Sometimes, when Google has one of those seasonal logos I click on it - to see what it’s about. Just now I noticed that the Google logo has a red planet with some little aliens representing the first “O” in Google. Out of curiosity I clicked on it, the image links to Google Mars, which is like Google Maps. Unfortunately it can’t be used with Google Earth client.

The blurb claims these are some of the most detailed maps of Mars ever made. I don’t know if the maps cover the whole planet, as admittedly my Martian geography is pretty poor.

Just for the fun of it I tried moon.google.com and it resolves. Go take a look and make sure to zoom in

The rest of the planets don’t resolve. You can aslo get to Google Mars via mars.google.com.

I seem to get a fair number of hits querying the basic formulas for velocity, so I’ve decided to post them.

Let [tex]\cal{O}[/tex] be an object moving along a path. The position of the object [tex]\cal{O}[/tex] is a function of the time [tex]t[/tex]. We denote the position vector [tex]\bold{x}(t)[/tex].

The time derivative of the position vector [tex]\bold{x}(t)[/tex] is the velocity vector [tex]\bold{v}(t)[/tex] of the object [tex]\cal{O}[/tex].

[tex]\bold{v}(t) \equiv \frac{d\bold{x}(t)}{dt} = \bold{\dot{x}}(t)[/tex].

The time derivative of the velocity vector [tex]\bold{v}(t)[/tex] is called the acceleration vector [tex]\bold{a}(t)[/tex] of the object [tex]\cal{O}[/tex] .

[tex]\bold{a}(t) \equiv \frac{d\bold{v}(t)}{dt} = \frac{d^2\bold{x}(t)}{dt^2} = \bold{\ddot{x}}(t) [/tex].

Integrate this last equation with respect to [tex]t[/tex],

[tex]\int \bold{\ddot{x}}(t) \ dt=\int \bold{a}(t) \ dt \implies \bold{\dot{x}}(t) =\bold{a}( t)t + \bold{C}[/tex],

where [tex]\bold{C}[/tex] is a constant vector to be determined by the initial conditions. Now, [tex]\bold{\dot{x}}(t) =\bold{v}(t)[/tex] so that

[tex]\bold{v}(t) =\bold{a}(t) t + \bold{C}[/tex]

We assume that at time [tex]t =0[/tex] the object [tex]\cal{O}[/tex] is moving with initial velocity [tex]\bold{v}(0) =\bold{v_0}[/tex]

[tex] \bold{v}(0) = \bold{v_0} =\bold{a}(0) \ 0 + C \implies C = \bold{v_0}[/tex]

and substitution gives the familiar formula for the velocity of an object

[tex]\bold{v}(t) = \bold{v_0} + \bold{a}(t) t[/tex].

We can integrate the equation

[tex]\bold{\dot{x}}(t) =\bold{v_0} + \bold{a}(t) t[/tex] with respect to [tex]t[/tex] a second time

[tex]\int \bold{\dot{x}}(t) \ dt = \int \bold{v_0} + \bold{a}(t) t \ dt[/tex] to give

[tex] \bold{x}(t) = \bold{v_0} t+ \frac{1}{2}\bold{a}(t) t^2 + \bold{K} [/tex], where [tex]\bold{K}[/tex] is a constant vector to be determined from the initial conditions.

We assume that at [tex]t =0[/tex] the object [tex]\cal{O}[/tex] has position vector [tex]\bold{x}(0)=\bold{x_0}[/tex]

[tex] \bold{x}(0) = \bold{x_0} = \bold{v_0} t+ \frac{1}{2}\bold{a}(0) 0^2 + \bold{K} \implies \bold{K}=\bold{x_0}[/tex]

and we have the familiar formula for the displacement (position) of the object at time [tex]t[/tex]

[tex] \bold{x}(t) = \bold{x_0} + \bold{v_0} t+ \frac{1}{2}\bold{a}(t) t^2 [/tex].

