How to Write (a_1+...+a_n)(b_1+...+b_n) in Einstein Notation?

[mnb96]

Hi!
I want to write the following expression in Einstein notation:

(a_1+\ldots+a_n)(b_1+\ldots+b_n)

I tried it by introducing the term I^{ij}=1 and writing:

a_ib_jI^{ij}

Are you aware of another, more convenient form for this?

 

[CFDFEAGURU]

According to the Einstein summation convention, when ever a expression contains one index as a superscript and the same index as a subscript a summation is implied over all values that the index can take. (As defined in Schutz "A First Course in General Relativity")

So what do you need the I super ij for?

I would just write it as

a (superscript i) * b (subscript i) for all values that i can take.

(My LaTex usage is horrible that is why I typed it out.)

Thanks
Matt

 

[Rasalhague]

So what do you need the I super ij for?

I would just write it as

a (superscript i) * b (subscript i) for all values that i can take.

In general

(a_1+\ldots+a_n)(b_1+\ldots+b_n) \neq \sum_{i=1}^{n}a_i\,b_i

 

[mnb96]

You probably read too fast: a^ib_i=a_1b_1+\ldots+ a_nb_n and that clearly does not produce all the combinations.
I'm trying to figure out if I really need to introduce that I^{ij}=1 term, or if I can express that by using the kronecker delta or the levi-civita symbol.

 

[CFDFEAGURU]

Yes, that is correct, but I said one is a superscript and one is a subscript. You have them written as both subscripts.

Yes, I did overlook the parenthesis.

Would it not then just be

a (superscript i) a (subscript i) * b (superscript 1) b (subscript i)

?

Thanks
Matt

 

[CFDFEAGURU]

Yes, thanks torquil.

Matt

 

[torquil]

You probably read too fast: a^ib_i=a_1b_1+\ldots+ a_nb_n and that clearly does not produce all the combinations.
I'm trying to figure out if I really need to introduce that I^{ij}=1 term, or if I can express that by using the kronecker delta or the levi-civita symbol.

The problem with this is that if you construct an expression out of a, b, the kronecker delta and/or the levi civita symbol, and it has no free indices, it will be rotationally invariant. Your original expression is not rotationally invariant, so you need something else, e.g. your matrix of 1's.

Torquil