# Measuring Geothermal (1)

In my previous article I promised to measure how much energy is produced from geothermal. I will make a first attempt today. I downloaded a big collection of borehole data from U of Michigan for us to analyze. I am interested in data as close to the surface as possible while being below 10 meters deep. Why 10 meters? It is commonly well known that the sun does not penetrate below 10 meters. Of the 1012 locations available, 312 have data for 50m and 100m depth. This is what we will be using.

Number of locations per country, and the country’s geographic middle latitude:

``````      1   AL    41.153332
29   AU    -25.274398
5   BR    -14.235004
3   BW    -22.328474
9   BY    53.709807
37   CA    56.130366
28   CN    35.86166
2   CU    21.521757
32   CZ    49.817492
1   IE    53.41291
2   IN    20.593684
2   IT    41.87194
22   JP    36.204824
20   KR    35.907757
2   MN    46.862496
1   NA    -22.95764
1   NE    17.607789
3   PE    -9.189967
3   PL    51.919438
27   RU    61.52401
8   SI    46.151241
4   TZ    -6.369028
4   UA    48.379433
2   UK    55.378051
42   US    37.09024
13   ZA    -30.559482
3   ZM    -13.133897
6   ZW    -19.015438``````

The average latitude weighted by number of locations is 30.7607. Here is all the data:

``````AL 02.50 18.55 19.25|CA 03.00 03.13 04.37|JP 01.31 16.10 17.21|SI 02.50 14.45 18.17
AU 03.10 21.36 21.88|CA 03.10 03.29 04.42|JP 01.35 15.87 16.60|SI 02.50 10.40 12.50
AU 03.87 21.33 22.04|CA 03.50 03.27 03.97|JP 02.47 18.30 19.10|SI 02.50 13.88 17.22
AU 02.42 24.65 25.01|CA 06.00 05.93 06.12|JP 01.36 16.05 17.05|SI 02.50 11.74 12.79
AU 04.00 21.16 21.66|CA 06.00 05.28 05.62|JP 01.36 16.27 16.94|TZ 02.50 23.67 24.37
AU 03.69 21.61 21.71|CA 03.00 06.07 07.29|JP 01.33 16.38 17.39|TZ 03.24 24.72 25.21
AU 03.37 14.46 15.71|CN 02.50 14.79 16.07|JP 01.36 15.58 15.70|TZ 02.63 24.12 25.13
AU 03.50 31.30 32.17|CN 02.10 09.51 09.73|JP 01.67 17.07 17.92|TZ 03.49 23.13 23.71
AU 02.17 31.20 32.18|CN 02.23 17.55 17.72|JP 01.32 17.53 18.57|UA 02.50 09.67 10.22
AU 02.50 26.75 27.32|CN 01.60 14.16 15.31|JP 01.36 15.41 15.68|UA 01.50 14.88 16.78
AU 02.80 24.42 25.34|CN 02.00 14.85 16.30|JP 01.78 15.80 17.10|UA 02.50 08.99 09.72
AU 01.90 19.07 20.91|CN 02.00 14.88 16.42|JP 02.47 18.73 20.27|UA 02.65 09.98 11.07
AU 04.26 21.53 22.48|CN 02.00 15.50 16.92|JP 01.78 15.76 17.05|UK 03.19 10.92 11.70
AU 03.01 18.47 19.86|CN 03.63 20.17 20.87|JP 02.00 16.84 18.11|UK 04.11 09.94 10.63
AU 03.05 18.73 19.92|CN 03.74 19.50 19.95|JP 01.96 17.26 19.29|US 02.75 15.84 16.35
AU 01.86 25.02 27.44|CN 03.38 21.02 21.24|JP 01.36 15.74 16.50|US 02.02 23.09 24.29
AU 03.01 13.78 14.68|CN 02.50 16.75 17.39|JP 01.32 15.65 16.75|US 02.02 23.05 24.20
AU 02.99 14.25 14.85|CN 02.50 16.68 17.39|KR 02.50 15.25 16.91|US 03.10 22.19 23.07
AU 02.68 18.80 19.30|CN 03.30 17.10 18.35|KR 02.50 15.22 16.32|US 02.60 26.17 27.72
AU 02.89 23.90 24.29|CN 02.50 13.05 14.90|KR 02.78 16.75 17.85|US 03.32 26.48 27.84
AU 02.77 24.80 25.13|CN 02.18 21.95 23.47|KR 02.78 14.17 15.24|US 03.32 25.90 27.31
AU 03.03 19.69 20.09|CN 02.38 21.42 22.36|KR 02.50 14.17 15.23|US 03.70 07.88 08.27
AU 05.54 22.87 23.25|CN 03.90 20.38 21.20|KR 02.21 13.75 15.02|US 02.91 17.03 17.72
AU 04.32 22.04 22.23|CN 02.75 18.84 19.89|KR 04.05 14.78 15.62|US 03.01 12.13 12.67
AU 03.99 22.45 22.65|CN 02.50 18.60 19.80|KR 02.78 14.90 16.10|US 03.23 10.41 11.68
AU 02.46 21.87 23.34|CN 02.50 20.70 21.10|KR 02.78 15.00 15.90|US 03.01 12.77 13.62
AU 03.44 20.54 20.61|CN 02.50 19.10 19.39|KR 02.78 14.00 15.20|US 02.95 04.47 05.68
AU 04.22 20.23 21.34|CN 02.50 18.39 18.75|KR 02.68 13.77 15.23|US 00.87 09.01 10.90
AU 03.51 24.32 24.46|CN 02.50 20.73 22.39|KR 02.68 13.67 14.14|US 01.33 11.21 13.75
AU 03.29 18.96 20.10|CN 02.50 20.73 22.39|KR 02.63 12.87 14.53|US 01.30 11.49 13.80
BR 02.50 27.35 28.24|CN 02.80 09.10 09.60|KR 02.68 13.93 15.32|US 01.20 14.90 18.88
BR 02.50 32.29 33.36|CN 02.20 16.75 18.30|KR 02.68 13.62 15.07|US 02.90 06.79 07.25
BR 02.50 31.48 31.84|CN 03.21 08.52 12.83|KR 02.68 13.76 14.80|US 02.48 08.50 09.34
BR 02.50 30.61 31.25|CN 03.82 18.96 19.75|KR 02.68 12.33 14.17|US 02.50 07.88 08.43
BR 01.37 21.08 22.73|CU 01.40 25.91 26.36|KR 02.58 16.00 17.50|US 01.20 07.31 08.57
BW 04.09 27.32 27.92|CU 02.60 24.42 24.52|KR 02.58 14.53 15.22|US 03.91 07.75 08.65
BW 04.00 26.23 26.91|CZ 02.21 11.67 13.33|KR 02.58 16.39 17.50|US 01.20 07.21 08.52
BW 03.60 26.14 26.52|CZ 02.75 09.61 10.56|MN 02.50 02.12 03.15|US 01.20 07.21 08.52
BY 02.50 09.50 10.56|CZ 02.24 11.67 12.67|MN 02.50 02.50 03.96|US 03.52 10.85 15.02
BY 02.50 07.83 08.38|CZ 02.24 09.27 10.64|NA 02.80 27.36 28.55|US 03.01 03.63 04.36
BY 02.50 08.86 09.60|CZ 02.34 05.76 06.87|NE 02.70 32.65 33.10|US 03.22 07.96 09.13
BY 02.50 07.83 07.97|CZ 02.50 08.01 09.66|PE 04.22 19.37 20.32|US 03.49 21.31 22.73
BY 03.00 07.39 07.54|CZ 02.50 06.80 08.10|PE 02.00 28.75 29.02|US 02.85 09.30 10.11
BY 02.50 08.09 08.68|CZ 02.50 10.51 11.84|PE 02.00 28.03 28.50|US 02.95 14.01 14.81
BY 02.50 07.37 07.63|CZ 02.97 08.22 09.09|PL 02.50 11.04 12.92|US 02.87 14.61 15.28
BY 02.50 07.58 07.90|CZ 02.50 09.05 09.89|PL 02.50 10.71 12.45|US 02.88 15.43 16.93
BY 02.50 07.42 07.83|CZ 02.50 08.81 09.47|PL 02.50 10.02 11.77|US 02.87 14.55 15.64
CA 03.31 04.26 04.55|CZ 02.50 09.30 10.00|RU 02.00 01.71 01.91|US 03.17 16.59 18.39
CA 03.37 04.50 04.94|CZ 02.50 10.40 11.20|RU 02.00 01.64 01.95|US 02.85 19.43 20.07
CA 03.03 08.69 09.28|CZ 02.50 11.17 12.45|RU 02.50 06.20 06.52|US 03.86 15.99 16.29
CA 03.03 08.69 09.28|CZ 02.50 08.64 10.41|RU 02.50 06.48 06.91|US 03.86 16.12 16.50
CA 03.13 05.83 06.39|CZ 02.34 09.19 09.50|RU 02.15 05.52 05.85|US 03.86 16.30 16.75
CA 02.49 04.03 04.43|CZ 02.31 10.35 11.10|RU 03.40 07.30 07.88|US 03.86 14.06 14.41
CA 02.22 03.98 04.70|CZ 02.50 11.47 13.64|RU 03.40 07.12 07.70|US 04.82 11.42 11.83
CA 01.60 06.71 07.86|CZ 02.50 11.34 13.96|RU 02.00 04.34 04.55|US 01.07 10.11 11.22
CA 04.58 06.63 07.01|CZ 02.31 11.00 12.20|RU 02.50 07.32 07.91|US 01.10 10.78 12.17
CA 03.60 05.01 05.09|CZ 02.50 08.53 09.40|RU 02.50 06.19 07.45|ZA 03.29 20.39 20.87
CA 03.30 03.45 04.13|CZ 02.50 09.56 10.80|RU 03.00 04.70 05.20|ZA 03.14 21.10 21.38
CA 03.00 03.32 04.15|CZ 02.50 11.28 12.59|RU 02.00 04.43 04.47|ZA 03.16 20.79 21.20
CA 03.10 03.37 03.73|CZ 02.50 07.85 08.67|RU 03.30 08.35 09.10|ZA 03.21 20.60 20.86
CA 03.60 02.97 03.57|CZ 02.50 08.85 10.17|RU 02.00 01.41 01.47|ZA 03.50 21.15 21.34
CA 03.30 03.30 03.95|CZ 02.50 08.27 09.65|RU 02.60 05.00 05.60|ZA 02.75 26.72 27.27
CA 02.50 09.23 10.11|CZ 02.72 11.76 13.92|RU 02.13 05.93 07.21|ZA 02.75 27.13 27.74
CA 04.00 07.36 07.44|CZ 02.50 04.70 05.70|RU 02.50 04.76 04.82|ZA 03.50 20.95 21.24
CA 04.00 08.61 09.07|CZ 02.50 06.52 08.46|RU 02.50 08.40 08.56|ZA 03.37 20.47 20.92
CA 04.30 04.38 05.05|CZ 02.50 07.55 09.51|RU 02.62 04.93 05.05|ZA 03.67 25.47 25.82
CA 04.20 04.83 05.27|CZ 02.50 07.37 09.24|RU 02.00 02.50 02.52|ZA 02.91 24.66 25.55
CA 03.10 04.15 04.46|CZ 02.50 07.52 09.38|RU 02.50 08.44 09.56|ZA 05.47 25.12 25.37
CA 03.00 04.12 04.50|IE 02.87 10.18 10.99|RU 02.00 05.05 05.38|ZA 02.21 22.35 23.56
CA 03.60 04.41 04.44|IN 03.00 29.38 29.53|RU 02.50 09.07 10.45|ZM 04.07 28.05 28.82
CA 03.60 04.28 04.48|IN 03.30 29.95 30.81|RU 02.06 04.50 05.50|ZM 02.81 26.47 27.59
CA 03.60 03.93 04.23|IT 01.80 15.38 16.43|RU 02.40 07.14 07.74|ZM 02.73 23.73 24.83
CA 03.60 03.29 03.89|IT 01.70 14.45 15.14|RU 02.50 06.55 07.25|ZW 02.81 23.16 23.91
CA 02.10 04.67 05.16|JP 01.38 15.14 15.51|RU 02.00 04.74 04.79|ZW 03.14 29.26 29.87
CA 02.00 04.49 04.72|JP 02.00 18.01 19.51|SI 02.50 11.82 12.83|ZW 02.38 22.53 23.79
CA 02.80 04.04 04.53|JP 01.31 15.44 15.45|SI 02.50 11.46 12.60|ZW 02.95 24.10 24.47
CA 02.20 02.13 02.62|JP 01.78 16.48 18.20|SI 02.50 12.74 15.08|ZW 03.10 24.40 24.63
CA 02.50 04.66 05.13|JP 02.74 16.86 17.39|SI 02.50 14.20 16.70|ZW 02.69 24.14 24.66``````