In a first treatment we often simplify things a bit. Firstly, to simplify notation we don’t write in the explicit time dependence for the position [tex]\bold{x}(t)[/tex] , so that it becomes [tex]\bold{x}[/tex] and similarly for [tex]\bold{v}[/tex] and [tex]\bold{a}[/tex]. Secondly we assume that the object starts from the origin rather than from some arbitrary position, thus we take [tex]\bold{x_0}=0[/tex], (which is really just a change of co-ordinates). This gives the familiar form of the equations of motion from high school text books:

[tex] \bold{x} = \bold{v_0} t+ \frac{1}{2}\bold{a} t^2 [/tex]
[tex]\bold{v} = \bold{v_0} + \bold{a} t[/tex].

These equations are commonly written as

[tex] s = u t+ \frac{1}{2} a t^2 [/tex]
[tex] v = u+ a t[/tex],

where we have used the alternate notation [tex]s \equiv x[/tex] for the position and [tex]u \equiv v_0[/tex] for the intial velocity. Note that we drop the bold face symbols [tex]\bold{v} \rightarrow v[/tex] which are used to denote vectors, since in a first treatment we consider only the one dimensional or scalar case.

Squaring the last equation gives
[tex]v^2 = (u+ a t)^2 = u^2 + 2 u a t + a^2t^2 = u^2 + 2a (ut+\frac{1}{2}at^2) [/tex]
[tex]\phantom{v^2}= u^2 + 2as[/tex]

another commonly used equation in high school physics texts.

The average velocity over the time interval [tex] (t_1,t_2) \equiv (t_i,t_f)[/tex] is given by

[tex]v_{av} = \frac{s_2 -s_1}{t_2-t_1} = \frac{s_f-s_i}{t_f-t_i} = \frac{\Delta(s)}{\Delta(t)}[/tex]

where, [tex] s_1 \equiv s_i[/tex] is the intial displacement and [tex]s_2 \equiv s_f[/tex] is the final displacement. Finally, in words: it is the change in position [tex]\Delta(s)[/tex] divided by the change in time [tex]\Delta(t)[/tex].

The first time I saw the axioms for tensor products of vector spaces I was confused to say the least. I kept thinking about why the particular axioms were chosen. Recently I have developed a better understanding of where the axioms come from.

We begin our considerations with 2 by 2 matrices [tex]A[/tex] and [tex]B[/tex] over a field [tex]\mathbb{F}[/tex], where
[tex] A= \left ( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right ) \quad \textrm{and} \quad B= \left ( \begin{array}{cc} b_{11} & b_{12} \\\ b_{21} & b_{22} \end{array} \right )[/tex]

The tensor (or Kronecker) product of [tex]A[/tex] and [tex]B[/tex] is defined as

[tex] A \otimes B = \left ( \begin{array}{cc} a_{11} B & a _{12} B \\\ a_{21} B & a_{22} B \end{array} \right ) [/tex]

[tex] = \left ( \begin{array}{cccc}a_{11} b_{11} & a_{11} b _{12} & a _{12} b_{11} & a _{12}b _{12} \\ a_{11} b_{21} & a_{11} b_{22} & a _{12} b_{21} & a _{12}b_{22} \\ a_{21} b_{11} & a_{21} b _{12} & a_{22} b_{11} & a_{22}b_{12} \\ a_{21} b_{21} & a_{21} b_{22} & a_{22} b_{21} & a_{22} b_{22} \end{array} \right )[/tex].

The tensor product of [tex]B[/tex] with [tex]A [/tex] is
[tex] B \otimes A = \left ( \begin{array}{cc} b_{11} A & b _{12}A \\ b_{21} A & b_{22}A \end{array} \right )[/tex]

[tex] = \left ( \begin{array}{cccc} b_{11} a_{11} & b_{11} a _{12} & b _{12} a_{11} & b _{12} a _{12} \\ b_{11} a_{21} & b_{11} a_{22} & b _{12} a_{21} & b _{12} a_{22} \\ b_{21} a_{11} & b_{21}a _{12} & b_{22} a_{11} & b_{22}a _{12} \\ b_{21}a_{21} & b_{21}a_{22} & b_{22} a_{21} & b_{22}a_{22} \end{array} \right )[/tex] .

