Radiation Pressure of Light

[WalterOrlov]

Hello!

Tony Rothman and Stephen Boughn discuss the difficulties in explaining
the radiation pressure: http://arxiv.org/abs/0807.1310

But the principle self is wrong. I contend that the mechanism can not
work at all. Due to the displacement of electrons through electric
field of light in the solid body the inner potentials are formed that
compensate the electric field of light.Therefore the electrons do not
move further inertional, but they are stopped and Lorentz-force is
equal to zero. My calculation:

http://sites.google.com/site/testsofphysicaltheories/English/radiation-pressure


Walter Orlov

[be...@iwaynet.net]

Interesting calculation. Too bad it's wrong. Not wrong in the
mathematical sense but wrong in that radiation pressure is demonstrated
to exist and can lift a glass bead on a light beam.

And if you talk to Treebert, you may also learn that it is the phenomena
of "radiation pressure" that explains gravity as well.

[WalterOrlov]

On 7 Feb., 09:35, "BJAC...@teranews.com" <b...@iwaynet.net> wrote:

> Interesting calculation. Too bad it's wrong. Not wrong in the
> mathematical sense but wrong in that radiation pressure is demonstrated
> to exist and can lift a glass bead on a light beam.

Of course, the radiation pressure of light exists. But the physicists
can not explain it properly.

[be...@iwaynet.net]

But a calculation that shows it can't exist certainly doesn't qualify as
a "proper" explanation, does it?

[WalterOrlov]

This calculation simply shows that the standart explanation is false.
Therefore we must seek proper interpretation, i.e. this chapter is not
yet complete.

[be...@iwaynet.net]

Ok. But instead of doing the Cat in the Hat and crossing out places it
is not, what is your idea FOR the "proper" explanation?

The idea of photon momentum transferring to matter is simple. And I'd
note that we already know that light is not electromagnetic radiation
even though it follows such laws in the macroscopic case. Hence to try
to apply Maxwellian principles at the microscopic scale isn't likely to
work (as you showed).

The real question then would be how exactly does the momentum of the
photon transfer to inert matter? Mechanical collisions are a comforting
thought but also highly unlikely in reality. But to answer the question
one would first have to discover exactly what a photon actually was.

[Jos Bergervoet]

On 2/8/2012 9:54 AM, BJA...@teranews.com wrote:
> On 2/8/2012 3:27 AM, WalterOrlov wrote:

..

>> This calculation simply shows that the standart explanation is false.
>> Therefore we must seek proper interpretation, i.e. this chapter is not
>> yet complete.
>
> Ok. But instead of doing the Cat in the Hat and crossing out places it
> is not, what is your idea FOR the "proper" explanation?
>
> The idea of photon momentum transferring to matter is simple. And I'd
> note that we already know that light is not electromagnetic radiation
> even though it follows such laws in the macroscopic case.

Where did you get the idea light is not EM radiation?

> Hence to try
> to apply Maxwellian principles at the microscopic scale isn't likely to
> work

Since when has it stopped working? Light has a
wavelength of c. 500nm and the microscopic structure
(atomic distances) is about 100 times smaller. And
therefore it still works the same as radiowaves.

> The real question then would be how exactly does the momentum of the
> photon transfer to inert matter? Mechanical collisions are a comforting
> thought but also highly unlikely in reality. But to answer the question
> one would first have to discover exactly what a photon actually was.

Why not use the Lorentz force of the incoming B
field on the current created by the incoming E-field?
That works for long waves and light is still a
long wave compared with atoms. Is it once again some
strange new theory of Jefimenko that forces you to
do otherwise? :-)

--
Jos

[WalterOrlov]

On 8 Feb., 12:35, Jos Bergervoet <jos.bergerv...@xs4all.nl> wrote:

> Why not use the Lorentz force of the incoming B
> field on the current created by the incoming E-field?

Because these forces will compensate each other every quarter period.
Around that it goes on my website.

[Jos Bergervoet]

Then you made an error. The correct computation can be found
in many EM books. The first order effect does indeed vanish, but
the second order effect does not (so the force is proportional
to the square of the wave amplitude!)

--
Jos

[WalterOrlov]

On 8 Feb., 19:32, Jos Bergervoet <jos.bergerv...@xs4all.nl> wrote:

> Then you made an error. The correct computation can be found
> in many EM books. The first order effect does indeed vanish, but
> the second order effect does not (so the force is proportional
> to the square of the wave amplitude!)
>

This is something new for me... Perhaps confuses you a force with the
Poyting vector?

[Benj]

On Feb 8, 6:35 am, Jos Bergervoet <jos.bergerv...@xs4all.nl> wrote:

> On 2/8/2012 9:54 AM, BJAC...@teranews.com wrote:

> Since when has it stopped working? Light has a
> wavelength of c. 500nm and the microscopic structure
> (atomic distances) is about 100 times smaller. And
> therefore it still works the same as radiowaves.

So you know the internal structure of a photons? The photoelectric
energy transfer clearly shows EM waves aren't it. Please enlighten us
with the knowledge we don't possess.

> > The real question then would be how exactly does the momentum of the
> > photon transfer to inert matter? Mechanical collisions are a comforting
> > thought but also highly unlikely in reality. But to answer the question
> > one would first have to discover exactly what a photon actually was.
>
> Why not use the Lorentz force of the incoming B
> field on the current created by the incoming E-field?
> That works for long waves and light is still a
> long wave compared with atoms. Is it once again some
> strange new theory of Jefimenko that forces you to
> do otherwise? :-)

An E field creates a current? Radiation pressure works with insulators
as well as metals. Orbital electrons may not be free to just blast
around as you seem to think. The theory of X-ray diffraction to the
contrary not withstanding. The question isn't wavelength but
intensity. The primary reaction has to be ONE photon with matter. How
does it impart momentum? Even if Maxwell waves can give a "correct"
answer we all know that if you lower the intensity enough there are
photons and not EM waves there. And that fundamental interaction
requires a knowledge of the internal structure of photons and atoms,
both of which are pretty vague at this point. Unless you can give us
that information otherwise all we have is the misapplication of a
known erroneous theory.

And if you'd actually bothered to read what Jefimenko has to say,
you'd know he is on your side in this one. He calculates EVERYTHING by
Maxwell from Gravity to Special Relativity. Actually I just looked
into what Jefimenko had to say just for fun. He just calculates the
radiation force on a object though a consideration of the momentum
light radiation. which of course, ignores all fundamental mechanisms.
He has also an interesting problem which is to calculate how two
spherical dust particles are attracted to each other in an isotropic
light field. Which as I alluded to in an early post is one theory of
gravity. (Feynman-Wheeler)

As to Walter Orlov's calculation, cripes, I hope you don't force us to
go through it in detail to find an error. Why don't YOU do that and
report back to us where he went wrong... :-)

[ti...@physics.uq.edu.au]

That's indeed the case for an atom in a dielectric. Not surprising, since
an atom in a dielectric that doesn't absorb or change the direction, and
hence doesn't change the momentum, of an incident plane wave shouldn't
experience a force.

Keep in mind that "radiation pressure" is _not_ a pressure. It's a force
density. If you didn't expect a zero result for a single dielectric atom
(neglecting re-radiation), then perhaps this is the cause.

The above zero result is not the case for conduction currents.

For an Ohmic conduction current, we have J=sE, where s is the
conductivity, and thus f=JB is the instantaneous force density in the
direction of propagation. So f=sEB. Take the time-average of that, rather
than hand-waving it away in the 3rd last paragraph (clearly, the
time-average of the force density will be proportional to the
time-averaged Poynting vector).

So radiation pressure due to absorption is easy - one gets the
time-averaged volume density of the force in a very straightfoward way.
Radiation pressure on a conductive reflector isn't too hard - don't assume
a perfect conductor, but a good conductor. Integrate the volume force
density in the direction of propagation to obtain the surface force
density. If you want, take the limit as the conductivity becomes very
large (and the reflection approaches complete reflection).

This is sufficient to unambiguously give the momentum flux density of an
electromagnetic wave in free space (known since 1884, courtesy of
Heaviside and Poynting, so hardly new). With that, it's easy to find the
radiation pressure on a dielectric body.

What's harder is to find the radiation pressure on a dielectric surface
(assuming we can treat a surface as a "thing") - else the
Abraham-Minkowski controversy would have ceased to be a controversy a
_long_ time ago. For some recent works with such calculation, see some of
the Abraham-Minkowski relevant work by, e.g., Mansuripur, or Kemp and
Grzegorczyk.

[Jos Bergervoet]

On 2/8/2012 10:52 PM, Benj wrote:
> On Feb 8, 6:35 am, Jos Bergervoet<jos.bergerv...@xs4all.nl> wrote:
>> On 2/8/2012 9:54 AM, BJAC...@teranews.com wrote:
>
>> Since when has it stopped working? Light has a
>> wavelength of c. 500nm and the microscopic structure
>> (atomic distances) is about 100 times smaller. And
>> therefore it still works the same as radiowaves.
>
> So you know the internal structure of a photons?

Not needed, the wavelength of light is still large
enough to treat its effect classically and get the
correct answer. (Read again what I wrote above!)

> Please enlighten us
> with the knowledge we don't possess.

Why would it be good for you to possess it? Some
people prefer fantasy over knowledge. Please first
give proof that you are not one of those. I would
not want to cause an allergic reaction, benj!

>> ... Is it once again some

>> strange new theory of Jefimenko that forces you to
>> do otherwise? :-)
>
> An E field creates a current?

This was discovered by Ohm, not by Jefimenko! But
there is no need for you to believe it.

..

> As to Walter Orlov's calculation, cripes, I hope you don't force us to
> go through it in detail to find an error.

I would advise you to ignore it completely.

> Why don't YOU do that and

I did!

