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I am reading the Colding and Miniczzi book "A course in Minimal Surfaces". In chapter 1, section 8 there is a derivation of the second variation formula, and the stability operator $L$, but just for the case of normal variations.

A normal vector fields $J$ on a minimal surface $\Sigma$ is said to be a Jacobi field if $LJ = 0$.

Then in the book they consider the case when $\Sigma \subset \mathbb{R}^3$ is the graph of a function satisfying the minimal surface equation. Then they consider the variation $F: \Sigma \times \mathbb{R} \to \mathbb{R}^3$ defined as: $$ F \colon (x_1, x_2, x_3, t) \mapsto (x_1, x_2, x_3 + t). $$ The variation field on the surface is $(0, 0, 1)$ and clearly $$ t \mapsto Vol\big(F(\Sigma, t) \big) $$ is constant.

Then in the book they conclude that the normal component of the variation field of this variation is a Jacobi Field.

I don't understand why this should be true. I mean, how can we assert that the normal component of the variation $F$ does not change the volume? In principle this could be false.

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If you're only interested in geometry (and not parametrization), then the normal component of a variation field is the only part that matters - any field can be written as the sum of a tangential part and a normal part, and variations of a map tangential to its image do not change its image.

Thus, the variation generated by the normal component of $(0,0,1)$ will have image coinciding with that of $F$, which clearly has constant volume.

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    Yes thanks! I actually just realized that it was a silly question. Because the first variation formula is: $$ \frac{d}{dt} Vol(F(\Sigma, t)) = - \int_\Sigma (F_t, H) $$ where F is a variation (not necessarily normal) with compact support. And since the mean curvature vector $H$ is normal, it's clear that the tangent component of the variation does not contribute. Thank you very much!! – 2017-01-19
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    There is one technical detail @Onil90, the variation $F$ (the translation) is not of compact support, so for your first variation formula, there would be a boundary term. – 2017-01-19
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    @JohnMa you are right. How can I solve this problem? – 2017-01-19
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    In fact you can just calculate $L f$, where $f = \vec n \cdot (0,0,1)$ is the orthogonal part of $(0,0,1)$ and check that to be zero. Of course it is not so geometric. @Onil90 – 2017-01-19
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    @JohnMa Yes, right, but it would be nice to avoid the computation.. :D mmmh – 2017-01-19