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From: Sue... on 8 Jun 2010 05:48
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.)
>
> Bounce a ball off a wall. KE_in = KE_out. No loss of KE. Change in
> momentum = 2 * momentum_in.

Is that ~consistent~ with the wikipedia page?
http://en.wikipedia.org/wiki/Radiation_pressure#Theory

If absorbed, the traversed volume is illuminated
once.

If reflected, the traversed volume is illuminated
twice.

Sue...


From: Timo Nieminen on 8 Jun 2010 06:26
On Jun 8, 1:38 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote:
> On Jun 7, 3:55 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
> > On Jun 7, 11:43 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote:
>
> > > This really suggests that what is regarded as radiation pressure might
> > > better be interpreted as photon momentum, whereas this figure that we
> > > are computing above is a fraud.
>
> > > > > being out in the sun all day. Rather than getting pushed over by the
> > > > > sunlight we're getting heated by it instead. There is no photon
> > > > > momentum in a literal sense.
>
> > > > There's a reason why I suggested the calculation:
>
> > > > > > Do the calculation, using Newtonian mechanics, for a stream of Newtonian
> > > > > > particles moving at c, with kinetic energy (for each particle) equal to hf
> > > > > > for, say, 500nm light, and total KE per second in the stream of 1300W.
>
> > > > Try this! Don't do this for real photons, do this for Newtonian particles
> > > > with the same kinetic energy as the photons.
>
> > > I understand that the
> > >    K = m v v / 2
> > > does not hold within relativity, and that there is another term, but
> > > on top of this we will not get any sense at c. Why move the chase over
> > > here?
>
> > Your main complaint with the result above is that you have difficulty
> > reconciling the large energy with the small momentum, the large power
> > with the small force.
>
> No, not at all. As I just stated in the 'twisted' paragraph it is that
> all of these figures are the same energy. To represent them as
> distinct things is breaking down under our careful analysis, which
> from my side was to attempt to isolate them.

Please, be more clear.

Is it OK for something (for example, a wave, a classical particle, or
a photon) to have a lot of energy and only a little momentum?

Is it OK for something (for example, a wave, a classical particle, or
a photon) to have a little energy and a lot of momentum?

Yes, for a given object, if the momentum is higher, the (kinetic)
energy is higher. Also, if the (kinetic) energy is higher, the
momentum is higher. This still doesn't mean that energy and momentum
are identical.

[moved]
> > The energy is still all there. We haven't
> > magicced the energy into becoming momentum; momentum and energy are
> > two different things.
>
> Well, to me this is becoming the most interesting part of the
> analysis.
> On the one hand the math we used to generate the photon momentum takes
> into account all of the photon's energy, yet upon absorbing the photon
> we do not see this come out as mechanical energy. This math actually
> does match experiment as you substantiate elsewhere, and so we should
> accept this one way paradigm, but this is not a completed theory then,
> for no discussion like this has been undertaken.
>
> To reverse the math we simply admit that when a photon hits a cold
> black surface that it's energy is transferred into that surface.
> Knowing the momentum of the photon we simple backtrack the math and
> get a small energy figure, and based on our earlier figures we will
> find 1300 watts of energy was absorbed. Because we put all of this
> energy in as momentum it follows that within the photon this became
> kinetic energy, and that the surface therefor receives 1300 watts of
> mechanical work. This is where you should falsify I believe.
[moved]
> I am so confused as to why you would deny any connection between
> momentum and energy, and it is clear that to uphold existing theory we
> cannot admit the 1300 watts of mechanical work, otherwise we'll really
> have high power solar mechanics going on, yet this is what the math
> would do. There must be a glitch imo.

[Big cut for brevity; here is the core issue for the moment.]

We can transfer - whether using classical particles, classical waves,
or quantum particles - momentum without transferring energy. You can
transfer energy without converting it all to work; that is, you can
transfer energy without it becoming KE. This works largely in the same
way in all three cases.

