Re: [tied] Re: Slavic palatalistions: why /c^/, /c/?

On Sat, 29 Oct 2005 08:51:30 +0000, tgpedersen
<tgpedersen@...> wrote:

>--- In cybalist@yahoogroups.com, Miguel Carrasquer <mcv@...> wrote:
>>
>> On Wed, 26 Oct 2005 11:44:30 +0000, tgpedersen
>> <tgpedersen@...> wrote:
>>
>> >
>> >
>> >BTW I had this thought:
>> >Suppose /c^/ ("/c´/") was once the palatalised partner of /c/ (they
>> >are both unmatched now, afaIk). Then original /i/, /e/ would have
>> >palatalised (result /c´/ > /c^/) and *ay > /E/ wouldn't
>(result /c/).
>>
>> There were more vowels in Slavic (to wit: a/o, a:/o:, u, u:,
>> au, aN), where the result is simply /k/.
>>
>
>True, but does it disprove the idea?
>AfaIk, PIE *i > Proto-Slav. *I which palatalises, so that would fit in
>with my idea, which goes something like:
>1) P.-Slav *k > c before front vowels (including *E)

I suppose by "E" you mean ê < *ai?

>2) P.-Slav. I palatalises the preceding consonant (standard theory)
>3) /c/ becomes /c^/ if palatalised.
>
>
>What does PIE *e become in Proto-Slavic?

/e/.

The actual developments can be given as follows:

PIE PS
*ka/*ko *ka > ko
*ke *ke > c^e 1st.pal.
*ki *ki > c^I 1st.pal.
*ku *ku > kU

*ka:/*ko: *ka: > ka
*ke: *ke: > c^a 1st.pal., ê > a
*ki: *ki: > c^i 1st.pal.
*ku: *ku: > ky
*kai/*koi *kai > cê 2nd.pal.
*kau/*kou *kau > ku
*kei *kei > c^i 1st.pal.
*keu *kjau > c^u j-pal. (au=au)

*kja/o *kja > c^e j-pal., a > e
*kju *kju > c^I j-pal., u > i > I
*kja:/o: *kja: > c^a j-pal., a: > e: > ê > a
*kju: *kju: > c^i j-pal., u: > i: > i
*kjai/oi *kjai > c^i j-pal., ai > ei > i: > i
*kjau/ou *kjau > c^u j-pal., au=au (and aN=aN).


=======================
Miguel Carrasquer Vidal
mcv@...