On Mar 9, 2:37 pm, Koobee Wublee wrote:
On Mar 9, 12:10 am, JanPB wrote:



On Mar 8, 10:10 pm, Koobee Wublee wrote:
Given

(ds)^2 = g_ij dq^i dq^j

Where

** i, j = 1, 2

Then,

(ds)^2 = g_11 dq_1 dq_1 + g_12 dq_1 dq_2 + g_21 dq_2 dq_1 + g_22 dq_2
dq_2

So, wise guy, go ahead and assign the numbers to g_11, g_12, g_21,
g_22, dq_1, and dq_2. Then, you will find a number for (ds)^2.
shrug

Where is the number? You said ds^2 was a scalar. Which number is ds^2
equal to on the sphere? Either write down the actual number or say
that you were wrong.

Oh, no. I am not wrong on this one.

If you are either incompetent in arithmetic or incapable of using a
calculator, as soon as you give me the numbers, g_11, g_12, g_21,
g_22, dq_1, and dq_2, we will give you an answer.

You said ds^2 was a scalar. If this is so, then why can't you tell us
what number is ds^2 equal to on the sphere?

There is no other exit.

In the meantime, keep whining in that fat castle in the air. It is
crumbling as we speak. shrug

I'm buying off Microsoft, did I tell you?

--
Jan Bielawski