Here is where it is worth sitting down and looking at these a little
systematically. (It can also be worthwhile putting some numbers in and
doing actual calculations; feel free to try, using a 500nm photon, a
classical particle of the same KE and speed (v=c), and a wave of power
1300W.)

Let us assume Newtonian mechanics, the energy of a photon E=hf,
momentum of a photon E=h/lambda. Let the aim be to (a) look at the
similarities between the 3 cases (classical particle, quantum
particle, classical wave), and (b) determine the momentum of a wave,
and consequent radiation pressure on reflection and absorption.
Galileian transformation between moving coordinate systems (so not
using special relativity).

I don't know your background knowledge, so let me speak about
momentum, force, and impulse first. Consider Newton's 2nd law: the
force applied to an object is the rate of change of momentum of the
object. With Newton's 3rd law, this means that force is the rate of
change of momentum. So, a force is a rate of transfer of momentum, and
is convenient when we're transferring momentum continuously. Not so
convenient to consider effectively instantaneous impacts. For this,
the impulse is better. Impulse is commonly defined as force * time
(for a constant force), but it's better to think of impulse as the
change in momentum, and force as the rate of change of momentum. For a
constant force, then we have force = impulse/time. If we have an
impulse of I per particle, and N particles per second, then F=I*N.
Similarly, if the transfer of energy per particle is E, the power is
E*N.

1a. Reflection by a stationary reflector.

(i) Consider a particle of mass m and speed v being reflected with no
loss of KE from a stationary plate (for the refletor to remain
stationary, there will need to be another force acting to hold it in
place). There is no change in energy, and no work is done on the
plate. But the momentum of the particle is reversed. This change in
the momentum, equal to double the particle momentum, will be the
impulse applied to the plate.

Thus,
KE = E = (1/2) mv^2,
p = mv = 2E/v,
I = 2mv = 4E/v.
With N particles per second, F = 4EN/v.

No work is done, the energy transferred per particle is zero, and the
power is zero.

(ii) A photon. As above, but we have:

E = hf,
p=h/lambda = E/c (similar to the energy/momentum relationship for a
particle, but without the factor of 2).
I = 2*E/c,
F = 2EN/c.

(iii) For a classical wave of power P, we have no work being done. If
we don't know the energy/momentum relation for a classical wave, we
don't know anything more, yet.

1b. Reflection by a moving reflector. Consider the above, as seen from
a moving coordinate system, moving at speed v_cs such that the
incident particle is faster or the incident photon or wave is
blueshifted.

(i) The particle speed is now v_2 = v + v_cs. The mass is unchanged.
The reflected particle is now at speed v_3 = v - v_cs.

E_in = (1/2)m v_2^2 = (1/2)m (v + v_cs)^2 = (1/2)m (v^2 + v_cs^2 +
2*v*v_cs)
E_out = (1/2)m v_3^2 = (1/2)m (v - v_cs)^2 = (1/2)m (v^2 + v_cs^2 -
2*v*v_cs)

E_in - E_out = 2m*v*v_cs.

By conservation of energy, this energy is transferred to the
reflector. In 1a(i), no energy was transferred to the reflector, and
therefore there was no heating of the reflector. Looking at this from
a different (moving) coordinate system doesn't change this; there is
still no heating. Therefore, the transferred energy must be the work
done on the reflector by the particle. (The restraining force stopping
the reflector from accelerating does work of the same magnitude, but
negative, so the total work on the reflector is zero, as expected,
since the KE of the reflector doesn't change.)

W = 2m*v*v_cs
P = 2m*v*v_cs * N

Since P = Fv,

F = P/v = W*N/v_cs = 2mvN = 4EN/v, where E is the original energy in
1a(i); this is exactly the same force as before. This is as expected,
since the change of coordinate system won't affect the force.

The change is momentum is still the same as before (the v_cs parts of
the momentum in and momentum out cancel), so from conservation of
momentum, we again have F = 4EN/v.

(ii) Photon.

