Channel Capacity - Heuristic proof

Quick recap of the channel capacity heuristic proof. Reference here: https://www.eecs.berkeley.edu/~dtse/Chapters_PDF/Fundamentals_Wireless_Communication_chapter5.pdf Assume we are transmitting a codeword in N dimensions. The dimension could represent time or any other signal characteristic. Assume it is corrupted with a noise vector (also N dimensions). Now each of the codewords can be represented as a point in N dimensional space. Assume also that we have a power constraint P (total power transmitted must not exceed P)- so all the received codewords must lie in an N-dimensional sphere of radius sqrt(N(P+sigma^2)). For error free decoding, all the received vectors (transmitted point + noise vector) should lie inside non-overlapping spheres inside this large sphere of radius sqrt(N(P+sigma^2)). The radius of this non-overlapping noise spheres has to be sigma^2 (noise variance) for a large N. Thus, max number of codewords that can be transmitted in this scenario can be geometrically deduced as sqrt(N(P+sigma^2))^N/sqrt(sigma^2)^N or equivalently the max number of bits that can be conveyed is log2(max_num_codewords). This yields the famous shannon expression for channel capacity. I.e no more information can be transmitted with the given power constraint on an AWGN channel with arbitrarily small error rate.