--
Jos

[Jos Bergervoet]

On 2/8/2012 10:06 PM, WalterOrlov wrote:
> On 8 Feb., 19:32, Jos Bergervoet<jos.bergerv...@xs4all.nl> wrote:
>
>> Then you made an error. The correct computation can be found
>> in many EM books. The first order effect does indeed vanish, but
>> the second order effect does not (so the force is proportional
>> to the square of the wave amplitude!)
>>
>
> This is something new for me...

That isn't surprising. If you don't read any books about
what has been done in the past few centuries but instead
only create websites with your own fantasies you know
almost nothing at all about science. Even if you are as
clever as Maxwell, you will in that case only know what
Maxwell would have known if he had not read books of
others and just thought about the matter for a few months.
And that isn't much..

Quick summary: wave hits matter, charges feel E-field
and move proportional to E, Lorentz force is proportional
to movement times B, so proportional to wave amplitude
times wave amplitude, i.e. amplitude squared. Proportional
to the power in the wave (which makes sense, since that
would also be proportional to the number of photons, if
you prefer that description).

The difficult part is to find the movement of the charges.
The E-field gives the force. You need the mass, resistance
and radiative effects that impede the motion to find how
much they move. For one single charge in vacuum you can
use the Abraham-Lorentz self-force approximation. For a
sheet of metal the sheet resistance and vacuum radiation
resistance. (For a perfect conductor only the latter.)

The essence is in the single-electron case: if you only
use the mass you get quarter-period cancellation (as
you perhaps have found, I did not read your webpage).
The Abraham-Lorentz self-force gives additional resistive
force which makes the movement _not_ exactly out of phase
with the E-field and cancellation is removed.

NB: also note the small amount of frequency doubling due
to the eight-shaped movement! (Already occurring without
A-L self-force..)

--
Jos

[WalterOrlov]

On 9 Feb., 02:34, Jos Bergervoet <jos.bergerv...@xs4all.nl> wrote:

> Quick summary: wave hits matter, charges feel E-field
> and move proportional to E, Lorentz force is proportional
> to movement times B, so proportional to wave amplitude
> times wave amplitude, i.e. amplitude squared.

If you interpret it that way - ok. But in the first quarter period the
force e.g. positive and in the next quarter period - negative, because
the direction of motion of the electrons changes, but the direction of
the magnetic field still retains.

[Jos Bergervoet]

I think we agree. And it happens if the velocity is 90 degrees
out of phase with the B-field. Which happens if you assume
frictionless movement of one single moving charge. Corrections:

a) For a single charge, as I already mentioned, the friction
of the radiative self-force (Abraham-Lorentz force) removes the
exact 90 degree phase difference between v and E (and B).

b) For many-body situations: the field from all moving
neighbors adds to the incoming beam and there E and B are
not in phase (it's near-field zone!) This also cancels the
cancellation, even without using the A-L force (which is
not the dominant effect in that case anyway).

c) Friction from Ohmic resistance or other loss mechanisms.

NB: For a perfect conductor c) will not help! You need b).

--
Jos

[WalterOrlov]

On 9 Feb., 12:05, Jos Bergervoet <jos.bergerv...@xs4all.nl> wrote:

> I think we agree. And it happens if the velocity is 90 degrees
> out of phase with the B-field.

Could you explain the reason for that? As far as I know, E-field and B-
field by light are synchronous, therefore they hit the surface of the
solid also simultaneously.

[Jos Bergervoet]

On 2/9/2012 1:21 PM, WalterOrlov wrote:
> On 9 Feb., 12:05, Jos Bergervoet<jos.bergerv...@xs4all.nl> wrote:
>
>> I think we agree. And it happens if the velocity is 90 degrees
>> out of phase with the B-field.
>
> Could you explain the reason for that?

I thought you already saw this (and that it was
the reason for your complaint which started this
thread).

> As far as I know, E-field and B-
> field by light are synchronous,

They are. So if the velocity, v, of the electrons,
is 90 degrees out of phase with E, then it also is
90 degrees out of phase with B. And _if_ v and B are
90 degrees out of phase the net result of the Lorentz
force cancels out. It changes every quarter period.

But the correct analysis shows that the situation
is different, as explained in the part you did not
quote. (Which is good. Interested readers should
retrace the thread to the original unaltered text.)

--
Jos

[WalterOrlov]

On 9 Feb., 14:03, Jos Bergervoet <jos.bergerv...@xs4all.nl> wrote:
> On 2/9/2012 1:21 PM, WalterOrlov wrote:
>
> > On 9 Feb., 12:05, Jos Bergervoet<jos.bergerv...@xs4all.nl>  wrote:
>
> >> I think we agree. And it happens if the velocity is 90 degrees
> >> out of phase with the B-field.
>
> > Could you explain the reason for that?
>
> I thought you already saw this (and that it was
> the reason for your complaint which started this
> thread).

Good. Why do you want enforce first the wrong interpretation?

[Jos Bergervoet]

On 2/9/2012 7:14 PM, WalterOrlov wrote:
> On 9 Feb., 14:03, Jos Bergervoet<jos.bergerv...@xs4all.nl> wrote:
>> On 2/9/2012 1:21 PM, WalterOrlov wrote:
>>
>>> On 9 Feb., 12:05, Jos Bergervoet<jos.bergerv...@xs4all.nl> wrote:
>>
>>>> I think we agree. And it happens if the velocity is 90 degrees
>>>> out of phase with the B-field.
>>
>>> Could you explain the reason for that?
>>
>> I thought you already saw this (and that it was
>> the reason for your complaint which started this
>> thread).
>
> Good.

Why is that good?

> Why do you want enforce first the wrong interpretation?

Why do you think I enforce an interpretation?

--
Jos

[WalterOrlov]

On 9 Feb., 21:52, Jos Bergervoet <jos.bergerv...@xs4all.nl> wrote:

> Why do you think I enforce an interpretation?
>

Those were your words:

"Why not use the Lorentz force of the incoming B
field on the current created by the incoming E-field?

That works for long waves and light is still a
long wave compared with atoms."

But you'd rather forget my question - this is actually irrelevant.

[Jos Bergervoet]

On 2/10/2012 7:47 AM, WalterOrlov wrote:
> On 9 Feb., 21:52, Jos Bergervoet<jos.bergerv...@xs4all.nl> wrote:
>
>> Why do you think I enforce an interpretation?
>>
>
> Those were your words:
>
> "Why not use the Lorentz force of the incoming B
> field on the current created by the incoming E-field?
> That works for long waves and light is still a
> long wave compared with atoms."

That clearly is a question: "Why not.."
So why do you call it an interpretation? Don't
you understand English?

--
Jos

[p.ki...@ic.ac.uk]

Your "Why not" question might not have been "enforcing an interpretation",
but it did read like a rhetorical question being used to (at the very
least) suggest an interpretation.

There is "understanding English", and understanding English: e.g.

"If that's a question, then I'm a Dutchman". :-)


#Paul

Ref: http://dictionary.cambridge.org/dictionary/british/i-m-a-dutchman

[Jos Bergervoet]

On 2/10/2012 4:41 PM, p.ki...@ic.ac.uk wrote:
> Jos Bergervoet<jos.ber...@xs4all.nl> wrote:
>> On 2/10/2012 7:47 AM, WalterOrlov wrote:
>>> On 9 Feb., 21:52, Jos Bergervoet<jos.bergerv...@xs4all.nl> wrote:
>>>
>>>> Why do you think I enforce an interpretation?
>>>
>>> Those were your words:
>>>
>>> "Why not use the Lorentz force of the incoming B
>>> field on the current created by the incoming E-field?
>>> That works for long waves and light is still a
>>> long wave compared with atoms."
>
>> That clearly is a question: "Why not.."
>> So why do you call it an interpretation? Don't
>> you understand English?
>
> Your "Why not" question might not have been "enforcing an interpretation",
> but it did read like a rhetorical question being used to (at the very
> least) suggest an interpretation.

Rhetorical questions might do that. But what interpretation
could it possibly be in this case?! The suggestion (if one
was intended) would be to use the theory of electromagnetics
to compute a force. Why would you (or Walter) call this an
"interpretation"? It's the most basic, unaltered use of the
theory!

> There is "understanding English", and understanding English: e.g.
>
> "If that's a question, then I'm a Dutchman".:-)

But you aren't, are you?! And I was not "interpreting"
anything! I may have been ignoring Walter's interpretation
(I didn't bother to read through it) and that may have
made him angry. But he now only is talking more nonsense!
Just using the theory that we currently have is *not* an
interpretation, not by any interpretation of interpretation
that I can think of..

--
Jos

[Salmon Egg]

In article
<cbcb446d-19b3-423c...@k6g2000vbz.googlegroups.com>,

You might be trying too hard. Forget electrons and and photons.
Radiation pressure id a CLASSICAL phenomenon dependent merely on
concepts well developed in the 19th century before Planck.

Work the model of a plane wave normally incident upon a semi-infinite
medium of conductivity sigma. You can calculate current in the slab as a
function of penetration into the medium. That allows you to calculate
the amount of the incident wave that gets reflected. If you do that
correctly. the total force on the medium is going to be as if the wave
is carrying a mass of power density divided by c^2. moving at a speed c.

For small values of sigma, there is no significant reflection, and if
the medium so that the momentum transfer (per unit area) from the
current set by the wave's electric field and the wave's magnetic field
is going to be (S/c^2)*c = S/c per unit area. S is the Poynting vector.

As sigma increases, the wave reflection approaches totality, and the
momentum exchange doubles.

That can and was derived without knowing anything about relativity,
photons, electrons, or quantum mechanics. It turned out hat when the
more modern ideas were developed, the classical calculations remained
valid.

--

Sam

Conservatives are against Darwinism but for natural selection.
Liberals are for Darwinism but totally against any selection.

[Jos Bergervoet]

Still, it is possible that Walter will now accuse you
of "enforcing an interpretation".

