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From: Tim BandTech.com on 13 Jul 2010 22:45
On Jul 13, 12:39 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
> On Mon, 12 Jul 2010, Tim BandTech.com wrote:
> > On Jul 11, 5:17 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
> > > On Jul 11, 10:05 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com>
> > > wrote:
>
> > > > >> They failed to assess the energy delay of redshifted light. Who else is
> > > > >> failing to do this?
>
> > > > > This is your conclusion from the article? Why so?
>
> > > > I see that we have overlooked this argument. I see that they are
> > > > overlooking this simple argument. Who else is overlooking this simple
> > > > argument? This is my question which you answer with a question. Why so?
>
> > > Tamara doesn't overlook this in the article. At least, not for my
> > > reading of "delay". I was wondering why you thought she did.
>
> > You are substantiating a claim without any evidence. Furthermore, you
> > are stating support for the delay paradigm. Care to form a quote here?
> > I only read the article in a local library, and in light of our
> > discussion see that the argument is flawed by overlooking the redshift
> > as energy delay. I seriously doubt that Tamara, whoever that is, has
> > cited the delay effect.
>
> (a) It's irrelevant. From the beginning when you introduced this paper
> into the discussion, I said it was irrelevant. In your previous message,
> you agreed that it was irrelevant. But since you insist, note how she
> wrote "the drop in energy is just a matter of perspective and relative
> motion".
>
> (b) She writes in terms of photons, so your "delay" isn't explicitly
> mentioned. What is mentioned is that the photon doesn't "lose" energy,
> which is photon-speak for essentially the same thing.
>
> (c) Tamara is the author, as you would note if you looked at the first
> page of the article.
>
> (d) Why did you think that she "failed to assess the energy delay of
> redshifted light" in any way that counts as "failing"?
>
> (e) Why do I answer your question ("Who else is failing to do this?") with
> a question? For clarification. You surely don't want the literal answer, a
> list of the people who haven't assessed it - a list which would be about 6
> billion names long, to just include living humans. You appear to think
> it's a Big Problem. Just what kind of problem?
>
> Anyway, it isn't relevant to radiation force. What is relevant is that the
> power is affected by redshift or blueshift. As of the previous pair of
> posts, we were agreed on that.
>
> > By stonewalling you are not furthering any argument. I ask you now
> > what are the stones in your wall?
>
> You're the one who's stonewalling:
>
>
>
> > > Let me instead return to a simpler point that you haven't answered
> > > yet, but which is very relevant. (Why is it relevant? Your argument
> > > has centered on "a = 0 means P = 0" for many iterations.)
>
> > > What is the dependence of the rate of work done on an object by a
> > > force on the velocity and acceleration of the object?
>
> > > What is the explicit formula for P(v,a)?
>
> > I don't honestly see this as a valid question. Why are you attempting
> > to put velocity and acceleration both in the equation? Is this at the
> > reflector? Whatever the velocity of the reflector there is no power
> > due to velocity, so this portion
> > d P(v,a) / dv = 0.
> > I guess this is a partial derivative, but I don't know a text format
> > for it so am using 'd'. Should we just consider P(x), with a position
> > x of the reflector? What I am stating is that the P(x) is zero, and so
> > the acceleration is zero, due to light hitting the reflector, because
> > all of the light is returned toward the source. But this whole
> > nomenclature is new, so I am not probably stating this clearly. What
> > is P()?
>
> P = rate of doing work. If a force is acting on some object, it might be
> doing work on the object.
>
> I've asked this a few times already, and you haven't answered. So, let me
> give you the answer:
>
> P = Fv.
>
> It doesn't depend on the acceleration. It depends on the velocity. Note
> well that when v = 0, P = 0.
>
> Yes, I know that you're saying that P=0 means that a=0. This is wrong.
>
> What this basic result from classical mechanics tells us is that when v=0,
> no work is being done on the object. Doesn't matter what the force is,
> what causes the force, or what the object is; the rate of doing work on a
> stationary object is zero. If the object is only instantaneously
> stationary, the rate is still zero at the instant when v=0.

