Captain Ron
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This may be a naive question from a new poster, but I'm curious about the equation E=mc^2 and I'm wondering if anyone could clear up my confusion.

From what I understand of the whole equation, it converts from units of mass to units of energy via the conversion factor c^2, which represents the energy density. I've gone over the basics of how it's derived, but don't really understand it completely.

But the work-energy equation, w = f * d looks very similar when you break it down to f = m * a, which gives you back

energy = m * a * d

Which looks more like you're just calculating the energy required to accelerate an object with a given mass from a state of rest to the speed of light in one second over a distance of 1 light second. But taken that way, accelerating an object to the speed of light over a shorter distance would actually require less energy.

Thankfully, I've long since completed my physics exams, so I don't even have to think about this for the next month, but I thought I'd share and see if someone could clear up my confusion or have a laugh.

Also...

If an electron is spinning at the velocity 'c', but is a point particle, if you plug it back into the F = m * a equation with the required centripetal force vector

a = c^2/r (where r is infinitely small)

you get:

F = m * c^2 / r (where r is infinitely small, making F infinite for any value of m.)

Which would mean that there is an infinite amount of force holding every electron together.

Discuss?

From what I understand of the whole equation, it converts from units of mass to units of energy via the conversion factor c^2, which represents the energy density. I've gone over the basics of how it's derived, but don't really understand it completely.

But the work-energy equation, w = f * d looks very similar when you break it down to f = m * a, which gives you back

energy = m * a * d

Which looks more like you're just calculating the energy required to accelerate an object with a given mass from a state of rest to the speed of light in one second over a distance of 1 light second. But taken that way, accelerating an object to the speed of light over a shorter distance would actually require less energy.

Thankfully, I've long since completed my physics exams, so I don't even have to think about this for the next month, but I thought I'd share and see if someone could clear up my confusion or have a laugh.

Also...

If an electron is spinning at the velocity 'c', but is a point particle, if you plug it back into the F = m * a equation with the required centripetal force vector

a = c^2/r (where r is infinitely small)

you get:

F = m * c^2 / r (where r is infinitely small, making F infinite for any value of m.)

Which would mean that there is an infinite amount of force holding every electron together.

Discuss?

0

## Posts

ismass. c^2 is just a conversion factor to make the units work, effectively.2) Don't try to use Newtonian physics to examine relativistic or quantum phenomena. It makes dumb things happen, like in your electron equation up there.

ElJeffeonMikeManon2. A point particle is a point particle. It has no internal structure, but it acts as a coherent packet of energy with a given mass energy and a pre-determined set of properties. I just think it's an oddity, thats all.

Captain RononIrond WillonLykouraghona(cp) = -ω^2*r

ω = v / r

a(cp) = v^2/r

F(cp) = m * v^2/r

And the value for v doesn't really matter either. If the r is vanishly small and the m is constant, F would have to be infinite.

Lykouragh - I thought electron spin was considered a real property, one thats required to explain things like magnetism and such.

Captain RononExcept that an electron doesn't have a classical radius so much as it has a probability distribution. It's not r = 0; it's r = Psi(r,omega,chi).

It is a real property - it's just not actual rotational velocity. It's a static property that acts kind of like rotational velocity, and cannot be increased or decreased, though it can be tilted.

Irond WillonIrond WillonI might be confused here, but I'm talking about the electron itself, not the probability mechanics of it's orbitals about the nucleus.

And that value would be static, just infinite.

I don't really find it that odd to think about I guess. QM has had a ton of problems in regards to dealing with the infinities that show up the equations. Hence Feynman's infinity - infinity = 0.

Captain RononE=γmc^2, where gamma is 1/sqrt(1-[u^2/c^2]), with u representing velocity.

From this, you can also derive E^2=(p^2)c^2+(m^2)c^4, where p is momentum.

These are the equations for the total relativistic energy of a particle.

As a rule, classical physics and relativistic physics are, for all intents and purposes, equivalent when dealing with the sizes and speeds of objects with which we generally interact here on earth. Classical physics, however, falls apart when dealing with highers speeds, and extremely small or large masses, or very high energy states. In those cases, we must use relativistic (or quantum) equations.

