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Timothy van der Linden writes:
> >And I don't disagree with that analysis; it's correct in a frame
> >where the ship is moving. In that frame, the photon has more
> >energy and momentum, and while it changes momentum by the same
> >amount as in the frame where the ship is "stationary", this
> >necessarily results in a different-looking scenario.
> I was wrong to compare it with the police radar, so forget that part. But I
> still don't see what was wrong with part preceding it.
> o you say the photon looses energy if it accelerates the ship or do you say
> that the energy of the photon stays the same?
> What does that different scenario look like?
In a frame where the ship is in motion, the photon changes energy
when it bounces off the ship; its reflected energy is less than
the incident energy if the ship receding, greater if the ship is
approaching. You actually were right to compare this to doppler
radar, because it's this effect that doppler radar measures to
> >> Assuming these mirrors have mass, they indeed will get some velocity that
> >> correspondends to the same momentum as the photon. By each reflection the
> >> photon will loose some momentum and lower its wavelength.
> >This is exactly the analysis I came up with. In the limit the
> >mirrors carry the momentum of the light pulse, and the light
> >pulse fades to nothing.
> Somehow we seem to agree on this, but I'm not sure why.
Because we're both thinking about it the right way? :-)
> >First you have to convince me that a moving object really does
> >exert more gravitational force.
> Let me quote a few sentences of "Introducing Einstein's relativity" by Ray
> d'Inverno in the paragraph "The principle of equivalence":
> Next, we wish to make explicit the assumption that matter
> both responds to, and is a source of, a gravitational
> field. However, we have seen in special relativity that
> matter and energy are equivalent, so the statement about the
> gravitational field applies to energy as well.
> This means that even photons excert and react to gravity as well as all
> other kinds of energy.
This is true, but it's more complicated than you make it out to
be. Note that I will continue to use Taylor and Wheeler's
terminology, so when I say "mass", I mean invariant mass.
A photon is massless, because it has energy equal to its
momentum, and the magnitude of its momenergy vector is 0.
However, multiple photons considered as a system may not be
massless! For example, two photons with energy/momentum p
travelling in opposite directions have mass 2p:
[ p p 0 0 ] + [ p -p 0 0 ] = [ 2*p 0 0 0 ]
magnitude [ 2*p 0 0 0 ] = 2*p
Two photons travelling at right angles to each other also have
[ p p 0 0 ] + [ p 0 p 0 ] = [ 2*p p p 0 ]
magnitude [ 2*p p p 0 ] = sqrt(2) * p
Two photons travelling in parallel have no mass:
magnitude [ 2*p 2*p 0 0 ] = 0
An absorber gains mass as well as momentum from a massless
[ m 0 0 0 ] + [ p p 0 0 ] = [ m+p p 0 0 ]
magnitude [ m+p p 0 0 ] = sqrt((m + p)^2 - p^2) = sqrt(m^2 + 2 * p * m)
Mass is a more subtle concept than energy; while energies add
linearly, masses as magnitudes of momenergy vectors do not.
> I would conclude from this that moving bodies excert greater gravitation
> either because of gain of mass or gain of energy. I feel both can be used,
> its a bit like the wave-particle duality.
What I really want to see is the general relativistic formula
that says whether spacetime curvature is the result of an
object's mass or its energy.
> >> Is it this translation of energy to mass that gives the trouble?
> >It is that my studies of relativistic kinematics do not allow for
> >the treatment of energy and mass as identical quantities.
> They are indeed not identical but equivalent, meaning they can be be
Not in Taylor and Wheeler's terminology. The only time they
allow mass and energy to be spoken of as equivalent is when you
are dealing with an object in its rest frame. In any other frame
the object's energy is not equivalent to its mass.