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The most simplified version of the energy equation (which is also the most known) is $E=mc^2$
However, I understand that this only applies to objects with non-zero mass and zero velocity. I also read that "relativistic mass" is equal to the "rest mass" multiplied by the Lorentz factor: $$ m=\gamma m_0 =\frac { { m }_{ 0 } }{ \sqrt { 1-\frac { { v }^{ 2 } }{ { c }^{ 2 } } } } $$ What exactly IS the "complete" equation for objects with non-zero mass that can be applied to objects with any velocity?

My knowledge would suggest that the equation is: $$ E=\sqrt { { (\gamma m{ c }^{ 2 }) }^{ 2 }+{ (\gamma pc) }^{ 2 } } $$ where E is energy, $\gamma$ is the Lorentz factor, $m$ is the mass, $c$ is obviously the speed of light, and $p$ is the momentum of the object (defined as $p=mv$ but multiplied by the Lorentz factor because the mass increases with velocity)

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$E^2 = p^2c^2 + m^2c^4$ - $p$ is momentum and $m$ is rest mass. No factors of $\gamma$ are required. –  John Rennie May 30 at 14:18

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First, a small correction: it should be $m_\text{rel} = \gamma m$, because $\gamma = (1-\frac{v^2}{c^2})^{-\frac12}$.

Your formula is close. The most general version, which is true all the time (even for zero mass) is

$$E^2 = (mc^2)^2 + (pc)^2.$$

Here $m$ is the rest mass and $p$ is the magnitude of the relativistic momentum, defined as $\mathbf{p} = \gamma m \mathbf{v}$. Note that the factor of $\gamma$ is already included in the momentum, and we use the rest mass, not the relativistic mass.

You can get another formula for this one: if you replace the definition of $\mathbf{p}$, you'll find

$$E = \gamma m c^2 = \frac{m c^2}{\sqrt{1-\frac{v^2}{c^2}}} = m_\text{rel}c^2.$$

This formula is also valid at all times. In older texts it usually to refer to the relativistic mass $m_\text{rel}$ as simply the mass $m$, and this is probably where the formula $E = mc^2$ comes from.

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I noticed the mistake in the formula a while after I posted it. 1) How can a formula that multiplies anything by $m$ be valid for an object where $m=0$ ? 2) Isn't momentum $\rho$ and not $p$ ? –  Arc676 May 31 at 3:36
    
@Arc676: Simple substitute $m = 0$. You'll get the energy-momentum relation for photons, $E = pc$. Also, we usually use the letter $p$ for momentum, not $\rho$, which is normally used for density of something. –  Javier Badia May 31 at 3:44
    
Isn't momentum calculated using $p=mv$ or $p=\gamma mv$ or is there a different formula for momentum that I don't know about that doesn't involve mass? –  Arc676 May 31 at 3:53
    
@Arc676: I see what you mean. It is indeed true that $p = \gamma mv$. Now it's not so simple to set $m=0$, because we would get $p = 0$. The trick is to notice that if we set $m = 0$ while at the same time setting $v = c$, we get $0/0$ in the formula, and that can equal anything. Griffiths wrote: "Personally, I would regard this 'argument' as a joke, were it not for that fact that photons are known to exist in nature". –  Javier Badia May 31 at 4:24
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@Arc676: All this says is that, in principle, a zero mass particle can have any energy and momentum as long as it travels at the speed of light. It is the job of quantum theory to tell us how to determine a photon's energy. It turns out that it is proportional to its frequency: $E= h\nu$. So only positive energies are allowed, but that is a further restriction imposed by quantum mechanics, and that is outside the scope of this discussion. –  Javier Badia Jun 3 at 1:53

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