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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?

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    $\begingroup$ Antiparticles are defined having positive energy $\endgroup$ – Triatticus May 18 '18 at 4:10
  • $\begingroup$ @Triatticus Aren't the antiparticles particles with negative energy? $\endgroup$ – mithusengupta123 May 18 '18 at 4:11
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    $\begingroup$ No. A way you can tell is to do a matter antimatter collision. There is energy that comes out. A lot. $\endgroup$ – AHusain May 18 '18 at 4:13
  • $\begingroup$ 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 $\endgroup$ – Triatticus May 18 '18 at 4:14
  • $\begingroup$ @Triatticus So what is the answer to my question? Two 4-momenta with nonzero components cannot add up to zero? $\endgroup$ – mithusengupta123 May 18 '18 at 7:47
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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.

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  • $\begingroup$ 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 $E=p_0$. $\endgroup$ – Arnold Neumaier Nov 23 '18 at 15:21
  • $\begingroup$ @ArnoldNeumaier What? The origin of energies is perfectly arbitrary both in relativistic and non-relativistic mechanics. $\endgroup$ – AccidentalFourierTransform Nov 23 '18 at 15:25
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    $\begingroup$ 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 $E$. $\endgroup$ – Arnold Neumaier Nov 23 '18 at 15:35
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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.

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  • $\begingroup$ You don't have translation invariance anymore, so what do you mean by 4-momentum? $\endgroup$ – AHusain May 23 '18 at 17:25
  • $\begingroup$ @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. $\endgroup$ – Michele Grosso May 24 '18 at 8:32
  • $\begingroup$ There is no well defined 4-momentum. Momentum is the charge for translation invariance. No invariable, no charge. $\endgroup$ – AHusain May 24 '18 at 18:00
  • $\begingroup$ +1, this is correct even though it may sound wrong to people who don't know GR. Related question here. $\endgroup$ – knzhou Nov 21 '18 at 21:59
  • $\begingroup$ @MicheleGrosso The question did have the "special relativity" tag. $\endgroup$ – jim Nov 21 '18 at 22:05

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