My URL is http://s.faculty.umkc.edu/shatzkinm/
my email address is: shatzkinm@umkc.edu
About Me
About the Medical Arts Symphony
About New Ear
ABOUT MY DATABASE
OF COMPOSERS' WHEREABOUTS
SOME MUSIC THEORY TOPICS
As a violinist, I have been a member of several orchestras--the Indianapolis, the Chattanooga(concertmaster),Rochester, Tulsa, and Kansas City. I was a member of resident string quartets at Pittsburg State University and the University of Missouri Kansas City. Also at the latter institution I was a member of the contemporary music ensemble and the baroque ensemble.
I am a former member of the contemporary ensemble, NewEar, and am presently the conductor of the Medical Arts Symphony.
I have written compositions for orchestra, chorus, voice, piano,
and ensembles.
TOC
The Medical Arts Symphony has been in existence for over 30 years. Its members are amateur musicians, many of whom are either active or retired members of the medical profession. Some are retired professional musicians. The orchestra gives two concerts yearly in Kansas City--usually in late November and late April. Professionals fill out the ranks for these concerts. Notable compositions performed were Berlioz's HAROLD IN ITALY, Rachmaninoff's RHAPSODY ON A THEME OF PAGANINI, and Brahms' Second and Third Symphonies. Several contemporary works were performed--several of them premiers of local composers'.
The orchestra for many years sponsored the Leopold Shopmaker Violin Competition. This awarded three prizes and a performance with the orchestra for the first-prize winner. A recent winner went on to win the Queen of Belgium Competition.
We welcome new, qualified members. There are openings in strings
and bassoon. If you would like to audition, call me at 816-523-7376. Rehearsals
are held at Battenfield Auditorium at the corner of Olathe Blvd. and Rainbow
Blvd. each Monday evening at 7:15. Our next concert (at the same location)
will be on November 16, 2002 at 8:00 PM. There is no admission charge.
The program consists of Beethoven's Symphony No. 7, and Liszt's Les
Preludes, Straus' Tales of the Vienna Woods, and Oklahoma. Our program
carries advertisements which help support the orchestra. If you wish to
advertise, please call me.
.
NewEar is a professional ensemble that performs contemporary music in Kansas City, Mo. It normally gives five concerts yearly at St, Mary's Episcopal Church at 1306 Holmes St. Its repertoire concentrates on works of the last two decades and has included several premieres, some by local composers.
They encourage the submission of scores, particularly those calling for some combination of clarinet, saxophone, bassoon, violin, piano, and percussion or a subset thereof. Address scores with return postage and envelope to: Jan Faidley, 5530 Roeland Drive, Mission, Kansas. 66205 (816-671-7638) (JFAID74702@AOL.COM).
A device I developed for analysis and teaching of SET THEORY and 12-TONE ANALYSIS, as well as composition is the MUSICAL SLIDE RULE. It consists of twelve slides mounted on a slab, held in place by two guides. There are twelve plywood slides that are 12 inches long and one centimeter wide. Each is covered with paper and marked off with squares of equal size throughout its length. The squares are filled with numbers as follows: from the top down, the numbers "0" through "11" twice in black ink, then the same numbers twice in reverse order in red ink. The slides rest on a slab of plywood that is 9 X 12 inches. They are held in place by two plywood guides, each 7 inches X 1 centimeter. These are screwed onto the slab 1 inch from each of the long sides of the slab, parallel with those sides, and centered. The slides should rest between them snugly enough to hold them in place when individually moved up or down, but also with freedom to move readily. A piece of clear plastic of about 7 X 5 inches is screwed into the guides so that it covers the central area over the slab(which is covered by the slides). With all slides in the same orientation, the display looks like this:
0 0 0 0 0 0 0 0 0 0 0 0 (in black) 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 101010101010101010101010 111111111111111111111111 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 etc. 0 0 0 0 0 0 0 0 0 0 0 0 111111111111111111111111 101010101010101010101010 9 9 9 9 9 9 9 9 9 9 9 9 etc. (IN RED)The Slide Rule has many applications for dealing with the set theory described by Allen Forte in his book THE STRUCTURE OF ATONAL THEORY(Yale University Press, 1973). It is assumed here that the reader is familiar with at least the basics of the theory.
