The Wikipedia article on momentum defines momentum as in classical mechanics:

… momentum is the product of the mass and velocity of an object.

However, an electromagnetic field has momentum, which is how solar sails work.

I would not suppose that this is a product of the ‘mass’ and ‘velocity’ of the field.

So, what is momentum, conceptually?

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    $\begingroup$ Potentially helpful: physics.stackexchange.com/a/216599/20427 $\endgroup$
    – user87745
    Commented Jun 22, 2016 at 7:46
  • $\begingroup$ @Dvij Thanks, the second link is helpful but I still want to know what momentum is at a conceptual level, because the answer speaks rather vaguely about "terms that remain constant in time" $\endgroup$
    – hb20007
    Commented Jun 22, 2016 at 7:51
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    $\begingroup$ momentum is the 'quantity of motion' that is conserved when you 'transfer motion' from one body to another; it turns out that not only is it possible to transfer motion between material bodies, but you can also do so using 'immaterial' fields $\endgroup$
    – Christoph
    Commented Jun 22, 2016 at 10:34
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    $\begingroup$ "I would not suppose that this is a product of the 'mass' and velocity of the field": Why not? Perhaps naively, I'd assume that a light quantum's momentum is exactly its mass (i.e. its energy's mass equivalent) times c. Is that not so? $\endgroup$ Commented Jun 23, 2016 at 15:19

4 Answers 4


Momentum / energy are the conserved Noether charges that correspond, by dint of Noether's Theorem to the invariance of the Lagrangian description of a system with respect to translation.

Whenever a physical system's Lagrangian is invariant under a continuous transformation (e.g. shift of spatial / temporal origin, rotation of co-ordinates), there must be a conserved quantity, called the Noether charge for that transformation.

We then define the conserved charges for spatial and temporal translation as momentum and energy, respectively; angular momentum is the conserved Noether charge corresponding to invariance of a Lagrangian with respect to rotation of co-ordinates.

One can derive the more usual expressions for these quantities from a Lagrangian formulation of Newtonian mechanics. When Maxwell's equations and electromagnetism are included in a Lagrangian formulation, we find that there are still invariances with the above continuous transformations, and so we need to broaden our definitions of momentum to include those of the electromagnetic field.

User ACuriousMind writes:

I think it would be good to point out that the notion of "canonical momentum" in Hamiltonian mechanics need not coincide with this one (as is the case for e.g. a particle coupled to the electromagnetic field)

When applied to the EM field, we use a field theoretic version of Noether's theorem and the Lagrangian is a spacetime integral of a Lagrangian density; the Noether currents for a free EM field are the components of the stress-energy tensor $T$ and the resultant conservation laws $T_\mu{}^\nu{}_{,\,\nu}=0$ follow from equating the divergence to nought. This includes Poynting's theorem - the postulated statement of conservation of energy (see my answer to this question here) and the conservation of electromagnetic momentum (see the Wiki article). On the other hand, the Lagrangian $T-U$ describing the motion of a lone particle in the EM field is $L = \tfrac{1}{2}m \left( \vec{v} \cdot \vec{v} \right) - qV + q\vec{A} \cdot \vec{v}$, yielding for the canonical momentum conjugate to co-ordinate $x$ the expression $p_x=\partial L/\partial \dot{x} = m\,v_x+q\,A_x$; likewise for $y$ and $z$ with $\dot{x}=v_x$. A subtle point here is that the "potential" $U$ is no longer the potential energy, but a generalized "velocity dependent potential" $q\,V-\vec{v}\cdot\vec{A}$. These canonical momentums are not in general conserved, they describe the evolution of the particle's motion under the action of the Lorentz force and, moreover, are gauge dependent (meaning, amongst other things, that they do not correspond to measurable quantities).

However, when one includes the densities of the four force on non-EM "matter" in the electromagnetic Lagrangian density, the Euler Lagrange equations lead to Maxwell's equations in the presence of sources and all the momentums, EM and those of the matter, sum to give conserved quantities.

Also note that the term "canonical momentum" can and often does speak about any variable conjugate to a generalized co-ordinate in an abstract Euler-Lagrange formulation of any system evolution description (be it mechanical, elecromagnetic or a even a nonphysical financial system) and whether or not the "momentum" correspond in the slightest to the mechanical notion of momentum or whether or not the quantity be conserved. It's simply a name for something that mathematically looks like a momentum in classical Hamiltonian and Lagrangian mechanics, i.e. "conjugate" to a generalized co-ordinate $x$ in the sense of $\dot{p} = -\frac{\partial H}{\partial x}$ in a Hamiltonian formulation or $p = \frac{\partial L}{\partial \dot{x}}$ in a Lagrangian setting. Even some financial analysts talk about canonical momentum when Euler-Lagrange formulations of financial systems are used! They are (as far as my poor physicist's mind can fathom) simply talking about variables conjugate to the generalized co-ordinates for the Black Schole's model. Beware, they are coming to a national economy near you soon, if they are not there already!

