In RD "images" we can observe real things like real stars, and real comets, and real planets. When are you going to figure out that it's not a graph and you can see real objects in these images? Flying stuff? Ya, flying stuff too!

And for the record, was that one loop or many?

 

[sol invictus]

Only thermal radiation, unless you've got some other light source.

Thermal, in a vacuum? At what temperature?

No, you'd only get thermal radiation if you did something like accelerate your detector at a constant rate. If it's inertial you'll just get quantum fluctuations with a spectrum determined by the characteristics of your detector and interactions of whatever you're detecting.

I don't see how it could. What kind of photodetector would you use?

Well, let's see. A static background B field conserves energy but not momentum (of charges passing through it), classically. So you'd be best off with something that measures the lateral acceleration of a charge, like (say) the scattering angle as an electron passes through the field. In a very weak field, you'd see that the electrons (all else being equal) change direction in discrete jumps as they pass through the field - or more precisely, the Fourier transform of a histogram of the angles they scatter through would be related to the FT of the field, but with the effects of discrete photon numbers visible (because the particles absorb integral numbers of photons from the field). For a strong field that discreteness is of course very hard to see, but it must be there nonetheless.

 

[Michael Mozina]

It starts as electricity and ends as heat. Or really kinetic energy(the field) added to mass. And since blackbody is the property of solid matter and solid matter does not exist at 6000K I think that extending the idea of blackbody to 6000K is in error. Unless it is taken into consideration that the spectrum is really an solid arc/plasma lamp spectrum.

You hit the nail on the head with that comment IMO. The smaller discharges (under the photosphere) tend to serve the role of the solid plasma lamp as these discharge filaments ionize heavier elements in loops all along the surface. Some of those ionized (and dusty) heavier elements make their way into the atmosphere and photosphere and that is what creates a more "white light' spectrum in the mostly neon photosphere.

The overall energy release of a Birkeland model would need to jive with standard theory, but in almost every other respect, it would work "very differently" than standard theory. There will be a "flow pattern" of particles moving away from the surface, and through the various plasma layers in the atmosphere. I would expect mass separation and thermoclines to form between and in plasma double layers, but the bulk of the heat is carried away, and contained in the flowing electrons and ions, not simply in the surface features.

I would also add here that nobody is claiming that the surface is "solid iron", it simply has a "crust". The only temperature that is 'critical' is the melting point of carbon which is closer to 4000K than 6000K. Anything below 4000K and solids could and would begin to form. The other debate over "optical depth" is really moot IMO since were simply talking about "how far" below the photosphere the surface might be located and that video of the flare demonstrates that it *MUST* originate far below the photosphere.

 

[Ziggurat]

Well, let's see. A static background B field conserves energy but not momentum (of charges passing through it), classically. So you'd be best off with something that measures the lateral acceleration of a charge, like (say) the scattering angle as an electron passes through the field. In a very weak field, you'd see that the electrons (all else being equal) change direction in discrete jumps as they pass through the field - or more precisely, the Fourier transform of a histogram of the angles they scatter through would be related to the FT of the field, but with the effects of discrete photon numbers visible (because the particles absorb integral numbers of photons from the field). For a strong field that discreteness is of course very hard to see, but it must be there nonetheless.

I don't think so. An electron moving in a magnetic field will emit synchrotron radiation, whose frequency (and hence momentum) is characterized by the field strength and not by the field size (which characterizes the photons forming the field). This is what changes the momentum of the electron, not absorption of photons in the field, whose frequency (and momentum) are continuous anyways in the case of an infinite volume. And even in a finite volume, the emitted radiation doesn't need to form standing waves, so the momentum transfer need not be discrete either.

 

[Ziggurat]

You hit the nail on the head with that comment IMO.

He summarized his misconceptions quite nicely, I'll give him that much. But he's still under the delusion (and so, apparently, are you) that only solids can act as blackbodies. This has already been falsified, Michael.

I would also add here that nobody is claiming that the surface is "solid iron", it simply has a "crust". The only temperature that is 'critical' is the melting point of carbon which is closer to 4000K than 6000K. Anything below 4000K and solids could and would begin to form.

And yet, 4000 K is still thermodynamically impossible. Your cathode refrigeration model doesn't work. Which is why you can't put any numbers on it.

 

[sol invictus]

I don't think so. An electron moving in a magnetic field will emit synchrotron radiation, whose frequency (and hence momentum) is characterized by the field strength and not by the field size (which characterizes the photons forming the field).