Every entry has a country code, the thermal conductivity (k) value, Temperature @50m depth in °C, and Temperature @100m depth in °C

For every country we will find an average value, and compute a conductive heat flux value using the following formula:

``CHF = k/50 * (T@100m - T@50m)``

Result:

``````AU 29 03.22 21.70 22.48 0.0502
BR 05 02.27 28.56 29.48 0.0418
BW 03 03.89 26.56 27.11 0.0428
BY 09 02.55 07.98 08.45 0.0240
CA 37 03.30 04.87 05.40 0.0350
CN 28 02.65 17.13 18.20 0.0567
CU 02 02.00 25.16 25.44 0.0112
CZ 32 02.48 09.12 10.43 0.0650
IE 01 02.87 10.18 10.99 0.0465
IN 02 03.15 29.66 30.17 0.0321
IT 02 01.75 14.91 15.78 0.0304
JP 22 01.67 16.46 17.42 0.0321
KR 20 02.70 14.44 15.64 0.0648
MN 02 02.50 02.31 03.55 0.0620
NA 01 02.80 27.36 28.55 0.0666
NE 01 02.70 32.65 33.10 0.0243
PE 03 02.74 25.38 25.94 0.0307
PL 03 02.50 10.59 12.38 0.0895
RU 27 02.42 05.54 06.04 0.0242
SI 08 02.50 12.58 14.73 0.1075
TZ 04 02.96 23.91 24.60 0.0408
UA 04 02.28 10.88 11.94 0.0483
UK 02 03.65 10.43 11.16 0.0533
US 42 02.71 13.55 14.73 0.0640
ZA 13 03.30 22.83 23.31 0.0317
ZM 03 03.20 26.08 27.08 0.0640
ZW 06 02.84 24.59 25.22 0.0358``````

The 2nd column is the # of locations per country. The 5th column is CHF. The average CHF for all locations is 48.9 mW/m² (0.0489157 W/m²).

Now we will calculate what the surface (Depth = 0m) temperature would be assuming no solar influence, and a continuation of the trend that existed between -100 and -50 meters. The formula is:

``T = T@50m - CHF*50/k``

Result:

``````AL 01 02.50 18.55 19.25 0.0350 17.850
AU 29 03.22 21.70 22.48 0.0502 20.920
BR 05 02.27 28.56 29.48 0.0418 27.639
BW 03 03.89 26.56 27.11 0.0428 26.010
BY 09 02.55 07.98 08.45 0.0240 07.509
CA 37 03.30 04.87 05.40 0.0350 04.340
CN 28 02.65 17.13 18.20 0.0567 16.060
CU 02 02.00 25.16 25.44 0.0112 24.880
CZ 32 02.48 09.12 10.43 0.0650 07.810
IE 01 02.87 10.18 10.99 0.0465 09.370
IN 02 03.15 29.66 30.17 0.0321 29.150
IT 02 01.75 14.91 15.78 0.0304 14.041
JP 22 01.67 16.46 17.42 0.0321 15.499
KR 20 02.70 14.44 15.64 0.0648 13.240
MN 02 02.50 02.31 03.55 0.0620 01.070
NA 01 02.80 27.36 28.55 0.0666 26.171
NE 01 02.70 32.65 33.10 0.0243 32.200
PE 03 02.74 25.38 25.94 0.0307 24.820
PL 03 02.50 10.59 12.38 0.0895 08.800
RU 27 02.42 05.54 06.04 0.0242 05.040
SI 08 02.50 12.58 14.73 0.1075 10.430
TZ 04 02.96 23.91 24.60 0.0408 23.221
UA 04 02.28 10.88 11.94 0.0483 09.821
UK 02 03.65 10.43 11.16 0.0533 09.700
US 42 02.71 13.55 14.73 0.0640 12.369
ZA 13 03.30 22.83 23.31 0.0317 22.350
ZM 03 03.20 26.08 27.08 0.0640 25.080
ZW 06 02.84 24.59 25.22 0.0358 23.960``````

The 7th column is our predicted temperature. The average value for all locations is 13.08 °C at an average latitude of 30.7607 N.

Mainstream climate science argues that the sun only provides the surface of the earth 163.3 W/m² (-41.5°C), and the rest of the energy is provided by greenhouse gases which boosts the global average temperature to ~15°C.

And yet here we have evidence that 312 locations don’t need greenhouse gases at all, because they get enough energy from geothermal.

https://phzoe.com

## 70 thoughts on “Measuring Geothermal (1)”

1. So Zoe, have you ever wondered how the underground temperature plot around 8 or 10Km extrapolates to the surface temperature? How do those underground regions “know” what temperature to “aim” at by the time the temperature plot gets to the surface? This happens in virtually all the scores of boreholes for which I have obtained data. The answer is of course due to the “heat creep” process that is explained at http://whyitsnotco2.com

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1. You don’t show your borehole data because it doesn’t support your claims. These underground regions miss their aim. I got ~13°C for 30N latitude, instead of the ~21°C I should have gotten. Makes perfect sense. Solar adds to Geothermal, it doesn’t set it.

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2. Blimey Zoe. You say you believe Prof Claes Johnson and yet you talk as if you think greenhouse gases warm some regions while other regions don’t need any !!!!!!!!!!!!

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1. I do not think that they warm some other regions. I’m simply saying that the evidence presented by boreholes is definitive for those areas. The reader can reach their own conclusion. Hopefully, they will understand that those other regions aren’t warmed by GHGs.

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3. Henrik says:

Brilliant!

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4. JaKo says:

Hi Zoe,
We talked about this elsewhere; however, there’s something I can’t dismiss so easily — recent human energy generation/consumption; how would that compare with the Geothermal:
IE at a rate — say @2013 ~18TW / (510*10^12m^2) yields some 35mW/m^2; well, a fraction would go to something else, while the bulk dissipates as heat, not from the Earth surface, but from buildings/structures, vehicles etc…
As you like to say: “Think about it!”
Cheers,
JaKo

Liked by 1 person

1. Well, you’re talking about a CSR type.
Geothermal has a small CHF, but a high CSR.
Human generation is a small CSR, and an even tinier CHF. Just divide that 35 mW/m^2 by the height of the atmosphere in meters, not to mention emission to space.
Cheers, Zoe

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5. patrick199405042 says:

Where do oceans get their energy from? Say, the surface of the oceans. They are kilometers deep, aren’t they?
How much energy do they get from the sun? And from other sources (?)

Apparently I am too dumb to use WordPress, so this is a different account. If you have any problem with me using two accounts, tell me. 😀

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1. Oceans get their energy from geothermal and insolation.

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1. patrick199405042 says:

Specifically, the surface. Geothermal can’t be it, can it? So why aren’t ocean surfaces freezing all over the world? Where does the energy come from? According to:
1) You
2) Mainstream science

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1. patrick199405042 says:

My bad, I have some more reading to do on your blog, your other article looks a bit scary 😉 but maybe I can make sense of it (not at 3AM 😀 ).

Liked by 1 person

2. 1) Geothermal and Insolation

Water in a gravity field convects. Heated water becomes less dense and travels up.

Boil a pot of water. First you get small bubbles – that’s the top evaporating, then you get bigger bubbles – that’s more top-down evaporating.

Heated water in a gravity field will first be noticed at the top, not the bottom.

Geothermal manifests itself upside-down in the ocean.

2) Insolation and Tidal sloshing

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6. Michael says:

Hey Zoe.

How does deep permafrost form then? I don’t understand with your model how permafrost could form to 300m deep, let alone the claimed Siberian record depth of over 1,400 m.

Liked by 1 person

1. Very simple. Geothermal is not above 0C. It needs solar power to be over 0C. Latitude still plays a role.

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1. Michael says:

“Now we will calculate what the surface (Depth = 0m) temperature would be assuming no solar influence, and a continuation of the trend that existed between -100 and -50 meters.”

Why is the geothermal at depth, beyond the reach of solar insolation, so different at different lattitudes if we are ruling out solar influence?
Sorry if it’s a stupid question.

Liked by 1 person

1. Boreholes are contaminated with inflowing air, and boreholes are not random, but dug for geothermal energy prospecting.

The only good thing about them is that they measure T at varying depth and provide a thermal conductivity constant (k).

An ideal experiment would use buried thermometers that can’t be contaminated with solar-heated descending air.

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2. P.S. There should be a little bit of latitudinal difference even in geothermal because the “inner sun”, so to speak, is even more oblate than the surface. The distance to same deep temperature should be smaller at equator than at poles. I read this in two papers, but I can’t be bothered to find them again. Eventually I will.

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3. I’d like to find out more about your thoughts on an inner sun. In my mental model, at the centre of the void (which is where the mainstream places its iron core) there is a very small sun, blazing away in gamma ray frequencies. Negative electrical energy is being created by charge separation and there is pristine thermal energy, pair production, and fusion going on. Fusion not due to heat or pressure but due to powerful negative electrical charge.

