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From: JohnM on 6 Aug 2010 04:06
On Aug 6, 12:40 am, Bill Ward <bw...(a)ix.REMOVETHISnetcom.com> wrote:
> [Hand edited]
>
> On Thu, 05 Aug 2010 09:09:42 -0700, JohnM wrote:
>
> On Aug 5, 8:45 am, Bill Ward <bw...(a)ix.REMOVETHISnetcom.com> wrote:> On Wed, 04 Aug 2010 23:12:58 -0700, JohnM wrote:
> > > On Aug 4, 9:41 am, Bill Ward <bw...(a)ix.REMOVETHISnetcom.com> wrote:
> > >> On Wed, 04 Aug 2010 00:14:37 -0700, JohnM wrote:
> > >> > On Aug 4, 1:58 am, Bill Ward <bw...(a)ix.REMOVETHISnetcom.com> wrote:
> > >> >> On Tue, 03 Aug 2010 00:31:46 -0700, Rob Dekker wrote:
> > >> >> > "Bill Ward" <bw...(a)ix.REMOVETHISnetcom.com> wrote in message
> > >> >> >news:EuqdnaQQKoUX5srRnZ2dnUVZ_gCdnZ2d(a)giganews.com...
> > >> >> >> On Mon, 02 Aug 2010 18:12:32 -0700, Rob Dekker wrote:
>
> > >> >> >>> "Bill Ward" <bw...(a)ix.REMOVETHISnetcom.com> wrote in message
> > >> >> >>>news:EIednfB4Dt7KdMvRnZ2dnUVZ_hCdnZ2d(a)giganews.com...
> > >> >> >>>> On Mon, 02 Aug 2010 01:22:56 -0700, Rob wrote:
>
> > >> >> >>>>> On Jul 30, 3:08 pm, Bill Ward
>
> <bw...(a)ix.REMOVETHISnetcom.com>
>
> > >> >> >>>>> wrote:
> > >> >> >>>>>> On Fri, 30 Jul 2010 11:45:01 -0700, Rob Dekker wrote:
> > >> >> >>>>>> > "Bill Ward" <bw...(a)ix.REMOVETHISnetcom.com> wrote in
> > >> >> >>>>>> > message