But I'm not talking about the radiation the charge emits - I'm talking about the radiation it absorbs from the field.

This is what changes the momentum of the electron, not absorption of photons in the field, whose frequency (and momentum) are continuous anyways in the case of an infinite volume.

It's synchrotron radiation which causes the acceleration of the electron, which caused the synchrotron radiation? No - it's the absorption of photons from the B field which causes the acceleration, which in turn can lead to synchrotron radiation. And the effect of the field is entirely accounted for by individual processes in which a field photon interacts with the electron - and the spectrum of those photons is determined by the field profile (and their quantity by its intensity).

One way to understand this is by conservation of momentum. A static B field with some spatial profile violates conservation of momentum (ignoring the backreaction of the particle on the field), because it violates translation invariance. How much momentum can it transfer because of that? One way to find out is to take its FT (come to think of it the FT of the vector potential is better, because it's more simply related to the momentum transfer).

And even in a finite volume, the emitted radiation doesn't need to form standing waves, so the momentum transfer need not be discrete either.

You seem to be mixing two things here - the momentum transfer due to radiation with the momentum transfer due to absorption. They're not the same. And even the absorbed radiation doesn't have a discrete FT - it's not a plane wave, it's a localized lump.

 

[Ziggurat]

You seem to be mixing two things here - the momentum transfer due to radiation with the momentum transfer due to absorption. They're not the same. And even the absorbed radiation doesn't have a discrete FT - it's not a plane wave, it's a localized lump.

If it's not a plane wave, then it doesn't have a well quantized momentum. Which means each such photon doesn't transfer discrete momentum either. So you can't measure the number of photons it absorbs by counting discrete changes in the electron's momentum, because they won't be there.

 

[sol invictus]

If it's not a plane wave, then it doesn't have a well quantized momentum.

Each photon carries a certain momentum determined by its frequency. The probability of absorbing a photon of a particular momentum is related to the FT of the classical field, which is a smooth function. By doing the experiment many times, you'll be able to reconstruct that function, and see the effects of the quantization (for example, the FT will fall off like a power high momentum, but quantization will mean all occupation numbers will be zero or exponentially small).

Which means each such photon doesn't transfer discrete momentum either.

Oh, but it does. It's the probability p(k) of absorbing a photon with momentum k that's continuous, not the k of any given photon.

So you can't measure the number of photons it absorbs by counting discrete changes in the electron's momentum, because they won't be there.

The change for any given electron is obviously discrete. A histogram of the changes for many electrons will be continuous in the limit of an infinite number of repetitions - is that what you mean? - but that's just as expected.

But I'm not sure what we're disagreeing over. Surely you agree that QED (plus the rest of the standard model) is the correct theory for electromagnetic interactions. Surely you agree that all QED processes arise from the cumulative effects of the only interaction that exists in the theory, which is the absorption or emission of a single photon by a charged particle.

But then the effects of a magnetic field on a particle can be understood that way - by the absorption of the photons making up that field. And the spectrum of those photons must be closely related to the FT of the field (that would be easy enough to check - just write down the Feynman rules in momentum space, and think about how the background field enters).

 

[Ziggurat]

Oh, but it does. It's the probability p(k) of absorbing a photon with momentum k that's continuous, not the k of any given photon.

But if you have a continuous distribution of momenta that can be absorbed, then the momentum change in the electron is continuous. There's no way to tell if you've absorbed four photons of one frequency, five of another, or some mix. All you will be probing is the magnetic field strength. Which is not enough to determine how many photons are in the field.

But I'm not sure what we're disagreeing over.

I'm saying that the only way to determine how many photons make up the field is to measure the field strength AND size, and then take the Fourier transform, because there's no way to directly measure those photons.

But then the effects of a magnetic field on a particle can be understood that way - by the absorption of the photons making up that field. And the spectrum of those photons must be closely related to the FT of the field (that would be easy enough to check - just write down the Feynman rules in momentum space, and think about how the background field enters).

The FT is a non-local property of the field. But the measurement you propose is essentially a local measurement. So how can it be sensitive to the differences? I don't think it can. I don't think it matters that one can do the calculations in terms of individual photons: I don't think you can extract the number of photons in the field from the measurement, because multiple scenarios will produce the same measured result.

 

[Michael Mozina]

But I'm not sure what we're disagreeing over. Surely you agree that QED (plus the rest of the standard model) is the correct theory for electromagnetic interactions. Surely you agree that all QED processes arise from the cumulative effects of the only interaction that exists in the theory, which is the absorption or emission of a single photon by a charged particle.

