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Bill Thmpson
2003-Apr-02, 03:18 PM
OK, here is the deal. since light, uh, travels at the speed of light, I cannot see how it could possibly have mass when I apply my understanding of the theory of relativity and the fact that inertia increases as something approaches the speed of light.

I am currently arguing with a co-worker, who insists that light has mass. Since I am getting paid to work and not argue, I would like to bring this "discussion" to an end. Please answer this question. Do photons have mass?

I will say this. I heard about a year ago that there was a discovery that photons were discovered to change some charasteristics and scientists were thinking that this was only possible if photons have mass. Can anyone elaborate?

Also, what about this whole thing about light being both particles and energy and Einstein being upset about this and some physicist proving that it is half and half in a strange way.

[I edited because of embarassing spelling errors]

Eta C
2003-Apr-02, 03:34 PM
In general relativity, anything moving at c must have a rest mass of 0. This includes photons and, until recent discoveries proved otherwise, neutrinos. Remember, however, that mass and energy are not different (E=mc2). So photons, despite being massless, are effected by gravity.

Some fringe theories do posit a possible mass for the photon, but if true would require some massive rebuilding of theories.

As to wave, particle duality it's more of a semantic problem. If asked if light is a particle or wave the correct response is neither. Rather it is something that we then decribe as having wave properties when appropriate (diffraction, interference) and particle properties (photoelectric effect) when appropriate. Remember that this duality applies to everything, electrons, protons and other particles can also display "wave" properties. So in short, the old, classical, definitions of wave and particle do not apply to the world as described by quantum mechanics.

patrioticamerican
2003-Apr-02, 03:40 PM
Photons have no mass - if they had mass they couldn't travel at the speed of light! They do exhibit what is called wave-particle duality, meaning that in certain experiments they exhibit the properties of waves, such as interference patterns, and in others particles, such as the photoelectric effect.

I believe what you're referring to is Einstein's disagreement with Heisenberg's Uncertainty Principle, from which comes his famous quote, "God does not play dice with the Universe".

Laser Jock
2003-Apr-02, 04:02 PM
I will say this. I heard about a year ago that there was a discovery that photons were discovered to change some charasteristics and scientists were thinking that this was only possible if photos have mass. Can anyone elaborate?


You're confusing photons with neutrinos. There is some evidence that some types of neutrinos have mass, but photons do not.

JS Princeton
2003-Apr-02, 05:01 PM
Photons do have an effective mass, though. One can see this by considering an experiment of a photon in a box bouncing back and forth. The box itself will expereince a change in momentum which will be equivalent to h{nu}/c^2.
That's the "effective" mass of a photon.

The rest mass of the photon is zero, but the effective mass does contribute to the gravitational field, interestingly enough.

snowcelt
2003-Apr-02, 05:30 PM
JS Priinceton wrote that "(t)he rest mass of the photon is zero, but the effective mass does contribute to the gravitational field, inerestingly enough". Incredible! I would be interested in knowing where this mass fits in; in the total mass of the universe. Thankyou

JS Princeton
2003-Apr-02, 05:34 PM
The "total mass" of the universe is a highly subjective measurement. Rather, what is generally discussed is an energy-density of the universe. Indeed, the photon energy density is a component of this, and was at earlier times an extremely important component. That was the so-called "radiation dominated" epoch of the universe. We then passed into matter domination and are now into dark-energy domination. The photon energy density is so low that it really doesn't contribute a substantial amount to shape the present universe.

snowcelt
2003-Apr-02, 06:49 PM
If we have passed into a "dark-energy" epoch, does this mean that we may eventually enter some other epoch in the far future? A place where photons will have no place?

DStahl
2003-Apr-02, 09:01 PM
Begorrah, a photon has kinetic energy, that's for certain--it spins those little thingamabobs with the vanes painted black on one side and white on the other, enclosed in glass globes. How can something without mass still have kinetic energy, your co-worker might ask? Usually when people say "That particle has mass" they mean "rest mass" and not "relativistic mass"--which, if am understanding JS Princeton and the others correctly, is what a photon has.

But in a table of particle characteristics, a photon is listed as massless: it has no mass in the way that particles of matter have mass. The so-far-unobserved graviton is also massless, and it is unclear to me whether all neutrinos are now thought to have a tiny mass or whether there is room in neutrino theory for a massless neutrino. The even more hypothetical and unobserved tachyon is supposed to have a mathematically imaginary mass, related to the square root of negative one. Sheesh!

JS Princeton
2003-Apr-02, 09:05 PM
Begorrah, a photon has kinetic energy, that's for certain--it spins those little thingamabobs with the vanes painted black on one side and white on the other, enclosed in glass globes.

"Kinetic" energy is really poorly defined for photons. Better to say that a photon has momentum


How can something without mass still have kinetic energy, your co-worker might ask? Usually when people say "That particle has mass" they mean "rest mass" and not "relativistic mass"--which, if am understanding JS Princeton and the others correctly, is what a photon has.

No, a photon has a zero restmass. That's part of the reason it can never be at rest, for then it would have zero energy and no longer be anything at all!



and it is unclear to me whether all neutrinos are now thought to have a tiny mass or whether there is room in neutrino theory for a massless neutrino.

If the neutrinos change flavor the way we think they do, all three of them have to have mass.

JS Princeton
2003-Apr-02, 09:06 PM
If we have passed into a "dark-energy" epoch, does this mean that we may eventually enter some other epoch in the far future? A place where photons will have no place?

Photons really have no place in our present universe. Locally, they can do all sorts of damage, but looking at the universe as a whole there's really no difference between a universe with photons and a universe without photons at this point.

DStahl
2003-Apr-02, 09:21 PM
JS Princeton--OK, momentum. That works for me.

I really love the conciseness of this: "No, a photon has a zero restmass. That's part of the reason it can never be at rest, for then it would have zero energy and no longer be anything at all!"

DaveC
2003-Apr-02, 10:13 PM
Begorrah, a photon has kinetic energy, that's for certain--it spins those little thingamabobs with the vanes painted black on one side and white on the other, enclosed in glass globes.

This would be a Crookes radiometer, and the motion isn't really a result of photon momentum at all. The globe is partially evacuated and light falling on the dark side of the vane causes radiative heating of the air at the vane's surface. As the air heats and expands it exerts a tiny pressure differential across the vane causing the mechanism to rotate.
If the bulb were completely evacuated, and the bearing were frictionless, it is possible that photons bouncing off the white (or silver) side of the vanes could cause movement in the opposite direction - that is the light colored side of the vane would move away from the light source. I'd guess you'd need a powerful source of photons to induce movement in the real world where frictionless bearings dont exist.

Bill Thmpson
2003-Apr-02, 10:31 PM
The neutrinos changing flavor is what what I think I heard about a year ago.

But, playing devils advocate -- and since I am pretty sure I am going to be asked this by my coworker -- how DO photons move those blades of those little toys that are painted black and white if they do not have a mass?

And, let's not forget those designs for a star ship using sails that reflect off of giant sails and push the craft slowly, over time, toward the speed of light. (maybe this is solar wind it is riding and this is somehow different) If this design relys on light pushing against it, does this not imply that photons have mass.

If Light does not have mass, how can it seemingly move things?

JS Princeton
2003-Apr-02, 11:00 PM
Well, DaveC explained the answer to the Crookes radiometer well, but in a vacuum, light still has momentum and in principle the ability to "push" things around.

A photon must have momentum since it has energy and no mass. Perhaps you've seen the relativistic expression for energy?

E^2=m^2c^4 + p^2c^2

For a photon, m=0 (there is a zero restmass) so all the energy must be in "p" which is the symbol for momentum. In effect, for a photon where m=0

E/c=p

Divide the energy of your wave by the speed of light and you get a momentum.

Incidetally, the equation also works for a massive body at rest. In this case, the massive body has no momentum (p=0), but it has a mass (let's call it "m"). This means the equation reduces to:

E=mc^2

Look familiar?

One can convert from one form of energy to another, and in point of fact, one can convert radiation energy (in the form of light) to kinetic energy (in the form of some particle moving). A dramatic example of this is the so-called "photoelectric effect" which is explained well here:

http://www.colorado.edu/physics/2000/quantumzone/photoelectric.html

In effect, the photon imparts a momentum on an electron in order to get it to move. In fact, sometimes you get reflected or scattered photons when you do the experiment that have a different energy. If you calculate the momentum difference associated with the change in energy ({delta}E/c) you will find that that's exactly the amount of momentum imparted on the electron. Thus the photons have the ability to exert a pressure (a force per unit area) on an electron, just like billiard balls.

More resources on the "mass" of light:

http://math.ucr.edu/home/baez/physics/Relativity/SR/light_mass.html

Bill Thmpson
2003-Apr-02, 11:37 PM
Very interesting.

I am impressed. This has been a very enlightening read!

Much thanks to everyone who contributed to this discussion !

:)

RickNZ
2003-Apr-03, 12:14 AM
Explain again to me please how is it that photons are effected by gravity?

Hey another question while were on the subject. Is it the wave or the particle aspect of the photon that travels at c?

DStahl
2003-Apr-03, 12:45 AM
DaveC: "This would be a Crookes radiometer, and the motion isn't really a result of photon momentum at all. The globe is partially evacuated and light falling on the dark side of the vane causes radiative heating of the air at the vane's surface. As the air heats and expands it exerts a tiny pressure differential across the vane causing the mechanism to rotate."

HO! Bad science on my part. Thanks for the correction. Let's see if I can put a mental post-it note on that one so I remember it next time!

JS Princeton
2003-Apr-03, 02:45 AM
Explain again to me please how is it that photons are effected by gravity?

Photons will always travel in geodesics which are paths that depend on the curvature of space (that's what gravity is due to). This is much like how planes take Great Circles when going from point A to point B. In effect, photons will travel the shortest distance between two points (principle of least action).



Hey another question while were on the subject. Is it the wave or the particle aspect of the photon that travels at c?

The wave and the particle both travel at the speed of light. The wave and particle are indistinguishable features.

RickNZ
2003-Apr-03, 03:15 AM
Hey another question while were on the subject. Is it the wave or the particle aspect of the photon that travels at c?

