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grav
2007-Oct-16, 06:54 PM
If a positron and electron annihilate, what is the frequency of light produced?
Is that frequency precise, depending upon the rest masses?

If we have a relative speed away from a light source, the light becomes redshifted.
How would this work out with a speed relative to the annihilation?
Would it depend upon the relative speeds of the particles before annihilation?
Would it depend upon the relative speed to the point of annihilation?
Since the annihilation would take some duration of time, no matter how small, there could be no actual pinpoint of annihilation, could there, maybe the halfway point?
Would it depend upon the relative speed to the light itself?
All three, or perhaps just the first two?

trinitree88
2007-Oct-16, 07:08 PM
If a positron and electron annihilate, what is the frequency of light produced?
Is that frequency precise, depending upon the rest masses?

If we have a relative speed away from a light source, the light becomes redshifted.
How would this work out with a speed relative to the annihilation?
Would it depend upon the relative speeds of the particles before annihilation?
Would it depend upon the relative speed to the point of annihilation?
Since the annihilation would take some duration of time, no matter how small, there could be no actual pinpoint of annihilation, could there, maybe the halfway point?
Would it depend upon the relative speed to the light itself?
All three, or perhaps just the first two?

grav. Electron /positron annihilation at rest gives an energy of 1.022 Mev. Plug that into E=hv as two photons of 511 kev each or three coplanar ones summing to that. v = c/lambda .... frequency = speed of light over wavelength. If you are considering extra kinetic energy due to the velocities of the annihilating species, then that adds to the energy, and hence the frequencies of the photons produced. pete.

grav
2007-Oct-16, 07:25 PM
grav. Electron /positron annihilation at rest gives an energy of 1.022 Mev. Plug that into E=hv as two photons of 511 kev each or three coplanar ones summing to that. v = c/lambda .... frequency = speed of light over wavelength. If you are considering extra kinetic energy due to the velocities of the annihilating species, then that adds to the energy, and hence the frequencies of the photons produced. pete.Thanks, trinitree. What exactly do you mean by "at rest". Certainly the particles wouldn't be at rest, but probably moving at c or close to it at the point of annihilation, in order to give E = mc2 = hv, right? If so, would that be c relative to each other? It would seem it would have to, since the frequency would differ to an observer depending upon their relative speed to the particles, adding or subtracting extra kinetic energy, as you said.

hhEb09'1
2007-Oct-16, 08:10 PM
Certainly the particles wouldn't be at rest, but probably moving at c ...I know you continue on with "or close to it" but it's probably worth noting that the particles don't move at c.

alainprice
2007-Oct-16, 08:56 PM
What about annihilation of virtual pairs?

They would appear briefly, and then annihilate, without any measurable kinetic energy. Obviously, that's not the answer you want.

Did you mean to say a beam of electrons going one way in a particle accelerator meets a beam of positrons going in the opposite direction? You need to state what is creating the anti-matter for a speed to be established. Are we talking about an stream of particles eminating from a black hole jet?

grav
2007-Oct-16, 09:14 PM
I know you continue on with "or close to it" but it's probably worth noting that the particles don't move at c.Yes. :) That's why I added it. I'm also wondering if the particles might get so close to c when attracting each other that they might break down on their own or something. I'm also wondering about collisions at much less than c, or if the particles might contain a magnetic component, perhaps due to the charge while spinning on their axes, or while revolving around each other, that builds quickly enough to keep them from colliding in the first place, but accelerating closer and closer to c as they revolve or something. In addition, I'm wondering about, with Relativity, their masses build up toward infinity as they approach c, according to a stationary observer, then why would we use the rest mass for E=mc2 to find the energy of annihilation, instead of the mass according to the observer at the time of annihilation?

grav
2007-Oct-16, 09:28 PM
What about annihilation of virtual pairs?

They would appear briefly, and then annihilate, without any measurable kinetic energy. The kinetic energy should depend upon the relative speed between the virtual pair and the observer, I would think. But that's another thing. If we are to say that a virtual pair arises in empty space with zero kinetic energy between them (their energies add up to zero), then that would define an absolute space, at that relative speed for an observer that makes the kinetic energy of each of two particles that are created exactly equal and opposite, wouldn't it? But that is only if they are said to arise from empty space. Since they would annihilate into discernable energy, such as light, and not back to nothingness, then that energy would have to have been there to create the virtual pair to begin with, and the space cannot be empty, so that clears up my perception of virtual pairs.


