## Tag Archives: Galileo

### Galilean relativity

Required math: algebra, vectors, matrixes

Required physics: basics

Although the theory of relativity is usually attributed to Einstein, beginning with his paper of 1905, the concept of relativity actually dates back 300 years earlier to Galileo. Considering that Galileo predated Newton (Newton was born in the same year that Galileo died), his views of how the world worked were nothing short of revolutionary. Although many modern students of physics regard Galileo’s theories as everyday common sense, they were anything but that in his day.

One of Galileo’s most striking proposals was that of the equivalence of all reference frames moving uniformly relative to one another. He stated that any two observers moving with a constant velocity relative to each other must formulate the laws of physics in exactly the same way. One consequence of this is that there is no such thing as absolute rest or absolute motion. There is no one special reference frame in the universe that can be regarded as more important than any other.

Coming at a time when the church still decreed that the Earth was the fixed centre of the universe and that all heavenly bodies revolved around it, this idea is far from a ‘common sense’ proposition. In everday life, it is natural to regard the surface of the Earth as a special reference frame, against which we can measure the motion of anything else. Even if the church’s doctrine had not prevailed at the time, such a theory would have been greeted with skepticism.

Einstein’s special relativity theory incorporates Galileo’s hypothesis as its first axiom (the constancy of the speed of light is the second axiom). The ‘special’ in special relativity refers to situations where there is no acceleration or gravity.

Mathematically, Galileo’s theory requires that the laws of physics are invariant under a particular kind of transformation of coordinates. An implicit assumption is that time is measured at the same rate by all observers (an assumption which is dropped by Einstein). Thus if we have two observers ${R}$ (who measures his coordinates using Roman letters such as ${t}$ for time and ${x}$ for distance along the ${x}$ axis) and ${G}$ (who uses Greek letters ${\tau}$ for time and ${\xi}$ for distance), then we always have

$\displaystyle t=\tau$

If we align the two observers so their respective axes are parallel, and if ${G}$ is moving in the ${+x}$ direction at a speed ${v}$ relative to ${R}$, then the transformation from ${G}$ to ${R}$ is

$\displaystyle x=\xi+v\tau$
That is, a point fixed at coordinate ${\xi}$ in ${G}$‘s frame is moving relative to ${R}$ so that after a time ${\tau}$, it will have moved ${v\tau}$ along the ${x}$ axis.

These two equations can be written in matrix form as

$\displaystyle \left(\begin{array}{c} t\\ x\end{array}\right)=\left(\begin{array}{cc} 1 & 0\\ v & 1\end{array}\right)\left(\begin{array}{c} \tau\\ \xi\end{array}\right)$

We can define the transformation matrix as

$\displaystyle S_{v}\equiv\left(\begin{array}{cc} 1 & 0\\ v & 1\end{array}\right)$

The transformation matrix ${S_{v}}$ is a linear map, in that it maps one vector onto another, and is a linear transformation, so that for two vectors ${\mathbf{A}}$ and ${\mathbf{B}}$ and scalar ${k}$:

 $\displaystyle S_{v}(\mathbf{A}+\mathbf{B})$ $\displaystyle =$ $\displaystyle S_{v}\mathbf{A}+S_{v}\mathbf{B}$ $\displaystyle S_{v}(k\mathbf{A})$ $\displaystyle =$ $\displaystyle kS_{v}\mathbf{A}$

The inverse transformation is found by moving at a velocity ${-v}$ so we get

$\displaystyle S_{-v}=\left(\begin{array}{cc} 1 & 0\\ -v & 1\end{array}\right)=(S_{v})^{-1}$

The fact that ${S_{-v}=(S_{v})^{-1}}$ is verified by direct multiplication$\displaystyle S_{-v}S_{v}=S_{v}S_{-v}=\left(\begin{array}{cc} 1 & 0\\ 0 & 1\end{array}\right)$

The transformation can be generalized to 2 or 3 dimensions by adding a couple of rows and columns to the matrix. Thus for a general 3-dimensional velocity ${\mathbf{v}=(v_{x},v_{y},v_{z})}$ we have

$\displaystyle S_{\mathbf{v}}=\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ v_{x} & 1 & 0 & 0\\ v_{y} & 0 & 1 & 0\\ v_{z} & 0 & 0 & 1\end{array}\right)$

Using this matrix to transform a general spacetime vector we get

$\displaystyle \left(\begin{array}{c} t\\ x\\ y\\ z\end{array}\right)=\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ v_{x} & 1 & 0 & 0\\ v_{y} & 0 & 1 & 0\\ v_{z} & 0 & 0 & 1\end{array}\right)\left(\begin{array}{c} \tau\\ \xi\\ \eta\\ \zeta\end{array}\right)$

Returning to the ${t-x}$ case, a general linear map can be defined by

$\displaystyle L=\left(\begin{array}{cc} a & b\\ c & d\end{array}\right)$

The area of a parallelogram spanned by two vectors ${\mathbf{X}}$ and ${\mathbf{Y}}$ is (can see this from drawing a diagram and using the fact that the area of the parallelogram is twice the area of the spanned triangle) ${|\mathbf{X}||\mathbf{Y}|\sin\theta}$ where ${\theta}$ is the angle between the vectors. This is the modulus of the cross product ${|\mathbf{X}\times\mathbf{Y}|=|x_{1}y_{2}-x_{2}y_{1}|}$. Since a vector transforms as ${\mathbf{X}'=L\mathbf{X}}$:

 $\displaystyle \mathbf{X}'$ $\displaystyle =$ $\displaystyle (ax_{1}+bx_{2},cx_{1}+dx_{2})$ $\displaystyle \mathbf{Y}'$ $\displaystyle =$ $\displaystyle (ay_{1}+by_{2},cy_{1}+dy_{2})$

The new area ${A'}$ is thus
$\displaystyle A'=|\mathbf{X}'\times\mathbf{Y}'|$

Writing this out as the determinant we get

$\displaystyle A'=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ ax_{1}+bx_{2} & cx_{1}+dx_{2} & 0\\ ay_{1}+by_{2} & cy_{1}+dy_{2} & 0\end{array}\right|$
Doing the sums, we get

 $\displaystyle$ $\displaystyle =$ $\displaystyle A'|(x_{1}y_{2}-x_{2}y_{1})(ad-bc)|$ $\displaystyle$ $\displaystyle =$ $\displaystyle |\mathbf{X}\times\mathbf{Y}||\det\; L|$ $\displaystyle$ $\displaystyle =$ $\displaystyle A|\det\; L|$

If ${|\det\; L|=0}$ this means that ${\mathbf{X}'\parallel\mathbf{Y}'}$ so the area spanned is zero. For a general linear map, the area spanned by the transformed vectors is not preserved, but for the special case of the Galilean transformation, ${L=S_{v}}$ and ${\det S_{v}=1}$, so the Galilean transformation preserves areas. The result extends to 3 spatial dimensions although the algebra is messier.