The problem is that I brought up the idea, not you. If you had said that photons were the carrier particle of the EM field, this conversation would probably be over already, and probably never questioned in the first place. I for one however have benefited from the discussion, so by all means, do continue.

 

[Ziggurat]

The problem is that I brought up the idea, not you. If you had said that photons were the carrier particle of the EM field, this conversation would probably be over already, and probably never questioned in the first place.

Don't flatter yourself. And the exchange we're having isn't about whether or not photons are the carrier particle for EM fields.

 

[sol invictus]

But if you have a continuous distribution of momenta that can be absorbed, then the momentum change in the electron is continuous.

You mean, the possible momentum changes? Then yes.

There's no way to tell if you've absorbed four photons of one frequency, five of another, or some mix.

I don't think that's the case. For one thing, you could use something more sophisticated than an electron - something with some structure, that's more sensitive to some frequencies than others. For another you can vary the field profile and strength, and do many experiments. And in reality you have to allow some input from theory to interpret the results, and the theory predicts a certain spectrum of momentum changes due to photon absorption, and you can go out and check the predictions of the theory.

All you will be probing is the magnetic field strength. Which is not enough to determine how many photons are in the field.

Not just the strength, because even classically the profile of the field (say the volume where it's non-zero) affects the momentum transferred. But yes, the strength is part of it as well.

I'm saying that the only way to determine how many photons make up the field is to measure the field strength AND size, and then take the Fourier transform, because there's no way to directly measure those photons.

So you agree (?) I have a theory in which I can assign a finite number of photons to the field configuration, and you agree that every single prediction of that theory - calculated using discrete interactions with finite numbers of photons - is correct, but you don't agree there's a way to measure that number or be sure it's accurate. If that's your position, I can't argue with it (nor am I very interested in trying).

The FT is a non-local property of the field. But the measurement you propose is essentially a local measurement.

Why? I can shoot electrons through from any angle, through the whole thing, through part of it, etc. How is that local?

I don't think you can extract the number of photons in the field from the measurement, because multiple scenarios will produce the same measured result.

Well, there is a significant subtlety here regarding infrared divergences. Any time you have massless particles like photons, the number of "soft" photons - photons with very low momentum and energy - gets hard to define. But that's a problem that's solved the minute you consider a detector with finite energy resolution, it goes away if the universe were huge but finite (which tells you right away it can't matter), and it's there even when the field is precisely zero. So I don't think it prevents one from answering this question, although I'll agree my argument is somewhat vulnerable on that point.

As an example more topical to this thread - if I asked you how many photons there are in the sun, or how many it emits per second, would you agree that's a sensible question with a finite answer?

 

[Ziggurat]

I don't think that's the case. For one thing, you could use something more sophisticated than an electron - something with some structure, that's more sensitive to some frequencies than others.

In other words, try to use multiple electrons. I don't see how that will change anything fundamentally. Nor is it clear what such a device might be like, nor why frequency selectivity will matter when the field is, in effect, static.

For another you can vary the field profile and strength, and do many experiments.

And if it's a local measurement, I don't think field profile differences will have any effect on your measurement.

And in reality you have to allow some input from theory to interpret the results, and the theory predicts a certain spectrum of momentum changes due to photon absorption, and you can go out and check the predictions of the theory.

That's not enough. If the theory will give you the same predicted experimental results for multiple field profiles (and I think it will), then you cannot determine the number of photons from the measurement.

So you agree (?) I have a theory in which I can assign a finite number of photons to the field configuration, and you agree that every single prediction of that theory - calculated using discrete interactions with finite numbers of photons - is correct, but you don't agree there's a way to measure that number or be sure it's accurate. If that's your position, I can't argue with it (nor am I very interested in trying).

You're missing a step, Sol. It is not enough to predict the experimental results, because what's in question isn't the theory. What's in question is the number of photons. And in order for the theory to calculate this, the theory needs the field profiles as an input. If I'm right that multiple field profiles will produce the same experimental outcome, then local experiments cannot determine the number of photons, because different field profiles will have different numbers of photons.

Why? I can shoot electrons through from any angle, through the whole thing, through part of it, etc. How is that local?

You can do a local measurement in multiple places, but each measurement is still local. What happens to the momentum of the electron doesn't depend upon the field in places the electron never goes. That's the sense in which it's local. But the Fourier transform, and hence the number of photons in the field, does depend upon the entire field profile. And if you want to do multiple measurements in multiple places to try to reconstruct all this stuff, well, all you're really doing is measuring the field itself, along with its profile, which you still need to take the Fourier transform of in order to get an answer for the number of photons. But the detector itself isn't counting photons. A single local measurement will not tell you the number of photons, even in a probabilistic sense.