Clearly, [tex]A \otimes B \ne B \otimes A[/tex] , so that the tensor product is non-commutative ^[1] (it is however associative).

We now consider the effect of multiplication by a scalar [tex] \nu \in \mathbb F[/tex] . We set

[tex] \nu A = \nu \left ( \begin{array}{cc} a_{11} & a _{12} \\\ a_{21} & a_{22} \end{array}\right ) = \left ( \begin{array}{cc} \nu a_{11} & \nu a _{12} \\\ \nu a_{21} & \nu a_{22} \end{array} \right) [/tex]

now
[tex] \nu A \otimes B = \left ( \begin{array}{cc} \nu a_{11} & \nu a _{12} \\\ \nu a_{21} & \nu a_{22} \end{array}\right) \otimes \left ( \begin{array}{cc} b_{11} & b _{12} \\\ b_{21} & b_{22} \end{array}\right) [/tex]
[tex] = \left ( \begin{array}{cc} \nu a_{11} B & \nu a _{12} B \\\ \nu a_{21} B & \nu a_{22} B \end{array} \right ) [/tex]

[tex] = \left ( \begin{array}{cccc} \nu a_{11} b_{11} & \nu a_{11} b _{12} & \nu a _{12} b_{11} & \nu a _{12}b _{12} \\ \nu a_{11} b_{21} & \nu a_{11} b_{22} & \nu a _{12} b_{21} & \nu a _{12}b_{22} \\ \nu a_{21} b_{11} & \nu a_{21} b _{12} & \nu a_{22} b_{11} & \nu a_{22}b_{12} \\ \nu a_{21} b_{21} & \nu a_{21} b_{22} & \nu a_{22} b_{21} & \nu a_{22} b_{22} \end{array} \right )[/tex]

[tex] = \nu \left ( \begin{array}{cccc} a_{11} b_{11} & a_{11} b _{12} & a _{12} b_{11} & a _{12}b _{12} \\ a_{11} b_{21} & a_{11} b_{22} & a _{12} b_{21} & a _{12}b_{22} \\ a_{21} b_{11} & a_{21} b _{12} & a_{22} b_{11} & a_{22}b_{12} \\ a_{21} b_{21} & a_{21} b_{22} & a_{22} b_{21} & a_{22} b_{22} \end{array} \right )[/tex]

[tex] = \nu (A \otimes B) = A \otimes \nu B [/tex] .

For arbitrary matrices [tex]A = a_{ij}, B = b_{kl}[/tex] the tensor product takes the form
[tex]A \otimes B = \left( \begin{array}{ccc} a_{11} B & \cdots & a_{1j} B \\ \vdots & \ddots & \vdots \\ a_{i1}B & \cdots & a_{ij}B\end{array} \right)[/tex]

For column vectors, [tex]u=\left( \begin{array}{c}u_1 \\\ \vdots \\\ u_j\end{array} \right), v= \left( \begin{array}{c} v_1 \\\ \vdots \\\ v_k \end{array} \right)[/tex] the tensor product is of the same general form.

A column vector is simply a [tex]k \times 1[/tex] matrix and the tensor product of [tex]u[/tex] with [tex]v[/tex] is

[tex]u \otimes v =\left( \begin{array}{c}u_1 v\\\ \vdots \\\ u_j v\end{array} \right)[/tex].

We can begin abstracting a bit now. Let [tex]V [/tex] and [tex]W[/tex] be two finite dimensional vector spaces. We define the tensor product [tex]V \otimes W[/tex] as the vector space spanned by all formal products [tex]v \otimes w[/tex] , [tex]v \in V[/tex] , [tex]w \in W [/tex] where [tex]\otimes [/tex] satisfies

(i) [tex] ( v_1 + v_2 ) \otimes w = v_1 \otimes w + v_2 \otimes w[/tex]

(ii) [tex] v \otimes ( w_1 + w_2 ) = v \otimes w_1 + v \otimes w_2 [/tex]

(iii) [tex] \nu ( v \otimes w ) = ( \nu v ) \otimes w = v \otimes ( \nu w ) [/tex],

for all [tex]v, v_1, v_2 \in V, w,w_1,w_2 \in W, \nu \in \mathbb{F}[/tex].

