TITLE: Are there as many real-closed fields of a given cardinality as I think there are?
QUESTION [17 upvotes]: Let $\kappa$ be an infinite cardinal.  Then there exists at least one real-closed field of cardinality $\kappa$ (e.g. Lowenheim-Skolem; or, start with a function field over $\mathbb{Q}$ in $\kappa$ indeterminates, choose an ordering and a real-closure).  
But I think there are many more, namely $2^{\kappa}$ pairwise nonisomorphic real-closed fields of cardinality $\kappa$.  This is equal to the number of binary operations on a set of infinite cardinality $\kappa$, so is the largest conceivable number.
As for motivation -- what can I tell you, mathematical curiosity is a powerful thing.  One application of this which I find interesting is that there would then be $2^{2^{\aleph_0}}$ conjugacy classes of order $2$ subgroups of the automorphism group of the field $\mathbb{C}$.   
Addendum: Bonus points (so to speak) if you can give a general model-theoretic criterion for a theory to have the largest possible number of models which yields this result as a special case.

REPLY [11 votes]: For real closed fields this is fairly easy.
First show that for any infinite cardinal k there are 2^k nonisomorphic linear orders
of cardinality k
For example if X is a subset of k let A_x be Q+2+Q if x is in k and Q+3+Q if x is not in X.  Let L_X be the sum of the A_x for x in k.  It is easy to see that L_X is isomorphic
to L_Y if and only if X=Y.
If F is a real closed and x and y are infinite element of R we say that x and y are 
comparable if and only if there are natural numbers m and n such that x is less than y^m and y is less than x^n.
The ordering of R induces a linear order L_R of the comparability classes, which we call the ladder of R.
Suppose L is a linear order. Let F be the real algebraic numbers. Let R_L be the real closure of the transcendental extension of the real algebraic numbers F(x_l:l\in L) ordered such that if i is less than j then x_i^n is less than x_j for all n. It's not hard to show that the ladder of R_L is isomorphic to L.
Thus if we start with nonisomorphic orders A and B then the fields R_A and R_B will be 
nonisomorphic.