E_in = hf(1+v_cs/c),
E_out = hf(1-v_cs/c),

E_in - E_out = 2hf v_cs/c. As for the particle case, there must still
be no heating.
P = 2hf N v_cs/c = 2EN v_cs/c

F = P/v_cs = 2EN/c, as before. Again, the change in coordinate system
can't have changed this, so this is as expected.

1/lambda changes by the same (1+ v_cs/c) factor due to the change in
coordinate system, the v_cs/c part cancels when we find the change in
momentum, so again,

I = 2*E/c,
F = 2EN/c.

(iii) Classical wave of (original) speed v_w and original power P0.

The power is proportional to the frequency. So,

P_in = P0 (1+v_cs/v_w)
P_out = P0 (1-v_cs/v_w)

P_in - P_out = 2P0 v_cs/v_w.

In 1a(iii), there was no absorption, and therefore no heating. The
change in coordinate system doesn't change that, so there is still no
heating. Thus, this is the rate of doing work on the reflector.

F = P/v = 2P0 v_cs/v_w = 2P0/v_w.

From conservation of momentum, the momentum flux of the original wave
(in the coordinate system where the reflector was at rest) is

p_flux = P0/v_w.

OK, that's enough for the moment. Note that the photon and wave case
are identical, apart from the stream of N photons per second coming in
discrete chunks, and the classical wave being continuous. Note that
the classical particle result is the same except for a factor of 2.
Further note that, in all 6 cases, there was no _absorption_ of
energy. That is, no energy went towards heating the reflector.

Let me know what you think.

Next will be: 2a. Absorption by a stationary absorber, and 2b.
Absorption by a moving absorber. If you want, you can try this one for
yourself. In all 6 cases, there will be heating; changing to the
moving coordinate system won't change the rate of heating. The change
in momentum of the particles/photons/wave will be approximately half.
For v_cs that is small compared to v, c, and v_w, the work done on the
absorber will be much small than the energy that heats the absorber.

There is a surprising result here! For the stationary absorber, the
change in momentum is exactly half that for the reflector. So, the
force is force is exactly half of the reflector force. The change in
coordinate system to the moving one can't change this force, but the
change in momentum in the moving coordinate system is only
approximately half, not exactly half (it's a little bit more than
1/2). If the extra momentum or momentum flux doesn't give us extra
force, what happens to this extra momentum. Answer: the absorbed
energy, now thermal energy, must have momentum when its moving!

OK, a few more quick points to reply to:

> > (3) Conservation of energy and momentum gives a prediction of
> > radiation pressure.
>
> Nah. Photon momentum is radiation pressure. Radiation pressure is a
> misnomer.

Not a misnomer. Newton 2: force = rate of change of momentum. If
photons don't have momentum, they can't produce a force. If photons
can produce a force, they must carry momentum. (Electromagnetic)
radiation pressure is just the effect of changing the momentum of
photons (or a classical EM wave), by reflection, refraction, or
absorption.

> I was just reading a wiki page on momentum and they claim that static
> electric fields have momentum. Do you balk at this? At this point I
> have no idea what to really believe. This is all a wake up call for
> me.

I don't trust it, the statement that static EM fields have momentum.
What we get out of the conservation laws is the Maxwell stress tensor
(or the 4x4 energy-momentum tensor, if we're doing a full relativistic
treatment; the Maxwell stress tensor is a 3x3 chunk of this). What is
meaningful then is either the divergence of the stress tensor (which
tells you the rate of change of momentum density), or its integral
over a closed surface (which tells you the rate of change of momentum
in the enclosed volume). It doesn't directly tell you the momentum
density.

There is similar funny business with the Poynting vector for certain
static fields. But again, it's the divergence of the Poynting vector,
or its integral, that are physically meaningful.

It's possible to calculate what the momentum density of a static field
should be. Start with no field, which means momentum density of zero,
and calculate the divergence of the stress tensor as the field builds
up to a final value. The integral of this over time should be the
momentum density. At first glance, this gives zero, which suggests
that a static field doesn't have momentum. I haven't checked this
properly.

