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From: NoEinstein on 10 Jun 2010 10:02
On Jun 9, 10:09 pm, Sam Wormley <sworml...(a)gmail.com> wrote:
>
Dear Sam: For someone with no history of making substantive posts,
you rate only as a jealous blow-hard. The thousands reading my words
every day don't do so because Sam Wormley replies. To prove that fact
to you, why don't you make one '+new post' (if you can) and see how
many readers leave me to follow you. You are sad, Sam, very sad.

Oh... The "correction" to Newtonian Physics in GR is to include the
Lorentz transformation "invented" to explain the nil results of the M-
M experiment. The latter gave us rubber rulers and... SR. Strictly
by chance, the math (but not the rationale) behind Lorentz's Beta is a
close analogy to having gravity forces increase, inversely, as the
distance; and to be subject to a "flow" that also increases,
inversely, as the square of the distance. So, without SR (which I've
disproved), there would be no GR (which I correctly define as being
caused by ether flow, NOT by space-time variance.) — NoEinstein —
>
> On 6/9/10 3:23 PM, NoEinstein wrote:
>
> > Dear Sam:  I've disproved SR.  There is no such thing as space-time
> > variance due to velocity or nearness to mass.
>
>    You even confuse SR and GTR! BTW-All you have demonstrated is
>    a lack of education in relativity and other branches of physics.
>    It's all in your posting record.

From: NoEinstein on 10 Jun 2010 10:34
On Jun 10, 7:43 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote:
>
Dear Tim: Again, your science is strong. As an architect, I had to
design bullet resistant enclosures for guard rooms and towers. The
early kinetic energy experiments were inspired to explain why wooden
plank bridges fractured more quickly when a person jumped up and down
in the middle. Those "scientists" made the erroneous assumption that
it must be the extra velocity causing the breaks. Since velocity
increased with the height of the jump, we've been stuck with
Coriolis's 1830 kinetic energy equation: KE = 1/2 mv^2 for much too
long.

What actually causes the planks to break is both a combination of the
velocity, AND the fact all materials respond very poorly to rapidly
applied loads. Hard materials, like steel, readily transfer
compressive stresses inward. When the point of a speeding bullet
touches the steel, the very high compressive stress is transferred in
a narrow cone through the entire thickness of the material. That
means that it is the sudden high pressure, not just the KE, that is
doing most of the damage. Steel could be only 10% as strong when
subjected to sudden loading. A huge compounding factor in strength
reduction is the "drilling" action caused by the spin of the bullet.
Combine the spin with the compressive strength reduction, and the
steel is, like, only .5% of normal strength. Very little of that
"penetrating power" was due to the KE of the bullet!

