text:
Special Relativity/Mathematical transformations

The teaching of Special Relativity on undergraduate physics courses involves a considerable mathematical background knowledge. Particularly important are the manipulation of vectors and matrices and an elementary knowledge of curvature. The background mathematics is given at the end of this section and can be referenced by those who are unfamiliar with these techniques.

The Lorentz transformation
The Lorentz transformation deals with the problem of observers who are moving relative to each other. How are the coordinates of an event recorded by one observer related to the coordinates of the event recorded by the other observer? The standard configuration used in the calculation of the Lorentz transformation is shown below:

There are several ways of deriving the Lorentz transformations. The usual method is to work from Einstein's postulates (that the laws of physics are the same between all inertial reference frames and the speed of light is constant) whilst adding assumptions about isotropy, linearity and homogeneity. The second is to work from the assumption of a four dimensional Minkowskian metric.

In mathematics transformations are frequently symbolised with the "maps to" symbol:

$$ (x,y,z,t) \mapsto (x^',y^',z^',t^') $$

The linearity and homogeneity of spacetime
Consider a clock moving freely. According to Newton's first law, objects continue in a state of uniform motion unless acted upon by a force; so, the velocity of the clock in any given direction ($$dx_i/dt$$) must be a constant.

If the clock is a real clock with readings given by $$\tau$$ then the rate of change with respect to these readings of the elapsed time at any other point in an inertial frame of reference, $$dt/d\tau$$, will be a constant. If the clock were to tick at an uneven rate compared with other clocks then the universe would not be homogeneous in time—at some times the clock would appear to accelerate. This would also mean that Newton's first law would be broken and the universe would not be homogeneous in space.

If $$dx_i/dt$$ and $$dt/d\tau$$ are constant then $$dx_{\mu}/d\tau\,(\mu=1,2,3,4)$$ is also constant. This means that the clock is not accelerating, i.e. that $$d^2x_{\mu}/d\tau^2 = 0$$.

Linearity is demonstrated by the way that the length of things does not depend on position or relative position; for instance, if $$x' = ax^2$$ the distance between two points would depend upon the position of the observer whereas if the relationship is linear ($$x' = ax$$) separations are independent of position.

The linearity and homogeneity assumptions mean that the coordinates of objects in the $$S'$$ inertial frame are related to those in the $$S$$ inertial frame by:

$$x_{\nu}' = (\sum \Lambda_{\nu\mu}x_\mu) + B_\nu$$

This formula is known as a Poincaré transformation. It can be expressed in indicial notation as:

$$x^{'\nu} = \Lambda_{\mu}^\nu x^\mu + B^\nu$$

If the origins of the frames coincide then $$B_\nu$$ can be assumed to be zero and the equation:

$$x_{\nu}' = \sum \Lambda_{\nu\mu}x_\mu$$

Those who are unfamiliar with the notation should note that the symbols $$x_1$$ etc. mean $$x_1=x$$, $$x_2=y$$, $$x_3=z$$ and $$x_4=t$$ so the equation above is shorthand for:

$$x' = a_{11} x + a_{12} y + a_{13} z + a_{14} t $$

$$y' = a_{21} x + a_{22} y + a_{23} z + a_{24} t $$

$$z' = a_{31} x + a_{32} y + a_{33} z + a_{34} t $$

$$t' = a_{41} x + a_{42} y + a_{43} z + a_{44} t $$

In matrix notation the set of equations can be written as:

$$\mathbf{x'}=\mathbf{\Lambda x}$$

The standard configuration (see diagram above) has several properties, for instance:

 * The spatial origin of both observer's coordinate systems lies on the line of motion so the x axes can be chosen to be parallel.

 * The point given by $$x=vt$$ is the same as $$x'=0$$.

 * The origins of both coordinate systems can coincide so that clocks can be synchronised when they are next to each other.

 * The coordinate planes ,y, y' and z,z', can be arranged to be orthogonal (at right angles) to the direction of motion.

 * Isotropy means that coordinate planes that are orthogonal at y=0 and z=0 in one frame are orthogonal at y'=0 and z'=0 in the other frame.

According to the relativity principle any transformations between the same two inertial frames of reference must be the same. This is known as the reciprocity theorem.

The Lorentz transformation
From the linearity assumption and given that at the origin $$y=0=y^'$$ so there is no constant offset then $$y^'=Ky$$ and $$y=Ky^'$$, therefore K=1. So:

$$y^'=y$$

and, by the same reasoning:

$$z^'=z$$

Now, considering the x coordinate of the event, the z and y axes can be assumed to be 0 (ie: an arbitrary shift of the coordinates to allow the event to lie on the x axes). If this is done then the linearity consideration and the fact that $$x=vt$$ and $$x^'=0$$ are the same point gives:

(1) $$x^' = \gamma (x - vt)$$

where $$\gamma$$ is a constant. According to the reciprocity theorem we also have:

(2) $$x = \gamma (x^' + vt^')$$

Einstein's assumption that the speed of light is a constant can now be introduced so that $$x=ct$$ and also $$x^'=ct^'$$. So:

$$ct^'= \gamma t(c - v)$$

and

$$ct= \gamma t^'(c + v)$$

So:

$$c^2tt^'= \gamma^2tt^'(c^2 - v^2)$$

and

$$\mathbf{\gamma = \frac {1}{\sqrt {1-v^2/c^2}}}$$

Therefore the Lorentz transformation equations are:

$$t^' = \gamma (t - vx/c^2)$$

$$x^' = \gamma (x - vt)$$

$$y^'=y$$

$$z^'=z$$

The transformation for the time coordinate can derived from the transformation for the x coordinate assuming $$x=ct$$ and $$x^'=ct^'$$ or directly from equations (1) and (2) with a similar substitution for $$x=ct$$.

The coefficients of the Lorentz transformation can be represented in matrix format:

\begin{bmatrix} c t' \\x' \\y' \\z' \end{bmatrix} = \begin{bmatrix} \gamma&-\frac{v}{c} \gamma&0&0\\ -\frac{v}{c} \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} c t\\x\\y\\z \end{bmatrix}. $$

A coordinate transformation of this type, that is due to motion, is known as a boost.

The Lorentz transformation equations can be used to show that:

$$c^2dt^{'2} - dx^{'2} - dy^{'2} - dz^{'2} = c^2dt^{2} - dx^{2} - dy^{2} - dz^{2}$$

Although whether the assumptions of linearity, isotropy and homogeneity in the derivation of the Lorentz transformation actually assumed this identity from the outset is a moot point.

Given that: $$c^2dt^{'2} - dx^{'2} - dy^{'2} - dz^{'2}$$ also equals $$c^2dt^{2} - dx^{2} - dy^{2} - dz^{2}$$ and a continuous range of other transformations it is clear that:

$$\mathbf{\Delta s}^2 = c^2\Delta{t}^{2} - \Delta{x}^{2} - \Delta{y}^{2} - \Delta{z}^{2}$$

The quantity $$\Delta s$$ is known as the spacetime interval and the quantity $$\mathbf{\Delta s}^2$$ is known as the squared displacement.

A given squared displacement is constant for all observers, no matter how fast they are travelling, it is said to be invariant.

The equation:

$$ds^2 = c^2dt^{2} - dx^{2} - dy^{2} - dz^{2}$$

is known as the metric of spacetime.

Another look at the Lorentz transformation
We have seen how length and duration both look different to a moving observer:

 * lengths are contracted by a factor γ, when measured at the same time.
 * durations are contracted by a factor γ, when measured in the same place.

We can extend these results to allow for measurements taking place at different times and places.

Lets consider two observers, O and O' such that O' is moving at velocity v along the x-axis with respect to O. We'll use primed variables for all the measurements O' makes.

We can assume for now both observers have the same origin and x-axis because we already know how to allow for observers being relatively rotated and displaced. We can put these complications back in later.

Now any length or duration can be written as the difference between two coordinates, for the two ends of the body or the start and end of the event, so it is sufficient to know how to change coordinates from one frame to the other.

We know how to do this in classical physics,

 * $$\begin{matrix} x' & = & x-vt \\ t' & = & t \end{matrix} $$

we need to extend this to relativity.

Notice that the in classical physics the relationship is linear; the graphs of these equations are straight lines. This makes the maths much simpler, so we will try to find a linear relationship between the coordinates for relativity, i.e equations of the general form

 * $$\begin{matrix} x' & = & mx + nt \\ t' & = & px + qt \end{matrix} $$

where m, n, p and q are all independent of the coordinates.

To begin with know that

 * when t=0, x′ = γx (Lorentz contraction)
 * when x=0, t′ = γt (Time dilation)

and that O' is travelling at velocity v. They measure their position to be at x′=0, but O measures it to be at vt so we must have

 * x′=0 when x-vt=0

The only relationship between x and x′ that satisfies these criteria is

 * $$x'=\gamma (x - vt) \, $$

Both observers must measure the same speed for light,

 * $$ x=ct \Rightarrow x'=ct' \,$$

or, substituting and rearranging,

 * $$ x=ct \Rightarrow t'=\gamma \frac{x-vt}{c}$$

The only linear relationship between t and t′ that satisfies these criteria is

 * $$t'=\gamma (-\frac{vx}{c^2} + t) \, $$

So the primed and unprimed coordinates are related by

 * $$\begin{matrix}

x' & = & \gamma (x-vt) \\ t' & = & \gamma \left(-\frac{vx}{c^2} + t \right) \end{matrix} $$

These equations are called the Lorentz transform.

They look simpler if we write them in terms of ct rather than t

 * $$\begin{matrix}

x' & = & \gamma \left( x-\frac{v}{c}ct \right) \\ ct' & = & \gamma \left( -\frac{v}{c}x + ct \right) \end{matrix} $$

Written this way they look much like the equations describing a rotation in three dimensions. In fact, once we allow for the different Pythagorean theorem, they are exactly like the equations for rotation.

 * If observers are moving relative to each other, their coordinate systems are rotated in the (x,ct) plane.

The derivation of the constancy of the speed of light from the first postulate alone
The following text was taken from a "Wikiversity" lecture.

The lorentz transformation can be deduced from the first postulate alone with no additional assumptions.

The Lorentz transformation describes the way a vector in spacetime as seen by an observer O1 changes when it is seen by an observer O2 in a different inertial system.

The postulate of special relativity states that the laws of physics are independent of the velocity of an observer. This is equivalent to the requirement that the mathematical formulation of the laws of physics must be invariant if the coordinates of space and time are altered using a lorentz transformation.

This is the reason why the Lorentz transformation is a central concept in special relativity.

Four-vectors are vectors that, subjected to the Lorentz transformation, behave like vectors from spacetime. The standard approach to find relativistic laws to the corresponding non-relativistic ones is to find four-vectors to replace space vectors. Surprisingly often this simple approach is successful.

Here, for simplicity, the space coordinates are chosen so that the x-axis points in the direction in which O2 is moving with respect to O1 and the y and z coordinates are not considered.

Here is a way to determine a matrix $$L(v)$$ that transforms coordinates from O1 to O2 where v is the velocity of O2 seen from O1.

Most generally, this matrix looks like $$L(v) =\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$$ and transforms $$\left(\begin{matrix}t \\ x\end{matrix}\right)$$ to $$\left(\begin{matrix}t^\prime \\ x^\prime\end{matrix}\right)$$ via $$\left(\begin{matrix}t' \\ x'\end{matrix}\right) = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)\left(\begin{matrix}t \\ x\end{matrix}\right) $$.

The first requirement on this matrix is that O2 will be at rest (in space) within its own coordinate system.

Therefore $$\left(\begin{matrix}t' \\ 0\end{matrix}\right) = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)\left(\begin{matrix} t \\ tv\end{matrix}\right) $$ for any time t.

This gives us $$ ct +tvd = 0 \Rightarrow c = -vd $$ and $$L(v) =\left(\begin{matrix}a & b \\ -vd & d\end{matrix}\right)$$.

