The Wikipedia article on momentum defines momentum as in classical mechanics:
[M]omentum 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, fundamentally?
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!
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 2v 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.
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").
The answer by @Selena Ballerina is physically correct, but I wanted to add a more intuitive answer about the geometry of momentum from the perspective of rigid body motion in Newtonian Mechanics.
Linear momentum $\boldsymbol{p}$ is defined by the cumulative impulse $\boldsymbol{J}$ needed to change a body from moving to being at rest. Think of having a huge hammer that you can swing to hit a moving object in order to make it come to rest. Momentum describes the quantity needed to do so.
$$ \boldsymbol{p} + \boldsymbol{J} = 0 \tag{1}$$
Angular momentum $\boldsymbol{L}$ also needs to be balanced $$ \boldsymbol{L} + \boldsymbol{r} \times \boldsymbol{J} = 0 \tag{2} $$ here angular momentum is summed about the origin.
A point particle of mass $m$ is moving, and at some instant it has a velocity vector $\boldsymbol{v}$ (magnitude and direction) and is located at position $\boldsymbol{r}$ relative to an inertial reference frame.
Linear Momentum, $\boldsymbol{p}$ is defined by the impulse $\boldsymbol{J} = \int \boldsymbol{F}\,{\rm d}t$ required to bring the particle to rest
$$\left. \begin{aligned} \boldsymbol{J} & = \int \boldsymbol{F}\,{\rm d}t \\ & = \int m (-\boldsymbol{a})\,{\rm d}t \\ & =- m \int \boldsymbol{a}\,{\rm d}t \\ \end{aligned} \right\} \; \boldsymbol{J} = -m \boldsymbol{v} $$ $$ \boldsymbol{p} = m \boldsymbol{v} \tag{3} $$
Angular Momentum, $\boldsymbol{L}$ needs to be balanced, which is trivial for a point mass since $\boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}$ $$\require{cancel} \begin{aligned} \boldsymbol{r} \times \boldsymbol{p} + \boldsymbol{r} \times \boldsymbol{J} & = 0 \\ \boldsymbol{r} \times \left(\cancel{ \boldsymbol{p} + \boldsymbol{J}} \right) & = 0 \end{aligned} $$
In summary, the total momentum of a point mass (defined by the combination of linear and angular momentum) is defined
$$\begin{Bmatrix}\boldsymbol{p}\\ \boldsymbol{L} \end{Bmatrix}=\begin{Bmatrix}m\boldsymbol{v}\\ \boldsymbol{r}\times m\boldsymbol{v} \end{Bmatrix}$$
which defines a line in space through the mass point $\boldsymbol{r}$, with direction parallel to $\boldsymbol{v}$ and magnitude $m \| \boldsymbol{v} \|$. This axis is called the axis of percussion.
The combination of the two vectors $\{ \boldsymbol{\hat{e}},\;\boldsymbol{r}\times \boldsymbol{\hat{e}} \}$ are called the Plücker coordinates of a line with direction $\boldsymbol{\hat{e}}$ through a point $\boldsymbol{r}$, because you can recover the properties of the line from the coordinates.
When it comes to mechanics, the geometry is that of lines, but with a magnitude value attached to them. This decomposes the momentum into $$\text{(momentum)} = \text{(magnitude)} \text{(line)}$$
similar to how vectors decompose into
$$\text{(vector)} = \text{(magnitude)} \text{(direction)}$$
A rigid body with mass $m$ and mass moment of inertia tensor about the center of mass ${\rm I}_{\rm cm}$ is in motion. The motion can be decomposed into the velocity of the center of mass $\boldsymbol{v}_{\rm cm}$ and a rotational velocity $\boldsymbol{\omega}$ about the COM.
Linear Momentum, $\boldsymbol{p}$ is defined by the impulse $\boldsymbol{J} = \int \boldsymbol{F}\,{\rm d}t$ required to bring the body to rest.
$$\left.\begin{aligned}\begin{aligned}\boldsymbol{J}\end{aligned} & =\int\boldsymbol{F}\,{\rm d}t\\ & =\int m\left(-\boldsymbol{a}_{{\rm cm}}\right)\,{\rm d}t\\ & =-m\int\boldsymbol{a}_{{\rm cm}}\,{\rm d}t \end{aligned} \right\} \boldsymbol{J}=-m\boldsymbol{v}_{{\rm cm}}$$
$$ \boldsymbol{p} = m \boldsymbol{v}_{{\rm cm}} \tag{4}$$
Angular Momentum, $\boldsymbol{L}$ is used to define the line of action of the impulse, which does not go through the center of mass. If it did go through the COM, then it would only bring the center of mass at rest, and the body would continue to rotate.
$$\begin{aligned}\boldsymbol{L} & =\boldsymbol{L}_{{\rm cm}}+\boldsymbol{r}_{{\rm cm}}\times\boldsymbol{p}\\ & ={\rm I}_{{\rm cm}}\boldsymbol{\omega}+\boldsymbol{r}_{{\rm cm}}\times m\boldsymbol{v}_{{\rm cm}} \end{aligned} \tag{5}$$
But we need to bring the above in the form of Plücker coordinates $\boldsymbol{L}=\boldsymbol{r}\times \boldsymbol{p}$ by finding an appropriate point $\boldsymbol{r}$ through the axis of percussion.
$$\begin{aligned}\boldsymbol{r} & =\frac{\boldsymbol{p}\times\boldsymbol{L}}{\|\boldsymbol{p}\|^{2}}\\ & =\frac{\boldsymbol{p}\times\left(\boldsymbol{L}_{{\rm cm}}+\boldsymbol{r}_{{\rm cm}}\times\boldsymbol{p}\right)}{\|\boldsymbol{p}\|^{2}}\\ & =\frac{\boldsymbol{p}\times\left(\boldsymbol{r}_{{\rm cm}}\times\boldsymbol{p}\right)}{\|\boldsymbol{p}\|^{2}}+\frac{\boldsymbol{p}\times\boldsymbol{L}_{{\rm cm}}}{\|\boldsymbol{p}\|^{2}}\\ & =\frac{\boldsymbol{r}_{{\rm cm}}\left(\boldsymbol{p}\cdot\boldsymbol{p}\right)-\boldsymbol{p}\left(\boldsymbol{p}\cdot\boldsymbol{r}_{{\rm cm}}\right)}{\|\boldsymbol{p}\|^{2}}+\frac{\boldsymbol{p}\times\boldsymbol{L}_{{\rm cm}}}{\|\boldsymbol{p}\|^{2}}\\ \boldsymbol{r} & =\boldsymbol{r}_{{\rm cm}}+\frac{\boldsymbol{p}\times\boldsymbol{L}_{{\rm cm}}}{\|\boldsymbol{p}\|^{2}}-\cancel{\boldsymbol{p}\left(\ldots\right)} \end{aligned}$$
Note that any component parallel to $\boldsymbol{p}$ and hence parallel to $\boldsymbol{v}_{\rm cm}$ can be ignored as it does not contribute to the angular momentum, due to the cross product.
$$ \boldsymbol{r} =\boldsymbol{r}_{{\rm cm}}+\frac{\boldsymbol{p}\times\boldsymbol{L}_{{\rm cm}}}{\|\boldsymbol{p}\|^{2}} \tag{6}$$
This means the line of action of momentum for a moving rigid body is beyond the center of mass, away from the center of rotation. This is exactly the definition of the axis of percussion. For example, for a swinging door, if you kick it at $2/3$ of its length away from the pivot, it is going to swing open without any reaction on the hinge. The factor of $2/3$ comes from equation (6) if you work out all the math.