<snip>

> > >> >> >>>>>> >> Energy cannot flow from cold objects to hot objects.
>
> > >> >> >>>>>> > That's a pretty bold statement that leads to quite absurt
> > >> >> >>>>>> > physics : Photons represent energy, but here you claim
> > >> >> >>>>>> > (absolute) that energy (thus photons too) cannot flow
> from
> > >> >> >>>>>> > a cold to a warm object.
>
> > >> >> >>>>>> > Think about that for a moment what would be involved for
> > >> >> >>>>>> > these photons to be so smart ? By which mechanism would
> > >> >> >>>>>> > they know the temperature of the object they originate
> from
> > >> >> >>>>>> > and then by which mechanism do they decide NOT to
> > >> >> >>>>>> > materialize when they arrive when they arrive on a warmer
> > >> >> >>>>>> > surface ? Even better, what would these photons do when
> the
> > >> >> >>>>>> > encounter a warmer object than the one they originated
> from
> > >> >> >>>>>> > ? Turnback? Whereto ?
>
> > >> >> >>>>>> > So the statement "Energy cannot flow from cold objects to
> > >> >> >>>>>> > hot objects" leads to completely bizzarre physics, which
> > >> >> >>>>>> > would require re-writing all basic physics books about
> > >> >> >>>>>> > photons and how they behave. Including finding an
> > >> >> >>>>>> > alternative for quantum physics.
>
> > >> >> >>>>>> > Best of luck with that Bill, but for now, I'll assume
> that
> > >> >> >>>>>> > the statement "Energy cannot flow from cold objects to
> hot
> > >> >> >>>>>> > objects" is only true in your mind, and not in the real
> > >> >> >>>>>> > world. In the real world, there are two energy flows. One
> > >> >> >>>>>> > from cold to warm, one from warm to cold.
>
> > >> >> >>>>>> Do you really think the mathematical shortcut of separating
> > >> >> >>>>>> the flow into forward and backward components is physical?
>
> > >> >> >>>>> As long as quantum physics and causality in this Universe
> are
> > >> >> >>>>> correct, there is no doubt that these two components are
> > >> >> >>>>> physical.
>
> > >> >> >>>>>> If so, describe an experiment that directly measures the
> > >> >> >>>>>> "backradiation".
>
> > >> >> >>>>> Point an IR detector at the hot object, and measure the
> power
> > >> >> >>>>> (W/m^2) radiating from it.
> > >> >> >>>>> Do the same thing for the cold object. Done.
>
> > >> >> >>>> Not quite.  How do you get the warm detector to absorb the
> cold
> > >> >> >>>> photons?
>
> > >> >> >>>> If the detector is cooled, you're just measuring plain old
> > >> >> >>>> "forward" radiation.
>
> > >> >> >>>> If you use a coherent detector, you're not measuring heat,
> > >> >> >>>> you're measuring EM energy.
>
> > >> >> >>> And EM energy (photons) do not heat up the object where it is
> > >> >> >>> received (including any coherent-or-not) IR detector ?
>
> > >> >> >> Coherent detection involves amplifying the incoming signal,
> which
> > >> >> >> always requires an external power source.
>
> > >> >> > Any 10 year old who built a crystal radio will prove you wrong.
>
> > >> >> In that case the amplitude is large enough no amplification is
> > >> >> required. Radio signals are not thermal radiation (heat).
>
> > >> >> >> No net heat needs to be transferred
> > >> >> >> to the detector by the signal.
>
> > >> >> > We were talking about "EM energy". Not "net heat". Don't change
> > >> >> > the subject please.
>
> > >> >> The S/B equation applies to thermal energy, not line emission, as
> in
> > >> >> radio transmitters.  If there's a change of subject, it's all
> yours.
>
> > >> >> >>> Also, it is then up to YOU
> > >> >> >>> to explain why you think that this measured EM energy
> (photons)
> > >> >> >>> would suddenly stop flowing once we remove the IR detector.
>
> > >> >> >> They don't.
>
> > >> > Progress! Bilbo's starting to get it. (Or else he's looking to
> > >> > redesign his cold-to-hot-heat-transfer straw man)
>
> > >> >> >>  They just can't transfer heat to a warmer object.
>
> > >> >> >> You're the one claiming cold photons can transfer energy to a
> > >> >> >> warmer object, not me.  So show me the transferred energy,
> don't
> > >> >> >> just assume it.
>
> > >> >> > The photon density (intensity) for any emitting (cold or warm)
> > >> >> > object is properly described with the Planck function. The
> > >> >> > transferred energy per photon is E=hv, with h being the
> Boltzmann
> > >> >> > constant and v the oscillation frequency of the photon.
> > >> >> > Preservation of energy states that the receiver of these photons
> > >> >> > will always heat up with energy E=hv per photon, regardless of
> the
> > >> >> > form of the receiver (coherent, hot object, cold object).
>
> > >> >> Cold photons can't heat a warmer object.  One way to think of it
> is
> > >> >> that all the lower energy states are filled, leaving none for
> > >> >> absorption.
>
> > >> > Nothing to do with filled and empty states. The energy of the
> > >> > incoming photon is absorbed by a transition between any two states
> > >> > having energy difference equal to 'hv' for the photon.
>
> > >> >> You are apparently confused about the difference between line
> > >> >> emission (radio)
> > >> >> and thermal emission (heat), so you may not understand that.
>
> > >> >> In a microwave oven, photons are generated by a magnetron as a
> line
> > >> >> emission, then converted to heat by absorption in water.  S/B is
> > >> >> valid only for thermal emission, not line emission, so the fact
> the
> > >> >> transmitter is cooler than the water has nothing to do with the
> > >> >> energy it transmits.
>
> > >> >> > So a coherent receiver (with or without a power source) is fine
> in
> > >> >> > determining the energy flow out of a cold object. Why is this so
> > >> >> > difficult for you to understand ?
>
> > >> >> Because it violates both the first and second laws of
> > >> >> thermodynamics. Other than that, it's a fine theory, just not
> > >> >> confirmable.
>
> > >> >> If you could show me thermal energy transferred from a cold
> "source"
> > >> >> to a warmer "sink" (heating it up), you could not only confirm
> your
> > >> >> theory, you could build a perpetual motion machine yielding free
> > >> >> energy.  Did you ever wonder why no one has ever done that?
>
> > >> > Ohh... not progress, after all. Bilbo was just re-designing his
> > >> > thermodynamic, straw man argument.
>
> > >> If it were a "strawman" argument, Morgan would easily be able to
> refute
> > >> it by showing us his free-energy machine.  Watch him dance and twist
> to
> > >> make excuses now.  He's getting right up there with Dawlish.
>
> > > Very clever twist by Bilbo here. Ditch the straw man completely after
> I
> > > debunked it, and switch to pure ad hom.
>
> > Pointing out the fact Morgan can't show a free-energy machine is hardly
> > ad hominem.
>
> Of course it's an ad hominem, as there is no such thing as a "free
> energy machine". Tell me bilbo, have you stopped beating your wife yet?
> - a rather similar form of personal attack, in which all possible
> answers implicate the recipient in those uniquely Islamic delights of
> wife-beating (approved by the Holy Koran, no less).
>
> [Morgan apparently doesn't even know the meaning of "reductio ad
> absurdum".]

Bilbo apparently doesn't even know the meaning of plain English. I've
rejoined the deviously separated parts of my original paragraph to
illustrate this.