Did you mean to say a beam of electrons going one way in a particle accelerator meets a beam of positrons going in the opposite direction? You need to state what is creating the anti-matter for a speed to be established. Are we talking about an stream of particles eminating from a black hole jet?Just a single positron and electron, like the process one would see in a particle chamber.

korjik
2007-Oct-16, 09:39 PM
Yes. :) That's why I added it. I'm also wondering if the particles might get so close to c when attracting each other that they might break down on their own or something. I'm also wondering about collisions at much less than c, or if the particles might contain a magnetic component, perhaps due to the charge while spinning on their axes, or while revolving around each other, that builds quickly enough to keep them from colliding in the first place, but accelerating closer and closer to c as they revolve or something. In addition, I'm wondering about, with Relativity, their masses build up toward infinity as they approach c, according to a stationary observer, then why would we use the rest mass for E=mc2 to find the energy of annihilation, instead of the mass according to the observer at the time of annihilation?

The most common electron-positron annihilation gives a pair of 511 MeV gammas.

If you take to slow moving particles and let them collide, they dont end up having any relativistic effects show in the annihilation. If I were to guess, I would say that it is a conservation of energy effect.

Start with two particles, far enough apart that the are very weakly interacting. That will give you a total energy in the system of

2mc^2 + a very small contribution from the field interaction.

At the point of collision the energy has to be the same.

You can also make a pseudoatom out of an electron and a positron. I think the term commonly used is positronium. Basically, the positron is acting like the proton in a hydrogen atom. Dosent last very long before it decays.

alainprice
2007-Oct-16, 10:04 PM
If you really want, set your own speed and use

E^2 = M^2C^4 + P^2C^2

where M is rest mass, P is momentum.


It is considered normal to choose a reference frame that puts the test object at rest, and then use special relativity to find the energy in other inertial frames.

grav
2007-Oct-16, 10:14 PM
The most common electron-positron annihilation gives a pair of 511 MeV gammas.

If you take to slow moving particles and let them collide, they dont end up having any relativistic effects show in the annihilation. If I were to guess, I would say that it is a conservation of energy effect.

Start with two particles, far enough apart that the are very weakly interacting. That will give you a total energy in the system of

2mc^2 + a very small contribution from the field interaction.

At the point of collision the energy has to be the same.

You can also make a pseudoatom out of an electron and a positron. I think the term commonly used is positronium. Basically, the positron is acting like the proton in a hydrogen atom. Dosent last very long before it decays.That's interesting. So a positron and electron set very close together and originally stationary to each other would produce rays of light with a total energy of just 2mc2 while the additional energy would depend upon how much they pick up while accelerating toward each other? I wonder how that would work out if they each accelerated toward each other from infinity.

So that's what a positronium is? Why isn't it as stable as an electron and proton?

grav
2007-Oct-16, 10:20 PM
If you really want, set your own speed and use

E^2 = M^2C^4 + P^2C^2

where M is rest mass, P is momentum.


It is considered normal to choose a reference frame that puts the test object at rest, and then use special relativity to find the energy in other inertial frames.Thanks. Is that P for the momentum between the particles upon colliding? It has nothing to do with the relative speed of the observer, right, or does it?

grav
2007-Oct-16, 10:25 PM
Does someone know how P would be expanded out in that formula alainprice posted?

grav
2007-Oct-16, 10:40 PM
Let's see. The equation E^2 = (mc^2)^2 + (pc)^2 for a single particle would actually describe the energy relative to an observer. So the energy of a positron and electron would be the sum of their individual energies to an observer, and this should match the energy of the light observed.

Okay, but are there two rays of light travelling in opposite directions? Do they have the same energy? If so, or if they can at least, then one observer in the forward direction should receive half of the energy according to the formula above and the other the other half, but the other half would be different to the other observer, depending upon their relative speed to the original particles as well, so one can find the difference in frequencies with relative speeds this way, by considering the two observers to be stationary to each other, and the original particles moving toward one of them.

grav
2007-Oct-16, 10:51 PM
According to that equation, a particle should have just as much extra energy moving toward the observer as it does moving away, so why wouldn't the frequency observed be the same as well? We can take the square root of (mc^2)^2 + (pc)^2 to be positive or negative, but the quantity would stay the same, and not provide for a frequency relative to the observer and source of light.

korjik
2007-Oct-17, 02:46 PM
That's interesting. So a positron and electron set very close together and originally stationary to each other would produce rays of light with a total energy of just 2mc2 while the additional energy would depend upon how much they pick up while accelerating toward each other? I wonder how that would work out if they each accelerated toward each other from infinity.

