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

[Frame Dragger]

I don't really object to the word spin being used, I just think there is a very significant fundamental difference between quantum spin and macroscopic spin. But this difference is normally made clear in textbooks, anyway, I think?Yeah, I know, it was just a U.S. vs British thing. :!)

The truly sad thing is that I'm American! First generation Greek, but that has nothing to do with it. I just happened to be living in Ireland during a critical moment in my education which has lead to pain, confusion and awkwardness since. THANKS FOR BRINGIG IT UP! *runs away sobbing*... yeah so there's that.

 

[Char. Limit]

Even so, I want to understand all of physics... even the parts that make no sense. Of course, the will to do something doesn't make it easy.

The photon disturbs me. It has no mass, yet it has momentum and energy... both of which have mass in their equations. It appears to me to simply be a packet (quantum?) of energy that...

Are photons one of the four almighty force-carrying particles?

Photon:
Has no mass
Energy determined by frequency
v=c at all times

What else is there?

 

[sylas]

Even so, I want to understand all of physics... even the parts that make no sense. Of course, the will to do something doesn't make it easy.

The photon disturbs me. It has no mass, yet it has momentum and energy... both of which have mass in their equations. It appears to me to simply be a packet (quantum?) of energy that...

Are photons one of the four almighty force-carrying particles?

Photon:
Has no mass
Energy determined by frequency
v=c at all times

What else is there?

Which equations do you mean? I've given the basic equation above, which is satisfied by every particle we know: massless, massive, relativistic, slow, all of them. The other thing photons have is quantum "spin".

Photons are the carrier particle for the electromagnetic force. (I don't know what you mean by "almighty".)

Other force carriers are the gluon (strong force), the W and Z bosons (weak force) and (undetected so far, but predicted) the graviton, for gravity.

Cheers -- sylas

 

[Frame Dragger]

Even so, I want to understand all of physics... even the parts that make no sense. Of course, the will to do something doesn't make it easy.

The photon disturbs me. It has no mass, yet it has momentum and energy... both of which have mass in their equations. It appears to me to simply be a packet (quantum?) of energy that...

Are photons one of the four almighty force-carrying particles?

Photon:
Has no mass
Energy determined by frequency
v=c at all times

What else is there?

They're the force carrying Gauge Bosons for the Electromagnetic force. You have Gluons for the Strong (nuclear) force that results in Colour Confinement of Quarks, the W and Z bosons (weak as Sylas said), and then maybe the Higgs Boson and/or a Graviton (speculative on both counts awaiting evidence in the LHC). The Higgs Boson specifically and the Higgs Mechanism in general are theorized to be the source of Mass in other particles.

It should be noted that ALL Gague Bosons are massless Spin 1 particles, and that is a central part of The Standard Model.

 

[Char. Limit]

The equations I was referencing are the only two I know for momentum, which are p=mv and p=\frac{d E_k}{dv}, and the three equations I know for energy, which are E=\frac{1}{2}mv^2, E=mgh, and E=\frac{pc+mc^2}{\sqrt{1-\frac{v^2}{c^2}}}. All of them depend on mass, don't they?

By almighty, I was just emphasizing their importance.

On the W and Z bosons, what's the difference, if they both carry the weak force?

 

[Frame Dragger]

The equations I was referencing are the only two I know for momentum, which are p=mv and p=\frac{d E_k}{dv}, and the three equations I know for energy, which are E=\frac{1}{2}mv^2, E=mgh, and E=\frac{pc+mc^2}{\sqrt{1-\frac{v^2}{c^2}}}. All of them depend on mass, don't they?

By almighty, I was just emphasizing their importance.

On the W and Z bosons, what's the difference, if they both carry the weak force?

In a single word: Charge. If you'd like some good reading, take a look at GUT theories in which the ElectroWeak force emerges from a union of the EM/Weak forces. Fascinating truly.

 

[sylas]

Here are your equations...

p=mv

This is an approximation which works only for massive particles and non-relativistic v. A more accurate equation for all velocities less than c is

p = \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}​

You don't need mass to have momentum. A particle like a photon has m=0 and v=c, and so the above is undefined. It works out to 0/0. A photon has momentum E/c.
​

p=\frac{d E_k}{dv}

I've never actually seen this before. It works for the approximations for slow particles, but is false in general. With a bit of calculus you can derive a more correct equation from E² = (pc)² + (mc²)² as follows.

