Since no particle acts on itself (i.e., F jj = 0 for 1 ≤ j ≤ N), we split the sum in terms below and above this diagonal and we add them together in pairs: ∑ ∑ k = 1 N F k ⋅ ⋅ r k = ∑ ∑ k = 1 N ∑ ∑ j = 1 N F j k ⋅ ⋅ r k = ∑ ∑ k = 2 N ∑ ∑ j = 1 k − − 1 ( F j k ⋅ ⋅ r k + F k j ⋅ ⋅ r j ) = ∑ ∑ k = 2 N ∑ ∑ j = 1 k − − 1 ( F j k ⋅ ⋅ r k − − F j k ⋅ ⋅ r j ) = ∑ ∑ k = 2 N ∑ ∑ j = 1 k − − 1 F j k ⋅ ⋅ ( r k − − r j ) where we have assumed that Newton's third law of motion holds, i.e., F jk = − F kj (equal and opposite reaction).