# How to derive: $\frac{E^2}{c^4} = \frac{p^2}{c^2} + m_0^2$?

My teacher wrote the following equation on the board a few days ago: $$\frac{E^2}{c^4} = \frac{p^2}{c^2} + m_0^2$$ And I got no idea how this is derived especially since we already derived that: $$m^2 = \frac{p^2}{c^2} + m_0^2$$ So I was hoping some one could help me out with this deriviation.

• What did your teacher say when you asked this question? – WillO Apr 21 at 22:48
• He is on vecation, so thats why i'm asking here :) – Jonas Wolff Apr 21 at 22:52

## 2 Answers

It's difficult to know how much detail to go in to without knowing what level you're learning the material at. I'll be assuming a slightly more advanced approach, that is also a little more rigorous. The downside is this post will be kind of long. The punchline is this: we want to make mechanics deal with something called the Lorentz transform in the same way Newtonian mechanics dealt with rotations. Doing that, and requiring that we get Newtonian mechanics when velocities are small, leads us to that form for the energy of a massive particle.

Note that I will stick with the more common usage that $$m$$ refers to the particle's rest mass, so there is not distinction between $$m$$ and $$m_0$$ in this answer. The idea of "relativistic mass" is a usage that is largely out of fashion as redundant. It comes from an attempt to continue to define dynamics in terms of ordinary velocity and acceleration. Such attempts are, IMHO, misguided, as they obscure the underlying simplicity of the physical laws by using the wrong quantities.

It is for that reason that we begin the derivation pretty deep down, briefly covering terminology that refers the underlying structures that the physical laws are simple in relation to, before we build back up to the definition of energy.

The dynamics of special relativity are concerned with events. 'Event' is the name we give to the idea of a point in space-time. In symbols, if we have a time $$t$$ and position $$(x,y,z)$$ then we call the collection $$(t,x,y,z)$$ an event, whether or not something existed or happened at that place and time.

From the postulates of special relativity we derive that all observers agree on the value of the expression $$(c\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2 \tag1$$ between two events separated by $$(\Delta t,\Delta x,\Delta y,\Delta z)$$. Quantities that have this property, that all observers will agree upon them, are regarded, in some sense, as being more real in physics because they are not contingent on the relationship between the observer and the system being observed.

Importantly, we also derive that if two inertial observers observe the same event, then the coordinates they assign to that event are related by something we call a Lorentz transformation. Think of it as the space-time version of a rotation, because it consists of combining rotations with something we call a boost. Specifically, if one inertial observer is moving with respect to another, but their coordinate axes are otherwise parallel, then we say their coordinate systems are related by a boost.

Because we have this new transformation law, we want to re-arrange Newtonian physics (classical mechanics) into chunks that behave cleanly under Lorentz transformation. The familiar $$\vec{F}=m\vec{a} = m \frac{\operatorname{d}^2 \vec{r}}{\operatorname{d}t^2}$$, for example, takes the form it does because of the way vectors and scalars transform under rotations. Notice that we can take two time derivatives of the vector $$\vec{r}$$ and get a new vector $$\vec{a}$$ because $$t$$ is a scalar under rotations. Under Lorentz transforms, though, $$t$$ becomes part of something we call a $$4$$-vector, so we won't want to construct our laws using derivatives with respect to $$t$$ because they'll be more complicated.

The new scalar we need for objects that have mass comes from the fact that if we take any two points on that objects path and dump them in to (1) we always get a positive number. That allows us to define something we call the proper time $$c \Delta \tau \equiv \sqrt{(c\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2}.$$ We call it the proper time because it is the time an inertial observer that travels between the two events will see go by. For Lorentz transformations $$\Delta \tau$$ is a scalar, so we can use it to take derivatives of $$4$$-vectors to get new $$4$$-vectors and construct our dynamical laws.

First, a bit of notation. We use lower-case Greek letters, super-script or subscript, to denote the components of a $$4$$-vector. Specifically, we identify an event with $$x^\mu$$ and say that $$(x^0, x^1, x^2, x^3) = (c t, x, y, z)$$. If we write a quantity with a Latin character in its subscript or superscript, for example $$x^i$$ or $$p_j$$, that is meant to be understood as the components of an ordinary vector (i.e. $$i$$ can be one of $$1$$, $$2$$, or $$3$$, and $$\mu$$ can be any of $$0$$, $$1$$, $$2$$, or $$3$$).

Notation out of the way, we construct something called the $$4$$-velocity by taking the derivative of an object's position with respect to its proper time $$u^\mu \equiv \frac{\operatorname{d} x^\mu}{\operatorname{d} \tau},$$ which, because $$x^\mu$$ are the components of a $$4$$-vector and $$\tau$$ is a Lorentz scalar, is also a $$4$$-vector. We can recover our ordinary notion of velocity $$v^i$$ from the $$4$$-velocity by taking the ratio $$v^i = c\frac{u^i}{u^0}.$$ Going the other direction ($$u^\mu$$ in terms of $$\vec{v}$$) we find that \begin{align} \left(u^0,u^1,u^2,u^3\right) &= \left(\gamma c, \gamma v^1, \gamma v^2, \gamma v^3\right) \text{ with} \\ \gamma &\equiv \frac{1}{\sqrt{1 - \frac{\vec{v}\cdot\vec{v}}{c^2}}}, \end{align} which is, perhaps, the more familiar relationship.

