Take a body having mass $M$ and velocity $v$, it has initial momentum $Mv$. The body radiates two photons of the same "mass" in directly opposite directions (one at the body motion direction, one - opposite direction). The momentum of the system gets reduced: $(M-2m)v + mc - mc = Mv - 2mv$. Does it mean, that the body velocity is increased or there is some other explanation for the missing momentum?

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    $\begingroup$ Photons have no mass. But why would a decrease in momentum equal an increase in velocity? $\endgroup$ – Omry Oct 10 '17 at 17:05
  • $\begingroup$ Photons have mass because E=mc2. If the body radiates photon with a mass m, it losses mass m. I doubt if the speed should increase. But I see some missing momentum in this setup if I use classical approach. $\endgroup$ – Hobbyist Oct 10 '17 at 17:10
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    $\begingroup$ Photons have no mass I'm afraid. They can carry momentum, given by $p = h/\lambda$, but they don't have any mass. Not that this makes any difference to your question. $\endgroup$ – John Rennie Oct 10 '17 at 17:11
  • $\begingroup$ No, photons do not have a mass. See physics.stackexchange.com/q/277173 and links therein. $\endgroup$ – Omry Oct 10 '17 at 17:12
  • $\begingroup$ The discussion about the mass is irrelevant. If a body radiates a photon it looses mass that is equal photon energy divided by the c^2. On the other hand the photon carries momentum that is not dependent on the body (initial) velocity. You can replace m with (E/c^2), it does not change the result. $\endgroup$ – Hobbyist Oct 10 '17 at 17:19

Whatever is emitting these photons is indeed faster at the end. It has to have the same momentum with less mass, and so its speed is higher after the emission.

Note that you have to be careful about what you mean by emit two photons with "the same momentum." Is this the same momentum in the original object's reference frame, or in the laboratory frame? These give quite different results.

Also note that you have some misconceptions about the relationship between mass and energy- we usually describe the photon as massless: it has energy and momentum but no mass. Rather than $E=mc^2$, a formula like $E=\sqrt{m^2c^4+p^2c^2}$ is more applicable here.

  • $\begingroup$ The question is not about a particle, but about a complex object (e.g. two lasers facing opposite directions plus some energy source) that emits two (or more) photons along the moving axis but in opposite directions. $\endgroup$ – Hobbyist Oct 10 '17 at 20:56

Check out this recent paper. It analyzes your problem (toward the bottom), using a quantized atom and field. The conclusion is: the emitter loses momentum, but this loss is not related to an acceleration. Rather, the emitter's mass changes.

In the emitter' rest frame, it cannot be accelerated because it is emitting photons of equal frequency in opposite directions. But two detectors of the photons that are moving relative to the emitter would see differently-Doppler-shifted photon frequencies, and would thus conclude that the emitter's momentum has changed. Indeed it has, but since it cannot have been accelerated, the change must come from mass.

\begin{equation} F=\frac{dp}{dt}=m\frac{dv}{dt}+\frac{dm}{dt}v \end{equation}

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    $\begingroup$ Thank you. Just as I thought the point is in mixing different reference frames. Only Doppler shift slipped from my mind somehow. $\endgroup$ – Hobbyist Oct 11 '17 at 16:30
  • $\begingroup$ @Hobbyist Yeah, and the most interesting thing pointed out in the paper is that the momentum change is proportional to the initial velocity, making it act in a friction-like way. This behavior would appear to conflict with the principle of relativity, except when you consider that the momentum can change without an acceleration. Thus, the emitter has the same acceleration in every frame (i.e. none). $\endgroup$ – Gilbert Oct 11 '17 at 16:43

You can't use Newtonian mechanics to explain this decay because, as you pointed out, it leads to unphysical results. Using special relativity everything makes sense: the photon has a four-momentum $p^{\mu}$ where $p^0 = E = c|\vec{p}| $ and $p^i$ is the $i$-th component of the momentum (as you see I have no mass involved in these equations).

