In relativistic mechanics with Lagranian $L(\dot q^i,q^i)$ of a particle, the conjugate momentum of the position coordinate $q^i$ is defined as (wiki)

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

As far as I understand, the position coordinate $q^i$ is considered to be contravariant and the conjugate momentum $p_i$ is covariant (here I am already not sure).

The position coordinate of the particle that we measure in nature is certainly corresponding to $q^i$. On the other hand, when we measure the momentum of the particle, do we measure the covariant component $p_i$ or the contravariant $p^i$ component of the momentum?

The signature of the metric is given by $(-,+,+,+)$ and $i=1,2,3$ are the indices corresponding to the spatial momenta $p_i$.

  • 3
    $\begingroup$ It depends on the details of the measurement and metric conventions. $\endgroup$
    – Qmechanic
    Mar 1, 2020 at 9:46
  • $\begingroup$ I am not struggling about sign conventions of $p_i$. But I think, we have to decide which one $p^i$ or $p_i$ is measured. $\endgroup$
    – user56224
    Mar 1, 2020 at 16:19
  • $\begingroup$ Concerning the "details of the measurement".. describe your measuring apparatus tensorially. Measurements of components are inner products. If your apparatus is described by a vector $v^i$, you are measuring $p_i$. $\endgroup$
    – robphy
    Mar 1, 2020 at 17:59
  • $\begingroup$ @robphy why do you think that measurements of components are inner products? $\endgroup$
    – user56224
    Mar 2, 2020 at 19:31

3 Answers 3


Them momentum vector that you wrote there $ p_i=\frac{\partial L}{\partial \dot q^i} $, which is a conserved quantity (If the theory respects translational symmetry), is associated with an Euler-Lagrange equation which is written for $ x_{\mu} $, say

$ \frac{\partial L}{\partial X^{\mu}}-\frac{d(\frac{\partial L}{\partial \dot{X^{\mu}}})}{dt} $.

Which in this case is a covariant vector. But you can write a similar equation for $ x^{\mu} $:

$ \frac{\partial L}{\partial X_{\mu}}-\frac{d(\frac{\partial L}{\partial \dot{X_{\mu}}})}{dt} $.

In this case, the conserved quantity is a contravariant vector. Now, these two equations are identical. Meaning that they contain similar information and they can be translated into each other via the metric tensor ($ g^{\mu\nu} $ or $ g_{\mu\nu} $). Now, when we are measuring a quantity (doesn't matter what), we are just measuring its components. So whether it is a covariant or contravariant vector (or tensor) is our own convention! Meaning that if we choose our coordinate vectors to be contravariant, then we would have a covariant metric tensor and a contravariant momentum. However, if we choose our coordinates to be covariant vectors, then the metric tensor is contravariant and the momentum vector is covariant! Since as a convention, the coordinates are contravariant vectors, then the measured momentum would also be a contravariant one!


I would like to make my own proposal for addressing my measurement question. Let $g_{\mu\nu}$ be a (pseudo) Riemannian metric of a manifold $M$ with signature $(-,+,+,+)$ and the contravariant coordiantes of the associated wordlines are given by $x^\mu(\lambda)$. Then the mass-shell condition of the 4-momenta is given by

$$p^\mu p_\mu=-(mc)^2.$$

Without loss of generality let the metric be diagonal. After some algebraic manipulations we can write (the positive root)

$$H\equiv p_0=\Big[ \frac{(mc)^2+g^{ij} p_i p_j}{-g^{00}} \Big]^{1/2}$$

where the summation of the momenta are for $i,j=1,2,3$. The metric components $g^{ij}$ on the right hand side are the inverse of $g_{ij}$. As far as I have seen in literature (in all applications), the prefactors in front of the momenta on the right hand side are always given by $g^{ij}$ but not by $g_{ij}$. Therefore it becomes obvious that the momenta $p_i$ are those which are associated with the experimental measurements.

Actually, such kind of distinction can not be done in ordinary flat Minkowski space with Euclidiean coordinates because there both $g^{ij}$ and $g_{ij}$ are the same. Here, one had to consider spherical coordinates at least.

Summarized, this result is correponding to the comments regarding my post that whenever the worldline is expressed in contravariant (covariant) coordinates then the assiociated momenta have to be of opposite type.


In ordinary geometry, one can regard $\hat x\cdot \_= \hat x^a g_{ab}$ as operator which picks out the $x$-component of a vector, as in $V_x=\hat x\cdot \vec V= \hat x^a g_{ab} V^b$. A similar situation happens in (say) phase space.

Quoting from Geroch's "Geometrical Quantum Mechanics":

  1. Observables; Poisson Bracket (p. 49...)

    By an observable of a system, we mean a scalar field on its phase space. Roughly speaking, observables are the things instruments measure. We think of an observable as a box having a dial and a little probe which sticks into the system. ...

    ...Two particular types of observables are of particular interest. Let $\alpha$ be a scalar field on configuration space. Then $\alpha \circ \pi$ is a scalar field on phase space (which ignores momentum, and looks only at configuration). Any observable of this form, $\alpha \circ \pi$, will be called a configuration observable. In elementary treatments of particle mechanics, "x", "y", and "z" are configuration observables. Now consider a contravariant vector field $\xi^a$ on configuration space. We introduce the following scalar field on phase space: it assigns, to the point $(q, p_a)$ of phase space, the number obtained by contracting $\xi^a$ at $q$ (a contravariant vector at $q$) with $p_a$ (a covariant vector at $q$). We write this scalar field $\xi^a p_a$. An observable expressible in this form will be called a momentum observable. In elementary particle mechanics, "$p_x$", "$p_y$", and "$p_z$" are examples of momentum observables (i.e., those defined by unit vectors pointing in the $x$, $y$, and $z$ directions).

From the quoted passage, the $p_a$ is a covector.


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