To work with a set, in most cases it is convenient to use the number of slides that corresponds to the cardinality of the set; e.g., the number of notes in the set. To illustrate, let us take 3-4 {0 1 5}.
THE MATRIX
If any three of the adjacent slides are arranged so the numbers
of the prime form are aligned at the top of the display, all 24 of the
equivalents of the set are shown:
0 1 5 (IN BLACK) 1 2 6 " 2 3 7 3 4 8 4 5 9 5 610 6 711 7 8 0 8 9 1 910 2 1011 3 11 0 4 0 1 5 etc. 011 7 (IN RED) 1110 6 " 10 9 5 9 8 4 8 7 3 7 6 2 6 5 1 5 4 0 4 311 3 210 2 1 9 1 0 8 011 7 etc.The matrix, arranged in this way--with the prime form at the top, shows all the equivalents in a manner that allows each equivalent to be quickly found. First of all, the transpositions of the prime form are in black and the inversions in red. Secondly, The left-most number of each equivalent is always the transposition number relative to the prime form: e.g., in black, the row (of the matrix) reading "8 9 1" is the prime form transposed up by 8 semitones, and in red the row reading "6 5 1" is the inversion of the prime form transposed up by 6 semitones.
FINDING PRIME FORM
The prime form of any set can quickly be found by aligning the numbers
of the unkown set at the top of the display. This still creates a matrix
of the set in the display, so, naturally, the prime form will be shown
in one or more of its rows. THE PRIME FORM IS THE ROW CONTAINING THE SMALLEST
NUMBER, READING THE DIGITS IN THE ROW IN ASCENDING ORDER AND AS A SINGLE
NUMBER.
Let us suppose we have the set [3 1 6] and wish to identify its prime form. We align three slides with these numbers at the top:
3 1 6 (in black) 4 2 7 5 3 8 6 4 9 7 510 8 611 9 7 0 10 8 1 11 9 2 010 3 111 4 2 0 5 3 1 7 etc. 0 2 9 (in red) 11 1 8 10 0 7 911 6 810 5 7 9 4 6 8 3 5 7 2 4 6 1 3 5 0 2 411 1 310 0 2 9 etc.By inspection, we find that the row containing "2 0 5" indicates the prime form, because these numbers, arranged in ascending order(025), form the smallest number of those that result when the numbers in each of the rows are so arranged. To use the matrix as it is now aligned to find transpositions, use the column(or slide)containing "0" in the prime form--in this case "0" appears in the second column. Thus, the numbers in the second column of the matrix can be used to identify transpositions of all equivalents.
I used the areas in the"margins" of the slide rule (to the left of the left guide and to the right of the right guide) to display all the prime forms of the set universe in ascending order--e.g., 012, 013, 014,--with the ordinal number of the short name in parentheses after the prime form; thus:
012(1)
013(2)
014(3>
etc.
Specifically, this shows, for example, that prime form {0 1 2} is
also called "3-1."
Transpositional relationships between equivalents can easily be determined by the index column--that is, the column containing "0" in the prime form. For example, in the matrix for {0 1 5} shown above, [6 4 9] has an index of 4 since that number is in the second column (the column containing ) in the prime form. Finding the transpositional relationship of this set to [1 11 4} is simply a matter of subtracting 4 from 11, 11 being the number in [1 11 4] that is in the second column. The difference, 7, is the number of semitones [1 11 4] is "above" [6 4 9].
Common tones, or invariances, between equivalents are easily found by inspection. In the partial matrix of {0 1 5 6} shown here, the presence of pc 6 is easily seen in three of the equivalents and the dyad [5 6] in two of them:
0 1 5 6
1 2 6 7
2 3 7 8
3 4 8 9
4 5 910
5 61011
It is easy to find common tones between members of different sets
simply by setting up the matrices of those sets side-by-side on the Slide
Rule. The matrices for {0 1 5 6} and {1 2 3 6}, shown in part below, reveal
that the dyad [5 6] is common to the prime form of the first set and of
t3 of the second:
0 1 5 6 0 2 3 6 etc. 1 3 4 7 2 4 5 8 3 5 6 9Although the interval vectors of each set can be accessed in Forte's Appendix I, in the absence (OR presence) of that appendix they can be derived from a matrix in the following manner:
To find the inverse of a set using the matrix, take any of
the numbers in that set and find its inverse on the same slide in the opposite
"side" of the matrix (i.e., in the other color). The row in which that
inverse occurs is the inverse of the whole set. e.g.:
In the matrix of {0 2 3 6} to find the inverse of the prime form
we can look for "0" in the leftmost slide in red (assuming that the prime
form is set up in black). The row containing "0" in red on the leftmost
slide indicates the set [0 10 9 6], which is the inverse of [0 2 3 6].