Suppose the gun weighs 10 pounds and impacts the shooter's shoulder
with a force of 30 pounds. The effective force from the bullet just
fired is only 40 pounds. Now, take that same bullet and push the
point against the steel plate with 40 pounds of pressure (the max
KE). You can't push the bullet into the steel, can you? I have some
ideas how to build bomb shelters strong enough to stop even the US's
bunker bombs. My understanding of what's allowing the penetrations
gives me that ability. — NoEinstein —
>
> On Jun 9, 3:40 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
>
> > On Jun 10, 4:10 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote:
>
> > > On Jun 8, 3:00 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
>
> > > > On Jun 8, 9:34 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote:
>
> > > > > On Jun 8, 2:30 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
>
> > > > > > On Mon, 7 Jun 2010, Tim BandTech.com wrote:
> > > > > > > On Jun 7, 3:55 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
>
> > > > > > [a very quick point, will return later]
>
> > > > > > > > Of course the energy hasn't disappeared! Momentum isn't energy, energy
> > > > > > > > isn't momentum! They're not the same thing!
>
> > > > > > > When the momentum is absorbed is not the energy likewise absorbed?
>
> > > > > > No! (I think "absorbed" is the wrong word here.)
>
> > > > > Is this your new careful use of language Timo? This 'no' is to mean
> > > > > that you wish to discuss the reflected case. Well, I am discussing the
> > > > > absorbed case. So please do not change the argument as a means of
> > > > > falsifying my argument.
>
> > > > So, be clear! We have been discussing the case of reflection as well,
> > > > not just absorption. If you're restricting yourself here to absorption
> > > > only, say so.
>
> > > > In the reflecting case, there can be a transfer of momentum with no
> > > > transfer of energy. You can have transfer of momentum without
> > > > absorbing the photon. If you absorb the photon, yes, you transfer all
> > > > of the momentum and energy. Most of the energy (all of the energy, for
> > > > a stationary absorber) _doesn't_ become work done on the absorber.
>
> > > > This is the next step (2a,b) in the other line of this thread, let us
> > > > deal with it there.
>
> > > > > This 'No' above can be miscontrued. I will
> > > > > take this wording as support of my argument, for there is no actual
> > > > > falsification content provided here.
>
> > > > That's very sloppy. A lack of falsification is not support (for
> > > > example, a complete lack of comment is also a complete lack of
> > > > support).
>
> > > > > > Bounce a ball off a wall. KE_in = KE_out. No loss of KE. Change in
> > > > > > momentum = 2 * momentum_in.
>
> > > I'm almost willing to accept this criticism, but I see that you are
> > > obfuscating the situation more than I am. I asked a simple question
> > > within a position on photon momentum and photon energy, and you
> > > invalidate the question, and then go on to say that I haven't been
> > > clear.
>
> > > Here at least we were discussing one simple point, rather than
> > > branching off into three or more methods. It is a point worth
> > > considering, since the computation that granted the photon momentum
> > > took its entire energy e=hf, or at least this is my best
> > > understanding. It then follows that when that photon is absorbed, that
> > > its entire energy is absorbed. This is merely the reversal of the
> > > equations which got us the momentum in the first place, and if we did
> > > not recover all of the energy, then we have a break with energy
> > > conservation.
>
> > But using the standard theory for photon momentum or radiation
> > momentum, if you know the photon energy, you know the photon momentum.
> > If you know the wave energy flux and speed, you know the momentum
> > flux. If you know the photon momentum, you know the photon energy; if
> > you know the wave momentum flux and speed, you know the wave energy
> > flux.
>
> > But the energy isn't momentum, and you can change the momentum - that
> > is, exert a force - without changing the energy. They are _different_.
>
> > More than that, if we use the energy to calculate the momentum, this
> > is just a _calculation_ that _we're_ doing. This doesn't affect the
> > energy, doesn't affect the momentum. The momentum is there even if we
> > don't calculate it, and the energy is there after we calculate it. We
> > aren't turning energy into momentum, or momentum into energy - to be
> > worried that we wouldn't "recover" the energy and violate conservation
> > of energy is meaningless in this context; we're not doing _anything+
> > to the energy in the _calculation_. If the photon/wave is reflected or
> > absorbed, then the reflection or absorption can do something to the
> > energy. And as you can see from the posted calculation, we use
> > conservation of energy to calculate what happens to the energy, if
> > anything.
>
> > > That this balance of energy is heat suggests within a cannonball model
> > > (which you are so fond of) that an incredible amount of that directed
> > > energy, all of which was used to compute a momentum, has turned to
> > > heat, with such a slight amount becoming kinetic as to be extremely
> > > difficult to notice. If we had, say, a large lead ball hanging on a
> > > string in stasis, and sent a tiny steel ball into the lead ball at