\begin{align*}<br /> p &amp; = \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}} \\<br /> \text{Hence} \; E^2 &amp; = \frac{m^2v^2c^2}{1-\frac{v^2}{c^2}} + m^2c^4 \\<br /> &amp; = \frac{m^2c^4}{1 - \frac{v^2}{c^2}} \\<br /> \text{Hence} \; E &amp; = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} \\<br /> \frac{dE}{dv} &amp; = \frac{mc^2(-2v/c^2)}{-2(1-\frac{v^2}{c^2})^{3/2}} \\<br /> &amp; = \frac{mv}{(1-\frac{v^2}{c^2})^{3/2}} \\<br /> &amp; = \frac{p}{1-\frac{v^2}{c^2}}<br /> \end{align*}​

E=\frac{1}{2}mv^2

This is an approximation which works only for massive particles and non-relativistic v; and which considers only "kinetic energy", not total energy.
​

E=mgh

This is for energy changes in a gravitational field. Photons change energy in a gravitational field in the same way as massive particles -- one of the early experimental confirmations of general relativity.​

E=\frac{pc+mc^2}{\sqrt{1-\frac{v^2}{c^2}}}

This looks like an error. Omit the pc on the top and it is correct.​

All of them depend on mass, don't they?

No. You've just give approximate forms that apply for massive particles. The general equations work fine whether m is zero or not. IMO the most important single equation is the one I gave earlier.

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

It is expressed in terms of momentum, rather than velocity.

On the W and Z bosons, what's the difference, if they both carry the weak force?

To be honest I'm not sure. I don't know much about the weak force.

Cheers -- sylas

 

[Char. Limit]

Actually, my last equation was an error. Thanks for noticing.

So... if the relativistic momentum equation for a photon is indeterminate... that would suggest to me that the photon's momentum cannot be described. Bah... quantum physics being strange and all, I should have expected that.

Next question: If a photon has no charge, and it can't, how does it carry an electromagnetic force?

Er... a photon has no charge, right?

 

[sylas]

So... if the relativistic momentum equation for a photon is indeterminate... that would suggest to me that the photon's momentum cannot be described.

No... it means photon momentum is not described in terms of mass and velocity. Photon momentum can be described just fine, and it is described as a function of frequency. p = hf/c

Next question: If a photon has no charge, and it can't, how does it carry an electromagnetic force?

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

 

[Char. Limit]

Can you describe massive object's momentum in the same terms, say, by using the de Broglie wavelength or something?

A photon has no charge. Thus, a proton is it's own antiparticle.

Two questions: Is the above statement correct, and if it is, is the reasoning correct?

 

[sylas]

Can you describe massive object's momentum in the same terms, say, by using the de Broglie wavelength or something?

Yes, I believe so.

A photon has no charge. Thus, a proton is it's own antiparticle.

Two questions: Is the above statement correct, and if it is, is the reasoning correct?

It is said that a photon is its own anti-particle; but you can't conclude that from lack of charge. A neutron is a counter example. It also has no charge, but it is not its own antiparticle.

Cheers -- sylas

 

[jtbell]

No... it means photon momentum is not described in terms of mass and velocity. Photon momentum can be described just fine, and it is described as a function of frequency. p = hf/c

And even in classical electrodynamics, the momentum density of an electromagnetic wave can be described just fine in terms of the E and B fields:

http://farside.ph.utexas.edu/teaching/em/lectures/node90.html

 

[Char. Limit]

So, if p=\frac{h\nu}{c} works for both massive and massless particles, why don't we use that equation instead? Sure, there's a slight hurdle when first learning about momentum, but the overall understanding would be greater in the long run.

 

[atyy]

So, if p=\frac{h\nu}{c} works for both massive and massless particles, why don't we use that equation instead? Sure, there's a slight hurdle when first learning about momentum, but the overall understanding would be greater in the long run.

The hurdle is relativistic quantum field theory. But I like your idea!

 

[Char. Limit]

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.

 

[atyy]

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.

 

[Char. Limit]

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).

 

[jacksonwalter]

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 e^{i\theta} = cos\theta + isin\theta, I think I'd be in a world of confusion.

 

[Char. Limit]

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!

 

[sylas]

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).

 

[Char. Limit]

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.

 

[sylas]

Apparently p really does equal hf/c.

Not for a massive particle, it doesn't.

 

[Char. Limit]

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 p=\frac{h}{\lambda} in high school? We learn more useless stuff than that in Am. Lit.

 

[Frame Dragger]

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 p=\frac{h}{\lambda} 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.

 

[DanRay]

Yes, it does follow the same rules for matter. The rule is:

E^2 = (pc)^2 + (mc^2)^2​

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?"

 

[Frame Dragger]

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?"

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

 

[sylas]

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

 

[Char. Limit]

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.

 

[DanRay]

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.

 

[Frame Dragger]

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.