Climate Sensitivity Reconsidered

The point of this post is to show a calculation by guest, Pochas, of the decay time that should be expected from the accumulation of heat in the mixed layer of the ocean.

I realized this prediction gives another test of the accumulation theory of climate change, that potentially explains high climate sensitivity to variations in solar forcing, without recourse to feedbacks, or greenhouse gasses, in more detail here and here.

The analysis is based on the most important parameter in all dynamic systems, called the time constant, Tau. Tau quantifies two aspects of the dynamics:

1. The time taken for an impulse forcing of the system, such as a sudden spike in solar radiation, to decay to 63% of the original response.

2. The inherent gain, or amplification. That is if the Tau=10, the amplification of a step increase in forcing will be x10. This is because at Tau=10, around one tenth of an increase above the equilibrium level will be released per time period. So the new equilibrium level must be 10 times higher than the forcing, before the energy output equals the energy input.

I previously estimated Tau from global temperature series, simply from the correlation between successive temperature values, a. The Tau is then given by:

Tau = 1/(1-a)

Pochas posted the theoretical estimate of the time constant, Tau, below, that results from a reasonable assumption of the ocean mixed zone depth of 100m.

The input – output = accumulation equation is:

q sin ωt /4 – kT = nCp dT/dt

where q = input flux signal amplitude, watts/(m^2 sec). The factor 4 corrects for the disk to sphere surface geometry.

k = relates thermal flux to temperature (see below) J/(sec m^2 ºK).

T = ocean temperature,

ºKn = mass of ocean, grams.

Cp = ocean heat capacity J/(g ºK)t = time, sec or years.

Rearranging to standard form (terms with T on the left side):

nCp dT/dt + kT = q sin ωt /4

Divide by k

nCp/k dT/dt + θ = q sin ωt /(4k)

The factor nCp/k has units of time and is the time constant Tau in the solution via Laplace Transform of the above.

n = mass of water 100 m deep and 1m^2 surface area = 10E8 grams.

Cp = Joules to heat 1 gram of water by 1ºK = 4.187 J/gram.

k = thermal flux equivalent to blackbody temperature, J/(m^2 sec ºK).

Solution after inverse transform, after transients die out:

Amplitude Ratio = 1/(1+ω²T²)^½

where ω = frequency, rad/yr

Derivation of k Stefan Boltzmann equation

q = σT^4k = dq/dt

Differentiating: dq/dt = 4σT^3

Evaluating at T = blackbody temp of the earth, -18 ºC = 256 ºK

k = 4 (5.67E-8) 256^3 = 3.8 J/(sec m^2 ºK)

Calculating Time Constant Tau

Tau = nCp/k = 10E8 (4.187) / 3.8 = 1.10E8 sec

Tau = 1.10E8 / 31,557,000 sec/yr = 3.4857 yr

_____________________________________

The figure of Tau=3.5 yrs is in good agreement with the empirical figures from the correlation of the actual global surface temperature data of 6 to 10. The effective mixed zone may be closer to 150m, and so explains the difference.

This confirms another prediction of the theory that amplification of solar forcing can be explained entirely by the accumulation of heat, without recourse to feedbacks from changing concentrations of greenhouse gases.

Advertisements

0 thoughts on “Climate Sensitivity Reconsidered

  1. Not sure if this is relevant to the issue above, but can someone tell me if out-gassing of the oceans is being directly measured anywhere?

    I have a reasonably accurate CO2 meter, used for testing enclosed spaces, as part of my work. Doesn’t have data logging unfortunately.

    I have recently been using it to monitor outdoor concentrations. In my location (19°S 146°E) the rural fire service conducts burn-off of accumulated fuel load. More needed this year because of several years of enhanced growth plus the place being trashed by Cyclone Yasi back in February. Yes, there is a distinct rise on CO2 down-wind of the burn-off.

    But there is also a rise when the meter is turned towards the north-easterlies that blow in from the Coral Sea in late afternoon. This is 200m away from the receptor, and there is very little development nearby, no industry in that direction, not much vegetation, no power stations, Defence aren’t bombing Rattlesnake Island – just the Great Barrier Reef and Pacific Ocean.

    http://people.aapt.net.au/jclark19/

    • ” troppo19 2 weeks ago

      Not sure if this is relevant to the issue above, but can someone
      tell me if out-gassing of the oceans is being directly measured
      anywhere?”

