Split the difference

Suppose that the free energy of a system is described by F = ( T T c ) m 2 + 1 4 m 4 \mathcal{F}=(T-T_c)m^2+\frac14 m^4 where T T is the temperature, and m m represents some property of the system, such as its overall magnetization. As T T rises and falls, but does not cross T c T_c , the allowed physical value(s) of m m , m m^* change somewhat (see Assumptions).

When T \displaystyle T crosses T c T_c , however, there is a qualitative shift in the allowed values of m m^* , and the system undergoes a phase transition.

Suppose we're working with the system at a temperature T \displaystyle T below T c T_c such that T T c = 10 T-T_c=-10 , find m \lvert m^*\rvert .

Assumptions

  • The system always resides in states m m^* which minimize its free energy. These are the "physical" states of the system.


The answer is 4.47214.

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1 solution

Ayush Verma
Oct 27, 2014

F = m 4 4 10 m 2 , f o r c r i t i c a l v a l u e , d F d m = 0 m 3 20 m = 0 m = 4.472135955 ( a s m > 0 ) F=\cfrac { { m }^{ 4 } }{ 4 } -10{ m }^{ 2 },\\ \\ for\quad critical\quad value,\\ \\ \cfrac { dF }{ dm } =0\Rightarrow { m }^{ 3 }-20m=0\\ \\ \Rightarrow m=4.472135955 (as\quad m>0)

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