The backward transformation from O2 to O1 is described by the inverse matrix $$L(v') = L(v)^{-1} = (\det(L(v))^{-1}\left(\begin{matrix}d & -b \\ vd & a\end{matrix}\right)$$ (where $$\, v'$$ is the velocity of O1 seen from O2) has to leave O1 at rest  in its own coordinates.

Assuming $$ v' = -v $$ (being strict this would have to be proved) we have $$\left(\begin{matrix}t \\ 0\end{matrix}\right) = (\det(L(v))^{-1}\left(\begin{matrix}d & -b \\ vd & a\end{matrix}\right)\left(\begin{matrix} t' \\ -t'v\end{matrix}\right) $$ for any time $$\, t'$$ resulting in $$ a=d \Rightarrow L(v) =\left(\begin{matrix}a & b \\ -va & a\end{matrix}\right)$$.

The coefficients are functions of v. For all v the transformation is invertible, so $$0 \neq \det L(v) = a(v)^2 + v a(v) b(v) \Rightarrow a(v) \neq 0$$.

For v=0 L(v) is the identity matrix, so b(0)=0. This allows to rewrite $$ b(v) = -v a(v) \varepsilon(v) $$ with a yet unknown function $$\varepsilon(v)$$.

Now $$ L(v) =a(v) \left(\begin{matrix} 1 & -v\varepsilon(v) \\ -v & 1\end{matrix}\right)$$.

Now for every $$v_0, v_1$$ there exists a velocity $$v_2$$ such that $$L(v_0) L(v_1) = L(v_2)$$. Using what we have proved up to now gives $$\begin{align} L(v_0)L(v_1) & = a(v_0) \left(\begin{matrix} 1 & -v_0\varepsilon(v_0) \\ -v_0 & 1\end{matrix}\right) a(v_1) \left(\begin{matrix} 1 & -v_1\varepsilon(v_1) \\ -v_1 & 1\end{matrix}\right) \\ & = a(v_0) a(v_1)\left(\begin{matrix} 1+v_0v_1\varepsilon(v_0) & -v_1\varepsilon(v_1)-v_0\varepsilon(v_0) \\ -v_0-v_1 & 1+v_0 v_1 \varepsilon(v_1)\end{matrix}\right) \\ & = a(v_2) \left(\begin{matrix} 1 & -v_2\varepsilon(v_2) \\ -v_2 & 1\end{matrix}\right) \\ & = L(v_2). \end{align}$$

Comparing the diagonal components, we see immediately that $$\varepsilon(v_0) = \varepsilon(v_1)$$ for arbitrary values of $$v_0, v_1$$, and so $$\varepsilon(v) = \operatorname{const.} = \varepsilon$$.

Using this we can rewrite the equation for the $$a(v_0)a(v_1)(1+v_0v_1\varepsilon) = a(v_2)$$ and $$\,a(v_0)a(v_1)(-v_0-v_1) = -v_2 a(v_2)$$ by comparing each of the components at indeces (1,1) and (2,1) respectively. Substituting $$a(v_2)$$ yields

$$(1) \ \ a(v_0)a(v_1)(-v_0-v_1) = -v_2 a(v_0)a(v_1)(1+v_0v_1\varepsilon) $$

from which we have $$v_2 = \frac{v_0+v_1}{1+v_0v_1\varepsilon}$$.

Reinserting this into the first equation we have $$ a(v_0)a(v_1)(1+v_0v_1\varepsilon) = a(\frac{v_0+v_1}{1+v_0v_1\varepsilon})$$.

In case $$ v := v_0 = -v_1$$ this simplifies to $$ a(v)a(-v)(1-v^2\varepsilon) = a(0) = 1$$ as $$\, L(0)$$ is the identity matrix.

Let $$\varphi(x) = \varphi(-x)$$ be an even function and $$\psi(x) = -\psi(-x)$$ an odd function such that $$a(v) = e^{\varphi(v) + \psi(v)}$$.

This gives us $$e^{\varphi(v) + \psi(v) + \varphi(v) - \psi(v)-2\pi\operatorname i k} = (1-v^2\varepsilon)^{-1}$$. Taking the logarithm and simplifying yields $$\varphi(v) = -\frac12\ln (1-v^2\varepsilon) + \pi\operatorname i k$$ from which we get $$a(v)=\pm\frac{1}{\sqrt{(1-v^2\varepsilon)}} e^{\psi(v)}$$ with any odd function $$\,\psi(-v)=-\psi(v)$$. The negative sign is ruled out by $$\, a(0)=1$$.

Substituting $$\, a(v)$$ in equation (1) verifies the solution but leaves us with $$\psi(v_0) + \psi(v_1) = \psi(\frac{v_0+v_1}{1+v_0v_1\varepsilon})$$.

To prove $$\psi(v) \equiv 0 $$ we flip the direction of the x-axis in the base system as well as in the transformed system. The choice of the direction of the axis must not lead to different results. In the flipped coordinates the Lorentz transformation takes the form $$L(v') = a(v') \left(\begin{matrix} 1 & -v'\varepsilon \\ -v' & 1\end{matrix}\right) = a(-v) \left(\begin{matrix} 1 & v\varepsilon \\ v & 1\end{matrix}\right) $$ for $$v' = -v$$.

On the other hand we can compute $$L(v')$$ by using a coordinate transformation on $$L(v)$$:

$$ L(v') = \left(\begin{matrix} 1 & 0\\ 0 & -1\end{matrix}\right) a(v) \left(\begin{matrix} 1 & v\varepsilon \\ v & 1\end{matrix}\right) \left(\begin{matrix} 1 & 0 \\ 0 & -1\end{matrix}\right) = a(v)\left(\begin{matrix} 1 & -v\varepsilon \\ -v & 1\end{matrix}\right)$$ from which we get $$\,a(-v) = a(v)$$ and hence $$\,\psi(-v)=\psi(v)$$. Together with $$\,\psi(-v)=-\psi(v)$$ this gives $$\psi(v) \equiv 0$$.

Looking at the eigenvectors of L(v) we see they are $$\left(\begin{matrix} 1 \\ \pm\frac{1}{\sqrt\varepsilon}\end{matrix}\right)$$ independent of v.

This is equivalent to a transformation invariant velocity $$\frac1\sqrt\varepsilon=:c_0$$ which experimentation shows to be equal to the speed of light. Altogether we have:

$$L(v)= \frac{1}{\sqrt{1-\frac{v^2}{c_0^2}}}\left(\begin{matrix} 1 & -\frac{v}{c_0^2} \\ -v & 1\end{matrix}\right) $$

Note that for $$\varepsilon=0$$ we would have the Galileian case with no finite invariant velocity.

Behold that if we did the same deduction not in the spacetime plane for (t, x) but in the space plane (x, y) with slope m=y/x instead of velocity v=x/t nothing would change except that, not by calculation but by experimentation alone, it would have turned out that $$\,\varepsilon = -1$$ and L(m) is just a rotation in the  euclidian space with no real valued eigenvectors.

It is an interesting exercise to compare this result with the assumptions of reciprocity, homogeneity and isotropy mentioned earlier. Do the linear equations contained in the matrix formulation of the Lorentz transformation above assume a particular spacetime from the outset?

It is further important to mention that the determinant $$\, \det L(v) = 1$$ independent of velocity. This directly implicates that spacetime volumes are conserved.

The geometry of space-time
SR uses a 'flat' 4-dimensional Minkowski space, which is an example of a space-time. This space, however, is very similar to the standard 3 dimensional Euclidean space, and fortunately by that fact, very easy to work with.

The differential of distance(ds) in Cartesian 3D space is defined as:

 * $$ ds^2 = dx_1^2 + dx_2^2 + dx_3^2 $$

where $$(dx_1,dx_2,dx_3)$$ are the differentials of the three spatial dimensions. In the geometry of special relativity, a fourth dimension, time, is added, with units of c, so that the equation for the differential of distance becomes:

 * $$ ds^2 = dx_1^2 + dx_2^2 + dx_3^2 - c^2 dt^2 $$

In many situations it may be convenient to treat time as imaginary (e.g. it may simplify equations), in which case $$t$$ in the above equation is replaced by $$i.t'$$, and the  metric becomes

 * $$ ds^2 = dx_1^2 + dx_2^2 + dx_3^2 + c^2(dt')^2 $$

Note that ds in this case is the distance, and not the interval. Caution should be exercised in the use of 'imaginary' time, it is not part of the modern theory of relativity. Blandford and Thorne (2004) in their "Applications of Classical Physics" (http://www.pma.caltech.edu/Courses/ph136/yr2004/ ), write the following about imaginary time: "(i) it hides the true physical geometry of Minkowski spacetime, (ii) it cannot be extended in any reasonable manner to non-orthonormal bases in flat spacetime, and (iii) it cannot be extended in any reasonable manner to the curvilinear coordinates that one must use in general relativity."

If we reduce the spatial dimensions to 2, so that we can represent the physics in a 3-D space

 * $$ ds^2 = dx_1^2 + dx_2^2 - c^2 dt^2 $$

We see that things such as light which move at the speed of light lie along a dual-cone:

defined by the equation

 * $$ ds^2 = 0 = dx_1^2 + dx_2^2 - c^2 dt^2 $$

or

 * $$ dx_1^2 + dx_2^2 = c^2 dt^2 $$

Which is the equation of a circle with $$r=c dt$$. The path of something that moves at the speed of light is known as a null geodesic. If we extend the equation above to three spatial dimensions, the null geodesics are continuous concentric spheres, with radius = distance = c×(±time).

 * $$ ds^2 = 0 = dx_1^2 + dx_2^2 + dx_3^2 - c^2 dt^2 $$

 * $$ dx_1^2 + dx_2^2 + dx_3^2 = c^2 dt^2 $$

This null dual-cone represents the "line of sight" of a point in space. That is, when we look at the stars and say "The light from that star which I am receiving is X years old.", we are looking down this line of sight: a null geodesic. We are looking at an event $$d = \sqrt{x_1^2+x_2^2+x_3^2} $$ meters away and d/c seconds in the past. For this reason the null dual cone is also known as the 'light cone'. (The point in the lower left of the picture below represents the star, the origin represents the observer, and the line represents the null geodesic "line of sight".)

The cone in the -t region is the information that the point is 'receiving', while the cone in the +t section is the information that the point is 'sending'.

Length contraction, time dilation and phase
Consider two inertial frames in standard configuration. There is a rigid rod moving along in the second frame at v m/s. The length of the rod is determined by observing the positions of the end points of the rod simultaneously - if the rod is moving it would be nonsense to use any other measure of length. An observer who is moving at the same velocity as the rod measures its "rest length". The Lorentz transformation for coordinates along the x axis is:

$$x^' = \gamma (x - vt)$$

Suppose the positions, $$x_1, x_2$$, of the two ends of the rod are determined simultaneously (ie: $$t$$ is constant):

$$(x^'_1 - x^'_2) = \gamma (x_1 - x_2) $$

Or, using $$L_0 = (x^'_1 - x^'_2) $$ for the rest length of the rod and $$L = (x_1 - x_2)$$ for the length of the rod that is measured by the observer who sees it fly past at v m/s:

$$\mathbf{L_0 = \gamma L}$$

Or, elaborating $$\gamma$$:

$$\mathbf{L = {L_0}\sqrt{1-v^2/c^2}}$$

In other words the length of an object moving with velocity $$v$$ is contracted in the direction of motion by a factor $$\sqrt{1-v^2/c^2}$$ in the direction of motion.

The Lorentz transformation also affects the rate at which clocks appear to change their readings. The Lorentz transformation for time is:

$$t^' = \gamma (t - vx/c^2)$$

This transformation has two components:

$$t^' = \gamma t - \gamma vx/c^2$$

and is a straight line graph (ie: $$t^' = mt + c$$).

