Prev: PNAC of course said it "wanted" the 9/11 attacks".
Next: Arctic sea ice extent
From: Timo Nieminen on 8 Jul 2010 18:02
On Jul 9, 7:10 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com>
wrote:
> Timo Nieminen wrote:
> > On Jul 7, 1:35 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com>
> > wrote:
> >> Yes, this is the lab table reference frame; the one we'd be able to do
> >> measurements from most easily. I see you've put lost in quotes here so I
> >> presume you do have further analysis within this frame, otherwise, when
> >> we remove these quotes we have a conservation of energy problem.
>
> > Energy before = energy after, no loss, no problem.
>
> You've deleted the most important part of my prior post.
> The redshift has been misinterpreted by both of us. It is merely an
> energy delay, for so long as the number of cycles is preserved then the
> energy is conserved. I've been discussing a 50% redshift, and a pulse 1
> ns long from the source will come back 2 ns long at 50% redshift. The
> pulse will have the same number of cycles and so this conservation of
> cycles is a pretty good paradigm.

You can call it a "delay". But it's a funny kind of delay. For
constant motion, at 50% redshift, 20 cycles go in, and 10 cycles come
out, in some time interval t. So, what is the delay of a given cycle?
Is the 10th cycle delayed by t/2? If so, the 20th cycle is delayed by
t, the 40th by 2t, etc. It isn't a constant delay.

That's in classical wave language. In quantum language, each photon
loses energy. The photons don't come as N photons per cycle, or with M
cycles per photon, so there isn't any delay per photon.

> I was just reading Scientific American, and they pose a conservation of
> energy problem in the cosmological redshift, but they never came up with
> this answer. This is a general answer and applies there too.

No, it doesn't apply here. The equivalent would be looking at the
energy of an object or a wave in an _accelerated_ frame. In a
uniformly accelerated frame, an initially stationary object has a
velocity of v=at. What is its KE? If we assume that KE = (1/2)mv^2
applies in an acceleated frame, we have KE appearing from nowhere.
Then we have a potential problem with either the conservation of
energy or, at least, with the question of what is energy in an
accelerated frame.

The radiation pressure thing can be analysed using inertial frames,
the redshift occurs in inertial frames. Different.

> Now, returning to the reflector, if a perfect reflector returns all of
> the energy from the source back to the source then by conservation of
> energy no work can have been done on the reflector. There is no redshift
>   energy loss. There is no acceleration, theoretically speaking. So I
> can still claim that there is a conflict between theory and experiment.
> Experiment shows a deflection, and supposedly even a doubling effect
> over an absorber, within photon momentum theory. Conservation of energy
> requires that the perfect reflector cannot accelerate.
>
> Energy before = energy after, no loss, no acceleration.
> I suspect you'll address this in your next post.

Energy before = energy after gives no acceleration if no work is done.
Work is done. We have power = Fv; the rate of doing work on an object
depends on the speed.

A perfect reflector is a _lossless_ reflector. For reflection of
particles, that means that no mechanical energy is lost, KE before -
KE after. This is not the same as KE of the reflected object not
changing - some of the KE can be transferred to the reflector. For
reflection a wave, similar. No energy is lost through absorption. You
say "if a perfect reflector returns all of the energy from the source
back to the source then by conservation of energy no work can have
been done on the reflector." Indeed, this is so. But in general, it
_doesn't_ return all energy from the source back to the source. It
does so when the reflector is stationary. Also in that case, there is
no work being done. Experimentally, nobody measures work being done at
the instant when the reflector is stationary. So, what conflict
between theory and experiment?

The reduction in energy due to a moving reflector, dependence of work
on speed, very small force despite a high KE, the factor of 2 with
reflection vs absorption, etc. - pretty much all of the points that
you complain about - occur in the calculation of force due to the
reflection of a continuous stream of high-speed matter using classical
mechanics.

