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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?

Qmechanic
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Steven Thomas Hatton
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  • Possible duplicate: https://physics.stackexchange.com/q/176555/2451 , https://physics.stackexchange.com/q/351903/2451 and links therein. – Qmechanic Sep 09 '20 at 17:07

1 Answers1

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To me, the most natural definition of momentum is via the Lagrangian formalism, which yields the one-form $p_\mu = \frac{\partial L}{\partial \dot x^\mu}$. 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 one-form. 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 $p_k = -a^2(t)m\dot x^k$.

J. Murray
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  • Just checking; you are saying it can be either, depending on whether the index is "up" or "down"? – m4r35n357 Sep 09 '20 at 19:31
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    @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. – J. Murray Sep 09 '20 at 20:13
  • OK cheers for that. – m4r35n357 Sep 09 '20 at 20:59
  • @J.Murray 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). Whether we express it covariantly or contravariantly, the expression refers to the same physical entity. I typically write momentum covariantly. But I don't have an ontological argument to consider that to be an inherent property of momentum. The identification of the covariant with the contravariant expression for momentum seems immutable. – Steven Thomas Hatton Sep 11 '20 at 20:22
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    @StevenThomasHatton I have edited my answer to address your comment. – J. Murray Sep 12 '20 at 16:56
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    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. – Zo the Relativist Sep 12 '20 at 18:42
  • The very last expression you wrote doesn't have matching up/downstairs indices. Is it meant to read $p_k = -a^2m\dot{x}_k$? – jacob1729 Sep 12 '20 at 23:47
  • @jacob1729 No. It's not a Lorentz-covariant expression, it's just shorthand for $p_1 = -a^2(t)m\dot x^1, p_2 = -a^2(t)m \dot x^2$, and $p_3 = -a^2(t) m\dot x^3$. – J. Murray Sep 12 '20 at 23:54
  • You are talking about generalized momentum. Also know as conjugate or cannonical momentum. That isn't (in general) the same thing as physical momentum. It is defined by analogy. – Steven Thomas Hatton Sep 13 '20 at 15:09
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    @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. – J. Murray Sep 13 '20 at 15:18
  • @J.Murray If I recall correctly you are talking about the generalized momentum conjugate to a cyclic variable. That would include angular momentum, and anything else whose generalized position coordinate doesn't appear in the Lagrangian. In fact, it's really a statement about individual momentum components. That is not, in general, physical momentum. – Steven Thomas Hatton Sep 14 '20 at 01:31
  • @J.Murray Im my previous comment, I should have written: it's really a statement about individual generalized momentum coordinates. – Steven Thomas Hatton Sep 14 '20 at 02:54
  • @StevenThomasHatton I don’t understand your objection, but this is now beyond the scope of a comment thread. Essentially all of the deep properties of momentum correspond to the canonical momentum, which is what appears e.g. in the Poincare algebra. If you want more clarification on this, I’d suggest asking a separate question. – J. Murray Sep 14 '20 at 04:57
  • @J.Murray I certainly don't mean to imply that generalized momentum is useless or nu-interesting. The more I think about this question the more it makes my head spin. There are simply too many caveats and considerations for me to succinctly state my objections. – Steven Thomas Hatton Sep 15 '20 at 01:50
  • @J.Murray I accepted the answer that momentum is a vector. – Steven Thomas Hatton Sep 15 '20 at 02:04
  • See my notes under the heading "Generalized Momentum Coordinates and Hamilton’s Equations" currently on page 49 (a moving target) https://drive.google.com/drive/folders/1pbWPPeQ7k4coa0Y7gjJspl9J7Ud91JZU Note that I am effectively lowering the generalized velocity index using a weighted metric to produce a generalized momentum component. The metric is, however, that of the manifold of unconstrained coordinates in configuration space. – Steven Thomas Hatton Sep 16 '20 at 22:13
  • @StevenThomasHatton I'm not sure to which document you're referring, but I don't have anything to add to my answer and with respect, I'm not particularly interested in parsing your notes. You should ask a new question which clarifies the concerns you're having and includes whatever context you'd like from your notes. – J. Murray Sep 18 '20 at 19:07
  • @J.Murray Sorry, I just clicked 'get link' and pasted what google gave me. I meant this. https://drive.google.com/file/d/1GLf_JbRGlbGJoeLK0dJmpYiAInjJYvrp/view?usp=sharing It shows that getting momentum from the Lagrangian, and 'lowering the index' on velocity while multiplying by mass are one in the same. It's not an insight that I have seen exposed elsewhere. – Steven Thomas Hatton Sep 20 '20 at 21:11
  • @StevenThomasHatton I fail to see how that differs from the first four lines of my answer, in which I wrote "[...] then it follows that $p_{\mu} = g_{\mu\nu} m \dot x^\nu$ [...] this is clearly the "brother" of the four-vector $\tilde{p}^\mu = m\dot x^\mu$ with the index raised/lowered via the metric." Furthermore, it is $p_\mu$, not $\tilde{p}^\mu=m\dot x^\mu$, which has all of the deep properties we associate with momentum, so your insistence that $\tilde p^\mu$ is the momentum and $p_\mu$ is not remains difficult for me to understand. – J. Murray Sep 21 '20 at 16:28
  • @J.Murray I guess, I'm equivocating. The vector and the 1-form are actually two different representations of the same geometric object. – Steven Thomas Hatton Sep 22 '20 at 21:12
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    @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 $p_\mu$. – J. Murray Sep 22 '20 at 21:37