The gradient of the graph is $$\gamma$$ so:

$$\Delta t^' = \gamma \Delta t$$

or:

$$t^'_1-t^'_2 = \gamma (t_1 - t_2)$$

Therefore clocks in the moving frame will appear to go slow, if $$T_0$$ is a time interval in the rest frame and $$T$$ is a time interval in the moving frame:

$$T = \gamma T_0$$

Or, expanding:

$$\mathbf{T = \frac{T_0}{\sqrt{1-v^2/c^2}}}$$

The intercept of the graph is:

$$-\gamma vx/c^2$$

This means that if a clock at point $$x$$ is compared with a clock that was synchronised between frames at the origin it will show a constant time difference of $$\gamma vx/c^2$$ seconds. This quantity is known as the relativistic phase difference or "phase".

The relativistic phase is as important as the length contraction and time dilation results. It is the amount by which clocks that are synchronised at the origin go out of synchronisation with distance along the direction of travel. Phase affects all clocks except those at the point where clocks are syncronised and the infinitessimal y and z planes that cut this point. All clocks everywhere else will be out of synchronisation between the frames. The effect of phase is shown in the illustration below:

If the inertial frames are each composed of arrays of clocks spread over space then the clocks will be out of synchronisation as shown in the illustration above.

It is interesting to see how the phase term cancels out when time differences are being considered. It cancels out because the phase term $$x^'v/c^2$$ is applied to the clock that is at a constant position in the primed frame. The Lorentz transform for time is:

 * $$t = \frac{t^' + (v/c^2)x^'}{\sqrt{(1 - v^2/c^2)}} \,$$

A difference between two times is:

 * $$t_1 - t_2 = \frac{t_1^' + (v/c^2)x^' - t_2^' - (v/c^2)x^'}{\sqrt{(1 - v^2/c^2)}} \,$$

The important thing to note is that the clock is at the same position $$x^'$$ in its own reference frame.

The result is that:

 * $$t_1 - t_2 = \frac{t_1^' - t_2^'}{\sqrt{(1 - v^2/c^2)}} \,$$

Where the phase terms have cancelled.

 * The Time Dilation Measurement through the Ives-Stilwell Experiment

Hyperbolic geometry
In the flat spacetime of Special Relativity:

$$s^2 = c^2t^{2} - x^{2} - y^2 - z^2$$

Considering the x-axis alone:

$$s^2 = c^2t^{2} - x^{2}$$

The standard equation of a hyperbola is:

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

In the case of spacetime:

$$\frac{(ct)^2}{s^2} - \frac{x^2}{s^2} = 1$$

Spacetime intervals separate one place or event in spacetime from another. So, for a given motion from one place to another or a given fixed length in one reference frame, given time interval etc. the metric of spacetime describes a hyperbolic space. This hyperbolic space encompasses the coordinates of all the observations made of the given interval by any observers.

It is possible to conceive of rotations in hyperbolic space in a similar way to rotations in Euclidean space. The idea of a rotation in hyperbolic space is summarised in the illustration below:

A rotation in hyperbolic space is equivalent to changing from one frame of reference to another whilst observing the same spacetime interval. It is moving from coordinates that give:

$$(ct)^2 - x^2 = s^2$$

to coordinates that give:

$$(ct^')^2 - {x^'}^2 = s^2$$

The formula for a rotation in hyperbolic space provides an alternative form of the Lorentz transformation ie:

$$\begin{bmatrix} c t \\x \end{bmatrix} = \begin{bmatrix} \cosh \phi & \sinh \phi \\ \sinh \phi & \cosh \phi \end{bmatrix} \begin{bmatrix} c t'\\x' \end{bmatrix}. $$

From which:

$$x = x^' \cosh \phi + ct^' \sinh \phi$$

$$ct = x^' \sinh \phi + ct^' \cosh \phi$$

The value of $$\phi$$ can be determined by considering the coordinates assigned to a moving light that moves along the x axis from the origin at $$v m sec^1$$ flashes on for $$t$$ seconds then flashes off.

The coordinates assigned by an observer on the light are: $$t',0,0,0$$, the coordinates assigned by the stationary observer are $$t, x=vt, 0, 0$$. The hyperbola representing these observations is illustrated below:

The equation of the hyperbola is:

$$(ct)^2 - x^2 = s^2 = (ct^')^2$$

but x=vt for the end of the flash so:

$$\tanh \phi = \frac {v}{c}$$

Now, from hyperbolic trigonometry:

$$\frac{1}{\sqrt{1-\tanh^2{\phi}}} = \sqrt{ \frac{\cosh^2\phi} {\cosh^2\phi-\sinh^2\phi} }= \cosh\phi$$

But $$\tanh\phi = \frac{v}{c}$$ so:

$$\cosh\phi = \frac{1}{\sqrt{1-v^2/c^2}} = \gamma$$

and, from the hyperbolic trigonometric formula $$\sinh\phi = \tanh\phi \cosh\phi$$:

$$\sinh\phi = {\frac{v}{c}}\gamma$$

Inserting these values into the equations for the hyperbolic rotation:

$$x = x^' \cosh \phi + ct^' \sinh \phi$$

$$x = \gamma x^' + ct^' \gamma v/c $$

Which gives the standard transform for x:

$$x = \gamma ( x^' + vt^')$$

In a similar way $$ct = x^' \sinh \phi + ct^' \cosh \phi$$ is equivalent to:

$$t = \gamma (t^' + vx/c^2)$$

So the Lorentz transformations can also be derived from the assumption that boosts are equivalent to rotations in hyperbolic space with a metric $$s^2 = c^2t^{2} - x^{2} - y^2 - z^2$$.

The hyperbolic angle $$\phi$$ is known as the rapidity of the boost.

Addition of velocities
Suppose there are three observers 1, 2, and 3 who are moving at different velocities along the x-axis. Observers 1 and 2 are moving at a relative velocity $$v$$ and observers 2 and 3 are moving at a relative velocity of $$u^'$$. The problem is to determine the velocity of observer 3 as observed by observer 1 ($$u$$).

It turns out that there is a very convenient relationship between rapidities that solves this problem:

If $$v/c = \tanh{\phi}$$ and $$u^'/c = \tanh{\alpha}$$ then:

$$u/c = \tanh(\phi + \alpha)$$

In other words the rapidities can be simply added from one observer to another ie:

$$\sigma = \phi + \alpha$$

Hence:

$$\tanh(\sigma) = \tanh(\phi + \alpha)$$

So the velocities can be added by simply adding the rapidities. Using hyperbolic trigonometry:

$$u/c = \tanh(\alpha + \phi) = \frac{\tanh{\alpha} + \tanh{\phi}}{1+\tanh{\alpha}\tanh{\phi}} = \frac{u^'/c + v/c}{1+u^'v/c^2}$$ Therefore:

$$\mathbf{u = \frac{u^' + v}{1+u^'v/c^2}}$$

Which is the relativistic velocity addition theorem.

The relationship $$u/c = \tanh(\phi + \alpha)$$ is shown below:

Velocity transformations can be obtained without referring to the rapidity. The general case of the transformation of velocities in any direction is derived as follows:

$$\mathbf{u^'}= (u^'_1, u^'_2, u^'_3) $$

where $$u^'_1$$ etc. are the components of the velocity in the x, y, z directions.

Writing out the components of velocity:

$$u^'_1 = dx^'/dt^'$$

$$u^'_2 = dy^'/dt^'$$

$$u^'_3 = dz^'/dt^'$$

But from the Lorentz transformations:

$$dx^' = \gamma (dx - v dt)$$

$$dy^' = dy$$

$$dz^' = dz$$

$$dt^' = \gamma (dt - vdx/c^2)$$

Therefore:

$$u^'_1 = dx^'/dt^' = \frac{\gamma (dx - v dt)}{\gamma (dt - vdx/c^2)}$$

$$u^'_2 = dy^'/dt^' = \frac{dy}{\gamma (dt - vdx/c^2)}$$

$$u^'_3 = dz^'/dt^' = \frac{dz}{\gamma (dt - vdx/c^2)}$$

Dividing top and bottom of each fraction by $$dt$$:

$$u^'_1 = \frac{\gamma (dx/dt - v)}{\gamma (1 - vdx/dt/c^2)}$$

$$u^'_2 = \frac{dy/dt}{\gamma (1 - vdx/dt/c^2)}$$

$$u^'_3 = \frac{dz/dt}{\gamma (1 - vdx/dt/c^2)}$$

Substituting $$\mathbf{u} = (u_1, u_2, u_3)$$

$$u^'_1 = \frac{u - v}{1 - uv/c^2}$$

$$u^'_2 = \frac{u_2}{\gamma (1 - uv/c^2)}$$

$$u^'_3 = \frac{u_3}{\gamma (1 - uv/c^2)}$$

The full velocity transformations are tabulated below:

Having calculated the components of the velocity vector it is now possible to calculate the magnitudes of the overall vectors between frames:

$$u = \sqrt{u_1^2 + u_2^2 + u_3^2}$$

$$u^' = \sqrt{u^{'2}_1 + u^{'2}_2 + u^{'2}_3}$$

Addition of velocities - another approach
In classical physics, velocities simply add. If an object moves with speed u in one reference frame, which is itself moving at v with respect to a second frame, the object moves at speed u+v in that second frame.

This is inconsistent with relativity because it predicts that if the speed of light is c in the first frame it will be v+c in the second.

We need to find an alternative formula for combining velocities. We can do this with the Lorentz transform.

Because the factor v/c will keep recurring we shall call that ratio β.

We are considering three frames; frame O, frame O' which moves at speed u with respect to frame O, and frame O" which moves at speed v with respect to frame O'.

We want to know the speed of O" with respect to frame O,U which would classically be u+v.

The transforms from O to O' and O' to O" can be written as matrix equations,

 * $$\begin{pmatrix} x' \\ ct' \end{pmatrix} $$$$= \gamma \begin{pmatrix} 1 & - \beta \\ -\beta & 1 \end{pmatrix}

\begin{pmatrix} x \\ ct \end{pmatrix} \quad$$
 * $$\begin{pmatrix} x \\ ct \end{pmatrix} $$$$= \gamma'

\begin{pmatrix} 1 & - \beta' \\ -\beta' & 1 \end{pmatrix} \begin{pmatrix} x' \\ ct' \end{pmatrix} $$

where we are defining the β's and γ's as

 * $$\begin{matrix}

\beta = \frac{u}{c} & \gamma = \frac{1}{\sqrt{1-\beta^2 }} \\ \beta^\prime = \frac{v}{c} & \gamma^\prime = \frac{1}{\sqrt{1-{\beta^\prime}^2 }} \end{matrix} $$

We can combine these to get the relationship between the O and O" coordinates simply by multiplying the matrices, giving

\begin{pmatrix} x \\ ct \end{pmatrix} = \gamma \gamma^\prime \begin{pmatrix} 1+\beta \beta' & - (\beta + \beta') \\ - (\beta + \beta') & 1+\beta \beta'  \end{pmatrix} \begin{pmatrix} x \\ ct \end{pmatrix} \quad (1)$$

This should be the same as the Lorentz transform between the two frames,

\begin{pmatrix} x \\ ct \end{pmatrix} = \gamma \begin{pmatrix} 1 & - \beta \\ -\beta'' & 1 \end{pmatrix} \begin{pmatrix} x \\ ct \end{pmatrix} \quad (2) \mbox{ where } \begin{matrix} \beta'' & = & \frac{U}{c} \\ \gamma & = & \frac{1}{\sqrt{1-{\beta}^2 }} \end{matrix} $$

These two sets of equations do look similar. We can make them look more similar still by taking a factor of 1+ββ' out of the matrix in (1) giving:

\begin{pmatrix} x \\ ct \end{pmatrix} = \gamma \gamma' (1+\beta \beta') \begin{pmatrix} 1 & - \frac{\beta + \beta'}{1+\beta \beta'} \\ - \frac{\beta + \beta'}{1+\beta \beta'} & 1  \end{pmatrix} \begin{pmatrix} x \\ ct \end{pmatrix} $$

This will be identical with equation 2 if

 * $$\beta''=\frac{\beta + \beta'}{1+\beta \beta'} \mbox{ (3a)  and }

\gamma'' = \gamma \gamma' (1+\beta \beta') \mbox{ (3b)}$$

Since the two equations must give identical results, we know these conditions must be true.

