Lazarus

Tom, you are one of the people whose opinion that I value the most so I appreciate your comments. ******* Swansont said: "An arbitrary number that works" disqualifies this as legitimate science. The Hubble constant, the Gravitational contant, c and other constants are arbitrary constants that work. They would all be different arbitrary constants that work in a different universe or if we used Klingon units. ******* Swansont said: How is it that we can get multiple redshift values from hydrogen? The choice of shells that the electron moves to and from determines the color of the light on emission and the velocity of the atom determines the shifted wave length. ******** The other objection that you have is that electrons in a curved path have to radiate. The "proof" that they radiate in a curved path is that the electrons in a Synchrotron radiate. That "proof" is suspect. The curvature of the path of the electrons is caused by the magnetic field from the coils at the corners of the Synchrotron. The approach of the electrons is at an angle to the coils so there is a component of the velocity straight toward the coil which would cause the electron to slow down, lose kinetic energy and radiate. Doesn't that bring the "proof" that electrons have to radiate when in a curved path into question?

Strange said: "It is like one of those mathematical tricks: "think of a number, now multiply by ... take away the number you first thought of ... your number was ..." ------------------------------------------------------------------------------------------------------------------------------------------ Thank you for reading this thread. It is hard to follow it around the loop of calculations but there really is a distinction between the result of the calculation and the testing of the result. The calculation from the electron ends with the new wave length. The calculated wave length is called wave and the intial wave length is called lamda. Lamda is the emitted wave length and wave is the observed wave length. Now to test it use the customary equations to see how the results compare. The program does the calculations for several different atom velocities..

1 Does this mean you have chosen a couple of arbitrary numbers in order to get the right answer? 2 Can you explain how you calculate the velocity of a galaxy from the wavelength? 3 How does the "standard method" calculate the velocity of a galaxy from a wavelength? ---------------------------------------------------------------------------------------------- 1 The velocity of the electron in shell 1 at the apogee was determined by trial and error.. A number different from it will give different results. 2 The post above shows the steps of the calculation. 3 The usual way of computing the velocity of galaxies is: z=wobserve/wemit-1 z = observed wave length over emitted minus 1 v=(((z+1)**2-1)/((z+1)**2+1)) * c Velocity of source

THE CONCEPT The velocity of stars and galaxies can be calculated from their Red Shift wave length. The wave length vs velocity can be calculated from the classical elliptical Bohr hydrogen atom. The results are similar. THE ASSUMPTIONS Using the classical elliptical Bohr hydrogen atom: The velocity of the electron at the apogee of shell 1 is 2,188,000 times .0030394. An arbitrary number that works. The velocity of the electron is inversely proportional to the shell number. The photon is released at the apogee. THE CALCULATION STEPS 1 Calculate the energy of a photon caused by an electron moving from shell 3 to shell 2 using the Rydberg constant.. This is the total energy change of the electron, kinetic and latent. 2 Calculate the kinetic energy change between shell 2 and 3 with the atom not moving. 3 Give some velocity to the atom. 4 Add the atom's velocity to the initial velocities for the electrons at the apogee. 5 Calculate the new difference of kinetic energies for the 2 shells. 6 Adjust the initial energy by the kinetic energy change. 7 Calculate the wave length associated with that energy. 8 Use the wave length to calculate the velocity of the galaxy that sent the photon to us. THE COMPARISON Now use the calculated wave length to calculate the velocity with the "Standard" method. The results are remarkably close.

Meanwhile, back at the ranch, can someone tell me whats wrong with the electron to stars calculation? Other than the electron is supposed to radiate.

Gref, my friend, I was wrong about the electron. If the current idea is correct a curving path for and electron would release a photon so the electron would lose some of it's kinetic energy. I don't think a larger body would lose kinetic energy. Your said: Posted 25 July 2013 - 12:52 PM Changes in momentum cause changes in kinetic energy by: where p is momentum m is mass of the body. Since momentum is mv so your equation becomes E=mmvv/2m or mvv/2 which is kinetic energy only. That doesn't prove much about momentum. In fact p=square root of 2mE. Since 2mE is 2mmvv/2 you have to know v and m to solve the equation for momentum. Also, to determine the results of a collision of 2 objects the total kinetic energy AND the total momentum must remain the same. Bob

My eating rule is only eat things that taste good. Works for me.

