What happens when one tries to work out a science problem when one does not know the math?
A poster asked about the effects of renmoving CO2 from the atmosphere.
Another poster who thinks he can understand dynamical systems with very limited math wrote back
"There would be no noticeable lag between CO2 reduction and H2O feedback kicking in and the cooling effect would be immediate... PS – No differential equations were used in the making of this answer (just a very simple 9 cell spreadsheet)."
He did not use differential equations but answer is clearly wrong though!
The Earth is in a disequilibrium state now. If CO2 remaimed constant, the Earth would heat up for many years to come. One can think of it as if the temperature has not yet caught up to the CO2 level. It is as if I keep raising the thermostat setting for a room, and then stop raising it. If the temperature of the room has not yet reached the thermostat level, the temperature will keep rising for awhile even if I do not rause the thermostat setting further.
So if we start decreasing the CO2 levels, what will happen is not that the temperature will immmediately drop as Trevor and his spreadsheet claim, but rather that *the rate of increase* in the temperature will drop. The temperature will rise for a while, but it would not rise as fast as it otherwise would have.
No doubt, most of you here have no idea about how what I just explained relates to differential equations.
You got it wrong, Ottawa Mike. You are not understanding the difference between "the derivative of temperature with respect to time" and "the second derivative of temperature with respect to time".
Trevor was claiming if CO2 would start decreasing, the derivative of temperature with respect to time would be immediately negative. (He of course did not use that wording--he is a non-calculus scientist of the highest magnitude) What I was claiming was that if CO2 started dropping, the second derivative of temperature with respect to time would be negative (it actually could be negative even other conditions) but that the derivative of temperature with respect to time would remain positive at first.
"I asked the question to which you refer http://answers.yahoo.com/question/index;…
so I'd be interested to see your answer there, preferably spelled-checked, and including any relevant differential equations (using any readable and unambiguous notation). "
Hey Dook, you would have gotten your answer "there" except you blocked me.
"I think this may be another case where you are primarily nitpicking someone's word choice. A cooling effect does not necessarily mean that temperatures will drop from what they were before... just that they will drop from what they otherwise would have been."
That is not what "cooling" means. You can create your own language in private, but it causes trouble if you use it in public.
One of the reasons why equations are often useful is because they are precise--there is no room to give a "new" meaning to something like dT/dt.
"I think this may be another case where you are primarily nitpicking someone's word choice. A cooling effect does not necessarily mean that temperatures will drop from what they were before... just that they will drop from what they otherwise would have been."
That is not what "cooling" means. You can create your own language in private, but it causes trouble if you use it in public.
One of the reasons why equations are often useful is because they are precise--there is no room to give a "new" meaning to something like dT/dt.