Writing the β's in terms of the velocities equation 3a becomes

 * $$\frac{U}{c}=\frac{\frac{u}{c} + \frac{v}{c}}{1+\frac{uv}{c^2}}$$

which tells us U in terms of u and v.

A little algebra shows that this implies equation 3b is also true

Multiplying by c we can finally write.

$$U = \frac{u+v}{1+\frac{uv}{c^2}} $$

Notice that if u or v is much smaller than c the denominator is approximately 1, and the velocities approximately add but if either u or v is c then so is U, just as we expected.

Acceleration transformation
This section is under development.

It was seen above that:

$$\frac {u}{c} = \tanh{\phi}$$

and, if $$\frac {v}{c} =\tanh \beta$$ and  $$\frac{u^'}{c} = \tanh \epsilon$$ the velocity addition theorem can be expressed as the sum of the rapidities:

$$\phi = \beta + \epsilon $$

If we differentiate this equation with respect to $$t$$ to investigate acceleration, then assuming $$v$$ is constant:

$$\frac {d\phi}{dt} = \frac{d\epsilon}{dt}$$

so

(1) $$\frac {d\phi}{dt} = \frac{d\epsilon}{dt^'}\frac{dt^'}{dt}$$

But $$\frac {d\phi}{dt} \!$$ is also equal to:

$$\frac {d\phi}{dt} = \frac {d\phi}{du}\frac{du}{dt}$$

But $$\phi = \tanh^{-1} (u/c) \!$$ and the derivative of an arctangent is given by:

$$\frac{d \tanh^{-1}(x)}{dx} = \frac{1}{1-x^2}$$

and hence:

$$\frac{d\phi}{du} = \frac{1}{c}\frac {1}{(1-u^2/c^2)}$$

For brevity we will use the notation:

$$\frac {1}{1-u^2/c^2} = \gamma^2(u)$$

ie: $$\gamma(u)$$ is gamma for observers moving at a relative velocity of $$u$$.

So:

(2)$$\frac {d\phi}{dt} = \frac{1}{c} \gamma^2(u) \frac{du}{dt}\!$$

The velocity of the object observed by the unprimed and primed observers is:

$$u^2 = {u_1}^2 + {u_2}^2 + {u_3}^2 \!$$

$$u^{'2} = {u_1}^{'2} + {u_2}^{'2} + {u_3}^{'2}\!$$

and:

$$u = \frac{dr}{dt}\!$$

$$u^' = \frac{dr^'}{dt^'}\!$$

So:

$$u^2dt^2 = dr^2 \!$$

$$u^{'2}dt^{'2} = dr^{'2} \!$$

and using the Minkowski metric:

$$ds^2 = dr^2 - cdt^2 = dr^{'2} - cdt^{'2} \!$$

So:

$$u^2dt^2 - cdt^2 = u^{'2}dt^{'2} - cdt^{'2} \!$$

Hence:

$$dt^2(c^2 - u^2) = dt^{'2}(u^{'2} - c^2) \!$$

Given that $$\gamma (u) = \frac {c}{\sqrt{c^2 - u^2}} \!$$ and $$\gamma (u^') = \frac {c}{\sqrt{c^2 - u^{'2}}}\!$$

(3)$$\frac{dt^'}{dt} = \frac {\gamma(u^')}{\gamma(u)}$$

Therefore, substituting (2) and (3) in (1):

$$\frac{1}{c} \gamma^2(u) \frac{du}{dt} = \frac{d\epsilon}{dt^'}\frac {\gamma(u^')}{\gamma(u)}$$

Applying the differential of arctanh as before to determine $$\frac{d\epsilon}{dt^'}$$:

$$\gamma^3(u^')\frac {du^'}{dt^'} = \gamma^3(u)\frac {du}{dt}$$

This is a different result from the Newtonian formula in which $$du/dt = du^'/dt^'$$. The proper acceleration, $$\alpha \!$$ is defined as the acceleration of an object in its rest frame. It is the instantaneous change in velocity for an observer for whom $$u^'=0 \!$$ and $$\alpha = du^'/dt^' \!$$. In these circumstances:

$$\alpha = \gamma^3(u)\frac{du}{dt}$$

The term $$\frac{du}{dt}\!$$ is called "coordinate acceleration".

Another approach to acceleration
Classically we would talk about a particle at x(t) with acceleration d²x/dt². In relativity we must treat time as just another coordinate, and use derivatives with respect to proper time, which means our notion of acceleration will change. We can assume the particle is moving slower than light.

Since τ increases monotonically with t (no time travel allowed) we can easily parameterise the path of the particle with τ rather than t. Derivatives with respect to these two variables then differ only by a factor of γ

 * $$\frac{d}{d\tau} = \gamma \frac{d}{dt} $$

which gives us the connection between the classical and relativistic formulae.

Lets suppose we have a particle moving on the path (x(τ),ct(τ)), where τ is the particle's proper time.

Its velocity four vector is

 * $$\underline{u} = \frac{d}{d\tau}(x, ct) = \gamma (\dot{\mathbf{x}}, c) $$

In the particles rest frame this is (0,c), which has constant magnitude -c2, but this scalar must be the same in all frames, thus

 * $$\underline{u} \cdot \underline{u} = -c^2 $$

That is, the magnitude of the velocity is always constant.

Differentiating this, we can immediately say

 * $$\underline{u} \cdot \underline{a} =

\frac{d}{d\tau} \underline{u} \cdot \underline{u}= 0 $$

so, velocity and acceleration are always perpendicular.

These two results are much simpler than in classical physics.

Differentiating the velocity gives us, after some algebra ,

 * $$\underline{a} = \gamma^4 \ddot{x} (1, \frac{v}{c}) $$

when the motion is along the x-axis. Since the spatial and temporal components must be 3-vectors and scalars respectively we immediately see that for motion in arbitrary directions,

 * $$\underline{a} = \gamma^4 \left( \ddot{\mathbf{x}}, \frac{\ddot{\mathbf{x}}\cdot \mathbf{v} }{c} \right) $$

The magnitude of this is

 * $$|\underline{a}| = \gamma^3 \ddot{x} $$

which is the classical magnitude, corrected by a factor dependent on γ. When the velocity is much less than c this factor is approximately 1, so in that limit the magnitudes are the same, as they should be.

Knowing this, we can work out the equation of motion for a particle with constant acceleration a along the x-axis. For simplicity, we'll assume the initial velocity is 0.

Since the acceleration is constant

 * $$ \frac{\dot{v}}{\left( 1-\frac{v^2}{c^2} \right)^{\frac{3}{2}}} = a $$

Integrating once we get

 * $$\frac{v}{\sqrt{1-\frac{v^2}{c^2}}} = at $$

or, on rearranging

 * $$ v = \frac{at}{\sqrt{1+a^2t^2/c^2}} $$

If at is much less than c this gives a velocity of approximately at, identical to the classical result, but as t tends to infinity the velocity tends to c.

The position is

 * $$ x = d+ \frac{c^2}{a}\sqrt{1+ \frac{a^2 t^2}{c^2}}$$

where d is some constant. We can choose our coordinate system to make d zero.

Calling the interval between the origin and the particle, I, we have

 * $$\begin{matrix}

I^2 & = & x^2 & - & c^2t^2 \\ & = & \frac{c^4}{a^2} \left( 1+ \frac{a^2 t^2}{c^2} \right) & - & c^2t^2\\ & = & \frac{c^4}{a^2} & & \end{matrix} $$

Thus the interval between the particle and the origin is constant. Notice, this is the equation for an hyperbola, so we know the particle's trajectory.

Momentum
Classically, momentum is velocity multiplied by mass. We can use the same definition in relativity, and see where it takes us.

 * $$\underline{p}=m_0 \underline{u}= m_0 \gamma (\mathbf{v}, c)$$

You may sometimes see the product m0γ which shows up here called the relativistic mass, but we will not be using this approach.

The mass, m0, is generally called the rest mass, to distinguish it from the relativistic mass.

The spatial component of the four-momemntum is clearly the classical momentum, scaled by a factor of γ. At speeds much less than c this will be approximately 1.

The temporal component is m0γc. To see what this means we can look at its value when v/c is small.

 * $$m_0 \gamma c = \frac{m_0 c}{\sqrt{1-v^2/c^2}}=

m_0 c \left( 1 + \frac{1}{2}\frac{v^2}{c^2} + \frac{3}{8}\frac{v^4}{c^4} + \cdots \right) $$

The first term in this expansion is a constant.

The second term is
 * $$\frac{m_0 v^2}{2c}$$

which we recognise as being the classical kinetic energy, divided by c.

Now, adding a constant to the definition of kinetic energy makes no real difference, since all that matters are changes in energy, so we can identify this temporal component of relativistic momentum with the energy over c.

 * $$\underline{p}= (\gamma \mathbf{p}, E/c)$$

We then have

 * $$E = m_0 c^2 + \frac{1}{2}m_0 v^2 + \frac{3}{8}\frac{m_0 v^4}{c^2}

+ \cdots $$

Even at rest, the particle has a kinetic energy,

 * $$E=m_0 c^2, \,$$

the most famous relativistic equation.

Force
Classically, we have

 * $$\mathbf{F}= \frac{d\mathbf{p}}{dt} $$

We can get the equivalent relativistic equation simply by replacing 3-vectors with 4-vectors and t with τ, giving

 * $$\underline{F} = \frac{d\underline{p}}{d\tau}$$

Provided the rest mass is constant, as it is for all simple systems, we can rewrite this as

 * $$\underline{F} = m \underline{a}$$

We already know about a so we can now write

 * $$\underline{F} = \gamma^4 \left( \mathbf{F},

\frac{ \mathbf{F} \cdot \mathbf{v} }{c} \right) $$

The temporal component of this is essentially the power, the rate of change of energy with time, as might be expected from energy being the temporal component of momentum.

Accelerated Frames and Event Horizons
Figure: Spacetime diagram showing the world line of the origin of a reference frame undergoing constant acceleration.

We know this world line is an hyperbola, with asymptotes x=±ct. Looking at the diagram show another property.

No light from the 'twilight zone', where x is smaller than ct, can ever reach the accelerated observer. Hence, no event in the twilight zone can affect the observer.

Conversely, there is a region, where x is smaller than -ct, which no light from the accelerated observer can ever reach. Hence the observer can not affect any events in this region.

Because of this we call the two asymptotes event horizons. In some ways, they are like one-way barriers in space-time. The accelerated observer can see objects vanishing behind an event horizon, but they never see any re-emerge from it.

These particular event horizons are observer dependent, as are all event horizons in special relativity. However, general relativity permits event-horizons which are observer independent.

We are not going to prove this, but we can some idea about how event horizons apply to general relativity by using its basic principle, called the equivalence principle by Einstein: that gravity is nothing more than an inertial force.

This means that gravity is locally indistinguishable from forces due to acceleration; if the elevator doors are closed, we can't tell whether we are stationary in a 1g gravity field or being accelerated at 1g in zero gravity.

This suggests that something can be learned about general relativity by examining the properties of accelerated reference frames

Now, since the gravitational force on the Earth points downward, it follows that we must be constantly accelerating upward as we stand on the surface of the Earth. The obvious problem with this interpretation of gravity is that we don't appear to be moving away from the center of the Earth, which would seem to be a natural consequence of such an acceleration.

However, the calculation of the previous chapter shows that even though the object associated with the curved world line in the figure above is accelerating away from the origin, it always remains the same distance (in its own frame) from the origin. In other words, even though we are accelerating away from the center of the Earth, the distance to the center of the Earth remains constant!

Therefore, because of the equivalence principle, we can expect to see the same things as an accelerated observer in special relativity, including an event horizon.