TOC
12-TONE ANALYSIS
Using all 12 slides, the Slide Rule can be used to display all 48
forms of a row (row-class). Each is easily identified by its initial pc
number. To illustrate, let us take the row of Schoenberg's Fourth String
Quartet: 0 11 7 8 3 1 2 10 6 5 4 9. The P transpositions are shown
as the rows descend--P1,P2,---P11. In red the I forms appear in "reverse"
order--I0, I11,---I1. The R and RI forms are read from right to left, their
transposition number being identified (as by many theorists)by the leftmost
pc number.
Hexachordal combinatoriality can be determined by inspection
of hexachords. For example, note the first hexachord of P0 and inspect
the hexachords of the remaining forms for the same pc content. In the matrix
of the Schoenberg row we find the pc content of the first hexachord of
P0 ( [0 11 7 8 3 1] ) in the second hexachord of I5 ( [3 7 11 0 1 8] ).
This not only reveals that the row is hexachordally combinatorial, but
also that it is so at the level of I5.
Unordered sets can be discovered within rows and row-forms
by thinking of the entire matrix as the union of submatrices--that is,
by using only some of the slides at one time.
Every unordered set is contained in every row-form since each form
contains all 12 pcs. To begin, let us determine the prime forms for the
tetrachords ( i.e., order numbers 0-3, 4-7, 8-11) in the Schoenberg row.
To do this, we treat the leftmost four slides as a submatrix, and the same
for f the four slides on which order numbers 4-7 and 8-11 are represented,
respectively. Using the method for determining prime form shown above,
we fing that the first tetrachord is {0 1 4 5} (this prime form is found
in I0), the seond is {0 1 2 5} (see I3) and the third is also {0 1 2 5}
(see P8].
If we do the same for trichords, we find the succession {0 1 5},
{0 2 7}, {0 4 8}, and {0 1 5}, but more interestingly, if we analyze EVERY
trichord formed by adjacent row members, we find {0 1 5} in the order-number
sets 0-1-2, 2-3-4, 7-8-9, and 9-10-11.
THE INTERVAL-CLASS SERIES(IC SERIES)
A twelve-tone row is most generalizably represented not by its P0
(the precompositional set), but by the ordered set of directed interval
classes that relate the successive members of the row. This can be represented
by ic numbers preceded by a + or - (except for pc 6, for which + and -
entail yield the same result). Thus the ic series for the P forms of the
Schoenberg row is:
-1-4+1-5-2+1-4-4-1-1+5.
Specifically, the first "-1" represents the directed ic relationship between order number 0 and order number 1 in any P form--in P0, these order numbers are pc 0 followed by pc 11 and 0-11 = -1. Similarly, from pc 11 to the following pc 7 in P0 the directed ic is -4, etc. This ic series holds for all P forms, thus is more general than P0 alone. For I forms, the series has the same order of numbers with signs reversed:
+1+4-1+5+2-1+4+4+1+1-5.
The series for R is I series in reverse:
-5+1+1+4+4-1+2+5-1+4+1
because when pitches are reversed, so are the directions they move from one to the next.
As one might expect, then, the series for RI is the series for P in reverse:
+5-1-1-4-4+1-2-5+1-4-1.
The ic series can be used (in the absence of a matrix or magic square) to construct any form, given its first pc. It can also be used to analyze statements of rows in 12-tone works. The ic series can also be used to discover ordered pc sets that are common to more than one form. This is done by using the following formulas and procedure:
FINDING ORDERED SETS
1. Examine the ic series for subsets whose elements are
1) the same,
2) the same, but with opposite signs,
3) the same, but reversed, or
4) the same, with opposite signs AND reversed.
