To make things more realistic and feasible, I changed the material of
the cavity for aluminum, which has a reflectivity of 99%. I figure out
the weight of each cavity as being 2 grams.
The aluminum coating on the inside has, on purpose, an irregular surface
in the order of nanometers, so the laser beam will disperse in each
bounce.
I suggest that the laser output be dispersed by a vibrating lens, prior
to its entry to the cavity. In this way, the beam will be continuously dispersed. Even more, 4 or 5 different lasers (total power = 5 W) be
placed in such a way that they cover the entire volume of the cavity, accelerating the process of energy spread in its volume.
New calculations, using aluminum:
h = 6.62607E-34 J.sec
λ = 5.50E-07 m
c = 3.00E+08 m/s
E(photon) = 1.21E-48 Joules
5 W = 5 Joules/sec = 4.12E+48 photons/sec
5 Joules/hour = 1.48E+52 photons/hour
Inner surface of cavity = 7.85E+03 mm²
Area of laser spot = 3.14159 mm²
2,500 spots
Time between hits = 3.333E-10 seconds (10 cm avg. trajectory)
3.00E+09 bounces/sec
photon hits/sec in the cavity = 1.37E+39
photons hit/sec x spot = 5.4879643E+35
Reflectivity with aluminum = 99.0%
photons reflected/sec x spot = 5.4330846E+35
photons absorbed/sec x spot = 5.49E+33
energy lost/sec x spot = 6.67E-15 J
energy lost/sec in the cavity = 1.67E-11 J
energy lost/Hr in the cavity = 6.00E-08 J
energy lost/Hr in the cavity = 1.43E-08 cal/Hr
energy lost in 72 Hr, in the cavity = 1.032E-06 calories
Even if I made errors in the order of 1,000:1, still the heat absorbed
is insignificant (0.00103 calories).
AS A REFERENCE:
1 calorie is the heat that raises 1"C to one gram of water.
With this new arrangement and materials, proving/disproving E=mc² should
be much easier and cheap. The weight increment is in the order of 9 μN (1.0368 μgrams of MASS) can be measured with EM balances of $ 10,000.
On Wed, 20 Nov 2024 14:46:05 +0000, rhertz wrote:
As I wrote before, the cavity NEVER reaches a steady state (thermal
equilibrium is just one possible goal to reach a steady state).
The stored energy increases constantly until the temperature of the
cavity
walls (not cancelled by any means) destroy the material that form the
cavity (either coatings or places on the cavity surface, making holes).
No, because the temperature of the foil rises until the power emitted
equals 5 watts. Once that happens, no further increase of stored
energy occurs.
P = ε * σ * T^4
Let us assume a 10x10x10 cm cube with surface area 0.06 m^3
So 5 watts total power emitted means 83 watts/m^2
Assume ε = 0.13
Let T_i be 273 K
83 = ε * σ * (T_f^4 - T_i^4)
11,260,344,593 = (T_f^4 - 5,554,571,841)
T_f = 274.8 K
In other words, the temperature of the foil rises to 1.8 degrees
above room temperature.
In physics, a steady state is a condition in which a system or process
does not change over time, or any changes are balanced out. This means
that the variables that define the system's behavior remain constant.
Some examples of steady states include:
Not bad for my estimation (2 gr) for the weight of the cavity,
with inner surface of 7,850 mm², and thickness of 0.1 mm.
THE SYSTEM WILL NEVER REACH INTERNAL EQUILIBRIUM!
That is IMPOSSIBLE because I'd be pumping ENERGY non-stop, forever if necessary. Can you get this, please?
THERE IS NO AMPLIFICATION OF LASER POWER, AND NEVER WAS MEANT TO HAPPEN.
WHAT HAPPEN IS A CONTINUOUS FEED OF ENERGY!. If you CAN'T SEE IT, then substitute the 5W laser by a hose POURING WATER INSIDE THE CAVITY. Some
water falls out, but most remain UNTIL THE CAVITY IS FULL OF IT.
When will it happen with the water analogy? Don't know/don't care.
The only reason by which I used three days to fill the cavity up is
because A LONGER PERIOD would accumulate much more perturbations and
external interferences, complicating the statistical processing of the electrical signal that IS LINEARLY PROPORTIONAL to the accumulation of
energy inside the cavity.
If you REFUSE to understand this, I advise you to go back to college or
high school, where you could re-learn elementary logic and arithmetic.
Say no more.
