Converting Kinetic Energy: Solving for Mass in E=mc2 Equation

• RuroumiKenshin
So apparently it is a measure of viscosity. But I don't think that's what pmb was talking about, because pmb said "momentum", which is a significantly different thing than viscosity. Momentum is a measure of the quantity of motion something has, while viscosity is a measure of how much resistance to flow a liquid has. There are a number of different symbols used to represent momentum, p happens to be one of them. I don't know why. Anybody else know?In summary, the conversation discusses the possibility of converting kinetic energy to mass by solving for "m" in the equation E=mc2. It is mentioned that this is only possible if "m" is less than or equal

RuroumiKenshin

Can you convert the kenetic energy of something to its mass by solving for m in the equation E=mc2?

Originally posted by MajinVegeta
Can you convert the kenetic energy of something to its mass by solving for m in the equation E=mc2?

Only if "m" is less than or equal to the rest mass!
Cheers, Jim

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

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

Does that help?

Originally posted by MajinVegeta
Can you convert the kenetic energy of something to its mass by solving for m in the equation E=mc2?

First let me suggest the web page I made regarding mass and energy

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

Originally posted by pmb
You can't take a single photon and change it into a particle with a non-zero rest mass.
Isn't this what happens with virtual particles; a photon becomes a particle/antiparticle pair?
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.
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?

Just curious.

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

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

Does that help?

Yes, immensely!
what's "p"? (I have an idea its poise?)

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

Virtual particles are not real, and are only mathematical formulations to explain something about quantum field theory.
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.

Poise? p is momentum.

Is poise a measure viscosity? I've been trying to find as much info. about it as I can, but I haven't gotten far. I guess this explains it! "p" always reminds me of poise! (fascinating, from a psychological perspective)

According to Wolfram, poise is "The cgs unit of dynamic viscosity, equal to 1 g cm-1 s-1, 1/10 Pa s."

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

P is momentum, where momentum is velocity*mass.
Viscosity I believe refers to friction inside a fluid. Very viscous fluids have little friction(like hot oil).

p is not poise!

Originally posted by MajinVegeta
Yes, immensely!
what's "p"? (I have an idea its poise?)

Hi Mr/Ms Vegeta,
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

1. What is the E=mc2 equation and why is it important?

The E=mc2 equation, also known as the mass-energy equivalence equation, is a fundamental equation in physics that relates energy (E) to mass (m) and the speed of light (c). It was derived by Albert Einstein and is important because it revolutionized our understanding of the relationship between mass and energy, showing that they are essentially interchangeable.

2. How does kinetic energy relate to the E=mc2 equation?

Kinetic energy is a form of energy that an object possesses due to its motion. In the E=mc2 equation, the "E" represents energy, which includes both kinetic energy and other forms of energy. The mass (m) in the equation refers to the rest mass of an object, which is its mass when it is not moving. Therefore, the E=mc2 equation shows that an object's kinetic energy is directly related to its mass.

3. What is the process for converting kinetic energy to mass using the E=mc2 equation?

The process for converting kinetic energy to mass using the E=mc2 equation involves rearranging the equation to solve for mass (m). This can be done by dividing both sides of the equation by the speed of light squared (c2), which gives the equation m = E/c2. Then, the value for kinetic energy (E) can be plugged in and the resulting mass will be in units of kilograms.

4. Can the E=mc2 equation be used for all types of energy?

Yes, the E=mc2 equation can be used for all types of energy. This is because energy is a broad concept that includes various forms such as kinetic energy, potential energy, thermal energy, and more. As long as the energy is measured in joules (J), it can be used in the equation to calculate the mass of an object.

5. How does the E=mc2 equation relate to Einstein's theory of relativity?

Einstein's theory of relativity is based on the concept that the laws of physics are the same for all observers in uniform motion. The E=mc2 equation is a direct consequence of this theory and demonstrates how mass and energy are interrelated. Additionally, the equation is used in the theory of relativity to explain the behavior of particles traveling at the speed of light.