Cosmic acceleration rediscovered

"greywolf42" wrote in message
...
George Dishman wrote in message
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My starting assumption was the same as you say above:

"The 'returning' is a local effect (at the point of fractional
return). dE = - const E, at the point of inspection."

Sorry, I did not consider that your starting assumption, because of the
large gap -- and because the next statement does not follow from the
above.

The E in that equation falls as the inverse square

This is your incorrect assumption. The E in the above equation does not
fall as the inverse square. The equation above is the pure linear form.
The only degradation is due to the fractional removal from every photon
(which continues in a straight line). Inverse square is the reduction in
the number of photons per unit area, as a spherical source spreads over
the
surface of larger spherical shells.

hence so does dE.

We are talking past each other so let me try another
way. Consider a area of 1m^2 at a distance of 1000AU
from the Sun. In one second an energy E passes through
that square. That will deposit a small fraction dE
into the aether.

Now consider an area of 1m^2 at a distance of 2000AU.
The energy passing through in one second will be
almost E/4 but very slightly less due to tired light.
Over this distance though the tired light loss will
be negligible compared to the r^-2 loss. The energy
deposited into the aether here will therefore be no
more than dE/4 and in fact slightly less.

If the temperature of the aether at 1000AU is T, that
at 2000AU should be very close to (but slightly less
than) T/sqrt(2) since the power from a black body
radiator is proportional to T^4.

I don't think so. It still seems to me that dE in
the equation you posted will follow an inverse square
from each (point) source of E.

As demonstrated above, you are incorrect. A tired light model will always
be (slightly) below a pure inverse square model.

Yes, that's what I'm saying, though I had ignored the
(slightly) part as it is negligible over short ranges
and it adds to the effect anyway. You are arguing that
the temperature is uniform aren't you?

George

George Dishman wrote in message
...

"greywolf42" wrote in message
...
George Dishman wrote in message
...


{snip abandoned attempt at description}

We are talking past each other so let me try another
way. Consider a area of 1m^2 at a distance of 1000AU
from the Sun. In one second an energy E passes through
that square. That will deposit a small fraction dE
into the aether.

Delta E. Yes.

Now consider an area of 1m^2 at a distance of 2000AU.
The energy passing through in one second will be
almost E/4 but very slightly less due to tired light.

Yes.

Over this distance though the tired light loss will
be negligible compared to the r^-2 loss. The energy
deposited into the aether here will therefore be no
more than dE/4 and in fact slightly less.

Yes.

If the temperature of the aether at 1000AU is T, that
at 2000AU should be very close to (but slightly less
than) T/sqrt(2) since the power from a black body
radiator is proportional to T^4.

Oops, here is your error.

The aether temperature is not merely a function of the local addition of
energy from starlight degradation. (Temperature is a function of E, not of
dE.) Starlight degradation is a miniscule contribution to the pre-existing
aether energy density. After all, the aether temperature (energy content)
is what drives gravitation. In fact, gravitation is a competing effect --
lowering the local aether temperature a tiny amount, in the vicinity of the
star.

I don't think so. It still seems to me that dE in
the equation you posted will follow an inverse square
from each (point) source of E.

As demonstrated above, you are incorrect. A tired light model will
always be (slightly) below a pure inverse square model.

Yes, that's what I'm saying, though I had ignored the
(slightly) part as it is negligible over short ranges
and it adds to the effect anyway. You are arguing that
the temperature is uniform aren't you?

Very close to uniform. Not uniform. Normal gravitation and light energy
are tiny variations on the pre-existing local energy density. (The sun's
surface gravitation is about 1 part in 10^8 of the maximum gravitational
acceleration.)

--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail}

"greywolf42" wrote in message
...
George Dishman wrote in message
...

snip
If the temperature of the aether at 1000AU is T, that
at 2000AU should be very close to (but slightly less
than) T/sqrt(2) since the power from a black body
radiator is proportional to T^4.

Oops, here is your error.