As an example more topical to this thread - if I asked you how many photons there are in the sun, or how many it emits per second, would you agree that's a sensible question with a finite answer?

Yes.

 

[GeeMack]

I don't have any running difference "graphs" on my website, only running difference "images". In RD "images" we can observe real things like real stars, and real comets, and real planets. When are you going to figure out that it's not a graph and you can see real objects in these images? Flying stuff? Ya, flying stuff too!

First, a running difference image is a graph. That has been shown time and again in this thread and on other forums. Each pixel graphically represents the difference in the brightness values between corresponding pixels in a pair of source images. And that's all. Matching corresponding bright pixels in the source images will create the exact same output value as matching corresponding dark pixels. There are no actual objects in a running difference image. The method of constructing the output makes that impossible. That is a fact and has not been refuted. Your unsubstantiated assertion to the contrary does not constitute a refutation in the real world of sane people and legitimate science.

Second, the LMSAL running difference image you tout as evidence for your insane fantasy was created using data obtained several thousand kilometers away from where you claim your physically impossible solid surface exists. That has also been clearly shown in this thread and in discussions on other forums. So not only is it physically impossible for that surface to exist, it is impossible that your interpretation of the graphic is correct. That is also a fact that has not been refuted. And again, your simple assertion to the contrary is not a valid refutation in the world of legitimate science and sane, intelligent people.

Third, even if we disregard the fact that a running difference graph can't possibly show any real things, and the fact that it is impossible to see anything thousands of kilometers from where the source data was obtained, there are a few thousand kilometers of opaque plasma between that location and where you claim to see that physically impossible surface. That has also been clearly shown in this thread and in many others on other forums. You can't see through that opaque layer. That also makes it impossible for your interpretation of the graphic to be correct. There is another fact which has not been refuted. Your repeated assertion that things are otherwise does not amount to a refutation in the eyes of legitimate science and sane, rational people.

Now, we know the above to be facts. Yet rather than address those facts, you choose to quibble with my using the term "graph" to describe a graph?

​

 

[sol invictus]

And if it's a local measurement, I don't think field profile differences will have any effect on your measurement.

I'm confused. If I shoot an electron through a B field, the momentum transferred will depend on a line integral of the field dotted into the electron's velocity along its path. That obviously depends on the field profile as well as its overall magnitude, and it's not very local. And by doing many such experiments, shooting from different angles etc., I can certainly reconstruct the full profile. And I can also reconstruct the number of photons that must be in there - it's like reconstructing the number of marbles in a jar by sending in some probe that bounces around for a while and comes out, and then doing it over and over again. Sure, a single such measurement isn't enough, and it might be hard to distinguish one bounce from two smaller bounces. But do it enough times, with enough different probes, and be clever enough, and you can count them all (that's more or less how vision works, actually).

That's not enough. If the theory will give you the same predicted experimental results for multiple field profiles (and I think it will), then you cannot determine the number of photons from the measurement.

Can you give an example of two field profiles that would have different numbers of photons (according to me) that will give the same results for all such experiments? Hell, even forgetting about having different numbers of photons? If the profiles are different in any way, even my simple electron scattering experiments are certainly going to detect it.

You can do a local measurement in multiple places, but each measurement is still local. What happens to the momentum of the electron doesn't depend upon the field in places the electron never goes.

You'll find that all works out fine when you do the calculation. For example if the field is very extended along some axis and very thin along another, and the electron flies through in the thin direction, the low momentum part of the FT coming from the extended direction won't affect it much, that's true. But that works just as well in momentum space as it does in position space, as I'm sure you know.

And if you want to do multiple measurements in multiple places to try to reconstruct all this stuff, well, all you're really doing is measuring the field itself, along with its profile, which you still need to take the Fourier transform of in order to get an answer for the number of photons. But the detector itself isn't counting photons. A single local measurement will not tell you the number of photons, even in a probabilistic sense.

Who said anything about a single local measurement? You'd need lots of these experiments, and they're really not very local in any sense I can think of.

Yes.

Then maybe we should discuss that instead. What is it about the sun, as opposed to a static magnetic field, that makes you think the number of photons in it is well-defined?

