With some trepidation I will try to describe why:

- monopole sound radiation is possible
- electro-magnetic radiation must be at least dipole
- gravitational radiation must be quadrupole.

First of all by ‘monopole’ I mean that there are solutions to the wave equation that are spherically symmetric; they are invariant under orthogonal transformations: SO(3).

The following is wrong! (By ‘dipole’ I mean that for any given axis (thru the source), there are solutions to the wave equation that remain invariant under rotations about that axis.
Such solutions are called ‘cylindrically symmetric’.)

Quadrupole radiation lacks these symmetries, but it may have bilateral symmetry about a plane thru the source.

A monopole wave source produces a spherically symmetric wave train. Sound waves including pressure waves in a solid may have spherical symmetry. The motion of material particles that constitute the sound wave are in the direction of propagation of the wave.

Shear waves in solids are different for the motion of material particles that constitute the propagation of a shear wave is perpendicular to the direction of propagation of the wave. Electromagnetic waves are like this too; the electric vector is at right angles to the direction of the light.

There is a mathematical theorem to the effect that you cannot comb the hair on a sphere. If there is a continuous vector field on a sphere that is everywhere tangent to the sphere, then the field will be zero somewhere on the sphere. At any moment there is somewhere on the Earth that the wind is not blowing. If there were spherically symmetric shear waves then at any point in time and any particular radius, the vector field would be everywhere zero, or everywhere non-zero, but the latter is precluded by the hair theorem. The constantly zero field is thus the only shear field with spherical symmetry.

Pressure waves are a scaler field since knowing the scalar pressure at each time and place is knowledge of the entire wave. Knowledge of the vector shear at every time and place constitutes knowledge of the entire shear wave and thus the shear wave is a vector wave.

The gravitational wave requires knowledge of a tensor at every place and time. Don’t panic; it is much simpler than the general case for several reasons. The tensor is the metric tensor which says how far apart places that you know of are. The only distances of interest to understanding the gravitational wave are the distance between two points in a plane perpendicular to the source of the waves. According to general relativity points that begin on a circle in a plane perpendicular to the direction of the wave, just floating in space, will find them selves briefly in an ellipse as the wave passes. After the wave has passed they will resume the circular shape and be once again at rest. Indeed it is a semantic game as to whether they can be said to have moved as the wave passed. The tensor is symmetric and only the components describing the geometry in our plane are of interest. This means that just three numbers describe the distortion of the ellipse at a particular time and place. Indeed Einstein’s equations say that the ellipse will distort, but not change its size and thus just two independent numbers remain to be observed as you measure the passing gravitational wave. (This assumes you already know the direction of the source.)

If there were dipole gravitational waves then you can imagine a large ring of cooperating observers symmetrically positioned in a large circle so that they observe the same effects of the wave. Each observer measures the distance to his neighbor by trading and timing light signals. A dipole wave would shrink or expand this distance simultaneously in the collective reference frame of the observers. I suggest here that even a brief simultaneous shortening of the circumference of the observer ring is physically objectionable. I would like to conclude that it is excluded by Einstein’s equations but I have certainly not done that.

Two black holes meeting head on have dipole symmetry and according to this account can not produce gravitational disturbances at a distance. Actually it implies only that the effects die off faster than r