Gadzukes, this is bad. The rate can still be zero when v=10 as well.
I'm sorry, but this v=0 condition is just plain wrong.
Power is energy divided by time. In electrical terms if we speak of
1000 watts then this is an instantaneous figure, and upon having a
steady 1000 watts for one hour then we would have one kilowatt hour.
We might for instance power a motor of a car with 1000 watts from
battery power, and we should expect that car to accelerate especially
if there is no loss to friction in its parts. By this simple example
power is tied to acceleration. The only construction that takes any
sense within your analysis is of a situation with friction and a
steady velocity, say a weight on a table being pushed accross the
table, but even under this situation if we increase the power the
weight accelerates and if we decrease the power the weight
decelerates.

As you say that redshift delay is overlooked... Aren't you willing to
admit that humans are capable of such? It really wouldn't surprise me
to find a prior analysis consistent with what I've stated. You've not
falsified the redshift argument.

- Tim


>
> What this means is that P=0 doesn't mean that you must have F=0.
>
> --
> Timo

From: Timo Nieminen on 14 Jul 2010 00:31
On Tue, 13 Jul 2010, Tim BandTech.com wrote:

> On Jul 13, 12:39 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
> >
> > I've asked this a few times already, and you haven't answered. So, let me
> > give you the answer:
> >
> > P = Fv.
> >
> > It doesn't depend on the acceleration. It depends on the velocity. Note
> > well that when v = 0, P = 0.
> >
> > Yes, I know that you're saying that P=0 means that a=0. This is wrong.
> >
> > What this basic result from classical mechanics tells us is that when v=0,
> > no work is being done on the object. Doesn't matter what the force is,
> > what causes the force, or what the object is; the rate of doing work on a
> > stationary object is zero. If the object is only instantaneously
> > stationary, the rate is still zero at the instant when v=0.
>
> Gadzukes, this is bad. The rate can still be zero when v=10 as well.
> I'm sorry, but this v=0 condition is just plain wrong.

Yes, the rate of doing work can be zero when v = 10 (in some units). Then
you must have F = 0.

But since P=Fv, you can't deduce that F must be zero because P is zero,
especially when you know that v = 0.

> Power is energy divided by time.

No, _average_ power is energy divided by time. Power is the rate of
transferring/transforming/moving energy.

> In electrical terms if we speak of
> 1000 watts then this is an instantaneous figure, and upon having a
> steady 1000 watts for one hour then we would have one kilowatt hour.
> We might for instance power a motor of a car with 1000 watts from
> battery power, and we should expect that car to accelerate especially
> if there is no loss to friction in its parts. By this simple example
> power is tied to acceleration. The only construction that takes any
> sense within your analysis is of a situation with friction and a
> steady velocity, say a weight on a table being pushed accross the
> table, but even under this situation if we increase the power the
> weight accelerates and if we decrease the power the weight
> decelerates.

Are you saying that P=Fv is wrong? If yes, say so explicitly, and
preferably give some support.

If you agree that it is correct, then we can proceed further.

If you don't agree, we're not going to be able to go any further without
dealing with this first. If you have an introductory physics textbook at
hand, preferably a calculus-based one, see what it says about
instantaneous power in the section on classical mechanics. (A
calculus-based book is better here since the calculus-avoiding
algebra-based books will introduce average power first, and then proceed
to instantaneous power.)

> As you say that redshift delay is overlooked... Aren't you willing to
> admit that humans are capable of such? It really wouldn't surprise me
> to find a prior analysis consistent with what I've stated. You've not
> falsified the redshift argument.

"Overlooked"? No, I wouldn't say so. Cycles don't disappear, they just
pass by at a different rate. This isn't a new revelation, it's known, and
known for a long time. So what is being overlooked?

The power is different (since the cycles pass by at a different rate) due
to redshift or blueshift, which is what is claimed, not that cycles
disappear or that energy is "lost".

As for falsifying your redshift argument (P=0, so a=0 and F=0), this is
the point of dealing with some elementary points of power, velocity, and
acceleration in classical mechanics first.

