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I was reading Peter Mann's Lagrangian & Hamiltonian Dynamics, and I found this equation (page 115):

$$p_i := \frac{\partial L}{\partial \dot{q}^i}$$

where L is the Lagrangian. I understand this is the definition of conjugate momentum, but I wanted to know if there is a particular reason for the momentum index to be a lower index and the coordinate index to be an upper index. Is it simply the author's preference or there is a deeper reason?

Qmechanic
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math-ingenue
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2 Answers2

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  1. The index of generalized coordinates $q^1, \ldots, q^n$, is conventionally$^1$ a superscript/upper index in physics.

  2. The Lagrangian momenta $p_i:=\frac{\partial L}{\partial \dot{q}^i}$ have a subscript/lower index since they transform under general coordinate transformations $q^i\to q^{\prime j}=f^j(q,t)$ as components of a co-vector/1-form $p=p_i\mathrm{d}q^i$, cf. e.g. this Phys.SE post.

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$^1$ Be aware that many authors don't bother to make such notational distinctions.

Qmechanic
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Usually, you would write your Lagrangian in some sort of form like:

$$L = {\dot q}^{i}{\dot q}_{i} = g^{ij}{\dot q}_{i}{\dot q}_{j}$$, because the lagrangian itself is a scalar. Then, if you took a variation with respect to the "downed" version, you'd be left with

$$\frac{\delta L}{\delta {\dot q}_i} = 2 g^{ij}{\dot q}_{j} = 2 {\dot q}^{i}$$

So, variation of a scalar with respect to a "downed" index leaves an "upped" index.

Zo the Relativist
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  • Sorry if this is a stupid question, but why are we bringing the metric tensor into the Lagrangian? – math-ingenue Mar 09 '21 at 22:00
  • @math-ingenue, it's just a shorthand for whatever inner product we're using to span the $q_{i}$'s. It's also how we map from upper indices to lower indices. It is often just the identity matrix, which makes a lot of this formalism seem absurd and overdone. – Zo the Relativist Mar 09 '21 at 22:03