--
Jos

[WalterOrlov]

On 12 Feb., 17:06, Jos Bergervoet <jos.bergerv...@xs4all.nl> wrote:

> Still, it is possible that Walter will now accuse you
> of "enforcing an interpretation".

Please listen to! As the representative of modern science, you have
failed!

Maxwell Maxwell is not the same. And with the radiation pressure he
was mistaken ;)

[WalterOrlov]

On 12 Feb., 12:55, Salmon Egg <Salmon...@sbcglobal.net> wrote:

> As sigma increases, the wave reflection approaches totality, and the
> momentum exchange doubles.
>

For the frequency range of light the sigma is constant. So, this
observation leads to a dead end.

[Szczepan Bialek]

"WalterOrlov" <wor...@yandex.ru> napisal w wiadomosci
news:30dee603-e9b8-4a17...@w9g2000vbv.googlegroups.com...

Maxwell wrote: http://en.wikisource.org/wiki/On_Physical_Lines_of_Force
"Undulations issuing from a centre would, according to the calculations of
Professor Challis, produce an effect similar to attraction in the direction
of the centre; but admitting this to be true, we know that two series of
undulations traversing the same space do not combine into one resultant as
two attractions do, but produce an effect depending on relations of phase as
well as intensity, and if allowed to proceed, they diverge from each other
without any mutual action. In fact the mathematical laws of attractions are
not analogous in any respect to those of undulations, while they have
remarkable analogies with those of currents, of the conduction of heat and
electricity, and of elastic bodies."

What is your opinion on the ""Undulations issuing from a centre produce
an effect similar to attraction in the direction of the centre" ?
S*

[Jos Bergervoet]

On 2/12/2012 6:33 PM, WalterOrlov wrote:
> On 12 Feb., 17:06, Jos Bergervoet<jos.bergerv...@xs4all.nl> wrote:
>
>> Still, it is possible that Walter will now accuse you
>> of "enforcing an interpretation".
>
> Please listen to! As the representative of modern science,

Yes, people should always listen to the representative
of modern science.

> you have failed!
>
> Maxwell Maxwell is not the same.

You think there were two Maxwells? And that they were
different? It might become a similar debate as about
Shakespeare then..

--
Jos

[Jos Bergervoet]

On 2/12/2012 6:49 PM, Szczepan Bialek wrote:
> "WalterOrlov"<wor...@yandex.ru> napisal w wiadomosci
> news:30dee603-e9b8-4a17...@w9g2000vbv.googlegroups.com...
>> On 12 Feb., 17:06, Jos Bergervoet<jos.bergerv...@xs4all.nl> wrote:
>>
>>> Still, it is possible that Walter will now accuse you
>>> of "enforcing an interpretation".
>>
>> Please listen to! As the representative of modern science, you have
>> failed!
>>
>> Maxwell Maxwell is not the same. And with the radiation pressure he
>> was mistaken ;)
>
> Maxwell wrote: http://en.wikisource.org/wiki/On_Physical_Lines_of_Force

Some scholars believe that others actually wrote wikisource.org

(I've also heard rumours here that there were actually two
Maxwells but I think that is just exaggeration..)

--
Jos

[WalterOrlov]

On 12 Feb., 20:42, Jos Bergervoet <jos.bergerv...@xs4all.nl> wrote:
> On 2/12/2012 6:33 PM, WalterOrlov wrote:
>
> > On 12 Feb., 17:06, Jos Bergervoet<jos.bergerv...@xs4all.nl>  wrote:
>
> >> Still, it is possible that Walter will now accuse you
> >> of "enforcing an interpretation".
>
> > Please listen to! As the representative of modern science,
>
> Yes, people should always listen to the representative
> of modern science.
>
> > you have failed!
>
> > Maxwell Maxwell is not the same.
>
> You think there were two Maxwells? And that they were
> different?

Jester

[Szczepan Bialek]

"Jos Bergervoet" <jos.ber...@xs4all.nl> napisal w wiadomosci
news:4f3816ff$0$6965$e4fe...@news2.news.xs4all.nl...

One of them is Heaviside.
Do you know that the "Maxwell equation" are wrote by Heaviside?
S*

[WalterOrlov]

On 12 Feb., 18:49, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:

> What is  your opinion on  the ""Undulations issuing from a centre  produce
> an effect similar to attraction in the direction of the centre" ?

My guess: It is about comparison with mechanical waves in elastic
body: take a rope and attach it to the wall. The other end take in
your hand and swing it - you are attracted to the wall. But that's
just a guess.

[Szczepan Bialek]

"WalterOrlov" <wor...@yandex.ru> napisal w wiadomosci

news:187dbba6-481f-47f9...@l1g2000vbc.googlegroups.com...

In XIX century most of scientists (including Faraday) were opinion that

"Undulations issuing from a centre produce
an effect similar to attraction in the direction of the centre"

An example:
"Wave models
Keller and Boisbaudran
In 1863 F.A.E. and Em. Keller[31] presented a theory by using a Le Sage type
mechanism in combination with longitudinal waves of the aether. They
supposed that those waves are propagating in every direction and losing some
of their momentum after the impact on bodies, so between two bodies the
pressure exerted by the waves is weaker than the pressure around them. In
1869 L. de Boisbaudran[32] presented the same model as Leray (including
absorption and the production of heat etc.), but like Keller he replaced the
particles with longitudinal waves of the aether."

From: http://en.wikipedia.org/wiki/Le_Sage%27s_theory_of_gravitation

But I do not know the details.

S*

[Salmon Egg]

In article
<6b4749c4-188d-4bb0...@hb4g2000vbb.googlegroups.com>,

I not know if I am trying to teach a dead dog a new trick. but here goes
anyway.

I do not know the relevancy of WalterOrlov's reply, but it certainly
factually incorrect. If it were correct, polished Ag, Au, Cu, Al, etc.
would all look the same. Brass, bronze, and many other alloys would look
like Cu. Various Au alloys also could not appear different from one
another.

[WalterOrlov]

On Feb 16, 5:20 pm, Salmon Egg <Salmon...@sbcglobal.net> wrote:

> I do not know the relevancy of  WalterOrlov's reply

Excuse me, I meant delta - the penetration depth. It varies greatly
for radio waves, but It is constant for the light.

[Timo Nieminen]

No, it isn't. For metals, you can make a case that it is
_approximately_ constant, but it is only _approximately_ constant, and
there are conductors other than metals.

If penetration depth/skin depth was constant for light, then
absorptivity would be constant for light. Since we see that it is not
constant across materials (different metals having different
reflectivities) nor constant across frequencies (as Sam already said,
we have coloured metals), it is not constant. Nor would we expect it
to be, theoretically.

Irrelevant anyway, since the original point was to do the radiation
pressure calculation for a finite conductivity, and then find out what
would happen for a perfect conductor by taking the limit as sigma-
>infinity. Which doesn't depend on having real materials with any
particular values of conductivity. Or even on having real perfect
conductors.

Why you bother arguing against a perfectly valid mathematical argument
by making irrelevant and incorrect claims about material properties, I
don't know.

[Salmon Egg]

In article
<3f8d1778-0ae3-4dbb...@z31g2000vbt.googlegroups.com>,

Still incorrect. Penetration depth is jointly and inversely proportional
to the frequency and conductivity. For short wavelengths of light as in
Cu, the conductivity drops. That accounts for the reddish color of Cu.

At some point, the wavelength gets to be so short that the plasma
properties of electron conductivity needs to be considered. Tell us
where in the spectrum that takes place. Meanwhile, I will stick to
classical thoughts until quantum properties are requirered for the
description.

[Jos Bergervoet]

On 2/18/2012 11:43 PM, Timo Nieminen wrote:
> On Feb 19, 6:12 am, WalterOrlov<wor...@yandex.ru> wrote:

...
...

> Why you bother arguing against a perfectly valid mathematical argument
> by making irrelevant and incorrect claims about material properties, I
> don't know.

On the other hand, some of us (occasionally) are arguing against
completely invalid, non-mathematical arguments, by making relevant
and correct claims! The question (why?) seems equally applicable..

--
Jos

[Timo Nieminen]

At least it makes logical sense. The "why?" in that case is
psychological/social, rather than scientific/logical, and if such a
response is futile, it's futile on psychological grounds, not logical
grounds.

A clear demonstration on the non-reversability of the logical "if"!

[WalterOrlov]

On Feb 18, 11:47 pm, Salmon Egg <Salmon...@sbcglobal.net> wrote:

> Tell us
> where in the spectrum that takes place.

http://books.google.de/books?id=nTKG4Mr35j8C&pg=PA234&dq=wellen+Bild+10.10

[WalterOrlov]

On Feb 18, 11:43 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:

> Why you bother arguing against a perfectly valid mathematical argument
> by making irrelevant and incorrect claims about material properties, I
> don't know.

General, in a mechanical sense, Salmon Egg is right. But the mechanism
self is not thereby explained. Sigma explains nothing.

[Jos Bergervoet]

Well, it _does_ explain the radiation pressure if you combine
it with the Lorentz force (unless of course you're a complete
imbecile in mathematics, but since you post from Russia we would
not immediately expect that..)

--
Jos

[WalterOrlov]

Have you still not understand anything?

[Jos Bergervoet]

> Have you still not understand anything?

I'm willing to change my mind, русский друг..

--
Jos

[WalterOrlov]

Yes, it's never too late, mijn vriend..

[Timo Nieminen]

On Feb 20, 2:37 am, WalterOrlov <wor...@yandex.ru> wrote:
> On Feb 18, 11:43 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:
>
> > Why you bother arguing against a perfectly valid mathematical argument
> > by making irrelevant and incorrect claims about material properties, I
> > don't know.
>
> General, in a mechanical sense, Salmon Egg is right.

So why argue about it? Especially incorrectly!

