# Sum of two 3-momenta that are equal and opposite is zero. Can the sum of two nonzero 4-momenta be zero?

The sum of two 3-momenta $\vec{p}_1$ and $\vec{p}_2$ can be zero if their magnitudes are equal and they are directed opposite to each other.

What about the sum of two 4-momenta $p_1^\mu$ and $p_2^\mu$?

Let's assume that the sum can be zero which implies that their individual components must add up to zero. In particular, $p_1^\mu+p_2^\mu=0$ implies for the zeroth components that $E_1+E_2=0$. Can this condition ever be satisfied other than the trivial case when $E_1=E_2=0$? Is this possible for a particle-antiparticle pair because antiparticles have negative energy?

• Antiparticles are defined having positive energy – Triatticus May 18 '18 at 4:10
• @Triatticus Aren't the antiparticles particles with negative energy? – mithusengupta123 May 18 '18 at 4:11
• No. A way you can tell is to do a matter antimatter collision. There is energy that comes out. A lot. – AHusain May 18 '18 at 4:13
• When Dirac first found the negative energy solutions, the idea then became to make an antielectron with positive energy as a negative energy particle would have no lower bound on its energy states – Triatticus May 18 '18 at 4:14
• @Triatticus So what is the answer to my question? Two 4-momenta with nonzero components cannot add up to zero? – mithusengupta123 May 18 '18 at 7:47

It is clear that $$p_1^\mu+p_2^\mu=0$$ if and only if $$E_1+E_2=0\qquad\text{and}\qquad \vec p_1+\vec p_2=\vec 0$$

OP correctly argued that $$\vec p_1+\vec p_2=\vec0$$ is perfectly possible, and asks whether $$E_1+E_2=0$$ is possible as well. The answer is yes. The origin of energies is irrelevant, and therefore if $$E_1+E_2=\mathcal E_0$$, it suffices to redefine $$E_i\to E_i-\mathcal E_0/2$$ to obtain $$E_1+E_2=0$$.

One could argue that the choice of origin we made above is not natural. If you choose, instead, the origin of energies such that $$E=m$$ when the particle is at rest, then $$E=\sqrt{\vec p^2+m^2}$$ is positive-definite. The sum of two positive-definite functions is also positive-definite, and therefore $$E_1+E_2>0$$, strictly. In that case, the sum of energies cannot vanish.

Finally, if you use massless particles (which have no rest frame), then $$E=|\vec p|$$, which is semi-positive definite. The only way to get $$E_1+E_2=0$$ is that $$E_i=0$$, that is, that $$\vec p_i=\vec 0$$. But a particle with no mass, no momentum, and no energy, is not really a particle at all: it is the vacuum. Whether this qualifies is a matter of opinion.

To clarify a misconception in the OP: anti-particles have positive energy. The expression $$E=\sqrt{m^2+\vec p^2}$$ is valid both for particles and anti-particles.

• Your answer is inconsistent. Your final sqrt expression for the energy proves that E is nonnegative for every particle and positive for a massive particle. In particular, the sum of two such energies can vanish only if both separately vanish. - Only nonrelativistic energy can be chosen to have an arbitrary origin. In relativistic theories, the origin cannot be displaced by Lorentz invariance since $E=p_0$. – Arnold Neumaier Nov 23 '18 at 15:21
• @ArnoldNeumaier What? The origin of energies is perfectly arbitrary both in relativistic and non-relativistic mechanics. – AccidentalFourierTransform Nov 23 '18 at 15:25
• What? Please prove your claim, and explain in which sense the relations $E=p_0$and $p^2=m^2$ for 4-momentum in the $+---$ metric, always assumed in the relativistic case, are invariant under shifting energies. You cannot shift, you need to Lorentz transform! And this preserves the sign of $E$. – Arnold Neumaier Nov 23 '18 at 15:35

On the event horizon of a static (Schwarzschild) black hole or on the ergosphere of a rotating (Kerr) black hole, a quantum fluctuation may produce a pair of particles one with positive energy and the other with negative energy. The particle with positive energy just before the horizon/ergosphere and the particle with negative energy just after. The negative energy particle is allowed by the time Killing vector switching to spacelike crossing the horizon/ergosphere.

It is a specific case, but in principle the 4-momentum addition of the two particles may give zero.

• You don't have translation invariance anymore, so what do you mean by 4-momentum? – AHusain May 23 '18 at 17:25
• @AHusain The Killing vector guarantees the energy conservation, but the change of its nature allows for the energy to be negative. The issue was how the temporal component of the 4-momentum addition could be zero. – Michele Grosso May 24 '18 at 8:32
• There is no well defined 4-momentum. Momentum is the charge for translation invariance. No invariable, no charge. – AHusain May 24 '18 at 18:00
• +1, this is correct even though it may sound wrong to people who don't know GR. Related question here. – knzhou Nov 21 '18 at 21:59
• @MicheleGrosso The question did have the "special relativity" tag. – jim Nov 21 '18 at 22:05