So that's what a positronium is? Why isn't it as stable as an electron and proton?

If all that is happening is the two particles accelerating one another, there is no additional energy.

You have to look at what is happening in the center of mass frame. if the center of mass is stationary, then all you have for energy is 2mc^2. If the center of mass is moving then there is another source of energy that makes the collision more complex. If you are talking about a relativistic center of mass, then what happens is beyond what I know.

Positronium is not stable because the two particles can annihilate. The collision is an allowed state. With an electron and proton, the collision is not an allowed state.

trinitree88
2007-Oct-17, 03:12 PM
If all that is happening is the two particles accelerating one another, there is no additional energy.

You have to look at what is happening in the center of mass frame. if the center of mass is stationary, then all you have for energy is 2mc^2. If the center of mass is moving then there is another source of energy that makes the collision more complex. If you are talking about a relativistic center of mass, then what happens is beyond what I know.

Positronium is not stable because the two particles can annihilate. The collision is an allowed state. With an electron and proton, the collision is not an allowed state.


korjik et al . Thanks. Kind of what I said initially. grav, if in principle you could create a positron/electron pair at rest. and in close proximity, all those equations fall out true. The look is at the momenta....two particles with no velocity have zero momenta....so they must travel at 180 degrees away from each other in order for plus momentum to equally, and oppositely cancel out minus momentum, and they MUST have exactly the same energy...(E=hv)...and photon momentum (E=hv/c)..except for the signs. Then the Law of Conservation of Momentum is upheld in the gamma photon-pair creation. Mastering that, you must also consider the possibility that three gammas emitted in a plane (coplanar), will also fill the bill nicely. No experiment has ever been done that violates conservation of momentum, it's near the top of the hierarchy of conservation laws.

In practice, particle/antiparticle pairs are created in collisions of a particle beam with a target. The pairs are separated out by charge, and a constant- momentum spectrometer. They are then collected in rings and "stored" until the beam density increases a lot to enhance the statistics in collisions. Early ring systems held semi-chaotic circulating particles or antiparticles. Later, the technique of stochastic cooling was invented....groups of particles are sensed as they circle the ring. "Slow" ones are speeded up, while "fast" ones are slowed down, like a bunch of jockeys fighting for the rail in a horserace. Since they travel at a substantial-fraction-of-light speed, the electronics must "sense" the bunch as they pass by, and a signal to modify (cool) them travels the shortcut across the ring to adjust the magnets and accelerating potential on the opposite side of the ring just before the "bunch" arrives. Tricky computing. The outcome is two countercirculating rings stacked above near each other, are abruptly diverted to an intersecting area where annihilations take place...(and a lot of interesting particle physics with implications for cosmology)...that's Fermilab's Tevatron in a nutshell.
Interestingly, annihilation of proton/anti-proton is quite "messy" with lots of energy and momenta distributed over the component quarks, and gluons interacting with antiquarks.antigluons....hence big messy jets of debris, but no free quarks. In electron/positron annihilations, they have no substructure to distribute the momenta/energy...so you can get a whopping big photon, or create a new particle anti-particle pair. However Bremstrahlung losses are greater in circulating electrons and positrons. What we need is a space-based LINAC...straight path, no Bremstrahlung, with stations set out over several light years............or use some incoming Tev cosmic rays instead. Nice. pete

alainprice
2007-Oct-17, 05:29 PM
You guessed, that equation I posted is for a single particle.

What you can do is take the event of annihilation as the origin. Use the formula for each particle to get the total energy.

BTW, if an electron and positron approach each other from infinity, the energy of annihilation is also infinite. You must set a limit on the distance between the particles. Think of the separation as potential energy, very much like the potential energy of a gravity well.

edit: P(momentum) is the usual mass x velocity . It's pretty much there for particles of no rest mass(photons).

grav
2007-Oct-20, 12:14 AM
If all that is happening is the two particles accelerating one another, there is no additional energy.

You have to look at what is happening in the center of mass frame. if the center of mass is stationary, then all you have for energy is 2mc^2.I'm thinking that only applies for the momentum. In the stationary center frame, the momentum of each particle is equal but opposite, so it cancels out to zero. As for the energy, however, they are both positive, becoming greater as they accelerate, and they would be added together, giving a positive value, regardless of the direction of motion, and the relativistic formula for energy gives the same value for the energy whether the particles are coming or going, as far as I can tell.

grav
2007-Oct-20, 12:15 AM
Thanks, everyone, for your replies. To simplify things quite a bit, let's say that a single particle is stationary with an observer, and then disintegrates into pure energy in the form of light, which then travels directly to the observer. Then the observer would measure the energy of that light as E=mc^2. And since E=hf also, whereas h should retain the same value in any frame if all things are truly relative, then the frequency of the light observed would be f = mc^2/h. So this is the frequency observed when stationary with the source of the light.