The wave and the particle both travel at the speed of light. The wave and particle are indistinguishable features.

Lets just assume im a pleb here ;) If the wave travels at c then would that not mean the photon particles need go much faster than c to travel in a wave pattern?

Unless the wave aspects are from the medium light travels thru. Woohoo ether does exist! LoL

Sigh i feel thick.

DStahl
2003-Apr-03, 03:35 AM
For a discussion of that, see this topic (http://www.badastronomy.com/phpBB/viewtopic.php?t=4410).

RickNZ
2003-Apr-03, 03:54 AM
Cheers for that, im on the right path now.


BTW is it possible for light to be made up by more than one particle type?

JS Princeton
2003-Apr-03, 06:49 AM
In as much as you can have different polarization of photons, light can be made up of two distinct kind of particles (photon polarizations). However, in order to be electromagnetic radiation, the fields that are coupled to each other (the E and B fields) are always the same "kind" of field. Therefore there is only one "kind" of photon.

DoctorDon
2003-Apr-03, 12:58 PM
My understanding is that it's not actually photon impact that drives those little spinning black/white light bulb things. I sat down with one once and looked at it closely and found that it actually spins in the reverse direction of what you'd expect from photon momentum transfer (you'd expect it to spin away from the white side, because reflection transfers twice as much momentum as absorption, but it spins away from the black). I did some poking about, and read that what's really going on is that the impact of light causes the black material to shed molecules, and it's the momentum of the ejection of the molecules that causes the thing to spin.

Seemed to make sense. 'Course, I have no way to experimentally verify it.

Don

kucharek
2003-Apr-03, 01:06 PM
The bulb with the light-mill isn't completely evacuated. The air above get's a little bit more warmed than the air over the silver side. Warm air gives more bumbs, so the mill begins to turn. Light-mills don't work in a perfect vacuum. Your first reasoning about your observation was correct, but not your conclusion.

Harald

DoctorDon
2003-Apr-03, 02:33 PM
The bulb with the light-mill isn't completely evacuated. The air above get's a little bit more warmed than the air over the silver side. Warm air gives more bumbs, so the mill begins to turn. Light-mills don't work in a perfect vacuum. Your first reasoning about your observation was correct, but not your conclusion.


Well, that's another reasonable-sounding theory that I have no way to confirm or falsify. :-) If I had one, I could pump all the air out and see if it still worked. If not, that would falsify the "black enamel emits more molecules" theory, and lend support to the "hot air near black side has increased pressure" theoy. :-) But I have neither mill nor pump, alas, so I must remain uncertain.

And it wasn't *my* conclusion -- as I said, it was an explanation I read somewhere.

Don

Reacher
2003-Apr-03, 03:13 PM
ok, 2 things. firstly, someone happened to mention that a photon at rest has no energy, and therefore is nothing. well, then, that recently invented machine( its in another strand) that slows down light, would cause a photon to lose energy... then the universe loses energy, etc. Secondly, i have read in one of my science books "the universe explained," that when a particle passes on a trajectory that eventually puts it on a rough "tangent" (ps. is this the correct word, considering that a black hole is spherical, or at least its event horison is, or something) with a black hole, then some electrons can be sheared off, and the universe looses mass, ore energy, or some such thing. well, by my understanding of what a black hole is - the singularity thing - then the black hole is a part of the universe, so the universe loses no mass, or energy.

Reacher
2003-Apr-03, 03:18 PM
pps: i cant work the quote thingy.

Bill Thmpson
2003-Apr-03, 04:08 PM
It seems my co-worker still insists that photons have mass. I have asked him to post his views here.

I hope he does.

Maybe he is shy.

:oops:

:roll:

JS Princeton
2003-Apr-03, 06:01 PM
ok, 2 things. firstly, someone happened to mention that a photon at rest has no energy, and therefore is nothing. well, then, that recently invented machine( its in another strand) that slows down light, would cause a photon to lose energy... then the universe loses energy, etc.

The "slowing down of light" can be described qualitatively as due to scattering interactions. In effect, what's going on is that the light travels at the speed of light and then has some "interaction time" with an atom or a molecule. This "interaction time" basically relies on changing the light's properties. We had a discussion earlier about whether one can say that the photon was "destroyed" and "recreated", but generally the rule is that when a photon is interacting it ceases to be a photon in the real sense of the word. Really a photon is only a photon when it isn't interacting. Thus it travels in its geodesic at the speed of light and doesn't really slow down in the machine you refer to with the high index of refraction. However, it appears to be that way due to the nature of scattering in a medium. The details are messy but we can go into them. The quick and dirty way to answer the question is that the rules about photon energy and propagation only apply when you are in a vacuum.


Secondly, i have read in one of my science books "the universe explained," that when a particle passes on a trajectory that eventually puts it on a rough "tangent" (ps. is this the correct word, considering that a black hole is spherical, or at least its event horison is, or something) with a black hole, then some electrons can be sheared off, and the universe looses mass, ore energy, or some such thing. well, by my understanding of what a black hole is - the singularity thing - then the black hole is a part of the universe, so the universe loses no mass, or energy.

Hmm, I don't exactly know what you are referring to here. It sounds to me like you are referring to a paraphrased version of Hawking Radiation (http://casa.colorado.edu/~ajsh/hawk.html), but the details of your description are quite different. A black hole really can't have electrons "sheared" off of it because there really can be no way of accelerating an electron past the event horizon. However, quantum mechanics has some interesting paradoxes, among them tunneling which allows for Hawking Radiation.

SeanF
2003-Apr-03, 07:25 PM
Secondly, i have read in one of my science books "the universe explained," that when a particle passes on a trajectory that eventually puts it on a rough "tangent" (ps. is this the correct word, considering that a black hole is spherical, or at least its event horison is, or something) with a black hole, then some electrons can be sheared off, and the universe looses mass, ore energy, or some such thing. well, by my understanding of what a black hole is - the singularity thing - then the black hole is a part of the universe, so the universe loses no mass, or energy.

Hmm, I don't exactly know what you are referring to here. It sounds to me like you are referring to a paraphrased version of Hawking Radiation (http://casa.colorado.edu/~ajsh/hawk.html), but the details of your description are quite different. A black hole really can't have electrons "sheared" off of it because there really can be no way of accelerating an electron past the event horizon. However, quantum mechanics has some interesting paradoxes, among them tunneling which allows for Hawking Radiation.

I think Reacher meant electrons would be sheared off the particle, not off the black hole . . . is that feasible?

JS Princeton
2003-Apr-03, 07:59 PM
I think Reacher meant electrons would be sheared off the particle, not off the black hole . . . is that feasible?

That might be interpreted just as the photoelectric effect, except conservation of momentum will demand that you see an energy change in the photon (it isn't unaffected, in other words).

Bill Thmpson
2003-Apr-04, 04:56 PM
My shy, misguided coworker has emailed me the following:


"If you want to calculate relative mass of photon you can use E = mc^2, where photon's energy can be calculated using
E = hv, where v is light frequency and h is Planck's constant."

Pshew! I have no idea where he is going with this. Is this something along the line of thought:

"If you cannot dazzel them with brilliance, baffle them with B.S"?

Is he resulting to trying to impress me? Can anyone elaborate on this? Does any of this make sense to anyone with a better background in Physics than I?

ToSeek
2003-Apr-04, 05:13 PM
My shy, misguided coworker has emailed me the following:


"If you want to calculate relative mass of photon you can use E = mc^2, where photon's energy can be calculated using
E = hv, where v is light frequency and h is Planck's constant."

Pshew! I have no idea where he is going with this. Is this something along the line of thought:

"If you cannot dazzel them with brilliance, baffle them with B.S"?

Is he resulting to trying to impress me? Can anyone elaborate on this? Does any of this make sense to anyone with a better background in Physics than I?

I think what he's saying is that if you have a photon with frequency v, you can calculate its mass (effectively, not its rest mass) as hv/c^2. I haven't checked this, but it sounds correct.

Bill Thmpson
2003-Apr-04, 05:24 PM
I think what he's saying is that if you have a photon with frequency v, you can calculate its mass (effectively, not its rest mass) as hv/c^2. I haven't checked this, but it sounds correct.

Okay... So, I need to look up this "Plank's Constant". But, while I do, does this mean we can say that a photon is "x grams"? where x is some real number?

I mean, "relative mass" isn't the same thing as "mass" as the commonly accepted difinition of the word, is it?

And is Plank's equation a real thing or something that is theoritical?

Wait a minute, if m = hv/c^2 then the mass would be determined on how great v is. So with the increase of, V the photon is more massive? This sounds like B.S. Am I wrong?

Bill Thmpson
2003-Apr-04, 06:05 PM
Before I respond to my good friend, I thought I would put my response in this forum and have you, experts, have a look at it:


E=mc^2 is the Theory of Relativity which illustrates how mass (not relativistic mass) and energy are interchangeable. Among other things, it is used to see how much energy would be generated if a certain amount of mass was converted into energy.

I think that Plank’s equation deals with quantum mechanics. So merging the two equations into one (using simple algebra), you get something strange m=hv/c^2 . But I think this is a misuse of the equation. The theory of Relativity will give you a specific mass for an amount of energy. But m=hv/c^2 gives you a specific mass for the frequency of light since both c and h are both constants.

So, using your logic, the mass of a photon increases as the frequency increases. Watch out, here comes a massive photon! Duck, everybody!!

This might be the quantum mechanics term “relativistic mass” but it is not mass. This is not mass in the commonly accepted definition of mass.

How does this sound?

ToSeek
2003-Apr-04, 06:08 PM
Wait a minute, if m = hv/c^2 then the mass would be determined on how great v is. So with the increase of, V the photon is more massive? This sounds like B.S. Am I wrong?

Yes. The higher the frequency of light, the more energy it has, and therefore the more mass-equivalent it has.

JS Princeton
2003-Apr-04, 06:10 PM
My shy, misguided coworker has emailed me the following:


"If you want to calculate relative mass of photon you can use E = mc^2, where photon's energy can be calculated using
E = hv, where v is light frequency and h is Planck's constant."



