Volney wrote:

>> Your attributions are screwed up. I wrote that to "Tadd Muraro".
>> "Stan Fultoni" displays excellent knowledge of SR, albeit from a
>> different perspective than me. Tom Roberts
>
> Apologies to Stan, I don't know how he got into the attributions by
> mistake. He does appear to know what he's talking about. My point
> remains,
> the nymshifter troll's (by whatever name it's currently using) current
> technique is to post just enough physics to keep people such as yourself
> replying to it.

you are not easy to fool. The mosts are not replying being unskilled,
having no idea.

On 2022-08-16 16:30:13 +0000, Ed Lake said:

> When the earth is moving toward the binary star system, the light
> from BOTH stars will be received at a higher frequency and a shorter
> wavelength due to the earth's movement. When the earth is moving
> away from the binary star system, the light from BOTH stars will be
> received at a lower frequency and a longer wavelength. Motion by
> the binary stars has no effect on what is observed on earth.

In the northen hemisphere in the winter one can observe the
spectrum of β Aurigae and see that every second night it is
a mixture of blueshifted and redshifted light.

Mikko

On 2022-08-15 19:35:02 +0000, Ed Lake said:

> On Monday, August 15, 2022 at 12:14:45 PM UTC-5, Mikko wrote:
>> On 2022-08-15 16:41:30 +0000, Ed Lake said:
>> > No, it just means that the CAUSE of the "spectroscopic" effect is not
>> > defined. It MUST be caused by the earth in its orbit moving toward
>> > or away from the binary star. It cannot be caused by the binary
>> > star moving toward or away from a STATIONARY EARTH because
>> > the earth is NEVER stationary.
>> That is obviously false. When one of the stars is seen redshifted
>> the other is observed blueshifted. This cannot be caused by Earths
>> motion as Earth cannot both away and towards the binary at the
>> same time.

> "Redshifted" means the energy frequency is slower than "normal"and
> "blueshifted" mean the energy frequency is faster than "normal."

Energy is not faster or slower. It is less or more.

But that was not how de Sitter and other astronomers understood
"redshift" and "blueshift". To them those words refer to the measured
wavelength of the light, i.e., how light is diffracted in the
spectrometer.

Anyway, whether you measure frequency or energy, most of the time
one of the stars is redshifted and the other is blueshifted from
their average spectrum. The mere existence of this difference and
variation refutes your opinion that the motion of the star has
no effect on its spectrum.

Mikko

Volney <volney@invalid.invalid> wrote:

> On 8/15/2022 4:04 PM, Tom Roberts wrote:
> > On 8/15/22 12:19 AM, Volney wrote:
> >> On 8/14/2022 10:37 PM, Tom Roberts wrote:
>
> >>> Yes, a monochromatic EM source has a frequency; individual photons
> >>> do not.
> >>
> >> If an individual photon has a specific energy E, determined by
> >> whatever produced it, and if E=hf, why can't we say the frequency f =
> >> E/h?
> >
> > You can say whatever you want, and make mathematical inferences using
> > whatever math you want -- that does not mean they are correct.
>
> OK, I am really asking *why* it is not considered correct.
>
> Is it because you can never *measure* the frequency (not energy) of a
> single photon, as that requires more than one measurement while a photon
> can only be measured once (as it is destroyed)?
>
> You say that f=E/h is not a measurement (although E is measured) so it's
> irrelevant?
>
> Because E can never be exact (due to HUP, ?E*?t > ?), or processes in
> the source which make E vary, perhaps ever so slightly?
>
> > In particular: YOU DID NOT MEASURE THE FREQUENCY OF THE PHOTON.
>
> I understand that. I measured the energy of the photon. The photon (now
> destroyed) had a certain amount of energy. I thought equations like E=hf
> were pretty much fundamental, not so much?
>
> > All else is subject to interpretation and may be wrong because the
> > English words you use do not necessarily correspond to physical
> > phenomena, or even to the mathematical quantities in the theory. That
> > happens in an attempt to claim "the frequency of the photon is E/h".
> >
> > Tom Roberts
>
> What *is* E=hf supposed to mean?

Don't worry, it is you who has it right, not Tom Roberts.
Nowadays \hbar has a defined value, which we may take as 1,
and E=hf is just saying the same thing in different units.

