# Is 4-momentum a vector or a 1-form?

This is a follow-on to https://physics.stackexchange.com/a/576885/117014. If we should not consider a vector and its "canonically" dual 1-form to represent the same object, then it seems that we should be able to say whether 4-momentum, for example, is a vector or a 1-form. I could ask the same thing about the electromagnetic field tensor, but momentum seems to be a good place to start. I tend to think of force as a 1-form because in pure classical mechanics it is the negative of the gradient of a potential (Ask Susskind.) Since force is also the time derivative of momentum, it seems reasonable to consider momentum to be a 1-form. But, we can also think of momentum as the product of mass with 4-velocity, which is a 4-vector.

So if a vector and its dual 1-form are not to be viewed as different representations of the same object, momentum must be one or the other. Either that, or we have two distinct geometric objects representing momentum.

Which is it?

Note: A $$k$$-form is a smooth, totally-antisymmetric $$(0,k)$$-tensor field. A one-form is therefore a covector field, but the momentum of a particle is not a field, so this answer has been revised to use the correct terminology.

To me, the most natural definition of momentum is via the Lagrangian formalism, which yields the covector $$p_\mu = \frac{\partial L}{\partial \dot x^\mu}$$. Taking the standard Lagrangian

$$L(x, \dot x) = m\sqrt{g_{\mu\nu} \dot x^\mu \dot x^\nu}$$ (where differentiation is taken with respect to the proper time), it then follows that $$p_\mu = g_{\mu\nu} m\dot x^\nu$$. That being said, this is clearly the brother of the 4-vector $$\tilde{p}^\mu = m \dot x^\mu$$, with the index raised/lowered via the metric.

From the Lagrangian standpoint, if we add a potential energy term then the Lagrangian equations of motion take the form

$$\frac{d}{d\tau} p_\mu = -\frac{\partial U}{\partial x^\mu} \equiv f_\mu$$ so as you say, from this perspective force is naturally a covector. But again, the metric provides us with an isomorphism, so solving

$$\frac{d}{dt}p_\mu = f_\mu$$ and $$\frac{d}{d\tau} \tilde{p}^\mu =g^{\mu\nu} f_\nu \equiv \tilde{f}^\mu$$

are ultimately equivalent.

If we consider the mythical classical point mass, it has a 4-momentum determined by its mass and its world-line. I call such a thing a "priority" object. It exists prior to any manifold parameterization, or metric (or observation).

Okay, that's fine. You're talking about $$p^\mu = m \dot x^\mu$$. This expression is perfectly well-defined with no additional structure needed.

Whether we express it covariantly or contravariantly, the expression refers to the same physical entity.

Without a metric (or some other structure which provides a similar isomorphism), you can't "express it covariantly." The momentum you referred to before is well-defined on its own, but you can't map it to a covector without implicitly making a choice of metric (or other index-lowering map).

I typically write momentum covariantly. But I don't have an ontological argument to consider that to be an inherent property of momentum.

For that, you'll need to be more specific about what you mean by momentum. If you're talking about the mass times the 4-velocity, that is a 4-vector. If you're talking about the canonical momentum which is conjugate to position in the Lagrangian or Hamiltonian pictures, and whose spatial components (i) act as the infinitesimal generators of spatial translations, and (ii) are conserved in the presence of spatial translation symmetry, then that object is a covector.

As a concrete example, consider the flat space FLRW spacetime in which

$$ds^2= c^2dt^2 - a^2(t)\big(dx^2+dy^2+dz^2\big)$$

This metric is homogeneous and isotropic, which implies 3-momentum conservation. However, it is not $$p^k = m \dot x^k,\ k=1,2,3$$ which is conserved, but rather $$p_k = -a^2(t)m\dot x^k$$.

• @m4r35n357 I'm saying that the canonical momentum of Lagrangian/Hamiltonian mechanics is a one-form, and that using the metric to raise its index gives you its 4-vector partner. Both are commonly referred to as momentum, and the placement of the index resolves any potential ambiguity. Commented Sep 9, 2020 at 20:13
• @StevenThomasHatton I have edited my answer to address your comment. Commented Sep 12, 2020 at 16:56
• As a note here, this business of "you can't link the vector and the one-form without a metric" business seems vague, but it becomes critically important if you're doing the intrinsic geometry on a null submanifold of a four space. In this case, the surface has a null tangent $\ell^{a}$ and a null normal vector $k^{a}$, and it will work out that the covariant three space is spanned by the two metric plus $\ell^{a}$, but the correct oneform to use to span the covariant space is $k_{a}$ plus the spanning oneforms on the two-space. Commented Sep 12, 2020 at 18:42
• @StevenThomasHatton I don't think I can be more clear with what I meant. If by "momentum" you mean "mass times 4-velocity", then that is a 4-vector. The momentum which is conserved via Noether's theorem is a covector. If your position is that only the former should be called momentum, then you have your answer, but you should be aware that the physics community at large does not typically abide by that restriction. Commented Sep 13, 2020 at 15:18
• @StevenThomasHatton I would say that the vector and the 1-form are different geometric objects, but (i) they are natural partners via the mapping induced by the metric, and (ii) to the extent that they differ, it is the latter rather than the former which carries the significance we attribute to momentum. In other words, the significance of $m\dot x^\mu$ is due to its partnership with $p_\mu$. Commented Sep 22, 2020 at 21:37