The particle before decaying in the "laboratory frame" has instead $p^0 = m \gamma c^2$ and momentum $p^i = m\gamma v_i$. If the particle has velocity along the $\hat{x}$ axis, then $p^1 = m\gamma v$ and $p^2 = p^3 = 0$.

Using the four-momentum conservation law (i.e. energy conservation + momentum conservation) you have:

$$ m \gamma c^2 = |\vec{p}|_{\gamma_1} c + |\vec{p}|_{\gamma_2} c $$ $$ m \gamma v = p^1_{\gamma_1} + p^1_{\gamma_2} $$ $$ 0 = p^2_{\gamma_1} + p^2_{\gamma_2}$$ $$ 0 = p^3_{\gamma_1} + p^3_{\gamma_2}$$

The physics you get from this is that the two photons don't have opposite momenta, neither the same energy (in the laboratory frame)! Moreover you have 6 variables but only 4 equations, which means you can't predict the direction of emission.

  • $\begingroup$ Not a decay, two well calibrated lasers emitting photons of the exact same energy in the opposite directions. Such a system has some energy source, draining it will reduce total energy of the body (2 lasers + battery) therefore it's mass. Aka hot watter have more mass than cold watter. $\endgroup$ – Hobbyist Oct 10 '17 at 18:43
  • $\begingroup$ I am sorry! I misunderstood your question... now I'll try to answer the true question $\endgroup$ – Matteo Oct 10 '17 at 18:47

Take a body having mass M and velocity v, it has initial momentum Mv. The body radiates two photons

The pi0 for example.

of the same "mass"

the mass of photons is zero and is the same for all photons.

in directly opposite directions (one at the body motion direction, one - opposite direction).

This is wrong, unphysical.

If the photons go into opposite directions you are at the center of mass of the decaying particle by definition.This means zero initial momentum for the decaying particle.

If the decaying particle has momentum the angle between the two photons will be less than 180 degrees and they will carry the original momentum , each photon having p=E/c, i.e. the energy(frequency) of the photons is increased over the energy in the center of mass frame.

The invariant mass of the two photons ( i.e. the four vector of the addition of the two fourvectors) , will be the mass of the original particle ( the pi0 mass for example) and the system carries the original momentum.

Here is the outline of a pi0 decay in a bubble chamber (go to the index for pi0 to see the full complicated picture, and a second example)


Edit after comment that it is a solid body emitting two photons in opposite directions:

Again the mass of each photon is zero. It is the energy that describes a photon and its momentum is E/c. for two photons in equal and opposite direction of the moving body,

E1/c-E2/c will be the momentum missing from the whole body momentum.

If E1=E2 in the center of mass system of the body, no momentum is transmitted. The energy of the body in the center of mass system ( relativistically calculated) will be less than the energy E1+E2. In a moving frameof reference the momentum of the photons will be doppler shifted with respect to the one in the center of mass .

  • $\begingroup$ Again, not a particle. See comments above. $\endgroup$ – Hobbyist Oct 10 '17 at 20:57
  • $\begingroup$ @annav If $E_1=E_2$ in the lab frame, the momentum of the body is unchanged in the lab frame. Since its energy decreased by $E_1+E_2$, does it mean that its rest mass decreased and so the velocity increases? $\endgroup$ – leongz Oct 11 '17 at 11:46
  • $\begingroup$ @leongz I guess so . If you do the experiment with two balls of equal and opposite momentum leaving a rocket forwards and backwards it would be the same classically too. the mass would be less the momentum conserved and the same ... $\endgroup$ – anna v Oct 11 '17 at 13:35

You seem to want that the two photons have the same energy, which I will denote by $E_\gamma$. Then the conservation of energy and momentum reads

$$\begin{align} E &= E' + 2E_\gamma\\ p &= p' \end{align}$$

where primed letters mean "after emission". With the relativistic expression of energy and momentum, these equation reads

$$\begin{align} m\gamma &= m'\gamma' + 2E_\gamma \tag{E}\\ m\gamma v &= m'\gamma'v', \tag{P} \end{align}$$

where $\gamma=(1-v^2)^{-\frac{1}{2}}$ and similarly for the primed one. Note that I am using units where the speed of light $c=1$.