Now, relativistic Doppler says that the frequency will change with a relativistic speed to the source of light by sqrt[(1+v/c)/(1-v/c)], where v is positive upon approach. That means that the frequency observed will be this much greater or smaller than the frequency emitted, depending upon v. So the frequency observed should be sqrt[(1+v/c)/(1-v/c)] mc^2/h upon approach and sqrt[(1+(-v)/c)/(1-(-v)/c)] mc^2/h = sqrt[(1-v/c)/(1+v/c)] mc^2/h upon recession. And so the original energy of the particle should be less or greater likewise, but according to the formula for relativistic energy, it is the same either way. Heck, according to the classical formula for energy, it is the same either way, and it wouldn't work out any better by applying classical Doppler either. I'm getting that in order for redshift or blueshift to vary with the direction of travel of the source, then the original energy must also vary with the direction, even though the energies of particles, at least those involved in collisions, as far as I can tell from the formulas and explanations for them, are said to be always positive and add together, and is the same regardless of direction directly toward or away, unlike momentum. Is there anything I'm overlooking here? How can we get an energy for the light that depends upon the direction of travel of the source when the energy of the source is the same either way?

grav
2007-Oct-20, 12:52 AM
From the above post, in order for the energy of a particle to equal that of the frequency shift observed according to relativistic Doppler, then the energy would have to be E = sqrt[(1 + v/c)/(1 - v/c)] mc^2, so that E^2 = [(1 + v/c)/(1 - v/c)] (mc^2)^2, where m is the rest mass of the particle. If m were to be considered the relativistic mass, whereas mrel = m/sqrt[1 - (v/c)^2], then E = sqrt[(1 + v/c)/(1 - v/c)] * sqrt[1 - (v/c)^2] mrelc^2, so E = (1 + v/c) mrelc^2. Just wanted to run through it real quick to see what it would amount to. That last result is interesting.

alainprice
2007-Oct-20, 04:57 PM
If a stationary particle of mass Mo spontaneously converts to light, the frequency of light observed is only equal to f = mc^2/h if a single photon is created.

trinitree88
2007-Oct-22, 06:24 PM
If a stationary particle of mass Mo spontaneously converts to light, the frequency of light observed is only equal to f = mc^2/h if a single photon is created.

alainprice....mathematically true. Physically, there's no validity to a system that creates a piece of momentum from out of nowhere. Conservation of momentum is always seen in a fixed reference frame. A single "stationary" particle with rest mass has zero momentum as v is zero, so it cannot in reality produce a single photon...or neutrino....emanating away in one direction,two opposing or three coplanar photons works, as does two Z0's as a Z/anti-Z pair at 180 degrees. It's how they make sense of collisions in particle physics. Every detector conserves momenta, energy, charge...etc..... in it's algorithms. pete.

alainprice
2007-Oct-22, 06:35 PM
AFAIK, the most probable scenario(involving photons only) are 3 co-planar photons. So yes, to confirm your reasoning, a single photon and therefore a known frequency is not possible.

grav
2007-Oct-22, 08:46 PM
Well, I was just trying to use that to simplify things, but let's go back to the original scenario, which would probably be better anyway, since it portrays what might actually take place.

Two particles of equal rest mass move toward each other with an observer stationary to their center of momentum, so that the particles move along the line of sight of the observer, one toward the observer and one away. When they annihilate, light is produced in opposing directions along the same line, one directly toward the observer and one away, each with the same energy, that of the rest mass of one of the particles plus that extra energy due to their motion, which is always positive, and the same extra kinetic energy for each particle to the stationary observer, being stationary to the center of momentum. So the total energy of each photon is the same, greater than that for just the rest mass, and in opposing directions.

We will now reflect one of the photons off of a mirror, also stationary with the center of momentum and observer, so that it is reflected back in the same direction as the other photon. We now have two photons travelling in the same direction. The observer still observes each with the same energy each since the mirror was stationary with the observer and center of momentum, so all measurements take place in the stationary frame. So now we have two photons travelling together toward the stationary observer, each with the same energy, that of their rest mass and that due to their relative speed to the center of mass at the point of annihilation.

