The reality of energy.
in [Relativity]
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From: Inertial on 1 Mar 2010 18:11 "ben6993" <ben6993(a)hotmail.com> wrote in message news:83a2bd7f-6a06-409c-bfed-7d668b4f069e(a)v20g2000yqv.googlegroups.com... > On Mar 1, 9:59 am, "Inertial" <relativ...(a)rest.com> wrote: > >> Photons don't have a point of view (in the sense that they don't have any >> inertial frame of reference where they are at rest). The energy of a >> photon >> is what happens to some other object when the photon 'hits' it. > > I accept that I can't use the photon's viewpoint. (You know I wrongly > keep trying to imagine what it is like from the photon's viewpoint.) :):) >> > by the time it reaches us it has redshifted quite a long way and to us >> > it appeard to have a reduced energy, > >> It always had reduced energy to us. > > OK, and > re-phrasing: We calculate it has less energy (because of redshift) > that an observer seeing a similar photon soon after its emission would > have calculated it. No .. we don't. The rephrasing doesn't help, you still have the same misconception. All similar (co-moving) observers record the SAME frequency. Red shift is only able to be used as a measure of distance on the assumption that objects further away from us are moving away faster than ones closer. It is NOT the case that the light from a distant star is more and more red shifts due to observers being further away .. its relativy velocity that does this. As far as we are concerned (ie in our frame of reference), the frequency of light leaving the star is the same as the frequency of light arriving here (ignoring such things as light slowing when it travels thru air etc) > And the greater the redshift, the less is the > energy we calculate. Yes .. because we are moving away from the source >> > can we think of that discrepancy between higher energy in its >> > framework and its lower energy in our framework as being the entropy >> > gain of the photon? > >> Nothing in the photon changed. The energy it 'has' in our frame is >> constant >> the whole time. The energy it 'has' in the frame of the star that >> emitted >> it is constant the whole time. > >> There is no change > > Agreed. Re-phrasing again: I was originally wondering if the lower > energy in our frame compared to the higher energy in the frame of > emission of the photon was comparable to a higher entropy state in our > frame. Nope .. as it is all symmetric. Unless entropy is frame dependent. > Ie is it really less energy Yes. Given Energy being the frame dependent value we use. 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. > or is it merely less useable > energy. But I can't see it now. It must be less energy. Energy, like velocity and momentum, is frame dependent. > It is odd that redshifted light gives less energy on impact, Not really > even > though the speed of the photon is still c But its momentum is less > despite the high speed of > separation of the two galaxies where emission and absorption occur. Not really that odd, once you get it :). Do some reading on relativistic dopler effect >> > 2. Assume that a rocket was using the entire mass of the universe as >> > fuel for its burners in order to accelerate closer and closer to speed >> > c. >> >> OK. A tad impractical, of course :) > > <snip> > >> I'm not sure how the mass gets into the rocket in this idea, if it wasn't >> there to start with. > > I was thinking of relativistic increase in mass through high speed. > > <snip> >> I'm not sure what the argument is. > > The idea was of doing work with the whole energy of (or most of) the > universe and so greatly increasing its entropy. But wondering if the > resultant high entropy system could still do work outside the system. > Ie energy can be different frames but can entropy also be different in > different frames. I'm not sure about entropy.
From: BURT on 1 Mar 2010 19:28 On Mar 1, 3:11 pm, "Inertial" <relativ...(a)rest.com> wrote: > "ben6993" <ben6...(a)hotmail.com> wrote in message > > news:83a2bd7f-6a06-409c-bfed-7d668b4f069e(a)v20g2000yqv.googlegroups.com... > > > On Mar 1, 9:59 am, "Inertial" <relativ...(a)rest.com> wrote: > > >> Photons don't have a point of view (in the sense that they don't have any > >> inertial frame of reference where they are at rest). The energy of a > >> photon > >> is what happens to some other object when the photon 'hits' it. > > > I accept that I can't use the photon's viewpoint. (You know I wrongly > > keep trying to imagine what it is like from the photon's viewpoint.) > > :):) > > >> > by the time it reaches us it has redshifted quite a long way and to us > >> > it appeard to have a reduced energy, > > >> It always had reduced energy to us. > > > OK, and > > re-phrasing: We calculate it has less energy (because of redshift) > > that an observer seeing a similar photon soon after its emission would > > have calculated it. > > No .. we don't. The rephrasing doesn't help, you still have the same > misconception. All similar (co-moving) observers record the SAME frequency. > > Red shift is only able to be used as a measure of distance on the assumption > that objects further away from us are moving away faster than ones closer.. > > It is NOT the case that the light from a distant star is more and more red > shifts due to observers being further away .. its relativy velocity that > does this. > > As far as we are concerned (ie in our frame of reference), the frequency of > light leaving the star is the same as the frequency of light arriving here > (ignoring such things as light slowing when it travels thru air etc) > > > And the greater the redshift, the less is the > > energy we calculate. > > Yes .. because we are moving away from the source > > > > > > >> > can we think of that discrepancy between higher energy in its > >> > framework and its lower energy in our framework as being the entropy > >> > gain of the photon? > > >> Nothing in the photon changed. The energy it 'has' in our frame is > >> constant > >> the whole time. The energy it 'has' in the frame of the star that > >> emitted > >> it is constant the whole time. > > >> There is no change > > > Agreed. Re-phrasing again: I was originally wondering if the lower > > energy in our frame compared to the higher energy in the frame of > > emission of the photon was comparable to a higher entropy state in our > > frame. > > Nope .. as it is all symmetric. Unless entropy is frame dependent. > > > Ie is it really less energy > > Yes. Given Energy being the frame dependent value we use. > > 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. > > > or is it merely less useable > > energy. But I can't see it now. It must be less energy. > > Energy, like velocity and momentum, is frame dependent. > > > It is odd that redshifted light gives less energy on impact, > > Not really > > > even > > though the speed of the photon is still c > > But its momentum is less > > > despite the high speed of > > separation of the two galaxies where emission and absorption occur. > > Not really that odd, once you get it :). Do some reading on relativistic > dopler effect > > > > > > >> > 2. Assume that a rocket was using the entire mass of the universe as > >> > fuel for its burners in order to accelerate closer and closer to speed > >> > c. > > >> OK. A tad impractical, of course :) > > > <snip> > > >> I'm not sure how the mass gets into the rocket in this idea, if it wasn't > >> there to start with. > > > I was thinking of relativistic increase in mass through high speed. > > > <snip> > >> I'm not sure what the argument is. > > > The idea was of doing work with the whole energy of (or most of) the > > universe and so greatly increasing its entropy. But wondering if the > > resultant high entropy system could still do work outside the system. > > Ie energy can be different frames but can entropy also be different in > > different frames. > > I'm not sure about entropy.- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - What about negative or anti energy proposed by Stephen Hawking? Mitch Raemsch
From: BURT on 1 Mar 2010 21:33 Mass is infinitely dense point particle of C squared Gamma energy. Mitch Raemsch
From: ben6993 on 2 Mar 2010 03:49 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. > 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. 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. > 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. 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. I will keep reading/learning.
From: Inertial on 2 Mar 2010 04:35
"ben6993" <ben6993(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. |