As to measurement: a measurement of E is competely equivalent
to a measurement of f.
You just measure it in the way that is most convenient and practical,

Jan

Mikko <mikko.levanto@iki.fi> wrote:

> On 2022-08-16 16:30:13 +0000, Ed Lake said:
>
> > When the earth is moving toward the binary star system, the light
> > from BOTH stars will be received at a higher frequency and a shorter
> > wavelength due to the earth's movement. When the earth is moving
> > away from the binary star system, the light from BOTH stars will be
> > received at a lower frequency and a longer wavelength. Motion by
> > the binary stars has no effect on what is observed on earth.
>
> In the northen hemisphere in the winter one can observe the
> spectrum of ? Aurigae and see that every second night it is
> a mixture of blueshifted and redshifted light.

Guess you mean Beta Aurigae
<https://en.wikipedia.org/wiki/Beta_Aurigae>
which is a 4-day spectroscopic double.
(in fact one of the first discovered, ever)

Hint: this is usenet.
Greek letters should be spelled out in ASCII,
rather than trying to get them from some non-ASCII character set,
(which will render as ? for some)

Jan

On Tuesday, August 16, 2022 at 2:15:41 PM UTC-4, tjrob137 wrote:
> On 8/16/22 8:41 AM, Ken Seto wrote:
> > On Monday, August 15, 2022 at 10:55:16 PM UTC-4, tjrob137 wrote:
> >> It is physically impossible to measure the frequency of a photon.
> >
> > It is possible to measure the frequency of a wave-packet.
> Sure.
> > That means that a photon is not a particle.
> Nope. You got it backwards (as usual).
>
> That means that a photon is not a wave packet. Indeed, for EM radiation,
> wave packets are made of myriad photons. So a photon

So a photon is much smaller than a wave? Can you define the size of a photon for me ?.
I can define the size (length) of a sodium wave.....it is 589 nm)

>
> Note that the word "photon" is defined in QED.
Yeah it is a virtual particle which mean that it can be anyhing (any size) you want it to be.

>You, and others around
> here, attempt to use other meanings for the same word, causing confusion
> in thought and word. YOU (and they) are confused. And wrong.
>
> Tom Roberts

On Tuesday, August 16, 2022 at 2:49:09 PM UTC-5, Paul Alsing wrote:
> On Tuesday, August 16, 2022 at 9:30:14 AM UTC-7, wrote:
> > On Monday, August 15, 2022 at 3:53:51 PM UTC-5, Paul Alsing wrote:
>
> > > Ed, what makes you think that the light from one star is approaching and the light from the other is receding?
>
> > It appears I made a mistake.
> Ya think?
> > In our expanding universe, nearly every binary star system is moving
> > away from the earth.
> Evidence for this extraordinary claim, Ed? In an expanding universe, except for the local group of galaxies, all the rest appear to be moving away from each other, but within our Milky Way galaxy, the same thing is NOT true of its stars! Wherever did you get such an idea? That is exactly what I mean when I tell you that you do not know what you do not know!

Yeah, somehow I was thinking that binary stars were outside of the
Milky Way and would therefore be moving away. It appears I have lost
interest in this discussion and I'm just not paying attention.

Ed

On Wednesday, August 17, 2022 at 2:21:19 AM UTC-5, Mikko wrote:
> On 2022-08-15 19:35:02 +0000, Ed Lake said:
>
> > On Monday, August 15, 2022 at 12:14:45 PM UTC-5, Mikko wrote:
> >> On 2022-08-15 16:41:30 +0000, Ed Lake said:
> >> > No, it just means that the CAUSE of the "spectroscopic" effect is not
> >> > defined. It MUST be caused by the earth in its orbit moving toward
> >> > or away from the binary star. It cannot be caused by the binary
> >> > star moving toward or away from a STATIONARY EARTH because
> >> > the earth is NEVER stationary.
> >> That is obviously false. When one of the stars is seen redshifted
> >> the other is observed blueshifted. This cannot be caused by Earths
> >> motion as Earth cannot both away and towards the binary at the
> >> same time.
>
> > "Redshifted" means the energy frequency is slower than "normal"and
> > "blueshifted" mean the energy frequency is faster than "normal."
>
> Energy is not faster or slower. It is less or more.
>
> But that was not how de Sitter and other astronomers understood
> "redshift" and "blueshift". To them those words refer to the measured
> wavelength of the light, i.e., how light is diffracted in the
> spectrometer.
>
> Anyway, whether you measure frequency or energy, most of the time
> one of the stars is redshifted and the other is blueshifted from
> their average spectrum. The mere existence of this difference and
> variation refutes your opinion that the motion of the star has
> no effect on its spectrum.
>
> Mikko

Yes, it appears I need to do more thinking about binary stars, but I've
simply lost interest. Sorry.