TarantioonDoesn't that have negative solutions?

Captain RononMaybe you're misreading it? Even if the momentum is negative (which it wouldn't be, because direction doesn't matter- energy is scalar) it's squared. Everything else in the equation is mass and the speed of light, and it's a sum.

TarantioonGetting the square root of E^2 should give you back two solutions, one positive and one negative. That was part of Dirac's hunt for the positron, I believe.

It doesn't matter for me really, I was just curious about certain ways that it's represented.

Captain RononWell the point is that you have two flaws in your reasoning regarding "spinning an electron at light speed":

1) Electrons can't really be "spun". They have a property called "spin" that kind of acts like they're being spun, but it's not actually being spun in the way we understand it in a classical sense.

2) You reference "r=0" in your "spinning an electron" equation. When dealing with a quantum particle, you don't assume "r=0". You integrate over the probability distribution.

Irond WillonIf it helps, that particular equation is derived from the other equation I showed you, and the one for relativistic momentum:

p=γmu

TarantioonDirac was working with a wave equation, not that energy equation. It's sort of a different

thang.ElJeffeonWhile E=mc^2 speaks specifically of particles at rest, the larger point is that energy is mass,

period. Rest energy isn't different from kinetic energy in this way. When you take a particle and speed it up, you're effectively giving it more mass. That's why the mass of particles goes up when you shoot them around at relativistic speeds - the mass is actually increasing, because you're pouring more energy into it, and energy = mass.Really, that's the best way to view E=mc^2. It's just saying, "energy = mass". Period.

ElJeffeon1. Spin angular momentum, you mean. Which acts just like regular spin momentum, implying a defined mass spinning with a given velocity on an axis, but because the electron doesn't appear to have a real physical structure its considered to be one of those wierd little intrinsic property that we don't try to extrapolate additional information from?

2. I didn't say r=0, I said r approaches 0. If r = 0 then the whole thing wouldn't work at all because you'd run into a nullility. I think we're talking about two different kinds of spin. I'm talking about the spin of a point particle in reference to itself, you're talking about it's orbital probability distribution about the nucleus maybe?

Eljeffe - I was trying to find a specific reference to rest mass in the derivation of the formula, but I can't seem to find it. I know one of the assumptions regarding it cancels certain other operators to zero, but I can't find the section I'm looking for.

Edit - Found it: He assumed that the universe was static and that time was infinite in both directions, and from this created the concept of a rest mass with zero momentum, as seen in the formula F * p = 0 stating: In the particle's rest frame, the momentum is (mc,0) and so for the force four-vector to be orthogonal, its time component must be zero in the rest frame as well, so F = (0,F). Since we've known the universe to be expanding since Hubble's discovery in 1929, and we've already observed a violation in cosmological isotropy from the WMAP observations of cosmic microwave background radiation, is there an information loss taking place by making these incorrect assumptions concerning the existence of the rest mass which begins to look like an absolute frame of reference?

I had it answered already, but it came down to treating small chunks of spacetime as though they were flat and static and getting back a 'good

enough' answer.

You might consider that a load of gibberish, but I'm just of a curious disposition.

Captain RononIt's rather the same situation with the orbital probability fields though as well - for whatever reason, the magnetic moment does in fact act somewhat like a conventional bar magnet ostensibly because the electron can be considered to be "spinning" through it's probability field. Of course this is all "to the first order" (waves hands).

electricitylikesmeonelectricitylikesmeonAnd I love physics, too.

JamesKeenanonEspecially the hand waving.

Captain RononelectricitylikesmeonYeah, but that leaves them sort of explained and unexplained at the same time.

ElJeffeonShhh... if no one knows, then the problem isn't there.

Captain RononRichyonThe "randomness" of quantum theory, for example, was never actually part of the theory. It was never "Our universe is

actuallycompletely random, unattached to physical law" In thought the random portion of quantum theory was to account for the secret key we're obviously missing.This key, once found, will bring quantum theory back into the realm of scientific predictability.

JamesKeenanonInterestingly, no. Dirac's delta function was designed to handle point particles. It's the kind of barbaric solution mathematicians hate, but it works.