Now let's give the observer a small relative speed to the center of momentum and follow the same procedure as above. That will give the particle that was originally travelling toward the observer a slightly lesser energy due to the lesser relative speed between them, and that of the particle that was travelling away from the observer a slightly greater energy. The energies of the two photons that result from the annihilation and are then made to move in the same direction toward the observer will still each have the same energy, then, since the annihilation itself doesn't take place any differently, but each of the photons will become redshifted due to the relative speed of the observer to the stationary frame. The sum of the energy for the redshifted photons, then, in order for the law of conservation to hold, must be the same as the sum of the energies of the individual particles to the moving observer.

I thought about this while I wrote it, and I realize now that I can probably work through that scenario to find a formula for it, so I'll get to work.

grav
2007-Oct-22, 10:07 PM
Okay, if the particles each have a speed of v relative to the stationary frame at the center of momentum upon annihilation, while the observer moves away from them at vo relative to the stationary frame as well, then the observer should measure the energy of each particle as

Ea^2 = (m c^2)^2 + m^2 (v - vo)^2 c^2
Eb^2 = (m c^2)^2 + m^2 (v + vo)^2 c^2

where a is the particle travelling toward the observer and b is travelling away. The ratio of the frequency of light observed to that emitted in the stationary frame, then, which is proportional to the energies, is

fo / fe = [ sqrt[(mc^2)^2 + m^2(v+vo)^2 c^2] + sqrt[(mc^2)^2 + m^2(v-vo)^2 c^2] ] / [ 2 sqrt[(mc^2)^2 + m^2 v^2 c^2] ]

Dividing numerator and denominator each by mc^2, we get

fo / fe = [ sqrt[1 + (v/c + vo/c)^2] + sqrt[1 + (v/c - vo/c)^2] ] / [ 2 sqrt[1 + (v/c)^2] ]

That certainly doesn't look like the formula for relativistic Doppler to me, though. In fact, it gets worse. As vo increases past vo>v, the energy of the particles according to the observer would increase as well, but as vo increases, the redshift should become greater also, resulting in a lesser and lesser energy observed according to the frequency of the light, not greater. So that is a conflicting result. If anybody sees anything I missed, please let me know. In the meantime, I'll keep working on it. It seems to me that in order for this to work, there would have to be some variation of energy according to direction, even though the formula for energy does not imply that, as far as I can tell. I'll look into that some more as well.

grav
2007-Oct-23, 11:45 PM
Okay, I think I've got it all figured out. The frequency for the reflected light would stay the same as that of the other photon in the stationary frame if they both started out that way, and so they should remain the same frequency as each other for a moving observer as well, but redshifted if the observer is moving away from the point of annihilation. That tells us right there that a moving observer still measures the energy of each of the original particles as the same as well, regardless of the direction of motion, since the resulting frequencies of the two photons will remain the same as each other at all times, although redshifted or blueshifted depending upon the relative motion of the observer. So a particle moving away from an observer at some relative speed will have exactly the same energy as a particle moving toward the observer at the same speed, and always positive.

But let's say that a stationary observer sees the particles collide at 5 m/sec and -5 m/sec. The energy of the two photons is the same. So now let's say that the observer is moving away from the point of annihilation (center of momentum) at 2 m/sec. Now the particles are travelling at 3 m/sec and -7 m/sec in respect to the observer, but the energies of the two photons still remain the same, but redshifted. So let's say another stationary observer sees them travelling toward each other at 7 m/sec and -7 m/sec. Those energies are the same also. So the moving observer sees the energy of the particles moving at -7 m/sec and 3 m/sec as the same, and the stationary observer sees the energies of the particles moving at 7 m/sec and -7 m/sec as the same, so, if all is relative, then the energies of the particles moving at 3 m/sec and 7 m/sec must also be the same, and we can do this for all relative speeds, so all energies must be the same, regardless of the relative speed or direction. At first, this might seem like a contradiction, but let's explore it further.

The energy emitted by a particle at rest is E=mc^2, so if all energies are the same regardless of relative speed or direction, then this must always be the same. In a way, that immediately makes sense, because nothing can travel faster than c, so a particle that gives up all of its energy can give no more energy than E=mc^2. So why does the relativistic formula for energy appear to add to this? Think about a black hole with a single photon orbitting around it at c perpendicularly to our line of sight. Now, if we were to obtain a relative speed away from the black hole, then from our point of view the photon is now travelling around the black hole at c plus its additional speed along the vector of our line of sight, giving a total speed for the photon of sqrt(c^2 + v^2). Now, relativity says an observer always measures the speed of light at c, so something must differ for this photon. So if the black hole is moving away from us at v, then the photon must be orbitting at sqrt(c^2 - v^2) = c*sqrt(1 - (v/c)^2), so that the overall speed we would measure for the photon is sqrt[(sqrt(c^2 - v^2))^2 + v^2] = c. now what would make the photon appear to be orbitting at a slower rate with a relative speed to the black hole? Time dilation. A time dilation of sqrt(1 - (v/c)^2) for the black hole and everything associated with it would make the photon appear to be orbitting at sqrt(c^2 - v^2).

