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Let's consider

$$E_f^2=(mc^2)^2+(pc)^2$$

where the $mc^2$ is the rest energy due to the rest mass -- in Finnish "lepomassa".

$$ \sqrt{(mc^2)^2+(pc)^2} - mc^2~=~(\gamma-1)mc^2$$

is the kinetic energy due to the movement because of momentum $p=\gamma mv$.

Now where is potential energy if $E_f=\gamma mc^2$ is the total energy?

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Related: physics.stackexchange.com/questions/69080/… –  jinawee Nov 18 '13 at 13:29

4 Answers 4

The formula you quote does not contain the potential energy, it is valid for a free particle (i.e. a particle which is not affected by external potential). You can link it to classical mechanics by evaluating it for small values of $p$ (more precisely: $ p \ll c$):

$$ E = \sqrt{\left(mc^2\right)^2 + p^2 c^2} = c \sqrt{m^2c^2 + p^2} = \cdots $$

$$ \cdots = mc^2 \sqrt{1 + \frac{p^2}{m^2 c^2}} \approx mc^2 \left( 1 + \frac{p^2}{2 m^2 c^2} \right) = \cdots $$

$$ \cdots = \text{constant} + \frac{p^2}{2m} = \text{constant} + \frac{1}{2} m v^2 $$

Here we see that the relativistic formula in the non-relativistic (i.e. small speeds) limit reduces to the classical one, apart for a constant energy associated to the mass of the object, which is a purely relativistic concept.

The constant is, by the way, $mc^2$, and that's explains why the formula $E=mc^2$ is so famous, as it catches one of the most astonishing concept of special relativity: an object just for existing and having mass $m$, has an energy $E=mc^2$, i.e. the rest energy.

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1  
-1 Everything you said is true, but it doesn't have anything to do with the question of the OP. –  JSQuareD Nov 13 '13 at 19:38
    
I think this is the answer. The expression given for E is just for a free particle, which means the only `potential energy' is the mass-energy. If there are other potential energies (gravitational, electromagnetic, etc.) they will be added to E. –  Zane Beckwith Nov 18 '13 at 2:24
    
Why does the potential energy have to be zero for a free particle? We can give a free particle any energy we want. It's free because the potential is constant (no forces). In general, a lot of discussions in physics benefit from keeping energy as an unreferenced quantity, because when we start to compare systems with already referenced energies, we can get into trouble if they use different zeros of energy. –  Nanite Nov 18 '13 at 21:04
    
@Nanite, special relativity actually kind of gets rid of the idea of energy being undetermined up to a constant. For example, if a free particle is at rest, physically we would say its total energy is the mass energy, $mc^2$; this means that if it were completely annihilated, $mc^2$ of energy would be released. But if we were free to add any constant energy to the particle's energy, that would mean any arbitrary amount of energy could potentially be released by annihilation. –  Zane Beckwith Nov 19 '13 at 15:05

The energy in your equation is for a free rigid body in the absence of a potential. We can see this if we start with a Lagrangian with a scalar function, $\Phi(q)$, and remember $\gamma$ is a function of $\dot{q}$, $$ L=T-V=-\gamma^{-1} (\dot{q}) \, mc^2-\Phi(q) $$ Then if we find the momentum $$ \pi=\frac{\partial L}{\partial \dot{q}}=\gamma^{-2}\frac{\partial \gamma}{\partial \dot{q}}mc^2=\gamma m\dot{q} $$ Thus, the Hamiltonian, $$ H=\pi\dot{q}-L=\gamma m\dot{q}^2+\gamma^{-1} (\dot{q}) \, mc^2 + \Phi(q) $$ which gives after factoring out $\gamma mc^2$, $$ H=\gamma mc^2(\frac{\dot{q}^2}{c^2}+\gamma^{-2})+\Phi(q)=\gamma mc^2+\Phi $$ The first term is the one you like, and the second one is the potential energy if you'd like.

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@hhh I started with the Lagrangian since it's a simple starting point. I then found the canonical momentum, and then preformed a Legendre Transform to the Hamiltonian, since we were talking about energy. It turned out to show that the energy you wrote in the OP doesn't include potential energy. We know this because we put in the potential energy ourselves in the Lagrangian, and in the end we got the energy, which was the original term, "total energy", and then more, the potential energy. –  kηives Jan 10 '13 at 18:31
    
Does Lagrangian or Lagrangian-Euler have anything to do with Lagrange's method? I see you have generalized things, cool, keep it -- taking me just more time to dig into this, like it :) –  hhh Jan 10 '13 at 18:33
    
@hhh no problem, I was just trying to think of a way to convince you to get energy, and still have the expression not be what you wrote in the OP, to demonstrate that potential energy is not included. The Euler-Lagrange equations and the Lagrangian are definitely related. The former is derived from the later. –  kηives Jan 10 '13 at 18:36
    
Lagrangian != Lagrange method's: please check the Wikipedia sites in the comment. You have used some continuous description with functionals to describe physics where the rigid body means some sort of trivial functional? –  hhh Jan 10 '13 at 18:39
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@hhh There are no functionals, only functions. 1) I have taken a Lagrangian which includes kinetic and potential parts. 2) I have found the conjugate momentum. 3) Using the momentum, I have preformed a Legendre transform to the Hamiltonian. 4) looking at the Hamiltonian I have deduced that your original expression does not contain potential energy since the Hamiltonian contains the OP expression and then some. –  kηives Jan 10 '13 at 19:21

Potential energy is the property of a system, not of individual particles. Even in classical mechanics this is true. The usual way of saying something's potential energy can be seen as an abuse of notation.

So for one particle $E=\gamma m c^2$ does not include potential energy, but the energy of the total system (a point charge and a capacitor, for example) include the potential energy. The potential energy shifts the rest mass of the system from the sum of the mass of individual components (in addition to the effect of relative motion).


Another viewpoint is that potential energy is stored in fields, so the shift in rest mass is due to energy of the field.

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In the standard formula given in the question posed, the potential energy is zero. The formula applies to a free particle only.

For a charged particle of charge Q in an electromagnetic field, the correct formula for the total (kinetic plus potential) energy is $$E= c\sqrt{(mc)^2 + (p+QA)^2} -QA_0,$$ (e.g., $Q=-e$ for an electron) where $A$ and $A_0$ are the space (vector) part and the time (scalar) part of the electromagnetic gauge potential. Here $A_0$ is the potential energy.

For gravitational forces, the correct formula is given by the solution $E=p_0$ of the equation $G(p,p_0)=const$, where $G$ is a Lorentzian quadratic form (whose coefficients define the metric tensor) in the space part $p$ and the time part $p_0$ of the relativistic 4-momentum vector. Here a potiential energy can be identified only in a nonrelativistic limit.

If both kinds of forces are present, the correct formula is given by the solution $E=cp_0$ of the equation $$G(p+QA,p_0+QA_0)=0$$. Specializing the quadratic form to $G(p,p_0)=const(p_0^2-p^2)$ gives the above formula.

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