Can matter be completely converted to energy as suggested by E=mc2?

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In summary: So the rest mass was eventually adopted as the standard, and the relativistic mass is now mostly used in theoretical physics.
  • #106
Char. Limit said:
Well, there's a saying: bad habits are hard to break. If you get fixed on the idea that momentum requires mass, as I did, it's hard to break.

I also support never telling students that you can't take the square root of a negative number. If they ask, don't waffle: tell them immediately about i.

I don't know if it has a name, but there's one of my philosophies.

Bad habits are very important, as long as they are good bad habits, like Newtonian physics, or special relativity, or general relativity, or quantum field theory.
 
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  • #107
Even so, shouldn't one at least introduce the Planck definition (I'm calling it that because Planck's constant is in it) of momentum in high school physics? Most of the class isn't listening to the teacher anyway (my physics class does, but we're an exception) and the ones who are listening can understand it, somewhat.

That's just my opinion, and whether you think it's right or not (it is), you must admit that it has some validity (I like parentheses).
 
  • #108
Char. Limit said:
Even so, shouldn't one at least introduce the Planck definition (I'm calling it that because Planck's constant is in it) of momentum in high school physics? Most of the class isn't listening to the teacher anyway (my physics class does, but we're an exception) and the ones who are listening can understand it, somewhat.

That's just my opinion, and whether you think it's right or not (it is), you must admit that it has some validity (I like parentheses).

My teacher in high school spent a full class period explaining how de Brolie's PhD thesis was only like 20 pages long and would muse about the fact that his body can be described by a wave, albeit one with a very short wavelength.

The problem with telling kids the generalized forms of every concept is that it seems to be much easier to start with the basic, specific case and then generalize later on. If I learned sin and cos by first learning the Taylor series expansions for them or [tex]e^{i\theta} = cos\theta + isin\theta [/tex], I think I'd be in a world of confusion.
 
  • #109
mv is not a specific case of hf/c though. They share none of the same units.

My wavelength is so small, I am a massive, thinking gamma ray. Fear me!
 
  • #110
Char. Limit said:
mv is not a specific case of hf/c though. They share none of the same units.

The units match, but you have the wrong formula. For momentum, it is p = h/λ. The energy of the particle is given by E = hf.

The velocity you obtain as fλ is not c, and it is not the velocity of the particle either. It is the phase velocity of the waves.

Using p = γmv and E = γmc2, you have λ = h/(γmv) and f = γmc2/h, and phase velocity is c2/v, which is greater than c. But that's okay, the phase velocity is not required to be less than c.

Cheers -- sylas

Postscript. Usain Bolt weighs 95 kg, and runs at about 10.44 m/s2. He has a wavelength of 6.68*10-37 m (about 4% of the Planck length) and a frequency of about 1.3*1052 Hz (about 700 million cycles per Planck time).
 
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  • #111
sylas said:
p = hf/c



It doesn't need charge; it needs to interact with charged particles. Beyond that, I'm not good on the details of carrier particles.

Cheers -- sylas

Apparently p really does equal hf/c.
 
  • #112
Char. Limit said:
Apparently p really does equal hf/c.

Not for a massive particle, it doesn't.
 
  • #113
I'm still not sure about that. Say I move at 1 m/s and weigh 50 kg (the truth, rounded). My de Broglie wavelength will thus be... Oh wait, it doesn't matter (it's h/50)... I read you wrong, sylas, and apologize. Still, can't we teach [tex]p=\frac{h}{\lambda}[/tex] in high school? We learn more useless stuff than that in Am. Lit.
 
  • #114
Char. Limit said:
I'm still not sure about that. Say I move at 1 m/s and weigh 50 kg (the truth, rounded). My de Broglie wavelength will thus be... Oh wait, it doesn't matter (it's h/50)... I read you wrong, sylas, and apologize. Still, can't we teach [tex]p=\frac{h}{\lambda}[/tex] in high school? We learn more useless stuff than that in Am. Lit.

If that had been taught in HS, my life woud have been a LOT easier. For some reason I found it much easier to work from abstraction within a definite framework, and THEN the arbitrary ****. Keep in mind that math and science are taught according to theories of learning that are far older than the Physics or Mathematics being taught!

However, in the bell-curve of life, more people would struggle with p=h/lambda than simplifications. Most people believe they never need to know that, and maybe they're right. *shrug*. High School and College are more about learning HOW to learn, than what you happen to pick up along the way. Post-Grad is where the real learning begins.
 
  • #115
sylas said:
Yes, it does follow the same rules for matter. The rule is:
[tex]E^2 = (pc)^2 + (mc^2)^2[/tex]​
p is momentum, and E is the total energy.

If the particle has no momentum, then this reduces to E = mc2.