Do this calculation. Do this calculation in the (instantaneous) rest
frame of the reflector, and in frames when the reflector is moving
towards and away from the direction of the source. Zero work, positive
work, and negative work. In all cases, you will see that energy is
conserved, with no absorption.
From: Tim Golden BandTech.com on 9 Jul 2010 09:15
Timo Nieminen wrote:
> On Jul 9, 7:10 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com>
> wrote:
>> Timo Nieminen wrote:
>>> On Jul 7, 1:35 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com>
>>> wrote:
>>>> Yes, this is the lab table reference frame; the one we'd be able to do
>>>> measurements from most easily. I see you've put lost in quotes here so I
>>>> presume you do have further analysis within this frame, otherwise, when
>>>> we remove these quotes we have a conservation of energy problem.
>>> Energy before = energy after, no loss, no problem.
>> You've deleted the most important part of my prior post.
>> The redshift has been misinterpreted by both of us. It is merely an
>> energy delay, for so long as the number of cycles is preserved then the
>> energy is conserved. I've been discussing a 50% redshift, and a pulse 1
>> ns long from the source will come back 2 ns long at 50% redshift. The
>> pulse will have the same number of cycles and so this conservation of
>> cycles is a pretty good paradigm.
>
> You can call it a "delay". But it's a funny kind of delay. For
> constant motion, at 50% redshift, 20 cycles go in, and 10 cycles come
> out, in some time interval t. So, what is the delay of a given cycle?
> Is the 10th cycle delayed by t/2? If so, the 20th cycle is delayed by
> t, the 40th by 2t, etc. It isn't a constant delay.

No. 20 cycles go in, and 20 cycles come out. The 20 that come out take
longer to return to the source, but this is accounted for by the Doppler
shift and space analysis. Whatever the redshift is determines the delay.
It is easiest to think in terms of a wave packet and so to consider a
packet of T second duration (wave interpretation) if it is redshifted by
a reflector travelling v away from the source the reflected frequency is
f ( 1 - v / c )
where f is the frequency of the source. The redshift as a ratio is then
( 1 - v / c )
The duration of the packet becomes
T / ( 1 - v / c )
There is no need to worry about which cycle got delayed by how much.
This is as one might pulse a laser by a clock in short bursts. The
redshifted bursts are longer in duration. So are the pauses between
bursts. There is no problem with this analysis. It is a conservation of
cycles, and I do believe that you will eventually admit that if the
pulse had 1E6 cycles hitting the perfect reflector that it will have 1E6
cycles returning to the source, no matter what frequency. This is a
conservation.

>
> That's in classical wave language. In quantum language, each photon
> loses energy. The photons don't come as N photons per cycle, or with M
> cycles per photon, so there isn't any delay per photon.

Yes, it is a wave interpretation, but so is much of the analysis we've
been doing. Certainly we'll have to admit that the photon's return to
the source is delayed. Since the photon is a packet of energy then the
delay will be consistent. It does seem to make a statement on the
photon: that photons do have a duration of their wavelength. This is a
good thing. Within their own frame the photons are like little rods
instantaneously travelling from source to destination, so by requiring
this duration some of the photon quality can be preserved in their frame.

>
>> I was just reading Scientific American, and they pose a conservation of
>> energy problem in the cosmological redshift, but they never came up with
>> this answer. This is a general answer and applies there too.
>
> No, it doesn't apply here. The equivalent would be looking at the
> energy of an object or a wave in an _accelerated_ frame. In a
> uniformly accelerated frame, an initially stationary object has a
> velocity of v=at. What is its KE? If we assume that KE = (1/2)mv^2
> applies in an acceleated frame, we have KE appearing from nowhere.
> Then we have a potential problem with either the conservation of
> energy or, at least, with the question of what is energy in an
> accelerated frame.

You'd have to read the article. They present redshift as an energy
conservation problem. They also maintain conservation of energy, but
this explanation, which is a clear interpretation is missing. So, is
this a new interpretation that I am offering? I really doubt it. It is
more a matter of awareness and corrective thinking in terms of the time
domain. Delay is a good context. The redshift of light in the universe
does not mean that the energy disappeared; it means that the energy is
still coming and will still be coming for quite some time. Even if a
star that is 1E7 light years away is extinguished instantaneously, the
light will still be coming at us for for 1E7 years.
To the SciAm people this is energy lost (under their first context), but
under my interpretation it is energy delayed.

