Restriction of a Lagrangian

Question

I'm wondering if anyone could help me with the following questions.

Let $M$ be the Minkowski spacetime, given $f\in C^{\infty}(M) ; f(m)=x^{0}(m)$, with $\{x^{\mu}\}$ being a global Cartesian coordinates system, given the 3-dimensional submanifold $M\supset F_{t}=f^{-1}(t)$ relative to a regular value $t\in\mathbb{R}$ of $f$, and given the Lagrangian: $$ \mathcal{L}\in C^{\infty}(TM) $$ $$ \mathcal{L}(x^{\mu},\dot{x}^{\mu})=-\sqrt{\eta_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}} $$ where $\eta$ is the Minkowski metric and $\{x^{\mu}\}$ a global Cartesian coordinates system; what is the coordinates espression of the Lagrangian on $F_{t}$: $$ (T\iota_{t})^{*}\mathcal{L}\in C^{\infty}(F_{t}) $$

I "know" from other "sources" that I should find: $$ (T\iota_{t})^{*}\mathcal{L}(x^{i},\dot{x}^{i})=\mathcal{L}\circ T\iota_{t}(x^{i},\dot{x}^{i})=-\sqrt{1 - \delta_{ij}\dot{x}^{i}\dot{x}^{j}} $$ Is it totally wrong?

Answer 1

Your suspicions are correct: It is wrong! At least as it is written presently.

Let us start form the embedding of manifold $$\imath_t : F_t \ni p \mapsto p \in M\:.$$ It induces and embedding of corresponding tangent bundles: $$T\imath_t : TF_t \ni (p,v) \mapsto (p, d\imath_t (v)) \in TM$$ The latter can only preserve the vectors tangent to $F_t$ seen as embedded submanifold in $M$. It cannot say anything about components non-tangent to $F_t \subset M$.

When one fixes a coordinate system $x^0,x^1,x^2,x^3$ adapted to $F_t$, i.e $F_t$ coincides with the set of points with $x^0=0$ (your $x^0$ is my $t+x^0$), then he/she also fixes a similar coordinate system referring to $TF_t$ and $TM$, passing in the naturally associated charts with coordinates, respectively, $x^0,x^1,x^2,x^3,\dot{x}^0,\dot{x}^1,\dot{x}^2,\dot{x}^3$ in $TM$ and $x^1,x^2,x^3,\dot{x}^1,\dot{x}^2,\dot{x}^3$ in $TF_t$.

With our choice of coordinates the bases turn out to be identical and thus $T\imath_t$ preserves the $3$ components $\dot{x}^i$. In other words, as said above, any vector transported from $F_t$ to $M$ keeps remaining tangent to $F_t$ seen as submanifold of $M$:

$$T\imath_t : TF_t \ni (x^1,x^2,x^3,\dot{x}^1,\dot{x}^2,\dot{x}^3) \mapsto (0, x^1,x^2,x^3,0,\dot{x}^1,\dot{x}^2,\dot{x}^3) \in TM$$

Therefore $$ (T\iota_{t})^{*}\mathcal{L}(x^{i},\dot{x}^{i})=\mathcal{L}\circ T\iota_{t}(x^{i},\dot{x}^{i})=-\sqrt{0 - \delta_{ij}\dot{x}^{i}\dot{x}^{j}} $$

which makes sense if you are allowed to consider complex values. Otherwise you should define the Lagrangian including an absolute value (the point is that as it stands the initial, unrestricted, $\mathcal{L}$ is not defined on $TM$, but only on the subset of causal elements $(p,v) \in T_pM$ with $v$ causal).

If you want to obtain the expression $\sqrt{1 - \delta_{ij}\dot{x}^{i}\dot{x}^{j}}$, you should fix the temporal component of vectors making use of a jet bundle over $x^0$ for instance... (However to be completely honest, all that seems to my like killing a fly with a gun.)

ADDENDUM: As I wrote in a comment now erased, every differentiable coordinate function like $x^0$ in a coordinate patch on a manifold is such that all their values are always regular. (In fact $dx^0|_p$ has to be an element of a basis $T_p^*M$ and thus it cannot vanish.) So it is not necessary to assume it separately, as you did in your question.

Answer 2

According to the regular value theorem we can find a chart on $M$ such that the immersion $\iota_t: F_t \rightarrow M$ takes the form $\iota_t (x^i) = (x^0_t, x^i)$. Note that $x^0$ is a fixed constant and only depend on the value of $t$. Thus, differentiation with respect to the first variable is the identity on $TM$ (and the wrt other coordinates results in $\dot x^i$) from which the wanted equation follows.