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Freeman
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According to the first law of thermodynamics

\begin{align}U=TS+YX+\sum_j\mu_jN_j.\end{align}

Where $Y$ is the generalized force, $dX$ is the generalized displacement.

Helmholtz Free Energy

\begin{align}F=U-TS=YX+\sum_j\mu_jN_j. \end{align}

Gibbs Free Energy

\begin{align}G=U-TS-YX=\sum_j\mu_jN_j\end{align}\begin{align}G=U-TS-YX=\sum_j\mu_jN_j.\end{align}

Therefore that

\begin{align}G=F-YX \end{align}\begin{align}G=F-YX.\end{align}

In your case, $Y=H$, $X=M$, so we get

\begin{align}G=F-HM.\end{align}

You can see the textbook:

A Modern Course in Statistical Physics by L. E. Reichl, 2nd, ed (1997), p23, 42, 45.

According to the first law of thermodynamics

\begin{align}U=TS+YX+\sum_j\mu_jN_j.\end{align}

Where $Y$ is the generalized force, $dX$ is the generalized displacement.

Helmholtz Free Energy

\begin{align}F=U-TS=YX+\sum_j\mu_jN_j. \end{align}

Gibbs Free Energy

\begin{align}G=U-TS-YX=\sum_j\mu_jN_j\end{align}

Therefore that

\begin{align}G=F-YX \end{align}

In your case, $Y=H$, $X=M$, so we get

\begin{align}G=F-HM.\end{align}

You can see the textbook:

A Modern Course in Statistical Physics by L. E. Reichl, 2nd, ed (1997), p23, 42, 45.

According to the first law of thermodynamics

\begin{align}U=TS+YX+\sum_j\mu_jN_j.\end{align}

Where $Y$ is the generalized force, $dX$ is the generalized displacement.

Helmholtz Free Energy

\begin{align}F=U-TS=YX+\sum_j\mu_jN_j. \end{align}

Gibbs Free Energy

\begin{align}G=U-TS-YX=\sum_j\mu_jN_j.\end{align}

Therefore that

\begin{align}G=F-YX.\end{align}

In your case, $Y=H$, $X=M$, so we get

\begin{align}G=F-HM.\end{align}

You can see the textbook:

A Modern Course in Statistical Physics by L. E. Reichl, 2nd, ed (1997), p23, 42, 45.

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Freeman
  • 226
  • 2
  • 6

According to the first law of thermodynamics

\begin{align}U=TS+YX+\sum_j\mu_jN_j.\end{align}

Where $Y$ is the generalized force, $dX$ is the generalized displacement.

Helmholtz Free Energy

\begin{align}F=U-TS=YX+\sum_j\mu_jN_j. \end{align}

Gibbs Free Energy

\begin{align}G=U-TS-YX=\sum_j\mu_jN_j\end{align}

Therefore that

\begin{align}G=F-YX \end{align}

In your case, $Y=H$, $X=M$, so we get

\begin{align}G=F-YX.\end{align}\begin{align}G=F-HM.\end{align}

You can see the textbook:   

A Modern Course in Statistical Physics by L. E. Reichl, 2nd, ed (1997), p23, 42, 45.

According to the first law of thermodynamics

\begin{align}U=TS+YX+\sum_j\mu_jN_j.\end{align}

Where $Y$ is the generalized force, $dX$ is the generalized displacement.

Helmholtz Free Energy

\begin{align}F=U-TS=YX+\sum_j\mu_jN_j. \end{align}

Gibbs Free Energy

\begin{align}G=U-TS-YX=\sum_j\mu_jN_j\end{align}

Therefore that

\begin{align}G=F-YX \end{align}

In your case, $Y=H$, $X=M$, so we get

\begin{align}G=F-YX.\end{align}

You can see the textbook:  A Modern Course in Statistical Physics by L. E. Reichl 2nd, ed (1997), p23, 42, 45.

According to the first law of thermodynamics

\begin{align}U=TS+YX+\sum_j\mu_jN_j.\end{align}

Where $Y$ is the generalized force, $dX$ is the generalized displacement.

Helmholtz Free Energy

\begin{align}F=U-TS=YX+\sum_j\mu_jN_j. \end{align}

Gibbs Free Energy

\begin{align}G=U-TS-YX=\sum_j\mu_jN_j\end{align}

Therefore that

\begin{align}G=F-YX \end{align}

In your case, $Y=H$, $X=M$, so we get

\begin{align}G=F-HM.\end{align}

You can see the textbook: 

A Modern Course in Statistical Physics by L. E. Reichl, 2nd, ed (1997), p23, 42, 45.

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Freeman
  • 226
  • 2
  • 6

According to the first law of thermodynamics

\begin{align}U=TS+YX+\sum_j\mu_jN_j.\end{align}

Where $Y$ is the generalized force, $dX$ is the generalized displacement.

Helmholtz Free Energy

\begin{align}F=U-TS=YX+\sum_j\mu_jN_j. \end{align}

Gibbs Free Energy

\begin{align}G=U-TS-YX=\sum_j\mu_jN_j\end{align}

Therefore that

\begin{align}G=F-YX \end{align}

In your case, $Y=H$, $X=M$, so we get

\begin{align}G=F-YX.\end{align}

You can see the textbook: A Modern Course in Statistical Physics by L. E. Reichl 2nd, ed (1997), p23, 42, 45.