I am so glad you asked. The discription of the calculations is at the head of this thread. The code to do the calculations and comparisons follows: ' shellsx.bas defdbl a-z ' Relate Bohr atom speed to energy of red or blue shifted photon ' Calculate Bohr electron energy change from one shell to another redblue=0' 0 = redshift, 1 = blueshift littlen=2' Lower number Shell bign=3' Higher number Shell velfact=.0030394' To adjust electron velocity at apogee of elipse incremt=0' Initial atom velocity increment delta=0' To change velocity print "Enter for red, 1 for blue" input redblue print "Little shell, Enter for 2 or 1 - 9" input a if a<>0 then littlen=a print "Big shell, Enter for 3 or 1 - 9" input a if a<>0 then bign=a print "Velocity factor, Enter for .0030394 or .001 to 1" input a if a<>0 then velfact=a ' Formulas ' c=f*l Speed of light = frequency times wave length ' e=h*f Energy of photon = frequency times Plank's ' 1/l=Rh*(1/n1*n1 - 1/n2*n2) l=wave length, Rh=Rydberg, n=shell # ' F=m*a Force = mass times acceleration ' Fc=v*v/r Centripital force = vel squared over radius ' ee=k/r Escape energy is proportional to inverse of r ' ke=m*v*v*.5 Kinetic energy=mass times vel squared times .5 ' z=wobserve/wemit-1' z = observed wave length over emitted minus 1 ' v=(((z+1)**2-1)/((z+1)**2+1)) * c' Yeah ' d=v/h' Distance = velocity divided by Hubble open "shells.txt" for output as #8 hubble=4.675e-17' Hubble constant rh=1.0973731e7' Rydberg constant plank=6.626e-34' Plank's constant c=3e8' Velocity of light in meters shell1v=2.188e6' Velocity of electron in shell 1 v1=shell1v*velfact' Modified velocity of electron in shell 1 m=.911e-30' Mass of an electron facte7=1e7' 10 to the 7th power facte9=1e9' 10 to the 9th power facte20=1e20' 10 to the 20th power z=0' Red shift factor vlittle=v1/littlen' Velocity of electron in smaller shell # vbig=v1/bign' Velocity of electron in bigger shell # r1sqr=1/(littlen*littlen)' Reciprical of little shell # squared r2sqr=1/(bign*bign)' Reciprical of big shell # squared rlamda=rh*(r1sqr-r2sqr)' Reciprical of wave length lamda=1/rlamda' Wave length of photon in meters, stopped freq=c/lamda' Frequency in Hz ienergy=freq*plank' Initial energy of photon in Joules vatom=0' Initial velocity of the Bohr atom kelittle=m*vlittle*vlittle*.5'Initial kinitic energy of little shell kebig=m*vbig*vbig*.5' Initial kinitic energy of big shell ideltake=kelittle-kebig' Initial kinetic energy change wave7=lamda*facte7' Adjust wave length for printing if redblue<>0 then print#8," Blueshift vs velocity" if redblue=0 then print#8," Redshift vs velocity" print#8," Eliptical Bohr Hydrogen atom print#8," Wave length =";using"###.##";wave7; print#8,"e-7" print#8," Atom Shell"; print#8,littlen; print#8," Shell"; print#8,bign; print#8," length energy" '''''print#8," velocity velocity velocity *e20 in nm %"; '''''print#8," Z Zv/c"; print#8," velocity velocity velocity Zv/c Z in nm"; print#8," *e20 %"; print#8, '''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''' for i=1 to 60 if redblue<>0 then' Doing blue shift av1=vlittle-vatom'Absolute velocity of little shell electron av2=vbig-vatom' Absolute velocity of big shell electron end if if redblue=0 then' Doing red shift av1=vlittle+vatom'Absolute velocity of little shell electron av2=vbig+vatom' Absolute velocity of big shell electron end if ake1=m*av1*av1*.5' Current kinetic energy of little shell electron ake2=m*av2*av2*.5' Current kinetic energy of big shell electron deltake=ake1-ake2' Current change of kinetic energy energy=ienergy-(deltake-ideltake)' Adjust energy for change in KE freq=energy/plank' Frequency wave=lamda*(ienergy/energy)' New wave length energy20=energy*facte20' Mult by 10 to the 20th wave9=wave*facte9' Mult by 10 to the 9th percent=(energy/ienergy)*100' Calc percent wave change if redblue<>0 then percent=(ienergy/energy)*100' Reverse for blue distance=vatom/hubble' distance z=wave/lamda - 1' calc z vz=(((z+1)*(z+1)-1)/((z+1)*(z+1)+1)) * c' Velocity from z in m/s print#8,using"####,###,###";vatom;av1;av2; print#8,using"####,###,###";vz; print#8,using"###.###";z; print#8,using"#######.###";wave9; print#8,using"#####.####";energy20 ''''' print#8,using"#####";percent ''''' print#8,distance delta=delta+100000 incremt=incremt+delta vatom=vatom+incremt' Speed up the atom ''''' vatom=vatom+100000' Speed up the atom if vatom>c goto c9999 next i c9999:

Don't you at least consider it a bit strange that the velocity of galaxies and stars can be computed by just assuming an electron orbiting a proton?