Ok, after all, if I went that far, it's stupid to pretend to leave the rest as an exercise. So, divide the second equation by the first to get


Therefore $v'\ge v$. The equality is for the case where $v=v'=0$, i.e. when the analysis is done in the rest frame of the source emitting the two photons.

By the way, note that $m'< m$. That's because of eqn (P) since $u\mapsto (1-u^2)^{-\frac{1}{2}}u$ increases with $u$. Well, in the rest frame just mentioned, this reasoning does not hold of course, since (P) is $0=0$ but then eqn (E) becomes


and since $E_\gamma>0$, we still have $m>m'$.

  • $\begingroup$ Photon energy in which reference frame? $\endgroup$ – Hobbyist Oct 11 '17 at 18:31
  • $\begingroup$ Your choice! I mean where do you want your two photons to have the same frequency? $\endgroup$ – user154997 Oct 11 '17 at 19:03
  • $\begingroup$ Same frequency in the body reference system. $\endgroup$ – Hobbyist Oct 11 '17 at 19:27
  • $\begingroup$ Then it's impossible: $p=p'=0$, so $v=v'=0$, and $E=E'$ and then $E_\gamma=0$. $\endgroup$ – user154997 Oct 11 '17 at 19:32
  • $\begingroup$ sorry, I forgot $m$ could be different from $m'$ in my last comment. So $E=m$ and $E'=m'$, and $2E_\gamma=m-m'$. $\endgroup$ – user154997 Oct 11 '17 at 19:49

In my opinion, an energy-momentum diagram (the energy-momentum analogue of a spacetime diagram) can clarify what is happening. I've drawn it on rotated graph paper so that the light-signals can easily be drawn. In addition, quantitative calculations (often done with various formula from relativity) can be read off from this diagram... but I won't discuss this point now.

Conservation of total 4-momentum requires: $$\tilde P_{fin}+\tilde {k_1}+\tilde {k_2}=\tilde P_{init}$$

Here is the situation in the rest-frame of the emitter (observer B).
In this frame, the photons have the same energy and equal-magnitude but oppositely-directed spatial-momentum.

The final 4-momentum of the emitter must be smaller than the initial 4-momentum of the emitter. However, note that the emitter's 4-velocity (the "normalized 4-momentum", which is the emitter's 4-momentum divided by the emitter's rest mass) is unchanged.

in the emitter-frame

Here is the situation in the lab-frame (observer A), who observes the emitter moving with $\beta=3/5$ [from O, count 5 diamonds up and 3 diamonds to the right to see this].
In this frame, the photons have the different energy and spatial-momentum, in accordance with the Doppler Effect (with $\beta=3/5$, the Doppler factor is $k=2$).
Of course, each photon still travels at the speed of light.

The forward photon in the lab frame has
twice the 4-momentum compared to the forward photon in the emitter-frame.
The backward photon in the lab frame has
half the 4-momentum compared to the backward photon in the emitter-frame.
(Geometrically, the area of the triangle is the same in the two frames, as required by a Lorentz transformation [whose determinant equals 1].)

Once again, we see that the final 4-momentum of the emitter must be smaller than the initial 4-momentum of the emitter. And again, the emitter's 4-velocity (which is the emitter's 4-momentum divided by the emitter's rest mass) is unchanged.

In the lab frame, the emitter travels with the same spatial-velocity after emitting the photons.

As @Luc said in the Oct 12 comment, if you want to change the spatial velocity of the emitter, emit photons of different energies in the emitter-frame.

in the lab-frame


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