How to set an adiabatic temperature gradient of 0.5K/km for the mantle below the lithosphere?

ASPECT
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Hello everyone!
I am designing a 2D mid-ocean ridge model. In the initial temperature model, my aim is that the lithosphere follows the half-space cooling model, and the mantle temperature below the lithosphere increases from the bottom of the lithosphere with an adiabatic temperature gradient of 0.5K/km to the bottom of the model (670km). I have written the alf-space cooling model, but I don’t know how to achieve a mantle temperature gradient of 0.5K/km? Thank you all.I would be very grateful if any friends could help me solve the problem
initial temperature model.txt (459 Bytes)

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Hi @sxj,

Thank you for posting the question to the forum and apologies for taking some time to get to it.

I have written the alf-space cooling model, but I don’t know how to achieve a mantle temperature gradient of 0.5K/km?

My suggestion would be to precompute the depth where you reach 1643 K, and then modify the function expression to include a conditional statement that if you are below that depth the temperature should increase by 0.5 K/km. A rough example is below.

subsection Initial temperature model
      set Model name = function
      subsection Function
        set Variable names = x,y
        set Function constants = ymax=670e3, vsub=9.512e-10, \
                                               Tm=1643, Ts=273, kappa=1e-6, y_Tm = 570e3
        set Function expression = if (y> y_Tm, \
                                                     Ts + (Tm-Ts)*(1-erfc((ymax-y)/(2*sqrt(kappa*(x/vsub)))), \
                                                     Tm + (y_Tm - y)*0.5/1.e3)
                                               
      end
end
1 Like
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Thank you for answering my question. I also have this idea, but when determining how deep 1643K can reach, my model is at a distance of 1670km from the mid-ocean ridge. When working backward from the half-space cooling model to y, the obtained y value is infinitely large, and it is impossible to determine how deep 1643K can be reached. What’s your opinion on this? Thank you very much!