Tired Light

If you check his website you will see that he is just expressing himself in an odd way and that you are missunderstanding his actual point.When he says per cubic metre he actually means 'n', which means number of free electrons. This 'n' should use the same lenght unit as used elsewhere in the formula, so if you use m^3, you use m^-3, if you use ft^3, you use ft^-3.I'm not sure if there is a previously established number for 'n', but Lyndon sets this as exactly 1 per cubic metre.On his webpage, he says that as a formula this should be expressed as H = 2nhr/m. Note that the 2 in there is probably a typo on his part, but I copied and pasted the formula directly from his webpage.This message has been edited by Melchior, 03-21-2005 12:34 PM

Sure you are. It only requires multiplication and division. I'll take you through what's in Message 123 step by step.Ashmore's Paradox is that the Hubble Constant of 2.06x10^-18s^-1 (this is Ashmore's value for the Hubble Constant, not any accepted value, but that's not important for now) is very close to the value hr/m per unit area that he calculates as 2.05x10^-18s^-1, and that this is no coincidence, that it tells us something about the nature of the universe, like that it is not expanding.Ashmore carries out his calculation in units of meters: The math is very simple. Get out your calculator. First we'll just do the math with the numbers. Multiple 6.626 times 2.82 and hit equals. Now divide the result (which hopefully was 18.68532, we'll round later) by 9.1 and you should get 2.0538461538, which we'll round to 2.05. So far, so good, now we'll calculate our powers of 10.For powers of 10 we have -34 and -15 in the numerator for a total of -49, and we have -31 in the denominator. Subtracting -31 from -49 we get -18, so the final answer is 2.05x10^-18.Now let's address the units. Putting all the units into a single expression we get: (m²kg/s)(m)/kg) = m³/s = m³s^-1So the final answer with units is 2.05x10^-18m³s^-1, just what Ashmore claims it is. He then divides by a cubic meter (m³) to get his final value of 2.05x10^-18s^-1. This last step is invalid, but that's not the point I'm focusing on in this message.Now let's repeat the exact same calculations, but this time in feet. These are the exact same constants as before, but converted from meters to feet using Google. For example to get Planck's constant in units of feet, I typed into Google "6.626x10^-34(m^2kg/s) in (ft^2kg/s)". The math is just as simple as before. Once again get out your calculator. First we'll again just do the math with the numbers. Multiple 7.132 times 9.25 and hit equals. Now divide the result (which hopefully was 65.971, we'll round later) by 9.1 and you should get 7.2495604395, which we'll round to 7.25. So far, so good, now we'll calculate our powers of 10.For powers of 10 we have -33 and -15 in the numerator for a total of -48, and we have -31 in the denominator. Subtracting -31 from -48 we get -17, so the final answer is 7.25x10^-17.Now let's address the units. Putting all the units into a single expression we get: (ft²kg/s)(ft)/kg) = ft³/s = ft³s^-1So the final answer with units is 7.25x10^-17ft³s^-1. We follow his procedure by dividing by unit volume, this time a cubic foot (ft³) to get a final value of 7.25x10^-17s^-1. Quite obviously, 7.25x10^-17s^-1 and 2.05x10^-18s^-1 are not equal, and 7.25x10^-17s^-1 is nowhere near Ashmore's choice for the value of the Hubble Constant, which is 2.06x10^-18s^-1.Because Ashmore's constant is actually a function of the units you begin with, it is not actually a constant and has no significance.Please let me know if I can help you out getting through any of the steps.--Percy

Your rage is understandable, but expressions of rage make it difficult to keep discussions on track, and also make it difficult for onlookers to tell the loons from the luminaries.--Percy

Thanks Parsominium, but apologies are nicer when given in a better smelling container.

