# Ex 1.28 Adding a total time derivative

Structure and Interpretation of Classical Mechanics

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An analogous result holds when the $f_\alpha$‘s depend explicitly on time.

b. Show that, by adding a total time derivative, the Lagrangian can be written in the form $L = A - B$, where $A$ is a homogeneous quadratic form in the generalized velocities and $B$ is independent of velocity.

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$L' = L + D_t F$

The addition of a total time derivative to a Lagrangian does not affect whether the action is critical for a given path.

Moreover, the additional terms introduced into the action by the total time derivative appear only in the endpoint condition and thus do not affect the Lagrange equations derived from the variation of the action, …

— 1.6.4 The Lagrangian Is Not Unique

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\begin{aligned} L &= T(t, q, v) - V(t, q) \\ &= \frac{1}{2} \sum_\alpha m_\alpha |\partial_0 f_\alpha (t,q) + \partial_1 f_\alpha (t,q) v|^2 - V(t, q) \\ \end{aligned}

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\begin{aligned} &L - \frac{1}{2} \sum_\alpha m_\alpha \left \{ [\partial_0 f_\alpha (t,q)]^2 + 2 \partial_0 f_\alpha (t,q) \partial_1 f_\alpha (t,q) v \right \} \\ &= \frac{1}{2} \sum_\alpha m_\alpha [\partial_1 f_\alpha (t,q) v]^2 - V(t, q) \\ \\ &L - \frac{1}{2} \sum_\alpha m_\alpha \left \{ g_\alpha(t,q) +[\partial_0 f_\alpha (t,q)]^2 + 2 \partial_0 f_\alpha (t,q) \partial_1 f_\alpha (t,q) v \right \} \\ &= \frac{1}{2} \sum_\alpha m_\alpha [\partial_1 f_\alpha (t,q) v]^2 - V(t, q) - \frac{1}{2} \sum_\alpha m_\alpha g_\alpha(t,q)\\ \end{aligned}

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\begin{aligned} D_t F &= - \frac{1}{2} \sum_\alpha m_\alpha \left \{ g_\alpha(t,q) +[\partial_0 f_\alpha (t,q)]^2 + 2 \partial_0 f_\alpha (t,q) \partial_1 f_\alpha (t,q) v \right \} \\ &= - \frac{1}{2} \sum_\alpha m_\alpha \left[ G_{\alpha 0} (t,q,v) + G_{\alpha 1}(t,q,v)v \right] \\ \\ G_{\alpha 0} &= g_\alpha(t,q) +[\partial_0 f_\alpha (t,q)]^2 \\ G_{\alpha 1} &= 2 \partial_0 f_\alpha (t,q) \partial_1 f_\alpha (t,q) \\ \\ \end{aligned}

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\begin{aligned} D_t F &= G_{0} (t,q,v) + G_{1}(t,q,v) v \\ \\ \end{aligned},

where

\begin{aligned} G_0 &= - \frac{1}{2} \sum_\alpha m_\alpha \left[ G_{\alpha 0} (t,q,v) \right] \\ G_1 &= - \frac{1}{2} \sum_\alpha m_\alpha \left[ G_{\alpha 1}(t,q,v) v \right] \\ \\ \end{aligned}

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Since $D_t F$ is a total time derivative,

\begin{aligned} \partial_1 G_0 &= \partial_0 G_1 \\ \\ \end{aligned}

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Then, by choosing a suitable function $g_\alpha(t,q)$, we can access a simple case that

\begin{aligned} \partial_1 G_{\alpha 0} &= \partial_0 G_{\alpha 1} \\ \\ \end{aligned}

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\begin{aligned} G_{\alpha 0} &= g_\alpha(t,q) +[\partial_0 f_\alpha (t,q)]^2 \\ G_{\alpha 1} &= 2 \partial_0 f_\alpha (t,q) \partial_1 f_\alpha (t,q) \\ \\ \partial_1 G_{\alpha 0} &= \partial_1 g_\alpha(t,q) + 2 [\partial_0 f_\alpha (t,q)] \partial_1 \partial_0 f_\alpha (t,q) \\ \partial_0 G_{\alpha 1} &= 2 \partial_0 \partial_0 f_\alpha (t,q) \partial_1 f_\alpha (t,q) + 2 \partial_0 f_\alpha (t,q) \partial_0 \partial_1 f_\alpha (t,q) \\ \\ \end{aligned}

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\begin{aligned} &\partial_1 g_\alpha(t,q) + 2 [\partial_0 f_\alpha (t,q)] \partial_1 \partial_0 f_\alpha (t,q) \\ &= 2 \partial_0 \partial_0 f_\alpha (t,q) \partial_1 f_\alpha (t,q) + 2 \partial_0 f_\alpha (t,q) \partial_0 \partial_1 f_\alpha (t,q) \\ \\ &\partial_1 g_\alpha(t,q) \\ &= 2 \partial_0 \partial_0 f_\alpha (t,q) \partial_1 f_\alpha (t,q) \\ \\ \end{aligned}

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\begin{aligned} L + D_t F &= A - B \\ \\ D_t F &= - \frac{1}{2} \sum_\alpha m_\alpha \left \{ g_\alpha(t,q) + [\partial_0 f_\alpha (t,q)]^2 + 2 \partial_0 f_\alpha (t,q) \partial_1 f_\alpha (t,q) v \right \} \\ \\ A &= \frac{1}{2} \sum_\alpha m_\alpha [\partial_1 f_\alpha (t,q) ]v^2 \\ - B &= - V(t, q) - \frac{1}{2} \sum_\alpha m_\alpha g_\alpha(t,q) \\ \end{aligned}

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This answer is not totally correct, since the generalized velocity, $v$, should be a vector.

— Me@2022-11-01 08:58:52 AM

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