Okay, so now, applying this to our energy formula, using all relativistic terms, where also the relativistic mass is m/sqrt(1 - (v/c)^2), then

E^2 = [(m/sqrt(1 - (v/c)^2))*[c*sqrt(1 - (v/c))]^2]^2 + [(m/sqrt(1 - (v/c)^2))*[c*sqrt(1 - (v/c)^2)]*v]^2
E^2 = [(mc^2)*(sqrt(1 - (v/c)^2)]^2 + (mvc)^2
E^2 = (mc^2)^2 * (1 - (v/c)^2) + (mvc)^2
E^2 = (mc^2)^2 - (mc^2)^2 * (v/c)^2 + (mvc)^2
E^2 = (mc^2)^2 - (mvc)^2 + (mvc)^2
E^2 = (mc^2)^2
E = mc^2

So the energy is always E = mc^2 regardless of the relative speed or direction. So where do we get the different redshifts and energies from? Imagine an observer receiving light at some frequency from an unknown source. Then the same observer obtains some relative speed to the initial point of reception, away from the direction of the light. Now the light has become redshifted. Nothing has changed with the source, but the observer has obtained a different relative speed. The light that is received was already in transit when it redshifted, so if all is relative, then we can say that it is due to a relative speed relative to the light itself, and leave the source out of it. But relativity also says that the speed of the light must remain at c, so if the pulses are now received at longer intervals of 1 - v/c, and we consider that the moving observer would see a time dilation for the original point of reception of sqrt(1 - (v/c)^2), so that the original frequency measured would appear greater by sqrt(1 - (v/c)^2), then the frequency now measured is redshifted to (1 - v/c)/sqrt(1 - (v/c)^2) = (1 - v/c) /sqrt[(1-v/c)(1+v/c)] = sqrt[(1-v/c)/(1+v/c)] times the originally measured frequency, the same as for relativistic Doppler. The difference in energies, then is all due to the relative speed in as far as a redshift or blueshift in respect to a motion to the light itself is concerned.

But there was something else about this that puzzled me. Let's say we have two observers that are side by side measuring frequencies from two different unknown sources in the same direction, but each measures the same frequency. The two observers then aquire a relative speed of v to the original point of reception at the same time in the same direction, together, away from the sources. If all is relative, then that means that each still measures the same frequency as the other, otherwise they would be able to tell the relation between the speed and emitted frequency of the sources, depending upon how each measurement now differs, which could only happen if an absolute frame of reference existed. So each measures the same frequency as the other at all times, but let's say that one of the sources is stationary with the original point of reception and the other is not, but moving away from it and the observers, but emitting a higher frequency so that the received frequency is redshifted to the same as the other stationary source. Using the formula for relativistic Doppler, then, the redshifted frequencies for the moving observers should also come out the same, regardless of the relative speeds to the sources. If we figure that the moving source was moving away at vs and the observers aquire a relative speed to their original point of reception of v, and if we just added vs and v to find the resulting relative speed for the moving source, it would not work out. But if we just consider this resulting relative speed to be v', and take the ratio of the original frequencies received to the final ones, which should be the same, and set them equal according to the relative speeds to the sources, then we get

[ f ] / [ f * sqrt((1 - v/c)/(1 + v/c))] = [f' * sqrt((1 - vs/c)/(1 + vs/c)] / [f' * sqrt((1 - v'/c)/(1 + v'/c))], where f is the frequency emitted and received from the stationary source, and the denominator on the left is the frequency measured with a relative speed of the observers, and f' is the emitted frequency of the moving source, so the numerator on the right is the original frequency received and the denominator is the frequency received with some new relative speed to the moving source, so

1/sqrt[(1 - v/c)/(1 + v/c)] = sqrt[(1 - vs/c)/(1 + vs/c)]/sqrt[(1 - v'/c)/(1 + v'/c)]
(1 - v'/c)/(1 + v'/c) = [(1 - vs/c)/(1 + vs/c)][( - v/c)/(1 + v/c)]
(1 - v'/c)/(1 + v'/c) = z
(1 - v'/c) = (1 + v'/c)z
(v'/c)(1 + z) = 1 - z
v'/c = (1 - z)/(1 + z)

z = (1 - vs/c - v/c + vs*v/c)/( 1 + vs/c + v/c + vs*v/c)