But this scenario cannot be what you are talking about with an oblate infrared internal star. Thought it could lead to that result I’m supposing. So if you can give me a better understanding of where you are going with this oblate internal infrared effect. Since I think it could reinforce me prejudices.

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4. “my” prejudices. Sorry don’t want to sound too antiquated here.

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7. Michael says:

Plasmoid in the centre driving earth’s geothermal?

I hadn’t thought of the idea of the earth’s “inner sun” being more oblate than the surface. Thanks for that. Would love to see some proxy evidence for this but i imagine it’s beyond our capacities at this stage.

Important to know what truly is energising the geothermal , and how this “inner sun” in turn is driven / powered. Too many hypotheses and naked speculation masquerading as supported theories ir even established facts post 1904 in all realms if science in my opinion.

Thanks Zoe.

Liked by 1 person

8. Michael says:

“The only good thing about them is that they measure T at varying depth and provide a thermal conductivity constant (k).”

I see now. Thanks.

Liked by 1 person

9. Zoe, I fear you’ve misunderstood the borehole data. The important variable is the heat flow from inside the earth this is the conductivity times the thermal gradient. The average heat flow of the 1,012 boreholes is 0.05 W/m2.

At the surface, this five-hundredths of a watt per square metre is added to the downwelling radiation. The surface temperature is NOT determined by the five-hundredth of one w/m2. It is determined by the radiation, and the geothermal heat flux is basically lost in the noise.

And if there were no radiation, the surface of the earth would only be heated by that 0.05 W/m2. In that case, the temperature would be down at 30 kelvin (-242°C). That’s all the geothermal warming would do.

Best regards,

w.

Liked by 1 person

1. Zoe, I asked a simple question. To move the discussion forwards, could you please answer it?

w.

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1. Why? You haven’t acknowledged what I’ve already said.

Your example is extremely irrelevant. 75C is not maintainable, and neither is 50C.

Assuming HSR is 16 W/m^2, the final steady state is:

Hot Side – 129.6K
Cold Side – 104.6K

CHF – 2.5 W/m^2
CSR – 6.11 W/m^2

* This assumes thermal conductivity (k) stays the same.

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2. Regarding my previous comment, let me do the calculations. Let’s assume a 1 metre cube in outer space, insulated on four sides, with an emissivity of 0.9 and a conductivity of 0.8. It receives 833 W/m2 on one end, and nothing on the other end. What are the eventual steady-state temperatures of the two opposite ends?

We know that at steady-state, the total amount emitted by the block has to be 833 W/m2, because that’s what it receives. We also know that the amount emitted by the cold end has to be 0.8 (Th – Tc), since this what is flowing through the cube and thus it is what is flowing out the end of the cube.

Solving these two equations gives us Th = 72.36°C and Tc = -59.8°C.

To check the answer, sigma epsilon (72.36+273.15)^4 + sigma epsilon (-59.8+273.15)^4 = 832.95 W/m2

and

(72.36 – -59.8) * .8 = sigma epsilon (-59.8+273.15)^4 = 105.7 W/m2

Best regards,

w.

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3. Willis,
Congratulations, you got CSR = CHF.
What was the point?

Your HSR = 832.95 – 105.7 = 727.25 W/m^2

This produces 63.4C, not 72.36C.

“We know that at steady-state, the total amount emitted by the block has to be 833 W/m2, because that’s what it receives.”

Terrible assumption.

Vibrational energy in = Vibrational energy out leaves no room for translational energy or bond energy.

Total energy is conserved, but strictly EM energy (vibrational) and heat flow need not be.

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4. Willis,
The cube will not reject 727.95 W/m^2.

You’re not going to have EM waves traveling back. This is junk science.

Boltzmann and Planck did not have two-way photon flow between two opposing walls. They had ONE wave per frequency. One!

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5. Zoe, thanks for the reply..

You say:

Willis,
Congratulations, you got CSR = CHF.
What was the point?

This seems to be the fundamental disagreement.

I say that the flow of heat through the block, the conductive heat flux (CHF) has to be equal to the amount flowing out of the far end of the block, the cold side radiation (CSR).

If not, then somewhere energy is either being created or destroyed. If energy is flowing through the block and it is NOT coming out of the cold side, what is happening to it?

And if more energy is flowing out of the end of the block than is flowing through the block, then where is that energy coming from?

What am I missing here?

Next, you say:

Your HSR = 832.95 – 105.7 = 727.25 W/m^2

This produces 63.4C, not 72.36C.

I fear that you’ve left the stated emissivity of 0.9 out of your calculation. My number is the correct one.

Finally, you say:

“We know that at steady-state, the total amount emitted by the block has to be 833 W/m2, because that’s what it receives.”

Terrible assumption.

Vibrational energy in = Vibrational energy out leaves no room for translational energy or bond energy.

Total energy is conserved, but strictly EM energy (vibrational) and heat flow need not be.

While you are 100% correct in general, in this example there is no change in translational or bond energy. As a result, vibrational energy in does have to equal vibrational energy out

My best to you as always, fascinating discussion.

w.

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6. Zoe Phin
February 9, 2020 at 12:33 pm

Willis,
The cube will not reject 727.95 W/m^2.

You’re not going to have EM waves traveling back. This is junk science.

Boltzmann and Planck did not have two-way photon flow between two opposing walls. They had ONE wave per frequency. One!

I don’t understand this claim at all. For example, the earth receives radiation at the surface. It also emits radiation at the surface. This is a two-way photon flow.

As another example, consider two stars that are far apart. If we bring them close together, both of them will warm up. Why?

Two-way photon flow.

Yes, there certainly are theoretical situations with one-way photon flow. But even in outer space, there’s microwave background radiation providing photon flow in the opposite direction to the thermal radiation coming from any solid object above absolute zero.

This comes up in discussions of the radiative heat transfer between two parallel plates. Every reference I can find to that question says there is a two-way photon flow, with each plate both absorbing and emitting radiation. See e.g. here, A google search for “radiation heat transfer between two parallel plates” will bring up many more, all of which will discuss the fact that each plate is both absorbing and emitting radiation, i.e., two-way photon transfer.

w.

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7. Willis,

How do you know?
If you have a satellite with a sensor facing Earth’s surface, it’s not facing the sun, is it?

“there’s microwave background radiation providing photon flow in the opposite direction to the thermal radiation coming from any solid object above absolute zero.”

All that was detected was photons flowing from the 4K reference source. Only ~1.25 flowed out, so it was assumed there is CMB. In reality, non-BB star light thermalized the sensor, at 2.75, so only 1.25 had to flow out. I’m not even mentioning the 2.75K BB expected from Earth at L2 distance.

“Every reference I can find to that question says there is a two-way photon flow, with each plate both absorbing and emitting radiation.”

That’s nice, but the fathers of radiation, Boltzmann and Planck didn’t do two-way photon flows to derive their equations.

Planck should have gotten DOUBLE the photon gas density in his chamber. Planck’s equation should have a 2x factor, if that was true. He doesn’t, and his experiment is well confirmed. There is only ONE wave per frequency between two opposing molecules.

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8. Willis,

A negative sign means the plate RECEIVES.

-X W/m^2 does NOT mean the plate sends X W/m^2 in the other direction.

NEGATIVE means reception, not direction.

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1. All solid objects emit thermal (longwave) infrared radiation.

