Tired Light

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

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

Just for clarification for others, Eta is talking about what is referred to in electrical engineering as drift velocity. If you push an electron into one end of a wire, another electron will fall out the other end almost instantly, the speed of propagation being very close to the speed of light. But it wasn't the same electron that fell out the other end. A loose analogy is ball bearings in a garden hose. If you push a ball bearing in one end, another will instantly fall out the other.But if you continually push electrons into one end of the wire and establish an electric current, the time it takes an individual electron to travel from one end of the wire to the other is nowhere near the speed of light. It isn't even near the speed of your car. You can even walk faster. Typical electron drift velocities are around 0.1 millimeters/second, depending upon the voltage and the nature of the conductor.Apparently, electrons also have a thermal velocity that at normal temperatures is an appreciable fraction of the speed of light. As one site puts it (http://c2.com/cgi/wiki?SpeedOfElectrons), "So they are buzzing about at random at high speeds, with a small superimposed drift velocity caused by the electric field."--Percy

These graphs from Eta from the paperHigh-Redshift Quasars Found in Sloan Digital Sky Survey Commissioning Data. V. Hobby-Eberly Telescope Observations can't be interpreted by the uninitiated, e.g.: I've just initiated myself with a short visit to Hyperphysics on Hydrogen and UCLA Cosmology Tutorial. Naturally, recent initiates are prone to error, so I'd appreciate any corrections.There are two spikes on each graph. Some simple calculations reveal that the line nearest to 6000 angstroms is the Lyman alpha line of 1220 angstroms red-shifted up to around 6000 angstroms. I don't know what the other spike is. The jaggedness of the graphs is caused by absorption at various red-shifts of intervening clouds.The lines shown on the graphs are limited by the bandwidth of the spectrograph used by the researchers, which they say has a range from 5000-10,000 angstroms. They Lyman alpha line is the only hydrogen line that fits within this spectrum for these z values around 3.8, so this particular study didn't have the capability to measure other hydrogen lines. For example, the lowest Balmer series line at 4102 angstroms would be red-shifted up to around 19,700 angstroms for z=3.8, way out of the range of this device.But this spectrograph could easily examine the Balmer lines by looking at objects with lower red shifts. The longest Balmer line at 6563 angstroms would be viewable by this device for any object within our galaxy, and within all nearby galaxies, where z is approximately 0. The next shorter Balmer line at 4861 angstroms would be viewable by this device for objects with z between .1 and .8. The Balmer line after that at 4341 angstroms would be viewable with z between .2 and 1. The shortest Balmer line at 4102 angstroms would be viewable with z between .3 and 1.2. It would be relatively easy to find large numbers of objects with such low z values, so it should be possible for this spectrograph to display all four Balmer lines on a single graph. Unless absorption by the interstellar medium is a problem at these wavelengths, I imagine this has already been done many times.I'm no astromer so I wouldn't know if this measurement has been done or not, but poking about on the web I found this spectra showing all the Balmer lines at Vspec - Line measurement: Is that what you were looking for?--PercyThis message has been edited by Percy, 04-12-2005 08:44 PM

I'm finding this stuff hard to interpret. Can you put it all in context for the rest of us?--Percy

Here's one of the spectra Eta is referring to from page 1235 of the paper at AJ Vol 121 pg 1232. The Lyman beta line is the little bump on the far left around 5200 angstroms: Don't both clicking on "Click to enlarge", it's already as big as it's going to get.--Percy

Thanks for all this information. An aside, and this will probably fall into your "How could he not know that" category, but one of the things I learned while trying to figure out what you were saying about spectra is that the absorption spectrum of intervening hydrogen clouds is a function of their red shift. It's obvious, I know, but I had never even thought about it before.--Percy

I agree, Eta was out of line, and I called him on it. I actually thought I could see the Lyman Beta lines while eyeballing the diagrams, but I agree they're very subtle, much more subtle than the Lyman Alpha lines.But I have the same question as Eta. Why are you so hot to prove the supposedly shifted spectra are incomplete? This isn't as bad as trying to prove a flat earth, but it's close. It's only a Doppler-type effect, after all, hardly anything startling or revolutionary. Even Halton Arp, the arch enemy of red shift, accepts that the spectra are all there. Why do you care so much? Do complete spectra have some kind of critical implication that hasn't been mentioned yet?--Percy

The most significant discussion in this long thread was between Sylas and Lyndon Ashmore. My view is that Sylas posted correct and useful scientific information in post after post while Lyndon made error after error and gradually descended into unintelligibility and finally ceased posting, after taking the obligatory parting shots, of course. Lyndon was resourceful in rebuttal, raising a variety of counterarguments that had to be carefully addressed, but once the correct information was out there his arguments were often revealed to be wrong or red herrings. His theory isn't as bad as perpetual motion machines, but it's close!--Percy