> But the mechanism
> self is not thereby explained. Sigma explains nothing.

As already posted earlier in this thread:

For an Ohmic conduction current, we have J=sE, where s is the
conductivity, and thus f=JB is the instantaneous force density in the
direction of propagation. So f=sEB. Take the time-average of that,
rather
than hand-waving it away in the 3rd last paragraph (clearly, the
time-average of the force density will be proportional to the
time-averaged Poynting vector).

By "nothing", do you mean "everything that was required to be
explained"?

[glen herrmannsfeldt]

Timo Nieminen <ti...@physics.uq.edu.au> wrote:

(snip)

> No, it isn't. For metals, you can make a case that it is
> _approximately_ constant, but it is only _approximately_
> constant, and there are conductors other than metals.

Well, there are complications due to the shape of the
Fermi surface. Find a good solid state physics book and for
some metals they will have a drawing of the shape of the
Fermi surface. That is, the momentum in 3D of the electrons
at the energy between the highest occupied states and lowest
unoccupied states. The colorful metals tend to have very
complicated Fermi surfaces.

> If penetration depth/skin depth was constant for light, then
> absorptivity would be constant for light. Since we see that it is not
> constant across materials (different metals having different
> reflectivities) nor constant across frequencies (as Sam already said,
> we have coloured metals), it is not constant. Nor would we expect it
> to be, theoretically.

Well, it is quantum mechanics in 3D that complicates things.

-- glen

[WalterOrlov]

On Feb 19, 8:57 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:

>
> As already posted earlier in this thread:
>
> For an Ohmic conduction current, we have J=sE, where s is the
> conductivity, and thus f=JB is the instantaneous force density in the
> direction of propagation. So f=sEB. Take the time-average of that,
> rather
> than hand-waving it away in the 3rd last paragraph (clearly, the
> time-average of the force density will be proportional to the
> time-averaged Poynting vector).
>
> By "nothing", do you mean "everything that was required to be
> explained"?

If the current flows, the potentials occur on the surface.If the
current is generated by the E-field of light, it flows only until the
surface potential completely compensated the E-field of light.Then no
current flows, i.e. the velocity of the electrons is equal to zero, v
= 0. According to the definition, the Lorentz force F = q . v x B and
it is equal to zero!

[Timo Nieminen]

Ah, you want to handwave it away! No, you are simply wrong. This can
be seen in two ways.

(1) Qualitatively. If we had J=0, we would have no absorption, and no
reflection, by a conductor. If we had J non-zero only for a negligible
time, we would have negligible absorption and negligible reflection.
Easy to see that polished metals are reflective, and easy to see that
conductors absorb (even polished metal - put a block of aluminium in
sunlight for a while).

(2) Quantitatively. There's an exact solution for the case of a
conducting sphere in a plane wave. Calculate it! For a sub-wavelength
sphere, you find that it doesn't absorb or reflect much at all. A
large sphere does. In the limit of a tiny, tiny sphere (like a metal
nanosphere), your handwaving dismissal of current becomes
approximately correct (never exactly). We also note, experimentally,
that radiation pressure on metal nanospheres is very small. Large
spheres, plenty of radiation pressure.

[Salmon Egg]

In article
<4e303e3f-e295-4612...@l1g2000vbc.googlegroups.com>,

Unfortunately, my German is not up to the task.

BTW, even at a short wavelength, it still likely that classical
mechanics can describe plasma reflectivity. The reddish color of Cu
arises if the electron density in Cu is too low to reflect blue light.
In a way, these electrons in the metal are equivalent to the electron
density in the iohnospher reflectin radio waves.

[WalterOrlov]

On 19 Feb., 22:24, Timo Nieminen <t...@physics.uq.edu.au> wrote:

> Ah, you want to handwave it away! No, you are simply wrong. This can
> be seen in two ways.
>
> (1) Qualitatively. If we had J=0, we would have no absorption, and no
> reflection, by a conductor.

The current exists as long as E-field changes. Therefore, both
reflection and absorption can naturally occur. During this time there
is also the Lorentz force. But it compensates itself every quarter
period, because the direction of the B-field remains the same, but the
direction of the current is changing.

[WalterOrlov]

On 20 Feb., 02:55, Salmon Egg <Salmon...@sbcglobal.net> wrote:
> In article
> <4e303e3f-e295-4612-86d0-d5b56a231...@l1g2000vbc.googlegroups.com>,

>
>  WalterOrlov <wor...@yandex.ru> wrote:
> > On Feb 18, 11:47 pm, Salmon Egg <Salmon...@sbcglobal.net> wrote:
>
> > > Tell us
> > > where in the spectrum that takes place.
>

> >http://books.google.de/books?id=nTKG4Mr35j8C&pg=PA234&dq=wellen+Bild+...

>
> Unfortunately, my German is not up to the task.
>

What should this be difficult? "Kupfer 300°K" is "Copper at room
temperature". 'Lambda' is the penetration depth and 'omega' is the
angular frequency. The angular frequency of light is ~ 10^15. This is
precisely the area (b) on the image.

[Timo Nieminen]

On Feb 20, 6:24 pm, WalterOrlov <wor...@yandex.ru> wrote:
> On 19 Feb., 22:24, Timo Nieminen <t...@physics.uq.edu.au> wrote:
>
> > Ah, you want to handwave it away! No, you are simply wrong. This can
> > be seen in two ways.
>
> > (1) Qualitatively. If we had J=0, we would have no absorption, and no
> > reflection, by a conductor.
>
> The current exists as long as E-field changes.

No. The current exists when E is non-zero. Which it is, almost always.
Ohm's law: J=sigma*E.

> Therefore, both
> reflection and absorption can naturally occur. During this time there
> is also the Lorentz force. But it compensates itself every quarter
> period, because the direction of the B-field remains the same, but the
> direction of the current is changing.

No. Ohm's law: J=sigma*E. J and B are in phase.

Note that for a time-averaged non-zero energy transfer - which is what
absorption is - the time-average of J.E must be non-zero. Since E and
B are in phase, absorption means that the time-average of JxB must
also be non-zero.

The maths is really simple to do. The results agree with experimental
observations of radiation pressure on absorbers. And conductive
reflectors. What is the problem?

[WalterOrlov]

On 20 Feb., 09:46, Timo Nieminen <t...@physics.uq.edu.au> wrote:
> On Feb 20, 6:24 pm, WalterOrlov <wor...@yandex.ru> wrote:

> > The current exists as long as E-field changes.
>
> No. The current exists when E is non-zero. Which it is, almost always.
> Ohm's law: J=sigma*E.
>

And where is this current flowing? If the current flows, takes place
the displacement of the load. This inevitably leads to the formation
of the potentials on the metal surface, which counteract the E-field
of light and compensate it completely, until it changes again.

[Timo Nieminen]

Have you ever played with RC circuits? Have you heard of a "time
constant" in RC circuits? What's the time constant for a sheet of good
conductor? The charge will accumulate on the edges normal to the plane
of polarisation, so approximate as two parallel wires. So C is
approximately pi*epsilon*L/arcosh(W/T) where L is the length normal to
the plane of polarisation, W is the width along it, and T is the
thickness. Resistance will be W/(sigma*L*T). Calculate the time
constant. T should be the penetration depth AKA skin depth. Don't
handwave - do the calculation.

For a small piece of conductor, you can treat a block as a parallel
plate capacitor. Then you find the time constant is epsilon/sigma. For
good conductors, the potential across the block due to accumulated
charge can follow the driving field. Then it is as you describe,
approximately, and there is very little energy transfer, very little
reflection, and very little radiation pressure. As experimentally
known for metal nanospheres. For poor conductors, even for small
objects, the charge moves too slowly, and you get absorption, and
radiation pressure.

[WalterOrlov]

On 20 Feb., 11:11, Timo Nieminen <t...@physics.uq.edu.au> wrote:

> For
> good conductors, the potential across the block due to accumulated
> charge can follow the driving field. Then it is as you describe,
> approximately, and there is very little energy transfer, very little
> reflection, and very little radiation pressure. As experimentally
> known for metal nanospheres. For poor conductors, even for small
> objects, the charge moves too slowly, and you get absorption, and
> radiation pressure.

So, basically, you admit that I'm right. Everything else is just the
evasive maneuver ;)

[Timo Nieminen]

No. Are you dishonest or stupid? Feel free to try to actually discuss
this seriously, if you want to find out where and why you are wrong.
If you're just out to display dishonesty and stupidity, then that is,
as they say, "a valid lifestyle choice", but it doesn't make you right
in any scientific way.

Enjoy your life choices!

[Jos Bergervoet]

On 2/20/2012 9:33 AM, WalterOrlov wrote:
> On 20 Feb., 02:55, Salmon Egg<Salmon...@sbcglobal.net> wrote:
>> WalterOrlov<wor...@yandex.ru> wrote:
>>> On Feb 18, 11:47 pm, Salmon Egg<Salmon...@sbcglobal.net> wrote:
>>
>>>> Tell us
>>>> where in the spectrum that takes place.
>>
>>> http://books.google.de/books?id=nTKG4Mr35j8C&pg=PA234&dq=wellen+Bild+...
>>
>> Unfortunately, my German is not up to the task.
>
> What should this be difficult? "Kupfer 300°K" is "Copper at room
> temperature".

I'm also puzzled by Sam's difficulty with it..
But you, Walter, were obviously wrong with your
remark about sigma. Since delta is constant in
this plot, sigma must decrease with the square
root of the frequency.

> precisely the area (b) on the image.

Precisely! Not that it matters much (and you
already wrote "excuse me" :-) ). But you were
wrong and sigma is not constant! And the real
question in the discussion was of course whether
the _reflectivity_ varies (to explain the color
of copper..) Reflectivity is approximately
proportional to:
delta * sigma
which in this case would give 1/sqrt(omega),
so it could explain the lower reflectivity for
blue light (for Cu and Au, but why Ag is
different then becomes the question.. Do you
have a plot for silver as well?)