2. All solid objects absorb thermal infrared radiation from the objects around them.

I truly don’t understand why you think that this is not two-way photon flow.

To put it another way:

Photons from object A hit and are absorbed by object B.

At the same time, photons from object B hit and are absorbed by object A.

Two-way photon flow.

Like I said, google ““radiation heat transfer between two parallel plates”. Every one of them will discuss the two-way photon flow between the two plates. It’s the only way to calculate the answer.

Now, to be clear, this is not HEAT flow. This is the flow of energy in the form of photons. Heat is the spontaneous flow of NET energy from hot to cold.

And heat flow is always one-way … but photon flow is assuredly two-way.

w.

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10. Willis,
Boltzmann and Planck did not derive their laws from two-way photon flow. They could not do so. Their laws only work with ONE wave per frequency per two opposing molecules.

I challenge you to derive their formulas using two-way photon flow.

Try it. When you fail …

“Now, to be clear, this is not HEAT flow. This is the flow of energy in the form of photons. Heat is the spontaneous flow of NET energy from hot to cold.”

LOL. And yet you want to equate internal heat flow rate per meter with cold side radiation. Nice!

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11. Zoe Phin
February 9, 2020 at 2:33 pm

Willis,
Boltzmann and Planck did not derive their laws from two-way photon flow. They could not do so. Their laws only work with ONE wave per frequency per two opposing molecules.

I challenge you to derive their formulas using two-way photon flow.

Try it. When you fail …

100% True. However, their equations are used constantly in situations of two-way photon flow. Without that, we couldn’t compute e.g. the radiative heat transfer between two parallel plates.

As you say … I challenge you to compute the radiative heat transfer between two parallel plates without using Planck, Boltzmann, or two-way photon flow.

Try it. When you fail …

“Now, to be clear, this is not HEAT flow. This is the flow of energy in the form of photons. Heat is the spontaneous flow of NET energy from hot to cold.”

LOL. And yet you want to equate internal heat flow rate per meter with cold side radiation. Nice!

Since there is no radiative energy impinging on the cold end, the cold side radiation IS the heat flow from the cold side. As a result, we can say that the heat flow through the block has to be equal to the radiative heat flow from the cold side. Otherwise, energy would have to be either created or destroyed, and that’s not possible, First Law of Thermodynamics.

Best regards,

w.

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12. Willis,

“However, their equations are used constantly in situations of two-way photon flow”.

That’s called imagination. You’re asserting that which you can’t prove.

“Without that, we couldn’t compute e.g. the radiative heat transfer between two parallel plates.”

Sure we can. The hotter plate generates waves that the cold plate lacks.

Show me where in Boltzmann’s and Planck’s cavity chamber there is a 2x factor for photon density. Do it.

“Since there is no radiative energy impinging on the cold end, the cold side radiation IS the heat flow from the cold side. As a result, we can say that the heat flow through the block has to be equal to the radiative heat flow from the cold side.”

Wrong. Conductive heat flux is translational energy derived from input. It is a subtraction from the input vibrational energy.

Input vibrational energy -> Internal translational energy + vibrational energy.

Output vibrational energy = Input vibrational energy – internal translational energy.

“Otherwise, energy would have to be either created or destroyed, and that’s not possible”

You destroyed 727 W/m^2. Or do you think it travels back and makes the source hotter?

Learn equipartition of energy and internal energy.

CHF and CSR have an inverse relationship.

You’re comparing profit/loss to assets/liabilities. It’s a silly thing to do.

You reject 727 W/m^2. You convert 105.7 W/m^2 into translational energy, which has to be used up by the time it gets to the cold side, and then you still think you have it to emit.

Imagine a solar power car. You charged it, you drove it, and then you think you can still emit the energy you spent on mechanically moving it. Where does this energy come?

Conductive heat flux is vibrational energy lost to mechanics, assuming radiation powered hot side.

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13. Willis,
“1. All solid objects emit thermal (longwave) infrared radiation.”

Only if there’s a disturbance in the electromagnetic field.

If there’s an object with same temperature next to it, there is 0 disturbance, there is 0 emission. No EM waves are formed.

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14. February 9, 2020 at 2:36 pm

Willis,

“1. All solid objects emit thermal (longwave) infrared radiation.”

Only if there’s a disturbance in the electromagnetic field.

If there’s an object with same temperature next to it, there is 0 disturbance, there is 0 emission. No EM waves are formed.

I begin to despair. They use night vision systems called “FLIR”, for “Forward Looking Infra-Red”, on helicopters. When they look down, they can see the entire landscape, people, cars, everything.

If objects don’t emit infrared until there is a disturbance in the force, how can FLIR work?

From West Texas A&M, emphasis mine:

Humans glow mostly in the infrared portion of the electromagnetic spectrum. Public Domain Image, source: NIST.
Yes, humans give off radiation. Humans give off mostly infrared radiation, which is electromagnetic radiation with a frequency lower than visible light. This effect is not unique to humans. All objects with a non-zero temperature give off thermal radiation.

From PhysicsNet UK:

All objects emit and absorb infrared radiation.The hotter an object is the more energy it radiates per second

From the Encyclopedia Britannica:

Virtually every object at Earth’s surface emits electromagnetic radiation primarily in the infrared region of the spectrum.

From MIT:

All macroscopic objects (above absolute zero) emit radiation.

From CalTech:

All objects that are not at absolute zero emit infrared radiation.

I could go on and on. This is not even a question. Not one of these sources makes your strange and totally untrue claim that objects only emit IR when there is a “disturbance in the electromagnetic field”. All of them say that if a solid object is above absolute zero, it emits thermal radiation. Period.

w.

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15. Willis,
If the object is hotter than the FLIR camera, it is detected by a positive voltage. If it is colder, then the FLIR camera heats the object, a negative voltage. There is only one wave per frequency per opposing molcules of camera and object.

All your references are just hustling the corpuscular theory of light. Tell me, what sense does wavelength mean to a corpuscle?

Now go ask your sources why photon gas density in a chamber doesn’t have a 2x factor due to two way flow.

Even your parallel plates show send and receive arrows at two different locations. Different molecules! A molecule can’t send and receive at the same time. Your plate can’t emit and receive X W/m^2 at the same time. Theoretically, half the plate can send X/2, and half can receive X/2.

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16. I give up. I can’t discuss physics with someone who thinks that an object in outer space a million miles from anything doesn’t emit IR until there is a “disturbance in the electromagnetic radiation”. I gave a half dozen references showing that is simply not true. You’ve ignored them completely and failed to produce even a single reference backing up your claim. How do you think infrared telescopes work?

Infrared telescope

All celestial objects with a temperature above absolute zero emit some form of electromagnetic radiation. In order to study the universe, scientists use several different types of telescopes to detect these different types of emitted radiation in the electromagnetic spectrum. Some of these are gamma ray, x-ray, ultra-violet, regular visible light (optical), as well as infrared telescopes.

One of the reasons that folks trust my scientific work is that when I’m wrong, I admit it, out loud and in public. Heck, I have a post called “Wrong Again”, and another called “Wrong Again Again”. Why? Because I was.

Until you learn to do that, there’s no point in me wasting my time.

My very best wishes to you, thanks for the interaction,

w.

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17. Willis,
How can you believe that infrared telescopes work and believe in downwelling infrared radiation from the atmosphere? One or the other.

