Scalar Product of Orthonormal Basis: Equal to 1?

Tony Stark
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What is the scalar product of orthonormal basis? is it equal to 1
why is a.b=ηαβaαbβ having dissimilar value
 
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There is no such thing as a scalar product of a basis. The only possible interpretation of your question is what the scalar product between two vectors in an orthonormal basis is. The answer is that if you take two different vectors of the basis, it is zero (this is the "normal" part of orthonormal) while if you take the inner product of one of the vectors in the basis with itself you get ±1 depending on whether you chose a timelike or spacelike basis vector.
 
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Orodruin said:
There is no such thing as a scalar product of a basis. The only possible interpretation of your question is what the scalar product between two vectors in an orthonormal basis is. The answer is that if you take two different vectors of the basis, it is zero (this is the "normal" part of orthonormal) while if you take the inner product of one of the vectors in the basis with itself you get ±1 depending on whether you chose a timelike or spacelike basis vector.
What do you mean by INNER PRODUCT
 
Orodruin said:
The answer is that if you take two different vectors of the basis, it is zero (this is the "normal" part of orthonormal)

Actually, it's the "ortho" part, correct? The "normal" part is the ##\pm 1## you get when you take the inner product of a basis vector with itself.
 
PeterDonis said:
Actually, it's the "ortho" part, correct? The "normal" part is the ##\pm 1## you get when you take the inner product of a basis vector with itself.

Indeed, writing faster than thinking. :rolleyes:
 
Scalar product, dot product, and inner product all mean the same thing AFAIK - wiki agrees, though I wasn't very fond of the rest of the wiki article, entitled "dot product".

I'd also interpret
What is the scalar product of orthonormal basis?

as "what is the scalar (or dot, or inner) product of the basis vectors in an orthonormal basis", though I'd stop short of insisting that that's what the OP must have meant in favor of attempting to try and find out if that's what they meant. I agree with the answers that have already been given - I hope they have cleared things up for the OP.

I also can't resist asking - doesn't Hartle cover this? I don't have the book, I would have thought it would have been covered in the text.
 

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