I don't understand whether something physical, like velocity for example, has a single correct classification as either a contravariant vector or a covariant vector. I have seen texts indicate that displacements are contravariant vectors and gradients of scalar fields are covariant vectors, but in A Student's Guide to Vectors and Tensors by Fleisch I found this statement:

[I]t's not the vector itself that is contravariant or covariant, it's the set of components that you form through its parallel or perpendicular projections. (p.121)

If physical concepts can be represented as either type of mathematical object, I don't understand how a displacement could be represented as a covector. Wouldn't its components transform the wrong way if the coordinates were changed?

If covariantness/contravariantness is part of the definition of a physical concept, I don't understand how force is classified. The gradient of a potential would have dimensions of length in the denominator, making it a covariant vector. Mass times acceleration has dimensions of length in the numerator, making it contravariant.

Edit: I read through the accepted answer to Forces as One-Forms and Magnetism. One thing I don't understand is whether in relativistic spacetime any vector quantity can also be represented as a 1 form (because there is a metric) or whether its classification as a 1 form or vector depends on how its components change under a coordinate transformation. Doesn't a displacement have to be a vector and not a 1 form?

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    $\begingroup$ I think every covariant vector defines a contravariant vector, because $E^{\star\star}=E$ and reversely every covector defines a vector. But in differential geometry, a position is a element of a manifold, the velocity is an element of the tangent space of the manifold at the considered position, and the momentum is an element of the cotangent space. It can be seen the following way: $\mathbf{p}=\mathbf{v}^T\mathbf{M}$ is row vector (i.e. a covector) and defines a linear form: $\mathbf{p}\,\mathbf{v}$ is an element of $\mathbb{R}$ (called energy when divided by 2). $\endgroup$
    – anderstood
    Commented Aug 17, 2014 at 4:03
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    $\begingroup$ That would depend on your choice of basis, wouldn't it? If one choses a "natural" basis, then some physical quantities like displacement will be contravariant, while others, like gradient, will be covariant. If one picks a dual basis, it would be the other way around. One does that in crystallography and solid state physics a lot, because working in Fourier space makes a lot of sense for problems in periodic lattices and potentials. Someone please correct me, if I am wrong. $\endgroup$
    – CuriousOne
    Commented Aug 17, 2014 at 6:57
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    $\begingroup$ @CuriousOne: A basis and a dual basis live in different spaces - a vector space and a dual space. While, in many cases, a natural notion for the isomorphism between them exists, expressing one as the other is not changing the basis, but applying the metric (or its inverse) to a given vector (which also produces the dual basis). On the topic of the question: I'll write an answer, but it's not gonna be authoritative. $\endgroup$
    – ACuriousMind
    Commented Aug 17, 2014 at 12:06
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    $\begingroup$ possible duplicate of Forces as One-Forms and Magnetism $\endgroup$
    – user4552
    Commented Aug 17, 2014 at 16:07
  • $\begingroup$ @BenCrowell I'm sorry. I don't fully understand the answer to that question. Is the following correct? Since my question concerns Newtonian 3-Space, forces can be represented with either contravariant or covariant vectors. Ditto for displacements, velocities, etc. But I thought contravariant vectors were defined as those that transform like displacements, so how could displacements be represented as covariant vectors. $\endgroup$
    – Noah
    Commented Aug 17, 2014 at 18:16

5 Answers 5


I understand force to be a 1-form, through the following reasoning. Given a time-independent, conservative lagrangian $L$, its differential (a 1-form in the purest sense) is $$ \mathrm{d}L = p_a ~\mathrm{d}\dot{x}^a + f_a~\mathrm{d} x^a $$ where $$ p_a = \frac{\partial L}{\partial \dot{x}^a},~f_a = \frac{\partial L}{\partial x^a}. $$ So the components of this 1-form are the force and the momentum. The momentum and the force are both interpreted as components of a covector for this reason. It shouldn't be surprising that they are the same type, given their relationship from Newton's second law. I also feel like it's sort of natural for momentum to be a 1-form, given it's "dual" nature to position.