As we've said earlier in this post, this is one definition you can give to mass. It is not, however, the definition physicists choose to use. physicists define the mass of the photon to be determined by the rest mass which is necessarily zero.

If a photon gets absorbed by an atom and you take the difference of the mass of the atom before and after the absorption, you will indeed find that the atom has increased in mass by h{nu}/c^2. However, this is an "after the fact" measurement of the "mass" of the photon. The photon itself is said to have all its energy tied up in momentum and none of it in rest mass.




Pshew! I have no idea where he is going with this. Is this something along the line of thought:

"If you cannot dazzel them with brilliance, baffle them with B.S"?



No, there actually is some content to his idea, it just isn't accepted as the definition for rest mass. Check the John Baez FAQ link I posted above for more.

ToSeek
2003-Apr-04, 06:14 PM
If a photon gets absorbed by an atom and you take the difference of the mass of the atom before and after the absorption, you will indeed find that the atom has increased in mass by h{nu}/c^2. However, this is an "after the fact" measurement of the "mass" of the photon. The photon itself is said to have all its energy tied up in momentum and none of it in rest mass.


Would the photon's gravitional effects also be based on a mass of h{nu}/c^2 (assuming you could measure something that small)?

JS Princeton
2003-Apr-04, 06:20 PM
Okay... So, I need to look up this "Plank's Constant". But, while I do, does this mean we can say that a photon is "x grams"? where x is some real number?

That's Planck, after the great quantum physicist Max Planck who discovered all sorts of things about the quantized nature of the universe.

As I said above, a photon's restmass is zero. However, one can do some experiments where one might infer that one had measured an effective mass of a photon that was "x grams" where x is a real number.



I mean, "relative mass" isn't the same thing as "mass" as the commonly accepted difinition of the word, is it?


No, and therein lies the problem.

There are so many definitions of mass that I'm not even going to attempt to get into the subject. Let's just say that we aren't quite sure what exactly "mass" is, but we have a pretty good idea how to measure it.

The best definition for mass we have right now is the following:

Sqrt[E^2-p^2c^2]/c^2=m

where c is the speed of light, p is the momentum and E is the energy. Now you're going to ask me what energy and momentum are and how they are measured. Well, the answer to that gets you rather deeply into quantum mechanics. A bit too deep probably for this discussion. Let's just say that they are well-defined quantities for any system moving at any speed (much better defined than mass, for example).



And is Plank's equation a real thing or something that is theoritical?


Very real indeed. Also part of theory, but theory isn't speculation, it's a way of describing reality.



Wait a minute, if m = hv/c^2 then the mass would be determined on how great v is. So with the increase of, V the photon is more massive? This sounds like B.S. Am I wrong?

Note that V here isn't velocity but rather frequency. What the statement is saying is that as the frequency increases the energy would increase (and therefore the effective mass associated with said photon). Indeed, higher frequency photons (x-rays, gamma rays) are more energetic. So it's not B. S.

JS Princeton
2003-Apr-04, 06:24 PM
Would the photon's gravitional effects also be based on a mass of h{nu}/c^2 (assuming you could measure something that small)?

This is a delicate question. The basic answer is yes, however it's not the same sort of curvature as is due to traditional mass. In particular, we are dealing with a photon that doesn't have anything to enter into the mass-density term of the stress-energy tensor, but does have some pressure. That's what really would be causing the gravitational curvature. At the most naive level, the answer is, however, that the effective mass is the quantity that contributes to gravitational effects.

JS Princeton
2003-Apr-04, 06:33 PM
[quote]E=mc^2 is the Theory of Relativity which illustrates how mass (not relativistic mass) and energy are interchangeable. Among other things, it is used to see how much energy would be generated if a certain amount of mass was converted into energy.


Yes, but the full story is that E^2=m^2c^4-p^2c^2. Only if your particle is at rest does E=mc^2 apply.



I think that Plank’s equation deals with quantum mechanics.

Not really. It only deals with the way one can measure the energy of light based on it's frequency. Higher frequency --> higher energy.


So merging the two equations into one (using simple algebra), you get something strange m=hv/c^2 . But I think this is a misuse of the equation. The theory of Relativity will give you a specific mass for an amount of energy. But m=hv/c^2 gives you a specific mass for the frequency of light since both c and h are both constants.

That's right. You are assuming all of the energy associated with the photon is somehow converted into "rest mass". This is in principle something that can be done by having the photon, for example, become absorbed by an atom.

The details of this are more complicated, however. Unless you destroy the photon out and out or have it interact with matter from your measuring device you cannot measure this associated rest mass.



So, using your logic, the mass of a photon increases as the frequency increases. Watch out, here comes a massive photon! Duck, everybody!!


No, the photon isn't massive, it's more energetic. There's a big difference. The photon has a zero mass and a lot of energy.

You better duck, though. Highly energetic photons can cause damage.



This might be the quantum mechanics term “relativistic mass” but it is not mass. This is not mass in the commonly accepted definition of mass.

Quantum mechanics only uses rest mass. In fact, there is no use for relativistic mass at all any more except in cocktail conversations about the peculiarities of relativity.

ToSeek
2003-Apr-04, 06:40 PM
You better duck, though. Highly energetic photons can cause damage.


Highly energetic photons, otherwise known as gamma rays (http://imagers.gsfc.nasa.gov/ems/gamma.html).

Bill Thmpson
2003-Apr-04, 10:52 PM
Interesting.

OK, I am considering the following as a reply to my co-workers comment.

Can anyone tell me if I the following statement is correct, or am I off the mark?

I know this will sound simplistic to all you Professors and Grad Students. But I think a simple responce to his comments is what is required:


The Theory of Relativity is not a theory of Equality. E=mc^2 states that mass and energy are interchangeable but it does not stipulate that they are equal in a sense that they are one and the same thing. If they were, they would obviously be measured the same way (for instance, mass is measured by grams but energy is not).

E=mc^2 uses a constant, c, which is the speed of light. Multiplying mass by the speed of light squared CHANGES the mass.

This equation can be written another way, m = E/c^2. Once again, mass and energy are different things. Dividing Energy by the speed of light squared CHANGES the property of Energy.

Similarly, the same can be said for the equation of light frequency and Energy (incidentally, this equation is about light frequency and not about individual photons) “using E = hv, where v is light frequency and h is Planck's constant” Once again, we have a conversion equation, not an equality.

So, nowhere in your email do you show how photons have mass.

daver
2003-Apr-05, 12:54 AM
Interesting.

OK, I am considering the following as a reply to my co-workers comment.

Can anyone tell me if I the following statement is correct, or am I off the mark?

I know this will sound simplistic to all you Professors and Grad Students. But I think a simple responce to his comments is what is required:


The Theory of Relativity is not a theory of Equality. E=mc^2 states that mass and energy are interchangeable but it does not stipulate that they are equal in a sense that they are one and the same thing. If they were, they would obviously be measured the same way (for instance, mass is measured by grams but energy is not).



I've only skimmed this thread; i apologize if i'm using old or sloppy terminology. Maybe someone who actually knows what he's talking about will clear up some of my misconceptions.

The E = m c**2 equation is usually broken out of the theory of reliativity. I think it can be derived from special relativity, but i've usually seen it presented separately.

Anyway, I disagree. Two molecules of hydrogen plus one molecule of oxygen mass more than two molecules of water. The difference is extremely small, but it is there. Or, if you'd rather, two protons plus two neutrons mass more than one helium nucleus. That difference is still small, but measurable.

A moving object seems to have more mass than a stationary object. A 1 kg weight moving at .6c would seem to mass about 1.25 kg--you could think of it as having 0.25 kg of kinetic energy added.

Energy and mass are measured in different units because the c**2 term is so large--an immeasurably small mass difference can result in a tremendous energy difference.





E=mc^2 uses a constant, c, which is the speed of light. Multiplying mass by the speed of light squared CHANGES the mass.



No, it changes the units. Like multiplying miles by 1.609344 changes the units to kilometers. c**2 is a conversion factor. Maybe that's what you were saying, and i just misunderstood.





This equation can be written another way, m = E/c^2. Once again, mass and energy are different things. Dividing Energy by the speed of light squared CHANGES the property of Energy.



ditto





Similarly, the same can be said for the equation of light frequency and Energy (incidentally, this equation is about light frequency and not about individual photons) “using E = hv, where v is light frequency and h is Planck's constant” Once again, we have a conversion equation, not an equality.

So, nowhere in your email do you show how photons have mass.

Individual photons carry energy. The energy carried by a photon is proportional to its frequency--a photon with twice the frequency will carry twice the energy. A photon of bllue light is nearly twice as energetic as a photon of red light. The energy of the photon can be measured in any of several units--mass units, eletron volts, frequency. Different applications would dictate the choice of which units to use.

JS Princeton
2003-Apr-05, 02:38 AM
The Theory of Relativity is not a theory of Equality. E=mc^2 states that mass and energy are interchangeable but it does not stipulate that they are equal in a sense that they are one and the same thing.

Oh yes it does. E=mc^2 means that mass is just another form of energy. It's stored in something we call mass, but it's still energy.


If they were, they would obviously be measured the same way (for instance, mass is measured by grams but energy is not).

Energy can be measured in grams simply dividing it by the appropriate conversion factor (1/c^2). One can measure energy in grams or in joules just as one can measure distance in miles or meters.



E=mc^2 uses a constant, c, which is the speed of light. Multiplying mass by the speed of light squared CHANGES the mass.

Incorrect. The mass stays the same. It's just a scaling factor. There is a LOT of energy in mass, that's all this statement means (because c^2 is a pretty large value).



This equation can be written another way, m = E/c^2. Once again, mass and energy are different things. Dividing Energy by the speed of light squared CHANGES the property of Energy.

This is closer, but is still slightly incorrect. It is talking about switching between points of view. There is a matter of needing a physical change to get an observational measurement of the change of a restmass energy into, say, a kinetic energy, but the math works out so that we don't have to measure it to know the properties.



Similarly, the same can be said for the equation of light frequency and Energy (incidentally, this equation is about light frequency and not about individual photons)

No, it is about individual photons. Individual photons each have a frequency.


"using E = hv, where v is light frequency and h is Planck's constant” Once again, we have a conversion equation, not an equality.

