Tchaikovsky's Snowflakes

An extreme example of how to deal with the Great Nineteenth century Rhythm Problem

The “problem” is an artistic and aesthetic one - in four or eight bar phrases, there is a tendency for a duple hypermetre to override the metre of the time signature. This hypermetric effect can pertain at higher levels too, so that eight bars of ¾ are perceived as two bars of 12/8, or even one bar of 24/16, thus giving the listener a metrical structure so unchallenged and regular that one is hardly aware of metre at all. This was the challenge, then, that faced the nineteenth century composer of salon pieces in the popular dance or folk-style – to stay within the metrical and phrasal limits of the genre, while allowing the listener to put their ears over the metrical parapet, as it were. This is achieved by compositional devices such as metrical dissonance; rhythmic interest; changes in dynamics, register or timbre; phrase extension or diminution; interleaving of declamatory passages with rhythmic ones.

It was Chopin who in sheer quantity achieved this more than any other nineteenth century composer. His waltzes, mazurkas and polonaises are text-book examples of how to avoid the “great nineteenth century rhythm problem”. However, probably one of the best examples of this avoidance is Tchaikovsky’s Snowflakes Dance in Nutcracker. The extremes of speed, timbre, dynamics and pitch, coupled with frequent caesuras make metre in the introduction almost imperceptible. The main theme at its entrance is an example of hypermetrical hemiolated additive metre in polymetric canon with an eight bar phrase of ¾ (!).

1   2   3   4   5   6   1   2   3 1 2 1 2 3 = 6+3+2+3
                                                   2 2 4 4
1   2   3   4   5   6   1   2   3 1 2 3 1 2 3 = 6+3+6*
                                                   2 2 4
1 2 3 2 2 3 3 2 3 4 2 3 5 2 3 6 2 3 7 2 3 8 2 3 = 3
                                                   4

* because of the melodic shape, more correctly 6+(2+2+2+3)+3
2 4 4
As if that weren't enough, because of the curious way Tchaikovsky has written the accompaniment, you can easily count 6 bars of 4/2 (from the main tune) and still end up together with the orchestra for when the next section begins. There is no reason why this shouldn't be so, for 4/2 contains 8 (quarter-note) beats, which multiplied by 6 bars makes 48 beats. 48 then divides nicely into 16 bars of 3/4

It will be seen that only point at which ¾ is reinforced as the metre is in the last bar of the eight bar phrase, and then only for one bar before returning to the prevailing hypermetrical organisation. It should be noted that the orchestra does not reinforce the ¾ time signature – the only person “in ¾” will be the conductor.

This is a metrical conundrum as complex as anything by Stravinsky in the context of an eight bar phrase in waltz time.