Now, to address your edit. Given a metric, any vector can be written as a 1-form. Given that the manifold you are in is affine, you can write displacements as vectors. However, nobody every writes down displacement 1-forms. You seem to think this is at odds with the fact that the components of 1-forms transform "covariantly" and those of vectors "contravariantly". Once you use the metric to "lower the index", a vector will transform as a 1-form. Say we go from coordinates $x\rightarrow y$. The metric transforms as $$ g'_{ab} = \frac{\partial x^c}{\partial y^a}\frac{\partial x^d}{\partial y^b} g_{cd}, $$ and a vector $v$ will transform as $$ v'^a = \frac{\partial y^a}{\partial x^b} v^b. $$ Putting these statements together, we see that $v$ with the lowered index transforms as a 1-form should: $$ v'_a = g'_{ab}v'^b = \frac{\partial x^c}{\partial y^a}\frac{\partial x^d}{\partial y^b} g_{cd} \frac{\partial y^b}{\partial x^e} v^e = \frac{\partial x^c}{\partial y^a} \delta_{de} g_{cd} v^e = \frac{\partial x^c}{\partial y^a} g_{cd} v^d = \frac{\partial x^c}{\partial y^a} v_c. $$

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    $\begingroup$ This is a cool answer. But is there any reason to demand that momentum is a one-form, other than that you want to take an exterior derivative of the Lagrangian? $\endgroup$ Commented Aug 21, 2014 at 8:27
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    $\begingroup$ @soulphysics: While trying to think of an additional example for momentum, I came up with one for force: the work-energy theorem, which can be cleanly expressed as $ \int_C f = \Delta K$. Often, expressions written as line-integrals can be written in a parametrization-independent way by recognizing the integrand as a 1-form. $\endgroup$ Commented Aug 21, 2014 at 12:49
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    $\begingroup$ Another example is in the Fourier transform, we take $\exp(i p_j x^j)$. $\endgroup$ Commented Aug 21, 2014 at 13:41
  • $\begingroup$ Can displacements be written as 1 forms in an affine space, or are displacements always necessarily vectors? $\endgroup$
    – Noah
    Commented Aug 27, 2014 at 3:25
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    $\begingroup$ Displacement is an object that connects the old position (point) to the new position (point). Thus, it is a vector in the affine space. (In other spaces the answer may be different, more or less general, but the basic idea is vector.) $\endgroup$
    – firtree
    Commented Aug 27, 2014 at 14:56

I have an answer here describing the general concept of true physical vectors and tensors which might be useful. (It also might be things you already know very well.)

This turned out to be superlong, I hope you don't die of age before finishing reading. I at least added some useful titles of sections so you can skip some parts. The conclusion is that it is way more plausible for force to be a covector (it's components covariant).

Co(ntra)variant vector or components?

So, we need equations of mathematical physics to be true, which means they are, indeed, invariant. We achieve this by postulating objects such as $\vec{v}$ or $\vec{a}$. We then have equations of this type: $$\vec{F} = m \vec{a} $$ These are expressible in words and talk about objects which are "real" and thus a notion of them varying with a description is truly absurd. But how do we gain quantitative hypotheses to test from them? We postulate they have certain relations with other objects allowing us to obtain a certain set of numbers completely characterizing such an object. This set of numbers are the components and can be summarized by a following relation (Einstein summation convention included): $$\vec{v} = v^{\,i} \vec{e}_i$$ That is, the components just give the coefficients of a sum of objects of the same type. We do not get past the veil of this mysterious arrowed object. Ever. Concepts such as direction or velocity cannot be reduced completely to components and there is always this "core" of physics. Vectors are vectors and nothing less. We can however choose a different set of core objects $\vec{e}'_j$. Since $\vec{v}$ is truly invariant, we have $$v'^j \vec{e}'_j = \vec{v} = v^i \vec{e}_i$$ From arguments sketched in the answer linked in the top of this post, we can deduce the coefficients of the relation $\vec{e}_i = (e_i)^j \vec{e}'_j$. Nonetheless, when we apply it to the previous equation we get $$v'^j = (e_i)^j v^i$$ $$\vec{e}_i = (e_i)^j \vec{e}'_j$$ Where I put the "core" relation a second time to understand the fact that the components of the vector transform exactly opposite to the basis components. This is also the meaning of the word "contravariance" of the vector components. They transform to counter the variance of the basis, so as a whole the vector will stay invariant. (When any set of object transforms the same as the basis, they co-vary, so they are covariant.)