Notice how we could have constructed $$u^\mu$$ by starting with the vector $$\vec{v}$$, sticking it into a false $$4$$-vector as $$(c, v^1, v^2, v^3)$$, and then dividing by the 'length' to get something that actually is a $$4$$-vector. The reason for pointing this out is that we have an important quantity in Newtonian mechanics called the momentum $$\vec{p}_N = m\vec{v}$$. It stands to reason that we can take $$u^\mu$$ and multiply it by another Lorentz scalar we get a new $$4$$-vector that is the special relativity version of the momentum $$p^\mu = m u^\mu.$$ We call this the $$4$$-momentum because its last three components correspond to ordinary momentum when an object's velocity is much smaller than $$c$$ (meaning $$\gamma \approx 1$$).

To get the special relativity version of $$\vec{F} = m\vec{a}$$ we start by noticing that we can also write it as $$\vec{F} = \frac{\operatorname{d} \vec{p}}{\operatorname{d} t}.$$ This suggests that the special relativity version of Newton's second law is $$F^\mu = \frac{\operatorname{d} p^\mu}{\operatorname{d} \tau} = m \frac{\operatorname{d}^2 x^\mu}{\operatorname{d} \tau^2}, \tag2$$ with $$F^\mu$$ the $$4$$-force.

Why go through all of this trouble? The Newtonian mechanics derivation of the kinetic energy is called the work-energy theorem. Generalizing this right is a little challenging — blindly changing ordinary vectors into $$4$$-vectors produces an identity that has a zero where we find kinetic energy in the old work-energy theorem was because $$u^0 u^0 - \sum_{i=1}^3 u^i y^i = c^2$$.

It turns out that the version that generalizes correctly is that $$K = \int \vec{v}\cdot \operatorname{d}\vec{p},$$ where $$\vec{v}$$ stays the ordinary velocity and $$\vec{p}$$ is the last three components of the $$4$$-momentum. Actually doing this integral is a touch involved, but doable by a strong student at the end of AP calculus or finishing their first year of college calculus. Here I will just show the setup and answer \begin{align} K & = \int v \operatorname{d} p \\ & = \int \frac{cp}{\sqrt{p^2 + (mc)^2}} \operatorname{d} p \\ & = c\sqrt{p^2 + (mc)^2} - m c^2 = c p^0 - m c^2. \end{align} We know that this is the correct approach because if we use calculus to expand $$K$$ in a Taylor series for when $$p \ll mc$$ we get the correct expression from Newtonian mechanics, $$K \approx \frac{p^2}{2 m}$$. We also know this to be the correct path for deeper theoretical reasons that are tied to Hamiltonian mechanics and Noether's theorem. The bare logic of that last one is: there is a relationship between position and momentum, and the same relationship ties time and energy, so if position and time are gathered into a $$4$$-vector, then so should energy and momentum be.

Finishing the derivation that $$E^2 = p^2 c^2 + m^2 c^4$$ requires a little logic. Kinetic energy is the energy you have to add to an object to change the velocity it travels at (equivalently, it's the energy you could extract from an object that is moving if you could slow it down without accelerating yourself). Thus, it must be zero when $$\vec{v}=0$$. The form of $$K$$ is almost like we're taking some more general expression for the energy and subtracting off that same expression when $$\vec{v}=0$$. Thus, we identify $$E=c\sqrt{p^2 + (mc)^2}$$ to say that $$K=E - [E]_{v=0}$$.

It also happens that $$E= cp^0$$, satisfying the deeper concerns I alluded to above.

After any derivation this long, it's a good idea to perform some sanity checks. Let us examine the zeroth component of (2)$$F^0 = \frac{\operatorname{d} p^0}{\operatorname{d}\tau}.$$ If we consider the ordinary power $$P = \frac{\operatorname{d}E}{\operatorname{d} t}$$, then that equation becomes $$F^0 = P \frac{u^0}{c^2}.$$ Unlike in ordinary Newtonian mechanics, not all of the components $$F^\mu$$ are independent. We have the identity $$p^0p^0 - p^1p^1 - p^2p^2 - p^3p^3 = m^2 c^2$$. If we take the derivative with respect to $$\tau$$ of both sides of that identity we get that $$F^0p^0 - F^1p^1 - F^2p^2 - F^3p^3 = 0.$$ In some sense, the $$4$$-force and $$4$$-momentum are always orthogonal to each-other. Thus, $$F^0 = \frac{\sum_{i=1}^3 F_i p_i}{p^0}.$$ Using this new identity we get that \begin{align} P & = \frac{c^2}{u^0} \left(\frac{\sum_{i=1}^3 F_i p_i}{p^0}\right) \\ & = \frac{c}{u^0} \sum_{i=1}^3 F_i v_i. \end{align} It takes only a little bit of calculus to see that $$\int P \operatorname{d}t = \int \sum_{i=1}^3 F_i v_i \operatorname{d}\tau,$$ the right hand side of which is the same as the integral we initially used for $$K$$. So, what we derived is, at least, internally consistent.

It is derived from the invariance of the norm of the 4-momentum $$p = (E,\mathbf pc)$$. With $$\eta$$ denoting the Minkowski metric with signature $$(+,-,-,-)$$,

$$m_0^2c^4 = \eta(p,p) = E^2 - p^2 c^2,$$

where $$m_0$$ is the invariant mass.