Can you give a simple, intuitive explanation (imagine you're talking to a schoolkid) of why mathematically speaking momentum is covector? And why you should not associate mass (scalar) times velocity (vector) and momentum (covector) in your head?

There were already answers to this question, but they all rely on rather abstract definition of a Lagrangian (difficult to explain to a schoolkid). Is there some motivation in the first place?

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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/176555/2451 Related: physics.stackexchange.com/q/172789/2451 $\endgroup$
    – Qmechanic
    Aug 14, 2017 at 21:42
  • $\begingroup$ No, I don't believe you can ( I would also like to thank you for frightening me on the level of complexity involved, I self study and had no idea before I searched : ), I think once you go out of the "normal" explanation, you can't avoid the rigor of the math involved, avoiding the Langranian , to me personally, makes things worse. The Langrangian is easier to understand than the what follows. $\endgroup$
    – user163104
    Aug 14, 2017 at 21:54
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    $\begingroup$ School kids don't know covectors $\endgroup$ Feb 21, 2018 at 14:34

2 Answers 2


Suppose you exert a force ${\bf F}$ on a particle of mass $m$ in order to move it along a path $\gamma$. The work required to do so is $$W = \int_\gamma {\bf F} \cdot d{\bf x} = \int_\gamma \frac{d{\bf p}}{dt} \cdot d{\bf x} = \frac{d}{dt} \int_\gamma {\bf p} \cdot d{\bf x}$$ (since the path $\gamma$ and therefore $d{\bf x}$ does not change over time).

Now the work done is a scalar; clearly it doesn't depend on your choice of spatial coordinates. You don't feel twice as tired after exerting the same force on the same physical system described using different coordinates. (More quantitatively, if you use a motor to push the mass, then you can measure the work done by the motor by seeing how much energy has been depleted from its battery. Clearly the battery doesn't "know" what coordinate system you used.) And the time derivative and the line integral also clearly do not depend on the choice of spatial coordinates. Therefore, the dot product ${\bf p} \cdot d{\bf x}$ must be coordinate-independent as well. So if you change coordinates in such a way that the numerical values of the components of $d{\bf x}$ double, then the numerical values of the components of the momentum ${\bf p}$ must halve in order to compensate. So the position and momentum transform oppositely under coordinate transformations (as a contraviant vector and covector, respectively).

If you don't like the integral or the time derivative in that explanation, you can alternatively note that the instantaneous power $P = {\bf F} \cdot {\bf v}$ that you apply to the mass is a scalar for the same reason, so force (the time derivative of momentum) and velocity (the time derivative of position) must transform oppositely under coordinate transformations.

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    $\begingroup$ I agree with the spirit but this feels a bit too fast. It sounds dangerously close to saying that in every dot product, one of the objects is naturally a covector, which is clearly false. $\endgroup$
    – knzhou
    Feb 22, 2018 at 23:30
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    $\begingroup$ @knzhou That's a very fair point. The point that I was trying to emphasize is that you can directly relate the work and power to something very obviously non-geometric, like the reading on a battery. Whereas if you, say, try to calculate the area of a parallelogram via $A = {\bf a} \cdot {\bf b}$, you also get a scalar quantity, but unlike in the previous case, this one is fairly clearly geometry-/coordinate-dependent (if you define "area" by "number of squares of unit coordinate length"). $\endgroup$
    – tparker
    Feb 23, 2018 at 1:36
  • $\begingroup$ @knzhou Although I can't think of a simple intuitive explanation for why my answer's argument doesn't also imply that the kinetic energy $\frac{1}{2} m {\bf v} \cdot {\bf v}$ is also geometry-independent. $\endgroup$
    – tparker
    Feb 23, 2018 at 1:41
  • $\begingroup$ Does it make sense to pull the time derivative outside the integral, as in $\int_\gamma (dp/dt)\cdot dx=(d/dt)\int_\gamma p\cdot dx$? The integral on the right is not time dependent. $\endgroup$
    – WillG
    Aug 19, 2022 at 6:01
  • $\begingroup$ @WillG I will admit that this explanation is rather hand-wavy and maybe not rigorous enough. But I think that the power version is a little more solid, because it only involves instantaneous quantities. $\endgroup$
    – tparker
    Aug 19, 2022 at 11:51

Since the initial posting of my question, a reformulation has appeared with the perfect answer. Although this answer does not address the issue of explainability to a school kid, it nevertheless should help you get the mathematical gist of the concept. Hopefully, a (very) motivated school kid can understand it :) Therefore, the question is closed now.


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