--
Timo
From: Timo Nieminen on 8 Jun 2010 06:31
On Jun 8, 7:48 pm, "Sue..." <suzysewns...(a)yahoo.com.au> 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.)
>
> > Bounce a ball off a wall. KE_in = KE_out. No loss of KE. Change in
> > momentum = 2 * momentum_in.
>
> Is that ~consistent~ with the wikipedia page?http://en.wikipedia.org/wiki/Radiation_pressure#Theory
>
> If absorbed, the traversed volume is illuminated
> once.
>
> If reflected, the traversed volume is illuminated
> twice.

Sure it's consistent. Energy density with reflection is double the
energy density with absorption, pressure is likewise double. (Not 1/3
since not omnidirectional.)
From: NoEinstein on 8 Jun 2010 06:32
On Jun 7, 3:55 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
>
Dear Timo: The 'velocity' in the momentum equation F = mv is actually
the top half of v / 32.174 feet/sec.—which is a simple unitless
fraction. Momentum is: An increase (or decrease) in the apparent mass
of an object due to its velocity. The 'units' of momentum is pounds.
Power is: Any force which is available to be used continuously, but
which may be used for any length of time. The 'units' of power is
pounds. A pound-second, would be a one pound force utilized for one
second. A KWH is 1,000 watts of power, or the force equivalent,
utilized for one hour. — NoEinstein —
>
> On Jun 7, 11:43 pm, "Tim BandTech.com" <tttppp...(a)yahoo.com> wrote:
>
> > On Jun 7, 1:41 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
> > > On Sun, 6 Jun 2010, Tim BandTech.com wrote:
> > > > On Jun 6, 9:02 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
>
> > > [big cut for brevity, the core seems to be:]
>
> > > [E=hf]
>
> > > > > > This expression takes what I believe
> > > > > > to be the entire energy of the photon and expresses a mechanical
> > > > > > momentum. Thus the 1300 watts of solar power should provide quite some
> > > > > > motive force if the translation is accurate. Thus far we do not see
> > > > > > any such mechanical force,
>
> > > > > Do the actual calculation, and you will see that the force is small. Note
> > > > > the 1/c in your expression; c is a large number.
>
> > > > > > and instead see the ability to absorb a
> > > > > > large amount of this as heat or with less efficiency to turn it into
> > > > > > eletricity with special diodes, called solar cells.
>
> > > > > Just as we would see with the collision of lightweight balls sticking to a
> > > > > massive object.
>
> > > > > Do the calculation, using Newtonian mechanics, for a stream of Newtonian
> > > > > particles moving at c, with kinetic energy (for each particle) equal to hf
> > > > > for, say, 500nm light, and total KE per second in the stream of 1300W.
>
> > > > OK. I just did set it up in a spreadsheet. Each photon at 500nm has
> > > > energy of
> > > >    5.03E-019 Joules
> > > > The total number of photons in one second that will stream is
> > > >    2.58E+021 photons
> > > > The momentum per photon is
> > > >    1.32E-027 kg m / s
> > > > The momentum for one seconds worth is
> > > >    3.42E-006 kg m / s
> > > > I agree that the momentum is small, but this does nothing to diminish
> > > > the power argument and the net energy of this momentum when absorbed.
>
> > > OK, this will do for a start. This _isn't_ the suggested calculation,
> > > but it _is_ a start. Note that E = hf = 3.98e-19J, so you underestimate
> > > photon flux by 20%, and this the momentum by 20%. With the right E, this
> > > brings you back to to the already-mentioned 4 or 9 micronewtons.
>
> > I had a typo on the constant c. Now I am in agreement. Nice to see you
> > are serious.
>
> > > > The momentum figure is operant at the speed of light and damn well
> > > > better be small or we'll all be vaporized. It's tough enough already
> > > > being out in the sun all day. Rather than getting pushed over by the
> > > > sunlight we're getting heated by it instead.
>
> > > Why would 4 or 9 (or 3 or 7) micronewtons push you over? Do you expect to
> > > feel this force?
>
> > Let's see now, I've got a momentum of 4.3E-6 kg m / s
> > but you've got a figure of 4E-6 kg m / s / s .
> > We have a units mismatch on momentum versus newtons,
>
> Momentum per second has the units of force. Force times time is
> impulse, same units as momentum. Per Newton 2, force is the rate of