If a particle has very low velocity, then p is close to mv, and E is close to mc2 + 0.5mv2. This is what you are probably used to as the rules for matter, but in fact it is this simplified matter rule which is the one that is incorrect.

If v = c, then you have a massless particle for which m = 0, and E = p/c, which is true for photons.

The momentum of photons is an experimental fact, observed in particle collisions every day.

Cheers -- sylas

I'm sorry that I don't have the time to keep up with this forum in a timely (in order) fashion but I do have additional thoughts on the comments you made in regard to my question. I am aware of the assumption that the energy recorded for photons in particle accelerators is regarded as momentum in the equations you Quote and as far as how the energy coming from photons affect the things they interact with it probably makes no difference what you call it. When I suggested that maybe a photon should be regarded as only energy what I mean is the photon is the energy not some amount of energy carried by some other unseen particle. The momentum question I brought up concerns the photons travel at c. The definable difference I am trying to consider is that c is not an imparted speed caused by adding energy to that imaginary particle but a feature of the energy called a photon that makes it very different from say the momentum of the billard balls they were discussing earlier. That momentum (for the cue ball) is imparted by the stick which receives its momentum from the muscle in my arm. The photon seems to require none of that to reach c. Part of its defination would have to be that it always moves at c unless it interacts. That it is said to give up its energy when it interacts is what my argument goes to. If the photon is the energy it still exists and still is the energy being measured regardless of how it seems to have changed. The key thought is the question I began with "what if photons are just energy?"
 
  • #116
DanRay said:
I'm sorry that I don't have the time to keep up with this forum in a timely (in order) fashion but I do have additional thoughts on the comments you made in regard to my question. I am aware of the assumption that the energy recorded for photons in particle accelerators is regarded as momentum in the equations you Quote and as far as how the energy coming from photons affect the things they interact with it probably makes no difference what you call it. When I suggested that maybe a photon should be regarded as only energy what I mean is the photon is the energy not some amount of energy carried by some other unseen particle. The momentum question I brought up concerns the photons travel at c. The definable difference I am trying to consider is that c is not an imparted speed caused by adding energy to that imaginary particle but a feature of the energy called a photon that makes it very different from say the momentum of the billard balls they were discussing earlier. That momentum (for the cue ball) is imparted by the stick which receives its momentum from the muscle in my arm. The photon seems to require none of that to reach c. Part of its defination would have to be that it always moves at c unless it interacts. That it is said to give up its energy when it interacts is what my argument goes to. If the photon is the energy it still exists and still is the energy being measured regardless of how it seems to have changed. The key thought is the question I began with "what if photons are just energy?"

:confused:

EDIT: *reads Sylas' post* Ahhhh... so it WAS nonsensical. I thought so, but the run-ons and blending of concepts made me wonder if I had missed some arcane branch of particle physics. :P
 
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  • #117
DanRay said:
I am aware of the assumption that the energy recorded for photons in particle accelerators is regarded as momentum in the equations you Quote and as far as how the energy coming from photons affect the things they interact with it probably makes no difference what you call it.

Momentum isn't some ambiguous quantity that can be whatever you want to call it. Momentum is a well understood physical concept, and photons most definitely have momentum.

When I suggested that maybe a photon should be regarded as only energy what I mean is the photon is the energy not some amount of energy carried by some other unseen particle.

That doesn't make any sense. A photon has energy, and momentum. No-one is attributing the energy of photons to other unseen particles.

The momentum question I brought up concerns the photons travel at c. The definable difference I am trying to consider is that c is not an imparted speed caused by adding energy to that imaginary particle but a feature of the energy called a photon that makes it very different from say the momentum of the billard balls they were discussing earlier. That momentum (for the cue ball) is imparted by the stick which receives its momentum from the muscle in my arm. The photon seems to require none of that to reach c. Part of its defination would have to be that it always moves at c unless it interacts. That it is said to give up its energy when it interacts is what my argument goes to. If the photon is the energy it still exists and still is the energy being measured regardless of how it seems to have changed. The key thought is the question I began with "what if photons are just energy?"

This is not sufficiently coherent to mean anything much. You are not even using words in ways that make sense; it's a classic case of "not even wrong". There's no point speculating about these things until you have a much firmer grasp of what we already know.

There's no particular mystery here. A photon has well defined energy and momentum, and this is as solidly confirmed as anything ever gets in physics. When a photon is absorbed, the energy and the momentum that was previously associated with the photon is passed on to other particles, consistent with conservation laws.

Cheers -- sylas
 
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  • #118
I think he means that a photon doesn't carry energy, a photon is energy.

Also, on FD's post, screw the bell curve, teach it right. If this includes two equations, so be it.
 
  • #119
sylas said:
Momentum isn't some ambiguous quantity that can be whatever you want to call it. Momentum is a well understood physical concept, and photons most definitely have momentum.



