E=Mc2

[qft123]

I think that E=MC^2 is slightly wrong.... I also want to add the fact that E=MC^2 was tested by atomic.bomb... According to me, 100percent of the mass is not converted inti energy..

1) Is this true??

2) It is a simplied version of E2= (MC2)2 p2c2..

I am not able to figure out how "p2c2" vanish, during the process of simplification???

[timo]

p -> 0.

[qft123]

How is the momentum Zero??

[timo]

Dunno. It's your equation and your simplification, and most particularly you haven't bothered properly describing the physical process you talk about.

[qft123]

For example, consider a simple hydrogen atom, basically composed of a single proton.

 

It has a tiny mass indeed. But in everyday quantities of matter there are a lot of atoms! For instance, in one kilogram of pure water, the mass of hydrogen atoms amounts to just slightly more than 111 grams, or 0.111 kg. Hence

E=(MC2)2 + p2c2

Simplified E = MC2

= 0.111 x 300,000,000 x 300,000,000 = 10,000,000,000,000,000 Joules

Why is the momentum always "0"?

[uncool]

Because the equation is about what happens when p is 0. E = m c^2 is the equation that relates rest mass with rest energy.

=Uncool-

[D H]

E=(MC2)2 + p2c2

The energy should be squared here: [imath]E^2 = (m_0c^2)^2 + (pc)^2[/imath]

 

Why is the momentum always "0"?

It's not. E=mc² is a simplification in the special case that momentum is zero. In other words, it is looking at things from the perspective of the rest frame of the object.

 

Another special case of the general mass-energy equivalence formula results for massless particles ([imath]m_0=0[/imath]) such as photons. There is no such thing as a rest frame for such particles; momentum is never zero in this case. The energy of a photon is given by [imath]E=pc[/imath].

 

 

Another way to look at [imath]E=mc^2[/imath] it is that the [imath]m[/imath] in this expression is the relativistic mass. The concept of relativistic mass however is a bit outdated and is no longer widely used.

[mississippichem]

I think that E=MC^2 is slightly wrong.... I also want to add the fact that E=MC^2 was tested by atomic.bomb... According to me, 100percent of the mass is not converted inti energy..

 

I think you are conflating the efficiency of a nuclear device (a question for engineers and bomb designers?) with the theoretical notion of mass energy equivalence (a very "physicsy" notion). Another example comes from chemical reactions. I can calculate how much anergy should be released in theory by such a reaction, but in reality I can never harness or even release 100% of that calculated energy. There are always loses or imperfections in a system. I don't know if those losses in a nuke are from engineering considerations or some theoretical notion from nuclear physics but I know that blowing up a nuke and measuring the Joules that come out is not a valid test of mass energy equivalence unless you account for all those other losses and imperfections in the system.

[swansont]

The problem is that this is looking at the situation kinda backwards. If you measured the mass before and after, the difference would tell you how much energy was released, in other forms (e.g. KE of massive particles, emitted photons). Since it's a reaction involving normal matter, the total number of baryons and leptons will be conserved, so there is neither a prediction nor an expectation that all the mass will be converted into energy.

 

I think you are conflating the efficiency of a nuclear device (a question for engineers and bomb designers?) with the theoretical notion of mass energy equivalence (a very "physicsy" notion). Another example comes from chemical reactions. I can calculate how much anergy should be released in theory by such a reaction, but in reality I can never harness or even release 100% of that calculated energy. There are always loses or imperfections in a system. I don't know if those losses in a nuke are from engineering considerations or some theoretical notion from nuclear physics but I know that blowing up a nuke and measuring the Joules that come out is not a valid test of mass energy equivalence unless you account for all those other losses and imperfections in the system.

 

The misconception here isn't even at the level of not achieving 100% of theoretical performance. (It's like expecting 100% thermodynamic efficiency, even though Carnot efficiency is the theoretical best one can do.) But it is a valid test of mass-energy equivalence, because total energy is still conserved.

[pmb]

I think that E=MC^2 is slightly wrong.... I also want to add the fact that E=MC^2 was tested by atomic.bomb... According to me, 100percent of the mass is not converted inti energy..

1) Is this true??

2) It is a simplied version of E2= (MC2)2 p2c2..

I am not able to figure out how "p2c2" vanish, during the process of simplification???

I never showe you where that expression came from. I posted a derivation which was similar to Einstein's in essense. It's at

http://home.comcast.net/~peter.m.brown/sr/mass_energy_equiv.htm

 

It goes like this. In frame S a body emits a two photon of equal enegy in opposite directions. In a frame moving relative to that frame we see the same thing. This time the photons don't have the same energy. They therefore don't have the same momentum. The sum of the momenta of the two photons yields a total momentum which is non-zero. This means that the momentum of the box decreases. This means that the body's mass decreased. Since the speed can be arbitrarily small the mass of the body decreases. Relativity is used to find a relationship between the energy lost E and the mass lost m. That relationship is E = mc². I think that there is another derivation which proves the generality of that expression.