What is the Invariant Mass of Two Perpendicular Photons?

[rashida564]

TL;DR

Two photons travels perpendicular to each other find their invariant mass

I think since Esystem=(PsystemC)^2 + (Minvariant C^2)^2. Then the invariant mass of the system should be zero, but I am hesitated with this is it always the case that photon that travels perpendicular to each other have zero invariant mass

 

[Orodruin]

Summary:: Two photons travels perpendicular to each other find their invariant mass

I think since Esystem=(PsystemC)^2 + (Minvariant C^2)^2. Then the invariant mass of the system should be zero, but I am hesitated with this is it always the case that photon that travels perpendicular to each other have zero invariant mass

Only a system of photons that travel in the same direction has zero invariant mass. As soon as you add a single photon traveling in a different direction, the invariant mass will be non-zero.

Edit: Your equation should also contain E^2, not E.

 

[Nugatory]

The general recipe for finding the invariant mass of a system works here: Choose an inertial frame in which the total momentum is zero; the energy in that frame is the rest energy; invariant mass is then calculated from $E=mc^2$.

If the two photons are moving in the same direction then there is no frame in which the momentum is zero and then this recipe cannot be used.

 

[Orodruin]

The general recipe for finding the invariant mass of a system works here: Choose an inertial frame in which the total momentum is zero; the energy in that frame is the rest energy; invariant mass is then calculated from $E=mc^2$.

I disagree, since it is not applicable when the invariant mass is zero, it cannot be a general recipe. Instead, I would compute the invariant mass from the relation provided by the OP (apart from the missing square), $E^2 = (\vec pc)^2 + m^2 c^4$. For the two-photon system, it is simple enough to compute the total energy and the momenta given the photons' directions and energies.

 

[vanhees71]

The total mass of a system is given by
$$m^2 c^2=P_{\mu} P^{\mu}=s,$$
where $P_{\mu}$ is the total momentum. For two photons you have
$$s=(q_1+q_2)^2=q_1^2 + q_2^2 + 2 q_1 \cdot q_2=2 q_1 \cdot q_2$$
From this you get
$$m^2=\frac{2}{c^2} q_1 \cdot q_2.$$
Now
$$q_1=(|\vec{q}_1|,\vec{q}_1), \quad q_2 = (\vec{q}_2,\vec{q}_2)$$
and thus [EDIT: Corrected in view of #6]
$$m^2 = \frac{2}{c^2} (|\vec{q}_1| |\vec{q}_2|-\vec{q}_1 \cdot \vec{q}_2)=\frac{2}{c^2} |\vec{q}_1| |\vec{q}_2| (1-\cos \vartheta),$$
where $\vartheta \in [0,\pi]$ is the angle between $\vec{q}_1$ and $\vec{q}_2$. Thus you always get a well defined $m^2>0$ always.

 

[PAllen]

The total mass of a system is given by
...
and thus
$$m^2 = \frac{2}{c^2} (|\vec{q}_1| |\vec{q}_2|-\vec{q}_1 \cdot \vec{q}_2)=\frac{2}{c^2} |\vec{q}_1| |\vec{q}_2| \sin^2 \vartheta,$$
where $\vartheta \in [0,\pi]$ is the angle between $\vec{q}_1$ and $\vec{q}_2$. Thus you always get a well defined $m^2>0$ always.

Shouldn’t sin² be (1 - cos ), i.e. not squared, in the last step? Note, with yours, parallel and anti parallel would be the same, which is incorrect.

Otherwise, love the derivation.

 

[SiennaTheGr8]

$\frac{4}{c^2}|\vec q_1||\vec q_2|\sin^2 \frac{\vartheta}{2}$?

 

[PAllen]

$\frac{4}{c^2}|\vec q_1||\vec q_2|\sin^2 \frac{\vartheta}{2}$?

That’s, of course, equivalent to what I wrote.

 

[SiennaTheGr8]

Yes, just guessing that that's what @vanhees71 meant to write.

 

[vanhees71]

Of course you are right. It's a typo (I've corrected it also in the original posting).