Conversion is another way of saying two concepts are equivalent... equal.



So, nowhere in your email do you show how photons have mass.

No, he does show how photons have mass, only this isn't the definition for mass that physicists use because it is subject to problems depending on what your frame of reference is. Rather, they define mass as I showed above:

m=Sqrt[E^2-p^2c^2]/c^2

ToSeek
2003-Apr-05, 03:09 AM
Energy can be measured in grams simply dividing it by the appropriate conversion factor (1/c^2). One can measure energy in grams or in joules just as one can measure distance in miles or meters.


And astronomers sometimes measure distance in terms of wavelength based on the different red shifts of the two objects, which really confused me until someone explained it.

kilopi
2003-Apr-05, 05:30 AM
If they were, they would obviously be measured the same way (for instance, mass is measured by grams but energy is not).

Energy can be measured in grams simply dividing it by the appropriate conversion factor (1/c^2). One can measure energy in grams or in joules just as one can measure distance in miles or meters.

One thing I don't think has been mentioned in this thread is the use of different conversion factors.

Bill, if you use the appropriate units, c2 is equal to one, with no units at all. That makes E = m. That makes the equivalence of mass and energy even harder to avoid.

The mass of the Sun is about 1.4 kilometers. :)

Bill Thmpson
2003-Apr-06, 05:46 AM
Conversion is another way of saying two concepts are equivalent... equal.


:-?

Wait a second, E=mc**2 uses a constant, c. A simpler equation that I will use to make a point NOW is one in geometry that also uses a constant, pi (gee, wouldn't it be great if we could use mathematical symbols?). The area of a circile is (pi)r^2 = A. Area and Radius are not equal.

So wouldn't you agree that you have made a blunder here in saying:




Conversion is another way of saying two concepts are equivalent... equal.


Just as Radius and Area are not equal in (pi)r^2 = a, Mass and Energy are not equal in E=mc^2.

E-mc^2
A=(pi)r^2

Isn't you statement that mass is really energy akin to saying that an area of a circle is really ...?

kilopi
2003-Apr-06, 09:37 AM
Wait a second, E=mc**2 uses a constant, c. A simpler equation that I will use to make a point NOW is one in geometry that also uses a constant, pi (gee, wouldn't it be great if we could use mathematical symbols?). The area of a circile is (pi)r^2 = A. Area and Radius are not equal.

So wouldn't you agree that you have made a blunder here in saying:
I would disagree.

It's not Radius that is being converted into Area, it's Radius Squared. I'm sure you'd agree that Radius Squared and Area share characteristics.

Bill Thmpson
2003-Apr-06, 02:29 PM
It's not Radius that is being converted into Area, it's Radius Squared. I'm sure you'd agree that Radius Squared and Area share characteristics.

I think you are missing the point to this discussion. Area and Radius share characteristics. But this is not the issue. Energy and Mass share characteristics but this is not the issue as well.

Actually, since Radius is a one dimensional measurement and Area is two dimensional support the point that I am making.

So does the fact that c in the Theory of Relativity is a meaurement that is light squared. So the outcome of E=m^2 would not be a measurement of mass -- as in grams -- but a measurement using both grams, meters squared, and seconds squared.

People are jumping into this discussion and making comments without reading my previous comments and are pulling some of my sentances out of context.

The point I am making is this.

Since I was about 12 years old, I have known that light could not have mass. If an object was accelerating through a vacuum traveling closer and closer to the speed of light, it would not and could not actually reach the speed of light. THis is simply because its inertia would increase exponentially and the amount of energy it would take for it to push itself a little faster would become too great.

So the speed of light cannot be reached (I think my co-worker and many others are influenced by popular science fiction)

Actually I saw an experiment which proves this concept of velocity influencing inertia. Two atomic clocks were at rest on the ground (please don't pull this paragraph out of context and go off on a tangent that if an object is sitting on the earth it is not really at rest. THis is irrevelant to the point I am making... I could be really precice and say that they are at rest relative to each other, but I am hoping this is not necessary. Oh, what the heck let me just say that they are relative to each other) -- what I mean is, they are relative to each other. The clocks were set so that they were each attached to a seperate strobe light so that the lights flashed in sync with each other. Now, one of the clocks with its light went for a ride in a supersonic jet for a while. WHen this jet returned to earth, the light attached to the clock, and the light attached to the clock that had remained on the ground were out of sync. THis proved that the inertia of the clock that went for a ride, had increased and so there was a slight time dialation.

I have known for years about this sort of thing.

Go back and read my origional posting.

A co-worker of mine is suggesting that photons have mass.

This is impossible. If something is traveling at the speed of light, it cannot have mass.

In an attempt to show that a photon really does have mass, my co-worker has said:


If you want to calculate relative mass of photon you can use E = mc^2, where photon's energy can be calculated using E = hv, where v is light frequency and h is Planck's constant.

But saying E=mc^2 is not the same as saying that E=m. Althought you can say that mass is a form of energy or mass contains energy, they are not one and the same as if E=m. You are leaving out an important constant from the equation just as you would if you were discussiong a formula for determining the area of a circle by its radius.

If people want to jump into the discussoin and use the very cool feature of extracting senatnces from my discussion, they should understand the context in which it was written.

The context in which it was written has been to show that photons do not have mass.

Objects traveling at the speed of light cannot have mass.

It is impossible for objects that have mass to reach the speed of light.

Now, please do not take the above sentances out of context to fly off on a tangent about quantum mechanics or theoritcal science fiction.

kilopi
2003-Apr-06, 04:20 PM
I think you are missing the point to this discussion.
The point is to try to prove your coworker wrong? :)

The context in which it was written has been to show that photons do not have mass.
I notice that the coworker uses "relative mass". I've used the term relativistic mass. In that sense, as JS and others have pointed out, the coworker is right.

JS Princeton
2003-Apr-06, 05:11 PM
It's not Radius that is being converted into Area, it's Radius Squared. I'm sure you'd agree that Radius Squared and Area share characteristics.

I think you are missing the point to this discussion. Area and Radius share characteristics. But this is not the issue. Energy and Mass share characteristics but this is not the issue as well.


kilopi is right on this matter. The area of the circle is directly parametrized by a radius. For any given radius, you can tell me the area. That means that the radius determines absolutely the area of the circle. It doesn't matter that radius and area are different measurements, all that matters is that they are intimately connected enough to have a simple conversion between the two concepts. Since that's the case, then radius and area must be speaking to a COMMON IDEA when it comes to the universe of all circles.

The same is true for mass and energy. If I give you a mass you can give me the energy associated with that mass. In the universe, mass and energy must be speaking to COMMON IDEA.

Get it?



Actually, since Radius is a one dimensional measurement and Area is two dimensional support the point that I am making.


All that shows is that the concepts inhabit different parameter spaces. What you fail to notice is that the two are inextricably linked.



So does the fact that c in the Theory of Relativity is a meaurement that is light squared.

No, c^2 doesn't change. It's a constant just like pi. Energy and mass have a one-to-one linear relationship in fact.


So the outcome of E=m^2 would not be a measurement of mass -- as in grams -- but a measurement using both grams, meters squared, and seconds squared.

Energy never equals mass squared.



Since I was about 12 years old, I have known that light could not have mass. If an object was accelerating through a vacuum traveling closer and closer to the speed of light, it would not and could not actually reach the speed of light. THis is simply because its inertia would increase exponentially and the amount of energy it would take for it to push itself a little faster would become too great.

So the speed of light cannot be reached (I think my co-worker and many others are influenced by popular science fiction)

This is all a decent analysis, but there remains the problem that mass and energy can be converted one to the other. This means that since a photon has an energy, one can envision a process where you convert that energy into a mass. In isolation and in point of fact when you are doing physics, you are correct, the photon has zero restmass.



Actually I saw an experiment which proves this concept of velocity influencing inertia. Two atomic clocks were at rest on the ground (please don't pull this paragraph out of context and go off on a tangent that if an object is sitting on the earth it is not really at rest. THis is irrevelant to the point I am making... I could be really precice and say that they are at rest relative to each other, but I am hoping this is not necessary. Oh, what the heck let me just say that they are relative to each other) -- what I mean is, they are relative to each other. The clocks were set so that they were each attached to a seperate strobe light so that the lights flashed in sync with each other. Now, one of the clocks with its light went for a ride in a supersonic jet for a while. WHen this jet returned to earth, the light attached to the clock, and the light attached to the clock that had remained on the ground were out of sync. THis proved that the inertia of the clock that went for a ride, had increased and so there was a slight time dialation.

The Lorentz frame changing didn't, however, change the masses of the clocks when they returned to the same reference frame, interestingly enough. In point of fact, you didn't actually observe a change in inertia, but you could, observing the time dilation and knowing a bit about the relationship with the speed of light, work out the fact that the inertia had to have changed while the clock was travelling at a higher speed.



But saying E=mc^2 is not the same as saying that E=m.

It depends on what units you are in. You can define a unit system that's perfectly reasonable where c^2=1. That would imply that E=m.


Althought you can say that mass is a form of energy or mass contains energy, they are not one and the same as if E=m.

That's true, they aren't one and the same, but you can switch from one form of energy to the other. This tells you how much mass you would get if you switched a photon's energy into restmass.


You are leaving out an important constant from the equation just as you would if you were discussiong a formula for determining the area of a circle by its radius.

The constant doesn't matter, what matters is actually the application. In certain situations, one can say that E=m, but that is only when you are rest. Otherwise E^2=m^2+p^2 (if c^2=1).

cable
2003-Apr-06, 06:22 PM
Duality means photon is a perticle AND a wave.
BUT we cannot see thse 2 aspects at SAME time in an experiment.

I think , photon is something else that our math cannot modelize as a whole.
a corpuscule, a wave, collapse of wave function .... all bad modelling.

Eta C
2003-Apr-06, 07:13 PM
Quite right Cable. As JPrinceton and I mentioned earlier in this thread, in modern physics no object can be defined as a particle or as a wave. All exhibit properties of both definitions depending on the effect being observed. To ask if an object (photon, electron, 747) is a particle or a wave is meaningless.

dlack
2003-Apr-06, 09:12 PM
I have a few things to add to this discussion.