--
Jos

[WalterOrlov]

On 20 Feb., 12:13, Timo Nieminen <t...@physics.uq.edu.au> wrote:

> Are you dishonest or stupid? Feel free to try to actually discuss
> this seriously,

I am merely the opinion that we should not simply ignore some physical
processes. They are present and must be considered carefully. But
Maxwell's authority affected your thinking skills - you think just
like in the books and call it a serious. Too bad ...

[Timo Nieminen]

On Feb 20, 10:01 pm, WalterOrlov <wor...@yandex.ru> wrote:
> On 20 Feb., 12:13, Timo Nieminen <t...@physics.uq.edu.au> wrote:
>
> > Are you dishonest or stupid? Feel free to try to actually discuss
> > this seriously,
>
> I am merely the opinion that we should not simply ignore some physical
> processes. They are present and must be considered carefully.

Then why do you ignore them? Why do you refuse to consider them
carefully?

[Jos Bergervoet]

On 2/20/2012 12:13 PM, Timo Nieminen wrote:
> On Feb 20, 8:43 pm, WalterOrlov<wor...@yandex.ru> wrote:
>> On 20 Feb., 11:11, Timo Nieminen<t...@physics.uq.edu.au> wrote:
>>
>>> For
>>> good conductors, the potential across the block due to accumulated
>>> charge can follow the driving field. Then it is as you describe,
>>> approximately, and there is very little energy transfer, very little
>>> reflection, and very little radiation pressure. As experimentally
>>> known for metal nanospheres. For poor conductors, even for small
>>> objects, the charge moves too slowly, and you get absorption, and
>>> radiation pressure.
>>
>> So, basically, you admit that I'm right. Everything else is just the
>> evasive maneuver ;)
>
> No. Are you dishonest or stupid? Feel free to try to actually discuss
> this seriously,

Maybe you and Walter can still discuss the case with
sigma for a long time, but actually the cases without
sigma are much more interesting! (And perhaps we can
all agree on them first?!) There we have frictionless
motion of the charges. Examples:

1) One point-charge in free space.
2) Finite-radius charged particle in free space.
3) Perfectly conducting sheet (sigma=0).
4) Plasma of free moving electrons (of mass m).
5a) Dielectric slab (much thinner than one wave).
5b) Dielectric slab of quarter lambda thickness.
5c) Dielectric slab of half lambda thickness.
5d) Dielectric slab of 78.25*lambda thickness.
5e) Dielectric half-space (ideal, loss-less).

For cases 2 and 3 I already explained it earlier
in this thread. Cases 5a..5e are examples of "the
current exists as long as E-field changes" as
Walter mentioned repeatedly (but for the wrong
examples), so he should be interested!

NB: In all cases here, except 1), 5c), 5e), there
*is* radiation pressure!

--
Jos

[Timo Nieminen]

On Feb 20, 10:12 pm, Jos Bergervoet <jos.bergerv...@xs4all.nl> wrote:
>
> Maybe you and Walter can still discuss the case with
> sigma for a long time, but actually the cases without
> sigma are much more interesting! (And perhaps we can
> all agree on them first?!) There we have frictionless
> motion of the charges. Examples:
>
> 1) One point-charge in free space.
> 2) Finite-radius charged particle in free space.
> 3) Perfectly conducting sheet (sigma=0).
> 4) Plasma of free moving electrons (of mass m).
> 5a) Dielectric slab (much thinner than one wave).
> 5b) Dielectric slab of quarter lambda thickness.
> 5c) Dielectric slab of half lambda thickness.
> 5d) Dielectric slab of 78.25*lambda thickness.
> 5e) Dielectric half-space (ideal, loss-less).
>
> For cases 2 and 3 I already explained it earlier
> in this thread. Cases 5a..5e are examples of "the
> current exists as long as E-field changes" as
> Walter mentioned repeatedly (but for the wrong
> examples), so he should be interested!
>
> NB: In all cases here, except 1), 5c), 5e), there
> *is* radiation pressure!

For 5e) you have radiation pressure. You will also have what we might
call an Abraham-Minkowski force, but that doesn't mean you won't have
any radiation pressure due to the reflection from the interface.

For 1), it depends on how you do your point charge. If you renormalise
so that it has a finite mass, then you'll have acceleration and
radiation, and radiation pressure. Basically the same as 2) in that
case.

3) is done as the limit sigma -> infinity. (Not zero!)

I don't know what the best approach to 5) is. Finite sigma and/or 3)
tells you the momentum of an EM wave in free space, and that's
sufficient (except for the full treatment of 5e)). That's the easiest
way, too, but whether it is the best way is another story. An atomic/
molecular approach works, and gives you 5e) fully as well, but is
somewhat more difficult.

[Jos Bergervoet]

Not on the particles that make up the dielectric.
"There is no radiation pressure on the dielectric"
is a defensible statement in that sense.

In all cases except the three I mentioned there is
radiation pressure force *on the particles.* In
those three there is not..

> For 1), it depends on how you do your point charge. If you renormalise
> so that it has a finite mass, then you'll have acceleration and
> radiation, and radiation pressure. Basically the same as 2) in that
> case.

Of course but then you have to assume a infinitely
increasing *negative* mass! And even then, it would
give unphysical results with runaway solutions. It
would be strange, if my objective is to take the
limit of case 2) for r->0, to assume that the mass
simultaneously becomes infinitely negative.. I might
just as well assume (while I'm busy) that the charge
also becomes zero in the process! But of course it's
all just a choice, for an unphysical situation..

> 3) is done as the limit sigma -> infinity. (Not zero!)

Oops.. I meant rho = 0 of course :-) We agree there
is pressure in that case, it seems..

> I don't know what the best approach to 5) is.

I don't see any problem there. Just use the force on
the charges in the material! (But see below..)

> Finite sigma and/or 3)
> tells you the momentum of an EM wave in free space, and that's
> sufficient (except for the full treatment of 5e)). That's the easiest
> way, too, but whether it is the best way is another story. An atomic/
> molecular approach works, and gives you 5e) fully as well, but is
> somewhat more difficult.

That's very nice, but for 5e) the result then remains
zero! Polarization currents are out of phase with E
while E and B are still in phase with each other (no
standing waves). Therefore current is out of phase
with B and no net Lorentz force on the current.

You get nonzero result only if you somehow consider
the EM field inside the region as part of the object!
Then you get the force at the interface. But why do it?

If, on the other hand, you don't do it, but you add
an infinitesimal loss (very small sigma) then the
force is spread out very deep inside the material
and is felt by the moving charge, which is then not
completely out of phase with the field any more..
So those two pictures are quite different!

If one wants the force on the material particles it
should be my picture. If one wants the abstractly
defined force on a region of space with everything
included that is inside it, then it might be yours!


JB

[Salmon Egg]

In article <SalmonEgg-569E8...@news60.forteinc.com>,

Salmon Egg <Salm...@sbcglobal.net> wrote:

> BTW, even at a short wavelength, it still likely that classical
> mechanics can describe plasma reflectivity. The reddish color of Cu
> arises if the electron density in Cu is too low to reflect blue light.
> In a way, these electrons in the metal are equivalent to the electron
> density in the iohnospher reflectin radio waves.

I was thinking a bit more about electron plasmas and their reflectances.
Oblique incidence on ionospheric layers allows reflection of frequencies
higher than the plasma frequency. Can the same apply to light reflecting
off of copper?

Does anyone know of measurements on polished copper that indicate a more
silvery reflectance as the incident angle increases?

[Jos Bergervoet]

On 2/20/2012 6:43 PM, Salmon Egg wrote:
> In article<SalmonEgg-569E8...@news60.forteinc.com>,
> Salmon Egg<Salm...@sbcglobal.net> wrote:
>
>> BTW, even at a short wavelength, it still likely that classical
>> mechanics can describe plasma reflectivity. The reddish color of Cu
>> arises if the electron density in Cu is too low to reflect blue light.
>> In a way, these electrons in the metal are equivalent to the electron
>> density in the iohnospher reflectin radio waves.
>
> I was thinking a bit more about electron plasmas and their reflectances.
> Oblique incidence on ionospheric layers allows reflection of frequencies
> higher than the plasma frequency. Can the same apply to light reflecting
> off of copper?
>
> Does anyone know of measurements on polished copper that indicate a more
> silvery reflectance as the incident angle increases?

I just tried shining a LED light on a golden watch.
It does seem to be like you say! With a grazing angle
the reflection becomes almost as white as the LED.
At about 70 degrees the color vanishes. But it is not
a reliable and quantitave measurement, of course..

Nevertheless, I repeated it with a Euro-cent coin and
I got the same impression. (This also shows how most
experiments can be done much cheaper if you give it
some thought!) It also works with the reflection of
the window in which I am typing this message on the
laptop screen. At about 70 degrees the incidence the
copper color seems to vanish, in my opinion..


-jrb

[Timo Nieminen]

If you do the calculation properly, one has radiation pressure on the
particles. More below.

> In all cases except the three I mentioned there is
> radiation pressure force *on the particles.* In
> those three there is not..
>
> > For 1), it depends on how you do your point charge. If you renormalise
> > so that it has a finite mass, then you'll have acceleration and
> > radiation, and radiation pressure. Basically the same as 2) in that
> > case.
>
> Of course but then you have to assume a infinitely
> increasing *negative* mass! And even then, it would
> give unphysical results with runaway solutions. It
> would be strange, if my objective is to take the
> limit of case 2) for r->0, to assume that the mass
> simultaneously becomes infinitely negative.. I might
> just as well assume (while I'm busy) that the charge
> also becomes zero in the process! But of course it's
> all just a choice, for an unphysical situation..

