Just for fun. Is KE=(1/2)mv^2 consistent with E=mc^2? What conditions & proof? This will be 50% of your grade …
Consistent in that the units are the same, and are valid units for energy, yes. But they are descriptions of entirely different forms of energy, and are not any more or less related than PE = mgh.
You bet it is! The condition is that v is much less than c.
The rest energy of a thing is E = mc[sup]2[/sup].
The total energy of a thing is gamma×mc[sup]2[/sup].
The kinetic energy is the difference of these, KE = (gamma - 1)mc[sup]2[/sup].
Recall that gamma = (1 - (v/c)[sup]2[/sup])[sup]-1/2[/sup]. Expanding this in Maclaurin series it’s:
gamma = 1 + 1/2 (v/c)[sup]2[/sup] - 3/8 (v/c)[sup]4[/sup] + …
So KE = (1/2 (v/c)[sup]2[/sup] - 3/8 (v/c)[sup]4[/sup] + … )mc[sup]2[/sup]
If v is much less than c, we can approximate this with just the first term, so
KE is approximately 1/2 (v/c)[sup]2[/sup] mc[sup]2[/sup] = 1/2 mv[sup]2[/sup]
In layman’s terms… a short fast little SOB on the football field hurts four times as much when he hits you as a guy twice his size going half his speed…
In glactic terms, that fragment of Shoemaker - Levy that hit Jupiter (fragment G) was only a couple of miles wide, going 26,000 miles per second, tore a hole in the atmosphere the size of Earth.
The underlying principle of this is stay the hell out of the way of really fast small objects.
D.
If the guy is twice the mass and half the speed, this should only hurt twice as much, if pain is proportional to energy.
TECHNICALLY it’s only twice as much, but it FEELS like 4x, especially if you’re playing outside. In the winter. And the windchill is 30 degrees.
For all you junior varsity football players out there - "Ever been ‘snotbubbled?’ " lol
D.
We’re you trying to translate KE = 1/2 mv2 to layman’s terms?
Slight nitpick: total energy that is independent of scalar fields and microscopic processes. Potential energy and thermal energy are not taken into account in your jaunt.
You’re basically saying that total energy is equal to rest mass energy plus kinetic energy. This is only true under certain circumstances. It is by no means universal.
In freshman physics the professor derived e=mc[sup]2[/sup] from f=ma. Classical physics is a special case within relativity, so by definition they must be consistent with one another.
Oh… okay. Well what do you call the E in E = gamma×mc[sup]2[/sup], then? Just “energy”?
You call it exactly what I outlined it as: kinetic energy plus rest mass energy.
The most general energy equation on the frame-dependent level is as follows:
E[sup]2[/sup] = m[sup]2[/sup]c[sup]4[/sup] + p[sup]2[/sup]c[sup]2[/sup]
Which in the limit where p = mv reduces to your equation.
Scalar field potentials and associated energies can be seen as artifacts as perspective most generally.
On the microscopic level Schrodinger’s Equation is the thing to use, which reduces in the time-indepedent case to
H|Psi> = E|Psi>:
Where Psi is your wavefunction, E is the energy eigen-value solution, and H is the Hamiltonian defined as the sum of the energy due to momentum and the energy due to any potentials.
… Well, maybe I’m doing it wrong, but I can’t get that to work. I can only get it to reduce to my equation when p = gamma mv.
Well if we’re getting nitpicky then H in the above eigenvalue equation is the Hamiltonian operator.
H = -(hbar[sup]2[/sup]/2m) (d[sup]2[/sup]/dx[sup]2[/sup]) + V
(In one dimension -I don’t know how to make a del)
Achenar you do have to use p = gamma mv. The only reason phycisists prefer JS’ equation is because momentum is considered to be more fundamental than kinetic energy. There’s no law of conservation of kinetic energy.
Sigh, that should be physicists
BTW, why don’t I have an exclamation point in the upper left of my posts
The exclamation points are to report a post to the moderator. Under the new vBulletin code you can’t report your own posts.
Search in ATMB for more info.
Yeah… I know that. I’m just thinking that the E in E[sup]2[/sup] = m[sup]2[/sup] + p[sup]2[/sup] is the exact same E as in E = gamma m. And I’ve never heard a physicist have a problem calling it “energy” or “total energy” rather than the unwieldy “kinetic plus rest energy”, although I acknowledge that it does leave out, say, potential energy.
I’ve always called it momentum-energy, meself. I figure that’s precise enough without being too unwieldy.
Thanks Exapno I was beginning to think that I’d been semi banned or something.
Achernar Most textbooks refer to it as the total energy of a particle but as g8rguy says it’s really the equation of the four-momentum. It says that a particle or a system of particles is on the mass shell. IOW the system’s real not virtual.