Note that Fleisch is just stating the just described. The object is invariant, the components transform. He is not saying you can get both covariant or contravariant, you get only one. (You can then use a trick with the metric to change the nature, but the resulting object isn't "the original one". This also addresses your edit concerning relativity.)

A note on the "displacement vector"

The "displacement vector" has it's components only coincidentally transforming as that of a vector in the case of Cartesian coordinates (and flat space) and only when the displacement is taken from the origin. Try moving the origin or switching to polar coordinates or any other curvilinear coordinate system. It's components will transform very differently from e.g. that of velocity.

It is good to remember that for a "true vector" in the sense of component transforms there always has to be something more "differential" about it rather than "difference".

So are the components of force contravariant or what?

As stated e.g. in the answer linked at the top of this post, the ultimate reference "contravariant" vector is the velocity and it's second derivative brother acceleration. In classical physics we definitely want the already mentioned equation $\vec{F} = m \vec{a} $ to hold.

Thus, it would seem straightforward to say that $\vec{F}$ must be a vector with contravariant components because the equation must be invariant (contravariant in components) and the right hand side is a vector. But is it?

There are two conventions. The first one says $m$ is just a number, so $\vec{F}$ must be a vector. But you can see in the answer of Zach McDargh and other comments that it can be a quite neat convention to actually have the force to be a covector. Otherwise, as you correctly observe, we would have to use the already mentioned trick with the metric to raise the indices of the gradient determining the force and other unnecessary junk. I am going to give a quick argument of further elegance elucidating the Newton's law with the so-called mass matrix.

The mass matrix

Consider the kinetic energy, it looks like this $$T = \frac{1}{2} m v^2$$ But the $v^2$ actually means $\vec{v}\cdot\vec{v}$. However, when we construct classical mechanics and pass to Lagrangian and Hamiltonian mechanics, we find out kinetic energy is the only place where the dot product necessarily enters the construction. We do not need it anywhere else, and everywhere it is, there is also $m$ and vice-versa.

The other funny thing is that with multiple variously constrained particles and with different masses it is practical to describe them all with a single set of coordinates $\mathbf{q} =(q_1,\dots q_N)$. Instead of writing out every kinetic energy term, it is practical to express the whole sum by $$T = \sum T_{\dots} = \frac{1}{2} \dot{\mathbf{q}}^T \mathbf{M} \dot{\mathbf{q}}$$ with $\mathbf{M}$ the "mass matrix" which is a mix of the metrics of the different particles times their masses, the $1/2$ is conventional, and the dot above the positions means the time derivative.

So even if we stick with just one particle, it seems natural to define the "mass matrix" as $M_{ij} = m g_{ij}$ with $g_{ij}$ the metric ($\delta_{ij}$ for the dot product in Cartesian coordinates in flat space). If we insist on the fact that the metric or mass never occur in dynamical equations separately, we reduce one element of our theory without changing it's physics. Occam's razor! It's simpler so it has to be done! (Until we want to measure angles and distances in a non-dynamical context...)

