In our previous article Introduction to Four-velocity vector, we have presented the spacetime velocity vector equivalent to the classical three dimensional velocity vector.

The next logical question should be: what could be the expression of the **spacetime momentum vector**?

##### Why the spacetime four-momentum could not be the old fashioned classical one

Let's be naive and suppose first that our spacetime momentum vector could be written as **p** = m**v** with v the classical newtonian velocity vector.

Let us consider what is commonly called an** inelastic collision**, i.e. two objects of the same kind, moving oppositely with equals speed v, hit each other and stick together, to become some new, stationary object (to help visualize this, we can imagine the two particules as being chewing gums...).

Let us estimate the momentum of the system before and after the collision in referentials R and R', with R' attached to the first left particule.

In referential R:

In referential R', the important thing to remember is that the relative speed of the right ball with respect to the speed of the left ball is NOT -v-v = -2v because in special relativity, as seen in our last article Introduction to Four-velocity vector, the velocity do not transform according to Galilean answer, but according to the following expression, with v being the relative speed of the referentials.

So in our case, in R' referential, we can write:

Obviously, as momentum must be conserved in all referentials^{[1]}, our hypothesis was wrong, and the relativistic momentum could not equal the particle's mass m multiplied by its ordinary spatial velocity v, as it is defined in Newtonian mechanics.

##### Four momentum definiton and space time representation

It would be certainly a good idea to multiply the four-velocity by the rest mass m of a particle to get our four-momentum definition:

Also we have seen previously that the scalar product of the four-velocity is given by **η _{μν}U^{μ}U^{ν}=c^{2}**

Therefore, as four-momentum is given by **P ^{μ}=mU^{μ}**, we can write:

**η**

_{μν}P^{μ}P^{ν}=m^{2}**η**

_{μν}U^{μ}U^{ν}=m^{2}**c**

^{2}and our four-momentum with norm ** mc **could be respresented with its two time and spatial components (respectively

*γmc*and

*γmv*) as follows on a spacetime diagram:

The proposed momentum arrow always has a length equal to *mc* and points off in the direction of travel of the object in spacetime.

The part of the momentum spacetime vector that points in the space direction has a length equal to *γmv *and that part of the momentum vector that points off in the time direction has a length equal to *γmc. *

^{}

**Remark 1**: In Newtonian limits, i.e if the speed *v* of our object is much less than the speed of light *c*, then *γ* is very close to one. In that case, we regain the old-fashioned momentum, namely the product of the mass with the speed p = mv.

**Remark 2: **With this new expression *γmv*, what happens if a constant force (which is the rate of change of momentum) acts on a body for a long time? As opposed to Newtonian mechanics where the body keeps picking up speed until it goes faster than light - the mass being constant-, in relativity, the body keeps picking up, **not speed, but momentum, which can continually increase because the mass is increasing**, and the velocity could never reach the speed of light.

In his book *SIx-Not-So-Easy pieces*^{[2]}, Richard Feynman illiustrates this form of 'inertia' when v is nearly as great as c, by the example of the synchrotron used in Caltech to deflect high-speed electrons: they need a magnetic field that is 2000 times stronger than would be expected on the basis of Newton's law.

##### Time component conservation law

The law of momentum conservation tells us that the total sum of all the new arrows must be exactly the same as the sum total of the original arrows. This in turn means that the sum total of the portions of each of the arrows pointing in the space direction must be conserved, as should the sum of the portions pointing in the time direction.

So we appear to have two new laws of physics: both *γmv *and *γmc *are conserved quantities.

So what does it mean exactly that *γmc *should be conserved?

Since c is a universal constant upon which everyone always agrees, then the conservation of *γmc *means first of all the conservation of *γm* (slightly new modified law of mass conservation with γ as new factor).

But if *γmc* is conserved, than so too is *γmc** ^{2} *simply because

*c*is a constant. Then in the limit of small speeds (v<<c), we can write:

We just have discovered that there is a thing that is conserved that is equal to something (mc^{2}) plus the kinetic energy 1/2mv^{2}. It makes sense to refer to "something that is conserved" as **the total energy of the body, which has now two bits**.

Thus, even if an object is standing still (v=0), it has still energy associated with it, and that energy is given by **Einstein's famous mass-energy equation E = mc^{2}**.

To be more precise, if we assume with Einstein, **that the energy of a body always equals mc^{2 } (refer to Einstein paper outlines E=mc2, November 21, 1905), or that the mass is equal to the total energy content divided by **

**, we should then write, where the**

*c*^{2}**"rest mass" m**

_{0 }represents the mass of a body that is not moving:

Total relativistic energy E consists of the particle's relativistic kinetic energy plus the second term m_{0}c^{2}, which is the particle's mass energy E_{0}. Providing that no external forces act, **total relativistic energy is conserved in all inertial frames**, irrespective of whether mass or kinetic energy are conserved. In high speed particle collisions for example, mass, kinetic energy, even the total number of particles may not be conserved, but the total relativistic energy of the system will be.

If we come back to our example of the collision of two lumps of clay, applying the conservation of the total relativistic energy gives:

Let's assume that each lump of clay has a mass of 5kg, each travelling at 3000m/s, the increase of mass of the system will be:

which is an exceedingly small amount!

##### Energy-momentum relation

We still have to elucidate a last question: the total relativistic energy's formula and the mass-energy equivalence equation we have established before can not apply to massless particule such as photons.

From above, we know that the norm of the four momentum is a constant in all inertial frames equals to rest mass x c = m_{0}c.

But we know also from the precedent paragraph that the **time component of the four-moment vector γm_{0}c could be expressed also as E/c** as we know that total relativistic energy E =

*γ*m

_{0}c

^{2}

And for a photon, which has zero 'rest mass' but still does have energy and momentum, we get

which describes the energy-momentum relation for a photon.

As the energy of a photon is given by **E = hc/λ**, where h = 6.63x10^{-34}Js (Planck's constant) and λ is the wavelength, we can easily calculate for example the momentum of a photon of blue light, which has a wavelength of 450nm

[1] Should it not be the case, then the fundamental Principle of Relativity would be violated, as we would have a way to distinguish a 'rest' frame where the conservation of momentum law would be valid (our referential R above) from a 'moving' frame where this law would not hold (our R' referential).

[2] Richard Feynman *SIx-Not-So-Easy pieces *Chapter 3 - The special Theory of Relativity §3.8 Relativistic dynamics