The first two parts of the defintion are straightforward, we want [tex]\otimes[/tex] to be linear in both the first and the second spaces (i.e. bilinear). The third axiom is the most confusing one. As you can see from the results derived from the Kronecker product of matrices, the third axiom makes sense if we are going to model tensor products algebraically.

Note that a general element of [tex]V \otimes W[/tex] is of the form [tex]\sum_{i,j} \alpha_{ij} v_i \otimes w_j[/tex] and cannot in general be expressed in the simpler form [tex]v \otimes w [/tex], for some [tex]v,w \in V[/tex].

Example, let [tex]V[/tex] be a 2-dimensionl vector space with basis [tex]\{e_1, e_2\}[/tex], then consider the space [tex]V \otimes V[/tex]. A typical element of [tex]V \otimes V[/tex] is
[tex]e_1 \otimes e_2 + e_2 \otimes e_1[/tex].
It cannot be expressed as a tensor product of two vectors, for suppose that there are vectors [tex]v, v’ \in V[/tex] such that
[tex] v \otimes v’ = e_1 \otimes e_2 + e_2 \otimes e_1 [/tex].
Now since [tex]\{e_1,e_2\}[/tex] is a basis of [tex]V[/tex] we can express [tex]v,v’[/tex] in terms of the basis vectors as
[tex]v = \alpha_1 e_1 + \alpha_2 e_2,\quad v’ =\beta_1 e_1 + \beta_2 e_2[/tex]
where [tex]\alpha_1,\alpha_2, \beta_1, \beta_2 \in \mathhbb{F}[/tex], so that

[tex]v \otimes v’ = (\alpha_1 e_1 + \alpha_2 e_2) \otimes (\beta_1 e_1 + \beta_2 e_2)[/tex]
[tex] =\alpha_1 \beta_1 \ e_1 \otimes e_1 + \alpha_1 \beta_2 \ e_1 \otimes e_2 + \alpha_2 \beta_1 \ e_2 \otimes e_1 + \alpha_2 \beta_2 \ e_2 \otimes e_2 [/tex]
[tex]= e_1 \otimes e_2 + e_2 \otimes e_1[/tex].

Now by equating coeffecients we must have [tex]\alpha_1 \beta_1=0[/tex], which implies that either [tex]\alpha_1 \beta_2 =0[/tex] or [tex]\alpha_2 \beta_1 =0[/tex], so that there is no solution, hence there are no vectors [tex]v,v’[/tex] such that [tex] v \otimes v’ = e_1 \otimes e_2 + e_2 \otimes e_1 [/tex].

The tensor product comes up a lot in quantum mechanics, for example in the addition of angular momenta.
It is also the basis of entanglement in quantum mechanics, a state such as
[tex] | \uparrow \ \rangle \otimes |\downarrow \ \rangle + |\downarrow \ \rangle\otimes |\uparrow\ \rangle [/tex]
is called an entangled state because it has no decomposition, just replace [tex]e_1[/tex] with [tex]| \uparrow \ \rangle[/tex] (spin up) and [tex]e_2[/tex] with [tex]| \downarrow \ \rangle[/tex] (spin down) in the previous argument.

A state such as
[tex] |\uparrow\ \rangle \otimes |\uparrow \ \rangle + |\downarrow\ \rangle\otimes |\uparrow \ \rangle [/tex] is not entangled because it can be expressed as [tex] (1|\uparrow \ \rangle + 1|\downarrow \ \rangle) \otimes ( 1|\uparrow \ \rangle + 0|\downarrow \ \rangle )[/tex].

[1]. To see how [tex]A \otimes B[/tex] and [tex]B \otimes A[/tex] are related, we introduce the permutation operator [tex]T[/tex] where,
[tex] T = \left ( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right )[/tex].

Note that [tex]T^2 =I[/tex], so that [tex]T = T^{-1}[/tex]. A quick calculation shows that
[tex] A \otimes B = T \cdot (B \otimes A ) \cdot T^{-1} [/tex].