LOG#018. Momenergy (III).

In this last article dedicated to momenergy we are going to learn:

1st) How momenergy transform under Lorent transformations, i.e., the Lorentz transformations of momenergy.

2nd) Some extra stuff on relativistic energy and other alternative deductions of some formulae we just derived in the last posts.


Momenergy is a spacetime vector. Indeed, we noted already that somehow energy is the four component of momenergy just in the same way time is the four component of spacetime events. Thus, suppose a mass point with rest mass m and it is moving at \mathbf{u}=(u,0,0) in the S-frame, and \mathbf{u}'=(u',0,0) in the S’-frame. The relative velocity between the two inertial frames will be given by \mathbf{v}=(v,0,0). In the S-frame, the total relativistic energy reads:

E=Mc^2=m\gamma c^2=\dfrac{mc^2}{\sqrt{1-\dfrac{v^2}{c^2}}}

and the relativistic momentum is

P_x=Mu=\gamma m u=\dfrac{E v}{c^2}

From the addition theorem of velocities, we also have:


Squaring this equation, we get the following result:


and if we make some algebra calculations


Inverting and taking the square root, we can deduce:


Thus, we have proved the useful relationship:

\boxed{\gamma (u')=\gamma (u) \gamma (v) \left(1-\dfrac{uv}{c^2}\right)}

In the case of arbitrary velocities, we could indeed generalize this last equation into:

\boxed{\gamma (\mathbf{u}')=\gamma (\mathbf{u}) \gamma (\mathbf{v}) \left(1-\dfrac{\mathbf{u}\cdot \mathbf{v}}{c^2}\right)}

We calculate the momenergy in the S’-frame mutatis mutandis. Firstly, the energy will be:

E'=\gamma (u')mc^2=\gamma (u)\gamma (v) \left(1-\dfrac{uv}{c^2}\right)mc^2=\gamma (u)\gamma (v) (mc^2-muv)=\gamma (v) \left(E-vP_x\right)

Secondly, the momentum is easily computed as well:

P'_x=\gamma (u') mu'=\gamma (u) \gamma (v) \left(1-\dfrac{uv}{c^2}\right) \left(1-\dfrac{uv}{c^2}\right) m\dfrac{(u-v)}{ \left(1-\dfrac{uv}{c^2}\right)}


P'_x=\gamma (u) \gamma (v)(mu-mv)=\gamma (v)\left(P_x-\dfrac{Ev}{c^2}\right)

In summary, we can completely write the Lorentz transformation (boost) of momenergy and its components from the next table:

\boxed{\mbox{Momenergy}\; S\rightarrow S' \begin{cases}P'_0=\dfrac{E'}{c}=\gamma (v) \left (\dfrac{E}{c}-\beta (v) P_x \right)\\ P'_x=\gamma (v) \left(P_x -\beta (v) \dfrac{E}{c}\right)\\ P'_y=P_y \\ P'_z=P_z\end{cases}}

\boxed{\mbox{Momenergy}\; S'\rightarrow S \begin{cases}P_0=\dfrac{E}{c}=\gamma (v) \left (\dfrac{E'}{c}+\beta (v) P'_x \right)\\ P_x=\gamma (v) \left(P'_x +\beta (v) \dfrac{E'}{c}\right)\\ P_y=P'_y \\ P_z=P'_z\end{cases}}

The general Lorentz transformation of momenergy is given (for the case of non-parallel motion to the x-axis) by:

\boxed{\mbox{Momenergy}\; S\rightarrow S' \begin{cases}P'_0=\dfrac{E'}{c}=\gamma (v) \left( \dfrac{E}{c}-\vec{\beta}(v)\cdot \mathbf{p}\right)\\ \mathbf{p}'=\mathbf{p}+(\gamma (v)-1)\dfrac{(\vec{\beta}(v)\cdot \mathbf{p})\vec{\beta}(v)}{\beta^2(v)}-\gamma (v)\vec{\beta}(v)\dfrac{E}{c}\\ \gamma (v)=\dfrac{1}{\sqrt{1-\beta^2 (v)}}=\dfrac{1}{\sqrt{1-(\beta^2_x+\beta_y^2+\beta_z^2)}}\\ \vec{\beta}(v)=(\beta_x,\beta_y,\beta_z)=(v_x/c,v_y/c,v_z/c)\end{cases}}