For acceleration in special relativity, inertial objects pass the event horizon at speed c at time infinity, relative to the observer.

Similarly, in general relativity, freely falling objects pass the event horizon at speed c at time infinity, relative to a static observer. By reversing the trajectories we can conclude that at the event horizon the escape velocity is c, i.e gravity is so strong light can not escape.

Thus, using the equivalence principle, and behaviour under acceleration in special relativity, we've predicted black holes.

There isn't actually an event horizon inside the earth, since the mass of the earth is not concentrated in a point and the earth can only be treated as a point mass while outside the earth's surface, which lies outside where the event horizon should otherwise have been.

The red shift
Spacetime diagram for explaining the gravitational red shift.

Light emitted at a lower level in a gravitational field has its frequency reduced as it travels to a higher level. This phenomenon is called the gravitational red shift.

We can see why this happens by using the principle of equivalence. Being in a gravitational field is equivalent to being in an accelerated frame, so knowing how the doppler shift works in such a frame will tell us how it works in a gravitational field.

We view the process of light emission and absorption from the unaccelerated or inertial frame, as shown in the figure above. In this reference frame the observer of the light is accelerating to the right, as indicated by the curved red world line, which is equivalent to a gravitational force to the left.

The light is emitted at point A with frequency $$\omega$$ by a source which is stationary at this instant. At this instant the observer is also stationary in this frame. However, by the time the light gets to the observer, they have a velocity to the right which means that the observer measures a Doppler shifted frequency $$\omega'$$ for the light. Since the observer is moving away from the source, $$\omega'<\omega$$, as indicated above.

The relativistic Doppler shift is given by

 * $$ \frac{\omega'}{\omega} = \sqrt{ \frac{1 - U/c}{1 + U/c}}, $$

so we need to compute U/c. The line of simultaneity for the observer at point B goes through the origin, and is thus given by line segment OB. The slope of this line is U/c, where U is the velocity of the observer at point B. From the figure we see that this slope is also given by the ratio X' /X.

Equating these, eliminating X in favor of L = √(X2 - X′2), which is the actual invariant distance of the observer from the origin, and substituting into the previous equation results in our gravitational red shift formula:

 * $$\begin{matrix}

\frac{\omega'}{\omega} & = & \sqrt{ \frac{X - X'}{X + X'} } & = & \sqrt{\frac{(L^2 + X'^2 )^{1/2} - X'}{(L^2 + X'^2 )^{1/2} + X'} }\\ & =& \frac{(L^2 + X'^2 )^{1/2} - X'}{L} & =& \frac{X-X'}{L} \end{matrix}$$

If X′ = 0, then there is no redshift, because the source is collocated with the observer. On the other hand, if the source is located at the origin, so X′=X, the Doppler shifted frequency is zero. In addition, the light never gets to the observer, since the world line is asymptotic to the light world line passing through the origin. If the source is at a higher level in the gravitational field than the observer, so that X′ < 0, then the frequency is shifted to a higher value, i. e., it becomes a blue shift.

To see how this doppler shift relates to the strength of gravity, g, and the distance h between the source and the observer, first note that

 * $$L=\frac{c^2}{g} \qquad X-X'=L-h $$

Making these substitutions gives

 * $$\frac{\omega'}{\omega} = 1-\frac{gh}{c^2} \qquad (1)$$

So the redshift is proportional to gravity.

Since this doppler shift doesn't depend on the type of wave we can conclude that it is actually caused by time dilation, just like the doppler shift due to relative motion.

That is, gravity slows down time.

Energy and frequency
In equation 1 gh is the change in gravitational potential energy so the change in frequency is proportional to change in potential energy, which suggests there might be a connection with energy conservation.

However, since we haven't yet established any connection between frequency and energy, we can't simply apply an energy conservation argument. Instead, we can argue in reverse, finding out what the energy-frequency relationship must be if energy is to be conserved.

Suppose we have two identical systems, both at rest in a uniform gravitational field g with initial energy E separated by a vertical distance h.

The system has mass E/c2, giving it potential energy, so the total energy of the two systems is initially

 * $$E_i=2E - \frac{gh}{c^2}E $$

where the second term is due to the lesser potential energy of the lower system.

The lower system emits a burst of waves, frequency ω, energy E(ω). When this reaches the upper system the waves have been red-shifted to frequency ω', energy E(ω'). This energy is absorbed by the upper system.

The total energy is now

 * $$E_f=E-E(\omega)+E+ E(\omega') - \left(E-E(\omega)\right)\frac{gh}{c^2}

$$

Since we want to preserve energy conservation, these two equations must give the same result. Equating them, we get

 * $$E(\omega') = E(\omega)(1-\frac{gh}{c^2})$$

Comparing this with the doppler shift we see that

 * $$\frac{E(\omega')}{E(\omega)} = \frac{\omega'}{\omega} $$

which can only be true if E(ω) is proportional to ω

I.e, energy conservation implies energy is proportional to frequency, which is one of the axioms of quantum theory.

We could equally well have started with the quantum result and proved the gravitational red shift must exist. Either theory requires the other for consistency.

Since energy and frequency are each the temporal components of a four-vector, their being proportional implies the four-vectors themselves, and their spatial components, are also proportional. So, for waves, momentum is proportional to k.

Remember too, we saw earlier, when we looked at Hamilton's equations, that classical mechanics would be equivalent to a theory of anisotropic waves, in the geometrical optics limit, if energy were proportional to frequency and momentum to wave number. This proportionality isn't just required for energy conservation, it would make possible a theory uniting waves and particles.

None of this actually proves the proportionality, doing that requires experiment, but it does make it a natural assumptiom, which is indeed confirmed by experiment.

Because of all this, from now on we'll assume that energy and frequency are related in this way, with the constant of proportionality being $$\hbar$$

 * $$ (c\mathbf{p}, E) = \hbar (c\mathbf{k}, \omega)$$

Gravity and curvature
The gravitational redshift also implies that space is curved. We can see this by considering a rectangle in space-time.

Without gravity, if we start at some point A, wait for time t, then move at light-speed to the right for a distance h, we get to the same place, B, as if we move at light-speed for a distance h then wait for time t at rest with respect to A.

With gravity, if we follow the first path, we rise a distance of ct then wait for time t. On the second path, we begin by waiting for time t, but this is dilated by gravity. To an observer at B we appear to be waiting for a time t(1+gh/c2) before we start, so we end up at B later than on the first path.

Thus, with gravity, it matters which order we add distance vectors in. This can't happen if space is flat, so space must be curved.

To describe how it's curved we'd need the techniques of General Relativity.

Potential Momentum
In classical physics we know that kinematics can often be described by a potential energy alone. Now we've seen that in relativity the energy is just the temporal component of the momentum 4-vector, so we should expect the same of the potential energy. To see how this works, we'll reason by analogy from the classical case.

For a free, non-relativistic particle of mass m, the total energy E equals the kinetic energy K and is related to the momentum Π of the particle by

 * $$ E = K = \frac{| \boldsymbol{\Pi} |^2}{2m} \qquad \mbox{(free, non-relativistic)} . $$

In the non-relativistic case, the momentum is Π= mv, where v is the particle velocity.

If the particle is not free, but is subject to forces associated with a potential energy U(x,y,z), then the equation must be modified to account for the contribution of U to the total energy:

 * $$ E - U = K = \frac{| \boldsymbol{\Pi} |^2}{2m} \qquad \mbox{(non-free, non-relativistic)} . $$

The force on the particle is related to the potential energy by

 * $$ \mathbf{F} = - \left( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y} , \frac{\partial U}{\partial z} \right) . $$

For a free, relativistic particle, we have
 * $$ E = ( | \boldsymbol{\Pi} |^2 c^2 + m^2 c^4 )^{1/2} \qquad \mbox{(free, relativistic)} . $$

The obvious way to add forces to the relativistic case is by rewriting this equation with a potential energy:

 * $$ E - U = ( | \boldsymbol{\Pi} |^2 c^2 + m^2 c^4 )^{1/2} \qquad \mbox{(incomplete!)} . $$

However $$\boldsymbol\Pi = (\Pi, E/c)$$ is a four-vector, so an equation with something subtracted from just one of the components of this four-vector is not relativistically invariant. In other words, this equation doesn't obey the principle of relativity, and therefore cannot be correct!

How can we fix this problem? One way is to define a new four-vector with U/c being its timelike part and some new vector Q being its spacelike part:
 * $$ \underline{Q} \equiv ( \mathbf{ Q}, U/c) \qquad \mbox{(potential four-momentum)} . $$

We then subtract Q from the momentum Π. When we do this, equation (13.5) becomes
 * $$ E - U = ( | \boldsymbol{\Pi} - \mathbf{ Q}|^2 c^2 + m^2 c^4 )^{1/2} \qquad \mbox{(non-free, relativistic)} . $$

The quantity Q is called the potential momentum and Q is the potential four-momentum.

If |Π-Q| is much smaller than mc, this becomes approximately

 * $$E = mc^2 + \frac{1}{2m}\left( \boldsymbol{\Pi} - \mathbf{Q} \right)^2 $$

This expression for the energy has the same form as the Hamiltonian we looked at for classical velocity dependent forces, so we know it predicts a force perpendicular to the velocity, when the condition is met. It turns out to be perpendicular even when the condition is met.

In classical physics the potential momentum is an optional extra. In relativity it is a necessary part of any potential field.

Some additional terminology is useful. We define
 * $$ \mathbf{ p} \equiv \boldsymbol{\Pi} - \mathbf{ Q} \qquad \mbox{(kinetic momentum}) $$

as the kinetic momentum since in the classical case it reduces to mv. In order to avoid confusion, we rename Π the total momentum. Thus, the total momentum equals the kinetic plus the potential momentum, in analogy with energy.

Conservation of 4 momentum
We earlier introduced the ideas of energy and momentum conservation. In other words, if we have a number of particles isolated from the rest of the universe, each with momentum pi and energy Ei, then particles may be created and destroyed and they may collide with each other.

In these interactions the energy and momentum of each particle may change, but the sum total of all the energy and the sum total of all the momentum remains constant with time:

 * $$ E = \sum_i E_i = \mbox{const} \qquad

\mathbf{p} = \sum_i \mathbf{p}_i = \mbox{const} $$

The expression is simpler in terms of four-momentum:

 * $$ \underline{p} = \sum_i \underline{p}_i = \mbox{const} $$

At this point a statement such as the one above should ring alarm bells. Just what does it mean to say that the total energy and momentum remain constant with time in the context of relativity? Which time? The time in which reference frame?

Suppose two particles exchange four-momentum remotely at the time indicated by the fat horizontal bar in the left panel. Conservation of four-momentum implies that

$$ \underline{p}_A + \underline{p}_B = \underline{p}'_A + \underline{p}'_B $$

where the subscripted letters correspond to the particle labels in the figure. Primed values refer to the momentum after the exchange while no primes indicates values before the exchange.

Now view the exchange from the reference frame in the right panel. A problem with four-momentum conservation exists in the region between the thin horizontal lines. In this region particle B has already transferred its four-momentum, but it has yet to be received by particle A. In other words, four-momentum is not conserved in this reference frame!

This problem is so serious that we must eliminate the concept of action at a distance from the repertoire of physics. The only way to have particles interact remotely and still conserve four-momentum in all reference frames is to assume that all remote interactions are mediated by another particle, or by a field.

If the force is being mediated by a particle then first, particle A emits particle C in a manner which conserves the four-momentum. Second, particle C is absorbed by particle B in a similarly conservative interaction.

If the force is being mediated by a field then first, particle A emits wave C, with momentum proportional to its wavenumber in a manner, which conserves the four-momentum. Wave C then travels at c or less until it is absorbed by particle B in a similarly conservative interaction.

We'll see that in quantum theory the difference between a particle and a field vanishes, so these two pictures actually both describe the same mechanism a different perspective. Which every picture we use, four-momentum is conserved at all times in all reference frames.

