This is probably an obvious question, but I don't see it answered anywhere.

Imagine we have an object in the Universe that is traveling, relative to the Milkyway at about .99999999999999999999999999c and I realize that's not possible, but imagine.

Speed is relative, so to the object, the Milkyway is flying past it and to us, the object is flying past us. That's all good and relative.

But the photons that our superfast object encounters are gonzo blue-shifted, and relativity says that's OK too. But lets say the object is traveling so fast, the photons it encounters are energetic enough to spontaneously undergo pair production. That seems difficult to explain. A relatively low energy visible light photon to our perception won't split into an Electron and a Positron, but a very energetic photon can, but it's the same photon just observed by 2 different reference frames?

I understand that observers won't agree on when things happen or on energy but the two observers should agree on events that take place and pair production for one but not the other seems impossible.

Is there a simple layman's way that this can be explained?

  • $\begingroup$ "...should agree on events that take place and pair production for one but not the other seems impossible." Not necessarily. To my understanding particle number of a field does not belong to its Lorentz invariants and so, yes, different observers could have perceptions of a slightly different version of the Universe if that's defined through the particles it contains. Also, pair production is not an event. Observing a photon in the past and an e-e+ pair in the present are. What happens inbetween is subject to all possible Feynman diagrams. $\endgroup$
    – The Vee
    Jul 6, 2016 at 10:59
  • 2
    $\begingroup$ "I realize that's not possible"... Actually it's been observed. $\endgroup$
    – user10851
    Jul 6, 2016 at 13:37

2 Answers 2


Yes, there is a simple explanation! The pair production process $\gamma \to e^+ + e^-$ is forbidden by energy-momentum conservation, so it doesn't happen in either frame.

One way to see this is that, as a massless object, the photon has "the most possible momentum for its energy", as it doesn't have the extra rest mass-energy. So if you try to make the momentum in this equation balance, the electron and positron will have too much energy, and vice versa.

However, a slightly modified version of your idea holds. Our universe is full of low-energy photons from the Cosmic Microwave Background. As you suggested, a very fast moving charged particle will see these photons as heavily blueshifted, and pair production is possible if the photons scatter off the object. (This is a $2 \to 2$ reaction, so the argument in the previous paragraph doesn't hold.)

The scattering steals some of the particle's energy, slightly slowing it down. This places a limit on the energy of cosmic rays, called the GZK bound.


Pair production cannot just happen. It requires some other particle/object to balance out the momentum. Lets suppose this object is a nucleus.

So the interaction looks to the fast moving observer like a blue shifted photon scattering off a stationary nucleus and producing an $e\bar{e}$ pair. To us it looks like a nucleus moving at relativistic speed scattering a photon and producing an $e\bar{e}$ pair. There is no fundamental disagreement between the observers as to what happens.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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