# Relativistic Quantum Mechanics?

1. Mar 4, 2014

### Zak

Just a thought...

Say that a stationary (relative to me) source of light emits a photon with the energy level (as observed by me) appropriate to pair produce an electron-positron pair.

Meanwhile, a second observer travels towards the light source, thus blue-shifting the light, and hence observes a photon with more energy than that of the one I observed when stationary.

Upon observing the same pair-production, surely it is not possible that both observers see an electron positron pair because if the observer in motion (relative to the light source) saw an electron-positron pair, there would have to be extra photons emitted to compensate for the surplus energy and these are not observed by the stationary observer as to me, the photon seems to have just enough energy for an electron-positron pair.

Is this where, to the observer in motion, the particles seem to be moving faster, hence to them the particle's mass dialates and accounts for the 'extra' energy of the photon they observe?

2. Mar 4, 2014

### tiny-tim

Hi Zak! Welcome to PF!
A photon cannot produce an electron-positron pair (with nothing else) …

in the centre of mass frame of the pair, the total momentum is zero, and it is not possible for a photon to have zero momentum.

3. Mar 4, 2014

### samalkhaiat

What do you mean by "appropriate"? The process
$$\gamma \rightarrow e^{ + } + e^{ - }$$
does not happen no mater how energetic the photon is. This is because it violates energy-momentum conservation.

4. Mar 4, 2014

### Zak

Ah ok, so then what are the conditions that allow pair production?
also, surely if the particles travelled forward diagonally, the net momentum would not be zero as there would be a net 'positive' momentum in, say, the y axis but not the x hence the initiall momentum of the photon travelling forward would be conserved?

5. Mar 4, 2014

### WannabeNewton

I would suggest reading a chapter on relativistic kinematics from a particle physics book. See, for example, chapter 3 of Griffiths "Elementary Particles". You really need to be comfortable with relativistic problem solving of scatterings and decays through conservation of 4-momentum and Lorentz-invariant quantities using both lab frames and center of mass frames before tackling the kind of questions you're attempting to tackle.

That's not the center of mass frame. The center of mass frame is defined as the frame in which the total momentum of the system vanishes. It is easy to prove that such a frame always exists for a finite collection of particles. The process $\gamma \rightarrow e^{+}e^{-}$ is impossible because in the center of mass frame post-scattering you get vanishing momentum for $\gamma$ from conservation of momentum which is physically impossible.

6. Mar 4, 2014

### Staff: Mentor

For an electron-positron pair you can always find a reference frame where the momentum is 0. That specific reference frame is known as the "center of momentum frame" and is the frame that tiny-tim was referring to.

You can have a pair of photons that produce an electron-positron pair. The likelihood of this interaction is very small, but it does happen. Search for two-photon physics or two-photon interaction.

7. Mar 4, 2014

### WannabeNewton

By the way, just as an aside, you don't really have creation and destruction of particles in relativistic QM. You need QFT for such processes.

8. Mar 4, 2014

### samalkhaiat

The centre of mass frame is no special. But if a process is impossible in one reference frame, it will be impossible in all reference frames. Simply, the process violates energy-momentum conservation in all reference frames.
Take for example, a photon with energy-momentum 4-vector $p^{ \mu }_{ ( \gamma ) } = ( | \vec { p }_{ ( \gamma ) } | \ , \ \vec{ p }_{ ( \gamma ) } )$ and consider the rest frame of the electron: $p^{ \mu }_{ ( - ) } = ( m \ , \ \vec{ 0 } )$. Let the positron energy-momentum 4-vector be $p^{ \mu }_{ + } = ( \sqrt{ m^{ 2 } + p^{ 2 }_{ ( + ) } } \ , \ \vec{ p }_{ ( + ) } )$. So, energy-momentum conservation for the process tells you that
$$( | \vec { p }_{ ( \gamma ) } | \ , \ \vec{ p }_{ ( \gamma ) } ) = ( m \ , \ \vec{ 0 } ) + ( \sqrt{ m^{ 2 } + p^{ 2 }_{ ( + ) } } \ , \ \vec{ p }_{ ( + ) } )$$
This implies
$$\vec{ p }_{ ( \gamma ) } = \vec{ p }_{ ( + ) } ,$$
and
$$| \vec { p }_{ ( \gamma ) } | = m + \sqrt{ m^{ 2 } + p^{ 2 }_{ ( + ) } }$$
Now, you can solve these two equations to find the photon momentum
$$\vec { p }_{ ( \gamma ) } = 0 .$$