LOG#232. Is it relativistic?

Today or not today. That is the point. Today. How to know if a given particle or sytem is (special) relativistic? That is a tricky question, since the reality is…Everythin is (special) relativistic. The question would be when you can use the usual newtonian aproximation. When you can and you can’t use the newtonian approximation? That is the subject today. </p>Firstly, newtonian physics or galilean relativity IS valid when you can safely say that

  • Linear momentum is linear in velocity AND mass, p=mv.
  • Kinetic energy is quadratic in velocity E_k=\dfrac{1}{2}mv^2.
  • Velocity is much lesser than the speed of light v<<c. Tricky: how much is much lesser? Without loss of generality, anything below 1\% of the speed of light is approximately relativistic but you can note special relativity in some examples, but galilean or newtonian physics is good enough for most of the cases.
  • Time is absolute an universal, t=t_0.
  • Galilean transformations hold.

Secondly, special (Einstein’s) relativity holds whenever:

  • Momentum is non-linear in velocity:

        \[P=MV=m\gamma v=\dfrac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\]

  • Kinetic energy is NOT quadratic in velocity, but the total energy E=Mc^2 minus the rest energy E_0=mc^2 is


  • Velocity is close (or equal, for massless particles/systems) to the speed of light:   

        \[v\sim c\;\;v\approx c\;\; v=c\]

  • Time is NOT universal, but relative to the observer, and time gets a dilation factor (time dilation): 

        \[t=\gamma t_0\]

  • Lorentz transformations hold.

Equivalently, in galilean relativity


for a free particle, and


for a particle/system under conservative/forces.

However, in special relativity, you get

    \[P^2=p_\mu p^\mu=-m^2c^2\]



in general systems. Here, E=E_k+E_0. Then,






The momentum is relativistic when E_k\sim 2mc^2. If you define \beta=v/c, and \gamma as above, then

  • A particle is galilean/newtonian iff v<<c, \beta<<1, \gamma\sim 1, E_k=\frac{mv^2}{2}.
  • A particle is special relativistic (einsteinian) iff v=c or v\sim c, or v\approx c, E_k=m(\gamma-1)c^2=E-E_0.

Mass measuremnts are general non-linear in velocity, but invariant mass definition is possible via scalar product in Minkovski spacetime. Particles moving at exactly the speed of light are massless. This is the case of gluons, photons, and gravitons (m_g=m_\gamma=m_g=0). particles without rest mass as these particles verify E=pc, p)E/c=h/\lambda. Massive particles satisfy a more general dispersion relationship, as above


and thus

    \[E=\pm\sqrt{(pc)^2+E_0^2}=\pm E_0\sqrt{1+\left(\dfrac{pc}{E_0}\right)^2}\]


    \[E=\pm pc\sqrt{1+\left(\dfrac{E_0}{pc}\right)^2}\]

The special case in which you have a massive particle with E_0<< pc is called ultrarelativistic case, and then you can approximate

    \[E\approx \pm pc\left(1-\dfrac{1}{2}\left(\dfrac{E_0}{pc}\right)^2\right)=\pm pc\mp\dfrac{E_0^2}{2pc}\]

Here we note the purely relativistic massless case and the ultrarelativistic case easily, and we can also distinguish the massive or almost massless case in the purely relativistic case or the ultrarelativistic case:

  • Ultrarelativistic: m\approx 0, m\neq 0, E_0<<pc, and v\approx c.
  • Relativistic regime: m\approx 0, m\neq 0, E_0\sim pc, or m=0, E=pc, v=c.  Here, T\geq 2mc^2 or E=pc=E_k.
  • Non-relativistic case: m arbitrary, E_0<<pc, v<<c. Here, E_k=T<< 2mc^2.

    \[E=\pm pc\mp\dfrac{E_0^2}{2pc}\]

In the massive ultrarelativistic case:

    \[E\simeq \pm pc\mp \dfrac{pc}{2}\]

 In general, relativistic particles are

  • Generally relativistic (v\sim c) with E^2=(pc)^2+(mc^2)^2, when roughly T=E_k\geq 2mc^2.
  • Ultrarelativistic almost massless or massive with v\sim c and E=\pm pc\mp E_0/2pc. This is the case of neutrinos. E_0<<pc.

See you in other blog post!







Liked it? Take a second to support amarashiki on Patreon!
Become a patron at Patreon!

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.