There is a real question to be asked - does an electron represent an infinite amount of energy? And the answer, if you apply the normal rules to integrating an electron's field over all space, is yes. But the answer is trivial because it doesn't predict anything. We could just as well pick the energy of everything in the universe to be infinite by choosing an infinite potential energy as our starting point. Like all physical properties, energy is only meaningful for what it predicts.

QM doesn't change this, btw. Under the conventional interpretation of collapse, a particle chooses a mathematically precise set of characteristics - a discrete location for a discrete time. In a collapse the radius of an electron does become 0. Another way of stating this is that an orbital isn't a smeared electron cloud, though that is an easy way to think of it. It's an area where a point particle might be with specified probability. The indeterminacy of that location doesn't change the energy inherent in the electrical potential of the particle.

zakkielonzakkielonEinstein also had a problem with entanglement, "spooky action at a distance," remember? I don't think he's wrong. I just don't think he had the equipment that we do now.

It's so counter-intuitive (and I know intuition has no place

reallyin science). To suggest particles act completely randomly destroys my strongly held deterministic beliefs. I therefore reject that idea entirely.And thus a church is born.

JamesKeenanonF = γ m (1 + γ2 v vt) a

and

a = (1 - v vt) F / (γ m)

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

stestaggonIt's not a shortcut, it's a scientific fact well demonstrated by Hiroshima and particle accelerators.

EDIT: You can't argue it's "mathematically defined to be conserved" because then you're arguing this fact isn't demonstrated. It is demonstrated however - energy is conserved through a constant to be equivalent to mass. The constant is mandated by gauge symmetry.

I don't know what you're trying to say here, but there's no solution where an electron represents an infinite amount of energy. Integrating it's electric field over all of space just gives you a potential energy for any particle at any point in space with relation to that electron because the electric field only represents a force and a force

is not energyin and of itself.EDIT: tl;dr energy is meaningless in relation to a single electron, since it's not going to act or be acted upon by any external body.

electricitylikesmeonanythinggoes, or how far away things are.Also: An electron on top of a potential well is not more massive than one at the bottom. Ergo, mass=/=energy. A photon has no mass. Ergo, mass=/=energy. Sorry if this is flip treatment of the subject, but this is a pretty obvious point, and you really have no excuse for missing it.

You're exceeding your expertise here. I'll try and give the short explanation.

Force fields are messy because they specify vectors for each point. It is often much simpler to work with a scalar field, and for certain kinds of vector fields

Fit is possible to specify a field U such that delU=F. In this case, the curl will be 0 and the closed integral of b]F[/b] dot d[b]s[/b will be 0.* If this force is not time-dependent, we can call itconservative. Given the above conditions and Newton's definition of force asF=m*d^2/dt^2(r), it is possible to prove that the quantity U(r)+m/2*(dr/dt)^2 will be conserved (for non-relativistic systems, of course. Relativistic systems are more fraught). Note that U is not actually fully determined byF; a constant of integration will always remain which may be chosen arbitrarily.So what does it mean to say energy is always conserved, if we can just define a potential energy to ensure that it will be? Does it tell us nothing about the universe? No, of course not. It tells us that the fundamental forces of the universe are all of such a nature as to permit us to define a potential energy, which turns out to tell us quite a lot about what sorts of things are possible. This is what it means to say that "conservation of energy has been demonstrated." But it is really very naive to suppose that energy was just out there waiting for us to discover it, or that it adds anything to our desciption of the universe. Physicists hunt for quantities that can be conserved because they make problems easy. That's what led Einstein to redefine energy - he needed a quantity that would be conserved relativistically. We don't imagine that the Hamiltonian or Lagrangian are directly observable properties of the world despite the fact that they are indispensible in dealing with many advanced problems.

*Magnetic fields are the exception to this - they have a curl but no divergence. It turns out that the highly specific properties of magnetic fields allow us to define a potential nonetheless, but magnetic potentials are a beast to work with.

If you're interested in a more rigorous treatment of this, I recommend advanced E&M and mechanics. My understanding is that nanotech people are often short-changed on the fundamentals in the rush to get microscopic.

