This thread (related to this problem) can be split into two questions, the first one being restricted to Ancient Greek, the second one being more general.

(1) Let's be, by example, two syllables, the first one having an onset and a nucleus, the second one beginning with a lengthened consonant due to syllabication. By example, Ancient Greek has γλῶττα, standing for [ glˈɔː . t̪ːa ] (from Sidney Allen, Vox Graeca, page 12 : "wherever the normal spelling writes a double consonant, it stands for a correspondingly lengthened consonant.")

My first question : is the first syllable a closed one ? I read in John Laver, Principles of Phonetics, p. 32 : "When a syllable ends in a vowel, with no final consonant, it is said to be an open syllable". So the question becomes : does the first syllable has a coda or not ?

Now, from Michel Lejeune (Phonétique historique du mycénien et du grec ancien), I read, page 70 (my translation) : "Geminate occlusives are the (phonemes) whose duration (=French 'tenue') is sufficient to be heard by ears and whose implosion and explosion, both audible, belong to two different syllables". Same statement in Laver's book (page 437) about another language (Nubian) having "consonants being geminated (being made long, across a syllable boundary)".

From Lejeune's statement it seems clear that [ glˈɔː . t̪ːa ] can be analysed as if the first syllable had [t̪] in its coda, the second syllable beginning with another [t̪], the first syllable being consequently a closed one.

(2) But I know that other languages have natural lengthened consonants at the beginning of a syllable. By example, (in Laver, p. 437) Pattani Malay has [lːabɔ], [sːiku], ... For these languages, if a syllable without a coda is followed by a syllable beginning with a lenghtened consonant (e.g. [pa . lːabɔ], an example I forged for the occasion), is the first syllable [pa] a short or a long one ? The word [lːabɔ] beginning with a long consonant, can this word somehow "share" its consonantal lenghth with the preceding word or is the syllabic boundary "impassable" ?

  • By "tenue" Lejeune presumably means "duration (of closure)"; "tense" isn't a correct translation here.
    – TKR
    Nov 7, 2013 at 17:48
  • I fixed my translation : thank you very much.
    – suizokukan
    Nov 7, 2013 at 22:05

1 Answer 1


To answer (1): in Greek, geminates syllabify across a syllable boundary, so the first syllable is closed, e.g. γλῶτ.τα. We know this for two reasons:

  1. We know that Greek doesn't allow a syllable to begin with a geminate, because if it did, there should be words where this occurs initially (e.g. *ττα) or after a consonant (e.g. *γλῶρττα), and such words don't exist.

  2. Vowels followed by a geminate make a heavy (a.k.a. "long") syllable, as can be seen from their scansion in poetry. E.g. in Iliad 1.3 πολλὰς δ’ ἰφθίμους ψυχὰς Ἄϊδὶ προΐαψεν, the first syllable must be heavy to satisfy the meter, so we know it's πολ- not πο-. A similar, non-metrical source of evidence is the comparatives and superlatives of o-stem adjectives, whose form varies according to the weight of the preceding syllable: -ότερος/-ότατος after a heavy syllable, -ώτερος/-ώτατος after a light syllable. Adjectives with a geminate before the -o- take the former set: περιττός, περιττότερος, περιττότατος. This shows that the syllabification is πε.ριτ.τός not πε.ρι.ττός.

  • A very detailed answer : thank you very much.
    – suizokukan
    Nov 8, 2013 at 5:45

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