# Is $E^2=(mc^2)^2+(pc)^2$ or is $E=mc^2$ the correct one?

I have been having trouble distinguishing these two equations and figuring out which one is correct. I have watched a video that says that $E^2=(mc^2)^2+(pc)^2$ is correct, but I do not know why. It says that $E=mc^2$ is the equation for objects that are not moving and that$E^2=(mc^2)^2+(pc)^2$is for objects that are moving. Here is the link to the video: http://www.youtube.com/watch?v=NnMIhxWRGNw

• are you maybe confused by the E=m*c^2 formula of relativistic mass fame ? In the formulas you are quoting m is the rest mass, usually denoted by m_0. The relation between the two is m=m_0*gamma, en.wikipedia.org/wiki/Mass_in_special_relativity Oct 29, 2014 at 13:01
• Oct 29, 2014 at 13:04
• See physics.stackexchange.com/q/1686 and physics.stackexchange.com/q/34008 and many other posts around the site about the professional take on "relativistic mass". Oct 29, 2014 at 18:20
• Nov 18, 2021 at 23:06

Let me clarify some confusions in the notation that other answers have alluded to but not clearly mentioned.

Historically, physicists liked to talk about two different definitions of mass

• The first is the rest mass of a particle $m_0$. This is the mass of the particle when it is at rest. For example, the rest mass of the electron is $(m_0)_{electron} = 9.1 \times 10^{-31}~Kg$. This is an absolute constant that is independent of the speed of the particle.

• The second is the relativistic mass $m$. This is the apparent mass of the particle when it is moving with speed $v$. It is related to the rest mass via the relation $$m = \gamma m_0 = \frac{m_0}{\sqrt{1-v^2/c^2}}$$ Note that the relativistic mass is NOT a constant. It depends on $v$.

In this historical notation, Einstein's famous formula that is completely correct in all frames is $$E = m c^2$$ However, it turns out via a series of algebraic manipulations that this equation also implies $$E^2 = ( m_0 c^2)^2 + (pc)^2$$ Let us prove this. $p$ is the momentum of the particle defined by $p = m v = \gamma m_0 v$. Thus $$(m_0 c^2)^2 + (pc)^2 = m_0^2 c^4 + \gamma^2 m_0^2 v^2 c^2 = m_0^2 c^4 \left( 1 + \frac{\gamma^2 v^2}{c^2} \right)$$ Now, we have the property $$1 + \frac{\gamma^2 v^2}{c^2} = 1 + \frac{\frac{v^2}{c^2}}{\left( 1 - \frac{v^2}{c^2} \right) } = \frac{1}{ \left( 1 - \frac{v^2}{c^2} \right) } = \gamma^2$$ Thus $$(m_0 c^2)^2 + (pc)^2 = m_0^2 c^4 \gamma^2 = (\gamma m_0)^2 c^4 = m^2 c^4 = (mc^2)^2 = E^2$$ Thus, in summary, in the historical notation, we have two completely equivalent formulae $$\boxed{ E^2 = (m c^2 )^2 = (m_0 c^2)^2 + (pc)^2}$$

In modern day notation, physicists have decided to drop discussion of the relativistic mass $m$ since it is not an absolute constant and depends on the speed of the particle. Nowadays, we only talk about the rest mass, $m_0$. However, in a confusing notational change physicists today decided to use $m$ for the rest mass (which in today's notation is not confusing at all, since we don't talk about relativistic mass, but it is often confusing to students who try to compare Einstein's original papers with books written today).

Following modern day notation then, we only have ONE equation, namely $$\boxed{ E^2 = (m c^2)^2 + (pc)^2 }$$ where in the above equation $m$ is now the rest mass.

• Nice post - very clear. so perhaps I am a bit old fashioned to use $m_0$ - maybe it is more common to use $m_0$ in Europe where I am based?
– tom
Oct 29, 2014 at 17:06
• @tom - maybe. Where I've studied (India and US) I've only ever seen usage of $m$ for rest mass (except maybe in high school when they tried to teach us both definitions of mass). In any case, I would be glad to go back to $m_0$ since that is so much clearer. Oct 29, 2014 at 17:15
• @Prahar could u please explain this term given by Nov 21, 2014 at 3:40
• p=mv=γm0v in ur answer. Nov 21, 2014 at 3:40
• I've used the equation $m = \gamma m_0$ which was written down before. Apart from that, the $p=mv$ part is the definition of momentum. Is that what you wanted clarified? Nov 24, 2014 at 14:10

Both are correct, within the domains for which they are correct.

More seriously, the general relation

$$E^2 = m^2c^4 + p^2c^2$$

holds for all objects, whether they have mass or not, whether they are moving or not.

The special case $E = mc^2$ is for $p = 0$, i.e. objects which do not move, as you said.

The special case $E = pc$ is for objects which have no mass, i.e., photons.

• The $m$'s in your answer, do they denote the relativistic mass or the rest mass? Oct 29, 2014 at 14:08
• @user929304: Rest mass, I consider relativistic mass an unnecessary concept. Oct 29, 2014 at 14:09
• @user929304: We cannot, but this relativistic mass does not transform properly under Lorentz transformations. There is nothing that could not be explained/calculated without it, and it just confuses many people. Oct 29, 2014 at 15:05
• @user929304 It's a purely geometrical phenomenon. It'd be like rotating a vector and asking it to trace out an ellipse instead of a circle. In Minkowski space, boosting a four-velocity traces out a hyperbola. Hyperbolas have asymptotes--lightlike rays, in this case. Oct 29, 2014 at 21:52
• @Muphrid interesting, where can I learn more about this point of view?(any recommended sources?) Unfortunately your summary was too short and dense to be understandable for me :( Oct 30, 2014 at 14:24

I agree with answer of ACuriousMind, but I think it might also help to think about it like this....