> > Comparing him to Dawlish should be a compliment, since they
> > share the Alinsky principles of argument.  Either way, he's toast, and
> > should be ashamed of himself for getting his obvious misstatement caught
> > so easily. There's no wiggle room.  No matter how he tries to tap dance
> > around the issue, the 2nd law disproves his silly contention.
>
> Clever red herring. No mention which of the thousand or so statements
> I have issued on alt. g-w over the years is the subject of his claim.
> Let alone why this mystery statement is a "misstatement"
>
> [See above, where he agrees with the claim that a cold object can
> transfer heat to a warmer one.]

No such agreement by me. Pure fabrication - or a lie, if you will.

> > Ignoring temperature seems to be the best he can do:
>
> Poor Bilbo seems not to know which Law of T.D. is which.
>
> > > But a very simple fact that every bit of matter is potentially a sink
> > > for any photon, however much or little energy it carries, refuses to
> go
> > > away just because we stop discussing it.
>
> > How droll.
>
> So what does one of those pesky, low-energy photons do when it finds
> itself on a collision course with a hot object? In bilbo's phantasy
> world it turns on its heel and scurries back to where it came. Neither
> Spencer nor Pielke offer this possibility for back radiation in their
> recent blogs, so where does bilbo get this absurd idea?
>
> Entropy.  Cold objects can't heat warm ones.  Otherwise CoE could be violated
> with a heat engine running on the delta T.

Bilbo imagines heat and energy are synonyms? What a loon!

So why doesn't he post his amazing 'discovery' in the comments at the
websites of Spencer and Pielke?

> > Morgan can't even lose gracefully.
>
> Only the stupidist of fuckwits could dream up a statement that
> idiotic. Once again, bilbo fails to surprise us.
>
> See what I mean?  Morgan phrases his surrender in a very strange way.

If I surrender, I will say so straight out. I have no need to beat
about the bush like bilbo.
From: Bill Ward on 6 Aug 2010 12:41
On Fri, 06 Aug 2010 01:06:42 -0700, JohnM wrote:

> On Aug 6, 12:40 am, Bill Ward <bw...(a)ix.REMOVETHISnetcom.com> wrote:
>> [Hand edited]
>>
>> On Thu, 05 Aug 2010 09:09:42 -0700, JohnM wrote:
>>
>> On Aug 5, 8:45 am, Bill Ward <bw...(a)ix.REMOVETHISnetcom.com> wrote:> On
>> Wed, 04 Aug 2010 23:12:58 -0700, JohnM wrote:
>> > > On Aug 4, 9:41 am, Bill Ward <bw...(a)ix.REMOVETHISnetcom.com> wrote:
>> > >> On Wed, 04 Aug 2010 00:14:37 -0700, JohnM wrote:
>> > >> > On Aug 4, 1:58 am, Bill Ward <bw...(a)ix.REMOVETHISnetcom.com>
>> > >> > wrote:
>> > >> >> On Tue, 03 Aug 2010 00:31:46 -0700, Rob Dekker wrote:
>> > >> >> > "Bill Ward" <bw...(a)ix.REMOVETHISnetcom.com> wrote in message
>> > >> >> >news:EuqdnaQQKoUX5srRnZ2dnUVZ_gCdnZ2d(a)giganews.com...
>> > >> >> >> On Mon, 02 Aug 2010 18:12:32 -0700, Rob Dekker wrote:
>>
>> > >> >> >>> "Bill Ward" <bw...(a)ix.REMOVETHISnetcom.com> wrote in
>> > >> >> >>> message
>> > >> >> >>>news:EIednfB4Dt7KdMvRnZ2dnUVZ_hCdnZ2d(a)giganews.com...
>> > >> >> >>>> On Mon, 02 Aug 2010 01:22:56 -0700, Rob wrote:
>>
>> > >> >> >>>>> On Jul 30, 3:08 pm, Bill Ward
>>
>> <bw...(a)ix.REMOVETHISnetcom.com>
>>
>> > >> >> >>>>> wrote:
>> > >> >> >>>>>> On Fri, 30 Jul 2010 11:45:01 -0700, Rob Dekker wrote:
>> > >> >> >>>>>> > "Bill Ward" <bw...(a)ix.REMOVETHISnetcom.com> wrote in
>> > >> >> >>>>>> > message
>
> <snip>
>
>> > >> >> >>>>>> >> Energy cannot flow from cold objects to hot objects.
>>
>> > >> >> >>>>>> > That's a pretty bold statement that leads to quite
>> > >> >> >>>>>> > absurt physics : Photons represent energy, but here
>> > >> >> >>>>>> > you claim (absolute) that energy (thus photons too)
>> > >> >> >>>>>> > cannot flow
>> from
>> > >> >> >>>>>> > a cold to a warm object.
>>
>> > >> >> >>>>>> > Think about that for a moment what would be involved
>> > >> >> >>>>>> > for these photons to be so smart ? By which mechanism
>> > >> >> >>>>>> > would they know the temperature of the object they
>> > >> >> >>>>>> > originate
>> from
>> > >> >> >>>>>> > and then by which mechanism do they decide NOT to
>> > >> >> >>>>>> > materialize when they arrive when they arrive on a
>> > >> >> >>>>>> > warmer surface ? Even better, what would these photons
>> > >> >> >>>>>> > do when
>> the
>> > >> >> >>>>>> > encounter a warmer object than the one they originated
>> from
>> > >> >> >>>>>> > ? Turnback? Whereto ?
>>
>> > >> >> >>>>>> > So the statement "Energy cannot flow from cold objects
>> > >> >> >>>>>> > to hot objects" leads to completely bizzarre physics,
>> > >> >> >>>>>> > which would require re-writing all basic physics books
>> > >> >> >>>>>> > about photons and how they behave. Including finding
>> > >> >> >>>>>> > an alternative for quantum physics.
>>
>> > >> >> >>>>>> > Best of luck with that Bill, but for now, I'll assume
>> that
>> > >> >> >>>>>> > the statement "Energy cannot flow from cold objects to
>> hot
>> > >> >> >>>>>> > objects" is only true in your mind, and not in the
>> > >> >> >>>>>> > real world. In the real world, there are two energy
>> > >> >> >>>>>> > flows. One from cold to warm, one from warm to cold.
>>
>> > >> >> >>>>>> Do you really think the mathematical shortcut of
>> > >> >> >>>>>> separating the flow into forward and backward components
>> > >> >> >>>>>> is physical?
>>
>> > >> >> >>>>> As long as quantum physics and causality in this Universe
>> are
>> > >> >> >>>>> correct, there is no doubt that these two components are
>> > >> >> >>>>> physical.
>>
>> > >> >> >>>>>> If so, describe an experiment that directly measures the
>> > >> >> >>>>>> "backradiation".
>>
>> > >> >> >>>>> Point an IR detector at the hot object, and measure the
>> power
>> > >> >> >>>>> (W/m^2) radiating from it.
>> > >> >> >>>>> Do the same thing for the cold object. Done.
>>
>> > >> >> >>>> Not quite.  How do you get the warm detector to absorb the
>> cold
>> > >> >> >>>> photons?
>>
>> > >> >> >>>> If the detector is cooled, you're just measuring plain old
>> > >> >> >>>> "forward" radiation.
>>
>> > >> >> >>>> If you use a coherent detector, you're not measuring heat,
>> > >> >> >>>> you're measuring EM energy.
>>
>> > >> >> >>> And EM energy (photons) do not heat up the object where it
>> > >> >> >>> is received (including any coherent-or-not) IR detector ?
>>
>> > >> >> >> Coherent detection involves amplifying the incoming signal,
>> which
>> > >> >> >> always requires an external power source.
>>
>> > >> >> > Any 10 year old who built a crystal radio will prove you
>> > >> >> > wrong.
>>
>> > >> >> In that case the amplitude is large enough no amplification is
>> > >> >> required. Radio signals are not thermal radiation (heat).
>>
>> > >> >> >> No net heat needs to be transferred
>> > >> >> >> to the detector by the signal.
>>
>> > >> >> > We were talking about "EM energy". Not "net heat". Don't
>> > >> >> > change the subject please.
>>
>> > >> >> The S/B equation applies to thermal energy, not line emission,
>> > >> >> as
>> in
>> > >> >> radio transmitters.  If there's a change of subject, it's all
>> yours.
>>
>> > >> >> >>> Also, it is then up to YOU
>> > >> >> >>> to explain why you think that this measured EM energy
>> (photons)
>> > >> >> >>> would suddenly stop flowing once we remove the IR detector.
>>
>> > >> >> >> They don't.
>>
>> > >> > Progress! Bilbo's starting to get it. (Or else he's looking to
>> > >> > redesign his cold-to-hot-heat-transfer straw man)
>>
>> > >> >> >>  They just can't transfer heat to a warmer object.
>>
>> > >> >> >> You're the one claiming cold photons can transfer energy to
>> > >> >> >> a warmer object, not me.  So show me the transferred energy,
>> don't
>> > >> >> >> just assume it.
>>
>> > >> >> > The photon density (intensity) for any emitting (cold or
>> > >> >> > warm) object is properly described with the Planck function.
>> > >> >> > The transferred energy per photon is E=hv, with h being the
>> Boltzmann
>> > >> >> > constant and v the oscillation frequency of the photon.
>> > >> >> > Preservation of energy states that the receiver of these
>> > >> >> > photons will always heat up with energy E=hv per photon,
>> > >> >> > regardless of
>> the
>> > >> >> > form of the receiver (coherent, hot object, cold object).
>>
>> > >> >> Cold photons can't heat a warmer object.  One way to think of
>> > >> >> it
>> is
>> > >> >> that all the lower energy states are filled, leaving none for
>> > >> >> absorption.
>>
>> > >> > Nothing to do with filled and empty states. The energy of the
>> > >> > incoming photon is absorbed by a transition between any two
>> > >> > states having energy difference equal to 'hv' for the photon.
>>
>> > >> >> You are apparently confused about the difference between line
>> > >> >> emission (radio)
>> > >> >> and thermal emission (heat), so you may not understand that.
>>
>> > >> >> In a microwave oven, photons are generated by a magnetron as a
>> line
>> > >> >> emission, then converted to heat by absorption in water.  S/B
>> > >> >> is valid only for thermal emission, not line emission, so the
>> > >> >> fact
>> the
>> > >> >> transmitter is cooler than the water has nothing to do with the
>> > >> >> energy it transmits.
>>
>> > >> >> > So a coherent receiver (with or without a power source) is
>> > >> >> > fine
>> in
>> > >> >> > determining the energy flow out of a cold object. Why is this
>> > >> >> > so difficult for you to understand ?
>>
>> > >> >> Because it violates both the first and second laws of
>> > >> >> thermodynamics. Other than that, it's a fine theory, just not
>> > >> >> confirmable.
>>
>> > >> >> If you could show me thermal energy transferred from a cold
>> "source"
>> > >> >> to a warmer "sink" (heating it up), you could not only confirm
>> your
>> > >> >> theory, you could build a perpetual motion machine yielding
>> > >> >> free energy.  Did you ever wonder why no one has ever done
>> > >> >> that?
>>
>> > >> > Ohh... not progress, after all. Bilbo was just re-designing his
>> > >> > thermodynamic, straw man argument.
>>
>> > >> If it were a "strawman" argument, Morgan would easily be able to
>> refute
>> > >> it by showing us his free-energy machine.  Watch him dance and
>> > >> twist
>> to
>> > >> make excuses now.  He's getting right up there with Dawlish.
>>
>> > > Very clever twist by Bilbo here. Ditch the straw man completely
>> > > after
>> I
>> > > debunked it, and switch to pure ad hom.
>>
>> > Pointing out the fact Morgan can't show a free-energy machine is
>> > hardly ad hominem.
>>
>> Of course it's an ad hominem, as there is no such thing as a "free
>> energy machine". Tell me bilbo, have you stopped beating your wife yet?
>> - a rather similar form of personal attack, in which all possible
>> answers implicate the recipient in those uniquely Islamic delights of
>> wife-beating (approved by the Holy Koran, no less).
>>
>> [Morgan apparently doesn't even know the meaning of "reductio ad
>> absurdum".]
>
> Bilbo apparently doesn't even know the meaning of plain English. I've
> rejoined the deviously separated parts of my original paragraph to
> illustrate this.
>
>> > Comparing him to Dawlish should be a compliment, since they share the
>> > Alinsky principles of argument.  Either way, he's toast, and should
>> > be ashamed of himself for getting his obvious misstatement caught so
>> > easily. There's no wiggle room.  No matter how he tries to tap dance
>> > around the issue, the 2nd law disproves his silly contention.
>>
>> Clever red herring. No mention which of the thousand or so statements I
>> have issued on alt. g-w over the years is the subject of his claim. Let
>> alone why this mystery statement is a "misstatement"
>>
>> [See above, where he agrees with the claim that a cold object can
>> transfer heat to a warmer one.]
>
> No such agreement by me. Pure fabrication - or a lie, if you will.

