Consider a small body of a given mass floating motionless in space. This body begins to emit a ray of light (say a laser) into deep space. Firstly, does the body begin to lose mass as it loses em waves to space and its surroundings?
So the longer the laser fires for, the more mass is coverted into energy, right? So the length of the laser beam at any given moment is proportional to the loss of mass, yes? And the entire length of the laser "contains" the exact amount of energy that the body has lost. Is that logical?
Then the laser hits another body with a given mass sitting stationary relative to the first body. At this point the second body “soaks up” all the laser as it is struck and does not reflect any light. So the length of the laser is now constant and I presume the second body starts to increase its mass. Am I still doing ok?
The other body gains the mass; in a natural transfer, part of the gain would show up as kinetic energy as CH suggests.
Hi dav57, You are thinking along the right lines, but I would highly recommend that you think of this in terms of the conservation of four-momentum rather than just an exchange between mass and energy. The reason is that not only does the laser carry energy, but it also carries momentum. So the first body will start to move one direction and the second body will start to move the other direction. So not all of the mass lost by the first will be recovered by the second as some of the mass is converted to KE of the two bodies. That is kind of hard to figure out without really considering the conservation of the four-momentum. -Dale
Ok , I get that bit. So the mass is converted to energy as you would expect and the two bodies start moving a little and A loses some mass and B gains some mass. Does B gain the same amount of mass as A lost when the laser has finished firing and has all been soaked up by B?
No, some of the mass will be "permanently" converted to KE of A and B even after all of the light is absorbed. -Dale
So A always loses more mass than B gains. Right? So the total mass of the two bodies at the finish has reduced by the amount which has been converted into KE of both bodies. The laser "held" the rest of the energy during travel - but some is converted to add to B's mass and B's KE.
What if body A were to fire two lasers in opposite directions such that there could be no conversion to KE for body A? Does this mean that ALL of the lost mass is contained as energy within the length of the two lasers?
Yes. Yes. Yes. Here is how you would analyze it mathematically: Considering a single spatial dimension and in units where c=1 the Four-Momentum is (E,p) for a massive body at rest this reduces to (m,0) in units where c=1. For light moving to the right this reduces to (E,E) and for light moving to the left this reduces to (E,-E), again in units where c=1. The mass is given by the norm of the four momentum: m^2=|(E,p)|^2=E^2-p^2. I will use A for mass A, B for mass B, L for light moving to the left, R for light moving to the right, and S for the conserved total system four momentum. First scenario: before laser A0=(m,0) B0=(m,0) S0=A0+B0=(2m,0) during laser R1=(E,E) A1=A0-R1=(m-E,-E) B1=B0=(m,0) S1=A1+B1+R1=(2m,0) after laser A2=A1=(m-E,-E) B2=B1+R1=(m+E,E) S2=A2+B2=(2m,0) Second scenario: before laser A0=(m,0) S0=A0=(m,0) during laser R1=(E,E) L1=(E,-E) A1=A0-R1-L1=(m-2E,0) S1=A1+L1+R1=(m,0) -Dale
Ok. Does a laser beam reflecting off of a highly reflective mirror cause the mirror to gain KE? If the laser travelling away from body B, that is, backwards before it were to bounce off a mirror and hence travel towards body B (as is the other laser) - does this means that body A has no added KE (as before with a twin laser) but now the loss in A's mass would be fully converted into the mass gain of B after the process has been completed?
Yes. M is mirror. Before reflection: M0=(m,0) R0=(E,E) S0=(m+E,E) After reflection: L1=(E,-E) M1=S0-L1=(m,2E) S1=(m+E,E) I don't understand where the mirror is in relation to A and B, but you should be able to work it out following the above examples. -Dale
Hold on, let me get this right. If the mirror fully reflects the light (I take it we can manufacture one that does a pretty good job) how does the light transfer energy to the mirror as KE when the light retains all the energy itself? Or is there some energy lost? If there is energy lost, I presume the light must reduce either intensity or frequency??? What is going on here?
Remember photons not only carry energy that must be conserved, but they also carry momentum that must be conserved. Momentum is a vector quantity, so when the photon reflects back the change in momentum is exactly twice the change in momentum when it is absorbed. It is the same as any perfectly elastic collision. -Dale
In which case the conservation equations are not as simple as I wrote because the E of the reflected photon is less than the E of the incident photon. -Dale