Thank you very much for the reference. Electrons radiate when passing through an magnetic field. That happens in a synchrotron and in space. The electron in a current loop must be in a zero magnetic field because just outside of the loop with the lines of force going north the inside of the loop will have the lines of force going south. If there is no detectable radiation that should mean that the curvature of the electron path is not a cause of radiation.

I see that the justification for being unable to detect radiation from a steady current loop is a mathematical cancellation of radiation. The physical loop has electrons that emit photons in the direction that the electrons are traveling. That should mean that the photons spread out like a sprinkler so the photons would not interfere with each other. Is the radiation really detectable or does the math match what actually happens with the photons?

You are right. I jumped the gun on that. If curature of the path of the eledtron causes a photon to be fired off in the forward direction the electron would loose kinetic energy. But doesn't that imply that a steady current in a closed loop would contantly be radiating? Greg, Let's move earth and the comet out to empty space and try again.

Try this. A comet goes by the earth. At a distance of on milliion miles coming and going it's kinetic and latent energy are the same. However, it's momentum is different.

A ball going east at 1 mile per hour and rebounded at 1 mile per hour going west would have the same kinetic energy but different momentum.

It would seem that an electron should only radiate on a change of kinetic or latent energy., not a change of momentum. Otherwise, there is creation of energy.

Everything else can orbit so why can't an electron?

Hi Tom, The Bohr atom is really irrevelant. The electron orbiting a proton is all that counts. I am fully aware that it goes against lots of things but I would appreciate any suggestions of the logical fallacy involved. Bob

Red shift of light can be calculated from the changing energy required to go from one atomic shell to another. Consider the electron in the elliptical Bohr hydrogen atom going from shell 2 to shell 3. With the atom at rest a photon released will have a wave length of 656 nm and energy of 30e-20 joules. The when the atom is moving the latent energy doesn't change but the kinetic energy of the electron does. That means the energy of a photon released at the apogee of the ellipse will diminish as the velocity of the atom increases. The energy of the photon is equal to the change in the latent energy minus the change in kinetic energy. The similarity of the computed energy and wave length to the currently used values is remarkable. The condition of the calculated table following is: The elliptical Bohr atom, shells 2 and 3 Atom velocity * * * * * * * >* * * * * * * down * *p * * >* e * e * * * * * <* * Electron clockwise * Photon velocity * * up * * * * * * * <* * * * * * The initial conditions are: The semi-major axis of the ellipse is the same as the circular Bohr radius The energy of the photon is the accepted one for change of level. The velocity of the electron shell 1 electron at the apogee was chosen to match the wave length of a photon from an atom at rest. The table below was generated by: 1 Calculating the energy required for an electron to go from the apogee of shell 2 to the apogee of shell 3, using various atom velocities. The light generated is red . 2 Calculating Z from the calculated wave lengths. 3 Calculating the atom velocity from the value of Z. The calculated velocities are quite similar to the accepted velocities for matching wave lengths. Redshift Calculation from Elliptical Bohr Hydrogen atom Red light Wave length = 656 nm Wave Photon Atom Shell 2 Shell 3 Calculated Length Energy Velocity Velocity Velocity Vel from Z Z in nm *e-20 0 3,325 2,217 0 0.000 656.112 30.2966 100,000 103,325 102,217 100,000 0.000 656.331 30.2865 400,000 403,325 402,217 400,201 0.001 656.988 30.2563 1,000,000 1,003,325 1,002,217 1,001,503 0.003 658.306 30.1957 2,000,000 2,003,325 2,002,217 2,006,338 0.007 660.515 30.0947 3,500,000 3,503,325 3,502,217 3,519,837 0.012 663.856 29.9432 5,600,000 5,603,325 5,602,217 5,651,326 0.019 668.591 29.7312 8,400,000 8,403,325 8,402,217 8,516,142 0.029 675.009 29.4485 12,000,000 12,003,325 12,002,217 12,237,762 0.042 683.446 29.0850 16,500,000 16,503,325 16,502,217 16,950,182 0.058 694.292 28.6306 22,000,000 22,003,325 22,002,217 22,800,477 0.079 708.026 28.0753 28,600,000 28,603,325 28,602,217 29,951,341 0.105 725.241 27.4088 36,400,000 36,403,325 36,402,217 38,583,267 0.138 746.697 26.6213 45,500,000 45,503,325 45,502,217 48,895,735 0.179 773.391 25.7024 56,000,000 56,003,325 56,002,217 61,106,276 0.229 806.665 24.6422 68,000,000 68,003,325 68,002,217 75,445,441 0.293 848.380 23.4305 81,600,000 81,603,325 81,602,217 92,144,384 0.374 901.198 22.0573 96,900,000 96,903,325 96,902,217 111,409,641 0.477 969.071 20.5124 114,000,000 114,003,325 114,002,217 133,376,657 0.613 1058.140 18.7858 133,000,000 133,003,325 133,002,217 158,029,811 0.796 1178.492 16.8673 154,000,000 154,003,325 154,002,217 185,073,456 1.054 1347.944 14.7469 177,100,000 177,103,325 177,102,217 213,740,068 1.440 1601.199 12.4144 202,400,000 202,403,325 202,402,217 242,537,915 2.073 2016.056 9.8598 230,000,000 230,003,325 230,002,217 268,988,479 3.283 2810.402 7.0730 260,000,000 260,003,325 260,002,217 289,497,775 6.492 4915.623 4.0438 292,500,000 292,503,325 292,502,217 299,620,447 38.747 26078.370 0.7622 So why do the electrons stay in their orbit? All that is needed for the distinct fixed orbits is for the nucleus to have a periodically changing or rotating magnetic field. I thought that would be highly unlikely until I found that the Handbook of Nuclear Properties of 1997 mentions nuclei with magnetic forces having a rotational character.

