Nth prime number

Created an array to hold all the discovered primes. For a given a number, check if it already is in the list or not. If yes, it is a prime. If not, check if any of the primes in the list, less than square root of the number, is a factor of this number or not.
             The solution proposed by the Project Euler authors improvised on this. They used an additional fact that any prime greater than 3 is of the form 6k+/-1.
              If you know the upper bound of the prime number, then the Sieve of Eratosthenes gives the answer much quickly.

import java.util.ArrayList;
import java.util.List;

public class NthPrimeNumber {
 public static void main(String[] args) {
  int index = 5;

  if (index == 1) {
   System.out.println(2L);
   return;
  }

  // An array of all the discovered primes
  List<Long> primes = new ArrayList<Long>();
  primes.add(2L);

  Long num = 3L;

  // Search for primes by dividing num with all the known prime numbers
  while (primes.size() != index) {
   boolean isComposite = false;
   double sqrt = Math.sqrt(num);
   for (Long p : primes) {
    if (p > sqrt)
     break;
    if (num % p == 0) {
     isComposite = true;
     break;
    }
   }
   if (!isComposite)
    primes.add(num);
   num++;
  }
  System.out.println(primes.get(index - 1));
 }
}