>
> The radiation pressure thing can be analysed using inertial frames,
> the redshift occurs in inertial frames. Different.
>
>> Now, returning to the reflector, if a perfect reflector returns all of
>> the energy from the source back to the source then by conservation of
>> energy no work can have been done on the reflector. There is no redshift
>> energy loss. There is no acceleration, theoretically speaking. So I
>> can still claim that there is a conflict between theory and experiment.
>> Experiment shows a deflection, and supposedly even a doubling effect
>> over an absorber, within photon momentum theory. Conservation of energy
>> requires that the perfect reflector cannot accelerate.
>>
>> Energy before = energy after, no loss, no acceleration.
>> I suspect you'll address this in your next post.
>
> Energy before = energy after gives no acceleration if no work is done.
> Work is done. We have power = Fv; the rate of doing work on an object
> depends on the speed.
>
> A perfect reflector is a _lossless_ reflector. For reflection of
> particles, that means that no mechanical energy is lost, KE before -
> KE after. This is not the same as KE of the reflected object not
> changing - some of the KE can be transferred to the reflector. For
> reflection a wave, similar. No energy is lost through absorption. You
> say "if a perfect reflector returns all of the energy from the source
> back to the source then by conservation of energy no work can have
> been done on the reflector." Indeed, this is so. But in general, it
> _doesn't_ return all energy from the source back to the source. It
> does so when the reflector is stationary. Also in that case, there is
> no work being done. Experimentally, nobody measures work being done at
> the instant when the reflector is stationary. So, what conflict
> between theory and experiment?

We're back to square one here. I will just stick with the current
context and explain that the interpretation above falsifies your own
statement:
" in general, it doesn't return all energy from the source
back to the source."
The redshift analysis that we are now focusing on does expose that the
energy is all returned. It just takes some time to return.

There is the additional conflict that the work by Nichols experiment
does operate at zero velocity and zero redshift. I can say therefor that
the reflector must not be perfect, but have not provided an explanation.
I've only falsified the usual claim. Also the title of this thread is
falsified, and I'm the one who wrote that.

>
> The reduction in energy due to a moving reflector, dependence of work
> on speed, very small force despite a high KE, the factor of 2 with
> reflection vs absorption, etc. - pretty much all of the points that
> you complain about - occur in the calculation of force due to the
> reflection of a continuous stream of high-speed matter using classical
> mechanics.
>
> Do this calculation. Do this calculation in the (instantaneous) rest
> frame of the reflector, and in frames when the reflector is moving
> towards and away from the direction of the source. Zero work, positive
> work, and negative work. In all cases, you will see that energy is
> conserved, with no absorption.

You keep saying to do this calculation. I have done it and come up with
an infinite mass at the reflector in order to return all of the energy
to the source. This implies zero acceleration and is inconsistent with
the Nichols experiment.

- Tim
From: Timo Nieminen on 9 Jul 2010 18:52
On Jul 9, 11:15 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com>
wrote:
> Timo Nieminen wrote:
> > On Jul 9, 7:10 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com>
> > wrote:
> > You can call it a "delay". But it's a funny kind of delay. For
> > constant motion, at 50% redshift, 20 cycles go in, and 10 cycles come
> > out, in some time interval t. So, what is the delay of a given cycle?
> > Is the 10th cycle delayed by t/2? If so, the 20th cycle is delayed by
> > t, the 40th by 2t, etc. It isn't a constant delay.
>
> No. 20 cycles go in, and 20 cycles come out. The 20 that come out take
> longer to return to the source, but this is accounted for by the Doppler
> shift and space analysis.

They take longer by a longer and longer amount of time. It isn't a
constant delay.

> There is no problem with this analysis. It is a conservation of
> cycles, and I do believe that you will eventually admit that if the
> pulse had 1E6 cycles hitting the perfect reflector that it will have 1E6
> cycles returning to the source, no matter what frequency. This is a
> conservation.

Something that is the same in all the reference frames we might
choose. Maybe not what we would call a conservation law. This is
invariance of phase, not conservation of energy.

