An Introduction to the Generation of Mass from Energy

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For point 2 I would refer you to the worked example where the pre-collision mass is that of the two colliding protons and the post-collision mass is that of the two protons plus the 'generated' kaons.

Sorry, I don't have seen your example yet, anyway, concerning the transformation from a ref. frame where both particle 1 and particle 2 have non zero velocity, I find, instead of equation (33) of "Introduction to Relativistic Collisions":

${E_3^0}^2={E_1^0}^2+{E_2^0}^2+2E_1^0 E_2^0+2T_1E_2^0+2T_1T_2+2T_2E_1^0-2c^2\vec{p_1}·\vec{p_2}$ (*)

Then we change frame on K', co-moving with particle 2; $\vec{v_1}'$ is now the relativistic relative velocity between particle 1 and particle 2 and:
$T_2'=\vec{p_2}'=0$
so (*) becomes equal to (33).
I apologize if you have already described this in your last example.

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Only a last comment on the title of this insight, Neil (by the way, from your nickname it seems we have the same age): if energy is always conserved, how can it "generate" something else ("mass")?

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[neilparker62]

Energy is neither created nor destroyed - simply changed from one form to another. I think that would apply to just about any form of energy 'generation'. In a nuclear power plant we generate energy from mass, here we generate mass from energy.

 

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Energy is neither created nor destroyed - simply changed from one form to another. I think that would apply to just about any form of energy 'generation'. In a nuclear power plant we generate energy from mass, here we generate mass from energy.

Energy cannot be "generated", only transformed or converted. Your examp
Anyway, just my opinion, of course.
Edit: I want to make another example: let's say a particle, in a ref. frame where it's still, could disintegrate completely into photons, in a spherically symmetric pulse of light. Which is the mass of this light pulse? It's not zero, it's the same m as before. But I'm sure a lot of untrained people would say it's zero.
Regards.

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Energy cannot be "generated", only transformed or converted. Your examp

Something went wrong here. The phrase should have been:
"Your example shows, in my opinion, that kinetic energy can be converted into rest energy or the reverse".

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[neilparker62]

Sorry, I don't have seen your example yet, anyway, concerning the transformation from a ref. frame where both particle 1 and particle 2 have non zero velocity, I find, instead of equation (33) of "Introduction to Relativistic Collisions":

${E_3^0}^2={E_1^0}^2+{E_2^0}^2+2E_1^0 E_2^0+2T_1E_2^0+2T_1T_2+2T_2E_1^0-2c^2\vec{p_1}·\vec{p_2}$ (*)

Then we change frame on K', co-moving with particle 2; $\vec{v_1}'$ is now the relativistic relative velocity between particle 1 and particle 2 and:
$T_2'=\vec{p_2}'=0$
so (*) becomes equal to (33).
I apologize if you have already described this in your last example.

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Not explicitly no. But I did indicate somewhere in the article that in the Lorentz factor $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ the 'v' was to be taken as a relative velocity in which case $T_2=0$.