The aether temperature is not merely a function of the local addition of
energy from starlight degradation. (Temperature is a function of E, not
of
dE.) Starlight degradation is a miniscule contribution to the
pre-existing
aether energy density. ...

Ah, the penny drops, thanks. Eddington showed energy
in starlight was equivalent to circa 2.8K but only a
small fraction of that transfers to the aether. Also,
even if the aether temperature is much higher than
2.8K, your previous comments would get round this if
the transfer of energy to electrons is slow compared
to the re-radiation as thermal energy by the electrons.
The total power transferred just needs to match that
radiated at 2.8K.

George

George Dishman wrote in message
...

"greywolf42" wrote in message
...
George Dishman wrote in message
...

snip

If the temperature of the aether at 1000AU is T, that
at 2000AU should be very close to (but slightly less
than) T/sqrt(2) since the power from a black body
radiator is proportional to T^4.

Oops, here is your error.

The aether temperature is not merely a function of the local addition of
energy from starlight degradation. (Temperature is a function of E, not
of dE.) Starlight degradation is a miniscule contribution to the
pre-existing aether energy density.

...

Ah, the penny drops, thanks.

You're welcome. One more side issue down.

Eddington showed energy
in starlight was equivalent to circa 2.8K but only a
small fraction of that transfers to the aether.

Actually, all of it would transfer back to the aether -- eventually. Only a
fraction from any one source would transfer at any given dV.

Also,
even if the aether temperature is much higher than
2.8K, your previous comments would get round this if
the transfer of energy to electrons is slow compared
to the re-radiation as thermal energy by the electrons.

The total power transferred just needs to match that
radiated at 2.8K.

You have the general idea.

Now I'm going to add one more bit of confusion. Thermal radiation from
non-fusing sources in thermal equilibrium, such as planetary bodies and
cooled collapsed objects will eventually be radiating just as much power as
they absorb from gravitation. And that thermal radiation will eventually be
re-absorbed into the aether. Which is the part of my model that I believe
you now understand.

But since we were mentioning starlight, we have to touch on the issue of
fusion inside stars. Fusion energy is a temporary additional source. So
the amount of heat leaving a fusing star will actually be above that
received from the gravitational force. Until it burns out and cools down.
The original source of the fusion energy would come from the orignal
formation of protons (and anti-protons) out of the original aether. And the
origin of the aether (and universe) is as yet undescribed by the model.

--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail}

"greywolf42" wrote in message
. ..
George Dishman wrote in message
...

snip
Eddington showed energy
in starlight was equivalent to circa 2.8K but only a
small fraction of that transfers to the aether.

Actually, all of it would transfer back to the aether -- eventually. Only
a
fraction from any one source would transfer at any given dV.

Yes, that's what I meant. You should be getting
used to my 'economical' style by now ;-)

Also,
even if the aether temperature is much higher than
2.8K, your previous comments would get round this if
the transfer of energy to electrons is slow compared
to the re-radiation as thermal energy by the electrons.

The total power transferred just needs to match that
radiated at 2.8K.

You have the general idea.

Now I'm going to add one more bit of confusion. Thermal radiation from
non-fusing sources in thermal equilibrium, such as planetary bodies and
cooled collapsed objects will eventually be radiating just as much power
as
they absorb from gravitation. And that thermal radiation will eventually
be
re-absorbed into the aether. Which is the part of my model that I believe
you now understand.

But since we were mentioning starlight, we have to touch on the issue of
fusion inside stars. Fusion energy is a temporary additional source. So
the amount of heat leaving a fusing star will actually be above that
received from the gravitational force. Until it burns out and cools down.
The original source of the fusion energy would come from the orignal
formation of protons (and anti-protons) out of the original aether. And
the
origin of the aether (and universe) is as yet undescribed by the model.

OK, I'll treat that as a side issue too. There may
be some problems with the details but as long as any
additional heating of bodies is at a low level then it
probably doesn't affect our conversaton significantly.

George