      The best thing to do is look at the Ar/N2 atmospheric ratio. The partitions of the two 9inert0 gasses between the oceans and the atmosphere are both temperature dependent, but have different slopes. Both Keeling and Bender have measured the saw-tooth change in the Ar/N2 ratio over the year, which partly tracks the saw-tooth of the CO2 steady state.
      Keeling or some of his kiddies have done a 5 year track of the Ar/N2 ration somewhere.

      http://bluemoon.ucsd.edu/publications/mip/Ar.pdf

      http://geoweb.princeton.edu/people/bender/lab/research_arn2.html

    • Yes, for a process with a period of 10 years and a capacitance of 100 m of water.  All other processes have different periods and time lags.  What we see in the temperature record is a superposition of processes with different time lags and amplitude ratios, giving the appearance of noise.  
      Here is a graph of integrated sunspot number vs temp anomaly I made some time ago (along the lines of Dave Stockwell’s Solar Accumulation Theory recently discussed here) which I believe, if you stare long enough, shows a solar influence at a time lag of 3 years.  Since ENSO dominates the record I wouldn’t attempt to calculate r^2.  If I’d known then what I know now, I would have made the lag 3.5 years.


      With the amplitude ratio calculated as given above the TSI signal should be vanishingly small, so the apparently visible effect is due to some parallel modality such a gravity, UV, or cosmic ray shielding.

      note:  the Amplitude ratio calculation should have Tau instead of T, i.e.
      Amplitude Ratio = 1/(1+ω²Tau²)^½
      Everywhere else T is temperature.

  2. Is this a derivation by thermal conductivity, rather than by overturning/mixing/eddies? Can imagine several modes that take heat from the surface down and up.
    Cohenite, it’s not as simple as out of synch by 3.5 years. That’s just a mathematical point in time after a step change. The effect goes on for some time. Think simple radioactive decay math.

    • The solution given is specifically for a sinusoidal input.  Its true that the time constant also gives the time for the dependent variable to reach 62% of its ultimate value after a step change. 

      The phase lag for a process like this that does not include reactance is always about 90 degrees. I calculate the phase 
      lag for the above process ϕ
      = arctan (ωTau) = 89.6 degrees.  The effect of thermal resistance would be to decrease the amplitude ratio still further.

  3. This looks interesting but is so riddled with errors that it does not inspire confidence, and I don’t have the time to redo it all to see if it can be corrected.

    q = σT^4k = dq/dt
    ? q=dt/dt ?
    watts/(m^2 sec).
    ?watt/sec?
    256 ºK
    ? the unit is kelvin, not degrees kelvin?
    ºKn = mass of ocean, grams.
    ???
    Cp = ocean heat capacity J/(g ºK)t = time, sec or years.
    ?it can’t be both, chose you units and stick to it?

    Also the std unit of mass is kg not g.

    I suggest you edit your post to make it more coherent. The idea is interesting.

  4. This looks interesting but is so riddled with errors that it does not inspire confidence, and I don’t have the time to redo it all to see if it can be corrected.

    q = σT^4k = dq/dt
    ? q=dt/dt ?
    watts/(m^2 sec).
    ?watt/sec?
    256 ºK
    ? the unit is kelvin, not degrees kelvin?
    ºKn = mass of ocean, grams.
    ???
    Cp = ocean heat capacity J/(g ºK)t = time, sec or years.
    ?it can’t be both, chose you units and stick to it?

    Also the std unit of mass is kg not g.

    I suggest you edit your post to make it more coherent. The idea is interesting.

  5. 2. The inherent gain, or amplification. That is if the Tau=10, the
    amplification of a step increase in forcing will be x10. This is
    because at Tau=10, around one tenth of an increase above the equilibrium
    level will be released per time period. So the new equilibrium level
    must be 10 times higher than the forcing, before the energy output
    equals the energy input.

    Sorry , that in nonsense. You don’t seem to understand the terms you are using.

  6. 2. The inherent gain, or amplification. That is if the Tau=10, the
    amplification of a step increase in forcing will be x10. This is
    because at Tau=10, around one tenth of an increase above the equilibrium
    level will be released per time period. So the new equilibrium level
    must be 10 times higher than the forcing, before the energy output
    equals the energy input.

    Sorry , that in nonsense. You don’t seem to understand the terms you are using.

    • Please explain why you think that.  For a system that loses 10% of a perturbation every time step, then the decay time Tau is 10 steps, as 0.9^10 = 0.35.  However, as above, as only 10% is lost every time step, the equilibrium level must increase until the output is x10 its original perturbation in order to balance the constant input.  So the units of Tau are both in time, and in gain.

  7. David, please change T to Tau as follows in the original posting:
    Amplitude Ratio = 1/(1+ω²T²)^½Amplitude Ratio = 1/(1+ω²Tau²)^½I would not want people to confuse T (temperature) with Tau (a time constant, units of time).  Leave T as is everywhere else.  Thanks for giving this problem some visibility

  8. Pingback: fijzoterapii i Dietetyki

  9. Pingback: zobacz tutaj

  10. Pingback: oferta

  11. Pingback: witryna

  12. Pingback: wynajem opiekuna wroclaw

  13. Pingback: zobacz

  14. Pingback: depilacja jaka

  15. Pingback: polecam link

  16. Pingback: strona firmy

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s