4

It's commonly posited that all PIE roots consist of two groups of consonants, neither of which can be empty. For example, the root *h₁ed- has the groups *h₁ and *d.

However, I'm not aware of any language that shows a reflex of initial *h₁. All the derivations I can think of would work just as well if the root was **ed-, with no initial consonant.

So: is there a language which shows a definite reflex of initial *h₁, which requires us to reconstruct *h₁ed- instead of **ed-? Or is the initial *h₁ posited just to make the root structure more elegant?

Improve this question
11
  • 3
    Fortson (in his famous textbook) very briefly addresses this issue on p. 77 (section 4.7), saying that e.g. there is no indirect evidence for h1es- (vs. -es) but in exercise 8 (on p. 86) he leaves it up to us to figure this out. He asks us to think about why we have long a in Vedic Sanskrit ā́sat- 'monster', which is a compound consisting of the prefix a- 'not, un-' and the pres.part. of the verb 'to be' (from PIE * h1es- in the zero-grade), bearing in mind that the a in these forms comes from PIE *n̥. Would you rather think about this exercise and post your thoughts first or?
    – Alex B.
    Sep 9, 2020 at 2:44
  • 2
    cf. Fortson 2010 "Specialists differ on whether to rewrite all traditionally vowel-initial roots in this way when direct evidence of a laryngeal is lacking; the tendency is to add the laryngeal regardless, for the sake of structural uniformity" (p. 77). Basically, it boils down to "Should we disallow vowel-initial root reconstructions in PIE?
    – Alex B.
    Sep 9, 2020 at 2:53
  • 2
    @AlexB. Ahh, so if we posit *n̥-h₁s- instead of *n̥-∅s-, the long ā can be explained as compensatory lengthening from the laryngeal being lost?
    – Draconis ♦
    Sep 9, 2020 at 2:57
  • 1
    Well done! This process is called compensatory lengthening or contraction, VHC > V̄C (also in -VH > -V̄), see §3.20 in Fortson 2010
    – Alex B.
    Sep 9, 2020 at 4:31
  • 3
    There are other such cases as well, where Greek shows a prothetic ε- in zero-grade forms, e.g., from the same verb, ἐσμέν, ἐστέ, εἰσί. Had the root been just *es-, the expected Greek outcome would have been *μέν, *στέ, *εἱσί (the final stress in 3pl is analogical in both cases). Also, there are still some roots where it is common not to reconstruct a laryngeal, for various reasons; e.g., *abōl ‘apple’, which is likely a loan word. So not all PIE roots necessarily had a mandatory initial consonant slot.
    – Janus Bahs Jacquet
    Sep 9, 2020 at 7:46

2 Answers 2

Reset to default
2

Very frequently, Tocharian has y- where *H1 is expected: for example "horse" is yuk. I suppose Fortson does not mention that kind of phenomenon because Tocharian is far from being completely understood.

Improve this answer
2

There is some evidence of initial *h₁ in roots like *h₁es- "be". Alex B in the comments quotes an exercise from Fortson's textbook, Vedic ā́sat "monster" from *n̥-h₁s-. If the root were **∅es-, we would expect **n̥-∅s- > **ás- with a short vowel. But reconstructing it with the laryngeal, *n̥-h₁s- > ā́s- due to compensatory lengthening.

Arnaud Fournet mentions some additional evidence from Tocharian, and Kloekhorst suggests that Anatolian shows an initial ʔ- as a direct reflex of *h₁-, but this is rejected by most other Anatolianists (like Melchert).

Still, though, there are a lot of roots reconstructed with initial *h₁- where there's no direct evidence from any of these sources. Regarding these roots, Alex B gives another quote from Fortson:

Specialists differ on whether to rewrite all traditionally vowel-initial roots in this way when direct evidence of a laryngeal is lacking; the tendency is to add the laryngeal regardless, for the sake of structural uniformity.

In other words, it's elegant to have all roots consist of two groups of consonants, so *h₁- is used when there's no evidence of any other consonant.

Improve this answer

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.