> It does seem to make a statement on the
> photon: that photons do have a duration of their wavelength. This is a
> good thing. Within their own frame the photons are like little rods
> instantaneously travelling from source to destination, so by requiring
> this duration some of the photon quality can be preserved in their frame.

No. Photons do not have a length of their wavelength They are not like
little rods. They are not like little classical particles. No, they
don't have "their own frame" (in the sense of having an inertial rest
frame). "Packet of energy" might be OK, but don't over-analogise from
it. [Irrelevant to the current topic, though, so perhaps let us ignore
this.]

> You'd have to read the article. They present redshift as an energy
> conservation problem.

Cosmological redshift. A "problem" for conservation of energy of the
universe. As the article concludes, "So the universe does not violate
the conservation of energy; rather it lies outside that law’s
jurisdiction."

No relative motion needed for cosmological redshift. Here, we're just
talking about conventional relative-velocity Doppler shift.

> >> Now, returning to the reflector, if a perfect reflector returns all of
> >> the energy from the source back to the source then by conservation of
> >> energy no work can have been done on the reflector. There is no redshift
> >>   energy loss. There is no acceleration, theoretically speaking. So I
> >> can still claim that there is a conflict between theory and experiment..
> >> Experiment shows a deflection, and supposedly even a doubling effect
> >> over an absorber, within photon momentum theory. Conservation of energy
> >> requires that the perfect reflector cannot accelerate.
>
> >> Energy before = energy after, no loss, no acceleration.
> >> I suspect you'll address this in your next post.
>
> > Energy before = energy after gives no acceleration if no work is done..
> > Work is done. We have power = Fv; the rate of doing work on an object
> > depends on the speed.
>
> > A perfect reflector is a _lossless_ reflector. For reflection of
> > particles, that means that no mechanical energy is lost, KE before -
> > KE after. This is not the same as KE of the reflected object not
> > changing - some of the KE can be transferred to the reflector. For
> > reflection a wave, similar. No energy is lost through absorption. You
> > say "if a perfect reflector returns all of the energy from the source
> > back to the source then by conservation of energy no work can have
> > been done on the reflector." Indeed, this is so. But in general, it
> > _doesn't_ return all energy from the source back to the source. It
> > does so when the reflector is stationary. Also in that case, there is
> > no work being done. Experimentally, nobody measures work being done at
> > the instant when the reflector is stationary. So, what conflict
> > between theory and experiment?
>
> We're back to square one here. I will just stick with the current
> context and explain that the interpretation above falsifies your own
> statement:
>     " in general, it doesn't return all energy from the source
>     back to the source."
> The redshift analysis that we are now focusing on does expose that the
> energy is all returned. It just takes some time to return.

An ever-increasing time.

> There is the additional conflict that the work by Nichols experiment
> does operate at zero velocity and zero redshift. I can say therefor that
> the reflector must not be perfect, but have not provided an explanation.
> I've only falsified the usual claim. Also the title of this thread is
> falsified, and I'm the one who wrote that.

Apply the same logic to the reflection of classical particles, and you
can say that there are no elastic collisions.

> > The reduction in energy due to a moving reflector, dependence of work
> > on speed, very small force despite a high KE, the factor of 2 with
> > reflection vs absorption, etc. - pretty much all of the points that
> > you complain about - occur in the calculation of force due to the
> > reflection of a continuous stream of high-speed matter using classical
> > mechanics.
>
> > Do this calculation. Do this calculation in the (instantaneous) rest
> > frame of the reflector, and in frames when the reflector is moving
> > towards and away from the direction of the source. Zero work, positive
> > work, and negative work. In all cases, you will see that energy is
> > conserved, with no absorption.
>
> You keep saying to do this calculation. I have done it and come up with
> an infinite mass at the reflector in order to return all of the energy
> to the source.

If so, then you've done it wrong, or haven't done enough of it. Yes,
with an infinite mass, the velocity of the reflector doesn't change.
But there is more:

(1) For the continuous stream case, at any instant when the speed of
the reflector is zero, no work is being done. Depending on the choice
of reference frame, you can have zero power, positive power, or
negative power.