> transfer of momentum.
>
> > but most
> > importantly each of these figures includes the dimension of kilograms,
> > and these kilograms are travelling at the speed of light, and
> > furthermore since we used the relativistic
> >    e = m c c
> > to get here there is no reason not to use it to get back a usual
> > amount of work, otherwise we are discussing an impossibility, right?
> > The math simply inverts, and we wind back up at the 1300 watt figure,
> > which is expressible in units through
> >    1 J = 1 N m = 1 W s
> > and so I must ask you what happened to all of the energy that we
> > started the calculation with? Where is it?
>
> We didn't do anything with the energy; it's still there.
>
> > We had 1.3 kJ at the
> > beginning of the calculation and now over one second you are claiming
> > to have just something tiny. Well, multiply by c and you'll discover
> > that the energy is all there in that momentum figure.
>
> Momentum _isn't_ energy, energy _isn't_ momentum.
>
> > When does the
> > usage of relativity end? only after the thing travelling at the speed
> > of light is absorbed; not before. Isn't this coherent?
>
> When does the usage of relativity begin? You get the same kind of
> result using Newtonian mechanics and tiny particles.
>
>
>
>
>
> > This really suggests that what is regarded as radiation pressure might
> > better be interpreted as photon momentum, whereas this figure that we
> > are computing above is a fraud.
>
> > > > being out in the sun all day. Rather than getting pushed over by the
> > > > sunlight we're getting heated by it instead. There is no photon
> > > > momentum in a literal sense.
>
> > > There's a reason why I suggested the calculation:
>
> > > > > Do the calculation, using Newtonian mechanics, for a stream of Newtonian
> > > > > particles moving at c, with kinetic energy (for each particle) equal to hf
> > > > > for, say, 500nm light, and total KE per second in the stream of 1300W.
>
> > > Try this! Don't do this for real photons, do this for Newtonian particles
> > > with the same kinetic energy as the photons.
>
> > I understand that the
> >    K = m v v / 2
> > does not hold within relativity, and that there is another term, but
> > on top of this we will not get any sense at c. Why move the chase over
> > here?
>
> Your main complaint with the result above is that you have difficulty
> reconciling the large energy with the small momentum, the large power
> with the small force.
>
> Do the calculation for Newtonian particles, and you get the same type
> of qualitative result (even similar numbers, just different by a
> factor of 2). Do you disbelieve this result?
>
> Note well, the _same kind_ of result. If you photon momentum
> calculation makes you think it's all rubbish, will getting the same
> kind of result with a Newtonian calculation make you think that
> Newtonian mechanics is all rubbish?
>
> Yes, of course (1/2)mv^2 is known to be incorrect at v=c. If this
> bothers you, do the calculation for v=c/2, where Newtonian mechanics
> is a reasonable (even if no longer very accurate) approximation.
>
> > We already are using
> >    E = m c c
> > to get momentum, and that is bad enough. To mix the two even more does
> > not seem appropriate. Already we are considering a Newtonian momentum
> > figure on something that is claimed to break Newtonian mechanics, but
> > then you don't care to perform the reversal do you? Again Timo, there
> > should be a more direct falsification.
> > You ask for yet more
> > construction here and fail to falsify my own argument.
>
> It's been falsified many times already. Once should be enough.
>
> Where is the logic in your claim? Your only basis for it above is that
> a large energy results in a small momentum, so must be wrong. But
> momentum and energy aren't the same thing! Qualitatively, one gets the
> same kind of result from Newtonian mechanics; it shouldn't be
> mysterious (unfamiliar, and offensive to our common sense is OK, since
> this is outside our common everyday experience of mechanics).
>
> The falsifications so far:
>
> (1) Radiation pressure is observed experimentally, for RF, light, and
> acoustic waves at least.
>
> (2) The result of high energy/small momentum is exactly as expected