Ed

On 2022-08-17 14:14:50 +0000, Ed Lake said:

> Yes, it appears I need to do more thinking about binary stars, but I've
> simply lost interest. Sorry.

It's OK to be uninterested about binary starts. Most peaople are.
Just don't write anything about Doppler shifts, radar guns, or
other related topics before you have properly thought about them.

Mikko

Ken Seto wrote:

>> That means that a photon is not a wave packet. Indeed, for EM
>> radiation,
>> wave packets are made of myriad photons. So a photon
>
> So a photon is much smaller than a wave? Can you define the size of a
> photon for me ?.
> I can define the size (length) of a sodium wave.....it is 589 nm)

quantum physics, idiot. You can't define sizes, not measure. You fucking
idiot. You are the reason they say relativity is right.

On 8/17/2022 10:52 AM, Bubba Romano wrote:
> Ken Seto wrote:
>
>>> That means that a photon is not a wave packet. Indeed, for EM
>>> radiation,
>>> wave packets are made of myriad photons. So a photon
>>
>> So a photon is much smaller than a wave? Can you define the size of a
>> photon for me ?.
>> I can define the size (length) of a sodium wave.....it is 589 nm)
>
> quantum physics, idiot. You can't define sizes, not measure. You fucking
> idiot. You are the reason they say relativity is right.

X X X

Please define the locations of the three X's above in only absolute
terms. Thanks.

On 8/17/22 3:49 AM, J. J. Lodder wrote:
> Volney <volney@invalid.invalid> wrote:
>> On 8/15/2022 4:04 PM, Tom Roberts wrote:
>>> On 8/15/22 12:19 AM, Volney wrote:
>> OK, I am really asking *why* it is not considered correct [to say
>> a photon has a frequency E/h].
>>
>> Is it because you can never *measure* the frequency (not energy) of
>> a single photon, as that requires more than one measurement while a
>> photon can only be measured once (as it is destroyed)?

That is one aspect of the reason.

The real reason is that a photon simply does not posses a frequency (or
wavelength). Speaking loosely, since the (real) photon propagates at c,
it has no elapsed proper time, and thus cannot vary over time, which is
the sine qua non of having a frequency. Note that the (real) photon is
simply transferring 4-momentum from one event to another (modulo spin,
and only for null-separated events); there's no frequency in that.

The equation E=hf applies to a monochromatic light SOURCE in its
inertial rest frame, relating the source and its photons; it simply does
NOT apply to the individual photons. Since the Doppler shifts for E and
f are the same, this equation can also be applied to a monochromatic
light beam in any inertial frame.

The fact that a photon's energy varies in different inertial frames
implies that its energy cannot possibly be intrinsic to the photon. So
even if E=hf applied to the photon, the frequency likewise could not be
intrinsic to the photon.

[If you examine the equations of QED, you'll find that
the photon propagator depends upon its 4-momentum --
there's no frequency in that. This is much more
complicated than my simple statement here.]

> Don't worry, it is you who has it right, not Tom Roberts.

Not true. See above.

> As to measurement: a measurement of E is competely equivalent to a
> measurement of f.

Nope. You are confusing a measurement of a light beam with a measurement
of a photon. See above -- E=hf simply does not apply to individual
photons, it applies to a monochromatic SOURCE (and by a mathematical
quirk to a monochromatic light beam).

Tom Roberts

On Wednesday, 24 August 2022 at 07:12:46 UTC+2, tjrob137 wrote:

> The real reason is that a photon simply does not posses a frequency (or
> wavelength). Speaking loosely, since the (real) photon propagates at c,
> it has no elapsed proper time, and thus cannot vary over time, which is
> the sine qua non of having a frequency. Note that the (real) photon is

It was obvious sure that you don't believe a word of
these wise things you sometimes write about the
reality and its models.