So he now says he agrees with me that cold objects cannot transfer heat
to a warmer one. Kind of makes one wonder what all his fuss was about.
I guess it was either that or come up with a free-energy machine.

>> > Ignoring temperature seems to be the best he can do:
>>
>> Poor Bilbo seems not to know which Law of T.D. is which.
>>
>> > > But a very simple fact that every bit of matter is potentially a
>> > > sink for any photon, however much or little energy it carries,
>> > > refuses to
>> go
>> > > away just because we stop discussing it.
>>
>> > How droll.
>>
>> So what does one of those pesky, low-energy photons do when it finds
>> itself on a collision course with a hot object? In bilbo's phantasy
>> world it turns on its heel and scurries back to where it came. Neither
>> Spencer nor Pielke offer this possibility for back radiation in their
>> recent blogs, so where does bilbo get this absurd idea?
>>
>[Entropy.  Cold objects can't heat warm ones.  Otherwise CoE could be
> violated with a heat engine running on the delta T.]
>
> Bilbo imagines heat and energy are synonyms? What a loon!

What an imagination Morgan has. He also projects a lot.

> So why doesn't he post his amazing 'discovery' in the comments at the
> websites of Spencer and Pielke?
>
>> > Morgan can't even lose gracefully.
>>
>> Only the stupidist of fuckwits could dream up a statement that idiotic.
>> Once again, bilbo fails to surprise us.
>>
>> See what I mean?  Morgan phrases his surrender in a very strange way.
>
> If I surrender, I will say so straight out. I have no need to beat about
> the bush like bilbo.

Note above, where he now claims he agrees with me that cold objects cannot
transfer heat to a warmer one. What's he been doing, if not beating around
the bush?