On Thu, 21 Nov 2024 21:33:35 +0000, Paul B. Andersen wrote:
Den 20.11.2024 02:02, skrev rhertz:
Not bad for my estimation (2 gr) for the weight of the cavity,
with inner surface of 7,850 mm², and thickness of 0.1 mm.
If the cavity is spherical, the diameter will be D = 5 cm.>> The time light will use to go from one surface to the opposite
surface is Δt = 16.33 ns
Even if the cavity may have another shape, I will use
Δt = 16.33 ns as the average time between the reflections
of the laser beam.
Den 20.11.2024 05:37, skrev rhertz:
THE SYSTEM WILL NEVER REACH INTERNAL EQUILIBRIUM!
That is IMPOSSIBLE because I'd be pumping ENERGY non-stop, forever if
necessary. Can you get this, please?
Power of laser P₀ = 5 W
Let's look at some facts:
Reflected power after n reflections
P(n) = P₀⋅Rⁿ (1)
where R is the reflectivity of the inner walls.
The energy stored in the cavity as laser light:
E = ∑{i = 1 to ∞}P₀⋅ Rⁱ⋅Δt = P₀⋅Δt ⋅∑{i = 1 to ∞} Rⁱ
∑{i = 1 to ∞} Rⁱ = R/(1-R) , a converging geometric array
E = P₀⋅Δt ⋅R/(1-R) (2)
Note that this is a constant.
Calculations with R = 0.99:
-------------------------
Let us consider that all the laser light is absorbed
when P(i) < 1e-10 W.
From (1) we find: P(2455) = 9.62e-11
t = 2455⋅Δt = 409 ns
This means that 409 ns after the laser light enters the cavity,
it will be absorbed by the wall.
The energy stored as laser light in the cavity will according to (2) be:
P₀⋅Δt ⋅R/(1-R) = 8.248e-8 J
Was it this energy you thought would increase indefinitely?
It won't. It is constant. And tiny.
Calculations with R = 0.999998:
-------------------------------
From (1) we find: P(12350000) = 9.37e-11
t = 12350000⋅Δt = 2.07 ms
This means that 2.07 ms after the laser light enters the cavity
it will be absorbed by the wall.
The energy stored as laser light in the cavity will according to (2) be:
P₀⋅Δt ⋅R/(1-R) = 4.166e-4 J
Generally we can say that the laser light will be absorbed
almost immediately after it enters the cavity, and the stored
energy in the form of laser light will be tiny and constant.
So all the 5 J that enters the cavity every second will heat
the inner wall of the cavity. Since the walls are only 0.1 mm
thick, the outer surface of the cavity will be approximately
the same temperature as the inner surface, so after a short time
the outer wall will radiate 5 W and the system will be in steady state.
You could equally well have heated the cavity with a Bunsen burner.
Do you think the mass increase due to the heat energy in the cavity
would be measurable? :-D
Paul, as usual you fucked it up.
CORRECTION 1:
If the cavity is spherical, the diameter will be D = 10 cm (not 5 cm).
NOTE 1: c= 3.0E+10 cm/s
The time light will use to go from one surface to the opposite
surface is Δt = 0.333 ns (NOT 16.33 ns).
NOTE 2: Δt = 0.333 ns IMPLY 3.00E+09 bounces/sec. With LIGO advanced technology, 1 photon per 5,000,000 photons is lost in every bounce.
This represents a reflectivity R = 0.9999998.
CORRECTION 2 (your fatal error:
The energy stored in the cavity as laser light:
E = ∑{i = 1 to ∞}P₀⋅ Rⁱ⋅Δt = P₀⋅Δt ⋅∑{i = 1 to ∞} Rⁱ
∑{i = 1 to ∞} Rⁱ = R/(1-R) , a converging geometric array
E = P₀⋅Δt ⋅R/(1-R) (2)
Your error here is to use only the energy that exist in the first P₀⋅Δt interval. This represents ONLY 1.37E+39 photons (550 nm).
You forgot that such quantity is supplied every 0.333 ns, being the
amount of photons/sec 4.12E+48 (3 billions times bigger).
PER SECOND, you have to account the impact of reflectivity on 3 billion chunks of 1.37E+39 photons EACH, and make the calculations for losses cumulative with each bounce + new feed of P₀⋅Δt photons.
The above means that you should count losses/sec IN THIS WAY:
Loss 1 = ∑{i = 1 to 3,000,000,000} Rⁱ, for the first pack of photons.
Loss 2 = ∑{i = 1 to 2,999,999,999} Rⁱ, for the 2nd. pack of photons.
