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# Vector calculus cylindrical polars

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1. Hi,

I'm studying vector calculus and have a question to ask regarding dot products in cylindrical polars.

In cartesian coordinates the dot product is represented as

A · B = AxBx + AyBy + AzBz

However, I was watching a lecture on the derivation of the divergence of a vector in cylindrical polars and instead of the above cartesian method, the brackets of the dot product were multiplied out.

I was wondering if anyone could explain why this is so?
2. (Original post by a nice man)
Hi,

I'm studying vector calculus and have a question to ask regarding dot products in cylindrical polars.

In cartesian coordinates the dot product is represented as

A · B = AxBx + AyBy + AzBz

However, I was watching a lecture on the derivation of the divergence of a vector in cylindrical polars and instead of the above cartesian method, the brackets of the dot product were multiplied out.

I was wondering if anyone could explain why this is so?
Not sure what you mean by the brackets multiplied out but hope it helps.
3. (Original post by tangotangopapa2)
Not sure what you mean by the brackets multiplied out but hope it helps.
If you watch this lecture

He multiplies out the brackets to give a total of 18 terms. There are twice the number of terms due to the product rule but even if this were not the case there would be nine terms.

Whereas in cartesian coordinates the dot product produces three terms as in

A · B = AxBx + AyBy + AzBz
4. (Original post by a nice man)
If you watch this lecture

He multiplies out the brackets to give a total of 18 terms. There are twice the number of terms due to the product rule but even if this were not the case there would be nine terms.

Whereas in cartesian coordinates the dot product produces three terms as in

A · B = AxBx + AyBy + AzBz
In Cartesian coordinate system with three basis vectors (linearly independent of course), there are nine terms (look at the derivation of the formula above). In special case where the three basis vectors are ortho-normal, six of the terms gets cancelled out as dot product of two perpendicular vector is zero (shown as Kronecker delta in figure above) and hence we are left out only with three terms out of the nine terms.

While in the video there are nine terms at first and each of the nine terms has been split into two applying the product rule of partial differentiation, hence giving eighteen terms.
5. (Original post by tangotangopapa2)
In Cartesian coordinate system with three basis vectors (linearly independent of course), there are nine terms (look at the derivation of the formula above). In special case where the three basis vectors are ortho-normal, six of the terms gets cancelled out as dot product of two perpendicular vector is zero (shown as Kronecker delta in figure above) and hence we are left out only with three terms out of the nine terms.

While in the video there are nine terms at first and each of the nine terms has been split into two applying the product rule of partial differentiation, hence giving eighteen terms.
Im not sure I completely follow...

So when a dot product is calculated it always gives nine terms, with each component of vector a being multiplied with each component of vector b?

However, in the case of orthonormal coordinates (such as i,j and k), six of the terms of the nine produced are equal to zero, leaving us with the three terms that make the familiar equation

A · B = AxBx + AyBy + AzBz ?
6. (Original post by a nice man)
Im not sure I completely follow...

So when a dot product is calculated it always gives nine terms, with each component of vector a being multiplied with each component of vector b?

However, in the case of orthonormal coordinates (such as i,j and k), six of the terms of the nine produced are equal to zero, leaving us with the three terms that make the familiar equation

A · B = AxBx + AyBy + AzBz ?
Yes, you are at the point. So, the following generalisation is presented without any proof.
Dot product in n-dimensions always gives n^2 terms. If orthonormal basis vectors are taken you are left with only n terms as (n^2 - n) terms are 0 which make the fimilar equations like
A · B = AxBx + AyBy + AzBz (In three dimensions) or P . Q = PxQx + PyQy (in two dimensions.)
7. (Original post by tangotangopapa2)
Yes, you are at the point. So, the following generalisation is presented without any proof.
Dot product in n-dimensions always gives n^2 terms. If orthonormal basis vectors are taken you are left with only n terms as (n^2 - n) terms are 0 which make the fimilar equations like
A · B = AxBx + AyBy + AzBz (In three dimensions) or P . Q = PxQx + PyQy (in two dimensions.)
Thanks for the help, I appreciate it!

Can't believe I made it this far without being taught that!

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