# LOG#016. Momenergy (I).

We have seen that space and time are merged into the spacetime in Special Relativity (SR). Morever, in a similar way, we have also deduced that momentum and energy are merged into an analogue concept: the momenergy.

Mathematically, it is easily understood looking at the spacetime and momenergy vectors:

$\mathbb{X}=X^\mu e_\mu=(x^0,x^1,x^2,x^3)=(ct,x,y,z)=(ct,\mathbf{r})=(ct,x,y,z)$

and

$\mathbb{P}=P^\mu e_\mu=(P^0,P^1,P^2,P^3)=\left(\dfrac{E}{c},\mathbf{P}\right)=\left(\dfrac{E}{c},P_x,P_y,P_z\right)=(Mc,P_x,P_y,P_z)$

or

$\mathbb{P}=P^\mu e_\mu=\left(m\gamma c, m\gamma v_x, m\gamma v_y, m\gamma v_z\right)$

Now, we are going to understand better these concepts using basic undergraduate Physics, showing that everything we get from SR is consistent with newtonian physics and the whole theory as a whole.

In non relativistic Mechanics, the notion of kinetic energy is introduced as the work necessary to change the speed of some object, i.e., the work necessary in order to change the velocity. From the general definition of “work” (or energy) through some path $\Gamma$:

$\boxed{\mbox{WORK}=W=\int_{\Gamma} F=\int_1^2 \mathbf{F}\cdot d\mathbf{r}}$

we can insert the Newton’s second law for the force, $\mathbf{F}=m\mathbf{a}$, and we obtain:

$W=\int_1^2 m\mathbf{a}\cdot d\mathbf{r}=\int_1^2 m \mathbf{a}\cdot \dfrac{\mathbf{dr}}{dt} dt=\int_1^2 m\mathbf{v}\cdot d\mathbf{v}$

We can now use the following formal differential to perform the previous integral:

$d(\mathbf{v}\cdot \mathbf{v})=d \vert \mathbf{v}\vert ^2= d v^2= 2 \mathbf{v}\cdot d\mathbf{v}\rightarrow \mathbf{v}\cdot d\mathbf{v}=\dfrac{1}{2}d\vert \mathbf{v}\vert ^2$

$W=\int_1^2 m\mathbf{v}\cdot d\mathbf{v}=m\int_1^2 \dfrac{1}{2}d\vert \mathbf{v}\vert ^2=\dfrac{1}{2}m\int_1^2 d\vert \mathbf{v}\vert ^2$

and thus, we get the classical result that the work is the variation of the kinetic energy function

$\boxed{W=\dfrac{1}{2}m v^2 \bigg| _1^2=\dfrac{1}{2}m v_2^2-\dfrac{1}{2}m v_1^2=\Delta E_k}$

with

$\boxed{E_k=\dfrac{1}{2}m v^2}$

What happens in SR? Indeed, something similar! Recall that the relativistic generalization of force is:

$\mathbf {F}_ {rel}=\dfrac{d(m\gamma \mathbf{v})}{dt}$

Therefore, the work is now:

$W=\int_{\Gamma} F_{rel}=\int_1^2 \mathbf{F}_{rel}\cdot d\mathbf{r}=\int_1^2 \dfrac{d(m\gamma \mathbf{v})}{dt}\cdot d\mathbf{r}=\int_1^2 d(m\gamma \mathbf{v})\cdot \mathbf{v}=\int_1^2 \mathbf{v}\cdot d(m\gamma \mathbf{v})$

To perform this integral we need to integrate by parts using the rule:

$\boxed{d(\mathbf{u}\cdot \mathbf{v})=d\mathbf{u}\cdot\mathbf{ v} + \mathbf{u}\cdot d\mathbf{v}}$

and the formal differential of the inverse relativistic dilatation factor $\gamma^{-1}$. Since,

$\gamma = \dfrac{1}{\sqrt{1-\dfrac{\mathbf{v}^2}{c^2}}}\rightarrow \gamma^{-1}=\sqrt{1-\dfrac{\mathbf{v}^2}{c^2}}$

it implies that

$d(\gamma^{-1})=-\dfrac{\mathbf{v}\cdot d\mathbf{v}}{c^2\sqrt{1-\dfrac{\mathbf{v}^2}{c^2}}}=-\dfrac{\gamma\mathbf{v}\cdot d\mathbf{v}}{c^2}$