1 - z = 2(vs/c + v/c)/(1 + vs/c + v/c + vs*v/c)

1 + z = 2(1 + vs*v/c)/(1 + vs/c + v/c + vs*v/c)

v'/c = (1 - z)/(1 + z) = (vs/c + v/c)/(1 + vs*v/c^2)

So this is the resulting relative speed, the same as relativity says as well. If a stationary observer sees a source moving away at vs and another moving observer travelling in the opposite direction at v, then the moving observer will see the source moving away at v'. And this is all the direct result of a time dilation and measuring c the same in all frames. The thing is, though, that time dilation doesn't just mess with the simple addition of relative speeds, then, but with the law of conservation of energy, and with the law of conservation of momentum as well, I'm sure, since it is also based upon relative speeds. They still work, but we would now have to use the original and final relative speeds, which don't add together as normal.

grav
2007-Oct-26, 11:24 PM
Imagine an observer receiving light at some frequency from an unknown source. Then the same observer obtains some relative speed to the initial point of reception, away from the direction of the light. Now the light has become redshifted. Nothing has changed with the source, but the observer has obtained a different relative speed. The light that is received was already in transit when it redshifted, so if all is relative, then we can say that it is due to a relative speed relative to the light itself, and leave the source out of it. But relativity also says that the speed of the light must remain at c, so if the pulses are now received at longer intervals of 1 - v/c, and we consider that the moving observer would see a time dilation for the original point of reception of sqrt(1 - (v/c)^2), so that the original frequency measured would appear greater by sqrt(1 - (v/c)^2), then the frequency now measured is redshifted to (1 - v/c)/sqrt(1 - (v/c)^2) = (1 - v/c) /sqrt[(1-v/c)(1+v/c)] = sqrt[(1-v/c)/(1+v/c)] times the originally measured frequency, the same as for relativistic Doppler. The difference in energies, then is all due to the relative speed in as far as a redshift or blueshift in respect to a motion to the light itself is concerned.This explanation of relativistic Doppler seems a little off somehow. Let me try that again. The observer that is moving relative to the original point of reception would say that another that remained at the point of reception is receiving the frequency 1 + v/c faster, so the moving observer is receiving it 1/(1 + v/c) slower in comparison. Also, the moving observer would say that the clocks are ticking slower for the other one that is stationary by sqrt(1 - (v/c)^2), so the other must be receiving the frequency 1/sqrt(1 - (v/c)^2) faster according to their own clock. So the moving observer is receiving it sqrt(1 - (v/c)^2) slower in comparison. So the resulting frequency is sqrt(1 - (v/c)^2)/(1 + v/c) = sqrt[(1 - v/c)/(1 + v/c)] slower for the observer moving away from the point of reception.

We could also do it by finding for what the stationary observer sees of the moving one. The stationary one would say that the moving one is receiving the frequency of the light 1 - v/c slower and that the moving observer's clock is also moving sqrt(1 - (v/c)^2) slower, so that they receive the frequency 1/sqrt(1 - (v/c)^2) faster, giving a frequency for the moving observer that is (1 - v/c)/sqrt(1 - (v/c)^2) = sqrt[(1 - v/c)/(1 + v/c)] slower than the stationary one. So it's the same difference either way, and the inverse of these for how the stationary observer receives the light compared to the moving one.