My reference is Boltzmann and Planck.
Your references are non-experimental. They are merely assertions.

There is NO electronic device that can detect two way photon flow. None. It’s an unfalsifiable metaphysical construct – not science.

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18. Willis,
What are your references’ references? What historic evidence do they all trace back to?

Am I wrong in this?

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19. Willis,
“It takes a great deal of effort to build an infrared space telescope. After many years of hard work and after overcoming several complications, IRAS was successfully launched on January 25, 1983. The telescope was housed in a dewar, filled with 127 gallons of liquid helium and contained 62 detectors. The entire telescope was cooled to a temperature of just a few degrees above absolute zero because otherwise the telescope itself would emit infrared radiation (heat) which would interfere with the observations. A space infrared telescope must be cooler than the objects in space that it will observe.”

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20. Willis,
Did you know that people emit based on their assets? When you want to buy something for \$10, you have to give all your assets. They then emit all your assets back to you minus \$10. Also, they give you all their assets, and then you return it all back.

This is a physics principle. You have to accept it.

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10. Zoe, I just took a look at your page. I fear that you’ve made a mathematical mistake. The problem is that you have overspecified the equation. Let me explain by a parallel example.

It is a physical impossibility for there to be more water flowing out of the end of a hose than there is flowing through the hose. Can’t happen. The flow through the hose must be equal to the flow out the end.

In the same way, It is a physical impossibility for there to be more energy flowing out of the end of a block of concrete than there is flowing through the block. It is logically impossible. The flow through the block must be equal to the flow out the end.

The problem in your equation is that you cannot specify both q, the heat flux, and Tc, the temperature at the end of the block. Instead, you have to solve it as two different equations.

After substitution of the known quantities, the first equation is:

[Equation 1] q == 0.08 (75 – Tc)

The second equation ensures that the amount of water flowing out of the end of the hose is equal to the amount flowing through the hose, viz:

[Equation 2] q == 5.67*10^-8 * 0.9 * (273.15` + Tc)^4

Substituting “q” from Equation 2 into Equation 1 we get:

Tc = -138°C

At that point, the flow through the block is equal to the flow out the end of the block at 17 W/m2, so the physically required condition is satisfied.

My best regards to you.

w.

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1. Willis,
You got two totally different concepts confused. Heat flux is a differential. Thermal conductivity (k) has a length component. Radiation is an absolute.

Boltzmann and Planck’s experiment is based on cavity experiments where Heat flux was nearly zero, i.e. thermal equilibrium. According to you the cavity should’ve emitted nearly zero.

Conductive heat flux is not like water flowing through a hose. It’s like a sponge that reduces water flow. Heat flux is a differential, not an absolute.

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2. SB Law is not
q = …

You’re confusing heat flux with energy flux density.

SB Law yields W/m^2

q is in Watts.

Heat flux already accounts for a difference between two energy levels. Equating it with cold side radiation is just nonsense.

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1. Thanks, Zoe. I fear I simplified it too far and lost clarity along the way. Let me take another shot. Divide through by A.

q/A (in W/m2) = k ∆T/L

For simplicity in dealing with abstract cases, A is typically set to one square metre, and L is set to one meter. I’ll use this for my example. This gives us

Flux in W/m2 = k (Th -Tc)

Note that since we’ve specified Th, and k is a property of the material, the energy flow and Tc are inversely related. The colder that Tc is, the more energy flows through the one metre cube of concrete.

Now, let’s assume we’re doing this in outer space. We have a cube with one side at an initial temperature of 70°C, with perfectly insulated sides so it can only gain or lose energy through the two opposite faces. To pick a number, let me specify that the flow into the hot side is 16 W/m2. This should give the same answer, since it’s eight times the energy and one-eighth the length of your example. This give us

16 W/m2 = k(70 – Tc)

or

Tc = 70 – 16/k

As in your example, that gives Tc = 50°C

But at that point, the cube is gaining energy at the rate of 16 W/m2, and it is losing energy at 550 W/m2.

What will happen to the temperature of the cube? Heat up, cool down, or stay the same?

w.

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1. Willis,
In my example, the hot side has a steady source of 833 W/m^2 (75C).

You can’t equate a conductive heat flux with radiation, they’re completely different.

For example:
The difference between surface and start of stratosphere temperature is ~70C. The height of stratosphere is 20,000 meters. The k value of air ranges by pressure, but we can use 0.025.

For 1 square meter surface:
q = 0.025 * 70 / 20000 = 0.0000875 W/m^2 for conductive heat flux.

Yes that hot sun produces a tiny CHF.

Conduction is semi order in random motion. Heat flux is just a differential of random motions between two locations. Inside matter, energy can go every which way, but at the edge … there is nothing to hold back its full energy.

My concrete block example is a standard high school/college example. I got Tc correctly, and …

SB Law will emit by TEMPERATURE (and emissivity) and not by Conductive Heat Flux.

SB Law was derived experimentally in a cavity chamber where CHF was nearly ZERO. A CHF of ZERO yields the highest surface radiation possible. But you want to equate a CHF of ZERO with the highest possible radiation. Makes no sense.

According to your view, Boltzmann should have got nearly ZERO radiative emission because CHF was nearly ZERO.

CHF is a DIFFERENTIAL, Radiation is an absolute.

An object with a CHF of ZERO does not mean it has no energy to emit at its surface. It’s already “emitting” this energy level on the inside.

Knowing the CHF doesn’t even tell you what it can emit.

A 1000C object with a CHF of 0 will emit a lot more than a 100C object with a CHF of 0.

You get it?

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2. Zoe, I left an answer but it ended up upstream in the wrong place.

In any case, you say:

“In my example, the hot side has a steady source of 833 W/m^2 (75C).”

Now, I assume that in your example the sides of the block are perfectly insulated.

So it is losing energy from only two places, the two opposite ends.

But if the hot end is at 75°C, then it is radiating back out the total amount entering the block … which leaves no energy to pass through the block and come out the other side.

So this cannot be a steady-state condition.

Best regards, and thanks for the discussion. Makes me think hard.

w.

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11. As another example of what the problem with this discussion is, I can’t discuss physics with someone who claims that an FLIR camera in a helicopter warms objects on the ground, viz:

Zoe Phin
February 9, 2020 at 5:08 pm

Willis,
If the object is hotter than the FLIR camera, it is detected by a positive voltage. If it is colder, then the FLIR camera heats the object, a negative voltage.

Seriously? An FLIR camera on a helicopter a half-mile in the air is warming anything on the ground that is cooler than the camera? Say what?

Plus which, if that’s happening, then the stuff on the ground is being warmed by IR from the camera, and the camera is imaging IR from the object on the ground … and of course, that means something is happening that you say can’t happen, two-way photon exchange.

Truly, Zoe, I thank you for the discussion, but I can’t get past claims like that, or that infrared telescopes wouldn’t work if there was downwelling longwave radiation from the atmosphere. Downwelling radiation from the atmosphere has been known about, discussed, and MEASURED BY INSTRUMENTS all over the planet for decades. But you gaily sail right on past that and declare it doesn’t exist.

I did like how you called the clear statements from MIT, CalTech, Encyclopedia Britannica, and other experts, all of which agreed 100% with each other that all objects above absolute zero radiate IR, as mere “assertions” … it gives me an insight into how you deal with being shown to be wrong by experts from all over the planet.

It’s simple.