(2) For the 2-object discrete case, in the centre of momentum rest
frame, no work is done. Depending on the choice of reference frame,
you can have zero work, positive work, or negative work.

(3) In either case, for an infinite-mass reflector, you can _still_
have zero work, positive work, or negative work.

In every case above, the force due to the reflection is the same. The
force is unaffected by the choice of reference frame (assuming that
you keep it simple and use Galileian relativity).

But since you went so far as you did, would you claim that there are
no elastic collisions? After all, for a supposedly elastic collision,
you can always choose a reference frame where the KE of the reflected
particle goes down.

For the radiation pressure thing, you are basically claiming that
there can't be any such loss of KE, and that no KE can be transferred
to the reflector without violating conservation of KE. Just because
the incident "thing" is radiation, not particles, doesn't make it
somehow magic. We're still talking about energy and momentum. The same
conservation laws still apply, both quantities still depend on our
choice of reference frame, and changes in both quantities still depend
on our choice of reference frame.

Also include the following, especially (5):

(4) Consider the effect of an additional restraining force stopping
the reflector from accelerating.

(5) For the reflection of a continuous stream of matter, what is the
dependence on the power on the reflector (i.e., the rate of doing work
on the reflector) on the acceleration? Write down the formula
explicitly. Then reconsider the statement below.

> This implies zero acceleration and is inconsistent with
> the Nichols experiment.

It implies zero velocity when there is zero redshift, that's all. See
formula from (5) above.

--
Timo
From: Tim Golden BandTech.com on 10 Jul 2010 08:32
Timo Nieminen wrote:
> On Jul 9, 11:15 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com>
> wrote:
>> Timo Nieminen wrote:
>>> On Jul 9, 7:10 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com>
>>> wrote:
>>> You can call it a "delay". But it's a funny kind of delay. For
>>> constant motion, at 50% redshift, 20 cycles go in, and 10 cycles come
>>> out, in some time interval t. So, what is the delay of a given cycle?
>>> Is the 10th cycle delayed by t/2? If so, the 20th cycle is delayed by
>>> t, the 40th by 2t, etc. It isn't a constant delay.
>> No. 20 cycles go in, and 20 cycles come out. The 20 that come out take
>> longer to return to the source, but this is accounted for by the Doppler
>> shift and space analysis.
>
> They take longer by a longer and longer amount of time. It isn't a
> constant delay.

For a constant velocity this is a linear problem. There is nothing wrong
here.

>
>> There is no problem with this analysis. It is a conservation of
>> cycles, and I do believe that you will eventually admit that if the
>> pulse had 1E6 cycles hitting the perfect reflector that it will have 1E6
>> cycles returning to the source, no matter what frequency. This is a
>> conservation.
>
> Something that is the same in all the reference frames we might
> choose. Maybe not what we would call a conservation law. This is
> invariance of phase, not conservation of energy.

Actually, it pretty directly transposes to conservation of energy via
e = h f .
10 cycles at one frequency are equivalent in energy to ten cycles at any
other frequency. This has nothing to do with phase until some strict
experiment is setup. This is a plane wave of monochromatic light, and
matches the Maxwell wave interpretation, though the energy can be shown
through the photon energy equation.

>
>> It does seem to make a statement on the
>> photon: that photons do have a duration of their wavelength. This is a
>> good thing. Within their own frame the photons are like little rods
>> instantaneously travelling from source to destination, so by requiring
>> this duration some of the photon quality can be preserved in their frame.
>
> No. Photons do not have a length of their wavelength They are not like
> little rods. They are not like little classical particles. No, they
> don't have "their own frame" (in the sense of having an inertial rest
> frame). "Packet of energy" might be OK, but don't over-analogise from
> it. [Irrelevant to the current topic, though, so perhaps let us ignore
> this.]