Some are quite satisfied by such renormalisation (e.g., see Rohrlich's
book, "Classical charged particles"). You don't renormalise
arbitrarily, but to give the experimentally observed charge and mass
and radius (i.e., zero) of the charged particle in question. IMO, it
isn't an entirely satisfactory procedure. But assuming a finite radius
that gives the correct mass doesn't work either, in the sense of
matching the experimental evidence.

> > I don't know what the best approach to 5) is.
>
> I don't see any problem there. Just use the force on
> the charges in the material! (But see below..)
>
> > Finite sigma and/or 3)
> > tells you the momentum of an EM wave in free space, and that's
> > sufficient (except for the full treatment of 5e)). That's the easiest
> > way, too, but whether it is the best way is another story. An atomic/
> > molecular approach works, and gives you 5e) fully as well, but is
> > somewhat more difficult.
>
> That's very nice, but for 5e) the result then remains
> zero! Polarization currents are out of phase with E
> while E and B are still in phase with each other (no
> standing waves). Therefore current is out of phase
> with B and no net Lorentz force on the current.

Treat the atoms/molecules as dipole. First approximation is that the
dipole moment is in phase with E. This gives no net force. But
reradiation means that the dipole moment will not be in phase, so one
does get a net force on an isolated atom/molecule, proportional to the
rate at which it extracts energy from the incident field and radiates
it as omnidirectionally as possible. It's the dielectric analog of
radiation by an illuminated electron changing it from no force to some
force.

I've not gone back to the original papers on this, only to where it's
used to correct the static polarisability of dielectric spheres (for
calculating radiation forces on small (Rayleigh limit) dielectric
spheres in optical tweezers, and for DDA scattering calculations). But
it can be had from antenna theory. Take a straight wire short dipole
antenna. 'Tis a capacitor. Put a capacitor in an AC circuit and it's
lossless. But an antenna also has its radiation resistance, even in
the absence of Ohmic resistance. So it's RC, not C. And not lossless
as far as the driving circuit goes.

> If one wants the force on the material particles it
> should be my picture. If one wants the abstractly
> defined force on a region of space with everything
> included that is inside it, then it might be yours!

There's another way, too. Don't have a sharp interface. Assume a
refractive index gradient instead. OK to assume all this happens over
d<<lambda. Then you have a standing wave component _and_ dielectric in
the same place.

[Jos Bergervoet]

On 2/20/2012 9:29 PM, Timo Nieminen wrote:
> On Feb 20, 11:43 pm, Jos Bergervoet<jos.bergerv...@xs4all.nl> wrote:
>> On 2/20/2012 1:30 PM, Timo Nieminen wrote:

...

>>> For 5e) you have radiation pressure. You will also have what we might
>>> call an Abraham-Minkowski force, but that doesn't mean you won't have
>>> any radiation pressure due to the reflection from the interface.
>>
>> Not on the particles that make up the dielectric.
>> "There is no radiation pressure on the dielectric"
>> is a defensible statement in that sense.
>
> If you do the calculation properly, one has radiation pressure on the
> particles. More below.

For convenience, once more the list:

1) One point-charge in free space.
2) Finite-radius charged particle in free space.

3) Perfectly conducting sheet (sigma=oo).

4) Plasma of free moving electrons (of mass m).
5a) Dielectric slab (much thinner than one wave).
5b) Dielectric slab of quarter lambda thickness.
5c) Dielectric slab of half lambda thickness.
5d) Dielectric slab of 78.25*lambda thickness.
5e) Dielectric half-space (ideal, loss-less).

..
>>> For 1), it depends on how you do your point charge. If you renormalise
>>> so that it has a finite mass, then you'll have acceleration and
>>> radiation,

..
>> .. But of course it's

>> all just a choice, for an unphysical situation..
>
> Some are quite satisfied by such renormalisation (e.g., see Rohrlich's
> book, "Classical charged particles").

> ..

r=0 gives problems, however you do it, but there's
no reason for taking the limit r=0 anyway. We know
that classical physics based on electromagnetism
alone is not the reality. (It doesn't even agree
with the fact that there are 10 dimensions!)

...

>> That's very nice, but for 5e) the result then remains
>> zero! Polarization currents are out of phase with E
>> while E and B are still in phase with each other (no
>> standing waves). Therefore current is out of phase
>> with B and no net Lorentz force on the current.
>
> Treat the atoms/molecules as dipole. First approximation is that the
> dipole moment is in phase with E. This gives no net force. But
> reradiation means that the dipole moment will not be in phase, so one
> does get a net force on an isolated atom/molecule,

One isolated dipole is an example that was not on
the list! But of course there you get the radiation
pressure with similar reasoning as for a single
charge.

I hope we can also quickly agree that without atomic
structure (i.e. bound charge as a continuum) we do
_not_ get any radiation pressure in case 5e).

But the question here is whether we get it from a
regular grid of polarizable molecules in a filled
half-space. The effect cancels in that case!

> I've not gone back to the original papers on this, only to where it's
> used to correct the static polarisability of dielectric spheres (for
> calculating radiation forces on small (Rayleigh limit) dielectric
> spheres in optical tweezers,

Now that you mention Rayleigh: (thinking about the
blue air) If the polarizable molecules are at random
positions in the dielectric, then I would expect
to get some radiation pressure in case 5e) by the
mechanism you mention. But it would be a very weak
force, related to small random fluctuations of the
coherent wave inside the dielectric.

For the ideal lattice, however, just like for the
continuum, the dielectric in case 5e) feels _no_
radiation pressure. The reason is that there are
no spherical outgoing waves around each dipole.
They all accumulate coherently in the wave that
has entered the dielectric (there is no additional
scattered field in this case). Comparing a few
of the cases now:

5b) Lorentz force explains the radiation pressure
perfectly. No force on the interfaces, a maximum
in the center of the slab. Net force as expected.

5c) Lorentz force explains it perfectly. No force
on the interfaces, a "pulling" force maximum in
the center of the first quarter lambda and an
equal pushing force in the center of the second
quarter (hence no net radiation pressure!)

5e) No radiation pressure on the charges if we
neglect lattice errors and vibrations (and if we
don't, the force on the interface would still be
negligible compared to the Poynting vector)

...

> There's another way, too. Don't have a sharp interface. Assume a
> refractive index gradient instead. OK to assume all this happens over
> d<<lambda. Then you have a standing wave component _and_ dielectric in
> the same place.

That also is a new example that was not yet in
the list! But I still think 5e) is one of the
most interesting ones..

-jrb

[ti...@physics.uq.edu.au]

The usual concrete reason for taking the r=0 limit is that it makes
structure-dependent terms go away. Otherwise, it's just a choice between
too-large electrons held together by magic and the infinite negative
"bare" mass required by renormalisation, whichever is less offensive.

>>> That's very nice, but for 5e) the result then remains
>>> zero! Polarization currents are out of phase with E
>>> while E and B are still in phase with each other (no
>>> standing waves). Therefore current is out of phase
>>> with B and no net Lorentz force on the current.
>>
>> Treat the atoms/molecules as dipole. First approximation is that the
>> dipole moment is in phase with E. This gives no net force. But
>> reradiation means that the dipole moment will not be in phase, so one
>> does get a net force on an isolated atom/molecule,
>
> One isolated dipole is an example that was not on
> the list! But of course there you get the radiation
> pressure with similar reasoning as for a single
> charge.
>
> I hope we can also quickly agree that without atomic
> structure (i.e. bound charge as a continuum) we do
> _not_ get any radiation pressure in case 5e).
>
> But the question here is whether we get it from a
> regular grid of polarizable molecules in a filled
> half-space. The effect cancels in that case!

I haven't seen the calculation done for a semi-infinite medium (or done it
myself for that case), since that would take infinite computer memory and
time. But for finite objects, it works OK. That is, for many dipoles, with
the dipoles small compared to the wavelength, the results approach the
continuum result.

WHy would the effect cancel? Sure, there will be no force on dipoles
within the medium, where "within" means far enough in so that the mean
field is the macroscopic continuum field. But this won't be true at the
edge.

Gordon, PRA 8, 14-21 (1973) is a good compact starting place.

You take the limit as the thickness of the region of which the change in
refractive index occurs goes to zero. This gives 5e).

[Salmon Egg]

In article <4f428f94$0$6964$e4fe...@news2.news.xs4all.nl>,

Jos Bergervoet <jos.ber...@xs4all.nl> wrote:

> I just tried shining a LED light on a golden watch.
> It does seem to be like you say! With a grazing angle
> the reflection becomes almost as white as the LED.
> At about 70 degrees the color vanishes. But it is not
> a reliable and quantitave measurement, of course..
>
> Nevertheless, I repeated it with a Euro-cent coin and
> I got the same impression. (This also shows how most
> experiments can be done much cheaper if you give it
> some thought!) It also works with the reflection of
> the window in which I am typing this message on the
> laptop screen. At about 70 degrees the incidence the
> copper color seems to vanish, in my opinion..

Very interesting.

I may have been premature dismissing the role of quantum mechanics. I
ran across some references to the band structure of metals affecting
reflectivity. My guess is that after you get the effective masses based
upon QM, you can calculate plasma frequency knowing the density of
electrons. Collision rates with the lattice and phonons will determine
the imaginary component of the refractive index.

[Jos Bergervoet]

On 2/21/2012 2:01 AM, ti...@physics.uq.edu.au wrote:
> On Mon, 20 Feb 2012, Jos Bergervoet wrote:

...

>> But the question here is whether we get it from a
>> regular grid of polarizable molecules in a filled
>> half-space. The effect cancels in that case!
>
> I haven't seen the calculation done for a semi-infinite medium (or done
> it myself for that case), since that would take infinite computer memory
> and time.

If you do it yourself you will not use computer
time! Otherwise I would consider it cheating. :-)

> ... But for finite objects, it works OK. That is, for many

> dipoles, with the dipoles small compared to the wavelength, the results
> approach the continuum result.