My shy, misguided coworker has emailed me the following:


"If you want to calculate relative mass of photon you can use E = mc^2, where photon's energy can be calculated using
E = hv, where v is light frequency and h is Planck's constant."


The correct response to this is "no, you cannot calculate the mass of a photon in that manner."

Your co-worker is not using the entire formula.

E = mc^2 only applies to massive particles whose speed is very small compared to that of the speed of light.

The whole formula, which applies to all particles (including photons) is:

(1) E^2 = (mc^2)^2 + (pc)^2

where E is the energy of the particle, m is the mass of the particle, p is the momentum of the particle, and c is the speed of light.

If you calculate the energy of a photon using its frequency, as in

(2) E = hf = hc/(lambda)

where lambda is the wavelength of the light associated with the photon

and calculate the momentum of the photon using

(3) p = h/(lambda),

and substitute (3) and (2) into (1), you will find that the mass of the photon is ZERO!

One can speak of the "relativistic mass" of a particle, but it is not a measureable quantity (as rest mass and velocity are for Newtonian particles), and it is NOT found from the formula E = mc^2.



Energy can be measured in grams simply dividing it by the appropriate conversion factor (1/c^2). One can measure energy in grams or in joules just as one can measure distance in miles or meters.


You can give energy in terms of c, or mass in terms of c. For example, it is common in relativistic calcultions to give a mass with units (electron-Volts / c). Where electron-Volts is a unit for energy, as is Joules.

HOWEVER, this is not the same as "measuring distance in miles or meters." Miles and meters are both units for the same quantity, distance. Joules and grams are units for different quantities, energy and mass.

What you have to remember is that c isn't really a unit or a constant of proportionality, it is a constant quantity, with the units (meters/second).

So, the Equation E = mc^2 has units:

J = kg * (m/s)^2.

which is precisely the definition of a Joule.

So, energy can not be measured in grams. Grams is the unit for mass. Energy is measured in Joules. Energy and mass are related, but they are not "the same thing."

JS Princeton
2003-Apr-07, 01:03 AM
HOWEVER, this is not the same as "measuring distance in miles or meters." Miles and meters are both units for the same quantity, distance. Joules and grams are units for different quantities, energy and mass.

I beg to differ. Of course, it depends on the regime you are using, but mass is fundamentally a kind of energy. As such, you can measure mass in grams or in joules. Likewise, you can measure energy in mass, as long as you are clear what you are doing is converting all the energy into restmass.

It's the same thing as meters and miles. One system says that you have the distance of 5280 feet (the size of some King of England's foot) as being a measure of distance. The other says that some multiple of the speed of light which determines the meter is a measure of distance. Both are standards that are derived from a fundamental idealization of the world. This is the same as giving a "mass" for an "energy" of vice versa.



What you have to remember is that c isn't really a unit or a constant of proportionality, it is a constant quantity, with the units (meters/second).


ALL conversion factors are constant quantities with units. The conversion between meters and miles is a constant with units. The are a few constants in physics that are unitless, but those are not conversion factors.



So, the Equation E = mc^2 has units:

J = kg * (m/s)^2.

which is precisely the definition of a Joule.

So, energy can not be measured in grams. Grams is the unit for mass. Energy is measured in Joules. Energy and mass are related, but they are not "the same thing."

You are arguing something along the lines of "Force is not the same thing as the product of mass and acceleration." I heartily disagree with that assessment.

They are same thing if you are in the correct regime.

For example, particle physics theorists tend to use Planck units where not only the speed of light but also Planck's Constant and the Gravitational Constant are all set to 1.

kilopi
2003-Apr-07, 09:01 AM
and substitute (3) and (2) into (1), you will find that the mass of the photon is ZERO!

Well, that's rest mass, which nobody seems to disagree is zero.


One can speak of the "relativistic mass" of a particle, but it is not a measureable quantity (as rest mass and velocity are for Newtonian particles), and it is NOT found from the formula E = mc^2.
I think JS gave a better answer to this earlier. The energy of a photon does contribute to gravitation, and in that sense can be said to have an "equivalent" mass. How would you calculate that?

dlack
2003-Apr-07, 04:36 PM
I beg to differ. Of course, it depends on the regime you are using, but mass is fundamentally a kind of energy. As such, you can measure mass in grams or in joules. Likewise, you can measure energy in mass, as long as you are clear what you are doing is converting all the energy into restmass.

It's the same thing as meters and miles. One system says that you have the distance of 5280 feet (the size of some King of England's foot) as being a measure of distance. The other says that some multiple of the speed of light which determines the meter is a measure of distance. Both are standards that are derived from a fundamental idealization of the world. This is the same as giving a "mass" for an "energy" of vice versa.

ALL conversion factors are constant quantities with units. The conversion between meters and miles is a constant with units. The are a few constants in physics that are unitless, but those are not conversion factors.

You are arguing something along the lines of "Force is not the same thing as the product of mass and acceleration." I heartily disagree with that assessment.

They are same thing if you are in the correct regime.

For example, particle physics theorists tend to use Planck units where not only the speed of light but also Planck's Constant and the Gravitational Constant are all set to 1.


I'm going to concede your point, because this is an argument of terminology. You're arguing that if someone on earth gives their weight in kilograms, they are correct because the mass they give is commonly thought of as "equivalent" to a certain weight.

I still don't like the idea of saying mass and energy are "the same." A certain amount of mass can be converted to a certain amount of energy under certain circumstances, but to me, that doesn't mean they are "the same."

Again, it's semantics. I don't think of F = ma as a conversion equation. I think of it as a law. I don't think of E = mc^2 as a conversion of mass into energy, I think of it as the law governing the relationship between mass and energy. I'm not going to say the way you think about it is wrong, I just don't like it.

(And I'm not saying force isn't the same as mass times acceleration, I'm saying force isn't the same thing as mass.)

dlack
2003-Apr-07, 04:47 PM
and substitute (3) and (2) into (1), you will find that the mass of the photon is ZERO!

Well, that's rest mass, which nobody seems to disagree is zero.


One can speak of the "relativistic mass" of a particle, but it is not a measureable quantity (as rest mass and velocity are for Newtonian particles), and it is NOT found from the formula E = mc^2.
I think JS gave a better answer to this earlier. The energy of a photon does contribute to gravitation, and in that sense can be said to have an "equivalent" mass. How would you calculate that?

I have class right now, but I'm going to come back with an example try to explain what I mean. Essentially, you can't just take an photon of energy E, plug that energy into E = mc^2, and then take that mass to be the mass (relativistic, rest, or otherwise) of a photon. E = mc^2 does not apply to a photon. The entire formula as given in my last post must be used, and momentum, energy, and charge all must be conserved.

kilopi
2003-Apr-07, 06:00 PM
I'm going to concede your point, because this is an argument of terminology. You're arguing that if someone on earth gives their weight in kilograms, they are correct because the mass they give is commonly thought of as "equivalent" to a certain weight.
I may be going out on a limb here in defending JS, but I'm willing to bet dollars to donuts that he definitiely did not mean that!

In fact, because that is such a common well-understood error, and such a gross mischaracterization of what he said, that I think you should apologize. :)

dlack
2003-Apr-07, 07:59 PM
I may be going out on a limb here in defending JS, but I'm willing to bet dollars to donuts that he definitiely did not mean that!

In fact, because that is such a common well-understood error, and such a gross mischaracterization of what he said, that I think you should apologize. :)

I'd rather hear what he has to say about it, but:

W = mg

E = mc^2

Force = mass

Energy = mass

What's the difference?

kilopi
2003-Apr-07, 09:00 PM
I'd rather hear what he has to say about it, but:

W = mg

E = mc^2

Force = mass

Energy = mass

What's the difference?
Until he gets done with class, we'll talk. What do you say about it? Do you agree with all four of those? Or, is there any of them that you definitely do not agree with?

dlack
2003-Apr-07, 09:35 PM
I'd rather hear what he has to say about it, but:

W = mg

E = mc^2

Force = mass

Energy = mass

What's the difference?
Until he gets done with class, we'll talk. What do you say about it? Do you agree with all four of those? Or, is there any of them that you definitely do not agree with?

I would say that the first two are true while I have issues with the second two.

I understand how someone could speak of the energy of a particle instead of its mass, because for a certain mass, we can calculate the energy that can be gotten from it in the proper circumstances. Likewise we could speak of the weight of a particle instead of its mass, because for a given mass having a given acceleration due to gravity, we could calculate its weight.

However, to me, that doesn't say that weight is the same as mass or that energy is the same as mass. That says that the weight of a particle is intimately related to its mass, and that the rest energy of a particle is intimately related to its rest mass.

For me, the statement that mass and energy are so intimately related that a certain amount of energy can be created from a certain amount of mass is a strong enough statement without trying to say mass and energy are "the same thing." They are closely related, but not "the same thing."

I do understand how the argument could be made that you could set c = 1 and have mass = energy, but then you have to admit that I can set g = 1 and have mass = weight.

Bill Thmpson
2003-Apr-07, 09:50 PM
I'm going to concede your point, because this is an argument of terminology. You're arguing that if someone on earth gives their weight in kilograms, they are correct because the mass they give is commonly thought of as "equivalent" to a certain weight.

I still don't like the idea of saying mass and energy are "the same." A certain amount of mass can be converted to a certain amount of energy under certain circumstances, but to me, that doesn't mean they are "the same."


I think everyone should be able to agree that to get energy from mass, you have to do something to the mass to convert it into energy.

I just thought I would run that up the flag pole and see if anyone salutes.

I also like this weight VS mass argument. stated here as an analogy. Weight and Mass are not one and the same thing. THis is a good point. It also shows how everyone can think something is true and yet what they think is completely wrong.

Everyone uses Kilograms or Pounds as a measurement for weight. But the correct measurement should be in Newtons.

Also, everyone use bathroom scales as a measurement of weight. Strangely, since it works on a spring mechanism, it really does measure weight... but the dial shows Pounds or Kilograms which is not correct. Kilograms is a measurement of mass. Take the bathroom scale up to the moon, and it would give you an incorrect measurement of mass.