I understand that this is a sidetrack, but still, since you've decided
to critique it, the photon's reference frame travels at c, and so due to
time dilation and length contraction is instantaneous in its own frame;
it ages zero seconds from transmission to reception. Also the spatial
dimension undergoes lengthening, bringing it toward infinity. I am not
saying that the length of the photon is the wavelength. I am saying that
the duration of time that this rod exists for, which instantly
propagates, corresponds to the wavelength, for this is the amount of
time that the energy transfers in. This is not standard relativity
theory, but the math does account for it. You say that they do not have
any inertial rest frame, but I just gave them one without conflict,
other than a division by zero for the spatial dimension.

>
>> You'd have to read the article. They present redshift as an energy
>> conservation problem.
>
> Cosmological redshift. A "problem" for conservation of energy of the
> universe. As the article concludes, "So the universe does not violate
> the conservation of energy; rather it lies outside that law�s
> jurisdiction."

They failed to assess the energy delay of redshifted light. Who else is
failing to do this?

>
> No relative motion needed for cosmological redshift. Here, we're just
> talking about conventional relative-velocity Doppler shift.
>
>>>> Now, returning to the reflector, if a perfect reflector returns all of
>>>> the energy from the source back to the source then by conservation of
>>>> energy no work can have been done on the reflector. There is no redshift
>>>> energy loss. There is no acceleration, theoretically speaking. So I
>>>> can still claim that there is a conflict between theory and experiment.
>>>> Experiment shows a deflection, and supposedly even a doubling effect
>>>> over an absorber, within photon momentum theory. Conservation of energy
>>>> requires that the perfect reflector cannot accelerate.
>>>> Energy before = energy after, no loss, no acceleration.
>>>> I suspect you'll address this in your next post.
>>> Energy before = energy after gives no acceleration if no work is done.
>>> Work is done. We have power = Fv; the rate of doing work on an object
>>> depends on the speed.
>>> A perfect reflector is a _lossless_ reflector. For reflection of
>>> particles, that means that no mechanical energy is lost, KE before -
>>> KE after. This is not the same as KE of the reflected object not
>>> changing - some of the KE can be transferred to the reflector. For
>>> reflection a wave, similar. No energy is lost through absorption. You
>>> say "if a perfect reflector returns all of the energy from the source
>>> back to the source then by conservation of energy no work can have
>>> been done on the reflector." Indeed, this is so. But in general, it
>>> _doesn't_ return all energy from the source back to the source. It
>>> does so when the reflector is stationary. Also in that case, there is
>>> no work being done. Experimentally, nobody measures work being done at
>>> the instant when the reflector is stationary. So, what conflict
>>> between theory and experiment?
>> We're back to square one here. I will just stick with the current
>> context and explain that the interpretation above falsifies your own
>> statement:
>> " in general, it doesn't return all energy from the source
>> back to the source."
>> The redshift analysis that we are now focusing on does expose that the
>> energy is all returned. It just takes some time to return.
>
> An ever-increasing time.
>
>> There is the additional conflict that the work by Nichols experiment
>> does operate at zero velocity and zero redshift. I can say therefor that
>> the reflector must not be perfect, but have not provided an explanation.
>> I've only falsified the usual claim. Also the title of this thread is
>> falsified, and I'm the one who wrote that.
>
> Apply the same logic to the reflection of classical particles, and you
> can say that there are no elastic collisions.
>
>>> The reduction in energy due to a moving reflector, dependence of work
>>> on speed, very small force despite a high KE, the factor of 2 with
>>> reflection vs absorption, etc. - pretty much all of the points that
>>> you complain about - occur in the calculation of force due to the
>>> reflection of a continuous stream of high-speed matter using classical
>>> mechanics.
>>> Do this calculation. Do this calculation in the (instantaneous) rest
>>> frame of the reflector, and in frames when the reflector is moving
>>> towards and away from the direction of the source. Zero work, positive
>>> work, and negative work. In all cases, you will see that energy is
>>> conserved, with no absorption.
>> You keep saying to do this calculation. I have done it and come up with
>> an infinite mass at the reflector in order to return all of the energy
>> to the source.
>
> If so, then you've done it wrong, or haven't done enough of it. Yes,
> with an infinite mass, the velocity of the reflector doesn't change.
> But there is more:
>
> (1) For the continuous stream case, at any instant when the speed of
> the reflector is zero, no work is being done. Depending on the choice
> of reference frame, you can have zero power, positive power, or
> negative power.
>
> (2) For the 2-object discrete case, in the centre of momentum rest
> frame, no work is done. Depending on the choice of reference frame,
> you can have zero work, positive work, or negative work.
>
> (3) In either case, for an infinite-mass reflector, you can _still_
> have zero work, positive work, or negative work.
>
> In every case above, the force due to the reflection is the same. The
> force is unaffected by the choice of reference frame (assuming that
> you keep it simple and use Galileian relativity).
>
> But since you went so far as you did, would you claim that there are
> no elastic collisions? After all, for a supposedly elastic collision,
> you can always choose a reference frame where the KE of the reflected
> particle goes down.
>
> For the radiation pressure thing, you are basically claiming that
> there can't be any such loss of KE, and that no KE can be transferred
> to the reflector without violating conservation of KE. Just because
> the incident "thing" is radiation, not particles, doesn't make it
> somehow magic. We're still talking about energy and momentum. The same
> conservation laws still apply, both quantities still depend on our
> choice of reference frame, and changes in both quantities still depend
> on our choice of reference frame.
>
> Also include the following, especially (5):
>
> (4) Consider the effect of an additional restraining force stopping
> the reflector from accelerating.
>
> (5) For the reflection of a continuous stream of matter, what is the
> dependence on the power on the reflector (i.e., the rate of doing work
> on the reflector) on the acceleration? Write down the formula
> explicitly. Then reconsider the statement below.