To illustrate, using the Schoenberg series:
-1-4+1-5-2+1-4-4-1-1+5
we find three two-number elements that have these similarities:
1. [-1-4] and its reverse [-4-1] 2. [-4+1] and its reverse [+1-4]
3. [+1-5] and with opposite signs [-1+5] If such similarities can be found
within an ic series, the same pcs can be found in the same order in more
than one row form.
Formulas for each of the four possibilities given above follow:
1. If the elements of a subset are the same, the pcs represented will appear in the order numbers of the second subset in the form Px-y, where x is the first pc of the first subset and y is the first pc of the second subset AND the pcs of the second subset will appear in the orcder numbers of the first subset in the form Py-x.
To illustrate, suppose that order numbers 0-2 in P0 consist
of pcs 0-5-4. The ic series for these pcs is therefore [+5-1]. Now let
us suppose that in order numbers 3-5 there is the progression 9-2-1, which
also has the series [+5-1]. Then according to the formula, x = pc 0, the
first pc of the first subset, and y = 9, the first pc of the second subset,
and the pcs of the first subset will appear in order numbers 5-7 in P3(0-9).
AND the pcs of the second subset will appear in the order numbers of the
first subset in P9(9-0). Let us show these three row forms as far as order
number 5:
P0: 0 5 4 9 2 1- - -
P3: 3 8 7 0 5 4- - -
P9: 9 2 1 61110- - -
2.If the elements of the second subset are the same, but with opposite
signs, the pcs of the first subset will appear in the order umbers of Ix+y
and the pcs of the second subset will appear in the norder numbers of the
first subset also in Ix+y, where x is the first pc of
the first subset and y is the first pc of the second subset.
To illustrate:
P0:0 5 4 7 2 910
I2:2 9 107 0 5 4
Here the first subset contains [0 5 4] and the second [2 9 10].
The ic series is +5-1+3-5+1. The formula results in I2(0+2).
3.If the elements are the same, but reversed, the pcs of the first
subset will appear in the order numbers of the second subset in the form
RIx+y where x is the first pc of the first subset and y is the last
pc of the second subset and the pcs of the second subset will appear in
the order numbers of the first subset in the same form.
To illustrate:
P0:0 4 1 6 3 8 5 9 RI9:- - - - - - - 0 4 1 6 3 8 5 9(Note that in this case the subsets have four elements.)
4. If the elements are the same, but with opposite signs and reversed, the pc of the first subset will appear in the order numbers of the second subset in Rx-y where x is the first pc of the first subset and y the last pc of the second subset. The pcs of the second subset will appear in the order numbers of the first subset in Ry-x:
P0:0 4 1 6 9 5 R7:- - - - - - - 0 4 1 8 711 R5: 10 211 6 9 5I have written a program in BASIC that accomplishes all of these and displays all the results. For a copy send me a formatted disc and a stamped, self-addressed envelope to 100 Morningside Drive/Kansas City/MO 64113
EXTRACTED UNORDERED SETS
It is well known that Schoenberg manipulated row statements to create
set-class unity by leading instrumental parts between nonadjacent row members,
using that freedom to allow a specific set-class to be heard several times.
A composer working with 12-tone procedures can prepare a chart that shows
all the possibilities for deriving a given set-class among nonadjacent
members of row that is used, using the matrix. The process is:
Make two columns. Label the first "row form" and the second "order
nos." Then examine each row of the matrix (in the sense of "row" vs. "column")
for the order position of the numbers in the prime form of the set-class.
The row is then entered in column 1 as "P0" or "P1"---"I11" and the order
numbers identified in the second column. To illustrate, let us begin a
chart that show all of the possibilities for deriving set-class {0 1 5}
in the Schoenberg row. For completeness, let us include possibilities where
the pcs are adjacent:
Examining P0---0 11 7 8 3 1 2 10 6 5 4 9 for the numbers in the
prime form of {0 1 5}, we find that pc0 is in order position 0, pc1 is
in order position 5, and pc 5 in order position 9; therefore, we enter
"0, 5, 9" in the second column . Moving to the second row of the matrix--P2,
we find that the pc numbers 0, 1, and 5 are found in order positions 1,
0, and 10. Thus the first column entry is "P2" and the second column entry
is "1, 0, 10". In this manner we make 24 entries in the two columns. This
completes the chart and we are now in a position to associate certain order
number groups in ANY of the 48 forms with the set-class {0 1 5}. For example,
no matter what row form is chosen, if orders no. 0, then 1, then 10 are
heard, the three pcs will be equivalent to {0 1 5}. The same is true for
order nos. 1, 0, and 10. Thus the ordering of the row is satisfied while
a desired unordered set-class is produced.
