## Monday, September 29, 2008

### Re-examing the Inner Product (Euclidean Space)

I'm auditing Kip Thorne's "Applications of Classical Physics" this year. Today was the first day of class, and Kip gave a preamble which, if I may be allowed to paraphrase in a wild, inaccurate, and completely unfair manner, (which is basically to say every word of this is existed only in my head) went something approximately like this:
Welcome to class. Here is the website. Please call me "Kip". I'm 68 years old and my signature lucky ponytail is a thing of the past. There's not enough time left for ridiculous formalities. I am about to retire. I'm going to start writing books and making movies and various other things that famous retired physicists can do. I'm too old to waterski, so this is the next best option. This is will be last class I ever teach. There will be no grades. I would prefer you actually learn some stuff. Now, let's suppose the laws of physics are frame-independent, and see what restrictions this places on force law dealing with a classical field in Minkowski space..."

Awesome.

One thing that interested me was our classes' definition of the inner product between two vectors. Since we wanted to do everything in a frame-independent manner, we couldn't simply define the inner product by

$\mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z$

because that presumes the existence of basis vectors. The other common definition is

$\mathbf{A} \cdot \mathbf{B} = A B \cos{\theta}$

which is good, but we wanted something that could generalize to the Minkowski space of special relativity. So the definition we came in two parts. First, we defined the inner product of a vector with itself. Then use that definition to bridge to the inner product of arbitrary vectors.

For a vector $\mathbf{A}$, define

$\mathbf{A} \cdot \mathbf{A} = - \Delta s^2$

where $\Delta s^2$ is the square of the physical invariant interval between two events. To measure this interval, have an unaccelerated clock move from the origin to the event. ($\Delta s$ should be real, meaning the square of the interval is positive, and $-\Delta s^2$ is negative. So the inner product of the vector between timelike events with itself is negative). In the case of spacelike events, you should instead find the inertial reference frame in which the events are simultaneous, and lay down a measuring stick between them to get the interval.

Now we know how to take the inner product of a vector with itself. Define the inner product of two vectors $\mathbf{A}$ and $\mathbf{B}$ by

$\mathbf{A} \cdot \mathbf{B} = \frac{1}{4}\left( (\mathbf{A} + \mathbf{B})^2 - (\mathbf{A} - \mathbf{B})^2 \right)$

This is more subtle than it might initially appear. You can't just go willy-nilly with the algebra and start simplifying that right hand side out. You don't know any properties of this inner product, so you cannot, for example, write

$(\mathbf{A} + \mathbf{B}) \cdot (\mathbf{A} + \mathbf{B}) = A^2 + \mathbf{A} \cdot \mathbf{B} + \mathbf{B} \cdot \mathbf{A} + B^2$

Instead you have to work with the actual sums. It was claimed in class that this definition of the inner product is bilinear in the arguments, which wasn't obvious to me. So I asked about it, and Kip suggested I try to work it out for myself in Euclidean space using the Pythagorean theorem. So here goes.

Break down the vector $\mathbf{B}$ into a component parallel to $\mathbf{A}$ and a component perpendicular to $\mathbf{A}$

$\mathbf{B} = B_\parallel + B_\perp$

The vector $\mathbf{A} + \mathbf{B}$ now becomes $A + B_\parallel \mathbf{\hat{\parallel}} + B_\perp \mathbf{\hat{\perp}}$, where $\mathbf{\hat{\parallel}}$ and $\mathbf{\hat{\perp}}$ are unit vectors in the directions parallel and perpendicular to $\mathbf{A}$.

Because these two vectors make a right angle, the Pythagorean theorem applies.

$\left| \mathbf{A} + \mathbf{B} \right|^2 = (A + B_\parallel)^2 + B_\perp^2$

similarly,

$\left| \mathbf{A} - \mathbf{B} \right|^2 = (A - B_\parallel)^2 + B_\perp^2$

This is great, because the right hand sides of those equations are just numbers. Expand that out, subtract the two equations, and divide by four to obtain

$\frac{1}{4} \left| \mathbf{A} + \mathbf{B} \right|^2 - \left| \mathbf{A} - \mathbf{B} \right|^2 = A B_\parallel$

That's just the normal definition of the dot product, and is obviously linear in both vectors. Now, how about Minkowski space?

Anonymous said...

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