You’re never wrong.

So as I said above, I’m gonna let this go. Be clear that I’m not going away mad, or angry.

I’m just gonna wish you all the warmest regards, and let this discussion go.

w.

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1. Willis,
“Seriously? An FLIR camera on a helicopter a half-mile in the air is warming anything on the ground that is cooler than the camera? Say what?”

Yes dear, the FLIR camera is registering voltage gain/loss at ~0.99 times the speed of light.

“Plus which, if that’s happening, then the stuff on the ground is being warmed by IR from the camera, and the camera is imaging IR from the object on the ground … and of course, that means something is happening that you say can’t happen, two-way photon exchange.”

No dear that is your imagination again. The Camera is not imaging colder radiation from the ground, it is imaging the differential it sends to the ground. If object is warmer than camera, then the camera records the difference gain.

FLIR cameras can’t work without a non-radiation based thermometer. IR telescopes can’t work without a KNOWN reference temperature.

Why? Because only the differential can get recorded. And if you have no absolute, you have nothing.

Those devices would not need thermometers or a reference temp if absolute radiation could be measured.

“Downwelling radiation from the atmosphere has been known about, discussed, and MEASURED BY INSTRUMENTS all over the planet for decades.”

Nonsense. What has been measured is the upwelling radiation from the detection instrument OR the detection instrument was super-cooled to get genuine DWLR; it doesn’t exist normally and at rates claimed.

Are you even aware of upwelling-from-instrument-IR? Why not?

“I did like how you called the clear statements from MIT, CalTech, Encyclopedia Britannica, and other experts, all of which agreed 100% with each other that all objects above absolute zero radiate IR, as mere “assertions” … it gives me an insight into how you deal with being shown to be wrong by experts from all over the planet.I did like how you called the clear statements from MIT, CalTech, Encyclopedia Britannica, and other experts, all of which agreed 100% with each other that all objects above absolute zero radiate IR, as mere “assertions” … it gives me an insight into how you deal with being shown to be wrong by experts from all over the planet.”

If they’re not merely assertions, surely you can provide where they get their idea from. What’s the origin of it? Are you saying statements from institutions are evidence in themselves? Science stops at these institutions? Say what?

I haven’t been shown to be wrong, I’ve only been told that I’m wrong.

I’ve read Boltzmann’s and Planck’s derivations and experiments. You haven’t. They could never have achieved what they did using two-way photon flow.

I will gladly concede and admit I was wrong WHEN you show me the 2x factor in Boltzmann’s and Planck’s derivation and experimental confirmation.

Do it! I don’t mind being corrected. I’ve spent 10 years on Wall Street. Have you ever speculated and invested? Constant readjustment, flip-flops, and timing. Nothing is static, and you spend half the time trying to prove yourself wrong. That’s how you succeed. And that’s how I succeeded.

Also, the economics taught in university is complete crap. Takes a smart person to realize that early. Hello!

I know academia is full of garbage. Now I apply that wisdom to physics. Not saying all of it is garbage. One must use discernment and critical thinking to wade through the mess.

P.S. Are you in anyway pressured to believe some things and not others?

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1. patrick199405042 says:

“No dear that is your imagination again. The Camera is not imaging colder radiation from the ground, it is imaging the differential it sends to the ground. If object is warmer than camera, then the camera records the difference gain.”

Hm, let’s apply this to backradiation:

Mainstream physics: Heat is transferred in both directions.
[NET heat transfer from hot object to cold] = [transfer hot to cold MINUS transfer cold to hot]

Vs. let’s call it “rational physics”:
[heat transfer from hot object to cold] = [transfer hot to cold which is weaker the smaller the difference in heat between hot and cold]

More CO2 means:
a) this heat exchange happens at more wavelengths
b) the temperature differential decreases (
the hot object remains the surface, unchanged
but the cold object is now the hotter, closer CO2
)

Therefore: Less cooling for the surface. Less cooling plus same sunshine equals warming. In both “models”.

Tbh, I’m confused as I often am. I came here for an inside into the skeptics’ minds but it is not always successful.

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1. Patrick,

“the temperature differential decreases (the hot object remains the surface, unchanged)”

Surface ONLY has temperature due to geothermal and solar.

“Less cooling plus same sunshine equals warming.”

Therefore this is a non-sequitur.
The CO2 gets warmed and the surface remains the same, as you just said.

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2. Hi Zoe!
Does CO2 ever cool down?
Who is its “partner” in that case? Earth will always be warmer because of altitude, right?
Does its “send temperature” increase on path to down to earth (should it find cooler “partners” there) and if so does a CO2 atom know its altitude to know its “arrival” temperature and hence evaluate who to “connect” with?
Or is it the “partners” that hook up after comparing local temp vs recieved?

I am confused…

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2. Steve Keppel-Jones says:

Willis, I know you’ve gone away, but for the benefit of others who may follow this trail to this point, Zoe is right. You seem to be thinking of photons as some sort of “bullet” that an atom can fire into the universe while wishing it “bon voyage!” That’s the old corpuscular theory. But photons don’t work that way.

It is more accurate to think of a photon as a “negotiation” between two atoms, one of which has an excess of energy within the allowable quanta for it to emit, and the other one of which can accept that extra quantum of energy. But _which_ two atoms isn’t determined until the photon’s wavefunction collapses, which, in our current understanding, actually requires an observer to observe it as well. (Unless you use the many-worlds interpretation.) Until that point, no energy has left the higher-energy atom. Its excess energy is just a wavefront propagating through space looking for a place to happen, and it can’t happen just anywhere. It’s very selective.

So a colder atom on the surface of the earth cannot transfer a quantum of energy to a warmer FLIR camera. It doesn’t matter how many “experts” describe radiation as something that all objects do. The quantum phenomenon isn’t quite that simple. In the singular, or in the aggregate (when many atoms are involved), heat (energy) always flows from higher-energy matter to lower-energy matter. So an infrared camera indeed cannot take a (positive) image of an object that is colder than it is. If you have an infrared detector at ambient surface temperature, it cannot measure the infrared flux from the colder atmosphere, because the flux is going the other way.

(Zoe’s comments seem to indicate that an IR camera may be able to take a negative image, of negative flux relative to its reference, which may be the case, I am not sure about that part. I have never seen IR images that have negative values relative to the sensor. But in that case, indeed, the camera is heating the object it is taking an image of.)

Liked by 1 person

1. I am maybe stuck in wrong thinking but how is the energy flow actually turned on? Is there enough information in the “wavefront”.
I am stuck thinking you can measure radiation from the sun the second it raises not 499×2 seconds after an “ACK signal” is transferred back. Even if the ACK is not needed the energy still has a 499 seconds journey.

Should I instead view the “wavefront” like a filled but closed hose that opens on “connection”? If so how does the frequency build up after the moment the “connection” happened.

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2. Steve Keppel-Jones
July 5, 2020 at 11:48 PM

Willis, I know you’ve gone away, but for the benefit of others who may follow this trail to this point, Zoe is right. You seem to be thinking of photons as some sort of “bullet” that an atom can fire into the universe while wishing it “bon voyage!” That’s the old corpuscular theory. But photons don’t work that way.

It is more accurate to think of a photon as a “negotiation” between two atoms, one of which has an excess of energy within the allowable quanta for it to emit, and the other one of which can accept that extra quantum of energy. But _which_ two atoms isn’t determined until the photon’s wavefunction collapses, which, in our current understanding, actually requires an observer to observe it as well.

