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Lagrangian is function of generalized co-ordinates, generalized velocities and time:

$$L=L(q,\dot{q},t)$$

Why the specific form $L=T-U$ is used as a definition of Lagrangian function? Here as usual $T,U$ are kinetic and potential energies respectively.

Why $L = \dfrac 1 2 m \dot{q}^2 + U(q)$ is not a Lagrangian function, even though this $L$ also is a function of $q,\dot q$?

Where did Lagrange write $L = T-U$ in his famous book "Mécanique analytique"? Or what justification in support of this definition, did he give there?

Qmechanic
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atom
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1 Answers1

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I'm not sure if Lagrange followed this derivation, but I think the timing works out. Starting from d'Alembert's principle of virtual work you can derive the lagrange equations of second kind
$$ \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial K}{\partial \dot{q}_j} - \frac{\partial K}{\partial q_j} = \sum_i \mathbf{F}_i \frac{\partial \mathbf{r}_i}{\partial q_j} $$ where $K$ is the kinetic energy. Rewriting the RHS for a konservative force $\mathbf{F}_i = -\nabla_i U(q_1, \dots, q_N)$ leads to \begin{align*} \sum_i \mathbf{F}_i \frac{\partial \mathbf{r}_i}{\partial q_j} = \sum_i -\nabla_i U \frac{\partial \mathbf{r}_i}{\partial q_j} = - \frac{\partial U}{\partial q_j} \end{align*} You can then move the term over to the LHS $$ \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial K}{\partial \dot{q}_j} - \frac{\partial}{\partial q_j}(K - U) = 0 $$ Now we also assume that the potential doesn't depend on the velocities $\dot{q}_j$ so we can include it in the first term for free since $\frac{\partial U}{\partial \dot{q}_j} = 0$ and you get $$ \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}_j} - \frac{\partial L}{\partial q_j} = 0 $$ with $L = K - U$.

Wihtedeka
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