 

[Ziggurat]

I'm confused. If I shoot an electron through a B field, the momentum transferred will depend on a line integral of the field dotted into the electron's velocity along its path. That obviously depends on the field profile as well as its overall magnitude, and it's not very local.

The momentum transfer only depends upon the field along the path. But the Fourier transform depends on more than the path, it depends upon the field everywhere.

And by doing many such experiments, shooting from different angles etc., I can certainly reconstruct the full profile.

Yes. You are thereby measuring the field strength and profile, and you can use that to take a Fourier transform and deduce the number of photons that the field is made of. But at no point can you take one of your measurements and say, "Ah, this means that there are X photons in the field". It doesn't work that way. There's no way to get the number of photons except by taking the Fourier transform of the field.

Sure, a single such measurement isn't enough, and it might be hard to distinguish one bounce from two smaller bounces.

Hard? I think it's impossible.

Can you give an example of two field profiles that would have different numbers of photons (according to me) that will give the same results for all such experiments?

All I need is for the fields to match along the path of the electron. If they differ in other locations, the experiment will produce the same outcome, but the Fourier transforms will not be the same, and the number of photons in the field will not be the same.

If the profiles are different in any way, even my simple electron scattering experiments are certainly going to detect it.

Yes, if you move your scattering experiment to different locations. In which case, you're doing exactly what I said before: you're measuring the field strength and profile, and reconstructing a Fourier transform from this. Your individual measurements are never counting photons.

Then maybe we should discuss that instead. What is it about the sun, as opposed to a static magnetic field, that makes you think the number of photons in it is well-defined?

Well, for starters the sun has a reasonably well-defined size. A 1 Tesla field does not have any defined size - that information was not given.

Furthermore, while it is possible to treat static macroscopic fields as a collection of photons, it's a bloody stupid way to treat the problem because it's needlessly complicated. But treating solar radiation that way doesn't have the same problem. The radiative aspects of it are the characteristics of primary interest to begin with.

 

[sol invictus]

Yes, if you move your scattering experiment to different locations. In which case, you're doing exactly what I said before: you're measuring the field strength and profile, and reconstructing a Fourier transform from this. Your individual measurements are never counting photons.

Well, as far as I can tell your position is identical to the one I said I wasn't interested in debating: sure. you have a theory that water is made of molecules, and your theory correctly predicts every aspect of water we've ever measured, but you can't really count the number of molecules in a drop even approximately.

Well, for starters the sun has a reasonably well-defined size. A 1 Tesla field does not have any defined size - that information was not given.

That's why I added that information in my very first post on this topic.

Furthermore, while it is possible to treat static macroscopic fields as a collection of photons, it's a bloody stupid way to treat the problem because it's needlessly complicated.

For most purposes, yes - but not for all. For example if you shine a laser beam (or a single photon for that matter) through a magnetic field, each photon in that beam will have a chance of scattering off the field due to the Euler-Heisenberg interaction. The easiest way to treat that is to draw Feynman diagrams in which the effect of the field arises from field photons interacting with the laser photons.

 

[Ziggurat]

For most purposes, yes - but not for all. For example if you shine a laser beam through a magnetic field, each photon in that beam will have a chance of scattering due to the Euler-Heisenberg interaction. The easiest way to treat that is to draw Feynman diagrams in which the effect of the field arises from field photons interacting with the laser photons.

That may be the easiest way to conceptualize the problem, but is that really the easiest way to calculate it? We're venturing into territory I'm not familiar with, but my intuition suggests that the final answer for the probability of a given laser photon scattering should just depend on some sort of path integral of the magnetic field along the beam's path. It would seem quite strange to me that you'd be better off taking the Fourier transform of the whole field in order to actually figure out how many photons you've got in the field.

 

[sol invictus]

That may be the easiest way to conceptualize the problem, but is that really the easiest way to calculate it?

It's pretty easy, at least if the field has a simple profile. If not it's hard, but so is any other method.

We're venturing into territory I'm not familiar with, but my intuition suggests that the final answer for the probability of a given laser photon scattering should just depend on some sort of path integral of the magnetic field along the beam's path.

Your intuition is correct to the extent that you could write an n-loop effective action, which will include the most important piece of the interaction that matters here, and then treat the B field classically as a source in that effective action. So long as the field is sufficiently classical that's a good approximation, but it remains an approximation.

It would seem quite strange to me that you'd be better off taking the Fourier transform of the whole field in order to actually figure out how many photons you've got in the field.

You don't need to figure out that number at any step - but if you work in momentum space (which is usually easier, because propagators are simple in momentum space) you do need to take the FT of the field.