And, everyone seems to overlook that the doctor's scale -- or the scale at a gym -- that uses a counter-balance system really does measure mass (while the bathroom scale measures weight because of the pull on a spring, the doctor's scale compares you against other object's graviataional pull, but the comparison would be the same on any gravity -- even on the moon, since it is a comparison of objects equally affected)-- take that mechanism (the doctor's scale) to the moon and it will give you the same result as on the Earth because the mass would not change. But the doctor says "step on the scale and let's see how much you weigh" not, "step on the scale and let's measure your mass" which would be more correct.

All this shows how people can get things all mixed up. The same is true for this discussion. And it is not clear who has the proper scientific credentials to back up what he or she is saying.

kilopi
2003-Apr-07, 10:32 PM
I would say that the first two are true while I have issues with the second two.
Kinda what I figured. But he didn't say the third one, so it's unfair to include it. That's called a straw man.


I understand how someone could speak of the energy of a particle instead of its mass, because for a certain mass, we can calculate the energy that can be gotten from it in the proper circumstances.
And how much would that be?

I do understand how the argument could be made that you could set c = 1 and have mass = energy, but then you have to admit that I can set g = 1 and have mass = weight.
It's not as arbitrary as that, though. It's c and G that are set equal to one. Since g = GM/r^2, so g has units of Kg/m^2, but G/c^2 would also be one, so meters and kilograms cancel to leave meters in the denominator. So, mass is measured in meters, but weight is not.


It also shows how everyone can think something is true and yet what they think is completely wrong.

Everyone uses Kilograms or Pounds as a measurement for weight. But the correct measurement should be in Newtons.

Well, clumsy as it is, pound is a measure of weight, so I guess that's a good example.


All this shows how people can get things all mixed up. The same is true for this discussion. And it is not clear who has the proper scientific credentials to back up what he or she is saying.
Credentials? We don't need no steenkin' credentials.

daver
2003-Apr-07, 11:33 PM
Everyone uses Kilograms or Pounds as a measurement for weight. But the correct measurement should be in Newtons.



Oh well. Nit picking time. "Pounds" is a legitimate unit of force; its associated unit of mass is a "slug". A "pound" is also a legitimate unit of mass; its associated unit of force is a "poundal". One of the many reasons people prefer to use the metric system.

Most people probably use "weight" when they're really concerned with "mass". And in most people's lives, the difference is insignificant. If the force of gravity in Europe were 10% higher than in the US there might be some incentive to disambiguate. However, for rather obvious reasons, the people who concocted the language weren't aware of the difference. I don't think you'll have much luck changing common parlance.

Sometimes people who know should know better get the two confused. Thrust being measured in kg is one example. Isp in 1/sec is another.

dlack
2003-Apr-08, 10:03 AM
I would say that the first two are true while I have issues with the second two.
Kinda what I figured. But he didn't say the third one, so it's unfair to include it. That's called a straw man.

My point is that, using his reasoning, if the fourth follows from the second, the third must follow from the first.




I understand how someone could speak of the energy of a particle instead of its mass, because for a certain mass, we can calculate the energy that can be gotten from it in the proper circumstances.
And how much would that be?

If you have two particles with given mass and momentum colliding to give you photons, the energy is given by E^2 = p^2c^2 + m^2c^4. (Notice their momentum matters).

However, you couldn't convert a single particle of given mass into a photon for the same reason you can't convert a single photon into a particle of given mass. (I show why below).


The energy of a massless photon can be converted to mass, but it's not a simple matter of E=mc^2.

The equation E^2 = p^2c^2 + m^2c^4 must be satisfied. Conservation of momentum must be satisfied. Conservation of Energy must be satisfied. Conservation of charge must be satisfied (start with a neutral photon, you better end up with a net charge of zero). Other conservations probably must be satisfied that I haven't studied yet.

A photon is never stationary (it moves at speed, um, c), so it never has zero momentum, so E = mc^2 does not apply.

For a similar reason you couldn't take one photon and convert it into one particle of a certain mass, because momentum will never be conserved.

The initial energy of the system is (for a photon):
E1 = p1c
The final energy (for a single massive particle):
(E2)^2 = (p2c)^2 + (mc^2)^2
Which means:
(E1)^2 = (E2)^2 = (p1c)^2 = (p2c)^2 + (mc^2)^2

But, remember momentum has to be conserved in the collision. The only possible way that p1 can equal p2 here (for a single particle) is if the mass of the new particle is zero! So a photon can't be changed into a single particle of given mass.

Now I think it would be possible to convert a single photon into two particles, but then the mass of the particles would depend on how much momentum each particle had, so you can't say a photon can give you a certain amount of mass.



I do understand how the argument could be made that you could set c = 1 and have mass = energy, but then you have to admit that I can set g = 1 and have mass = weight.
It's not as arbitrary as that, though. It's c and G that are set equal to one. Since g = GM/r^2, so g has units of Kg/m^2, but G/c^2 would also be one, so meters and kilograms cancel to leave meters in the denominator. So, mass is measured in meters, but weight is not.

I'm not sure what you're doing here. I never said weight was measured in meters (which is what you're trying to disprove, apparently). And what's the significance of the fact that G/c = 1 here? They don't ever appear in the same equation (of the ones you've used, anyways). Your argument looks like an argument to show that Force can be measured in (kg^2)/(m^2).

W = mg = (kg)(kg/m^2)

Which goes back to what I said before. If you want to think of it that way, you can, and apparently it's useful sometimes, but what you are doing is playing with units. You aren't proving that mass is the same thing as energy or that force is the same thing as mass or whatever is measured in (kg^2/m^2).

kilopi
2003-Apr-08, 11:01 AM
I would say that the first two are true while I have issues with the second two.
Kinda what I figured. But he didn't say the third one, so it's unfair to include it. That's called a straw man.

My point is that, using his reasoning, if the fourth follows from the second, the third must follow from the first.

No, that's using your reasoning. Of course, you think you're duplicating his reasoning, but I don't think you are. That's what makes it a straw man.

The energy of a massless photon can be converted to mass, but it's not a simple matter of E=mc^2.
Well, try the other way then. What if you had a body of mass M in space that was gravitationally affecting other bodies, and you converted it entirely to energy. (1) How much energy would you get? (2) What would the effect be on the other bodies, gravitationally?

I'm not sure what you're doing here. I never said weight was measured in meters (which is what you're trying to disprove, apparently).
No, I wasn't claiming that you said it was. I was merely following the scheme laid out in physics textbooks. For instance, the sidebar on page 29 of Misner, Thorne, and Wheeler's huge Gravitation.

dlack
2003-Apr-08, 11:30 AM
I would say that the first two are true while I have issues with the second two.
Kinda what I figured. But he didn't say the third one, so it's unfair to include it. That's called a straw man.

My point is that, using his reasoning, if the fourth follows from the second, the third must follow from the first.

No, that's using your reasoning. Of course, you think you're duplicating his reasoning, but I don't think you are. That's what makes it a straw man.

I'd still like to know if he agrees with my use of his reasoning.



The energy of a massless photon can be converted to mass, but it's not a simple matter of E=mc^2.
Well, try the other way then. What if you had a body of mass M in space that was gravitationally affecting other bodies, and you converted it entirely to energy. (1) How much energy would you get? (2) What would the effect be on the other bodies, gravitationally?

I already addressed this:


If you have two particles with given mass and momentum colliding to give you photons, the energy is given by E^2 = p^2c^2 + m^2c^4. (Notice their momentum matters).

However, you couldn't convert a single particle of given mass into a photon for the same reason you can't convert a single photon into a particle of given mass. (I show why below). A single particle can't be converted into a single photon of energy because momentum won't be conserved. If a couple particles are used, or however you do it (provided you conserve energy, momentum, etc.), the energy is given by the formula.

I'd like to see some sources about the idea that photons have gravitational attraction. I don't think gravity is part of this discussion, but you keep bringing it up. Photons don't have mass. A quick google search gives:
http://www.usatoday.com/weather/resources/basics/wonderquest/photonmass.htm



I'm not sure what you're doing here. I never said weight was measured in meters (which is what you're trying to disprove, apparently).
No, I wasn't claiming that you said it was. I was merely following the scheme laid out in physics textbooks. For instance, the sidebar on page 29 of Misner, Thorne, and Wheeler's huge Gravitation.

You said:


So, mass is measured in meters, but weight is not.
I took this to mean that you were trying to show that weight isn't measured in meters.

I still don't understand what you were trying to do.

I don't have your particular textbook, so I don't know what exactly they are doing on that page, but I'm looking at p. 324 of Physics by Tipler, and Newton's Law of Gravity is laid out fairly simply. Nowhere does it say anything about measuring mass in meters. In no formula does G/c appear.

(edit: particluar. blah.)

kilopi
2003-Apr-08, 12:20 PM
I'd still like to know if he agrees with my use of his reasoning.
Well, if he did, I wouldn't, so we can continue. :)



Well, try the other way then. What if you had a body of mass M in space that was gravitationally affecting other bodies, and you converted it entirely to energy. (1) How much energy would you get? (2) What would the effect be on the other bodies, gravitationally?

I already addressed this:
OK, since the momentum of the mass M is relatively zero, your equation reduces to E=mc^2, as you'd think. So your answer to (1) would be Mc^2?

I'd like to see some sources about the idea that photons have gravitational attraction. I don't think gravity is part of this discussion, but you keep bringing it up. Photons don't have mass. A quick google search gives:
http://www.usatoday.com/weather/resources/basics/wonderquest/photonmass.htm
That link discusses rest mass, and no one in this thread disputes that the photon has zero rest mass. That doesn't seem to be of any contention. Since you're unaware that photons contribute to gravitational attraction, I guess your answer to (2) would be zero?

I still don't understand what you were trying to do.

::snip::

Nowhere does it say anything about measuring mass in meters. In no formula does G/c appear.
They're called geometrized units (http://www.alcyone.com/max/physics/laws/g.html), and I apologize, they appear on page 36, not page 29, of MTW.

