- #1

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What is the total energy of a particle in a potential? Is it

$$E=\gamma m_0 c^2+E_pot$$

or is it still

$$E=\gamma m_0 c^2$$

where ##m_0## is a bigger mass than the particle would have in absence of the potential?

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- Thread starter greypilgrim
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- #1

- 432

- 23

What is the total energy of a particle in a potential? Is it

$$E=\gamma m_0 c^2+E_pot$$

or is it still

$$E=\gamma m_0 c^2$$

where ##m_0## is a bigger mass than the particle would have in absence of the potential?

- #2

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[itex]E = \gamma mc^2 + q \Phi[/itex]

where [itex]\Phi[/itex] is the electric potential.

- #3

Jonathan Scott

Gold Member

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(For gravity, where energy acts as a source so everything is non-linear, this gets much more complicated and as far as I know there isn't any satisfactory answer to where the equivalent of potential energy resides, not even in GR).

- #4

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But how does this work with

[itex]E = \gamma mc^2 + q \Phi[/itex]

where [itex]\Phi[/itex] is the electric potential.

$$E^2=c^2\cdot \mathbf{p}^2+m^2\cdot c^4 \enspace ?$$

If we look at two identical particles with the same velocity where one is in an electric potential and the other is not, the right sides of this equation are the same, but not the energy squared on the left?

- #5

Jonathan Scott

Gold Member

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In contrast, gravitational potential energy (which is negative relative to the local rest mass) is part of the energy of the particle and is assumed to contribute its inertia, but to get the usual conservation laws to work (at least for a weak field approximation) there also has to be positive energy in the field which compensates for the double effect of each particle having the whole potential energy.

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