From: tgpedersen
Message: 30927
Date: 2004-02-10
> Torsten:I was only trying to cheer you up with a word with many /a/'s, you
> > Bla-bla-bla. Examples?
>
> As much as "bla-bla-bla" is an annoying response, I admit
> to deserving it in this instance, at least in part.
> TheThs is how I see it:
> vowel *a most certainly needs to be reconstructed in IE
> on typological grounds as well as simply comparative
> grounds. It's not true that all instances of *a are
> the product of uvular lowering. A perfect example would
> be *kad- (traditional k^ad-). This root does not show
> evidence of a laryngeal, yet there's *a!
>
> I just made a long, perilous journey out to the SFU to
> check up on some much-needed reading and I specifically
> looked for ablaut on *a-grade verbs. Well... it was slim
> pickin's, lemme tell ya. However, I did find some tiny
> morsels of glosses like Gr /ke-kas-mai/ (< *kad-) which
> seems to show that *a alternated with neither *e nor *o.
> It didn't alternate at all! With this type of reduplication,
> you'd expect *ke-kos-mai.
>
> Then, I started reviewing what I'd already proposed about
> Mid IE and that rule makes sense. So far, I view IE *a
> as being mostly derived from *e due to colouring, with
> a few true instances being resistant to shifting to *o
> during the Late IE vowel shift of *a > *o neighbouring
> labial phonemes. The rest may have once been coloured
> by uvulars but the uvulars were replaced by plain sounds,
> as perhaps with *kad-. So in total, regardless of anything,
> *a is underlyingly derived from the same vowels that would
> produce *e or *o, the ones that participate in ablaut.
>
> One would expect that coloured *a, or that is "phonetic"
> rather than "phonemic" *a, does still participate
> in ablaut, and lo and behold it does: *stax- vs *ste-stox-
> (but only because *stax- is in fact *stex-). Whereas,
> *kad- and *mad-, not being coloured by uvulars, are TRUELY
> *kad- and *mad-. Thus they don't participate in ablaut
> because *a is treated as a "second *o". At least in
> regards to *mad-, this is what *a should have been if
> it weren't for neighbouring *m. So *mad-, being like **mod-,
> is already ablauted and therefore doesn't participate in
> ablaut.
>