Basically, to address the first comment at the top of this post, I was illustrating why the reasoning could not be extended to include the formula that you thought it could be, using your reasoning. That is, speaking for myself and not necessarily JS, the reason for our conclusions is not just a simple matter of re-interpreting any old equal sign.

dlack
2003-Apr-08, 01:00 PM
OK, since the momentum of the mass M is relatively zero, your equation reduces to E=mc^2, as you'd think. So your answer to (1) would be Mc^2?


NO!

If the momentum of the initial mass is zero, it can't be converted into a single photon of given energy!

Conservation of momentum must be obeyed, and a photon never has zero momentum. So if you try to convert a mass M with momentum 0 into a photon of energy, you get . . . a mass M with momentum 0.

I think you could possibly have a particle change into several photons moving in opposite directions. Then momentum is conserved, and the total energy would be mc^2. So the energy of each photon depends on its individual wavelength.



I'd like to see some sources about the idea that photons have gravitational attraction. I don't think gravity is part of this discussion, but you keep bringing it up. Photons don't have mass. A quick google search gives:
http://www.usatoday.com/weather/resources/basics/wonderquest/photonmass.htm
That link discusses rest mass, and no one in this thread disputes that the photon has zero rest mass. That doesn't seem to be of any contention. Since you're unaware that photons contribute to gravitational attraction, I guess your answer to (2) would be zero?

I'm saying that I haven't studied the effects on nearby particles of converting massive particles (with gravitational attraction) into photons of energy (which don't have gravitational attraction). My guess would be that the effects of gravity are pretty small for this case since we're dealing with quantum particles with extremely small mass. No one's talking about converting entire planets into energy, right?

dlack
2003-Apr-08, 01:08 PM
They're called geometrized units, and I apologize, they appear on page 36, not page 29, of MTW.
Basically, to address the first comment at the top of this post, I was illustrating why the reasoning could not be extended to include the formula that you thought it could be, using your reasoning. That is, speaking for myself and not necessarily JS, the reason for our conclusions is not just a simple matter of re-interpreting any old equal sign.

According to your link:


As a result of converting to geometrized units, all quantities are expressed in terms of a unit of distance

So if you're using geometrized units, all quantities are in centimeters, so energy can't be expressed as kilograms, as was stated initially to start this particular aspect of the debate.

As I said, I'm not saying this isn't a legitimate way of playing with units to make calculations easier, I'm saying this doesn't show that energy is the same thing as mass. They're intimately related and can be converted into one another, but they aren't the same thing.

kilopi
2003-Apr-08, 01:09 PM
OK, since the momentum of the mass M is relatively zero, your equation reduces to E=mc^2, as you'd think. So your answer to (1) would be Mc^2?

NO!

I'm surprised at the vehemence of this answer.


If the momentum of the initial mass is zero, it can't be converted into a single photon of given energy!

Especially if that mass M is even as large as a baseball maybe. But why must the energy be just a single photon?


Conservation of momentum must be obeyed, and a photon never has zero momentum. So if you try to convert a mass M with momentum 0 into a photon of energy, you get . . . a mass M with momentum 0.

I think you could possibly have a particle change into several photons moving in opposite directions. Then momentum is conserved, and the total energy would be mc^2.

But, that's what I said, Mc^2. Now I'm even more suprised at your first answer.

I'm saying that I haven't studied the effects on nearby particles of converting massive particles (with gravitational attraction) into photons of energy (which don't have gravitational attraction). My guess would be that the effects of gravity are pretty small for this case since we're dealing with quantum particles with extremely small mass. No one's talking about converting entire planets into energy, right?
Sure, why not? :)

OK, let's start with a baseball. So, your answer to (2) would be zero?

kilopi
2003-Apr-08, 01:17 PM
So if you're using geometrized units, all quantities are in centimeters, so energy can't be expressed as kilograms, as was stated initially to start this particular aspect of the debate.
No, energy can be expressed in kilograms, it's just not the standard way of it. If ergs can be expressed in centimeters, and grams expressed in centimeters, it should be easy to see how the transition from ergs to grams could be made.

Ring
2003-Apr-08, 01:49 PM
Mass is equal to the energy of a system that cannot be transformed away. In other words if the system has a zero momentum frame then the system has mass. So a single photon cannot have mass, but a system of photons can.

This is also why mass cannot be converted to energy. If a closed system initially has zero momentum then it must always have zero momentum and therefore the mass cannot change. A vault containing a nuclear weapon weighs the same both before and after detonation.

Even though a single photon doesn’t have mass it does have energy and energy effects the geometry of spacetime. However if you travel along with a beam of light you can reduce its energy to any value you please and this is why curvature is covariant not invariant.

kilopi
2003-Apr-08, 02:16 PM
Mass is equal to the energy of a system that cannot be transformed away. In other words if the system has a zero momentum frame then the system has mass. So a single photon cannot have mass, but a system of photons can.
That's still "rest mass." Everybody seems to agree that a photon has zero rest mass.


This is also why mass cannot be converted to energy. If a closed system initially has zero momentum then it must always have zero momentum and therefore the mass cannot change. A vault containing a nuclear weapon weighs the same both before and after detonation.
There is no energy released in a nuclear explosion?

Actually, that's probably an illustration of the point. The gravitational effect doesn't change, even after the mass has been converted to energy.


Even though a single photon doesn’t have mass it does have energy and energy effects the geometry of spacetime.
I think that's what we're talking about.

However if you travel along with a beam of light you can reduce its energy to any value you please and this is why curvature is covariant not invariant.
You can't travel along with a beam of light. You can set a reference frame to do so. Or am I wrong?

Ring
2003-Apr-08, 03:10 PM
]
There is no energy released in a nuclear explosion?

Fusion/fision isn't about converting mass to energy its about converting potential energy to kinetic and electromagnetic energy. In other words it's about changing one type of energy into another. The remnant of the nucleus has less internal energy and as a result of this decrease in internal energy it has less mass, not vice versa. Mass and energy are properties of a system, and not things in their own right



You can't travel along with a beam of light. You can set a reference frame to do so. Or am I wrong?

Well, if you move towards a light source it will appear blueshifted and vice versa, I think that's what I was actually trying to say.

dlack
2003-Apr-08, 05:19 PM
Kilopi, my insistence that the energy of a mass can't be released in a single photon is due to the fact that I thought the discussion was about assigning a "mass" to a photon. You can't assign any kind of mass to a single photon. That's where my vehemence comes from. Apparently the discussion has shifted from there, but it seemed like you were trying to get back to the "a photon has some kind of mass associated with it" issue.

My other point has been that when you transform energy into mass, the mass does not come from the (partial) equation E = mc^2. The whole equation I've typed several times must be used.

Ring is absolutely wrong. Mass can be converted into energy, and energy into mass. This occurs in the explosion of a nuclear weapon, and it occurs in the formation of molecules.

As for the effect this has on gravity, I don't know the answer. I'm sure there would be some dramatic results if you took an entire planet and converted it into energy. I don't think that loss of gravity would be the biggest concern in such a scenario, however. I think the resulting enormous amount of energy would be the biggest concern.

As for the issue with the units, I've conceded that you can play with units in order to make calculations simpler. I don't understand why you won't concede that if you do it with c, I can just as easily and legitimately do it with g.

daver
2003-Apr-08, 06:04 PM
As for the issue with the units, I've conceded that you can play with units in order to make calculations simpler. I don't understand why you won't concede that if you do it with c, I can just as easily and legitimately do it with g.

c is a universal constant, it's the same on the earth, the moon, or omicron ceti 5. g varies, even on earth. I memorized 9.80665 m/sec/sec a while back, i don't know if that is currently accepted as the standard or not.

Ring
2003-Apr-08, 06:21 PM
Ring is absolutely wrong. Mass can be converted into energy, and energy into mass. This occurs in the explosion of a nuclear weapon, and it occurs in the formation of molecules.


Use your own equation:

m^2 = E^2 - p^2 (c=1)

If the system has zero momentum before detonation then it must have zero momentum after detonation. So m = E = no conversion of mass to energy.

Nucleons have orbitals similar to electrons in an atom. And the nucleons in these orbitals have potential energy. When a reaction occurs the nucleus rearanges itself and the potential energy decreases (binding energy increases) and this decrease in potential energy results in kinetic and electromagnetic energy. (This is a gross simplification). There's a local mass defect but the system energy and mass remains the same.

dlack
2003-Apr-08, 06:55 PM
Ring is absolutely wrong. Mass can be converted into energy, and energy into mass. This occurs in the explosion of a nuclear weapon, and it occurs in the formation of molecules.


Use your own equation:

m^2 = E^2 - p^2 (c=1)

If the system has zero momentum before detonation then it must have zero momentum after detonation. So m = E = no conversion of mass to energy.



Suppose a system of mass m has momentum p. The energy of this system is found using the equation.

If this mass is converted into energy, the total momentum of the (massless) photons of energy will be equal to p, the initial momentum. The total energy of the photons will be equal to the initial energy.

Thus, momentum is conserved and energy is conserved. MASS is NOT conserved.

I'm taking a break from this thread for the day. If someone could help Ring out, that would be great.

dlack
2003-Apr-08, 06:56 PM
As for the issue with the units, I've conceded that you can play with units in order to make calculations simpler. I don't understand why you won't concede that if you do it with c, I can just as easily and legitimately do it with g.

c is a universal constant, it's the same on the earth, the moon, or omicron ceti 5. g varies, even on earth. I memorized 9.80665 m/sec/sec a while back, i don't know if that is currently accepted as the standard or not.

Alright, I see how that difference can be argued. I concede this point.

Ring
2003-Apr-08, 08:39 PM
Suppose a system of mass m has momentum p. The energy of this system is found using the equation.

If this mass is converted into energy, the total momentum of the (massless) photons of energy will be equal to p, the initial momentum. The total energy of the photons will be equal to the initial energy.

Thus, momentum is conserved and energy is conserved. MASS is NOT conserved.

Question: What physical system are we talking about?
Answer: A stationary nuclear weapon

Question: What’s the momentum of a stationary nuclear weapon?
Answer: Zero.

Question: What’s the momentum of this system after it detonates?
Answer: Zero. It’s a closed system so momentum must be conserved.

Question: Given m^2 = E^2 – p^2 what’s the mass of the detonated system?
Answer: The same as the pre detonated system, since p =0, m = E!