On Tuesday, August 23, 2022 at 10:12:46 PM UTC-7, tjrob137 wrote:
> On 8/17/22 3:49 AM, J. J. Lodder wrote:
> > Volney <vol...@invalid.invalid> wrote:
> >> On 8/15/2022 4:04 PM, Tom Roberts wrote:
> >>> On 8/15/22 12:19 AM, Volney wrote:
> >> OK, I am really asking *why* it is not considered correct [to say
> >> a photon has a frequency E/h].
> >>
> >> Is it because you can never *measure* the frequency (not energy) of
> >> a single photon, as that requires more than one measurement while a
> >> photon can only be measured once (as it is destroyed)?
> That is one aspect of the reason.
>
> The real reason is that a photon simply does not posses a frequency (or
> wavelength). Speaking loosely, since the (real) photon propagates at c,
> it has no elapsed proper time, and thus cannot vary over time, which is
> the sine qua non of having a frequency. Note that the (real) photon is
> simply transferring 4-momentum from one event to another (modulo spin,
> and only for null-separated events); there's no frequency in that.
>
> The equation E=hf applies to a monochromatic light SOURCE in its
> inertial rest frame, relating the source and its photons; it simply does
> NOT apply to the individual photons. Since the Doppler shifts for E and
> f are the same, this equation can also be applied to a monochromatic
> light beam in any inertial frame.
>
> The fact that a photon's energy varies in different inertial frames
> implies that its energy cannot possibly be intrinsic to the photon. So
> even if E=hf applied to the photon, the frequency likewise could not be
> intrinsic to the photon.
>
> [If you examine the equations of QED, you'll find that
> the photon propagator depends upon its 4-momentum --
> there's no frequency in that. This is much more
> complicated than my simple statement here.]
> > Don't worry, it is you who has it right, not Tom Roberts.
> Not true. See above.
> > As to measurement: a measurement of E is competely equivalent to a
> > measurement of f.
> Nope. You are confusing a measurement of a light beam with a measurement
> of a photon. See above -- E=hf simply does not apply to individual
> photons, it applies to a monochromatic SOURCE (and by a mathematical
> quirk to a monochromatic light beam).
>
> Tom Roberts

So, it's attached to the wavepacket, description?

In quantum physics "every particle has a frequency and a wavelength".

Or rather that's what I gather: the entire basis of Heisenberg uncertainty
principle is particle/wave duality, ....

Here though "photons", ..., "the emissive as tachyonic, for example the
radiative", if not "photons are tachyonic" is "photons are not bradyonic".

The balance of 1-1 photon-electron, energy in the electro-optic,
or as it were, besides "electron/photon detector", frequency and wavelength
are reciprocals, their units in energy's.

This is still for resolving "ultraviolet catastrophe". Also blackbody radiation vis-a-vis
optical and thermal, photon's still attached to the description, called part of wave-poacket.

For that photons result over the dimensionless in units, energy, I think all
that is subsumed in calling them wave-packets.

On August 23, 6 PM UTC-7, tjrob137 wrote:
>>> OK, I am really asking *why* it is not considered correct [to say
>>> a photon has a frequency E/h]
>
> The real reason is that a photon simply does not posses a frequency.
> Speaking loosely, since the (real) photon propagates at c,
> it has no elapsed proper time, and thus cannot vary over time, which is
> the sine qua non of having a frequency. Note that the (real) photon is
> simply transferring 4-momentum from one event to another (modulo spin,
> and only for null-separated events); there's no frequency in that.
> The equation E=hf applies to a monochromatic light SOURCE in its
> inertial rest frame, relating the source and its photons;

Given a large number of photons, traveling as a 'packet' (whatever that
means). That forms a classical light beam. Does the frequency spectrum
of that beam, as defined by Maxwell's model, correspond exactly to the
spectrum of the photons?

Quantum mechanics models everything as a wave in Hilbert space.
Waves interfere, therefore 'photon waves' must interfere. But to
compute the interference, one requires frequency and phase
information. E = hf is a scalar; where's the phase?

