Vector Cubed: Understanding the Expansion of (a_x + a_y + a_z)^3

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Homework Help Overview

The discussion revolves around the mathematical interpretation of the expression (a_x + a_y + a_z)^3, specifically in the context of vector operations and their properties. Participants explore whether this expression can be simplified or interpreted meaningfully within vector algebra.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants question the validity of "squaring" or "cubing" vectors and discuss the implications of dot products and scalar products in this context. There is exploration of whether the expression could represent the magnitude cubed or if it holds any mathematical meaning at all.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants suggest that the expression may not be mathematically defined, while others propose that it could refer to the magnitude of the vector raised to the third power. Clarifications regarding notation and context are also being sought.

Contextual Notes

There is mention of confusion arising from notation and the context in which the expression was encountered, indicating that participants are grappling with the implications of their calculations and the definitions of vector operations.

MadMax
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Ok so if I have (a_x + a_y + a_z)^2

I get a_x^2 + a_y^2 + a_z^2 as the answer because the cross terms go to zero. For example a_x . a_y = 0 etc...

What if I have (a_x + a_y + a_z)^3 ?

Will the answer be a_x^3 + a_y^3 + a_z^3 , or will it be a_x^3 + a_y^3 + a_z^3 + a_x^2 a_y + a_x^2 a_z + a_y^2 a_x + a_y^2 a_z + a_z^2 a_x + a_z^2 a_y ?

My motivation for thinking it might be the first one is that a term like a_x^2 a_y could be expanded into one like a_x . a_x . a_y and then would go to zero if we take the dot product a_x a_y . On the other hand the factor a_x . a_x is a scalar and a scalar a_x^2 times a vector a_y will remain as a_x^2 a_y . On the.. third.. hand if we take the supposed commutative properties of a dot product then (a_x . a_x) . a_z should be the same as a_x . (a_x . a_z) which I seem to argue is not the case. We cannot argue that a_x . (a_x . a_z) != 0 but we can argue that (a_x . a_x) . a_z = a_x . (a_x . a_z) = 0 ... So I guess the first anwer is the right one.

Confirmation or help in the right direction would be much appreciated. Thanks.
 
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I don't think you can actually "square" or "cube" a vector. What you can say is that the dot or scalar product of a vector with itself is equal to square of the modulus of the vector, i.e.,

[tex]\vec{r}.\vec{r} = \left|\vec{r}\right|^2[/tex]​

But [tex]\vec{r}.\left(\vec{r}.\vec{r}\right)[/tex] is meaningless, since the term within the parentheses is a scalar, and scalar products are defined only between vectors.

The multiplication defined for vectors is different from the multiplication defined for real numbers.
 
Ya. Basically, a dot product of two vectors is a scalar. If you try to take another dot product with a dot product, you end up with a vector dotted with a scalar which is meaningless unless you are merely multiplying the vector by the factor of the scalar in which you would be using two different types of multiplication from squaring to cubing.
 
How would you interpret [tex](\mathbf{a}_x + \mathbf{a}_y + \mathbf{a}_z)^3[/tex], where [tex]\mathbf{a}_l = |a_l| \hat{i}[/tex], ?
 
MadMax said:
How would you interpret [tex](\mathbf{a}_x + \mathbf{a}_y + \mathbf{a}_z)^3[/tex], where [tex]\mathbf{a}_l = |a_l| \hat{i}[/tex], ?


Sum of three vectors raised to the power 3? That's not defined mathematically.
 
So if I have [tex]\mathbf{a}^3[/tex], where a is a vector, the only thing it could mean is [tex]|\mathbf{a}|^3[/tex]?
 
Last edited:
MadMax said:
So if I have that, the only thing it could mean is [tex]|\mathbf{a}|^3[/tex]?

It can't be...it's not a meaningful expression.

Maybe you came across this

[tex]\left(\mathbf{a}.\hat{i} + \mathbf{a}.\hat{j} + \mathbf{a}.\hat{k}\right)^3[/tex]​

or something similar?

EDIT: I just read your edited post. Maybe what you read was just the magnitude cubed. In what context did you come across this term?
 
Last edited:
MadMax said:
How would you interpret [tex](\mathbf{a}_x + \mathbf{a}_y + \mathbf{a}_z)^3[/tex], where [tex]\mathbf{a}_l = |a_l| \hat{i}[/tex], ?

Did you actually encounter this in your reading or in a problem or are you just curious?

There are only two mathematically defined ways of combining three vectors. That's because the multiplication operations are binary, so you have to select two of the three for the first operation. That first operation has to produce a vector in order to perform the second binary operation with the third remaining vector. So you only get these possibilities:

A · (B X C) , the so-called scalar triple product or

A X (B X C) , the so-called vector triple product.

I can't recall having ever seen an A^3 notation involving a vector. If you tried to apply either of these to the situation A=B=C, you'd get zero since A X A = 0.
 
MadMax said:
So if I have [tex]\mathbf{a}^3[/tex], where a is a vector, the only thing it could mean is [tex]|\mathbf{a}|^3[/tex]?

I think the consensus is that the writer must mean the magnitude cubed. Was the name of the vector notated in boldface or not? If the vector name was just in plain face, it *would* mean the magnitude cubed and is the only way this would make sense. (If it was in boldface, the writer probably just slipped up...)
 
  • #10
Quite possibly. I'm basically doing a calculation and checking the result is the same as what somebody else got. We've both been lazy with notation and now I'm confused. Along the way we had somehting like [tex]\int d^3 \mathbf{b} f(\mathbf{a},\mathbf{b})[/tex]. After integration we both got terms with a^3 in them, and neither of us have indicated what that means exactly... Mind you at one point all the a terms where written as a^2 and then after simplification we had these terms with a^3. I know for sure that a^2 means |a|^2 ... so I guess a^3 must mean |a|^3 .
 
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  • #11
Having read the new posts and considering it must mean |a|^3. Thank you all very much for your help. :)
 
  • #12
Some physics textbooks denote the volume element in a volume integral as [tex]d^3\vec{r}[/tex], but that usually represents the usual dxdydz (in the Cartesian system, of course.)
 

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