> > > high velocity then we would observe some heating, but too achieve the
> > > level of heating that light achieves will be quite some trick to mimic
> > > in the terms of massive collisions.
>
> > Not at all. Wikipedia tells me that the energy of a typical 5.56mm
> > NATO bullet when fired is 1.7kJ. Shoot them into a massive target at a
> > little under 1 round per second, and you achieve approximately the
> > same heating. (Not the same force as with light! Just the same
> > heating.)
>
> Here again I see your obfuscatory tactic. Firstly you falsify and in
> your conclusion you agree. The amount of heating that light is capable
> of when absorbed versus the work that can be done mechanically due to
> that absorption are remarkable in comparison to your NATO bullet. It
> is a fine argument, except that where you write 'Not at all.' at the
> beginning we may as well replace these words with 'Yes, exactly!'. So
> why am I now spending time here? Even your falsifications are
> affirmations. Your argumentation has become very weak, and as you
> yourself admit the amount of mechanical energy that those NATO bullets
> will provide is appreciable. I will presume that as you discuss
> heating you are comparing to one square meter of sunlight since that
> has been our most used figure. Still, what the NATO bullet's target
> enters as awfully important to the analysis. If it were to hit a
> receptacle capable of not failing struturally which could propen a
> magnet through a coil of wire then we'd be generating electrcity with
> some of its energy. Still, keeping things simple we could simply put
> the target on some on Einstein's railway bed and shoot inline with the
> tracks and start to see the impressive mechanical energy; if we can
> make a light enough target that will not fail. Here we could scale
> down to a Nichols radiometer or a Crooks radiometer, and the
> mechanical energy will be way out of whack with the sunlight. I have
> no idea why I write so much on such a poor refutation.
>
>
>
>
>
>
>
> > > The fact that much of this energy
> > > could have been turned into electricity as well poses an even more
> > > intricate model. Likewise, that we might have chosen to reflect the
> > > energy with so little interaction goes in complete opposition to this
> > > idea of the absorber's collision, and the fact that matter varying
> > > from a sooted up plate to a polished plate could affect such a change
> > > in behavior requires much more dynamics.
>
> > Which means what? Electromagnetically, the polished plate is a good
> > conductor, and presents a very high impedance mis-match to the wave/
> > photon - high reflectivity results. The sooted plate is very different
> > electromagnetically. High enough conductivity to give high losses, but
> > much smaller impedance change, so not so reflective. We can model
> > reflection from metal or soot reasonably well (at least for isolated
> > bits of soot - there are some practical difficulties for a real sooty
> > surface, but an ideal flat smooth sheet of soot would be simple).
>
> > If you care about the dynamics, it can be described,
> > electromagnetically. In either case, you can find the Lorentz force
> > acting on the induced currents in the material, and the total force is
> > equal to the radiation pressure. So, we have a choice, we can either
> > use the details of the dynamics, or use conservation of energy and
> > momentum, and we get the same answer.
>
> > And to comment on something in your reply to Sue: "Well, if the photon
> > momentum is taken to be coming from the entire photon energy, then
> > there is no room for the rotational quality to contain more energy."
> > We're not turning the energy into the momentum, calculating the
> > momentum doesn't get rid of the energy, the energy is still there. It
> > hasn't been used up and become unavailable for turning into angular
> > momentum - especially since we don't turn the energy into angular
> > momentum any more than we turn it into momentum.
>
> > Take a pulse of light with energy A, momentum B, angular momentum C,
> > and combine it with a pulse of energy A, momentum B, angular momentum -
> > C, the final pulse has a total energy of 2A, total momentum 2B, and
> > total angular momentum 0. The final absence of angular momentum
> > doesn't change the energy at all.
>
> This is a poor argument above because you've merely inverted the
> angular momentum. Let's consider two photons
>    A, B
> and grant them the same total energy
>    e(A) = e(B)
> and now go on to consider granting A some angular momentum, and leave
> B's angular momentum at zero:
>    L(A) = a , L(B) = 0 .
> Isn't it fairly obvious that this requires some energy relationship
> and that  A's linear momentum must be less than B's under this
> situation? Doesn't this refute your repetitive argument of isolation
> between momentum and energy?
>
> We can even reuse your NATO bullet argument; for if the barrel of one
> gun is strongly rifled so as to spin the bullet(analogous to A above)
> and another gun is not so rifled but has straight grooving(B above)
> then we must accept that there is a balance of energy based upon how
> equally metered the explosive which propelled each bullet is, and
> grant the work done on each bullet to be identical from stasis; before
> they are fired. You see how carefully I must ...
>
> read more »- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -

From: Tim BandTech.com on 10 Jun 2010 12:57
This is a fascinating story. Thanks for sharing it.
How well it will apply to light is trouble.
If anything the heating effect of light could suggest it is doing
quite some shredding, eh?
In large enough quantity it will shred just about anything.
I had no idea that the kinetic energy formula was from Coriolis.

To me, as we try to take freedoms to find a better theory, we should
realize that as we shuffle things they will hopefully cohere into a
simpler structure. Within the fundamentals there may be multiple
parallel theories treating different behaviors as fundamental. It
seems that radiation pressure and photon momentum are two such
concepts. They do not each deserve fundamental treatment from within
one theory. It is possible that I have badly digested the information,
or digested bad information. These sorts of possibilities are why we
should head for the fundamentals and treat those as open problems,
rather than try to work too much on top of them.

I do accept your experimental evidence, but so much is a matter of
interpretation. We want to know why, and if the answer doesn't make
perfect sense then we will attempt a new answer. Should we eat
particle/wave duality? No. Waves dissipate as they propagate;
particles do not. Under relativity the photon is time dilated such
that in the photon's own frame of reference it passes instantaneously
from source to destination. This is somewhat a zero dimensional
interpretation, and that light behaves as a ray is so simplistic an
understanding that we call it 'ray tracing' when we do simple optics.

The fact that we happily go about projecting lines as entities in
planes and in 3D space suggests that we may likewise impose a zero
dimensional ray. This concept is born from polysign numbers at my
website. Time is P1, and light somehow is nearby in the
interpretation. Still, I'm not claiming these realizations as a
complete theory. They are consistencies. I respect the relative
reference frame, but this alone is sufficient to recover an isotropic
claim. The basis within each frame need not be isotropic. If each
frame is of a structured basis as
P1 P2 P3
or
a11
a21 a22
a31 a32 a33
then we no longer have an isotropic space, but this interpretation
needs to be on a per particle basis, and is consistent with spin
concepts, whereby a unique frame can be applied per particle. We
recover the isotropic sensibility firstly by allowing these frames to
interact relatively, and from our own human sensibility, because we
are conglomerates of these fine instances. This would be a fair
particle theory, or at least a basis for one. This is actually fairly
old information for me, but I share it again hoping to wake something
up in somebody. I suppose it sounds cryptic, but if you study my site
http://bandtechnology.com/PolySigned
it could become less so.

- Tim

On Jun 10, 10:34 am, NoEinstein <noeinst...(a)bellsouth.net> wrote:
> On Jun 10, 7:43 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote:
>
> Dear Tim: Again, your science is strong. As an architect, I had to
> design bullet resistant enclosures for guard rooms and towers. The
> early kinetic energy experiments were inspired to explain why wooden
> plank bridges fractured more quickly when a person jumped up and down
> in the middle. Those "scientists" made the erroneous assumption that
> it must be the extra velocity causing the breaks. Since velocity
> increased with the height of the jump, we've been stuck with
> Coriolis's 1830 kinetic energy equation: KE = 1/2 mv^2 for much too
> long.
>
> What actually causes the planks to break is both a combination of the
> velocity, AND the fact all materials respond very poorly to rapidly
> applied loads. Hard materials, like steel, readily transfer
> compressive stresses inward. When the point of a speeding bullet
> touches the steel, the very high compressive stress is transferred in
> a narrow cone through the entire thickness of the material. That
> means that it is the sudden high pressure, not just the KE, that is
> doing most of the damage. Steel could be only 10% as strong when
> subjected to sudden loading. A huge compounding factor in strength
> reduction is the "drilling" action caused by the spin of the bullet.
> Combine the spin with the compressive strength reduction, and the
> steel is, like, only .5% of normal strength. Very little of that
> "penetrating power" was due to the KE of the bullet!
>
> Suppose the gun weighs 10 pounds and impacts the shooter's shoulder
> with a force of 30 pounds. The effective force from the bullet just
> fired is only 40 pounds. Now, take that same bullet and push the
> point against the steel plate with 40 pounds of pressure (the max
> KE). You can't push the bullet into the steel, can you? I have some
> ideas how to build bomb shelters strong enough to stop even the US's
> bunker bombs. My understanding of what's allowing the penetrations
> gives me that ability. — NoEinstein —
From: Timo Nieminen on 10 Jun 2010 15:19
On Jun 10, 9:43 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote:
> On Jun 9, 3:40 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
> > On Jun 10, 4:10 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote:
>
> > > If we had, say, a large lead ball hanging on a
> > > string in stasis, and sent a tiny steel ball into the lead ball at
> > > high velocity then we would observe some heating, but too achieve the
> > > level of heating that light achieves will be quite some trick to mimic
> > > in the terms of massive collisions.
>
> > Not at all. Wikipedia tells me that the energy of a typical 5.56mm
> > NATO bullet when fired is 1.7kJ. Shoot them into a massive target at a
> > little under 1 round per second, and you achieve approximately the
> > same heating. (Not the same force as with light! Just the same
> > heating.)
>
> Here again I see your obfuscatory tactic. Firstly you falsify and in
> your conclusion you agree.