In other words, momentum and energy are transferred from particle A to particle B in a two step process. In between the momentum resides in a particle or field.


text:
User talk:Pluke

A-Level Computing
Hi, I'd be more than happy to help out, I also despise the Heathcote book, and have personally found it useless in some cases. I'll add bits and pieces to it, since computing is one the most enjoyable and easiest subjects for me. --Krackpipe 11:46, 11 January 2006 (UTC)

Hi, Thanks for the offer of help with the CIE section of the A-level Computing book. I'll be adding to it over the course of the year as I teach the course, but the more people there are working on it the better, so do feel free to dip in as you feel like it! Hoping that the book will become a great resource for any exam board. --Pastbury (discuss) 14:04, 27 June 2012 (UTC)

Print version
I completely don't understand where's the problem. The proper use of print version is: you go to A-level Computing/Print version and click "Press Refresh this page before you print to have the latest changes.". You will then see true print version and I can assure you won't see unneccessary templates there.

Anyway, because print version notice seemed to introduce some confusion, I changed it to be more clear. --Derbeth talk 09:04, 24 June 2006 (UTC)

 * Page A-level_Computing/Print_version is no true print version, you should not print it in this shape. Besides print version notice, there are also "edit" links, left navigation bar and lots of elements that should not appear on print. This page is not ment to look like it would afer being printed, because it simply cannot. You have to click the bolded "Printable version" link or "Refresh" link in order to get page which really is ready to be printed.
 * To sum up, you don't have to get rid of all templates like the print version notice, from A-level_Computing/Print_version, because this is not page that will be printed. It's just an overview of how print version looks like. --Derbeth talk 09:24, 24 June 2006 (UTC)

Wikistudy?
What is wikistudy, I ask for you becouse you have edit it. Is it like wikijunior but for older. Does it have own Wikistudy books--Tigru 05:38, 5 October 2006 (UTC)

Wikibooks Newsletter, Volume 1
(Wikibooks gazette home | Discuss | Bulletin board | Subscription list)

This is a short newsletter that is being distributed to all active wikibookians. You are getting this message because you are recognized as an established contributor to the project. This newsletter will be distributed on a regular basis to help share news, information, and tips. It comes from a bot account, User:The Staff. User:The Staff is currently operated by a team of wikibooks admins, the complete list of which is available on the user page of the bot. If you would like to not receive this newletter anymore, please remove your name from the list at Active wikibookians.

The work you do at Wikibooks is greatly appreciated. However there are plenty of other opportunities for you to get involved and help us to create a thriving Wikibooks community. We are sure that there are things we can do to help you and your understanding of Wikibooks and similarly there are certainly things you could do to help Wikibooks become a better place.

We would like to ask all wikibookians to add the Bulletin Board to your watchlists. The Bulletin Board is a fast and easy way for wikibookians to communicate important news and events to the entire community. If you have important news to share with the community, you can feel free to add your own entry to that page.

If you have general questions or comments about Wikibooks, you are welcome to post a message on The Staff Lounge, a free discussion area. Your input would also be welcomed in the Votes for Deletion and Requests for Adminship discussion pages. These pages are all active discussion areas that help to shape the Wikibooks community as a whole.

Sometimes it is easy to forget that the Wikibooks community is much larger and more diverse then the people who work in a single book, or on a single bookshelf. Hopefully, together we can all make Wikibooks a better place, and a more valuable educational resource.

The Staff 04:17, 19 December 2006 (UTC)

A-Level computing
Thanks. I'm currently studying the AQA course so I'll be able to update as I glide (hopefully!) through it. TheGenius1 17:50, 8 February 2007 (UTC)

Hello
Hi I´m Oceancetaceen, want to ask you something about Wikistudy. I come from Germany Duisburg Rheinhausen and want to ask you: Are you the main author and did you need a Germanbook for it, you could be look on the german wikibook pages, there I work mainly and write books for school, please reanswere on my dicussion page.

I sorry that I cant so many english, do you know germans in english Wikibooks?--Oceancetaceen (talk) 15:29, 24 January 2008 (UTC)

defining tables w/ integrity rule
Hi Pluke. I've been looking at your book - our curricula are not the same as the UK's, but similar - and I came across the DDL page. The primary key definition seems to be a bit different from what I've learnt at school. I'm wondering if this

What I learnt at school was:

I've tested yours and it works well, but I wonder if your code tells the RDBMS that the field is supposed to be not null and unique, or if that's already 'implied' in this code. Of course, RDBMSes are supposed to alert the data entry operator if the key field is null or non-unique, but I think it's better to define it explicitly. Thanks Kayau (talk · contribs) 12:31, 25 January 2013 (UTC)

Can you explain your edit revertion?
Hi ,

Why did you revert my revertion on a page? I reverted that page because the content was quite mixed-up, attribution content was changed without reason(names) and there did not seem to be anything proper with that set of edits. Of course I may be wrong, so I'm interested to hear your response. --Leaderboard (discuss • contribs) 18:45, 10 March 2015 (UTC)

ː My mistake, reverting

Adding BBC BASIC answers
A-level Computing/AQA/Problem Solving, Programming, Data Representation and Practical Exercise/Skeleton code/2011 Exam/Section B

I added some alternative answer's using BBC BASIC (which is still highly regarded in the UK). It's also something that's provided on the Raspberry Pi (as Brandy).

If it assists there is a manual for the Windows version (which for most practical purposes is the same as BASIC V on Arm/RISC OS based systems (the exception being related to the host platform) here http://www.cix.co.uk/~rrussell/bbcwin/manual/index.html.

Many many years ago there was a British publication called 30 Hour Basic, which may also prove useful.

Sfan00 IMG (discuss • contribs) 11:26, 30 April 2015 (UTC)

A-level Computing/AQA/Paper 1/Skeleton program/2017
I see you reverted my reversion on that page. This is, well, unexpected. I saw those edits as vandalism, in fact tbh I still see them as vandalism; that IP was vandalizing at the time, there and elsewhere, and I gave them a one-week block for it. I understand that page to be a skeleton, deliberately, meant to not have specific questions filled in. Perhaps I'm missing something. How do you see it? --Pi zero (discuss • contribs) 13:50, 23 November 2016 (UTC)

 * Hi pi zero. The edits elsewhere might be vandalism, but the edits on this page are contributing to filling in the skeleton outline. Hope that makes sense. Pluke (discuss • contribs) 15:52, 23 November 2016 (UTC)

 * That's... thought-provoking. If someone at that IP was being deliberately helpful, it may suggest my block was too long.  I'll have to reassess that action.  --Pi zero (discuss • contribs) 16:27, 23 November 2016 (UTC)

Unexplained revert
Hi Pluke, could you explain why you reverted 's edits without giving a reason why? It would be more helpful in the future if you used edit summaries to explain your reversion. --SHB2000 (discuss • contribs) 10:57, 26 January 2023 (UTC)

 * @SHB2000 sure, the edit removed an answer to a question. The previous structure of these pages had the question as the title to make it easier to students to navigate, rather than Question 1, Question 2, Question 3 etc. Sorry I should have made my reasoning clearer Pluke (discuss • contribs) 11:50, 26 January 2023 (UTC)
 * Thank you for the clarification! I apologize for the original revert. —Kittycataclysm (discuss • contribs) 13:06, 26 January 2023 (UTC)
 * no need to apologise @Kittycataclysm, I should have been better with my rationale! Pluke (discuss • contribs) 13:10, 26 January 2023 (UTC)
 * Understandable, and no need to apologise. We all forget to include edit summaries at times. SHB2000 (discuss • contribs) 23:11, 26 January 2023 (UTC)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User talk:Pluke


text:
User talk:206.45.65.20

Please stop adding nonsense to Wikibooks. It is considered vandalism. If you would like to experiment, use the sandbox. Thank you.

title: User talk:206.45.65.20


text: User talk:69.40.239.20

title: User talk:69.40.239.20


text:
User talk:62.171.194.8

Please stop adding nonsense to Wikibooks. It is considered vandalism. If you would like to experiment, use the sandbox. Thank you.

-- Herby talk thyme 14:36, 10 October 2007 (UTC)

title: User talk:62.171.194.8


text:
Georgia Water/Best Practices/Conservation

Best Practices in Water Conservation

Water Audits

 * System Audits (note: method to be developed for Georgia EPD )
 * End Use Audits (note: method to be developed for Georgia EPD )

Reuse Feasibility Studies

 * (note: method to be developed for Georgia EPD )

Education and Outreach Programs

 * (note: method to be developed for Georgia EPD )

Formulating Water Conservation Plans
Ten Key Steps to Planning a Successful Water Conservation Program*
 * Identify conservation goals
 * Develop a water-use profile and forecast
 * Evaluate planned facilities
 * Identify and evaluate conservation measures
 * Identify and assess conservation incentives
 * Analyze benefits and costs
 * Select conservation measures and incentives
 * Prepare and implement the conservation plan
 * Integrate conservation and supply plans, modify forecasts
 * Monitor, evaluate and revise program as needed

Source: *Handbook of Water Use and Conservation, Amy Vickers, 2001, pages 2 and 3 (as quoted from powerpoint presentation, Georgia DNR Pollution Prevention Assistance Division.

Water Conservation Techniques
(List of water conservation techniques, organized by who can implement, with following information for each technique: description, link to best example, estimating water saved, estimating costs, implementation guidance.)

Water Utilities

 * Water loss and leakage management
 * Water pricing

Identifying Water Conservation Techniques
(need to organized this list according to who implements) office sanitation)
 * Require 1.6 gal/flush toilet to be installed at the time of sale
 * Rebates for 6/3 dual flush or 4-liter toilets
 * Rebates for high efficiency clothes washers
 * Low income home leak detection and repair
 * Distribute retrofit kits w/low flow showerheads
 * Increase school education programs
 * Require high efficiency clothes washing machines
 * Insulate hot water piping
 * Rebates for 6/3 dual flush or 4-liter toilets
 * Require 6/3 dual flush or 4-liter toilets for new homes
 * Residential water audits
 * Guide to a self-water audit on North Georgia Retailer’s web site or as a mail-out
 * Permanent irrigation restrictions
 * Regulations for rain sensor/shut-offs on automatic systems
 * ET controller service (w/weekly updating) by vendor
 * Increase public education (expand web site, videos, CD’s)
 * Xeriscape demonstration gardens
 * Xeriscape education and staff training at retail garden/irrigation supply houses
 * Homeowner irrigation classes
 * Trigger shut-off valves and hose end timers
 * New home efficiency rating
 * Require model homes be landscaped with low water use landscaping
 * New home efficiency awards
 * Pressure regulation design for new areas
 * Promote water efficient plantings at new homes
 * Landscape requirements for new homes (turf limitations/regulations)
 * Regulations for rain sensor/shut-offs on automatic systems
 * Developer financed off-site conservation projects
 * /Rainwater harvesting/ (tanks) and irrigation systems
 * Offer incentives for replacement of clothes washers in coin-operated laundries
 * Distribute retrofit kits w/low flow showerheads
 * Incentives for retrofitting sub-metering
 * Regulations on sub-metering procedures (to protect tenant)
 * Require 1.6 gal/flush toilet to be installed at the time of sale
 * Require sub-metering multifamily units
 * Require efficient (such as horizontal axis) clothes washers
 * Require 6/3 dual flush or 4-liter toilets for new homes
 * Public relations campaign to set irrigation timers back in the fall
 * Residential (HOA) water audits
 * Rebates for high-tech irrigation controllers
 * ET controller service (w/weekly updating) by vendor
 * Rain-sensor retrofits on existing controllers
 * Regulations for rain sensor/shut-offs on automatic irrigation systems
 * New home efficiency ratings
 * New home efficiency award
 * Pressure regulation design for new areas
 * Landscape requirements for new landscaping systems (turf limitations/regulations)
 * Require efficient irrigation system design standards
 * Developer financed off-site development conservation projects
 * Commercial water audits and feasibility reports
 * Rebates for replacing high use commercial toilets with 1.6 gal/flush
 * Rebates for replacing high use commercial urinals with 0.5 gal/flush
 * 1.6 gal/flush toilet direct installation for high use commercial sites
 * Require 1.6 gal/flush toilet to be installed at the time of sale
 * Offer incentives for replacement or lease of clothes washers in coin-operated laundries
 * Require car washes to recycle water
 * Require or offer rebates for meters on cooling towers
 * Cooling tower regulations
 * Restaurant low flow spray rinse nozzles
 * Focused water audits for hotels/motels (showers and toilets only)
 * WAVE Program (US EPA) for hotels
 * Hotel retrofit (w/financial assistance)
 * Employee education program
 * Award program for water savings by businesses
 * Capacity buy-back for process improvements
 * 1.6 gal/flush toilet replacement incentives
 * Rebates for X-Ray recycling units
 * Replace inefficient water using equipment
 * Require car washes to recycle water
 * Require efficient (such as horizontal axis) clothes washers
 * Rebates for waterless urinals
 * Promotion and/or rebates for laundry recycle systems at commercial laundries
 * Self-closing faucets
 * Require efficient process equipment for selected businesses (restaurants, hotels/motels,
 * Prohibit once through cooling and non-recycling fountains, other non efficient water features
 * Require 0.5 gal/flush urinals in new buildings
 * Irrigation audits of large turf areas
 * Rebates for high-tech irrigation controllers for irrigation accounts
 * Surcharge on water bill for irrigation accounts to fund rebate programs
 * Rain-sensor retrofits on existing controllers
 * Assignment of irrigation budgets, tracking and water bill feedback to irrigation customers
 * Financial penalties for water use exceeding budget
 * Training landscape maintenance persons
 * ET controller service (w/weekly updating) by vendor
 * Financial incentives, rebates for irrigation upgrades
 * Regulations for rain sensor/shut-offs on automatic systems
 * Require dedicated irrigation meters
 * ET controller service (w/weekly updating) by vendor
 * Load shedding rate option for accounts w/ET controller service
 * Landscape requirements for new landscaping systems (turf limitations/regulations)
 * Require efficient irrigation system design standards
 * Assignment of irrigation budgets, tracking and water bill feedback
 * Financial penalties for water use exceeding budget
 * Replacement of inefficient toilets, showers, faucets
 * Installation of waterless urinals, dual flush toilets
 * City/County Department water reduction goals
 * City/County building facility water audits
 * Increase Xersicape landscaping of City/County & Utility facilities where appropriate
 * Public swimming pool water audits
 * ET controller service (w/weekly updating) by vendor
 * Irrigation scheduling with ET data
 * System water audits
 * Leak detection
 * Set water rates based on conservation pricing