The weights to be moved during exercise are placed on a weight bar--I
use five pounds on each side. Nylon cords are tied at one end near the
weights on the bar and run up to, behind, and through the U-shaped rings.
A single cord is used for each arm. The cord is run through holes drilled
near the ends of a tubular piece of wood. The wood, which serves as a handle,
is about 4 inches long and an inch wide. The cord, which goes "in" one
hole and "out" the other, is led back through the ring and down to be fastened
to the bar near the other end of the cord.
Thus, the machine consists of a support that allows weights
on a bar to be raised and lowered by means of cords attached at one end
to the bar and that run through two handles. The amount of weight, of course,
can be changed at will. Other factors, such as the height of the support
and the lengths of the cords can be changed to suit the desires of the
user.
EXERCISING
The machine allows gentle and smooth resisted pulling and pushing
in almost any direction while the user is standing (it ought to be possible
also to sit while exercising--I haven't tried it). My routine consists
of the following:
DAY 1: Standing very near the support and facing it, I lift
straight up with my palms toward the wall (8 repetitions), then, bending
down to accomodate the length of cord, I push down (avoiding collisions
between my hands and the rising bar) 8 times, then standing up again, lift
8 times with palms facing me and finally, bending down, push down
with palms facing inward. I then repeat the routine. I vary the routine
by standing with my back to the machine and lifting with palms facing away
from me. This is like the overhead press using the bar directly.
DAY 2: 1. Facing the machine and about three feet away, I
pull the handles all the way into my body, keeping palms down--ten repetitions
with my left foot forward and ten with my right foot forward.
Interspersed with these lifts is some chinning (not on the machine!),
then some pulls to the side,using the machine: facing it, about
two feet from it, I hold the handles with palms facing each other and twist
my whole body to one side as far as I can 8 times, then to the other side
8 times. More chinning, then back to the twist for 5 repetitions on each
side.
2.Turning around, my back to the machine, I push away from my body
horizontally as before (10 with left foot forward, 10 with right foot).
I vary this by pushing with one hand at a time.
3.I stand closer to the machine, facing it, and pull outward with
my hands moving in opposite directions, then inward. The hands cross over/under--I
change the hand that goes over each time. 15 repetitions out and in. I
vary this by pulling with one hand at a time.
4.I stand a little further away, with palms down, and alternately
pull down and toward my sides and lift up and over my shoulders. This is
varied by one hand at a time and by pulling down with one hand while pulling
up with the other.
Having gone through these routines, I repeat them, this time with
palms facing in the direction opposite to the earlier one. In addition,
for the "facing-away" portion of 1. I vary by holding my arms out straight
to my side at various starting positions, varying from straight down to
straight up, and pull forward, like a baseball pitcher using both hands
(one hand at a time is also good).
After these lifts I do some chinning (not on the machine!), interspersed
with some pulls to the side,using the machine: facing it, about
two feet from it, I hold the handles with palms facing each other and twist
my whole body to one side as far as I can 8 times, then to the other side
8 times. More chinning, then back to the twist for 5 repetitions on each
side. (I finish this session with some lifting directly with another weight
bar.)
TOC
ORCHESTRATION ANALYSES
Analysis #1
RIMSKY-KORSAKOV, SCHEHERAZADE, mm. 1-13
The opening 6 measures is a "unison" passage that descends. Its
highest pitch is E4, which makes it dark, and it uses all instruments
that can play the soprano notes of the line except the horns and trumpets.
The horns and trumpets are omitted after the first note possibly
because they would not in be in their best playing registers.
The line of the first 6 measures is doubled in three octaves. The
tuba leaves the lowest octave in m. 2, again, possibly because the composer
thought that, if it continued, it would not sound its best that low, and
that it would sound more effective in the middle octave from that point.
In m. 3 the trill is not given to to the trombones, tuba, or bass.