Loss 3 = ∑{i = 1 to 2,999,999,998} Rⁱ, for the 3rd. pack of photons. .....
Loss n = ∑{i = 1 to 3,000,000,001-n} Rⁱ, for the nth. pack of photons.
THEN CALCULATE THE SUM OF LOSSES FOR THE 3 BILLIONS PACKS OF PHOTONS.
You'll understand that such is not an easy task, in particular when calculating the losses per hour on in 72 hours.
YOUR NUMBERS ARE WAY OFF!! (3,000,000,000 off maybe?).
Now smile, asshole.
On Fri, 22 Nov 2024 20:30:39 +0000, Paul B. Andersen wrote:
<snip all the history of your life. You didn't read my disclaimer>
Bottom line:
It doesn't matter how you put energy into a closed cavity.
In the cavity there will be a black body radiation with
temperature equal to the temperature of the walls.
The reflectivity, albedo or colour of the inner surface
of the cavity are irrelevant. The radiation in the cavity
will always be black body radiation.
Make a hole in your cavity, and you have a perfect
black body radiation source.
Its temperature will not be very high, though-
It is not a BB radiation source!
You ignored the fact that the heat goes away from the cavity. It doesn't remain neither inside nor outside. It's eliminated by cooling
mechanisms, as I wrote as an initial condition. Read all the posts.
There is no light energy left within the cavity, nor heat energy outside
it. You better think again about it.
You also ignored my post apologizing to all people that participated in
this thread. That makes you a bigger ASSHOLE than what I thought.
Now, start thinking in my NEXT IDEA:
Willing to try to prove/disprove E=mc² at a macroscopic level, I'll
think of an experiment that incorporates electromagnetic oscillations
passing through the cavity, which will be converted in a CAPACITOR, by cutting it in halves and isolating them with a thin ring.
What I propose to MEASURE is the changes in the frequency of the LC oscillator, within a time window of about 3 msec, which repeats
permanently.
I'll use a relationship between mass and capacitance for the cavity,
with frequency around 1 Mhz or greater.
It was a failed idea for an experiment, but there are OTHER WAYS to
check E=mc² at a macroscopic level, without resorting to nuclear energy
crap (Kg evaporated vs. energy provided), or else.
Ask ChatGPT:
any other non-relativistic means to prove E=mc^2?
or
Is there any way to prove E=mc^2 at higher level than quantum?
or
can I use electrostatic energy?
The last one gave interesting insights. Ask to it sequentially the above lines.
Keep smiling, asshole.
Prokaryotic, I was thinking about what you wrote on the cavity behaving
as a black body and, as I wrote before, I completely disagree to take it
as a black body radiating energy, once equilibrium has been reached.
My main doubt was that, once in equilibrium and having gained as heatall the energy supplied by the 5W laser, the aluminum cavity HAD TO
radiate using the external surface AS WELL AS the internal surface. I
thought that almost HALF of the heat was going to be radiated INTO THE CAVITY.
So the stored radiation energy within the cavity would be:
E = 4⋅σ⋅V⋅T_f⁴ = 1.15e-08 Joules
Quite far from the 477.5 Joules you fantasised about.
In fact, the heat energy stored in the aluminium would be much higher,
E = 0.06 Joules. (If my calculation is correct.)
Not much to weight, is it? :-D
I'm going to tell this one more time, because it's the center of the
problem:
Using a modified Stefan's formula (by 4/c) to calculate the internal temperature of a small aluminum cavity IS AN ABERRATION OF COMMON SENSE.
In your calculation of 1,000,000+ K inside the cavity, YOU SHOULD HAVE STOPPED at 660.3"C (930.3 K) when ALUMINUM MELTS.
Why did you persist in using such stupid value? I can't figure it out.
This means that half of the accumulated 955 Joules remain within the
cavity.
Assuming that the environmental temperature is 293 K and that the
experiment is conducted in vacuum,
P = ε σ A_e (T_f^4 - T_i^4)
5 = 0.13 * 5.67e-8 * 0.03144 * (T_f^4 - 293^4)
5 = 0.13 * 5.67e-8 * 0.03144 * (T_f^4 - 7,370,050,801)
21,575,509,276 = T_f^4 - 7,370,050,801
T_f^4 = 28,945,560,077
T_f = 412.5 K (actually, the numbers only justify 1-2 sig figures)
A realistic temperature when we know that the cavity
must get rid of 5W is:
|Den 26.11.2024 03:25, skrev ProkaryoticCaspaseHomolog:
Assuming that the environmental temperature is 293 K and that the experiment is conducted in vacuum,
P = ε σ A_e (T_f^4 - T_i^4)
With 1 mm thick wall, the temperature difference between
the inner wall and the outer wall would be ΔT = 0.0008⁰K which
is negligible in this context.