(**** Edited because I now see that he mentions Bremsstrahlung in his paper as one of two mechanisms involved for radiating photons ****)Bremsstrahlung (free-free) scattering is radiation from an electron being decelerated in the potential of an ion.What he is talking about is some "invented" mechanism he calls double Mossbauer where the plasma as a whole absorbs most of the energy of the incoming photon and allows a low energy photon to be passed on. (**** I see he has Bremsstrahlung as part of this ****)Thermodynamics is a problem for his model but not in the way you are saying.Here is a quick run down of terms:Compton scattering - incoming photon scatters of electron where the electron is high energyThomson scattering - the low photon energy equivalent of the aboveRayleigh scattering - the equivalent to Thomson scattering where the electron is a bound atomic electronBremsstrahlung - electon scattering of an ion and emitting radiation as it decelerates, hence the term Bremsstrahlung which is German for breaking radiationMossbauer effect - scattering of radiation from atoms in a lattice where the macrostructure absorbs and redistributes the energy of the photon and allows a lower energy photon to be emittedThe problems with his model are that the IG plasma is too rarefied to allow a Mossbauer effect to occur. This is really at the heart of his thermodynamic problem. The plasma cannot perform this redistribution of momentum and energy because it cannot itself be in thermodynamic equilibrium because the mean free path of the electrons is so great they cannot act as a coherent structure inside of many many Hubble Times. I might calculate this later and post it.So what a rarefied plasma, at high T, which is what is really out there will do is actually to INVERSE Compton scatter incoming photons. This is, for CMB photons called the Sunyaev-Zeldovich effect.Inverse Compton scattering is Compton scattering where the electron has greater energy than the incoming photon so the photon actually gains energy at the expense of the electron. THIS is the effect that usually is in play in astrophysics - as contrasted with lab Compton experiments where the electrons are the free valence electrons in a metal and so the photon has the greater energy. Now of course Lyndon cannot have the SZ effect occurring because he wants his plasma to produce the CMB. So he postulates this silly Mossbauer type mechanism to get low energy photons.His problems with this are many fold. The key ones being:1) Plasma cannot act in the coherent manner he requires since their mean free paths are too high and the density way too low. The standard equations of plasma physics CANNOT be applied because the assumptions in their derivation are violated2) The Mossbauer type effects would NOT produce a Blackbody spectrum anyway. This is also a thermodynamical problem for him too.3) Even if you allow the foregoing violations of physics then the IG plasma varies enough in the local Universe - even away from the clutter of the galactic plane that fluctuations from the CMB isotropy would be at a much greater level than the observed fluctuations.and so on and so on....This message has been edited by Eta_Carinae, 03-21-2005 12:54 PMThis message has been edited by Eta_Carinae, 03-21-2005 02:50 PMThis message has been edited by Eta_Carinae, 03-21-2005 02:52 PM

Thanks very much for going to all that work, Percy, but I don't think either of us has the time for you to educate me to the point of comprehending all this math, not that it couldn't be done.I don't think your time has been wasted though. Since it's Ashmore's I'll let him be my advocate if he cares to.

I'm confused. The math involved is addition, subtraction, multiplication and division. These operations are not a problem for anyone armed with a calculator, so I don't get it.I can simplify this even more. I've taken the values in units of meters for the variables in the expression hr/m and divided by a cubic meter, and I've put it in a form suitable for cut-n-pasting into Google. Cut-n-paste this expression into Google's search box, click the search button, and you'll get 2.05x10^-18m³s^-1: I've done the same thing for the same variables for the same expression hr/m, but this time in units of feet. Just cut-n-paste this expression into Google's search box, click the search button, and you'll get 7.25x10^-17s^-1: You've just proven that Ashmore's constant is not a constant and is actually dependent upon choice of units. That's all there is to it!If you doubt that the two values are actually just the same value in different units, then just cut-n-paste this into Google: You'll get 2.05x10^-18m^3s^-1. Ashmore's mistake is dividing by what he calls unit space. He does this in order to make the units come out. The step is invalid, of course, the units equivalent of a fudge factor.--PercyThis message has been edited by Percy, 03-21-2005 01:40 PM