From: JohnM on 6 Aug 2010 13:35
On Aug 6, 6:41 pm, Bill Ward <bw...(a)ix.REMOVETHISnetcom.com> wrote:
> On Fri, 06 Aug 2010 01:06:42 -0700, JohnM wrote:
> > On Aug 6, 12:40 am, Bill Ward <bw...(a)ix.REMOVETHISnetcom.com> wrote:
> >> [Hand edited]
>
> >> On Thu, 05 Aug 2010 09:09:42 -0700, JohnM wrote:
>
> >> On Aug 5, 8:45 am, Bill Ward <bw...(a)ix.REMOVETHISnetcom.com> wrote:> On
> >> Wed, 04 Aug 2010 23:12:58 -0700, JohnM wrote:
> >> > > On Aug 4, 9:41 am, Bill Ward <bw...(a)ix.REMOVETHISnetcom.com> wrote:
> >> > >> On Wed, 04 Aug 2010 00:14:37 -0700, JohnM wrote:
> >> > >> > On Aug 4, 1:58 am, Bill Ward <bw...(a)ix.REMOVETHISnetcom.com>
> >> > >> > wrote:
> >> > >> >> On Tue, 03 Aug 2010 00:31:46 -0700, Rob Dekker wrote:
> >> > >> >> > "Bill Ward" <bw...(a)ix.REMOVETHISnetcom.com> wrote in message
> >> > >> >> >news:EuqdnaQQKoUX5srRnZ2dnUVZ_gCdnZ2d(a)giganews.com...
> >> > >> >> >> On Mon, 02 Aug 2010 18:12:32 -0700, Rob Dekker wrote:
>
> >> > >> >> >>> "Bill Ward" <bw...(a)ix.REMOVETHISnetcom.com> wrote in
> >> > >> >> >>> message
> >> > >> >> >>>news:EIednfB4Dt7KdMvRnZ2dnUVZ_hCdnZ2d(a)giganews.com...
> >> > >> >> >>>> On Mon, 02 Aug 2010 01:22:56 -0700, Rob wrote:
>
> >> > >> >> >>>>> On Jul 30, 3:08 pm, Bill Ward
>
> >> <bw...(a)ix.REMOVETHISnetcom.com>
>
> >> > >> >> >>>>> wrote:
> >> > >> >> >>>>>> On Fri, 30 Jul 2010 11:45:01 -0700, Rob Dekker wrote:
> >> > >> >> >>>>>> > "Bill Ward" <bw...(a)ix.REMOVETHISnetcom.com> wrote in
> >> > >> >> >>>>>> > message
>
> > <snip>
>
> >> > >> >> >>>>>> >> Energy cannot flow from cold objects to hot objects.
>
> >> > >> >> >>>>>> > That's a pretty bold statement that leads to quite
> >> > >> >> >>>>>> > absurt physics : Photons represent energy, but here
> >> > >> >> >>>>>> > you claim (absolute) that energy (thus photons too)
> >> > >> >> >>>>>> > cannot flow
> >> from
> >> > >> >> >>>>>> > a cold to a warm object.
>
> >> > >> >> >>>>>> > Think about that for a moment what would be involved
> >> > >> >> >>>>>> > for these photons to be so smart ? By which mechanism
> >> > >> >> >>>>>> > would they know the temperature of the object they
> >> > >> >> >>>>>> > originate
> >> from
> >> > >> >> >>>>>> > and then by which mechanism do they decide NOT to
> >> > >> >> >>>>>> > materialize when they arrive when they arrive on a
> >> > >> >> >>>>>> > warmer surface ? Even better, what would these photons
> >> > >> >> >>>>>> > do when
> >> the
> >> > >> >> >>>>>> > encounter a warmer object than the one they originated
> >> from
> >> > >> >> >>>>>> > ? Turnback? Whereto ?
>
> >> > >> >> >>>>>> > So the statement "Energy cannot flow from cold objects
> >> > >> >> >>>>>> > to hot objects" leads to completely bizzarre physics,
> >> > >> >> >>>>>> > which would require re-writing all basic physics books
> >> > >> >> >>>>>> > about photons and how they behave. Including finding
> >> > >> >> >>>>>> > an alternative for quantum physics.
>
> >> > >> >> >>>>>> > Best of luck with that Bill, but for now, I'll assume
> >> that
> >> > >> >> >>>>>> > the statement "Energy cannot flow from cold objects to
> >> hot
> >> > >> >> >>>>>> > objects" is only true in your mind, and not in the
> >> > >> >> >>>>>> > real world. In the real world, there are two energy
> >> > >> >> >>>>>> > flows. One from cold to warm, one from warm to cold.
>
> >> > >> >> >>>>>> Do you really think the mathematical shortcut of
> >> > >> >> >>>>>> separating the flow into forward and backward components
> >> > >> >> >>>>>> is physical?
>
> >> > >> >> >>>>> As long as quantum physics and causality in this Universe
> >> are
> >> > >> >> >>>>> correct, there is no doubt that these two components are
> >> > >> >> >>>>> physical.
>
> >> > >> >> >>>>>> If so, describe an experiment that directly measures the
> >> > >> >> >>>>>> "backradiation".
>
> >> > >> >> >>>>> Point an IR detector at the hot object, and measure the
> >> power
> >> > >> >> >>>>> (W/m^2) radiating from it.
> >> > >> >> >>>>> Do the same thing for the cold object. Done.
>
> >> > >> >> >>>> Not quite.  How do you get the warm detector to absorb the
> >> cold
> >> > >> >> >>>> photons?
>
> >> > >> >> >>>> If the detector is cooled, you're just measuring plain old
> >> > >> >> >>>> "forward" radiation.
>
> >> > >> >> >>>> If you use a coherent detector, you're not measuring heat,
> >> > >> >> >>>> you're measuring EM energy.
>
> >> > >> >> >>> And EM energy (photons) do not heat up the object where it
> >> > >> >> >>> is received (including any coherent-or-not) IR detector ?
>
> >> > >> >> >> Coherent detection involves amplifying the incoming signal,
> >> which
> >> > >> >> >> always requires an external power source.
>
> >> > >> >> > Any 10 year old who built a crystal radio will prove you
> >> > >> >> > wrong.
>
> >> > >> >> In that case the amplitude is large enough no amplification is
> >> > >> >> required. Radio signals are not thermal radiation (heat).
>
> >> > >> >> >> No net heat needs to be transferred
> >> > >> >> >> to the detector by the signal.
>
> >> > >> >> > We were talking about "EM energy". Not "net heat". Don't
> >> > >> >> > change the subject please.
>
> >> > >> >> The S/B equation applies to thermal energy, not line emission,
> >> > >> >> as
> >> in
