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Re: one question

Kevin C. Houston writes:
 > >  > The wires are clearly dissipating a momentum equal to 2 E/C.
 > >  > that is what i was trying to say with my bb analogy.
 > > 
 > > I'm not sure "dissipating momentum" is quite the term for it.
 > > After all, the laser with momentum -E/c is tugging through the
 > > wires against the plate with momentum E/c.  The momenta cancel.
 > YES!  That's what I've been saying all along!  that a physical structure 
 > can transmit (through tugging) opposite momenta that can then cancel!

I thought about deleting that paragraph ("I'm not sure
'dissipating momentum' . . . ") because it's not really
consistent with my position that you only have momentum if you
have motion.  If the things don't move, they don't have momentum,
or absorb momentum, or dissipate momentum.

What I was really objecting to in your parasail design was not
the idea that you had cancelling momenta, but that you thought
that creating sideways momentum meant a decrease in forward
momentum.  For example, a device like this:

   E/c --v                         :::: p = [ 0 E/(2*c) ]
|######~~~~~~~~~~~~~~~~~~~~~~~~~~~~::/ |
|######~~~~~L~A~S~E~R~~~~~~~~~~~~~~:/  | beam splitter
|######~~~~~~~~~~~~~~~~~~~~~~~~~~~~:\  |
|######~~~~~~~~~~~~~~~~~~~~~~~~~~~~::\ |
                                   :::: p = [ 0 -E/(2*c) ]

# = laser  ~ = laser light  -- = wires 

also won't move.  The beam splitter "absorbs momentum" just like
the black absorbing plate did, even though it splits the beam
into two beams traveling in the +y and -y directions.  This setup
also has the advantage of not requiring an unobtainium heat sink,
as long as you get a Perfect Mirror (tm) from Acme Physics

Without wires to hold the beam splitter in a fixed position with
respect to the laser, it too would move away from the laser.
It's critical to understand that momentum is a vector quantity to
understand how the beam splitter can move, even though it is
emitting two beams whose momentum magnitudes are both p/2 while
it is picking up momentum with magnitude p from the laser.
Although it is intuitive to think in terms of the magnitudes of
the momentum components and then try to add those to account for
system momentum, it's unfortunately wrong.

Although the laser light carries momentum [ p 0 ] to the unbound
beam splitter and the splitter, as a result of reflecting halves
of the beam, produces two smaller beams with momentum [ 0 p/2 ]
and [ 0 -p/2 ], those two beams add to momentum [ 0 0 ] -- they
carry no forward momentum whatsoever; as a pair, they carry no
momentum at all.  Yet conservation of momentum requires that the
total momentum remain [ p 0 ], and the only place this can show
up is in the beam splitter itself.  Since the beam splitter
stopped all forward motion of the beam, it's not even that
unintuitive that it should get the forward momentum.

The ultimate case of absolutely needing to think of momentum as a
vector quantity is when you have an object with zero momentum
that splits into two or more objects with nonzero momenta.  The
vector sum of momenta of all the objects will still be zero, but
in a sense momentum did come from nowhere since an unmoving
object gave rise to motion.

 > > Like I said before, momentum means motion.  A non-moving object
 > > has no momentum.  If the laser and the plate are held together
 > > with wires and don't move relative to each other or your
 > > observer, then you can't say that they have momentum.
 > The laser imparts momentum when it leaves, and when it strikes the black 
 > plate.

Once, during the time that the beam is in progress to the absrber

For much the same reasons that I don't like "relativistic mass",
I don't like the notion of imparting momentum to a non-moving
object.  Clearly it has lead to some incorrect physical
thinking.  If something can't be put into uniform linear motion
relative to something else, then it can't "absorb momentum"
relative to that thing.

 > > The wires are indeed under tension, because there is a force
 > > between the laser and the plate.  This tension was created in the
 > > first instant the laser was turned on, and a small amount of its
 > > energy went into stretching the wires before it was all spent on
 > > heating the plate.
 > so you're saying that the plate heats more if it is held in place than if 
 > it was free to move?

Yes.  If it was free to move, it would acquire velocity; as it
acquired velocity, the light it received would become weaker due
to doppler-shifting, and so the plate would receive less energy.
If it is held in place, then it never has velocity relative to
the laser, which always transmits the same energy per unit time
to the plate.