If two infinitely long sets of deep-sea waves, having slightly different wave-lengths, and therefore slightly different velocities, are superimposed, we obtain a resultant wave-train which exhibits a variation in wave-amplitude along its course periodically. If we were to look along the train, we should see the wave-amplitude at intervals waxing to a maximum and then waning again to nothing. These points of maximum amplitude regularly arranged in space constitute, as it were, waves on waves. They are spaced at equal distances, and separated by intervals of more or less waveless or smooth water. These maximum points move forward with a uniform velocity, which we may call the _velocity of the wave-train_, and the distance between maximum and maximum surface-disturbances may be called the _wave-train length_. Let _v_ and _v′_ be the velocities, and _n_ and _n′_ the frequencies, of the two constituent wave-motions. Let λ and λ′ be the corresponding wave-lengths. Let V be the wave-train velocity, N the wave-train frequency, and L the wave-train length. Then N is the number of times per second which a place of maximum wave-amplitude passes a given fixed point. Then we have the following obvious relations:— _v_ = _n_λ, _v′_ = _n′_λ′, N = _n_ - _n′_ = _v_/λ - _v′_/λ′ Also a little consideration will show that— L/λ′ = λ/(λ - λ′) since λ is nearly equal, by assumption, to λ′. Hence we have— 1/L = 1/λ - 1/λ′; and also V = NL Accordingly— V = N/(1/L) = (_v_/λ - _v′_/λ′)/(1/λ - 1/λ′) Let us write 2π/_k_ instead of λ, and 2π/_k′_ instead of λ′; then we have— V = (_vk_ - _v′k′_)/(_k_ - _k′_) (i.) And since _k_ and _k′_, _v_ and _v′_ are nearly equal, we may write the above expression as a differential coefficient; thus— V = _d_(_vk_)/_d_(_k_) (ii.) Suppose, then, that, as in the case of deep-sea waves, the wave-velocity varies as the square root of the wave-length. Then if C is a constant, which in the case of gravitation waves is equal to _g_/2π, where _g_ is the acceleration due to gravity, we have— _v_^2 = Cλ, or _v_^2 = (_g_/2π) × λ But λ = 2π/_k, hence— _vk_ = 2πC/_v_ Hence if we differentiate with respect to _v_, we have— _d_(_vk_)/_dv_ = -2πC/_v_^2 Again, _k_ = 2π/λ = 2πC/_v_^2; therefore— _d_(_k_)/_dv_ = -2(2πC/_v_^3) Hence, dividing the expression for _d_(_vk_)/_dv_ by that for _d_(_k_)/_dv_, we have— V = _d_(_vk_)/_d_(_k_) = _v_/2 In other words, the wave-train velocity is equal to half the wave-velocity. This is the case with deep-sea waves. Suppose, however, that, as in the case of air waves, the wave-velocity is independent of the wave-length. Then if two trains of waves of slightly different wave-length are superposed, we have _k_ and _k′_ different in value but nearly equal, and _v_ and _v′_ equal. Hence the equation (i.) takes the form— V = _v_