> >> > >> >> radio transmitters.  If there's a change of subject, it's all
> >> yours.
>
> >> > >> >> >>> Also, it is then up to YOU
> >> > >> >> >>> to explain why you think that this measured EM energy
> >> (photons)
> >> > >> >> >>> would suddenly stop flowing once we remove the IR detector..
>
> >> > >> >> >> They don't.
>
> >> > >> > Progress! Bilbo's starting to get it. (Or else he's looking to
> >> > >> > redesign his cold-to-hot-heat-transfer straw man)
>
> >> > >> >> >>  They just can't transfer heat to a warmer object.
>
> >> > >> >> >> You're the one claiming cold photons can transfer energy to
> >> > >> >> >> a warmer object, not me.  So show me the transferred energy,
> >> don't
> >> > >> >> >> just assume it.
>
> >> > >> >> > The photon density (intensity) for any emitting (cold or
> >> > >> >> > warm) object is properly described with the Planck function.
> >> > >> >> > The transferred energy per photon is E=hv, with h being the
> >> Boltzmann
> >> > >> >> > constant and v the oscillation frequency of the photon.
> >> > >> >> > Preservation of energy states that the receiver of these
> >> > >> >> > photons will always heat up with energy E=hv per photon,
> >> > >> >> > regardless of
> >> the
> >> > >> >> > form of the receiver (coherent, hot object, cold object).
>
> >> > >> >> Cold photons can't heat a warmer object.  One way to think of
> >> > >> >> it
> >> is
> >> > >> >> that all the lower energy states are filled, leaving none for
> >> > >> >> absorption.
>
> >> > >> > Nothing to do with filled and empty states. The energy of the
> >> > >> > incoming photon is absorbed by a transition between any two
> >> > >> > states having energy difference equal to 'hv' for the photon.
>
> >> > >> >> You are apparently confused about the difference between line
> >> > >> >> emission (radio)
> >> > >> >> and thermal emission (heat), so you may not understand that.
>
> >> > >> >> In a microwave oven, photons are generated by a magnetron as a
> >> line
> >> > >> >> emission, then converted to heat by absorption in water.  S/B
> >> > >> >> is valid only for thermal emission, not line emission, so the
> >> > >> >> fact
> >> the
> >> > >> >> transmitter is cooler than the water has nothing to do with the
> >> > >> >> energy it transmits.
>
> >> > >> >> > So a coherent receiver (with or without a power source) is
> >> > >> >> > fine
> >> in
> >> > >> >> > determining the energy flow out of a cold object. Why is this
> >> > >> >> > so difficult for you to understand ?
>
> >> > >> >> Because it violates both the first and second laws of
> >> > >> >> thermodynamics. Other than that, it's a fine theory, just not
> >> > >> >> confirmable.
>
> >> > >> >> If you could show me thermal energy transferred from a cold
> >> "source"
> >> > >> >> to a warmer "sink" (heating it up), you could not only confirm
> >> your
> >> > >> >> theory, you could build a perpetual motion machine yielding
> >> > >> >> free energy.  Did you ever wonder why no one has ever done
> >> > >> >> that?
>
> >> > >> > Ohh... not progress, after all. Bilbo was just re-designing his
> >> > >> > thermodynamic, straw man argument.
>
> >> > >> If it were a "strawman" argument, Morgan would easily be able to
> >> refute
> >> > >> it by showing us his free-energy machine.  Watch him dance and
> >> > >> twist
> >> to
> >> > >> make excuses now.  He's getting right up there with Dawlish.
>
> >> > > Very clever twist by Bilbo here. Ditch the straw man completely
> >> > > after
> >> I
> >> > > debunked it, and switch to pure ad hom.
>
> >> > Pointing out the fact Morgan can't show a free-energy machine is
> >> > hardly ad hominem.
>
> >> Of course it's an ad hominem, as there is no such thing as a "free
> >> energy machine". Tell me bilbo, have you stopped beating your wife yet?
> >> - a rather similar form of personal attack, in which all possible
> >> answers implicate the recipient in those uniquely Islamic delights of
> >> wife-beating (approved by the Holy Koran, no less).
>
> >> [Morgan apparently doesn't even know the meaning of "reductio ad
> >> absurdum".]
>
> > Bilbo apparently doesn't even know the meaning of plain English. I've
> > rejoined the deviously separated parts of my original paragraph to
> > illustrate this.
>
> >> > Comparing him to Dawlish should be a compliment, since they share the
> >> > Alinsky principles of argument.  Either way, he's toast, and should
> >> > be ashamed of himself for getting his obvious misstatement caught so
> >> > easily. There's no wiggle room.  No matter how he tries to tap dance
> >> > around the issue, the 2nd law disproves his silly contention.
>
> >> Clever red herring. No mention which of the thousand or so statements I
> >> have issued on alt. g-w over the years is the subject of his claim. Let
> >> alone why this mystery statement is a "misstatement"
>
> >> [See above, where he agrees with the claim that a cold object can
> >> transfer heat to a warmer one.]
>
> > No such agreement by me.  Pure fabrication - or a lie, if you will.
>
> So he now says he agrees with me that cold objects cannot transfer heat
> to a warmer one.  Kind of makes one wonder what all his fuss was about.
> I guess it was either that or come up with a free-energy machine.