\boxed{\mbox{Momenergy}\; S'\rightarrow S \begin{cases}P_0=\dfrac{E}{c}=\gamma (v) \left( \dfrac{E'}{c}+\vec{\beta}(v)\cdot \mathbf{p}\right)\\ \mathbf{p}=\mathbf{p}'-(\gamma (v)-1)\dfrac{(\vec{\beta}(v)\cdot \mathbf{p})\vec{\beta}(v)}{\beta^2(v)}+\gamma (v)\vec{\beta}(v)\dfrac{E}{c}\\ \gamma (v)=\dfrac{1}{\sqrt{1-\beta^2 (v)}}=\dfrac{1}{\sqrt{1-(\beta^2_x+\beta_y^2+\beta_z^2)}}\\ \vec{\beta}(v)=(\beta_x,\beta_y,\beta_z)=(v_x/c,v_y/c,v_z/c)\end{cases}}

Indeed, we could have obtained these equations, of course, simply plugging the momenergy vector to the general Lorentz transformation in matrix form. That is, the above Lorent transformations are:


and the inverse transformations are thus



The work done when a force F_x is applied on a point mass along a path length dx is given by dW=F_xdx=dK. That is, as a force is applied, the supplied energy is stored as kinetic energy dK. When the point particle moves with \mathbf{u}=(u,0,0) and it has rest mass m, we get

dK=F_xdx=\gamma^3 (u)m\dfrac{du}{dt}dx=\gamma^3 (u) m\dfrac{dx}{dt}du=\dfrac{mudt}{\left(1-\beta^2(u)\right)^{3/2}}=mc^2\dfrac{\beta (u)d\beta (u)}{\left(1-\beta^2(u)\right)^{3/2}}

If the particle begins the motion with u=0 ( null velocity), integrating this last equation provides

K=mc^2\int_0^{\beta} \dfrac{\beta (u)d\beta (u)}{\left(1-\beta^2(u)\right)^{3/2}}=mc^2\dfrac{1}{\sqrt{1-\beta^2 (u)}} \bigg| _0^{\beta(u)}

and then

K=\dfrac{mc^2}{\sqrt{1-\beta^2(u)}}-mc^2=\left(\gamma (u)-1\right)mc^2=Mc^2-mc^2=(M-m)c^2=\delta M c^2


\delta M \equiv M-m

being the difference between the total relativistic mass and the rest mass. This remarkable equation was derived before, and it says that, in special relativity and relativistic theories based on the Lorentz group of transformations, the kinetic energy is the difference of two energies: the total energy and the rest energy. The total energy is, of course:

E=Mc^2=m\gamma c^2=\dfrac{mc^2}{\sqrt{1-\beta^2 (u)}}

and the rest energy is


If we differentiate the function K(u), the relativistic kinetic energy, with respect to time, we obtain:


Writing down the relativistic definition of force


Plugging the kinetic energy:


but we know that M=m\gamma and \mathbf{a}=\dfrac{d\mathbf{u}}{dt}, so we can recast the equation in an equivalent form as follows

\gamma m\mathbf{a}=\mathbf{F}-\dfrac{1}{c^2}\dfrac{dK}{dt}\mathbf{u}

and we get the previously obtained equation


In conclusion, the rate of change of the kinetic energy of a body with respect to time, i.e., the power P it absorbs, is equal to the scalar product of the force and the velocity. This is a well known result from classical mechanics and special relativity does not change it at all, since the dot product is a good invariant.


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