Melchior brings up the fact that later in his paper lyndonashmore multiplies the h*Re/Me by the electron density which he says is approx. 1/m^3 and thus restores his paradox.This is true.BUT a couple of points need to be made about this:1) Lyndon starts of with using this and puts this in later. His initial point is that h*Re/Me when divided by his unit volume has a numerical equivalence to the Hubble constant.2) The value for the electron density is similarly a free parameter and can be used in an ad hoc manner to restore the eqivalence.In fact Lydon himself at this later stage uses a value for n of 0.1/m^3 which spoils his paradox. His paradox would remain if n is exactly 1/m^3 since this would scale appropriately for feet or cm or whatever length unit.3) His value of 1 or 0.1 per m^3 for the electron density is only appropriate for a certain regime of space.This all becomes a question of where in the Universe his 'process' (which cannot possibly occur for other reasons I mentioned in an earlier post) occurs.If he wants it in the IGM then his density is way too high. n is more likely in the 10^-3 to 10^-5 area. This destroys his "paradox"
If he wants it more nearby then the value is very variable from say 0.1 to 10 per m^3 - this gives him the problem that it would vary tremendously over the sky - this is not observed.If he wants it really local then n is say 10 - 1000 per m^3 - again totally destroying his "paradox". Also it would be very variable too.
No matter what, even if it could occur (which it cannot anyway) it would either totally destroy his "paradox" or if the source is in just the right range of n values then it would be ruled out by inhomogeneity in the signal.Also think about this:His initial premeise is that h*Re/Me when divided by 1 m^3 has a relation to the Hubble constant.What is the motivation from h*Re/Me to begin with. Just because something has the the dimensions of m^3 s^-1 and then divide by m^3 doesn't mean it has physical significance.Note that his h*Re/Me is a hodgepodge to begin with. He has a classical concept (Re) which corresponds to nothing when it comes to quantum effects yet he multiplies this by Plancks constant. WHY? Just because the units come out to something he wants doesn't mean this is the correct thing to do.

No. I've read the whole paper, and have explained this in various posts. Look at Message 136, for example.The H = 2nhr/m is a units correct formula; and the 2 is not a typo. The "derivation" is in the paper, in section 4. The equality in this case is not calculated, but obtained by fudging the factor n to be 0.5. This value is not obtained by measurement, but by assuming his equation in the first place. See section 5 of the paper.The equation H = hr/m is a simple units error. There is therefore no equality here at all. He discusses this in section 1 of his paper, as an unexplained "coincidence". The coincidence in values vanishes in other units. He uses the nonsensical phrase "hr/m per cubic meter" to express a so-called paradox in values that actually depends on units chosen.The word "per" is wrong as well. That would make sense if h/rm gave a value with units m^-3. For example, n does measure quantities "per" cubic meter because the units are m^-3. h/rm, however, has units m³s^-1. You could describe it as "cubic meters per second", but "per cubic meters" is just another error.H in Hertz is close to hr/m in cubic meters per second, but it is nowhere near hr/m in cubic inches per second.Cheers -- SylasPS. Added in edit. Regarding fudge values of n... if you read the paper you see he also fudges to 0.57. It depends which estimates of H he wants to match. He can match any value of H just by picking a value for n; so again... no paradox.This message has been edited by Sylas, 03-21-2005 09:57 PM

Ashmore is wrong. I have not used curvature to explain where energy went. But this is off topic here. I have started the thread at Message 1 for this subject, and if Ashmore repeats the above nonsense in the appropriate thread I will explain -- again -- how these things really work in modern physics.

There are other major problems some of which I have hinted at in earlier posts.Let's ask him, since earlier he bandied the term, what the Debye radius is for his plasma and why he thinks he can treat his plasma as a macroscopic fluid. (i.e. it can absorb the photons energy/momentum and reradiate at 2.73K)You see, his mechanism relies upon using the equations of magnetohydrodynamics for the intergalactic plasma. This is why he mentions restoring forces and the like.But the plasmas in interstellar or especially intergalactic space are way to tenuous for this to apply. You have to treat this as a particle scattering event NOT a macroscopic plasma.So Lyndon what is your Debye radius?This message has been edited by Eta_Carinae, 03-21-2005 06:09 PM

Smell? O, that would be the stench of a pair o' dogs. The poo is all over this thread. Sorry about that.

"We" don't have a problem, mr. Ashmore, as far as I can see. Your students have, though. You are a physics teacher. Do your students know that your theory ranks as "cranky" on Crank Dot Net?As you may have seen, I suggested a new name for it: Ashmore's Real Paradox. Let's make that nice and short: ARP. Now, here's another suggestion: how about a HALT ON ARP?

HALT ON ARP!

Can we keep the general chatter down in this thread. It is difficult enough for us non-physics types to understand what is going on without the extraneous comments.

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