Since Cartesian coordinates and space are different things, a curved 3 dimensional space can be represented by 4 Cartesian dimensions. Throw in time and that makes 5 dimensions. If space is curved in 3 dimensions it takes 7 Cartesian coordinates. The shortest distance between any 2 points on a "grid" line may be along the line. Consider a grid fine enough that a misshapen rectangle is nearly flat and the grid line around it nearly straight. With 4 adjacent "rectangles" the grid line across the center is the shortest path between the 2 ends. Now fold it along the vertical centerline to match the curvature of the well. The crooked line is still the shortest distance as long as we stay in our space. --------------------- ! ! ! ! ! ! ! ! ! --------------------- ! ! ! ! ! ! ! ! ! --------------------- ^ Fold here The sun should be at the bottom of the well. I still can't see any reason why the photon should turn towards the center of the well. The only way it can get closer to the center of the well is to go deeper into the well. If it did that it should continue down the well just like the penny does in the science displays.

I think everybody is in agreement that the photon really changes direction as an asteroid does when it passes the earth. My problem was that I was under the impression that the curvature of space was the only thing affecting the path. I looks like the proper picture is that, regarding time as a forth dimension, space curves into a fifth dimension and there is a gravity like force parallel to the axis of the fifth dimension. That means Spyman's pool table analogy is pretty good.

The pool table analogy has a problem. It introduces a force that is not in the situation that we are discussing, namely the gravitational force towards the center of the earth. Einstein and Newton both considered the motion of a photon to be along what it thought was a straight line unless affected by an external force. If you put a dent in the space containing the line, the photon should still continue along the line and eventually be going in the same direction that it was originally. What would cause the photon to leave the line?

Geodesic, parabola or ellipse, close enough for government work. Whoops. Sorry, some of the very knowledgeable people who post on this forum are government people. Your comment is the very thing that I have the question about. Why does it come out in a different direction rather than follow a "grid"?

Krash, Some of the Astronomy Magazine articles are available online but that article doesn't seem to be. Perhaps you can explain the essence of the difference in a short summary.

Krash, That is exactly the grid that I see everywhere. --------------------------------------------------------------- Janus, Both I and Isaac would be happy to agree with you that the photon appears to follow a geodesic because of gravity. The curvature of space is supposed to replace gravity as the mechanism for determining the path of a photon. So the photon should follow the grid line, not the geodesic. If the proton follows the geodesic didn't it get out of it's place in curved space? Naughty photon.

A photon thinks it is following a straight line. If you put a dip in the line by bending space the photon will still follow the line. The line slopes down then slopes up so the photon should do the same. A penny put into one of those relativity cones in science shows goes down because of gravity. Gravity is replaced by curvature of space in the current model. If the photon did the same thing as the penny, photons would be captured by the mass. If the "analogy" is not correct, what is the real path of the photon?