> from either relativistic or Newtonian mechanics. That energy and
> momentum are different things is also clear in both mechanics. Both
> are also observed to give a good description of reality (Newtonian
> mechanics is it's regime of applicability).
>
> (3) Conservation of energy and momentum gives a prediction of
> radiation pressure.
>
> (4) Direct calculation of the forces gives electromagnetic radiation
> pressure, using classical electromagnetic theory (which is also
> observed to give a good description of reality). In particular, the
> same physics that is used to (successfully) describe electric motors
> etc. results in a prediction of radiation pressure.
>
> The results of (3) and (4) are the same, and the same as (2) using
> relativistic mechanics of zero rest-mass photons. The theoretical
> models agree with experiment (with 10% for Nichols and Hull, 4% for
> Bell and Green, 0.1% for R. V. Jones, within 5% for our own absolute
> measurements).
>
> If you choose to outright reject relativistic and Newtonian mechanics,
> the conservation of energy and momentum, and electric motors, and
> ignore a diverse range of thousands of experiments, what can be said?
> (Other than you _not_ being open to falsification!)
>
> > Your argument is claiming somewhat that the energy has disappeared. It
> > has not, and the math which has generated the argument holds, and its
> > reversal should be considered, since otherwise the conservation of
> > energy will be destroyed.
>
> Of course the energy hasn't disappeared! Momentum isn't energy, energy
> isn't momentum! They're not the same thing!
>
> Seriously, it's time to do the Newtonian calculation (at v=c, or v=c/2
> if you prefer).
>
>
>
> > > > So long as the momentum is derived from the power then there is
> > > > nothing wrong with my own logic. The power is still there, even in
> > > > this slight figure, because momentum is not the same as energy or
> > > > power. It's still 1300 watts, and a one second pulse of this is quite
> > > > some energy, though I was just looking at lasers and most can only
> > > > pulse this kind of power for shorter durations. Energy takes many
> > > > forms, and we see that they can be converted, but this does not mean
> > > > that they actually are converted, though we can still write the
> > > > equation. Somehow the physics culture has come to take the photon
> > > > momentum as fundamental but instead it is a trick, and we may as well
> > > > give them mass from another trick by the same trickster.
>
> > > > Is my logic a trick?
> > > > No, I do not think so. If it is then I do wish you would expose it as
> > > > such via falsification rather than via more construction. We've
> > > > already constructed the things that we need. There is no need for any
> > > > more.
>
> > > Power isn't force, energy isn't force, work isn't force.
>
> > > As far as I can tell, your argument here is solely that 1300W is a lot of
> > > power, and if light has momentum, 1300W of light must have lots of
> > > momentum, and since we're not pushed around (as far as we can notice) by
> > > sunlight,
>
> ...
>
> read more »- Hide quoted text -
>
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From: NoEinstein on 8 Jun 2010 06:40
On Jun 7, 4:04 pm, Sam Wormley <sworml...(a)gmail.com> wrote:
> On 6/7/10 2:47 PM, NoEinstein wrote:
>
Dear Sam: 1. Photons are emitted by all masses whether charged or
not. 2. Photons propogate at 'c' plus or minus the velocity of the
source. 3. Photons are absorbed by all masses whether charged or
not, and such exchange will cause a gravitational attraction between
the masses.

Photons are energy, and thus have no momentum. And, lastly, photon
energy is identical throughout the Universe regardless of the
velocity. Higher energy light merely has the photons spaced closer
together in the trains. — NE —
>
> > I had intended to include the following links that relate to the time
> > of travel of photons around light courses defined by mirrors.  — NE —
>
>    From the quantum mechanical perspective,
>
>      1. photons are emitted (by charged particles)
>      2. photons propagate at c
>      3. photons are absorbed (by charged particles)
>
>    Photon momentum
>      p = hν/c = h/λ
>
>    Photon Energy
>      E = hν

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