There are equations named recursion relations. We supposed to draw a diagram that shows renormalization group theory trajectories with recursion relations.

Fox exp: Recursion Relation:

$$ (x')^4 = x^2 + y^{-1}, y' = x^{-1} + y^2. (\text{not real one, just for exp}) \\ x = \mathrm e^J, y = \mathrm e^H $$

$ x' $ and $ y' $ are decimated $ x $ and $ y $, not derivatives.

I should show dots: fixed point, sinks, etc. Especially I don't know how to show points and make all flows attract to a point or run away from them.

enter image description here

image source

Can I plot this with Mathematica, can you suggest anything for me?


StreamPlot[{ very long recursion relation for J'(J,H) , very long recursion relation for H'(J,H) }, {J,0,5}, {H,-5,5}] 

I'm very new with Mathematica (~1 days) so I just wrote only this one. This code gives me some trajectories. I wonder can I lead those trajectories to a point that I want, like in this diagram.

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    $\begingroup$ Could you clarify what are the quantities k1, k2 you want to plot in terms of x, y, x', y' ? $\endgroup$ Commented Jun 14, 2020 at 18:20
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    $\begingroup$ A short answer to your question is YES. For more help you need to provide some Mathematica code for us to start with. $\endgroup$
    – yarchik
    Commented Jun 14, 2020 at 19:09
  • $\begingroup$ Thanks for the answers. Sorry, I forgot about K's. x = e^J -> J and K same -> x = e^K. J's (or K's) are coupling energies between two nearest neighbors. (Also: x'=e^J') Instead of K's, we can think of the diagram as the J-H plane. (H is an external field) The points in the diagram will change but it doesn't matter, I just put it as an example. And I edited my question. $\endgroup$
    – nara
    Commented Jun 15, 2020 at 6:54
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    $\begingroup$ Some minimal example would be nice to have. What is very long recursion relation? $\endgroup$
    – yarchik
    Commented Jun 15, 2020 at 9:15
  • $\begingroup$ In terms of J: J'=log{[(e^J+H)+(e^J-H)]/(e^H+e^-H)^2}, H'=log{(e^J+H)/(e^-J-H)} something like that. $\endgroup$
    – nara
    Commented Jun 15, 2020 at 9:24

1 Answer 1


It could be a bit late but maybe could be useful for other people.

So if you have a recursion relation similar to:

$k' = 2k (\frac{e^{4k}+1}{e^{4k}+3})^2 \\ b' = 3b \frac{e^{4k}+1}{e^{4k}+3}$

The ' is referred to the next step k, not to a derivative. To plot the RG flow you can use the following code:

rg flow recursion

Instead if you have a RG flow defined using differential equation similar to:

$ \frac{dk}{dl} = \pi^3 a^4 y^2 \\ \frac{dy}{dl} = (2- \frac{\pi}{k})y$

To plot the RG flow in this case you can use a code similar to:

enter image description here

You can also make the graph fancier adding the critical point and the other stuff!

  • 1
    $\begingroup$ Welcome to Mathematica StackExchange! Instead of posting screenshots of the code, it is more preferable to copy&paste the code directly into your answer. This way others can reuse the code without rewriting it again :) $\endgroup$
    – Domen
    Commented Jul 19, 2023 at 14:35

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