--
Timo
From: Tim BandTech.com on 14 Jul 2010 09:31
On Jul 14, 12:31 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
> On Tue, 13 Jul 2010, Tim BandTech.com wrote:
> > On Jul 13, 12:39 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
>
> > > I've asked this a few times already, and you haven't answered. So, let me
> > > give you the answer:
>
> > > P = Fv.
>
> > > It doesn't depend on the acceleration. It depends on the velocity. Note
> > > well that when v = 0, P = 0.
>
> > > Yes, I know that you're saying that P=0 means that a=0. This is wrong.
>
> > > What this basic result from classical mechanics tells us is that when v=0,
> > > no work is being done on the object. Doesn't matter what the force is,
> > > what causes the force, or what the object is; the rate of doing work on a
> > > stationary object is zero. If the object is only instantaneously
> > > stationary, the rate is still zero at the instant when v=0.
>
> > Gadzukes, this is bad. The rate can still be zero when v=10 as well.
> > I'm sorry, but this v=0 condition is just plain wrong.
>
> Yes, the rate of doing work can be zero when v = 10 (in some units). Then
> you must have F = 0.
>
> But since P=Fv, you can't deduce that F must be zero because P is zero,
> especially when you know that v = 0.
>
> > Power is energy divided by time.
>
> No, _average_ power is energy divided by time. Power is the rate of
> transferring/transforming/moving energy.
>
> > In electrical terms if we speak of
> > 1000 watts then this is an instantaneous figure, and upon having a
> > steady 1000 watts for one hour then we would have one kilowatt hour.
> > We might for instance power a motor of a car with 1000 watts from
> > battery power, and we should expect that car to accelerate especially
> > if there is no loss to friction in its parts. By this simple example
> > power is tied to acceleration. The only construction that takes any
> > sense within your analysis is of a situation with friction and a
> > steady velocity, say a weight on a table being pushed accross the
> > table, but even under this situation if we increase the power the
> > weight accelerates and if we decrease the power the weight
> > decelerates.
>
> Are you saying that P=Fv is wrong? If yes, say so explicitly, and
> preferably give some support.
>
> If you agree that it is correct, then we can proceed further.
>
> If you don't agree, we're not going to be able to go any further without
> dealing with this first. If you have an introductory physics textbook at
> hand, preferably a calculus-based one, see what it says about
> instantaneous power in the section on classical mechanics. (A
> calculus-based book is better here since the calculus-avoiding
> algebra-based books will introduce average power first, and then proceed
> to instantaneous power.)
>
> > As you say that redshift delay is overlooked... Aren't you willing to
> > admit that humans are capable of such? It really wouldn't surprise me
> > to find a prior analysis consistent with what I've stated. You've not
> > falsified the redshift argument.
>
> "Overlooked"? No, I wouldn't say so. Cycles don't disappear, they just
> pass by at a different rate. This isn't a new revelation, it's known, and
> known for a long time. So what is being overlooked?

It seems to me that some posts ago you were in disagreement with the
energy delay concept, and now you are in support of the argument. If
anyone has bothered to read along with your own writings there has
been a lack of concession on many points, and now we are nearly to the
bottom of it. If I were to pick some points out from twenty posts ago
or so then we would likely just repeat the loop.

I have no problem with instantaneous power and do have an appreciation
of it. Power can come in many forms, as energy does too. When we speak
of electromagnetic radiation we are speaking of a different type of
power than mechanical systems, but they can be translated. Still, it
seems that the actual translation takes place via electricity. The
sole exception to this is the radiation pressure, which we've now
strayed far away from. By admitting that the redshift energy delay is
valid you've admitted that your own argument for a redshift in concord
with the radiation pressure is flawed.

If you have a clean argument that denies this conclusion then why
don't you build it completely here? By tiptoeing along I see evidence
that you are putting stones in a wall, and the wall has grown quite
long here on this thread. If you'd like I'll go to the trouble of
producing a series of falsification. These are not falsifications that
I have constructed; they are conflicts within your own words, which I
must extract. The language that we use, particularly as math or
science type people should be strong. The mind should be even stronger
so that to concede a flawed argument allows one to move on. Otherwise
we are dealing in a pile of accumulation that includes much fecal
matter mixed with our food.

- Tim

>
> The power is different (since the cycles pass by at a different rate) due
> to redshift or blueshift, which is what is claimed, not that cycles
> disappear or that energy is "lost".
>
> As for falsifying your redshift argument (P=0, so a=0 and F=0), this is
> the point of dealing with some elementary points of power, velocity, and
> acceleration in classical mechanics first.
>
> --
> Timo

From: Timo Nieminen on 14 Jul 2010 19:24
On Wed, 14 Jul 2010, Tim BandTech.com wrote:

> If you have a clean argument that denies this conclusion then why
> don't you build it completely here?