SYSTEM MASS IS a universally conserved quantity.

dlack
2003-Apr-08, 10:17 PM
Ok, I couldn't stay away for a whole day.

MOMENTUM IS A VECTOR!

The magnitudes of the individual momenta of the photons doesn't have to be zero, the vector sum of the momenta must be zero.

Suppose we have a massive particle at rest at the origin.

Initial momentum is zero.

Now suppose this particle is converted into massless photons of energy.

As long as the momentum in the +x direction is the same as the momentum in the -x direction, momentum is conserved.

The sum of the momenta of the photons is zero.

Mass has been converted into energy.

The algebra may be a little more complicated for a 3-D space, but the idea is the same. The vector sum of the momenta must be zero.

Ring
2003-Apr-08, 10:45 PM
Ok, I couldn't stay away for a whole day.

MOMENTUM IS A VECTOR!

So is the energy monentum four-vector and its magnitude is the system mass, its time component is energy and its space component is momentum.

m^2 = e^2 - p^2


The magnitudes of the individual momenta of the photons doesn't have to be zero, the vector sum of the momenta must be zero.

That's correct and since the sum of the momenta equals zero: m = E


Suppose we have a massive particle at rest at the origin.

Initial momentum is zero.

And therefore the final momentum of the system must also equal zero and: m =E


As long as the momentum in the +x direction is the same as the momentum in the -x direction, momentum is conserved.

The sum of the momenta of the photons is zero.

That's correct and since p = 0: m = E.


Mass has been converted into energy.

Incorrect. m = E

dlack
2003-Apr-08, 11:14 PM
When I say mass, I mean rest mass.

We started out with a particle of mass m.

We got photons with mass zero.

There was less mass after the "event" than there was before the "event."

Mass was not conserved.

Ring
2003-Apr-08, 11:43 PM
When I say mass, I mean rest mass.

So do I. That's what the magnitude of the energy momentum four-vector is.


We started out with a particle of mass m.

We got photons with mass zero.

The individual photons have zero mass but the system comprised of two photons with anti parallel momentum has a mass of m = E.


There was less mass after the "event" than there was before the "event."

No, there is exactly the same amount of mass both before and after the event because p of the system = 0 and therefore m = E.


Mass was not conserved.

Not true. m = E.

dlack
2003-Apr-09, 12:10 AM
The individual photons have zero mass but the system comprised of two photons with anti parallel momentum has a mass of m = E.


I have real issues with this statement. Do you have some sources as to how a system comprised of massless particles can be said to have mass?

Bill Thmpson
2003-Apr-09, 01:03 PM
So if you're using geometrized units, all quantities are in centimeters, so energy can't be expressed as kilograms, as was stated initially to start this particular aspect of the debate.

Is this a thowback to my comment about radius and area of a circle?

If so, then it is wrong.

Area would be centimeters sqared, not centimeters.

dlack
2003-Apr-09, 02:28 PM
So if you're using geometrized units, all quantities are in centimeters, so energy can't be expressed as kilograms, as was stated initially to start this particular aspect of the debate.

Is this a thowback to my comment about radius and area of a circle?

If so, then it is wrong.

Area would be centimeters sqared, not centimeters.



::sigh::

It had to do with Princeton's comment that energy could be measured in kilograms and Kilopi's link to geometrized units (wherein all quantities are measured in cm).

But I've already given up this part of the argument.

JS Princeton
2003-Apr-09, 04:38 PM
The individual photons have zero mass but the system comprised of two photons with anti parallel momentum has a mass of m = E.


I have real issues with this statement. Do you have some sources as to how a system comprised of massless particles can be said to have mass?

Ring is absolutely right. Momentum that is equal an opposite in a system will sum to zero. This means that all the energy has to be bound up in mass.

This is the reason that an electron and a positron colliding creates two photons travelling in opposite directions in the center of momentum frame.

Mass and energy are inextricably connected. Mass IS an energy, and while it's true that energy is not necessarily in the form of mass, one can in principle convert all the energy you want into mass up to the amount of energy divided by c^2. This, then, is how you might express energy in kilograms. Call it "potential restmass energy of the system" if you like.

dlack
2003-Apr-11, 12:28 AM
Ring is absolutely right. Momentum that is equal an opposite in a system will sum to zero. This means that all the energy has to be bound up in mass.


I discussed this with my physics professor.

He says a system comprised solely of photons has no mass.

He says the formula:

E^2 = p^2c^2 + m^2c^4

can't be applied to an entire system, only to each individual particle (individual photons, electrons, positrons, etc.).

So, for our example of a mass at rest transforming into two photons:

(1)Momentum is conserved for the system.

The initial momentum is the momentum of the mass, zero.

The final momentum is the sum of the two (vector) momenta of the photons, which is zero if they are moving in opposite directions.

(2)Energy is conserved for the system.

Initial energy is E = mc^2.

Final energy is E = (p1c)^2 + (p2c)^2 , where p1 is the momentum of one photon, and p2 is the momentum of the other. (Obviously, p1 = p2 because of (1), so final energy is E = 2(p1c)^2).

So, energy is conserved, momentum is conserved, and mass need not be conserved. In the beginning, the system has mass. In the end, it has none.

JS Princeton
2003-Apr-11, 02:13 AM
Ring is absolutely right. Momentum that is equal an opposite in a system will sum to zero. This means that all the energy has to be bound up in mass.


So, energy is conserved, momentum is conserved, and mass need not be conserved. In the beginning, the system has mass. In the end, it has none.

And these statements are equivalent. If the sum of the momenta are zero, then the energy that cannot be transformed away is left to mass.

dlack
2003-Apr-11, 03:15 AM
And these statements are equivalent. If the sum of the momenta are zero, then the energy that cannot be transformed away is left to mass.

I don't know what you mean by "transformed away."

The energy of the system is always the same. None gets "transformed away."

In this case, the energy starts out as rest energy of the mass, and finishes up as momentum of the individual photons. There is no "extra" energy to be "left to mass." All of the energy is accounted for in the massless photons.

There is no mass in a system comprised solely of photons.

JS Princeton
2003-Apr-11, 01:37 PM
And these statements are equivalent. If the sum of the momenta are zero, then the energy that cannot be transformed away is left to mass.

I don't know what you mean by "transformed away."

The energy of the system is always the same. None gets "transformed away."


Depends on the system. If you are measuring the momentum of a group of massive particles than this statement is exactly wrong. However, it is right for photons. The issue is then one of reference frames and expectation.



In this case, the energy starts out as rest energy of the mass, and finishes up as momentum of the individual photons. There is no "extra" energy to be "left to mass." All of the energy is accounted for in the massless photons.

There is no mass in a system comprised solely of photons.

The photons themselves have zero rest mass. However the system can have non-zero total rest energy:

www.phys.ufl.edu/~hill/teaching/ 2003/3063/notes/Class6.pdf

dlack
2003-Apr-11, 03:12 PM
And these statements are equivalent. If the sum of the momenta are zero, then the energy that cannot be transformed away is left to mass.

I don't know what you mean by "transformed away."

The energy of the system is always the same. None gets "transformed away."


Depends on the system. If you are measuring the momentum of a group of massive particles than this statement is exactly wrong. However, it is right for photons. The issue is then one of reference frames and expectation.
You still haven't explained what exactly you mean by "transformed away."




In this case, the energy starts out as rest energy of the mass, and finishes up as momentum of the individual photons. There is no "extra" energy to be "left to mass." All of the energy is accounted for in the massless photons.

There is no mass in a system comprised solely of photons.

The photons themselves have zero rest mass. However the system can have non-zero total rest energy:

www.phys.ufl.edu/~hill/teaching/ 2003/3063/notes/Class6.pdf

Are you talking about this?:
http://www.phys.ufl.edu/~hill/teaching/2003/3063/notes/Class6_files/v3_master03_image008.jpg

The line that says "Strange result: total rest energy of the 2 photons ? 0, whereas each photon has mc2 = 0" ?

I'm going to have to look at this closer when I have more time, but everything else on that page seems to agree with what I've been saying.

For example, here:
http://www.phys.ufl.edu/~hill/teaching/2003/3063/notes/Class6_files/v3_master03_image005.jpg

Says mass does not have to be conserved. They say mass is "invariant" when looking at a given system from different reference frames. But that's not what we've been talking about. We're talking about staying in one reference frame and watching mass transform into energy. (At least that's what I've been talking about.)

kilopi
2003-Apr-11, 03:34 PM
Can we agree that there is a difference between "rest mass" and "relativistic mass", and can we recast the claims and statements in those terms?

I notice that the OP was about relativistic mass.

JS Princeton
2003-Apr-11, 03:39 PM
You still haven't explained what exactly you mean by "transformed away."


Simply the fact that momenta is dependent on the reference frame you choose. Rest mass is not dependent on that.



The line that says "Strange result: total rest energy of the 2 photons /= 0, whereas each photon has mc^2 = 0" ?

I'm going to have to look at this closer when I have more time, but everything else on that page seems to agree with what I've been saying.

I think we're in agreement on the basics, it's just a matter of coming to terms with the fact that rest energy of a system of two photons does not have to be equal to zero.



For example, here:
http://www.phys.ufl.edu/~hill/teaching/2003/3063/notes/Class6_files/v3_master03_image005.jpg

Says mass does not have to be conserved. They say mass is "invariant" when looking at a given system from different reference frames. But that's not what we've been talking about. We're talking about staying in one reference frame and watching mass transform into energy. (At least that's what I've been talking about.)

Well, I've been talking about something slightly different. The fact that mass is invariant when looking at a given system from different reference frames is exactly what is meant by mass being the energy that you cannot transform away. You can't simply stay in one reference frame and "watch the mass transform into energy" because that's not what happens in something like an annihilation. Rather, what's going on is ultimately a conservation of energy. There is an equivalency of type: if you look at any given photon the rest mass is zero, but the rest energy of the system isn't equal to zero.

In reality, this is all basically semantics. It has to do with how you DEFINE your coordinates and your four vector energies. We are basically in agreement, there's just a little sticking point in terms of modifiability of the reference frames (how such a modification affects your four-vector energy).