grav
2007-Oct-26, 11:50 PM
Think about a black hole with a single photon orbitting around it at c perpendicularly to our line of sight. Now, if we were to obtain a relative speed away from the black hole, then from our point of view the photon is now travelling around the black hole at c plus its additional speed along the vector of our line of sight, giving a total speed for the photon of sqrt(c^2 + v^2). Now, relativity says an observer always measures the speed of light at c, so something must differ for this photon. So if the black hole is moving away from us at v, then the photon must be orbitting at sqrt(c^2 - v^2) = c*sqrt(1 - (v/c)^2), so that the overall speed we would measure for the photon is sqrt[(sqrt(c^2 - v^2))^2 + v^2] = c. now what would make the photon appear to be orbitting at a slower rate with a relative speed to the black hole? Time dilation. A time dilation of sqrt(1 - (v/c)^2) for the black hole and everything associated with it would make the photon appear to be orbitting at sqrt(c^2 - v^2). I used a photon travelling perpendicular to the line of sight for this, but I was wondering how this would work out if the line of sight lay upon the same plane as the orbit of the photon around the black hole. Well, at the points in front and in back, where the photon crosses the line of sight directly perpendicularly, it is still travelling at c*sqrt(1 - (v/c)^2) relative to the black hole from the observer's point of view, becoming sqrt[(sqrt(1 - (v/c)^2))^2 + v^2] = c relative to the observer. But on the left and right sides of the orbit, since it still travelling at c all the way around to the observer while the black hole moves at v along the line of sight, the observer would say that the photon is travelling at c-v and c+v relative to the black hole. And an observer stationary with the black hole would say the same thing of the moving observer. But let's see what the formula for rrelativistic speeds says about this. For c-v, it becomes c' = [(c-v) + c]/[1 + (c-v)c/c^2] = (-v)/[1 + (c-v)/c] = (-v)c/[c + (c-v)] = (-v)c/(-v) = c. So even though each thinks the other might see the photon travelling at c-v, depending upon the side of the orbit, since the speeds don't add, the other really still sees it travelling at c. Now let's try it for c+v. We get c' = [(c+v) + c]/[1 + (c+v)c/c^2] = (2c + v)/[1 + (c+v)/c] = (2c + v)c/[c + (c+v)] = (2c + v)c/(2c + v) = c. Same thing.

grav
2007-Oct-27, 12:12 AM
I found that the energy of particles must remain the same regardless of relative speed or direction, but redshifted or blueshifted by relative motion between points of reception in regard to the light itself, without concern for the motion of the source or emitted frequency. Of course, this can still translate to a redshift or blueshift due to the relative motion to the source in a roundabout way as well, as found by using the addition for relative speeds. But no matter what the relative speed of the particles, to each other or to an observer, they should always annihilate with E=mc^2. So to an observer in the stationary frame of the center of momentum, if I have done this right, there will be no redshift or blueshift, and so the total energy of annihilation of the two particles will always be measured at E=2mc^2, regardless of how they were moving before the collision.

But this is using all relativistic terms, so I was wondering if there was any other way it could work out otherwise. Well, the equal energies in any frame thing had nothing to do with relativity, so it applies no matter what. If the equal energy in every frame was simply zero, then the internal energy of the particles would have to be negative, in order to cancel out the positive kinetic energy. But there is no such thing as negative energy, as far as I know. So if the total energy is always nonzero and positive, then there should be something about it that remains constant in every frame, otherwise the energy would become arbitrary between frames. It is difficult to explain exactly what I mean by that, though, but one might see what I mean if they think about it a minute. Something about the energy should remain constant in all frames, either the total energy altogether or one of its components, allowing an observer in the same frame to measure it in the same way as any other frame. Of course, the mass of all particles is not constant, and neither can the energy be if we measure for less than the entire particle, since any mass can be broken down into two or more parts, and each of those parts cannot have the same energy as the original whole, because that would upset the conservation of energy to begin with. So that just leaves the speed component, which must be constant, then. The speed of light.

So it would seem that E=mc^2 is a definite, then, and so is the constancy of the speed of light. But this wouldn't necessarily mean that the speed of light is a constant in all frames in itself, but only relative to the source of emission, or in the same frame as the particles themselves that are then annihilated. But it does seem that in order for c to be constant to the particles in each frame, that it would also remain constant when going from one frame to another, so always c to the receiver as well, regardless of the frame or relativistic speed. I will have to look into this some more, as well as how the relativistic mass is affected between frames, to see what the specific case should be.

Eta C
2007-Oct-29, 03:51 AM
Grav et al. This is really a simple question. You need to boost to the center of mass frame and determine the total available energy. A simple example is in a e+ e- collider such as SPEAR, the one I did my dissertation work on. There the colliding beams are in the CM frame, and the available energy is more or less the beam energy as it outweighs the mass energy of the electrons. Also, if you sit at a resonance energy, the result of the annihilation may not be two photons. You may produce a particle out of the available energy.

To give an example. If you run the two beams at 1.55 GeV you have a center of mass energy of about 3.1 GeV, right on the J/Psi resonance. The 1.2 MeV of mass energy from the electrons is negligable comapred to the beam energy. If you up the beam energy to about 1.885 GeV you'll be at a CM energy of 3.77 GeV, the Psi'' resonance.

This probably muddies the waters but you should realize it's not really a hard question. The total energy in the CM frame determines what the available energy for the collision is.