Bilbo's disingenuous turn-of-phrase style seems to be deserting him.
Any one with half-a-brain+ could see through that paragraph. Hint:
photons are neither hot nor cold. They are merely energetic.

> >> > Ignoring temperature seems to be the best he can do:
>
> >> Poor Bilbo seems not to know which Law of T.D. is which.
>
> >> > > But a very simple fact that every bit of matter is potentially a
> >> > > sink for any photon, however much or little energy it carries,
> >> > > refuses to go away just because we stop discussing it.
>
> >> > How droll.
>
> >> So what does one of those pesky, low-energy photons do when it finds
> >> itself on a collision course with a hot object? In bilbo's phantasy
> >> world it turns on its heel and scurries back to where it came. Neither
> >> Spencer nor Pielke offer this possibility for back radiation in their
> >> recent blogs, so where does bilbo get this absurd idea?
>
> >[Entropy.  Cold objects can't heat warm ones.  Otherwise CoE could be
> > violated with a heat engine running on the delta T.]

My question has not been answered, neither by bilbo nor anyone else.
What do those pesky photons do that escape from a low temperature body
and find themselves on a collision course with a hot body?

> > Bilbo imagines heat and energy are synonyms? What a loon!
>
> What an imagination Morgan has.  He also projects a lot.

Is bilbo saying heat and energy *are* synonyms? Can anyone else make
out what his response is supposed to be?

> > So why doesn't he post his amazing 'discovery' in the comments at the
> > websites of Spencer and Pielke?
>
> >> > Morgan can't even lose gracefully.
>
> >> Only the stupidist of fuckwits could dream up a statement that idiotic..
> >> Once again, bilbo fails to surprise us.
>
> >> See what I mean?  Morgan phrases his surrender in a very strange way..
>
> > If I surrender, I will say so straight out. I have no need to beat about
> > the bush like bilbo.
>
> Note above, where he now claims he agrees with me that cold objects cannot
> transfer heat to a warmer one.  What's he been doing, if not beating around
> the bush?

He's been explaining how 'back radiation' in the atmosphere is a real
phenomenon, despite bilbo's attempts to shout down the discussion

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