This is what I said. It's not, so near as I can tell, anything like what you said.

zakkielonIn regards to the r -> 0, although Dirac's equation involved point particles, when dealing with reality is it better to assume that the point particle represents an indivisible that still has a non-zero radius?

Captain RononAre you arguing that mass is not total energy? Because I don't think that's what he's trying to say. Just that mass is a component of energy, just the same as momentum. And that mass is composed of energy.

I'm not seeing any real conflicts between this and what you're saying, but it's entirely possible I'm missing something.

TarantioonHey, you guys wanna hear a joke I made up when I was in college?

- Why is it so windy in the physics department?

- From all the hand-waving!

IreneDAdleronAnd nonetheless light does in fact have a speed, in the sense that it will cover a particular distance in any inertial frame in a finite time. Hence the speed can change. That it's more convenient to define things in terms of the speed of light does not necessarily mean light must be absolutely constant - there's no particular reason that it cannot represent some underlying set of parameters with the actual speed of light a variable.

This is such a misnomer there's no excuse for you using it. The

of the electron as defined by E=mc^2 will never change because the only way to know the electron's mass is to pull it out of the system and measure it. The mass of the system with electron and potential well does.rest massConsider a void with a hydrogen atom in it. By adding a photon of sufficient energy (13.6 eV for example), the hydrogen atom (i.e. proton-electron system) absorbs the photon and the electron is pulled apart. At infinity (i.e. a very large distance) we still will not have recovered the original photon, but we will have increased the total energy of the system, which is how E=mc^2 came about of course - you approach

cbut never reach it, yet keep adding energy so where's it going?And this is observed in practice - nuclear binding forces show a mass defect due merely to the creation of potential energy in the nuclear strong force between the nucleons, which is why energy is released during nuclear fusion.

What your example tries to use as a trick is to imply that an electron can never weigh

lessthen it's rest mass, which is not the case as exemplified by the nucleon example.EDIT: Choosing an integrating factor is not nearly as arbitrary as you make it seem, since it carries through to the entire integration. Your contention seems to be that because you can choose an integrand that gives a striking result, clearly you can simply choose to get a striking result and then launch into asking questions about why maths should tell us anything about the universe at all. Valid questions, but attacking mass == energy isn't the way to go about them because it's just so solid. Of course it's not surprisingly when things are equivalent - because if they're unrelated variables we just say "when that's 1, and that's 1, they equal some

constant* 1". The constant is of course just a result of the fact that our units of measurement are rather arbitrary.EDIT 2: I mean, you're not wrong - one of the big issues is always predictive power, and making any mathematical transform always needs a compelling argument for being able to do it hence the issue of inertial mass and gravitational mass (why are they apparently the same?) but your argument against the energy/mass equivalence is just kind of baffling.

electricitylikesmeonAnd this shows that mass=energy... how? What are you trying to prove here, exactly?

I'll put it this way. Mass is a rational quantity. It has an absolute 0. Energy is not a rational quantity. We can pick the 0 wherever we like. The two are not the same. I cannot simply tell you the energy of a particle and enable you to deduce its mass, nor can I tell you the mass of a particle and enable you to deduce its energy.

zakkielonParticles don't act completely randomly - they act according to a probability distribution. There's not a simple mechanism underlying this - fundamentally, the variance in the position and velocity (or time and energy) distributions are linked in such a way that narrowing the distribution in one exacerbates the other, and these intrinsic probabilistic properties are necessary to explain physical properties like Bose-Einstein condensates, superfluids and, hell, conductors.

Irond Willonrealitywhen we speak of probabilities. Our inclusion of this probabilistic variance allows now us to make fairly accurate predictions, sometimes even really fucking accurate predictions.However, I still maintain a belief in the idea that sooner or later, we'll identify exact, quantitative and mathematically finite causes, actions, everything.

In the years to come, supercomputers will be built which can perfectly predict the actions of a small number of molecules a few seconds ahead. This number will grow, the time ahead the computer can perfectly predict will grow. In time, laboratory chemical reactions will be modeled exactly,

exactlyhow they'll play out before the experiment is ever done. Soon, entire weather systems and gas flows can be near perfectly predicted. Then truly perfectly predicted.Planetary AC!JamesKeenanon