Which means: radiation pressure is found as
Lorentz force in the bulk. And for lossless
dielectric it only occurs in a standing wave
pattern. It vanishes for a pure forward wave.
And therefore these forces sum to zero for
multiple half-wavelength thickness. And they
are absent if there is no standing wave.

There is never any "extra" force on the
interfaces! (I wonder where you got this
idea.. There are no delta-functions here.)

> WHy would the effect cancel? Sure, there will be no force on dipoles
> within the medium, where "within" means far enough in so that the mean
> field is the macroscopic continuum field. But this won't be true at the
> edge.

??
A radiation pressure force in an infinitely
thin sheet at the interface is never present! At
least not if we are talking about the Lorentz
force on the charges. There are no other forces
we need to include. And there is no way to give
them a delta-function behavior at the interface!
There is no delta-function singularity in the
current at the interface, and also not in the
B-field. So forget it, Timo! It is impossible.

In addition:

- If there was an interface force it would
corrupt the correct result that you already
get from the bulk forces, for the cases of
half-wave and quarter-wave dielectric slab.

- We never used it to explain the radiation
pressure in the lossy-conductor case, where
the force was clearly a bulk effect in the
skin-depth layer, which can be a deep as you
want for a low sigma. No surface term needed,
indeed not even allowed! For it would corrupt
the result which was already correct.

- Likewise, a slightly lossy dielectric
would give the correct force purely from the
bulk effect in the penetration depth of the
wave (which could be very deep, certainly
not a delta-function!)

>> ...
>>> There's another way, too. Don't have a sharp interface. Assume a
>>> refractive index gradient instead. OK to assume all this happens over
>>> d<<lambda. Then you have a standing wave component _and_ dielectric in
>>> the same place.
>>
>> That also is a new example that was not yet in
>> the list! But I still think 5e) is one of the
>> most interesting ones..
>
> You take the limit as the thickness of the region of which the change in
> refractive index occurs goes to zero. This gives 5e).

There is no reason at all to treat this as
a limiting case since there are no infinities
involved in a step function of the refractive
index. It would be as unnecessary as computing
2+2 as a limiting case..

There are no delta-functions here, Timo.

-jrb

[Jos Bergervoet]

That is for normal incidence (I forgot to make
this clear). For other angles there is a surface
charge because there is a normal E-field and it
moves over the surface with the wave pattern, so
this also creates a surface current. But as the
angle approaches normal incidence this vanishes
and only bulk forces are then possible.

-jrb (sorry for responding to my own message..)

[Timo Nieminen]

On Feb 21, 8:42 pm, Jos Bergervoet <jos.bergerv...@xs4all.nl> wrote:

> On 2/21/2012 2:01 AM, t...@physics.uq.edu.au wrote:
>
> > On Mon, 20 Feb 2012, Jos Bergervoet wrote:
>   ...
> >> But the question here is whether we get it from a
> >> regular grid of polarizable molecules in a filled
> >> half-space. The effect cancels in that case!
>
> > I haven't seen the calculation done for a semi-infinite medium (or done
> > it myself for that case), since that would take infinite computer memory
> > and time.
>
> If you do it yourself you will not use computer
> time! Otherwise I would consider it cheating. :-)
>
> > ... But for finite objects, it works OK. That is, for many
> > dipoles, with the dipoles small compared to the wavelength, the results
> > approach the continuum result.
>
> Which means: radiation pressure is found as
> Lorentz force in the bulk. And for lossless
> dielectric it only occurs in a standing wave
> pattern. It vanishes for a pure forward wave.
> And therefore these forces sum to zero for
> multiple half-wavelength thickness. And they
> are absent if there is no standing wave.
>
> There is never any "extra" force on the
> interfaces! (I wonder where you got this
> idea.. There are no delta-functions here.)

There is no "interface as a thing"; there is only a mathematical
boundary that you can place, where there are no dipoles on one side.
In this kind of practical dipole approximation, one is no longer
dealing with a continuum, but with microscopic fields. See Gordon (op.
cit) for some of the differences, as far as forces go. (More on ths
right at the end.)

> > WHy would the effect cancel? Sure, there will be no force on dipoles
> > within the medium, where "within" means far enough in so that the mean
> > field is the macroscopic continuum field. But this won't be true at the
> > edge.
>
> ??
> A radiation pressure force in an infinitely
> thin sheet at the interface is never present!

No infinitely thin sheet. No "interface as thing". So, not a valid
objection. Justify it for an atomic/molecular medium. (Apart from
Gordon, this might also be done by de Groot & Suttorp, who do look at
atomic/molecular media and microscopic fields, but I don't have a copy
to check.)

> >>> There's another way, too. Don't have a sharp interface. Assume a
> >>> refractive index gradient instead. OK to assume all this happens over
> >>> d<<lambda. Then you have a standing wave component _and_ dielectric in
> >>> the same place.
>
> >> That also is a new example that was not yet in
> >> the list! But I still think 5e) is one of the
> >> most interesting ones..
>
> > You take the limit as the thickness of the region of which the change in
> > refractive index occurs goes to zero. This gives 5e).
>
> There is no reason at all to treat this as
> a limiting case since there are no infinities
> involved in a step function of the refractive
> index. It would be as unnecessary as computing
> 2+2 as a limiting case..

For a real medium, made up of atoms/molecules, there is no step
function of the refractive index. To assume the change in refractive
index happens over some small non-zero is more realistic than the
sudden change we have in the usual continuum approximation. It's a
better average of what happens if we consider the actual microscopic
fields.

Dipole approximations as discussed here usually use dipoles much
larger than atoms/molecules, to avoid overstraining ourpoor feeble
computers. But since the wavelength provides a scale that we should be
safe if much-smaller-than, this works in practice, and in theory.

[Jos Bergervoet]

On 2/21/2012 9:15 PM, Timo Nieminen wrote:
> On Feb 21, 8:42 pm, Jos Bergervoet<jos.bergerv...@xs4all.nl> wrote:

...
...

> There is no "interface as a thing"; there is only a mathematical
> boundary that you can place, where there are no dipoles on one side.

Let's again look at the other side where the dielectric is.
Everything works fine if we add an infinitesimal dissipation
in the dielectric. The polarization current then has a tiny
in-phase component with the B-field and the time-averaged
bulk Lorentz force gives the correct radiation pressure.
No extra interface force is present. It is just the very
weak bulk force in a very large volume (the penetration
depth following from this very small dissipation).

If we go from infinitesimal losses to zero losses we get
the limit of zero force over an infinite volume. Asking for
trouble, of course, but that does NOT mean that suddenly
a completely different mechanism will pop up to give us an
interface force that wasn't there earlier!

For a better way to resolve it (and in the process find out
whether Abraham was right or Minkowsky) see below..

...

> For a real medium, made up of atoms/molecules, there is no step
> function of the refractive index. To assume the change in refractive
> index happens over some small non-zero is more realistic than the
> sudden change we have in the usual continuum approximation. It's a
> better average of what happens if we consider the actual microscopic
> fields.

It doesn't change the outcome for normal incidence of a
wave on loss-less dielectric. Once you make the transition
layer thin w.r.t. the wavelength the result smoothly goes
back to the stepped-index result. The latter already has
all fields continuous at the interface.

The best resolution is to look at the wavefront! Suppose it
enters at t=0, then in the medium the wave has shape:

sin(wt-kx) theta(wt-kx)

where theta is the Heaviside step function. The bulk
Lorentz force then will have this shape:

F ~ B(x,t) * dE(x,t)/dt
~ sin(wt-kx) cos(wt-kx) theta(wt-kx)
~ sin(2wt-2kx) theta(wt-kx)

Indeed at each point its time-average is zero. But as the
wavefront passes, the force always starts with a _positive_
sign, and subsequently averages out to 0. If we look at the
total force over the region already filled by the wave:

integrate(F dx) = sin^2(wt) theta(t)

A non-zero result! So it is the first quarter wavelength
of the penetrating wave train, somewhere deep inside the
medium, that gives a Lorentz force equal to the radiation
pressure. And again, no interface force should be added.
(and none is present of course). Physically, this means
that where the wavefront passes, some positive momentum
is locally transferred to the crystal by the Lorentz force
and then the local momentum keeps on oscillating between
this positive value and zero. For the total momentum in
the crystal we have to integrate the total force over time
and this gives:

integrate(F dx dt) ~ (t/2 - sin(2wt)/4w) theta(t)

Apart from a small oscillating term it is linearly rising
with time, as it should! So now we know the momentum that
is given to the crystal, which means we also know how much
is still in the EM field! So, after a century of doubt,
finally the Abraham-Mikowsky controversy has been resolved.
(Unless one disagrees..)

NB: Any other, possibly non-sinusoidal, wave train has the
same property of always starting with a positive force and
giving a positive integral over the medium. Basically
because integral(f(x) f'(x) dx) = 1/2 [f(x)^2 - f(0)^2]
= f(x)^2 if f(0) = 0.

--
Jos

[John Polasek]

Here is an interesting application of radiation pressure: the
adjustment of the acceleration of the Pioneer 10 satellite. It
experienced a mysterious (measured) acceleration back toward the Sun
of 8.74E-10ms^-2

The native acceleration was deduced/measured to be 7.88E -10 m/s
squared, but it was adjusted to 8.74E -10 m/s squared. This change was
attributed to the one-way radiation of its 8 W(!) narrow-cast
transmitter, which supposedly "pushed back from" the Sun by the thrust
of the radiation, thus biassing the value downward.