You were quite specific: "level of heating that light achieves will be
quite some trick to mimic". This level of heating is easy to mimic.

> The amount of heating that light is capable
> of when absorbed versus the work that can be done mechanically due to
> that absorption are remarkable in comparison to your NATO bullet.

So, you want to change "level of heating" to "amount of heating versus
work"? If you meant this in the first place, you weren't clear enough.
To criticise my reply to your actual original words on the basis of
your _changed_ version lies somewhere on the scale from weaseling to
complete bullshit.

If we can't mimic the ratio of heating:work that we get with light
using massive objects, it's because we can't get objects of low enough
mass, some orders of magnitude less massive than an electrons.

But you never specified any numbers in your argument, only that it is
somehow wrong for almost all of the energy to go into heating, and not
work, when the object is "absorbed".

99% is almost all. 99.9% is almost all. 99.99% is almost all. The
Newtonian particle is as offensive to your argument as a photon,
unless there is something magic about some tiny fraction of work/
heating somewhere between what we get with electrons and photons.

> > Take a pulse of light with energy A, momentum B, angular momentum C,
> > and combine it with a pulse of energy A, momentum B, angular momentum -
> > C, the final pulse has a total energy of 2A, total momentum 2B, and
> > total angular momentum 0. The final absence of angular momentum
> > doesn't change the energy at all.
>
> This is a poor argument above because you've merely inverted the
> angular momentum.

The final "zero" isn't an inversion. Energy, momentum, and angular
momentum are all conserved quantities. Is there anything wrong with
the above, other than you not liking it?

> Let's consider two photons
>    A, B
> and grant them the same total energy
>    e(A) = e(B)
> and now go on to consider granting A some angular momentum, and leave
> B's angular momentum at zero:
>    L(A) = a , L(B) = 0 .
> Isn't it fairly obvious that this requires some energy relationship
> and that  A's linear momentum must be less than B's under this
> situation?

No, it isn't obvious.

You're claiming your pre-conceptions as the ultimate truth. Therefore,
doesn't look like you are open to any kind of re-evaluation of your
argument, or any disproof.

Some more could be said about photons, angular momentum, and left- and
right-circularly polarised photons (note: "photon" wasn't consider
above). But, of what use, given your attitudes above? Perhaps you will
be happier discussing science fiction with NoEinstein, who doesn't
believe in KE or momentum, so won't bring these inconvenient issues
into a discussion of radiaton forces.
From: Tim BandTech.com on 10 Jun 2010 18:03
On Jun 10, 3:19 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
> On Jun 10, 9:43 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote:
>
>
>
> > On Jun 9, 3:40 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
> > > On Jun 10, 4:10 am, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote:
>
> > > > If we had, say, a large lead ball hanging on a
> > > > string in stasis, and sent a tiny steel ball into the lead ball at
> > > > high velocity then we would observe some heating, but too achieve the
> > > > level of heating that light achieves will be quite some trick to mimic
> > > > in the terms of massive collisions.
>
> > > Not at all. Wikipedia tells me that the energy of a typical 5.56mm
> > > NATO bullet when fired is 1.7kJ. Shoot them into a massive target at a
> > > little under 1 round per second, and you achieve approximately the
> > > same heating. (Not the same force as with light! Just the same
> > > heating.)
>
> > Here again I see your obfuscatory tactic. Firstly you falsify and in
> > your conclusion you agree.
>
> You were quite specific: "level of heating that light achieves will be
> quite some trick to mimic". This level of heating is easy to mimic.
>
> > The amount of heating that light is capable
> > of when absorbed versus the work that can be done mechanically due to
> > that absorption are remarkable in comparison to your NATO bullet.
>
> So, you want to change "level of heating" to "amount of heating versus
> work"? If you meant this in the first place, you weren't clear enough.
> To criticise my reply to your actual original words on the basis of
> your _changed_ version lies somewhere on the scale from weaseling to
> complete bullshit.