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    $\begingroup$ I think it would be good to point out that the notion of "canonical momentum" in Hamiltonian mechanics need not coincide with this one (as is the case for e.g. a particle coupled to the electromagnetic field) $\endgroup$
    – ACuriousMind
    Commented Jun 22, 2016 at 14:15
  • $\begingroup$ @ACuriousMind See update. Yes, it was a bit silly of me to forget this, given that the central theme of my answer - conservation - doesn't even hold for many canonical momentums. I guess whenever I see people groping for "meaning" in momentum, my answer is the idea I focus on, because I think it is a very deep insight to understand that conservation can come from symmetry. There are those in physics (I get the impression you're not one of them) who say "meh, you're just replacing one why with another", but I think there is a genuine and deep difference: conservation of abstract .... $\endgroup$ Commented Jun 23, 2016 at 3:58
  • $\begingroup$ ....quantities is something you just have to accept like the Israelites receiving Moses's transcriptions of the behests of a control freak in the sky, whereas the notion that our World doesn't depend on our descriptions of it is something even tiny children begin to grasp: it's something one naturally sees for oneself and is a much more "visceral" notion. $\endgroup$ Commented Jun 23, 2016 at 3:58
  • $\begingroup$ @WetSavannaAnimal, I would treasure the opportunity to furnish you with some bundaberg overproof rum and/or concentrated caffeine and simply sit there and bask in your awesomeness! $\endgroup$ Commented Jan 13, 2020 at 3:19

Without touching on electromagnetism, I'd like to bring up this construction from mechanics (it's in the Feynman lectures).

Consider two equal particles approaching each other with equal speed.

A---->           <----B

You can argue from first principles that if they stick together they will not be moving afterwards -- any argument you could make that the composite particle moves to the left is also an argument for making it move to the right; the symmetry of the situation doesn't allow for any nonzero movement afterwards. Simple enough.

Now consider a stationary particle A being approached by an equal particle B with velocity $2v$.

B------>            A

From Galilean invariance you can move the reference frame so that it is travelling to the right at constant speed $v$. Then this situation becomes the original situation, and the composite particle AB is stationary with respect to the new reference frame which is moving at speed $v$. Now translate back to the original reference frame and the composite particle is now travelling at speed $v$.

Here you can define $p=mv$ and observe that it is conserved.

You can extend this idea to systems of multiple equal particles and multiple collisions, constructing situations equivalent to composite masses in any ratio you like. $p=mv$ is conserved by inductive argument.

  • $\begingroup$ "You can extend this idea..." Feynman does, IIRC. $\endgroup$
    – spraff
    Commented Jan 26, 2021 at 15:25
  • $\begingroup$ Ah, sorry, I missed the part about composite masses. $\endgroup$
    – tparker
    Commented Jan 26, 2021 at 16:10
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    $\begingroup$ This "thought experiment" using Galilean invariance is also at the heart of a very intuitive argument that kinetic energy is quadratic in velocity and linear in mass physics.stackexchange.com/a/14752/222958 (one of my favorite answers ever) $\endgroup$
    – D.R
    Commented Apr 22, 2023 at 4:17

As the wiki article you quote states, momentum is defined as the product of the velocity times the mass of an object.

Classical mechanics developed theoretically on the lines explained by WetSavanna in the other answer, the conservation of momentum and energy being cornerstones of the theory. Classical mechanics is a very successful theory, and conservation of momentum is a law.

Then comes classical electrodynamics with Maxwell's equations. It can be shown that the electromagnetic wave carries energy. Then using the law of conservations of momentum, the momentum carried by the electromagnetic wave can be derived, as shown here. I.e. momentum conservation will be violated if the electromagnetic wave in addition to energy does not carry momentum. Please see the link for details.

So the concept of momentum comes from the classical mechanics definition and is extended into relativistic mechanics, classical electrodynamics is fully relativistic, so as to include the electromagnetic wave as a four vector too.


Momentum ($p$) is "really" $mv$, even for light and EM fields. This can be proven by the use of $E = mc^2$.
The momentum for a photon (EM) is $p = mv$. Where the mass is given by $m = E/c^2$ and $v = c$. Substituting these into the equation, one obtains, $p = E/c$. Although this equation "looks" different from $p = mv$, because it was derived using the basic definition, it is equivalent (and applicable to objects that have energy - but no "rest mass").

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    $\begingroup$ The sheer violence that $E=mc^2$ has done in the name of $E_0=mc^2$ puts the inquisition to shame. $\endgroup$
    – user87745
    Commented May 19, 2020 at 23:32

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