It could have been given to the tuba and bass, but Rimsky might have thought
their trills would be unclear. In the two scores I've seen, the marking
for the clarinets shows a trill sign above the staff only, which could
mean that the second clarinet does not trill. I assume that it does. The
effect of some instruments trilling while others do not is, to my ear,
a bit muddy or "messy."
The violins do not play the last note before the GP, of course,
because they can't. This is not noticed because there enough instruments--especially
the trombones--that do play it with strength.
After the GP, the trombones are omitted, with a considerably softening
effect that is more effectively achieved than a simple "diminuendo" mark
would be, and it enforces a change of color. The low clarinets emerge here.
The next passage is a woodwind group that sets a simple, chorus-like
phrase in a very non-straightforward manner. This passage surely was inspired
by the beginning of Mendelssohn's Midsummer Night's Dream overture, which
it resembles in obvious ways.
M. 8: The middle register clarinets enclose the low register
flutes.
M. 9: fl. 1 crosses over cl. 1 to continue the soprano while
cl. 1 crosses over both flutes to take the tenor. The clarinets are in
their weakest register and the vibrato of the flutes (as opposed to the
non-vibrato of the clarinets) gives greater warmth to the passage. Otherwise,
the change in color from clarinet to flute in the soprano is hardly noticeable
because of their respective registers.
M. 10: As the texture expands in contrary motion of the outer
parts, bn. 1 enters on a relatively weak G4, taking over the tenor line
from cl. 1 and avoiding parallel fifths in the clarinets. Entering in the
middle of the texture, its new sound (double reed) is less noticeable than
it would be as the bass. (In the Mendelssohn overture, instruments are
added with each chord, and always BELOW the previous notes.)
M. 11: Here is a considerable change in scoring, one
that matches the change in tonal implication (the F# major triad, which
implies V in B minor, is surprising after the modal E minor beginning):
fl. 1 drops out (if it continued, it might be superflous?) and is
replaced by ob. 2 while ob. 1 doubles fl.1 (the first unison doubling).
It is in this pitch area that the flutes and oboes match most in quality
and strength; thus the change in color is the least it could be for these
instruments. But the passage becomes suddenly more intense with the introduction
of three more double reeds--bn. 2 joins in, like bn. 1, as an inner voice
to minimize intrusiveness, while cl. 2 continues as a warm and increasingly
resonant bass.
M. 12: In the final chord the flutes double the oboes an octave
higher in the soprano and alto. Along with that, the cl. 1 suddenly jumps
up over the oboes and flutes (of m. 11) to double the 2nd alto note A while
cl. 2 likewise crosses the bassoons to double the bass an octave higher.
The bassoons unite on the bass note that "should have been given to cl.
2," while hn. 1 enters to take the tenor that "should have been given to
bn. 1," again, disguising its presence by being surrounded, pitch-wise.
The register of the horn is borderline in terms of balance--very much higher,
and it would become rather intense.
The final chord completes the gradual increase in the number of
instruments while keeping the spacing between instruments vertically and
reinforces the implied crescendo by the increased strength of the flutes
and cl. 1, which are now in stronger registers, adding to the greater intensity
of more double reed instruments. The manipulation of the lower instruments
avoids parallel fifths in every chord change! The sudden octave doubling
of the soprano prepares the surprising and effective entrance of the solo
violin on E6, which, with the harp (playing the whole chord an octave lower),
suggests an abrupt scene change to the heroine, singing to the harp!
The crossings of instrumental lines are less noticeable than they might
be in other scorings because the instruments are all woodwinds (including
horn?) and mostly in inner parts and because the overall voice-leading
(irrespective of scoring choices) is mostly stepwise.
Copyright c by Merton Shatzkin 1999
TOC
ANALYSIS #2
WAGNER, TRISTAN
UND ISOLDE, mm. 1-25
This excerpt illustrates two techniques very well--highlighting
(using one or more instrument(s) to double only one or a few notes of an
ongoing line in other instruments) and linking (the transfer of a line
from one instrument to another).
mm. 1-2: The vc. are in their strongest and most singing register.
