# Vectors and Scalars

1. Mar 2, 2010

### Char. Limit

If someone takes a vector, and squares it, does it become a scalar?

Also, is it possible to take the square root of a vector, and would the result be a vector or a scalar?

Lastly, the logarithm of a measurement is dimensionless. However, if you raise the base of that logarithm to the power of the logarithm, are the units recovered?

2. Mar 2, 2010

### CRGreathouse

How do you square a vector?

Probably depends on the definition of squaring a vector, above. Usually I'd say neither is defined, but there are cases where it's useful to define them; in those cases it would depend on the definition. If you define vector multiplication as
v1 * v2 = {[n n], v1 = v2 = [0 0 ... (n zeros) ... 0]
{ apricot, otherwise
for any vectors v1, v2 then sqrt([a b]) = [0 0 ... (a zeros) ... 0 0] if a = b and undefined otherwise.

No.

3. Mar 2, 2010

### Char. Limit

Well, I was considering the standard equation for kinetic energy:

$$E_k=\frac{1}{2}mv^2$$

Velocity is a vector, and you are squaring it.

Also, if the logarithm of measurement isn't dimensionless, than what is it? Take for example pKa... if not dimensionless, what are its units?

4. Mar 2, 2010

### CRGreathouse

Yeah, funny thing that... you're really squaring speed in that equation -- direction plays no part. Don't ask me why it's described as velocity.

Edit: If you interpret squaring as the dot product of a vector with itself, then squaring a vector *does* give a scalar. Of course norming it and squaring gives the same result... at least classically.

Ka is dimensionless, and so is its logarithm pKa.

Last edited: Mar 3, 2010
5. Mar 2, 2010

### Char. Limit

OK, let me present a different example...

From first-order kinetic rate laws...

$$ln[A]_t - ln[A]_0 = -kt$$

What is the unit $$ln[A]_t$$, given that the unit for $$[A]_t$$ is M, or molar?

Also, from what I've heard of kinetic energy, v is described as velocity. I'm also trying to think of another equation where a vector is squared, and all that I'm coming up with is $$a=\frac{v^2}{r}$$...

6. Mar 2, 2010

### Tao-Fu

When you have vector quantities that get squared in physics, it usually means taking the dot product (scalar product) of the vector with itself. This gives the square of the magnitude of the vector, since it is always parallel with itself and gives a directional cosine of 1. Similarly, a vector (cross) product of a vector with itself always gives zero, so there's no sensible reason for including it in an expression.

7. Mar 3, 2010

### HallsofIvy

There are three kinds of products for vectors:

Scalar product, which is a product of vector and scalar doesn't,apply here- you can't take the scalar product of a vector with itself.

You can take take the cross product of a vector with itself but the result is always the 0 vector so $\vec{v}\times\vec{v}$ is not interesting and we don't use $\vec{v}^2$ to mean that.

That leaves only the dot product. $\vec{v}^2= \vec{v}\cdot\vec{v}= |\vec{v}|$, a scalar.

8. Mar 3, 2010

### Rasalhague

$$\vec{v}\cdot\vec{v}=|\vec{v}|^2$$