From first principles I calculated a force for radiation just from the
wattage:
Power equals force times c.
F = 8W/c F = 2.669E-8N M = 310 Kg
dA = 8.74 - 7.88 = .86E-10 m/s squared (by subtraction)
dAcalc = F/M = 0.86E-10 m/ss
These numbers are from my best memory; the 310Kg was 'back calculated'
but I seem to recall it might have been 284Kg (couldnt find the mass
on the web).
There's no proof inherent here; it's simply noting that the radiation
pressure was given 100% credulity. (8W is a nightlight!)
John Polasek

[Timo Nieminen]

On Feb 23, 8:35 pm, Jos Bergervoet <jos.bergerv...@xs4all.nl> wrote:
> On 2/21/2012 9:15 PM, Timo Nieminen wrote:> On Feb 21, 8:42 pm, Jos Bergervoet<jos.bergerv...@xs4all.nl>  wrote:
>
> > There is no "interface as a thing"; there is only a mathematical
> > boundary that you can place, where there are no dipoles on one side.
>
> Let's again look at the other side where the dielectric is.
> Everything works fine if we add an infinitesimal dissipation
> in the dielectric. The polarization current then has a tiny
> in-phase component with the B-field and the time-averaged
> bulk Lorentz force gives the correct radiation pressure.
> No extra interface force is present. It is just the very
> weak bulk force in a very large volume (the penetration
> depth following from this very small dissipation).

I strongly suspect that this won't give the right answer. For a semi-
infinite medium with low reflectivity, this looks like it gives
approximately F=I/c, the force on a perfect absorber.

Given that we _know_ that the reflection isn't sudden at an interface
for an atomic medium, why look further? Especially since the continuum
version of such a distributed reflection model works.

> For a better way to resolve it (and in the process find out
> whether Abraham was right or Minkowsky) see below..
>
>   ...
>
> > For a real medium, made up of atoms/molecules, there is no step
> > function of the refractive index. To assume the change in refractive
> > index happens over some small non-zero is more realistic than the
> > sudden change we have in the usual continuum approximation. It's a
> > better average of what happens if we consider the actual microscopic
> > fields.
>
> It doesn't change the outcome for normal incidence of a
> wave on loss-less dielectric. Once you make the transition
> layer thin w.r.t. the wavelength the result smoothly goes
> back to the stepped-index result.

Yes, that's the point. And we can calculate, without undue difficulty,
the force on the region over which the change in index happens. And we
can see that the total force, integrated over this region remains
constant as we reduce the thickness of this region. So we shouldn't
assume that it suddenly goes to zero as the thickness becomes zero; we
should assume that the transition is smooth.

So, we _can_ assume a short, but continuous transition, and get the
correct answer.

> So, after a century of doubt,
> finally the Abraham-Mikowsky controversy has been resolved.
> (Unless one disagrees..)

The controversy isn't about the total force. Both Abraham and
Minkowski give the same total force, so calculating the force doesn't
resolve anything. There is no physics in Abraham-Minkowski, only the
philosophy of what part of the force to call "electromagnetic" (in the
simple case of a medium completely characterised by a constant
permittivity and permeability, Minkowski says "It's all
electromagnetic!").

An atomic medium strongly suggests that Abraham is correct (or at
least more correct), but that's not new. Again, Gordon 1973. Or our
newest paper on this (Journal of Optics, 13(4), 044017, 2011).

Assuming an atomic medium makes a bigger difference in the angular
momentum controversy, since it (apparently) changes the angular
momentum density. Which means that situations where the "carried at
the edge" picture requires superluminal transport of angular momentum
don't require such in an atomic-medium carried-in-the-bulk picture.

[Jos Bergervoet]

On 2/23/2012 9:26 PM, Timo Nieminen wrote:
> On Feb 23, 8:35 pm, Jos Bergervoet<jos.bergerv...@xs4all.nl> wrote:
>> On 2/21/2012 9:15 PM, Timo Nieminen wrote:> On Feb 21, 8:42 pm, Jos Bergervoet<jos.bergerv...@xs4all.nl> wrote:
>>
>>> There is no "interface as a thing"; there is only a mathematical
>>> boundary that you can place, where there are no dipoles on one side.
>>
>> Let's again look at the other side where the dielectric is.
>> Everything works fine if we add an infinitesimal dissipation
>> in the dielectric. The polarization current then has a tiny
>> in-phase component with the B-field and the time-averaged
>> bulk Lorentz force gives the correct radiation pressure.
>> No extra interface force is present. It is just the very
>> weak bulk force in a very large volume (the penetration
>> depth following from this very small dissipation).
>
> I strongly suspect that this won't give the right answer. For a semi-
> infinite medium with low reflectivity, this looks like it gives
> approximately F=I/c, the force on a perfect absorber.

What else would you want for this "medium with low
reflectivity"? It then is approximately a perfect
absorber so shouldn't this be the force you expect?

Anyhow, the general case with reflectivity R gives
F = (1+R^2) I/c, i.e. all transferred momentum is
(eventually) transferred to the crystal, because no
wave is left after absorption has taken place..
But completely without absorption there would
be less force on the crystal because the Abraham
momentum remains in the wave. In that case:
F = 2 R^2 I/c [force on the crystal]
F = (1-R^2) I/c [creating Abraham momentum]
+ -------------
F = (1+R^2) I/c [total force incoming radiation]

> Given that we _know_ that the reflection isn't sudden at an interface
> for an atomic medium, why look further?

My objective is to compute the Lorentz force from
the B-field on the polarization current, wherever
it occurs! And if it occurs deeper in the medium
then I have to look there. In the case of normal
incidence there is no Coulomb force, so the Lorentz
force is all I need to find the complete momentum
transfer to the crystal.

> Especially since the continuum
> version of such a distributed reflection model works.

But the continuum description is exactly what I'm
looking at! (I prefer to avoid the microscopic
description if possible). The Lorentz force in
the bulk of the dielectric, described by the
continuum limit. That's all. (That's all that's
needed!)

>> It doesn't change the outcome for normal incidence of a
>> wave on loss-less dielectric. Once you make the transition
>> layer thin w.r.t. the wavelength the result smoothly goes
>> back to the stepped-index result.
>
> Yes, that's the point. And we can calculate, without undue difficulty,
> the force on the region over which the change in index happens. And we
> can see that the total force, integrated over this region remains
> constant as we reduce the thickness of this region.

It goes to zero! (The Lorentz force in this transition
layer.) To have a nonzero limit the force would have to
diverge if the thickness shrinks. But it cannot do so
because neither the current nor the B-field are diverging.
So I'm puzzled what remaining force you expect to see!

> So we shouldn't
> assume that it suddenly goes to zero as the thickness becomes zero; we
> should assume that the transition is smooth.

Anyhow, it goes to zero, so we could just as well
have started without the transition region, which
also gives zero. And it hass continuous fields,
so they already _are_ smooth in that case.

> So, we _can_ assume a short, but continuous transition, and get the
> correct answer.

But also without it we get the correct answer
As I said: we can compute 2+2 as a limiting
case (of 2+2+epsilon, for instance). But why?
The correct answer is: "No force in the transition
layer if its thickness goes to zero."

>> So, after a century of doubt,
>> finally the Abraham-Mikowsky controversy has been resolved.
>> (Unless one disagrees..)
>
> The controversy isn't about the total force. Both Abraham and
> Minkowski give the same total force, so calculating the force doesn't
> resolve anything.

It resolves it completely if you accept that the
force gives the momentum transfer to the crystal
and the remaining momentum per volume area is the
momentum in the wave. i think that is a reasonable
point of view. (And the result is the Abraham
momentum!)

I find it surprising why there is a "controversy"
since the whole thing is as simple as described
above!

Also interesting: this Lorentz force only is present
in regions where there either are standing waves, or
the medium is lossy, or the amplitude of the passing
wave is increasing! (The last one is crucial, since
for the crystal momentum you need the history of
forces from the time before the wavefront arrives
to the steady state. The integral gives the acquired
momentum.

> There is no physics in Abraham-Minkowski, only the
> philosophy of what part of the force to call "electromagnetic" (in the
> simple case of a medium completely characterised by a constant
> permittivity and permeability, Minkowski says "It's all
> electromagnetic!").

As I explained, I was just trying to base it all on
the Lorentz force. Obviously the momentum transfer
to the lattice by this force would mean that only
the "left over" momentum in the wave should be
called "electromagnetic". Unless of course there
are mixed terms messing it up (momentum and energy
are quadratic in the amplitude, after all..)

> An atomic medium strongly suggests that Abraham is correct (or at
> least more correct), but that's not new. Again, Gordon 1973. Or our
> newest paper on this (Journal of Optics, 13(4), 044017, 2011).

My simple "Lorentz force" view gives the same! At
least for case 5e) of the earlier list, it leaves
exactly the Abraham momentum in the wave. And my
analysis is not based on an atomic medium but on the
continuum description. Of course my usage of a force
acting on the polarization current is ultimately
justified by saying "these are physically moving
charges at the microscopic level". Anyhow, this does
not "strongly suggest" but it "simply proves"
that Abraham is correct! (More difficult would be
to prove that Minkowsky is necessarily wrong..)

> Assuming an atomic medium makes a bigger difference in the angular
> momentum controversy,

Let's please save that subject for another thread! :-)
(Actually we had that other thread already with Radi.
There I also was only interested in whether I could
understand the forces, not the philosophy.)

-jrb

[Jos Bergervoet]

On 2/24/2012 10:17 AM, Jos Bergervoet wrote:
> On 2/23/2012 9:26 PM, Timo Nieminen wrote:

..

>> So, we _can_ assume a short, but continuous transition, and get the
>> correct answer.
>
> But also without it we get the correct answer
> As I said: we can compute 2+2 as a limiting
> case (of 2+2+epsilon, for instance). But why?
> The correct answer is: "No force in the transition
> layer if its thickness goes to zero."

Again, I should explicitly add: for normal
incidence! For any other angle you get a surface
charge and a Coulomb force on the surface. But
the normal beam only gives a bulk force on the
dielectric and it is only Lorentz force.

-jrb