Is there a situation in which two photons/particles violate Lorentz transformation?

It does not necessarily have to be a pair production or anything with a name. Consider two photons moving away from or moving towards each other in the void. If imagining photons in relativity is inconvenient, take two particles with $$v_1=v_2\gg 0.99\,c$$ instead. Either way, the relative velocity of the two inertial reference frames $$v$$ is more than $$c$$ (the speed of light), which gives an imaginary number for $$\gamma = \sqrt{1-\frac{v^2}{c^2}}.$$

Is this a violation of Lorentz transformation? Why?

If Lorentz transformation is not explaining the nature as it is, why are we not upgrading the transformation?

• Why would you put the rate at which the distance is decreasing between two particles into $\gamma$? That doesn't really make sense. Commented Jul 19 at 16:15
• @Triatticus I did not. It can be decreasing, as it can be increasing. I let the reader choose either of the situations. Should I derive $\gamma$ here? I thought it was a very well-known factor in physics.
– user366915
Commented Jul 19 at 16:30
• Photons don't have (inertial or otherwise) rest frames. Commented Jul 19 at 16:36

What you are not taking into account is the relativisic velocity addition.

When calculating the gamma factor we are calculating it for a single reference frame moving relative to our reference frame. You have two objects moving away from each other. Let's call them A and B and our reference frame E for Earth, velocities $$v_{AE} =-0.8c$$ and $$v_{BE} =0.6c$$, where E indicates the velocity is measure relative to the Earth frame. We can calculate the gamma factor for A from the point of view B reference frame or from our own reference frame. To calculate the velocity of A from B's point of view ($$v_{AB'}$$) we use the relativistic velocity addition formula. $$v_{AB'} = \frac{v_{AE} - v_{BE}} {1-\frac{v_{AE} v_{BE}} {c^2}} = \frac{(-0. 8) - 0.6}{1-(-0.8)\times0.6} = -0.945c$$.

What if B is a photon going to right at v = c = 1? What is the speed of B from A's point of view? The calculation is: $$v_{BA'} = \frac{v_{BE} - v_{AE}} {1-\frac{v_{BE} v_{AE}} {c^2}} = \frac{1.0 - (-0.8}{1-(-0.8)\times1.0} = c,$$.

Which is as it should be because the speed of a photon is always c with respect to any observer.

When B is moving relative to us his clocks are time dilated and his rulers are length contracted and his notion of what events are simultaneous are different to ours so it is not too surprising that he measures the velocity of A differently to what we measure. The relativistic addition equation is derived from the Lorentz transformations and takes all these effects into account. The Lorentz transformation have at their heart that the speed of light of is c in reference frame and the relativistic velocity addition formula is consistent with that. When you say the gamma factor is Y, you have to be clear what that gamma factor applies to. You can't simply say the gamma factor of A and B is Y. It applies either to A from some other observers point of view or to B from some other observer's point of view.

What you did was introduce a third inertial frame in which the relative velocity between the two particles is indeed higher than $$c$$. This does not violate Special Relativity. If you go in the reference frame of one of the particles you get the velocity of the other to be actually smaller than $$c$$.

And please, don't jump to conclusions. Since physicists use the theory as it is there is good reason for that. The Lorentz transformations are an accurate description of nature in the context of Special Relativity.

• Could you please go in the reference frame of one of the particles and show the velocity of the other to be actually smaller than $c$?
– user366915
Commented Jul 19 at 16:46
• I can but I won't, you just have to use Lorentz transformations. You can do it yourself or look it up in any Special Relativity textbook Commented Jul 19 at 16:49
• @Raymond if the particles are massive they have a rest frame, lorentz transform to one of the particle rest frames and use the velocity addition formula to see that the particle at rest does not observe another approaching it at a velocity greater than $c$. Commented Jul 19 at 16:50
• I have done that, and the result is not as you described.
– user366915
Commented Jul 19 at 16:51