Has anything changed since Planck derived E = hf?
--
Rich

On 8/25/22 3:20 PM, RichD wrote:
> On August 23, 6 PM UTC-7, tjrob137 wrote:
>>>> OK, I am really asking *why* it is not considered correct [to
>>>> say a photon has a frequency E/h]
>>
>> The real reason is that a photon simply does not posses a
>> frequency. Speaking loosely, since the (real) photon propagates at
>> c, it has no elapsed proper time, and thus cannot vary over time,
>> which is the sine qua non of having a frequency. Note that the
>> (real) photon is simply transferring 4-momentum from one event to
>> another (modulo spin, and only for null-separated events); there's
>> no frequency in that. The equation E=hf applies to a monochromatic
>> light SOURCE in its inertial rest frame, relating the source and
>> its photons;
>
> Given a large number of photons, traveling as a 'packet' (whatever
> that means). That forms a classical light beam. Does the frequency
> spectrum of that beam, as defined by Maxwell's model, correspond
> exactly to the spectrum of the photons?

Approximately, yes; exactly, no.

> Quantum mechanics models everything as a wave in Hilbert space.

More relevant is QFT, in which photons are represented by a propagator
in the myriad Feynman diagrams that comprise the perturbation
approximation to the amplitude for the physical process observed.

FYI: a Feynman diagram is not really a picture, it is an
integral with a definite value. The diagrams are a convenient
way to manage the combinatorial complexity of calculating
the amplitude for a given observable physical process in
the model represented by the Lagrangian used (the Lagrangian
determines the rules by which the diagrams are drawn).

> Waves interfere, therefore 'photon waves' must interfere.

You are inappropriately intermixing two different levels of description.
Photons are from QED, while the sort of "waves" you imagine are from
classical electrodynamics, or non-relativistic QM. Photons are neither
particles nor waves, they are excitations of the photon field.

> But to compute the interference, one requires frequency and phase
> information.

In CE and QM, yes. In QED it happens automatically in the summation over
Feynman diagrams, and in the integrals of the individual diagrams.

> E = hf is a scalar; where's the phase?

As I said before, E=hf applies to the SOURCE, not the individual
photons. Go look up how Planck derived it. So your question makes no sense.

> Has anything changed since Planck derived E = hf?

Yes, a whole host of things about this are now known. But it remains
true that E=hf applies to the source, not the individual quanta (photons).

Tom Roberts

On Thursday, August 25, 2022 at 7:57:23 PM UTC-7, tjrob137 wrote:
> On 8/25/22 3:20 PM, RichD wrote:
> > On August 23, 6 PM UTC-7, tjrob137 wrote:
> >>>> OK, I am really asking *why* it is not considered correct [to
> >>>> say a photon has a frequency E/h]
> >>
> >> The real reason is that a photon simply does not posses a
> >> frequency. Speaking loosely, since the (real) photon propagates at
> >> c, it has no elapsed proper time, and thus cannot vary over time,
> >> which is the sine qua non of having a frequency. Note that the
> >> (real) photon is simply transferring 4-momentum from one event to
> >> another (modulo spin, and only for null-separated events); there's
> >> no frequency in that. The equation E=hf applies to a monochromatic
> >> light SOURCE in its inertial rest frame, relating the source and
> >> its photons;
> >
> > Given a large number of photons, traveling as a 'packet' (whatever
> > that means). That forms a classical light beam. Does the frequency
> > spectrum of that beam, as defined by Maxwell's model, correspond
> > exactly to the spectrum of the photons?
> Approximately, yes; exactly, no.
> > Quantum mechanics models everything as a wave in Hilbert space.
> More relevant is QFT, in which photons are represented by a propagator
> in the myriad Feynman diagrams that comprise the perturbation
> approximation to the amplitude for the physical process observed.
>
> FYI: a Feynman diagram is not really a picture, it is an
> integral with a definite value. The diagrams are a convenient
> way to manage the combinatorial complexity of calculating
> the amplitude for a given observable physical process in
> the model represented by the Lagrangian used (the Lagrangian
> determines the rules by which the diagrams are drawn).
> > Waves interfere, therefore 'photon waves' must interfere.
> You are inappropriately intermixing two different levels of description.
> Photons are from QED, while the sort of "waves" you imagine are from
> classical electrodynamics, or non-relativistic QM. Photons are neither
> particles nor waves, they are excitations of the photon field.
> > But to compute the interference, one requires frequency and phase
> > information.
> In CE and QM, yes. In QED it happens automatically in the summation over
> Feynman diagrams, and in the integrals of the individual diagrams.
> > E = hf is a scalar; where's the phase?
> As I said before, E=hf applies to the SOURCE, not the individual
> photons. Go look up how Planck derived it. So your question makes no sense.
> > Has anything changed since Planck derived E = hf?
> Yes, a whole host of things about this are now known. But it remains
> true that E=hf applies to the source, not the individual quanta (photons).
>
> Tom Roberts

So you're giving the wave ascription to the photons not the wave?