We will also need this auxiliary result:

$\gamma^2=1+\dfrac{\gamma^2\mathbf{v}^2}{c^2}$

Then, we can now integrate by parts the relativistic version or work, obtaining firstly:

$W_{rel}= m\gamma \mathbf{v}^2 \bigg| _1^2-\int_1^2 (m\gamma \mathbf{v}) \cdot d\mathbf{v}=m\gamma \mathbf{v}^2 \bigg| _1^2+\int_1^2 (-m\gamma \mathbf{v}) \cdot d\mathbf{v}$

or equivalently

$W_{rel}=m\gamma \mathbf{v}^2 \bigg| _1^2+mc^2\int_1^2 \dfrac{1}{c^2}(-\gamma \mathbf{v}) \cdot d\mathbf{v}$

that is

$W_{rel}=m\gamma \mathbf{v}^2 \bigg| _1^2+mc^2\int_1^2 d(\gamma^{-1})=\left[m\gamma \mathbf{v}^2+\dfrac{mc^2}{\gamma}\right] \bigg| _1^2$

But after some algebra we get:

$W_{rel}=\left[m\gamma \left( \mathbf{v}^2+\dfrac{c^2}{\gamma^2}\right)\right] \bigg| _1^2=\left[m\gamma \left( \dfrac{\gamma^2\dfrac{\mathbf{v}^2}{c^2}+1}{\dfrac{\gamma^2}{c^2}}\right)\right] _1^2=\left[mc^2\gamma \left(\dfrac{1+\dfrac{\gamma^2\mathbf{v}^2}{c^2}}{\dfrac{\gamma^2}{1}}\right)\right] _1^2$

Using the last auxiliary result from the previous discussion, i.e.,

$\gamma^2=1+\dfrac{\gamma^2\mathbf{v}^2}{c^2}\rightarrow \dfrac{1+\dfrac{\gamma^2\mathbf{v}^2}{c^2}}{\gamma^2}=1$

we have

$W_{rel}=\left[ m\gamma c^2 \right] _1^2= \Delta E_{rel}^{total}$

where

$\boxed{\Delta E_{rel}^{total}=W_{rel}=\gamma c^2 \Delta m= \gamma c^2 (m_2-m_1)= \Delta M c^2 = (M_2-M_1)c^2}$

and of course, the total relativistic energy is the celebrated equation

$\boxed{E=Mc^2=m\gamma c^2 \equiv E_{rel}^{total}=\mbox{TOTAL RELATIVISTIC ENERGY}}$

A interesting question is now, what is the kinetic energy in the relativist domain? Well, we can define the relativistic kinetic energy $K=K(v)$ as some complex function but how can we determine it? Indeed, it is easy. We can define the relativistic kinetic energy as the total relatistic energy MINUS the rest energy. The rest energy is:

$\boxed{e=mc^2 \equiv \mbox{RELATIVISTIC REST ENERGY}}$

Sometimes it is written, with $m_0$ as the invariant mass we called “m” in our notation, as follows:

$E_0=m_0c^2$

The relativistic kinetic energy and the total energy are then:

$\boxed{K=K(v)=E-mc^2=\gamma mc^2-mc^2=(\gamma - 1)mc^2}$

$E=mc^2+K(v)$

In the small velocity limit, we have the following approximations:

$\boxed{K(v)=\dfrac{1}{2}mv^2+\dfrac{3}{8}\dfrac{mv^4}{c^2}+\mathcal{O}\left( \dfrac{v^6}{c^4}\right)\approx \dfrac{1}{2}mv^2}$

$\boxed{E=mc^2+\dfrac{1}{2}mv^2+\dfrac{3}{8}\dfrac{mv^4}{c^2}+\mathcal{O}\left( \dfrac{v^6}{c^4}\right)\approx mc^2+\dfrac{1}{2}mv^2}$

Therefore, the relativistic kinetic energy approaches the classical result in the limit of small velocities, as we expected, and the total relativistic energy agrees with the classical result, excepting a constant we called the rest energy and further corrections we neglect if $v<.