Reference

 * Georgia DNR Water Conservation Program


text:
User:The Doc

Sysop on it:Wb. To contact me: it:utente:The Doc

title: User:The Doc


text:
User talk:The Doc

Hello and welcome to Wikibooks!

Here are some tips to help you get started:
 * You can sign your name on Talk and vote pages using four tildes, like this: ~~~~.
 * Remember to conduct any editing experiments in the sandbox.
 * Check out our help page if you need additional help.
 * Eventually, you might want to read the Manual of Style and Policies and Guidelines.
 * Templates can save your time
 * Remember to maintain a Neutral point of view.
 * Explore, be bold in editing pages, and have fun!

You will find more resources in Community Portal. If you want to ask a question, visit the Study help desk, the Staff lounge, or ask me personally on my talk page.

Good luck! -- LV (Dark Mark) 17:25, 11 January 2006 (UTC)

title: User talk:The Doc


text:
User talk:163.248.81.90

Welcome to Wikibooks, collection of open-content textbooks! You don't have to log in to read or edit articles on Wikibooks, but creating an account is quick, free and non-intrusive, requires no personal information, and gives you many benefits, including:
 * The use of a username of your choice
 * The ability to view all your contributions via a "My contributions" link
 * Your own user page
 * Your own talk page which, if you choose, also allows users to send you messages without knowing your e-mail address
 * The use of your own personal watchlist to which you can add articles that interest you
 * The ability to rename pages
 * The ability to upload images
 * The ability to customize the appearance and behavior of the website
 * The eligibility to become an administrator
 * The right to be heard in votes and elections
 * Your IP address will no longer be visible to other users.

We hope that you choose to become a Wikibooks and create an account. We hope you enjoy your time here on Wikibooks as a Wikibookian! -- LV (Dark Mark) 17:35, 11 January 2006 (UTC)

title: User talk:163.248.81.90


text:
User talk:216.239.1.4

Welcome to Wikibooks, collection of open-content textbooks! You don't have to log in to read or edit articles on Wikibooks, but creating an account is quick, free and non-intrusive, requires no personal information, and gives you many benefits, including:
 * The use of a username of your choice
 * The ability to view all your contributions via a "My contributions" link
 * Your own user page
 * Your own talk page which, if you choose, also allows users to send you messages without knowing your e-mail address
 * The use of your own personal watchlist to which you can add articles that interest you
 * The ability to rename pages
 * The ability to upload images
 * The ability to customize the appearance and behavior of the website
 * The eligibility to become an administrator
 * The right to be heard in votes and elections
 * Your IP address will no longer be visible to other users.

We hope that you choose to become a Wikibooks and create an account. We hope you enjoy your time here on Wikibooks as a Wikibookian! -- LV (Dark Mark) 17:36, 11 January 2006 (UTC)

title: User talk:216.239.1.4


text:
Tcl Programming/Working with files

Files and channels
In addition to the functionalities known from C's stdio, Tcl offers more commands that deal with files, similar to what shells provide, though often a bit more verbose. Here are some examples:

glob *.tcl

List all files in the current directory that match the pattern *.tcl.

file copy /from/path/file.ext /to/path . file delete /from/path/file.ext . file rename before.txt after.txt . cd /an/other/directory . pwd To let code temporarily execute in another directory, use this pattern: set here [pwd] cd $someotherdir cd $here
 * 1) ... code here

More precisely, many "file" operations work in general on "channels", which can be
 * standard channels (stdin, stdout, stderr)
 * files as opened with open ...
 * pipes as opened with 'open |...
 * sockets (TCP)

File names
Files are often addressed with path names, which indicate their position in the directory tree. Like Unix, Tcl uses "/" as path separator, while on Windows "\" is the native way - which brings trouble not only to Tcl, but even to C, because "\" is the escape character on both, so that e.g. \t is parsed as horizontal tab, \n as newline, etc. Fortunately Windows accepts natively "/" as well, so you can use forward slash in both Tcl and C programs as path separator without worries. However, you still have to take care of escape sequences. One stopgap measure is But Tcl allows to use the "normal" separator / in almost all situations, so you're safer off with that. Unfortunately, things are sad for Mac users, since MacOS (before X) accepts only ":" as file separator.
 * to escape the escape character, i.e. write \\ for \, or
 * brace backslashed names, e.g. {\foo\bar\grill.txt}

If you need to, here's ways to convert between the two forms: % file normalize \\foo\\bar\\grill.txt C:/foo/bar/grill.txt % file nativename /foo/bar/grill.txt \foo\bar\grill.txt

You can even use file join command: ''file join arg1 arg2 ... argN''

Tcl will then take care of all platform dependent details to create platform independent path. For example:

set somepath [file join foo bar grill.txt]

will result in following path (on windows machine): foo/bar/grill.txt

Input and output
Tcl's input/output commands are pretty closely based on those from C's stdio (just strip off the leading f):

 * set handle [open filename ?mode?]
 * set data [read $handle ?int?]
 * tell $handle
 * seek $handle offset ?from?
 * puts ?-nonewline? ?$handle? content
 * gets $handle ?varname?
 * close $handle

C's printf functionality is split in two steps: puts $handle [format "%05d %s" $number $text]
 * format the data into a string with format (which is very much like sprintf)
 * output the resulting string with puts. For example,

To process a text file line by line, you have two choices. If the file is smaller than several megabytes, you can read it just in one go: set f [open $filename] foreach line [split [read $f] \n] { # work with $line here ... } close $f For files of any big size, you can read the lines one by one (though this is slightly slower than the above approach): set f [open $filename] while {[gets $f line] >= 0} { # work with $line here ... } close $f Finally, if you can format your file so that it is executable Tcl code, the following reading method is fastest: source $filename

To "touch a file", i.e. create it if not exists, and in any case update its modification time, you can use this: proc touch name {close [open $name a]}

"Binary" files
All files are made of bytes, which are made of bits, which are binary. The term "binary" with files relates mostly to the fact that they can contain bytes of any value, and line-ends (Carriage Return+Newline in the DOS/Windows world) are not to be translated. Tcl can handle "binary" files without a problem -- just configure the channel as binary after opening: set fp [open tmp.jpg] fconfigure $fp -translation binary set content [read $fp] close $fp Now the variable content holds the file's contents, byte for byte.

To test whether a file is "binary", in the sense that it contains NUL bytes: proc binary? filename { set f [open $filename] set data [read $f 1024] close $f expr {[string first \x00 $data]>=0} }

The file command
Many useful operations with files are collected in the file command. The first argument tells which operation to do:

 * file atime name ?time?
 * file attributes name
 * file attributes name ?option?
 * file attributes name ?option value option value...?
 * file channels ?pattern? - returns the handles of currently open files
 * file copy ?-force? ?- -? source target
 * file copy ?-force? ?- -? source ?source ...? targetDir
 * file delete ?-force? ?- -? pathname ?pathname ... ?
 * file dirname name - e.g. [file dirname /foo/bar/grill.txt] -> /foo/bar
 * file executable name
 * file exists name
 * file extension name - e.g. [file extension /foo/bar/grill.txt] -> .txt
 * file isdirectory name
 * file isfile name
 * file join name ?name ...?
 * file link ?-linktype? linkName ?target?
 * file lstat name varName
 * file mkdir dir ?dir ...? - creates one or more directories (folders)
 * file mtime name ?time?
 * file nativename name
 * file normalize name
 * file owned name
 * file pathtype name
 * file readable name
 * file readlink name
 * file rename ?-force? ?- -? source target
 * file rename ?-force? ?- -? source ?source ...? targetDir
 * file rootname name - e.g. [file rootname /foo/bar/grill.txt] -> /foo/bar/grill
 * file separator ?name?
 * file size name
 * file split name - e.g [file split /foo/bar/grill.txt] -> {foo bar grill.txt}
 * file stat name varName
 * file system name
 * file tail name - e.g. [file tail /foo/bar/grill.txt] -> grill.txt
 * file type name
 * file volumes - Windows: returns your "drive letters", e.g {A:/ C:/}
 * file writable name

title: Tcl Programming/Working with files


text:
User talk:Whitenoise~enwikibooks

Hello and welcome to Wikibooks!

Here are some tips to help you get started:
 * You can sign your name on Talk and vote pages using four tildes, like this: ~~~~.
 * Remember to conduct any editing experiments in the sandbox.
 * Check out our help page if you need additional help.
 * Eventually, you might want to read the Manual of Style and Policies and Guidelines.
 * Templates can save your time
 * Remember to maintain a Neutral point of view.
 * Explore, be bold in editing pages, and have fun!

You will find more resources in Community Portal. If you want to ask a question, visit the Study help desk, the Staff lounge, or ask me personally on my talk page.

Good luck! -- LV (Dark Mark) 17:38, 11 January 2006 (UTC)

VfD
Was this your comment, or someone trying to impersonate you? -- LV (Dark Mark) 17:39, 11 January 2006 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Whitenoise. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Whitenoise~enwikibooks that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:51, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 05:38, 19 April 2015 (UTC)

title: User talk:Whitenoise~enwikibooks


text:
User talk:Brian Brondel

Wikibooks Newsletter, Volume 1
(Wikibooks gazette home | Discuss | Bulletin board | Subscription list)

This is a short newsletter that is being distributed to all active wikibookians. You are getting this message because you are recognized as an established contributor to the project. This newsletter will be distributed on a regular basis to help share news, information, and tips. It comes from a bot account, User:The Staff. User:The Staff is currently operated by a team of wikibooks admins, the complete list of which is available on the user page of the bot. If you would like to not receive this newletter anymore, please remove your name from the list at Active wikibookians.