So, the mass matrix has two lower indices and is doubly covariant, the same as the metric. If we then contract it with a vector in one index we get e.g. $$p_i = M_{ij} v^j$$ Momentum is thus a covector whose components transform covariantly. It is now obvious that we have $$\vec{F}=m\vec{a} \to F_j = M_{ij} a^i$$ Or even better for systems with varying mass $$F_j = \frac{d}{dt}p_j$$

So there is no problem with Newton's equation and it's covariance. By this argument, it is much more natural to define force to be a covector.

A feinschmecker note: You might argue that the concept of a mass matrix is not needed for the whole argument. But I am assuming we are not Landau, and we want to find an elegant form of the Newton's law without an arbitrarily placed metric instead of "deriving" the Newton's law from Lagrangian mechanics.

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    $\begingroup$ Thank you for this -- it dispelled a long-standing worry that had been keeping me (a geometer with no physics education) awake at nights. $\endgroup$
    – helveticat
    Commented Jul 10, 2018 at 8:53

There are two kinds of mathematical situations considering vectors and covectors. They are those:

  1. In the space without scalar product (dot product) they are very different entities and one should never confuse them.
  2. In the space with scalar product they are the same thing viewed from two perspectives. One can freely convert one entity to another with metric tensor, whether that does make sense or not.

Now, when we talk about physics, we find many different theories with different spaces, some are given scalar product and some are not. For example, the $(x,y,z)$ space of Newtonian mechanics is given the Euclidean dot product, and the $(P,V)$ space of states of some quantity of gas is given none. Force does appear only in all kinds of mechanics, be that Newtonian, Lagrangian, SR, GR, quantum mechanics, quasiparticle mechanics in solids, and so on. And here is my humble observaion:

The natural definition of force is used only in those mechanical theories where the spaces are given dot product. These are Newtonian mechanics, SR, GR, Newtonian with restraints, some cases of quantum mechanics and such. In those cases, you can freely think of force as being a vector or a covector.

And in the theories without some implicit dot product, the concept of force is used in some abstract extended sense (if used at all). Such theories include Lagangian and Hamiltonian mechanics, more general cases of quantum mecanics, quasiparticle mechanics and such. In those cases, it is explicitly stated what does the term force mean, and often the notion of force is not used at all. Then you just follow the definitions of a particular book, and do not think of these forces as of usual forces in everyday life. [Usually the force is thought of as a covector, and the acceleration is connected to it with a tensorial mass, $f_a=m_{ab}w^b$.]


Not disagreeing with other (older) answers. But it seems that they ignore the most obvious/compelling reason why force is represented as a covector: in analytical mechanics, the force field is derived as the spatial gradient of a scalar potential energy, $F_i=-\partial V / \partial x^i$. And the gradient of a scalar is a covector (a.k.a. differential of the function $V$).

Now, in the case where the force is not derived from an energy gradient (like the force of poking a billiard ball with a cue or dragging a brick by a string), then it's not at all clear that you'd benefit from the covector representation. But then you'd also not be working on a manifold in which the invariance of tensors under coordinate transformations is useful, anyway.

To summarize, we represent force as a covector in many physics applications (orbital mechanics, E&M, SR, GR, QM) because the "force" is a spatial gradient of a scalar energy. This goes for non-mechanics applications too, such as the wavenumber covector of a traveling wave, $\exp[i(\mathbf{k}\cdot\mathbf{x}-\omega t)]$, where $k_i=\partial\phi/\partial x^i$ is the gradient of the scalar phase function $\phi(\mathbf{x})$. See, e.g., here.


I would argue that a force is a contravariant vector. Consider a rock falling along Earth's gravity. The gravity potential mgz is horizontal and its differential is a 1-form which kernel is a horizontal plane.

But the rock falls vertically, which is a favored direction outside the kernel of the 1-form. A 1-form doesn't have such a notion of favored direction outside its kernel. This is where the scalar product comes in and defines a perpendicular direction for the rock to fall.

So you either agree that the 1-form asks the scalar product a perpendicular direction every time it puts a body in motion, or that the force is already a contravariant vector, with a definite direction of motion.


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