$E^2=m_0^2c^4+p^2c^2 =m^2c^4$

where $m_0$ is the rest mass and $m$ is the relativistic mass (or inertial mass), defined as $m = \gamma m_0 = m_0 / \sqrt{1 - v^2/c^2}$.

The relatavistic mass increases as the momentum of the mass increases. At rest the two are equal to each other. As the speed of an object and its momentum increase the mass of the object increases.

so I think about it as

$E^2=m_0^2c^4+p^2c^2$

and

$E=mc^2$

• Yes, and relativistic mass is equal to $E_{r}/c^2 + E_{k}/c^2$, where $E_{r}$ is the energy due to rest mass $m_0 c^2$ and $E_{k}$ is the relativistic kinetic energy $(\gamma - 1) m_0 c^2$. More generally, the inertial mass of any bound system is equivalent to (it resist acceleration the same way as) a point particle with rest mass equal to the sum of the rest-mass energies, kinetic energies, and internal potential energies of all the parts of the system--the inclusion of potential energy is why, for example, a hydrogen atom has less mass than the sum of masses of an electron and a proton. Oct 29, 2014 at 12:41
• I should add that when I said the inertial mass of a bound system was determined by "the sum of the rest-mass energies, kinetic energies, and internal potential energies of all the parts of the system", I meant for the non-rest-mass energies to be evaluated in the center of mass frame of the system. Also, since energy can be stored in fields which don't have scalar potentials like the magnetic field, I assume in that case you'd have to include the energy in the magnetic field created by all parts of the bound system. Oct 29, 2014 at 16:57
• @Hypnosifl - thanks for the comments and edit - good comments.
– tom
Oct 29, 2014 at 22:54

The equation $$E^2=(mc^2)^2+(pc)^2$$ represents the correct energy-momentum relationship. It gives the total energy $E$ for an object of invariant mass (rest mass) $m$ that is observed to move with momentum $p$. This equation is applicable regardless whether the object is observed to be in motion ($p \ne 0$), or is observed to be at rest ($p = 0$). In the latter case, the energy-momentum equation simplifies into the well-known $E=mc^2$.

As an aside (some might call it nitpicking), when discussing the generics of the energy-momentum equation, it is good form to write the equation such that both sides of the equation are independent of the frame of observation chosen: $$E^2 - (pc)^2 = (mc^2)^2$$ Same math, different physics. (Note that this relativistically invariant relationship is simply the expression for the square norm of the energy-momentum four-vector.)

• +1, could you expand on your last point? "same math, different physics". Thanks Oct 29, 2014 at 15:51
• @user929304 - added a final remark to my answer (text in brackets). Does this answer your question? Oct 29, 2014 at 16:03

All of the other answers are great, and I highly recommend reading them. However, I think something is missing if you don't attempt to get an intuitive understanding with geometry: This triangle shows that the equation $$E^2=(mc^2)^2+(pc)^2$$ can be represented via a sort of reverse Pythagorean theorem. The special case of $$E=mc^2$$ can be found by setting $$p$$ to zero, and appears as if the $$pc$$ side of the triangle is zero in size, changing the shape to a line with $$E$$ on top and $$mc^2$$ on the bottom. Likewise, for light, we can show the special case of $$E=pc$$ by setting the rest mass $$m$$ to zero, which transforms the triangle into a vertical line with $$E$$ on the left and $$pc$$ on the right.

Much of this is hard-won insights after a couple of decades of independent study. Look into it, and I think you might find there's something pretty useful here.

Newton's Second Law can be written:

$$\frac{\text{Impulse}}{\text{mass}} = \text{Change in Velocity}.$$

But in relativistic mechanics we have

• Impulse/mass (in ls/s) = $\sinh(w)$,
• Change in Velocity (in ls/s) = $\tanh(w)$
• Time Dilation/Length Contraction Factor (in s/s or ls/ls) = $\cosh(w)$

where $w$ is the rapidity. When rapidity is small, $\sinh(w)= \tanh(w) = w$

You can see part of this in $$E^2 = p^2 c^2 + (mc^2)^2$$

So this equation is essentially the hyperbolic trig equivalent of Pythagoras theorem.

$$(mc^2)^2 \cosh^2(w) = (mc^2)^2\sinh^2(w) + (mc^2)^2$$

or

$$(mc^2)^2 \gamma^2 = (mc^2)^2 \left(\frac{\text {impulse}}{\text{mass}} \text{(in ls/s)}\right)^2 + (mc^2)^2$$

You can also get the kinetic energy out of this equation by subtracting 1 from the time-dilation factor, and multiplying the result by $mc^2$. The equation is not terribly useful for that purpose at low velocities, though, since the time-dilation factor, $\gamma$, will be something like 1.00000000004 and it won't fit into your calculator.

Once you confirm that all this is really hyperbolic trig, if you can find a calculator with easy access to hyperbolic trig functions, you'll find it much easier to put things into rapidities.