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Lost mass is converted into kinetic energy after nuclear reactions but what happens to potential energy? (in the case of a non isolated system)

I don't know a lot of potential energy and in my idea it is related do mass and distance from other masses (but this should be wrong because I don't consider massless interactions and I don't see how to integrate it with the total energy conservation, including to this the potential energy) and that potential energy belongs to a system rather than to a single object, is for me something related to the "environment".

Someone can enlight me?

If some mass is transformed in energy what happens to potential energy related to that mass?

Example: I have a system A before the nuclear reaction and a system B composed of masses, there is a potential energy for the system A connected to the interactions and masses Then we have the nuclear reaction for A Some mass of A is transformed in energy, before the reaction there was a gravitational force on A from B (ignore for a moment the remaining interactions) and a related potential energy of this mass connected to the system B , but when transformed in energy where does this potential energy go?

Qmechanic
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Anto
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1 Answers1

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You use of potential energy is a bit off here. Potential energy remains constant through the calculation. Also the gravitational forces here are negligible. So we stay in special relativity. This makes the calculations easier.

We start with the relativistic energy equation $E=\gamma mc^2$ where $\gamma$ is the Lorentz factor. So in a nuclear reaction start with $\gamma=1$ which corresponds to $v=0$. Energy being conserved means that $E_o=E$ subsequently $\gamma_o\cdot m_o\cdot c^2=\gamma\cdot m\cdot c^2$ since $\gamma_o$ is $1$ and eliminating $c^2$ we get $m_o=\gamma\cdot m$. After the nuclear interaction we have a decrease of mass between the original from $m_o$ to $m$ where $m<m_o$. Subsequently $\gamma>1$. If we divide by $m$ and expand gamma we get $\frac{m_o}{m}=1+\frac{1}{2}v^2 + ...$. So the converted mass goes into the second term which is the kenitic energy of the resultant particle(s).

Michael
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