I'll have to own here that I should have used the word 'relative'
within the context, but I see it is fairly easy to interperet since
the context of the whole argument is still present. Hell, I can match
the heat of sunlight rubbing some steel on a rock. The lead of the
bullet will melt on impact. Your interpretation of my writing is
clearly not coherent at many levels. I am still in the context of
photon energy and photon momentum; the qualities which you say are
different, and yet which mathematically we've come to see are tied in
exactitude together, and now we are considering an additional
energetic attribute of angular momentum. Now, it seems to me that the
angular momentum will outstrip the linear momentum by far, and this
should put a hiccup into the photon energy derivation. I'm not
claiming to have resolved this, but I'm ready to get into it, and I
admit I don't have nearly the background that you have. My only real
power is in approaching the problem as an open problem.

>
> If we can't mimic the ratio of heating:work that we get with light
> using massive objects, it's because we can't get objects of low enough
> mass, some orders of magnitude less massive than an electrons.
>
> But you never specified any numbers in your argument, only that it is
> somehow wrong for almost all of the energy to go into heating, and not
> work, when the object is "absorbed".
>
> 99% is almost all. 99.9% is almost all. 99.99% is almost all. The
> Newtonian particle is as offensive to your argument as a photon,
> unless there is something magic about some tiny fraction of work/
> heating somewhere between what we get with electrons and photons.
>
> > > Take a pulse of light with energy A, momentum B, angular momentum C,
> > > and combine it with a pulse of energy A, momentum B, angular momentum -
> > > C, the final pulse has a total energy of 2A, total momentum 2B, and
> > > total angular momentum 0. The final absence of angular momentum
> > > doesn't change the energy at all.
>
> > This is a poor argument above because you've merely inverted the
> > angular momentum.
>
> The final "zero" isn't an inversion. Energy, momentum, and angular
> momentum are all conserved quantities. Is there anything wrong with
> the above, other than you not liking it?

The context of the discussion for me revolves around the photon energy
and how we can come to attribute the photon momentum to the photon
energy without concern for such things as angular momentum. If we
allow for some angular momentum then we should address the photon
energy and it's relation to this new mechanical attribute. This is as
I've already suggested a matter of splitting the energy. Arguing that
it doesn't matter is like arguing that the spin on the bullet of the
NATO gun doesn't matter, which I see you've deleted. I'm not honestly
too keen on the quantum spin and angular momentum. The photon as an
oscillation already allows it to be rotational inherently. I am
looking at this as the major energetic component of the photon; not
the momentum, since the ability to capture the photon energy as
something other than heat does actually exist.

- Tim

>
> > Let's consider two photons
> > A, B
> > and grant them the same total energy
> > e(A) = e(B)
> > and now go on to consider granting A some angular momentum, and leave
> > B's angular momentum at zero:
> > L(A) = a , L(B) = 0 .
> > Isn't it fairly obvious that this requires some energy relationship
> > and that A's linear momentum must be less than B's under this
> > situation?
>
> No, it isn't obvious.
>
> You're claiming your pre-conceptions as the ultimate truth. Therefore,
> doesn't look like you are open to any kind of re-evaluation of your
> argument, or any disproof.
>
> Some more could be said about photons, angular momentum, and left- and
> right-circularly polarised photons (note: "photon" wasn't consider
> above). But, of what use, given your attitudes above? Perhaps you will
> be happier discussing science fiction with NoEinstein, who doesn't
> believe in KE or momentum, so won't bring these inconvenient issues
> into a discussion of radiaton forces.

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