That doesn't make any sense. A photon has energy, and momentum. No-one is attributing the energy of photons to other unseen particles.



This is not sufficiently coherent to mean anything much. You are not even using words in ways that make sense; it's a classic case of "not even wrong". There's no point speculating about these things until you have a much firmer grasp of what we already know.

There's no particular mystery here. A photon has well defined energy and momentum, and this is as solidly confirmed as anything ever gets in physics. When a photon is absorbed, the energy and the momentum that was previously associated with the photon is passed on to other particles, consistent with conservation laws.

Cheers -- sylas

I know I'm not making sense to you but its not because I don't understand the basics and what you are talking about. So let me try to make my question clear another way. What kind of energy is it that the photon carries that is referred to as momentum. When it is involved in a photoelectric effect its energy becomes electric energy, but when I hear the word momentum I think of kenetic energy.

So I guess the question becomes if it is just ordinary kenetic energy as with the billiard ball how does that convert to electric energy. I am also saying that I perceive a difference between the statement "A photon has energy" and the statament "A photon is a form of energy." Which is what would result if the answer to my question were yes. I know the energy equations are well understood and much verified and all of my questions may for most purposes boil down to a not too useful semantics argument but I still can't get over the fact that whenever light (and other photons) are propagated the seem to instantaneously take off at c (or whatever speed they can attain in their current enviornment that is near c) without the need to accelerate.
 
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  • #120
Char. Limit said:
I think he means that a photon doesn't carry energy, a photon is energy.

Also, on FD's post, screw the bell curve, teach it right. If this includes two equations, so be it.

Teaching, as an enterprise, is meant to be the communication of knowledge. If you're not achieving that goal, you're not an effective teacher. That doesn't make things right the way they are, but your view ignores the reality which is pretty gray.
 
  • #121
DanRay said:
I know I'm not making sense to you but its not because I don't understand the basics and what you are talking about. So let me try to make my question clear another way. What kind of energy is it that the photon carries that is referred to as momentum. When it is involved in a photoelectric effect its energy becomes electric energy, but when I hear the work momentum I think of kenetic energy. So I guess the question becomes if it is just ordinary kenetic energy as with the billiard ball how does that convert to electric energy. I am also saying that I perceiv e a difference between the statement "A photon has energy" and the statament "A photon is a form of energy." Which is what would result if the answer to my question were yes. I know the energy equations are well understood and much verified and all of my questions may for most purposes boil down to a not to useful semantics argument but I still can't get over the fact that whenever light (and other photons) are propagated the seem to instantaneously take off at c (or whatever speed they can attain in their current enviornment that is near c) without the need to accelerate.

No let me be clear for everyone else here: GO do your homework on the basics FIRST. That's how it works. You need a knowledge-base before you start finding paradoxes that don't exist because you don't grasp the basics.
 
  • #122
Are you sure that a photon IS not the energy "carried"?
 
  • #123
Frame Dragger said:
gray

Yeah, you are pretty confused! :smile: I personally spell "esophagus" for dull scientific stuff, but "colour", "flavour" and "amoeba" for more exciting stuff.
 
  • #124
All I'm saying is give h/lambda a chance in high school. I've already learned about h and lambda the normal way... So it's a simple extension to connect the two.
 
  • #125
DanRay said:
What kind of energy is it that the photon carries that is referred to as momentum. When it is involved in a photoelectric effect its energy becomes electric energy, but when I hear the word momentum I think of kenetic energy.

That's where you are going wrong. Momentum and energy are different things. We don't refer to the energy of a photon as momentum.

A photon is just a quantum of light, or a "packet" of light. Light carries energy and it carries momentum. You can call it "light energy" if you like.

I still can't get over the fact that whenever light (and other photons) are propagated the seem to instantaneously take off at c (or whatever speed they can attain in their current enviornment that is near c) without the need to accelerate.

I'm not sure why you can't "get over it". This usually means there's some assumption which you need to drop. In this case, it seems that when you think "momentum" you are thinking only in terms of moving balls. But momentum is not limited to moving balls.

Think rather of a wave in the ocean. A wave carries momentum, and it doesn't accelerate either. A photon is very much like a little wave packet, and just like is normal for waves, there's no acceleration as a wave is formed.

Cheers -- sylas
 
  • #126
Char. Limit said:
Are you sure that a photon IS not the energy "carried"?

Energy is a property of a photon, just as momentum, wavelength and frequency are properties. It is just as incorrect to say that a photon is energy, as it is to say that a photon is momentum, or a photon is wavelength, etc.

I think laymen and beginning students get led to this idea by the frequent of "pure energy" in science fiction. There is no such thing as "pure energy." Energy is always a property of something.
 
  • #127
But... but I want it to be.
 
  • #128
Char. Limit said:
But... but I want it to be.

Now THAT was one of the most honest statements I've heard in my life. Keep that sense of humour and open mind and you'll be alright.
 
