Log-likelihood of Normal Distribution: Why the term $\frac{n}{2}\log(2\pi \sigma^2)$ is not considered in the minimization of SSE?

Question

Reading some books and papers like the great one: ''Bundle Adjustment - A Modern Synthesis'' (page 10), I found that the cost function weigthed Sum of Squared Error (SSE):

$SSE = \frac{1}{2} \sum_i \Delta z_i(x)^T\,W_i\,\Delta z_i(x)$ $\,\,\,\,\,\,\,\,\,$(respecting the notation from the article linked above)

Represents as well the negative log-likelihood of the Normal Distribution from where the ground-truth data was obtained (considering that $W_i$ aproximates the inverse of the covariance matrix). Thereby, minimizing $SSE$, we will obtaine the parameters $x$ that best fit this Normal Distribution.

However, looking at some posts like this one form Wikipedia, they state that the log-likelihood for the Normal Distribution is given by:

$\log(\mathcal{L}(\mu,\sigma))= -\frac{n}{2}\,\log(2\pi\sigma^2)-\frac{1}{2\sigma^2}\sum_{i=1}^n(x_i-\mu)^2$

So, Why the term $\frac{n}{2}\,\log(2\pi\sigma^2)$ is not considered in the previous reasoning of minimizing $SSE$ = maximizing the likelihood?

Thanks in advance!

Answer 1

Because that part of the log likelihood is constant (with respect to $\mu$). Leaving it out saves some computation, but does not affect the ML estimate.

If you are also estimating $\sigma$ then you would need to include that part as well.

Answer 2

I want to add that this gist of code implements NLL https://gist.github.com/sergeyprokudin/4a50bf9b75e0559c1fcd2cae860b879e