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RuroumiKenshin

^{2}?

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RuroumiKenshin

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Only if "m" is less than or equal to the rest mass!Originally posted by MajinVegeta

^{2}?

Cheers, Jim

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And this is the way you get energy in any nuclear reaction !

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Well.... the full equation is E^2 = p^2*c^2 + m^2*C^4

Does that help?

Does that help?

- #5

pmb

First let me suggest the web page I made regarding mass and energyOriginally posted by MajinVegeta

^{2}?

http://www.geocities.com/physics_world/mass_energy.htm

Depends on what that something is and what you have to work with I suppose. For example: You can't take a single photon and change it into a particle with a non-zero rest mass. However you can take something with a non-zero rest mass and change it into two (or more photons. As far as kinetic energy and mass goes - sure. Suppose there is an inelastic collision - E.g. two balls of putty. Each of which has the same mass and heading toward each other along a straight line (i.e. a headon collision). When they collide they stick together. Then the rest energy of the final lump of clay will be greater than the sum of the two rest masses of the original lumps of clay. The difference in mass being dM = K/c^2 where K = sum of kinetic energies of the two lumps of clay.

Pete

- #6

LURCH

Science Advisor

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Isn't this what happens with virtual particles; a photon becomes a particle/antiparticle pair?Originally posted by pmb

You can't take a single photon and change it into a particle with a non-zero rest mass.

Something I've always wondered about that; doesn't it require that the temperature of the final "lump" must be higher than that of the two original lumps? And that this extra energy accounts for the gained mass, and as the heat radiates off, and the lump returns to its original temp (the temp it had before the collision), the mass of the final lump becomes equal to the sum of the masses of the original two?Suppose there is an inelastic collision - E.g. two balls of putty. Each of which has the same mass and heading toward each other along a straight line (i.e. a headon collision). When they collide they stick together. Then the rest energy of the final lump of clay will be greater than the sum of the two rest masses of the original lumps of clay. The difference in mass being dM = K/c^2 where K = sum of kinetic energies of the two lumps of clay.

Just curious.

- #7

pmb

Hi LURCH

re - "Isn't this what happens with virtual particles; a photon becomes a particle/antiparticle pair?"

Not that I'm aware of. When it comes to virtual particles you're talking quantum field theory - And that's what I refer to as "High Guru Physics" and until I formally learn it I don't touch it. It's far too easy to take things the wrong way if you don't know the full theory.

However, that said, suppose a photon decayed into a particle/antiparticle pair each of which has a non-zero rest mass. Then there would be a frame of referance in which the total momentum was not zero. But look at it from the point of view of an observer in the frame of referance for which the total momentum vanishes. Before the decay there was one photon which had momentum - therefore in that frame the total momentum was *not* zero. Then after the decay the the total momentum *was* zero. Hence a violation of momentum.

However if there is a heavy nuclei nearby then you can have this process happen and the nuclei absorbs the residule momentum - in effect acting as a catlyst.

re - "Somthing I've always wondered about that; doesn't it require that the temperature of the final "lump" must be higher than that of the two original lumps? And that this extra energy accounts for the gained mass, and as the heat radiates off, and the lump returns to its original temp (the temp it had before the collision), the mass of the final lump becomes equal to the sum of the masses of the original two?"

Depends on the specifics of the situation. I'd say for a "real" ball of clay - yes. But in general - no. Temperature is related to the kminetic energy of the particles which make up the body. However if the energy goes into potential energy then there need not be an increase in temperature - the increase in mass is then due to an increase in potential energy. Think of two balls which collide with a third body - a spring. The end effect is the spring is compressed and all motion stops. The energy is all in rest energy and the potential energy of the spring.

Hope that helps.

Pete

- #8

RuroumiKenshin

Yes, immensely!Originally posted by FZ+

Well.... the full equation is E^2 = p^2*c^2 + m^2*C^4

Does that help?

what's "p"? (I have an idea its poise?)

- #9

RuroumiKenshin

Virtual particles are not real, and are only mathematical formulations to explain something about quantum field theory."Isn't this what happens with virtual particles; a photon becomes a particle/antiparticle pair?"

Anyway, what you said doesn't quite make sense; if they are an anitparticle/particle pair, they'd have to be compatible (virtual particles are compatible, right?). That sort of a pair won't last very long; antiparticles annihilate each other.

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Poise? p is momentum.

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RuroumiKenshin

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RuroumiKenshin

what's dynamic viscosity (I have an idea, but I'm not going to even mention it..)?

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emu

Viscosity I believe refers to friction inside a fluid. Very viscous fluids have little friction(like hot oil).

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Hi Mr/Ms Vegeta,Originally posted by MajinVegeta

Yes, immensely!

what's "p"? (I have an idea its poise?)

Please ignore the six posts preceding this one that are a little off topic.

Also please look back to my response of your original post where I went, succinctly, to the bottom line, namely; that any iota of mass enhancement that is attributable to a particle's velocity is recoverable only in the form of momentum or kinetic energy imparted to the matter into which it is invading. The momentum/kinetic energy that is imparted results in a like amount sacrificed by the speeding "bullet". The rest-mass is retained until the last iota of velocity imposed physical motion is converted. Consider Fz's formula"

Well.... the full equation is E^2 = p^2*c^2 + m^2*C^4

Does that help?

Perhaps the confusion with the momentum term could have been avoided if it had been replaced by its equivalent:

E^2 = m^2*v^2*c^2 + m^2*c^4

I wonder: why has kinetic energy not been included? - must be because that would mean that mass would have been figured-in three times; the wonderment is why two masses have been allowed. The pertinent point remains that when the particle has reached its rest mass, its velocity has become zero and the term vanishes. Cheers, Jim

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