Re: A small conservation of energy puzzle

On Thu, 02 Feb 2006 17:30:59 -0700, N:dlzc D:aol T:com (dlzc) wrote:

Dear sal:

"sal" <pragmatist@xxxxxxxxxx> wrote in message
news:pan.2006.02.02.19.27.41.825015@xxxxxxxxxxxxx
On Thu, 02 Feb 2006 10:54:52 -0800, PD wrote:
...
On a spaceship, an astronaut converts a small mass, "dM", into a
single photon of frequency "f", which zips away from the ship at C.

A single chunk of mass cannot convert into a single photon without
violating conservation of momenergy. This is why that doesn't happen.

You're thinking of the case where there is an isolated mass which is
completely converted into photons. That's not what we have here.

"Dude", the ship has a certain *momentum* before dM is converted. You
posit that the energy is all

"all" in the EMITTER's frame. In the observer's frame some of the energy
is absorbed by the ship when it recoils. Energy is, of course, extremely
frame-dependent.

converted to a photon propagating normal to the flight path of the
ship...

.... normal to the path, in the OBSERVER's frame. In the SHIP'S frame the
photon is emitted at a different angle -- its path points somewhat toward
the back. This is important!

The angle it makes with the right-angle line, in the _ship's_ frame, is

arctan (v/sqrt(1-v^2))

When v=0 that's zero -- it's at right angles in both frames. As v->1 the
angle tends to 90 degrees -- the photon is shot almost straight out the
back in the emitter's frame.

If you don't believe me, run the photon's 4-momentum through the Lorentz
transform -- either way -- and see that the angles actually do what I'm
claiming.


Guess what? The ship either has to speed up
ever so slightly, and alter course very slightly away from this normal or
violate conservation of momentum.

I agree completely. If you wade through the 200 lines of solution I
posted earlier, that's the whole point of the exercise -- the ship
"recoils" in its CM frame.

When you Lorentz-transform the recoil into the observer's frame, it
turns out to have a component along the line of flight, even though the
photon's momentum in that frame does _not_. This is counter intuitive but
true in this case none the less. Part of the reason is that the recoil is
_not_ the inverse of the photon's 4-momentum -- it's the _difference_
between the 4-momentum which the parcel of mass "dM" had, and the photon's
4-momentum. Hence, when we Lorentz-transform the photon's momentum and
the recoil to another frame -- which is basically a rotation -- the
3-momentum components of the photon's momentum and the recoil are no
longer necessarily opposites, even though they were in the emitter's frame.

The upshot is that the ship does, indeed, _both_ speed up slightly and
alter course slightly.


Alternatively, such conversions *require* an even number of photons be
emitted due to conservation of spin (and/or even what Paul cites).

If one photon is emitted normal to the path, one or more photons must be
emitted to carry off the balance of the momentum, presumably mostly
rearward and away from the stationary observer. And then you are stuck
as to knowing how much energy is going to be received at the one you
"catch".

Again, let's say an astronaut shines a flashlight out the window. (Energy
is consumed from the batteries, which lose a microscopic increment of mass
as a result; there's our dM.) An observer sees that the beam appears to
come out traveling perpendicular to the line of flight ... as measured by
the observer.

Surely you're not saying that's impossible -- that would be a really
absurd argument. Just sweep the beam from back to front; at some point
it'll look to the observer like it's at 90 degrees to the line of flight!

Note that we do not need a "second photon" to conserve momentum; the
ship just recoils. That's where the balancing momentum goes.
Keep in mind that flashlights exist, and real astronauts can use them.
They do _not_ have to shine a separate beam out the back window to get the
momentum equation to balance. :-)

In the two-photon scenario, I think you may also have overlooked
something: If the two photons each come out at 90 degrees to the path in
the EMITTER's frame, they _won't_ appear that way to the observer. To get
them to each come out 90 degrees "off the bow", 180 degrees to each other,
in the observer's frame, you need to have them angled to the back in the
emitter's frame. And then their 3-momentum won't cancel in the emitter's
frame; to make it work you need to carry away the extra momentum with
something -- a spaceship, say -- and that object will speed up.

But finally, if you absolutely insist that flashlights must emit even
numbers of photons, due to spin conservation, I will say "Heck I didn't
know that" and we can use two photons instead of 1. But that changes very
little; it just leaves us with an extra factor of 2 in the photon count
and factor of 1/2 in the per-photon energy, which we can ignore because
they cancel. In any case that's a QM issue which has nothing to do
with this problem -- we don't require QM to solve this beyond the basic
h*f formula for the energy of a photon.


Your mathematical gyrations are nice, but I think you are missing
stuff, as I said before.

Let's go over the angles one last time, because I think this is part of
the problem in picturing what's going on. (And then I will stop repeating
myself.)

In the SHIP'S frame the photon is emitted at angle from line of flight of:

pi/2 + arctan (v/sqrt(1 - v^2))

In the OBSERVER's frame that looks like 90 degrees flat. And that's at
the heart of the counterintuitive recoil behavior.

Work it through for yourself if you don't believe it. The only
"mathematical gyration" I used is the Lorentz transform, which isn't much
a gyration. :-)

And then, if you still think I'm missing something, please try to
determine what else it might be.


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