Whatever acceleration is imparted to the reflector will be drained from
the source. The analogy to all of this mechanical contraption is only
partial. I've just shown you that the redshifted perfect reflector
returns all of the energy to the source. Therefore the reflector does
not accelerate. I understand that this is in conflict with experiment.

>
>> This implies zero acceleration and is inconsistent with
>> the Nichols experiment.
>
> It implies zero velocity when there is zero redshift, that's all. See
> formula from (5) above.

No. I have just shown that the redshift is only a delay in the return of
energy. All of the energy is returned by a perfect reflector, even at
relativistic velocity. Whether zero redshift, or 50% redshift. There is
no accounting for reflector acceleration here.

- Tim

>
> --
> Timo
From: Timo Nieminen on 10 Jul 2010 17:38
On Jul 10, 10:32 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com>
wrote:
> Timo Nieminen wrote:
> > They take longer by a longer and longer amount of time. It isn't a
> > constant delay.
>
> For a constant velocity this is a linear problem. There is nothing wrong
> here.

It still isn't a constant delay. The energy in during time T equals
the energy out during time 2T. Whether or not the cycles eventually
get back doesn't matter - the power matters, not eventual conservation
of energy.

Deficit spending at twice one's income is still deficit spending,
getting more and more in debt, even if every borrowed dollar is paid
back (with an every increasing delay).

> > Something that is the same in all the reference frames we might
> > choose. Maybe not what we would call a conservation law. This is
> > invariance of phase, not conservation of energy.
>
> Actually, it pretty directly transposes to conservation of energy via
>     e = h f .

OK, use E=hf if you want. The photons are countable, the count must
stay the same despite the change in reference frame, the energy of the
photons depends on the choice of reference frame. The redshifted wave
has less power. End of story, yes?

> 10 cycles at one frequency are equivalent in energy to ten cycles at any
> other frequency.

Does it matter? For the 50% redshift case, our wavelength is doubled.
If the total energy in the cycle is the same, then we have E', the E
field in the redshift frame, given by E' = E/sqrt(2). If the total
energy of the cycle is the same, and the cycle now occupies twice the
spatial extent, the energy density must be halved, and thus E^2 must
be halved. Likewise, H'=H/sqrt(2). The energy flux, the Poynting
vector, ExH, is halved as well.

Since the EM wave travels at the same speed in both frames, the energy
flux is proportional to the energy density. The energy flux, the
power, is what matters.