The tempo is probably too slow for the slur to be executed literally--the
players probably need to change bow--I recommend that they change in different
places, to retain the effect of the slur.
mm. 2-3: The e.h. links from the vc. on D#4. Ob. 1 and the
bassoons are highlighted by ob.2 and the clarinets, respectively, creating
an emphasis on the downbeat. Each of the four pitches is played by two
"instruments" (counting the vc as "one"). After the clarinets and
vc drop out, only double reeds are left, which should make a good blend.
mm. 4-7: This is a sequence fo mm. 1-3 with similar format: the
oboe and clarinet roles are reversed. This causes a mixture of single-reed
(clarinet) and double reeds. The clarinet, now in a stronger register,
is more effective in the soprano than it would have been in the first phrase.
mm. 8-11: The vc. are now quite high and somewhat strained-sounding.
The pitch relationship between the three opening phrases is such that the
vc. leap up to the first note of the previous woodwind phrase: e.g., to
G#4 in m. 5, which retakes that note in m.2 and to B4 in m. 8, which retakes
that note in m. 6, thus creating a sort of delayed, or implied, linking.
This phrase is another quasi-sequence, but with different harmony.
The instrumental assignments are the same as in mm. 1-3, with the addition
of hn. 2, which (together with cl. 2, which is in a weak register) highlights
C4.
We see that in m. 10 cl. 1 highlights the soprano ob. 1, whereas
in m. 2 ob. 2 is used for the same purpose. This "change" is possibly because
the clarinet is now in a stronger register.
mm. 12-13: This is an echo, 8va, of mm. 10-11. The phrases,
from the beginning, exhibit an alternation in the sopranos of double-reed
and non-double-reed (oboe, clarinet,oboe, flute). Fl. 1 is in a brighter
register than it would have been for any of sopranos of the previous phrases.
Now, Wagner apparently did not want emphasis on the downbeat--there is
no highlighting.
mm. 14: The violins (reinforced by the lower octave) echo
the soprano of m. 13 and m. 11. This resumes the alternation
between woodwinds and strings.
m. 15: The alternation and echoing continues. The upper octave
is given to only two flutes while there are two clarinets and e.h. on the
lower octave. For balance purposes, the stronger register of the flutes
requires more instruments below since these instruments are in weaker registers.
Also, including e. h. in the mix retains the instrumental sound of the
previous woodwind phrase.
mm. 16-17: The adding of instruments helps the notated crescendo
from m. 14. The lower-string pizzicato highlights the notes marked "sf",
and is resonantly full, covering many notes of the harmony, stretching
from the bass to just below the soprano, A secondary line, G#4-B4-A4,
important because it recalls those notes (in rearranged order) in mm, 2
and 3, is brought out by doubling in ob. 1 and hn. 3, both in strong registers.
The surprising F major triad ("deceptive" resolution of V) in m.
17 is emphasized also by fuller texture, but evaporates when the vc. return
melodically, taking the violins' last note an octave lower. The bass
clarinet, unobtrusively entering, takes the bass line of a sustained woodwind
group.
mm. 18-21: The vc. are accompanied by woodwinds continuing
from mm. 16-17, now articulated by the addition of vla.-cb. pizzes (this
resumes the pizz. of m. 16 in a softer manner). In m. 20 hn. 3 holds the
G5 that the vc. have arrived at while the vc. move away and back to it
while hn. 4 reinforces bn. 2 below.
The vc. are in the same register as they were in mm. 1-10 and again
rise sequentially, again reaching Bb 5 as part of the climax (as they did
in m. 9). In m. 21, on the surprising Neapolitan sixth chord (beat two)
on a hushed "p" marking, the pizzes are replaced by sustained arco and
the warmer-sounding strings replace the bassoons.
In m. 22, second beat, the vn. 2 cross over the vc. to take over
the soprano, having entered as if they were continuing the viola part.
They play on the G string with its darker, huskier sound--otherwise, they
might have sounded thin after the vc. In m. 23 vn. 1 takes over the line,
linking from vn. 2's D#4, while the vc. move to a unison C#4 with vn. 2
and then drop a seventh, anticipating the same notes an octave higher they
will play when they resume the lead in m. 25. Vn. 1 move between the same
notes in reverse order (from D#4 up to C#5) as they ascend, with increasing
intensity and straining on the G string. At the high point, after
the horns and bassoons come in to fill out the lower part of the texture,
the vc. take over the line again, linking with the vn. 1.
Copyright c by Merton Shatzkin 1999
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