  • #129
jtbell said:
Energy is a property of a photon, just as momentum, wavelength and frequency are properties. It is just as incorrect to say that a photon is energy, as it is to say that a photon is momentum, or a photon is wavelength, etc.

I think laymen and beginning students get led to this idea by the frequent of "pure energy" in science fiction. There is no such thing as "pure energy." Energy is always a property of something.

You have to admit, "...Fired a crackling lance of pure energy; blasting Gorzingo to its componant parts." Sounds better in a fictional novel than, "...He knew he only had a megawatt... maybe 2 from his Airplane-sized chemical laser. So, with careful aim from the help of an entire flight crew, a beam of electromagetic energy rapidly heated Gorzingo until his sunburn was UNBEARABLE!"

I think Gambit is the only fictional figure I've ever heard of who specifies the kind of potential energy they work with... from nowhere... presumably sourced FTL... lol.
 
  • #130
The photon momentum

In 1916 A. Einstein published a paper in 'Mitteilungen der Physikalischen Gessellschaft Zurich', called 'On the Quantum Theory of Radiation'. In this he shows that the absorption/emission of a photon ( light quantum) by an atom involves an exchange of momentum. Unless momentum is conserved, Planck's black-body spectrum is altered.

A translation of this this paper is in van der Waerden's 'Sources of Quantum Mechanics' (1967). As v d Waerden says 'All subsequent research on absorption, emission and dispersion of radiation was based on [this] paper'.

So, when a photon is absorbed by an atom. it's energy goes into the atom, which also experiences a 'kick'. In fact, this kick is the reason why doppler laser-cooling of atoms works.
 
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  • #131


Mentz114 said:
In 1916 A. Einstein published a paper in 'Mitteilungen der Physikalischen Gessellschaft Zurich', called 'On the Quantum Theory of Radiation'. In this he shows that the absorption/emission of a photon ( light quantum) by an atom involves an exchange of momentum. Unless momentum is conserved, Planck's black-body spectrum is altered.

A translation of this this paper is in van der Waerden's 'Sources of Quantum Mechanics' (1967). As v d Waerden says 'All subsequent research on absorption, emission and dispersion of radiation was based on [this] paper'.

So, when a photon is absorbed by an atom. it's energy goes into the atom, which also experiences a 'kick'. In fact, this kick is the reason why doppler laser-cooling of atoms works.

Fair enough... but what that "going in" entails, and the exact nature of that "kick" still has the problem of facing Classical and then Quantum (probabilistic) models and is a little weird. The "sucking out" as a result of DLC is also fascinating, understandable, but doesn't really make the photon or any gauge boson less... odd.
 
  • #132


Frame Dragger said:
Fair enough... but what that "going in" entails, and the exact nature of that "kick" still has the problem of facing Classical and then Quantum (probabilistic) models and is a little weird. The "sucking out" as a result of DLC is also fascinating, understandable, but doesn't really make the photon or any gauge boson less... odd.
I should point out that the photons we are discussing are not gauge bosons. These arise when a charge interacts with a magnetic or electric field. Because it's a quantum phenomenon, it is argued that the force must be carried by virtual quanta, the so-called gauge bosons. I don't think they can be detected as themselves.
 
  • #133


Mentz114 said:
I should point out that the photons we are discussing are not gauge bosons. These arise when a charge interacts with a magnetic or electric field. Because it's a quantum phenomenon, it is argued that the force must be carried by virtual quanta, the so-called gauge bosons. I don't think they can be detected as themselves.

Good point. I stand corrected.
 
  • #134


Frame Dragger said:
Good point. I stand corrected.

You were right the first time.
 
  • #135


atyy said:
You were right the first time.

Damn it! I thought I was, but everyone is open to the fallacy of an appeal to authority... I thought I must have been missing something.
 
  • #136
Happens to everyone, FD... Even me, and I usually AM wrong.
 
  • #137
Char. Limit said:
Happens to everyone, FD... Even me, and I usually AM wrong.

Very gracious Char. Limit, thank you.
 
  • #138
If photons always travel at c, why does light get slower through a medium (and don't tell me it doesn't, I've seen rainbows)?

Even if the photon is bouncing off things, its velocity shouldn't change, because it's massless...
 
  • #139
Char. Limit said:
If photons always travel at c, why does light get slower through a medium (and don't tell me it doesn't, I've seen rainbows)?

Even if the photon is bouncing off things, its velocity shouldn't change, because it's massless...

I'm not sure I know what you mean by the photon "bouncing off things", or why being massless would change how it propogates through a prism vs. vacuum out of line with observational evidence.
 
  • #140
Well, if light is traveling through a medium, it seems to me that a photon would keep either "bouncing off" or getting absorbed by atoms. So, if photons always travel at c, why do mediums slow
light down?
 

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