One can do the full SR calculation, too. For a plane EM wave, when
transforming to a reference frame moving along the direction of
propagation such that we have a redshift, we have (ref., e.g.,
http://scienceworld.wolfram.com/physics/ElectromagneticFieldTensor.html)

E = gamma (E - vB)
B = gamma (B - v/c^2 E)

> This has nothing to do with phase until some strict
> experiment is setup.

No, "phase" tells us where the cycles begin and end. N cycles in
meaning N cycles out is invariance of phase. But irrelevant.

> This is a plane wave of monochromatic light, and
> matches the Maxwell wave interpretation, though the energy can be shown
> through the photon energy equation.

E = hf tells us the power is reduced. Maxwell tells us the power is
reduced. Surely this is enough?

> I understand that this is a sidetrack, but still, since you've decided
> to critique it, the photon's reference frame travels at c [...]
> This is not standard relativity
> theory, but the math does account for it. You say that they do not have
> any inertial rest frame, but I just gave them one without conflict,
> other than a division by zero for the spatial dimension.

Invalid mathematics is not "without conflict". The math _doesn't_
account for it. The equivalent choice of reference frame in Galileian
relativity is one at infinite velocity. Doesn't work either,
mathematically.

> >> You'd have to read the article. They present redshift as an energy
> >> conservation problem.
>
> > Cosmological redshift. A "problem" for conservation of energy of the
> > universe. As the article concludes, "So the universe does not violate
> > the conservation of energy; rather it lies outside that law’s
> > jurisdiction."
>
> 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?

How is this at all relevant to radiation pressure?

> >> There is the additional conflict that the work by Nichols experiment
> >> does operate at zero velocity and zero redshift. I can say therefor that
> >> the reflector must not be perfect, but have not provided an explanation.
> >> I've only falsified the usual claim. Also the title of this thread is
> >> falsified, and I'm the one who wrote that.
>
> > Apply the same logic to the reflection of classical particles, and you
> > can say that there are no elastic collisions.

Doesn't the same logic imply that there can't be elastic collisions?

> > (5) For the reflection of a continuous stream of matter, what is the
> > dependence on the power on the reflector (i.e., the rate of doing work
> > on the reflector) on the acceleration? Write down the formula
> > explicitly. Then reconsider the statement below.
>
> Whatever acceleration is imparted to the reflector will be drained from
> the source. The analogy to all of this mechanical contraption is only
> partial.

If the motion of the reflector can be described using classical
mechanics, the motion of the reflector can be described using
classical mechanics. If we have a reflector of mass m, with a force of
F being exerted on it, what is the dependence on the rate of change of
its KE on its acceleration? On its velocity?

What is the rate of change of its KE when v=0?

What is the rate of transfer of energy from the source of the force to
the reflector when v=0?

> I've just shown you that the redshifted perfect reflector
> returns all of the energy to the source. Therefore the reflector does
> not accelerate. I understand that this is in conflict with experiment.

When handwaving "theory" is in conflict with experiment, the
handwaving explanation is probably wrong. Acceleration is not
"drained" from the source. Not in classical mechanics, not in
relativistic mechanics, not in wave theory. You could re-state (badly,
IMHO) the concept of conservation of energy as "change in energy due
to acceleration is drained from the source". Note the difference.
Acceleration is not energy; we don't have a law of conservation of
acceleration.

Returning all energy back to the source with ever-increasing delays
doesn't help. Power matters. A difference between power in and power
out matters.

> >> This implies zero acceleration and is inconsistent with
> >> the Nichols experiment.
>
> > It implies zero velocity when there is zero redshift, that's all. See
> > formula from (5) above.
>
> No. I have just shown that the redshift is only a delay in the return of
> energy. All of the energy is returned by a perfect reflector, even at
> relativistic velocity. Whether zero redshift, or 50% redshift. There is
> no accounting for reflector acceleration here.

So, what is the dependence of the rate of doing work on the reflector
on the acceleration of the reflector?

--
Timo
First  |  Prev  |  Next  |  Last
Pages: 1 2 3 4 5 6 7 8 9
Prev: PNAC of course said it "wanted" the 9/11 attacks".
Next: Arctic sea ice extent