The whole argument has been presented; what has followed has been trying
to find out where you disagree with it.

It doesn't help that you will - for post after post - refuse to answer
direct questions intended to find out. It doesn't help that you will post
multiple paragraphs of waffle instead of a simple "yes" or "no".

For example, do you agree that P=Fv is correct? (Here, P is the rate of
doing work on an object moving at v by a force F. Since this rate of
doing work is a power, I've used the traditional "P".)

If yes, we can move on to the next step.

If no, can you provide a falsification, an opposing derivation, or such?

We have, so far, agreed:

(a) The power of an EM beam (i.e., the energy flux) is affected by
redshift or blueshift.

(b) The cycles don't disappear. (I don't see why you think this is
relevant, but let me leave it here on the list.)

Can we add (c) P=Fv to the list?

--
Timo

From: Tim BandTech.com on 15 Jul 2010 09:46
On Jul 14, 7:24 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
> On Wed, 14 Jul 2010, Tim BandTech.com wrote:
> > If you have a clean argument that denies this conclusion then why
> > don't you build it completely here?
>
> The whole argument has been presented; what has followed has been trying
> to find out where you disagree with it.
>
> It doesn't help that you will - for post after post - refuse to answer
> direct questions intended to find out. It doesn't help that you will post
> multiple paragraphs of waffle instead of a simple "yes" or "no".
>
> For example, do you agree that P=Fv is correct? (Here, P is the rate of
> doing work on an object moving at v by a force F. Since this rate of
> doing work is a power, I've used the traditional "P".)

I agree that under some conditions that the equation you use is
correct. A more careful expression is:
P(t) = F(t) dot v(t)
where F and v are vectors, generally in (x,y,z) space. However, if we
take a beam of light whose intensity over some small area is 1000
watts, we should not expect to receive 1000 watts of mechanical work
from that light source. In this way I will be able to falsify your own
claim here. You have not constructed any scenario, nor have you
expressed any motivation for why you discuss this formula. Here I am
jumping through a hoop for you, and suspect that you are merely going
to present a long series of such hoops, rather than provide a complete
argument.

I stand by my original argument that a perfect reflector should not
accelerate under the exposure to light simply because in order for it
to accelerate it must have absorbed some of the light's energy, which
then means that it was not a perfect reflector.

When such an argument is made and the argument is incorrect there is
generally a literal falsification of the grammar, or at the very least
some part of the grammar requires qualification. I am open to a deeper
analysis that allows for motion of the mirror, but seriously doubt
that it will come from a classical physics argument. Supposedly the
Nichols experiment has been repeated in strong vacuum and is
consistent with the theory of radiation pressure, yet where this
additional energy comes from is troubling. By falsifying my own simple
argument above we could get somewhere. I'm fairly comfortable with the
idea that there are no perfect reflectors, but then the mechanism of
the translation of this small portion of the light's energy will need
to be addressed.

That there would be a charge differential built up on the reflector I
find believable. Then when we discuss the photon momentum we have a
mechanism other than a massive photon. The directed energy of the
electromagnetic wave interacts with electrons in such a way that when
they are travelliung sympathetically (the electron is orbiting away
from the source) the relative speed difference is low, and so the
possibility of a standing wave synching up here is a pretty good loose
paradigm. Anyway, this sort of mechanism I find more interesting than
abstracting.

There is a conflict in photon/wave theory that we are taught to eat
called particle-wave duality. This discussion is one way to attack
that paradigm, or at least to transform the conflict. The photon
momentum discussion just dodges photon mass, but still makes use of
the kilogram within it's units. From our earlier discussion I realize
that photon momentum is the particle equivalent of the wave theory's
radiation pressure. I have yet to see the radiation pressure as
sensible.

- Tim

>
> If yes, we can move on to the next step.
>
> If no, can you provide a falsification, an opposing derivation, or such?
>
> We have, so far, agreed:
>
> (a) The power of an EM beam (i.e., the energy flux) is affected by
> redshift or blueshift.
>
> (b) The cycles don't disappear. (I don't see why you think this is
> relevant, but let me leave it here on the list.)
>
> Can we add (c) P=Fv to the list?
>
> --
> Timo

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