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Here's an example to describe the issue

Supossed a high power laser (eg a 100 kW laser, ie, electromagnetic weapons) is fired to a target, then it will receive energy and move. (and likely to burn or explode, but that's not the point)

My question is, statting that photons have no mass

Would the laser transmitter get a kickback/recoil like in a "normal" gun?

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3 Answers 3

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Dear HDE, The laser beam obviously has energy and momentum so the laser transmitter gets recoiled due to the conservation of energy and momentum. See also:

http://en.wikipedia.org/wiki/Poynting_vector

The property we call mass is expressed by:

$m ~=~ \sqrt{\frac{E^2}{c^4} - \frac{p^2}{c^2}}$

which is zero for photons even though the energy $E$ and momentum $p$ are not zero. The mass m is zero if energy relates to momentum as $E^2=p^2c^2$. This shows you that photons can't be at rest because in that case both $p$ and $E$ are zero.

Regards, Hans

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Thanks for the answer, I will tell you one source of my doubt, I had in mind that "center of mass can't be changed only by internal forces", then as photons have no mass, then the system must include the photons in the center of mass calculus!, how could a photon (without mass) affect the center of "mass"? that's a strange thing, if the target mass enter to play then again it has sense, but it's hard to figure out how center of mass can be affected without a mass interaction, perhaps photon are obliged to interact sooner or later, but that's just a philosophic thought –  HDE Jun 1 '11 at 14:51
    
@HDE a lazer beam is zillions of photons and does have an effective mass, from the total four vector of the photons comprising the beam. Now I think of it, it is the heisenberg uncertainty principle, or the dispersion of the beam momentum in the perpendicular to the axis direction that will be definitive for the mass. –  anna v Jun 1 '11 at 14:58
    
Hi HDE, The confusion is understandable because there are two different definitions of mass. If you say that the photon has no mass then we are talking about the "rest" mass. When you are referring to the center-of-mass then a different definition of mass is used: The "inertial" mass where all kinetic energy is also considered "mass" physlink.com/Education/AskExperts/ae161.cfm –  Hans de Vries Jun 1 '11 at 16:09
    
de Vires I see, thanks.. Why your equation differs from @Zassounotsukushi answer $m=\frac{h}{\lambda c}$ ? (leaving aside the N) –  HDE Jun 1 '11 at 16:32
    
Please read up on four vectors en.wikipedia.org/wiki/Four-vectors. The formula that Hans de Vries has given holds also for the sum of energies and the summed three momenta when there are many particles.If there were no dispersion in the beam nor a heisenberg position/momentum uncertainty then the total invariant mass would be 0 because for each individual photon E=p in units of c=1, and the sum in the direction of motion would be equal. –  anna v Jun 1 '11 at 16:38

I asked a related question recently, Explain how (or if) a box full of photons would weigh more due to massless photons. Both of our questions are concerned with the accounting for physical properties when a photon is in-flight.

The transmitter is pushed back as the photons are emitted. A given force $F$ comes from the laser at all times it is on, which can be calculated from the momentum (addressed in other answers). If the laser is on, pointing in a single direction for a given time $t$, a quantifiable impulse is imparted to the laser apparatus.

Mass-Energy Balance

The challenging part comes when you consider that the laser apparatus (obviously including the energy source) looses mass through this process. If you take the laser apparatus to be at rest, no energy is imparted to it since power is force times the velocity $P=F v$, and $v=0$. That means that all mass that leaves the laser apparatus system ($\Delta m$) is accounted for by energy in the laser beam.

$$E=\Delta m c^2 = N p c$$

$N$ is the number of photons emitted in $t$ time.

Center of Mass (CM) and Net Velocity

Now consider an even more challenging component - the center of mass of the total system. Consider the laser and connected devices to be at some point A, isolated in empty space. Then consider that the laser fires directly toward another heavy isolated system in empty space B for time $t$. Take the distance between them to be $d$ and assume that $d>>c t$. That assumption is to make the beam like a "batch" single emission. The A system is moving away from B after it is fired. Both A and B are at rest before firing, $m_A=m_B$, and the system including both masses, system AB, has a CM at $d/2$.

If you include the photon in your definition of system AB after firing but before absorption by B, the system MUST have a net velocity of zero and the CM must not move. This is an extremely challenging concept, and to the extent of my understanding, you must simply consider the photon's relativistic mass. it has no rest mass, but the location of the photons must have some weighting in calculation of the system CM for the typical principles to still hold, this is $m=\frac{N h}{\lambda c}$. The momentum of the photons balance the recoil momentum of system A, and the position of the photon balances the recoil movement of system A. In this way, the photon is very very similar to if the process were instead a bullet. The main difference is that it has no rest mass and moves at the constant speed of light.

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+1 thanks for addressing the center of mass issue and the "before absorption moment" that's the source of the question as I put in other comments.. relativistic mass sounds as a convincing answer –  HDE Jun 1 '11 at 16:30

Yes, recoil makes part of such a process. You may consider this as a decay of an excited laser into two parts: a photon beam goes in one direction, a deactivated laser goes in the opposite one. This happens during emitting photons. The beam fate is a quite another story.

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When? at the moment of the shot? or does the reaction "waits" till it hit the target?, what if it doesn't hit the target till many years?, then there would be no action force...so why would be a reaction? still confusing for me, can you explain it further? thanks –  HDE Jun 1 '11 at 14:08
1  
@HDE I added an explanation to my answer. –  Vladimir Kalitvianski Jun 1 '11 at 14:14

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