So the stored radiation energy within the cavity would be:
E = 4⋅σ⋅V⋅T_f⁴/c = 1.15e-08 Joules
In fact, the heat energy stored in the aluminium would be much higher,
You have barely written an OP in the last 5 years, and when you did, you fucked it up big time.
Ashamed of those who mocked you, you went into
the "silence cone" for months.
But you were reading every single post,
looking for what? Consolation? A friendly hand to rescue you from exile?
I did that for you, remember? I LIED to save your ass(face), and you
thank me.
Den 21.11.2021 08:23, skrev Richard Hertz:Nov. 18, 1915)
The formula (8) on the paper is equal to Einstein's Eq. 13 (paper
https://tinyurl.com/yzzbratp
(Equation 13) ε = + 3π α/[a (1 – e²)]
with an error lower than 1.6E-07
Or equation (14): ε = 24π³a²/T²c²(1−e²)
This is the equation you will find in many books and papers.
T² = 4π²a²/GM (1)
where T is the period of a test particle in orbit
around a mass M. a is the semi-major axis.
My equation (8): ε = 6π(GM)²/G(M+m)a(1−e²)c²
If we in this equation set (GM)²/G(M+m) ≈ GM when m/M << 1
and from (1): GM = 4π²a²/T²
we get the equation: ε ≈ 24π³a²/T²c²(1−e²)
which is the same as Einstein's equation (14)
The difference is that Einstein's equation (the equation commonly used)
is the perihelion advance of a test particle in orbit around a mass M,
while my equation (8) is the perihelion advance of a mass m in orbit
around a mass M.
The relative difference is (as you said):
((GM)²/G(M+m) - GM)/GM = -1.66E-7
which obviously is negligible, probably less than the precision
of G, M and m.
That's why Einstein's equation (14) safely can be used.
BTW, thanks for a sensible post.
You haven't produced many of those lately.
--
Paul
I LIED because I took pity of your "exile", so I wrote that your
calculations were correct, within an error margin.
After that, you started to writing posts again, but NEVER tried again to write an OP (Original Post). You should keep this in mind.
Instead, you choose to continue with your usual M.O., which is to read everything and insert comments, but not an OP again.
My behavior is exactly THE OPPOSITE ONE: I mostly write OP, even with
the most ridiculous idea that I come up with, BECAUSE I'm here to have
fun. FUN, not scientific research.
To do that, I think about different ideas (as a hobby), and if I find something that can stir the waters, creating controversy to animate this circus, I wrote the OP with my best criteria and my best thoughts.
I pretend to animate the circus, not to gain a Nobel Prize. And, of the
most importance, I INVENT most of the content (original thinker), taking
care to make the most of sense and the lowest amount of ridiculous
equations and assessments. To do so, I first think, then I do some
research to validate the ideas. Even if I can't find any support, but
I'm convinced the idea is good, I create the OP and wait for the hell
broke loose.
Your behavior is mostly parasitic, as you live from the ideas of others.
You live citing and quoting papers from others that support relativity.
This behavior of you, playing safe from your comfort zone shows, among
other things, that you are an intellectual coward. You are afraid that
you might publish something wrong or stupid.
I don't care about what others think about me, but you CAN DIE if you
fuck it up, which you did many times in the last decade and a half.
I also learn by successive refinements of the ideas in the OP, through interactions with others. And if I make a mistake, and I realize it, I APOLOGIZE without any shame.
Now, think about what I wrote, and keep smiling, asshole.
[...] Of all the schemes I could think of whereby a DIY amateur scientist
can verify E=mc², the only feasible one that I can think of is to
measure the energy of electron-positron annihilation.
I'd need
the help of an expert former EE to put things together (hint, hint).
Sealed sodium-22 positron sources are readily available for sale
online.
Calibrating the gamma-ray detectors will be the hard part.
Once
calibrated, I'd place two gamma-ray detectors 180 degrees apart
equidistant from an appropriate target. Connected to the detectors
will be a whole slew of equipment for coincidence counting, energy measurement, etc. that the expert former EE will be responsible for.
With luck, the DIY experimenter will be able to confirm the
simultaneous emission of two 511 keV photons from each annililation.
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