Those are some strong photons, ....

What is this "beams", rays.

I suppose the beam is well enough the front of the wave, ....

My only usual understanding of light is rays from the sun.

On Tuesday, August 16, 2022 at 11:24:33 AM UTC-4, Volney wrote:
> On 8/16/2022 9:41 AM, Ken Seto wrote:
> > On Monday, August 15, 2022 at 10:55:16 PM UTC-4, tjrob137 wrote:
> >> On 8/15/22 3:44 PM, Tadd Muraro wrote:
> >>> Would win a noble prize able to measure the frequency of a single
> >>> photon.
> >>
> >> No. It is physically impossible to measure the frequency of a photon.
> >
> > It is possible to measure the frequency of a wave-packet. That means that a photon is not a particle. It is a wave-packet as shown on page 33 of my book in the following link:
> Nobody cares about your make-believe Muddle Mechanics, Stupid Ken. Go
> back to the little kids' table, this is an adult discussion about the
> frequency of the photon and whether that exists, not a made-up garbage
> "theory".

Hey Moron_y, Are you assuming that you are everybody?
Model Mechanics gives a realistic description of a photon.
BTW how does the source emits particle (photons) at infinitum?
Does it means that every source contains an infinite amount of energy?
I think not.
A source can cause a wave-packet in the aether (the E-Matrix) because
of its absolute motion in the E-Matrix.
This means that current physics failed to understand the nature of light.

On 8/28/2022 1:34 PM, Ken Seto wrote:
> On Tuesday, August 16, 2022 at 11:24:33 AM UTC-4, Volney wrote:
>> On 8/16/2022 9:41 AM, Ken Seto wrote:
>>> On Monday, August 15, 2022 at 10:55:16 PM UTC-4, tjrob137 wrote:
>>>> On 8/15/22 3:44 PM, Tadd Muraro wrote:
>>>>> Would win a noble prize able to measure the frequency of a single
>>>>> photon.
>>>>
>>>> No. It is physically impossible to measure the frequency of a photon.
>>>
>>> It is possible to measure the frequency of a wave-packet. That means that a photon is not a particle.

Non sequitur.

>>> It is a wave-packet as shown on page 33 of my book in the following link:
>> Nobody cares about your make-believe Muddle Mechanics, Stupid Ken. Go
>> back to the little kids' table, this is an adult discussion about the
>> frequency of the photon and whether that exists, not a made-up garbage
>> "theory".
>
> Hey Moron_y, Are you assuming that you are everybody?

Prove me wrong, Stupid Ken. Show me one person, other than yourself, who
cares about your make-believe Muddle Mechanics. Or admit that "nobody
cares" is correct.

And tell us, how many copies of your book did you sell (not counting any
sock puppets).

> Model Mechanics gives a realistic description of a photon.

No, it is merely a package of assertions and is worthless for physics.

> BTW how does the source emits particle (photons) at infinitum?

Who said it does?

> Does it means that every source contains an infinite amount of energy?

Who said it does? A finite source will emit a finite (albeit large)
number of photons, each with a finite amount of energy, totaling a
finite amount of energy.

> I think not.

You finally got something correct! Is this the very first time you got
something correct?

> A source can cause a wave-packet in the aether (the E-Matrix) because
> of its absolute motion in the E-Matrix.

Gobbledygook word salad assertions.

> This means that current physics failed to understand the nature of light.

No, a pack of assertions does not mean anything. Your inability to
understand physics shows YOU failed to understand the nature of light.

On Wednesday, August 10, 2022 at 10:44:38 AM UTC-7, Athel Cornish-Bowden wrote:

> I did have a brief and cursory look. The thing that struck me is that
> it contains no literature references apart from one to Einstein (1905).
> Does he imagine that science is advanced by geniuses sitting in their
> armchairs and thinking, without trying to build their ideas on those of
> their predecessors?
>
>
> --
> Athel -- French and British, living mainly in England until 1987.
Funny, that's just what Einstein did in his 1905 paper. It contained no references.