The reality of energy.
in [Relativity]
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From: ben6993 on 2 Mar 2010 06:44 On Mar 2, 9:35 am, "Inertial" <relativ...(a)rest.com> wrote: > "ben6993" <ben6...(a)hotmail.com> wrote in message > > Light always travels at speed c compared to any observer, but the > > effort of keeping up this speed when an observer is retreating fast > > makes the photon seem tired to the retreating observer. Ie energy is > > lost (somewhere in this process) in the maintenance of speed c. That > > doesn't sound so odd. > It may not sound odd .. but it is not what happens. There is no tiring of > light. It doesn't take any 'effort' for light to travel at c. That energy > 'loss' is from the moment the light is emitted. It is not something that > happens over distance. That's where you are getting confused. I know that I have much to learn but I did not think that the 'distance' aspect was confusing me. I think it was again a problem with my previous wording. I knew that distance has nothing to do with causing a redshift except inasfar as an observer (or the observer's galaxy) needed distance and time to build up a high speed worthy of calling it a Hubble size redshift. I agree that a minute redshift could be obtained instantly by just stepping back, away, from a torch shone at you, and that would have done just as well as an example. > Have a look at relativistic Doppler shift to see how it works. Will do. Particularly trying to understand the reason why the photon energy is reduced in the observer's framework.
From: Y.Porat on 2 Mar 2010 10:10 On Mar 2, 1:44 pm, ben6993 <ben6...(a)hotmail.com> wrote: > On Mar 2, 9:35 am, "Inertial" <relativ...(a)rest.com> wrote: > > > "ben6993" <ben6...(a)hotmail.com> wrote in message > > > Light always travels at speed c compared to any observer, but the > > > effort of keeping up this speed when an observer is retreating fast > > > makes the photon seem tired to the retreating observer. Ie energy is > > > lost (somewhere in this process) in the maintenance of speed c. That > > > doesn't sound so odd. > > It may not sound odd .. but it is not what happens. There is no tiring of > > light. It doesn't take any 'effort' for light to travel at c. That energy > > 'loss' is from the moment the light is emitted. It is not something that > > happens over distance. That's where you are getting confused. > > I know that I have much to learn but I did not think that the > 'distance' aspect was confusing me. I think it was again a problem > with my previous wording. I knew that distance has nothing to do > with causing a redshift except inasfar as an observer (or the > observer's galaxy) needed distance and time to build up a high speed > worthy of calling it a Hubble size redshift. I agree that a minute > redshift could be obtained instantly by just stepping back, away, from > a torch shone at you, and that would have done just as well as an > example. > > > Have a look at relativistic Doppler shift to see how it works. > > Will do. Particularly trying to understand the reason why the photon > energy is reduced in the observer's framework. -------------------- what is so diffficult for you to understand??: if the target is 'running away from the moving photons then LESS PHOTON UNITS * PER SECOND* ARE HITTING THE TARGET !!! less photon units = less enrgy !!! no need to stumble on entropy questions and that again shows you that photon energy emission is TIME DEPENDENT !! Y.Porat ----------------------
From: ben6993 on 2 Mar 2010 10:28 On Mar 2, 3:10 pm, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > what is so diffficult for you to understand??: > if the target is 'running away from the > moving photons > then > LESS PHOTON UNITS * PER SECOND* > ARE HITTING THE TARGET !!! > less photon units = less enrgy !!! > no need to stumble on entropy questions > and that again shows you that > photon energy emission is > TIME DEPENDENT !! Yes, the target is running away from the photon, but the photon is still moving at relative speed c no matter fast the observer retreats. Also, it is less energy even when only one photon is involved. I can understand (I trust) a ball being caught by a retreating catcher, and also doppler for sound waves, but the fact is that you cannot reduce your relative speed wrt a photon no matter how fast you retreat. Even though you do not reduce the photon's relative speed, the momentum is reduced and the energy is reduced. So the retreat works wrt energy and momentum (p) yet fails to work with the speed. I want to understand why E and p are reduced, and I am not sure that knowing all the maths formula will help me to know 'why'. But I will try.
