What Happens When You Divide a Vector by Its Zero Magnitude?

[yungman]

Let \vec{A} be a vector with length |\vec{ A}|

\hat{A} \;=\; \frac{\vec{A}}{|\vec{ A}|}

1) What is \frac{\vec{A}}{|\vec{ A}|} If |\vec{A}| = 0?

2) What is |\frac{\vec{A}}{|\vec{ A}|}| If |\vec{A}| = 0?

My answer for 1) is \hat{A} and 2) equal to 1

Please tell me why?

 

[CompuChip]

Well, the only vector whose length is zero, is the zero vector (\vec 0).
And the one vector that you can't stretch or shrink to unit length, is the unit vector.
So IMO, \hat 0 is not defined and the answers to both your questions are basically the same as the answer to "what is x / 0?".

By the way, how do you mean "please tell me why." Should we tell you why you answered A and 1?

 

[yungman]

This is from PDE book by Strauss p194. On the Green's function on a sphere.

The equation is:

G(\vec{x},\vec{x_0}) = -\frac{1}{4\pi \left\|\vec{x} -\vec{x_0}\right\|} + \frac{1}{4\pi \left\|\frac{r_0}{a} \vec{x} -\frac{a}{r_0}\vec{x_0}\right\|}

The book gave

G(\vec{x},0) = -\frac{1}{4\pi \left\|\vec{x} \right\|} + \frac{1}{4\pi a}

when \vec{x_0}=0

If what you said is true, this will not be correct.

Please help.

 

[Mark44]

This is from PDE book by Strauss p194. On the Green's function on a sphere.

The equation is:

G(\vec{x},\vec{x_0}) = -\frac{1}{4\pi \left\|\vec{x} -\vec{x_0}\right\|} + \frac{1}{4\pi \left\|\frac{r_0}{a} \vec{x} -\frac{a}{r_0}\vec{x_0}\right\|}

The book gave

G(\vec{x},0) = -\frac{1}{4\pi \left\|\vec{x} \right\|} + \frac{1}{4\pi a}

when \vec{x_0}=0

If what you said is true, this will not be correct.

Please help.

I don't see what the problem is. G(x, 0) doesn't involve x_0, so it doesn't matter that x_0 happens to be zero.

 

[yungman]

I don't see what the problem is. G(x, 0) doesn't involve x_0, so it doesn't matter that x_0 happens to be zero.

But you can clearly see the first term of the denominator equal zero and the second term is a. This means:

|\frac{\vec{x_0}}{r_0}|=1

r_0=|\vec{x_0}|

 

[HallsofIvy]

But that is NOT in what you quoted orginally. And, surely, where it the text says
|\frac{\vec{x_0}}{r_0}|=1
it also has a provision for the case that |\vec{x_0}|\ne 0.

 

[yungman]

But that is NOT in what you quoted orginally. And, surely, where it the text says
|\frac{\vec{x_0}}{r_0}|=1
it also has a provision for the case that |\vec{x_0}|\ne 0.

The textbook did not say anything, but if you look at the formula given by the book, |\frac{\vec{x_0}}{r_0}|=1 is the only way for the forumlas to be true. That is where I am confused. Here is what the book gave:

If x_0\neq 0:

G(\vec{x},\vec{x_0}) = -\frac{1}{4\pi \left\|\vec{x} -\vec{x_0}\right\|} + \frac{1}{4\pi \left\|\frac{r_0}{a} \vec{x} -\frac{a}{r_0}\vec{x_0}\right\|}

The book gave when \vec{x_0}=0


G(\vec{x},0) = -\frac{1}{4\pi \left\|\vec{x} \right\|} + \frac{1}{4\pi a}

 

[yungman]

Anyone please? This is on p194 of the PDE book by Strauss equation (11).
Thanks

Alan

 

[Tac-Tics]

I don't know the book you're talking about, and it really doesn't matter for your question.

Division by zero, even when we're talking about vectors, is undefined.

If you arrive at a statement \hat{A} \;=\; \frac{\vec{A}}{|\vec{ A}|} and |A| = 0 in your calculations, you have made a mistake earlier.

It could very well be that the mistake is in your textbook.

Resolving this invalid division is simply a matter of restricting the domain appropriately. Many authors will take this for granted and perform operations which are invalid at a handful of points.

For example, if I have two functions f and g, and f(x) g(x) = 1, I CANNOT say that f(x) = 1/g(x), because g(x) might evaluate to zero at some places. More correctly, I must say f(x) = 1/g(x) for all x where g(x) /= 0... but that's a lot to write out, and often clutters the more important point being made by the author.

Note that \hat{A} \;=\; \frac{\vec{A}}{|\vec{ A}|} is how you "normalize" a vector (find a vector in the same direction, but whose length is 1). The zero vector is the only vector that cannot be normalized, because it doesn't have a direction.