The work you do at Wikibooks is greatly appreciated. However there are plenty of other opportunities for you to get involved and help us to create a thriving Wikibooks community. We are sure that there are things we can do to help you and your understanding of Wikibooks and similarly there are certainly things you could do to help Wikibooks become a better place.

We would like to ask all wikibookians to add the Bulletin Board to your watchlists. The Bulletin Board is a fast and easy way for wikibookians to communicate important news and events to the entire community. If you have important news to share with the community, you can feel free to add your own entry to that page.

If you have general questions or comments about Wikibooks, you are welcome to post a message on The Staff Lounge, a free discussion area. Your input would also be welcomed in the Votes for Deletion and Requests for Adminship discussion pages. These pages are all active discussion areas that help to shape the Wikibooks community as a whole.

Sometimes it is easy to forget that the Wikibooks community is much larger and more diverse then the people who work in a single book, or on a single bookshelf. Hopefully, together we can all make Wikibooks a better place, and a more valuable educational resource.

The Staff 04:16, 19 December 2006 (UTC)

How is this image not PD-Gov?
not PD-Gov...evidence? --Iamunknown 20:10, 31 January 2007 (UTC)

 * Ah, I see. "The Government is granted for itself and others acting on its behalf a paid-up, nonexclusive, irrevocable worldwide license in these documents to reproduce, prepare derivative works, and perform publicly and display publicly by or on behalf of the Government." Sad. --Iamunknown 20:11, 31 January 2007 (UTC)

title: User talk:Brian Brondel


text:
Cookbook:Homemade Seitan

Seitan is a food made from gluten protein that has been extracted from wheat flour. It has a springy texture, with protein strands similar to those in chicken running through it, making it a reasonable replacement for meat in vegan or vegetarian dishes.

Ingredients

 * 10 oz (280 g) high-gluten wheat flour (such as bread flour)
 * Water

Procedure

 * 1) Add 2 cups water to the flour. Knead together until well-combined and elastic. It is advisable to keep hands wet so that the dough doesn't stick to the hands.
 * 2) Cover dough with water, and place in refrigerator for 1 hour.
 * 3) Gently knead the dough underwater until water is cloudy; dump cloudy water and replace with clear, cold water. In the beginning the dough can still easily dissolve underwater so it is advisable to only cautiously squeeze it against the bowl. Later you might find it easier to knead over the water, squeezing out the water, and only to dip the mass into the water to wash off the starch, if the mass contains too much water you easily wash out the gluten as well. The enriched water can be used to gain the insoluble starch, which settles on the bottom of the container, and what remains can serve as a base for grain milk.
 * 4) Continue kneading and replacing water until water remains clear after kneading. At the end, you should be left with a mass of gluten.
 * 5) Divide gluten mass into loaf- or roll-shaped halves. At this point, there are several things you can add to the mass:
 * 6) * Soy sauce (recommended): traditional in the making of seitan.
 * 7) * Spices: seitan can simulate Italian sausage by adding the correct spices, or adding poultry seasoning can make the seitan more similar to chicken or turkey.
 * 8) * Nutritional yeast (recommended): B12 found in some brands of nutritional yeast is a vitamin that is usually obtained through meats, eggs and/or dairy products.
 * 9) Place gluten loaves in pot, and cover with vegetable broth or dashi.
 * 10) Bring to a boil, then reduce heat to simmer; simmer for 1–1.5 hours.
 * 11) Remove from heat and serve as desired or use in place of meat.

Notes, tips, and variations

 * To avoid the washing process you can also buy 100% gluten flour in some stores. This flour would require only enough water to form it into a solid mass.
 * Seitan created using just gluten flour can be rubbery and has a very "gluteny" taste. Adding 1 part garbanzo bean flour to 4 parts gluten flour makes for a less chewy texture and a better flavor.
 * Because of their high protein content, gluten "steaks" can be grilled and fried to good effect.

Warnings

 * A disposable respirator mask can help to avoid inhaling the hygroscopic gluten flour, which becomes very sticky in combination with liquid. The lungs do have a certain capacity to resorb organic substances inside the lungs, but one shouldn't try to test that ability.

Kochbuch/ Seitan

title: Cookbook:Homemade Seitan


text:
User:Gubbubu

I'm a maths and culture fanatic, workaholic, and an admin in Oldwikisource and Hungarian Wikibooks. My hungarian wikipedia user page is: here.

hu:User:Gubbubu

title: User:Gubbubu


text:
Cookbook:Seitan Burger

| Vegan cuisine

Seitan burgers have a consistency similar to that of fillet and are easy to cover with breading. For the breading, the usual spices for meat can be used.

Ingredients

 * 1¼ cup water
 * Soy sauce
 * 1 cup gluten flour (see seitan for how to gain gluten from wheat flour)
 * Salt
 * Spices (e.g. garlic powder, optional)
 * Onions (optional), chopped

Breading (optional)

 * Breadcrumbs
 * Salt
 * Pepper
 * Cumin
 * Oregano
 * Savory

Procedure

 * 1) Combine water and soy sauce.
 * 2) Combine gluten, salt, and spices.
 * 3) Combine the water and gluten mixtures. Knead until smooth.
 * 4) If using, knead in chopped onions.
 * 5) Shape the seitan dough into patties.
 * 6) Combine all the ingredients for the breading, and use it to coat the patties on all sides.
 * 7) Fry the patties in an oiled pan. Don't use too much oil but enough to let the breading pick up some oil on both sides.

Kochbuch/ Seitan-Bratling ספר המתכונים הטבעוני/סייטן מהיר הכנה

title: Cookbook:Seitan Burger


text:
Common Lisp/First steps/Installation

In order to start programming in Common Lisp, you need to obtain a Common Lisp implementation. There are many of them available for most operating systems. There are both free and commercial implementations.

Emacs
The easiest way to get started with Common Lisp is to use Emacs. Emacs, or you'll hear people sometimes refer to it as "vanilla Emacs", is the preferred way to write lisp by a majority of lispers. If you're also new Emacs then the below might be helpful, but they might also feel overwhelming for new users. If you're going to use Emacs, you should probably also install slime. Slime will help you edit common lisp; it comes included with the environments below.

Portacle
Portacle is a portable and multi-platform Common Lisp environment, which is usually recommended to beginners nowadays. It offers SBCL, Emacs (with nice customization), Quicklisp and Git packaged together. There is no installation needed, so it really is a straightforward way to go.

Doom Emacs
Doom Emacs is a popular pre-configured version of Emacs that is suitable for beginners and advanced users alike.

spacemacs
spacemacs is another popular pre-configured version of Emacs that comes fully equipped for use with Common Lisp.

Vim
vim, or neovim, is a very common text editor for programmers. If you're already familiar with vim, you can still use it to edit Common Lisp as well.

slimv
slimv is probably the most popular vim plugin for editing Common Lisp.

vlime
vlime is a newer alternative to slimv that is mostly written in viml (vim's extension language).

Free implementations
Note: Even if you don't want to install Portacle, it's still highly recommended to use the Emacs/SLIME combination to interface with these implementations. The reason is that the text console which they provide is quite awkward to work with, so most users prefer a nicer environment like the one SLIME provides.

GNU CLISP
GNU CLISP is a popular free implementation which can run on Windows and most Unix-like systems. If you're using Windows, you don't have much choice but to use CLISP. CLISP provides all ANSI Common Lisp features and is quite easy to install.

Features

 * Full Unicode support
 * Localised interface and compiler messages
 * High portability: runs practically anywhere
 * Compiles to portable bytecode (most or all other Common Lisp implementations compile to native code)
 * Small memory footprint
 * Arbitrary precision floating point arithmetic

Shortcomings

 * Slow at number crunching

CMUCL
CMU Common Lisp is a high performance Common Lisp implementation.

Features

 * Non-native threading (“green” threads; x86 only)
 * Produces fast code due to its optimizing compiler that produces native code
 * Fast compiler

Shortcomings

 * Relatively low portability (see the CMUCL portability webpage for a list of supported platforms)

SBCL
Steel Bank Common Lisp, a fork of CMU Common Lisp, is a portable implementation featuring native code compilation. It's easily the most popular Common Lisp compiler on platforms which neither CMUCL nor CCL will run on and an excellent choice even on those mainly because of its better debugging support (for instance, its compiler yields more warnings and is said to be stricter regarding ANSI compliance).

Features

 * Unicode support
 * Quite portable: runs on Unix-like systems including Mac OS X and Microsoft Windows.
 * Native threading support on Intel-based platforms
 * Good speed because of the optimizing compiler that produces native code

Shortcomings

 * Slow compilation speed (SBCL performs *many* optimizations, which can be slow)

Clozure CL
Clozure CL is a Common Lisp implementation the PowerPC and AMD64. There is a pre-release x86 version. It is easily the most popular choice on Mac OS X, but it runs equally well on GNU/Linux and FreeBSD/amd64 systems; there is also a beta for Windows. On Mac OS X, CCL features a Cocoa interface.

Features

 * Full Unicode support
 * A fast compiler that produces equally fast code
 * A Cocoa interface for Mac OS X GUI development
 * Native threading

Embeddable CL (ECL)
ECL is a Common Lisp implementation focusing on portability and embeddability. These goals are obtained by translating CL into C and passing that to a C compiler (typically gcc). It runs on Linux, FreeBSD, NetBSD, OpenBSD, Solaris and Windows, running on top of the Intel, Sparc, Alpha and PowerPC processors.

Features

 * Unicode support
 * Translates CL into C, which produces relatively fast native code
 * Interpreter which circumvents the slow build times
 * Load on demand. ECL defers loading modules until needed
 * Ability to output portable C source files that can be compiled on other architectures
 * Call C from Lisp and call Lisp from C (i.e. ECL can be used as an extension language for any program that can call a C function)
 * Native threading

Shortcomings

 * Relatively slow compilation method (gcc is slow, maybe if you use a faster C compiler)
 * Relatively fragile error handling (i.e. you can seg-fault ECL fairly easily)

Armed Bear Common Lisp
ABCL is a Common Lisp implementation that runs on top of the Java Virtual Machine. It provides a runtime system, a compiler that compiles Lisp source to JVM bytecode, and an interactive REPL for program development.

Features

 * Produces portable Java bytecode
 * Java libraries are transparently available

Shortcomings

 * Quite slower than other implementations

Commercial implementations

 * Allegro Common Lisp (ACL)
 * LispWorks
 * Corman Lisp
 * Scieneer Common Lisp

There are free trial versions of Allegro, LispWorks, and Corman Lisp, which can be handy for learning the language. All of them work on Windows.

title: Common Lisp/First steps/Installation


text:
User talk:Grue

Hello and welcome to Wikibooks!

Here are some tips to help you get started:
 * You can sign your name on Talk and vote pages using four tildes, like this: ~~~~.
 * Remember to conduct any editing experiments in the sandbox.
 * Check out our help page if you need additional help.
 * Eventually, you might want to read the Manual of Style and Policies and Guidelines.
 * Templates can save your time
 * Remember to maintain a Neutral point of view.
 * Explore, be bold in editing pages, and have fun!

You will find more resources in Community Portal. If you want to ask a question, visit the Study help desk, the Staff lounge, or ask me personally on my talk page.

Good luck!

And it's always good to see a familiar name from en.wikipedia. -- LV (Dark Mark) 20:34, 11 January 2006 (UTC)

title: User talk:Grue


text:
Talk:Super Mario 64/Bob-omb Battlefield

Screenshot
Figured out why it wasn't working :) Mason11987 20:46, 11 January 2006 (UTC)

Bob-omb battlefield
This level is easy, it only takes two tries to do it without getting any coins

title: Talk:Super Mario 64/Bob-omb Battlefield