From: BURT on 2 Mar 2010 15:22 On Mar 2, 1:35 am, "Inertial" <relativ...(a)rest.com> wrote: > "ben6993" <ben6...(a)hotmail.com> wrote in message > > news:baac47b7-a6a8-4267-9d7c-a636fe3e1e06(a)e7g2000yqf.googlegroups.com... > > > > > > > On Mar 1, 11:11 pm, "Inertial" <relativ...(a)rest.com> wrote: > >> "ben6993" <ben6...(a)hotmail.com> wrote in message > > > <snip> > > >> > And the greater the redshift, the less is the > >> > energy we calculate. > > >> Yes .. because we are moving away from the source > > > Sorry, I was not clear enough about what I meant. I used the word > > redshift to imply that the observer was moving rapidly away from the > > emission source. I knew that you don't get redshift simply through > > distance alone. But as Hubble related the distance of separation to > > the speed of separation, it is the size of the speed of separation > > which gives the amount of redshift. > > That really makes no difference here .. in our frame of reference the light > left the start red-shifted .. it does not become redshifted along the way.. > > >> EG. if you throw a ball, in your frame the moving ball now has a lot of > >> kinetic energy. In the frame of the thrown ball, it is at rest (ignoring > >> gravity etc for now) and has no kinetic energy. > > > The catcher of a hard, fast ball yields or retracts the hands to > > reduce the relative velocity of the ball. That works and seems easy > > to understand. > > Yes .. but not really relevant to what we're talking about > > > But with light you can't actually reduce its relative velocity. No > > matter how fast you yield your hands or how fast the galaxy is > > retreating from the emission galaxy, the photon still moves at > > relative constant speed c. Yet the light does have less energy in > > our retreating frame. > > That's right. Have a look at relativistic Doppler shift to see how it works > > >> But its momentum is less > > > Yes, I have checked, E=pc so p= h/wavelength which reduces with > > redshift. > > It does seem odd, but tracing back to where the oddness starts, it > > seems to start with the existence of the redshift. Though, of course, > > the redshift does exist. > > That's right. Have a look at relativistic Doppler shift to see how it works > > > Light always travels at speed c compared to any observer, but the > > effort of keeping up this speed when an observer is retreating fast > > makes the photon seem tired to the retreating observer. Ie energy is > > lost (somewhere in this process) in the maintenance of speed c. That > > doesn't sound so odd. > > It may not sound odd .. but it is not what happens. There is no tiring of > light. It doesn't take any 'effort' for light to travel at c. That energy > 'loss' is from the moment the light is emitted. It is not something that > happens over distance. That's where you are getting confused. > > > I will keep reading/learning. > > Have a look at relativistic Doppler shift to see how it works.- Hide quoted text - > > - Show quoted text - Light has no kinetic energy because C is a constant. Mitch Raemsch
From: Mahipal7638 on 2 Mar 2010 16:25
On Mar 2, 10:28 am, ben6993 <ben6...(a)hotmail.com> wrote: > On Mar 2, 3:10 pm, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > > > what is so diffficult for you to understand??: > > if the target is 'running away from the > > moving photons > > then > > LESS PHOTON UNITS * PER SECOND* > > ARE HITTING THE TARGET !!! > > less photon units = less enrgy !!! > > no need to stumble on entropy questions > > and that again shows you that > > photon energy emission is > > TIME DEPENDENT !! > > Yes, the target is running away from the photon, but the photon is > still moving at relative speed c no matter fast the observer retreats. > Also, it is less energy even when only one photon is involved. Somewhere you apologized for imaging traveling at the speed of a photon. Why?! Is that thought experiment restricted to a teenager Einstein only? Consider this Ben, there far more potential frames of references than any theoretical model can claim to have considered. I used to call this the Infinite Perspectives Machine(s) -- plural or otherwise. > I can understand (I trust) a ball being caught by a retreating > catcher, and also doppler for sound waves, but the fact is that you > cannot reduce your relative speed wrt a photon no matter how fast you > retreat. Even though you do not reduce the photon's relative speed, > the momentum is reduced and the energy is reduced. So the retreat > works wrt energy and momentum (p) yet fails to work with the speed. I > want to understand why E and p are reduced, and I am not sure that > knowing all the maths formula will help me to know 'why'. But I will > try. There is no "try" only "I will do." Perhaps you might learn what I mean from this link: http://www.hep.ucl.ac.uk/opal/gammagamma/gg-tutorial.html While electrons collide, photons are absorbed or released. Except for the low-lower-lowest collision cross-sectional area graphed measures at the site above for photon-photon collisions. Google, The truly DoNoEvil, is your friend if you know what to search. But shhhhh... don't giveaway your search strings/spellings to anyone by the physical act of typing. I know, they are not watching the keystrokes. I must hanson laugh now! Ahahahahahahaha.... Enjo(y)... -- Mahipal Just My GPS And Me At The Edge Of The Universe Just My GRB And Me At The Edge Of The Universe Just My GOD And Me At The Edge Of The Universe Just My DOG And Me At The Edge Of The Universe -- Bee Gees. |