So you are telling us that before the sun emits a photon towards some star that is 300,000 light years away, it has to “negotiate” with that star as to relative energy levels.

So the sun sends out a request for the “negotiations” you describe. It takes 300,000 years for the request to get there, and another 300,000 years to get the answer back … and you are seriously claiming that this process goes on in real time for every photon emitted?

Pass …

w.

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1. Willis,
All the objects in the universe have always been “tethered”.

You can’t use your corpuscular theory to derive Planck’s or Boltzmann’s Law. They can only be derived by two-ended wave theory.

Can you explain what photon “wavelength” means to a corpuscle? What property of a corpuscle (a dot) is a wavelength? There seems to be a lack of physical sense there.

Have you considered that even in your corpuscular framework that photons can only charge toward where there is a potential for them to charge, such as another object?

-Z

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12. Steve Keppel-Jones says:

Willis, perhaps “negotiation” was the wrong term, because it’s not a two-way process. No answer is required, so maybe “agreement” or “congruence of states” is better. In any case, before an atom in the sun can lose a quantum of energy, another atom has to accept it. It doesn’t matter where that other atom is in the universe, but if it is 300,000 light years away, it will take that long for the wavefunction to collapse and be observable at the other end, depending of course on which frame of reference you’re sitting in. (In the photon’s frame of reference it’s instantaneous.) Quantum physics sure is weird.

MrZ, in the aggregate case, multiple photons can of course be propagating at the same time. Each one will need to arrive somewhere and collapse before a measurement can be recorded (more precisely, at the instant the measurement is recorded). I don’t think a hose is very good analogy. Each photon has to make its own connection for its wavefunction to collapse, so there is no “link” that stays open afterward for energy to flow through.

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1. It is fascinating and engaging to think about those things.
Still, for me impossible to grasp. If not a hose can you instead give me a queue like thing. I am struggling to understand how the photon can arrive at the same time it got the “get to go” signal. Especially when the warmer atom is 499 seconds away unless there is something that has infinit speed.

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1. gbaikie says:

If you going a the speed of light you go across the universe in a second.
But someone not going speed of ;light, will experience it as it taking you 13 billion years to go across
the universe.
A problem with you going the speed light is the acceleration needed reach such a velocity.
Light attains such speed when on turn on a light switch.
Back in spaceship somehow going speed of light, before take one step, you crossed the universe if somehow
turn around and head back at speed of light, before take one step you would cross the and universe and back where started. Matter can’t do that. Matter can not do 1/10th the speed of light in a second. Or 1/100th of speed light in a second.
Another aspect going speed of light is one gains mass- a brick going speed light will increase it’s graviational force- increases it’s mass.
No one knows what mass/gravity or light is. Instead it’s the measuring it’s effects.

One effect of heat is that flows from hot to cold.
In terms of radiant energy. At earth distance from Sun, sunlight which not magnified can heat to certain temperature and in terms of sunlight’s effect upon blackbody. It can heat to temperature of about 120 C.
If magnify sunlight, sunlight can heat a blackbody to the temperature of the sun itself, 5,778 K [5504.85 C}, but no further magnification can exceed this temperature.
So sun at Earth distance from the Sun acts as heat source which about 120 C.
But after going thru 1 atm of Earth atmosphere, the temperature of sunlight is about 80 C. Or it become about 1050 watts of direct sunlight. Or it’s 1050 watts of direct sunlight and if include indirect sunlight it’s 1120 watts per square meter.
And one uses only the direct sunlight in terms of how hot sunlight can heat a blackbody surface..

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13. Steve Keppel-Jones says:

MrZ, it sure is a fascinating topic. Light (photons) do have infinite speed in their own frame of reference. But in the frame of reference of a stationary observer, like you or me, it takes a bit of time to travel. Either way, we would not be able to detect a loss of the energy quantum in the higher-energy (warmer) atom before the corresponding gain of energy in the lower-energy (colder) atom, no matter how far apart they are. The situation is even less like a queue than a hose, unfortunately 🙂

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14. Hmmm
Most good ideas have a simple explanation. This one, in contrast, rely on very complex maters.
“Infinite speed, but It takes a bit of time to travel vs a stationary observer”. Seams you are saying we are in different dimensions.

Maybe you can explain something very simple in our world where this has an impact. Should be plenty of examples since this virtually touches everything.

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15. Steve Keppel-Jones says:

MrZ, it is not quite a case of different dimensions, but accelerating does change your frame of reference, and all the dimensions around you change as you do. So that’s a reasonably close analogy, as far as it goes.

For a simple example in our world, the relativistic effects within Earth itself are pretty small. But if you compare, say, the clocks of astronauts on the International Space Station vs. stationary clocks on Earth, the astronauts’ clocks will lose about 0.01 seconds per year. That’s small, but still measurable!

The next best example I can think of would be geosynchronous satellites, which are about 1/4 of a light-second away. So it takes about half a second to get a signal there and back. From the point of view of the photons, that’s instantaneous, but it’s a bit tricky to see what difference that makes to us. You could set up a diffraction slit experiment reflecting photons off a mirror on a geosynchronous satellite, then you could see that the photon’s wavefunction doesn’t collapse until it arrives at the detector in your experiment (based on the interference pattern), rather than when it left your source half a second earlier – but that’s not really a real-world example per se, and it’s the same result no matter how far the photons travel, too, so it’s not that exciting 🙂

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16. MrZ says:

Thanks for trying so hard.
Can I use sound as analogy? If we skip the fact sound is pushing air it is still not noticeable until the air bounces against matters and matters hears. Similarly maybe I could think that the photons leaves its parent. A wave and hence energy is not created until the “first” photon bounces of something. From the receivers point of view it has an endless amount of sources. As photons bounces On it they delay the following photons if ever so little. This delay causes a queue that builds a standing wave that in turn represent the energy in terms of a frequency and amplitude.

A crazy analogy?

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1. Jarle says:

MrZ, “From the receivers point of view it has an endless amount of sources”. How could anything have an endless amount of sources?

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17. Steve Keppel-Jones says:

MrZ is basically right that a receiver of a photon has an endless number of possible sources, and vice versa. Which atom transfers a photon of energy to which other atom isn’t determined until the corresponding wavefunction collapses, and until then, everything overlaps, like waves on the sea.

Sound is not too bad of an analogy, but “bouncing” is not the most common type of interaction. Photons can bounce (the incident wave is reflected) at interfaces of materials of significantly different propagation characteristics, like mirrors. But most of the time, a photon will be absorbed rather than bounce. This has no effect on following photons, so there is no delay and no “queue” per se. But there can certainly be multiple photons following each other, if atoms with excess energy emit more than one photon. They don’t queue up, though. Each one propagates independently, although their wavefunctions do interact with each other (to form interference patterns etc.) Maybe that is what you were thinking of when you thought of standing waves.

I think waves on the sea is a better analogy than sound waves, but they’re pretty similar, I suppose.

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18. Zoe, I managed to find the geothermal gradients of two Tasmanian mines in my Archive folder. They show a kink of increasing rock temp at around 1000m+, which correlates with the anomalous warming data from the bottom of Lake Tanganiyka.
I’ll have to email them to you as they won’t copy from my clipboard.

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