Recently, I learned how to derive $E_0 = m_0c^2$. However, to do so, one must accept the relativistic mass equation. How does one derive the Lorentz factor $\gamma$?

  • $\begingroup$ I'm pretty sure you just let $p=\gamma m_0v$ and the rest sorts itself out from there $\endgroup$
    – Jim
    Mar 30, 2015 at 22:35
  • $\begingroup$ @Jimnosperm This might sound dumb, but I never learned how to derive the Lorentz factor. So, how do you? $\endgroup$
    – Jimmy360
    Mar 30, 2015 at 22:38
  • $\begingroup$ this might answer your question: physics.stackexchange.com/a/171121/75518 $\endgroup$
    – image357
    Mar 30, 2015 at 22:43
  • $\begingroup$ Or this: physics.stackexchange.com/q/80511/23473 $\endgroup$
    – Jim
    Mar 30, 2015 at 22:46

3 Answers 3


After some searching, I found a very simple way to do so.

enter image description here

The blue disks shows show light traversing up a rod. If the rod were stationary then the light would reach the top in ct where t starts at t' = 0. However, with the horizontal motion of the rods, the light takes a diagonal path, indicated by the red line. By the Pythagorean Theorem, we get $$d^2 = l^2 + (vt)^2$$ (d being the distance of the first half of the red line)

Since l = ct, we get $$d^2 = c^2t^2 + v^2t^2$$

Since d = ct', we get $$c^2t'^2 = c^2t^2 + v^2t^2$$

Solving for t', we get $$t' = \dfrac{t}{\sqrt{1-\dfrac{v^2}{c^2}}}$$

Divding by t, we get $$\frac{t'}{t} = \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$$

Which is the Lorentz Factor

  • $\begingroup$ Small typo: 'Since l = ct', we get' should be 'Since l = ct, we get' $\endgroup$
    – HelloWorld
    Jun 23, 2017 at 19:51
  • $\begingroup$ Where from do you suppose that the length of the moving rod (normal to its motion) is the same as the length of the rest rod $\ell'=\ell$ ??? $\endgroup$
    – Frobenius
    Sep 30, 2017 at 12:06
  • $\begingroup$ This explanation is lame and gives the correct answer just by coincidence (just to 'fit' into Pythagorean Theorem). Why rod is dragging light within itself? Speed of light doesn't depend on motion of its source. In this example it DOES. $\endgroup$
    – m8labs
    May 9, 2021 at 16:20
  • $\begingroup$ @m8labs c doesn’t change with the speed of the source. That’s why this derivation works in the first place. The thing that must change is time, not c. $\endgroup$
    – Jimmy360
    Jun 14, 2021 at 18:25
  • 1
    $\begingroup$ @Jimmy360 This is incorrect. At the [Since d=ct'] step you made a mistake. The distance is not proportional to t, but to t' like this: (ct′)^2 = (ct)^2 + (vt')^2 $\endgroup$ Jun 28, 2021 at 18:57

We can use the fact that the speed of light is the same in every inertial reference frame: $$c=\frac{dx}{dt}=\frac{dx'}{dt'}$$ From that we can construct the space-time interval: $$(ds)^2=c^2dt^2-dx^2$$ The space-time interval is invariant in every inertial reference frame. $$(ds)^2=(ds')^{2}$$ In proper frame, the space-time interval has the following form: $$(ds)^2=c^2dt^2$$ where $dt$ is usually written as $d\tau$ and it is called the proper time. We know that the space-time interval is the same in every inertial reference frame, so we can write that the space-time interval in our proper frame is equivalent to any other inertial reference frame: $$c^2d\tau^2=c^2dt^2-dx^2$$ Divide both sides by $c^2dt^2$: $$\frac{d\tau}{dt}=\sqrt{1-\frac{v^2}{c^2}}$$ so: $$d\tau=dt\sqrt{1-\frac{v^2}{c^2}}$$ or: $$dt=\frac{d\tau}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma\cdot d\tau$$

  • $\begingroup$ This should be accepted answer. $\endgroup$
    – m8labs
    May 9, 2021 at 16:21

I think the Lorentz factor is wrong because it's based on Einstein's 2nd postulate which implies that the speed of light is constant irrespective of the motion of the emitting body. However, most people apply it as if it were constant for all observers.

In any case, it's a postulate i.e. an assumption. Einstein used the speed of light as constant after Maxwell calculated its value from the permeability and permittivity of free space which are constants and cannot vary under any circumstances.

If you move an electric charge in front of a magnetic field detector, it would detect the presence of a magnetic field. More importantly, if you move a magnetic field detector in front of a static charge, it'll also detect a magnetic field. Hence, electromagnetism is subject to relativity as Galileo implied. Well, he said that everything is relative.

Now light is made of photons which are electromagnetic particles. That means that velocity will interfere with the strength of the magnetic field. And it's either the velocity of light relative to the observer or the observer's velocity to that of light - they're equivalent. Maybe that's how the Doppler shift happens. We don't know what wavelength refers to. Maybe it's inversely proportional to the strength of the magnetic field component of the electromagnetic field.

Keeping the strength of the magnetic component of the electromagnetic field invariant across all frames of reference, may be a mistake. This means that all forms of relativity may be wrong. I'm certain of it. It could turn out that Galilean transformation was the right one to use after all.

  • 4
    $\begingroup$ If you really believe that Einstein was wrong you need to propose the alternate hypothesis and show how it explains all the same phenomena that SR and GR do. Personal speculation with no justification has no place in answers at this site. $\endgroup$
    – Bill N
    Dec 18, 2016 at 22:57
  • $\begingroup$ If a photon is ahead of you and you try to accelerate to catch up with it, you'll fail because, according to SR, the photon will always be travelling at the speed of light in front of you and the distance between you anf the photon will always be greater than 300,000 km and will keep increasing at the rate of 300,000 km per second. $\endgroup$ Dec 19, 2016 at 21:35
  • $\begingroup$ @KasimMuflahi relativity is a bit about different phenomena, which is most people have no idea of, but still trying to wrap their heads around (including you and me). As I understand it, you will MEASURE it still 300000 km/s, but outside observer will of course see you (say take video of you) travelling with the same as light speed. And you will be frozen on that "video", making your measurements for infinite amount of time. From your POV it's ok, it's just the observer live its life 'fast'. $\endgroup$
    – m8labs
    May 9, 2021 at 16:31

Not the answer you're looking for? Browse other questions tagged or ask your own question.