User:Benedek/Sandbox

just testing out the math features... and recording snippets from discussion on some physics topics that I found helpful. (for those that stumble across this, this is where I throw stuff while trying to get it to display right, or for saving discussion snippets for myself. Don't expect it to be coherent, and don't worry ... this thing won't end up copy-pasted into an article.)

However, if you do come across this, here are somethings I still haven't figured out with the formatting:
 * How do you force latex process (sometimes it seems to convert to html?). I've seen many pages that don't seem to have this problem, so there must be a way to fix it.
 * horizontaolly spacing subscripts, superscripts correctly
 * adding extra vertical space between "accents" on numbers like $$\bar{0},\hat{0},\tilde{0}$$ for it is sometimes hard to read in superscripts $$x^\bar{0},x^\hat{0},x^\tilde{0}$$. Oh well, at least it is more readible for letters.

If you know how to fix those issues, please drop a note on my talk page.

Hey, found a nice help document: http://en.wikibooks.org/wiki/Wikibooks:TeX_markup

Geometric view of SR
Special Relativity has shown that many of the things we refer to in everyday experiences are actually coordinate system dependent, but because "everyday" speed differences are such that $$\frac{v}{c} \ll 1$$, we often don't notice this. We say "let's meet for lunch at 12:15" and there are no discussions or worries about how we synchronize our clocks and keep them synchronized. We quote flying times and distances at the airport without worrying about "according to who? The passengers or the ground crews?". But SR showed indeed that distances and times to events are but mere convenient labels, and have no fundemental meaning.

In 1907, a mere two years after the initial publication of SR, Hermann Minkowski showed that it may be more enlightenning to view SR as a geometric theory of a four dimensional space-time. As he was famously quoted:
 * Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality.

The reason this view is so much more convenient is that geometrical objects are independent of a coordinate system. So now you are speaking of things that are true regardless of the arbitrary labelling system one wishes to use. Geometric objects however can still be (and often are) described by coordinate systems. But it is important to note here that since they are independent of a coordinate system, if you choose to use one, you are free to choose any coordinate system you wish. For the flat space of SR, inertial coordinate systems are the most convenient choice. The components of the metric (a description of the geometry of space) in these coordinate systems is actually named the Minkowski metric after the man that introduced it, and may look familiar to some of you:
 * $$g_{\alpha\beta} = \begin{bmatrix}

-1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Einstein eventually embraced this geometric view, which became quite important in his formulation of General Relativity in 1916. As the geometric view can often help one look "past the messy coordinates", this is the modern way SR is used. This is how you will see it used in almost all publications, and also how it is usually taught to physics undergraduate students during their upper level classes. So there are many good reasons to learn to think with this "geometric view".

As it sometimes confuses people initially, it should be noted that while this "geometric view" may involve different mathematical techniques, it isn't a new theory. It is still SR, and mathematically equivalent to all the coordinate based methods. Actually, when Minkowski first introduced it in 1907, many complained about his grandiose statements, for he 'had done nothing new'. The convenience and power of this method soon won everyone over though, and is why this is currently the modern view of SR.

Vectors
When using a coordinate system, a vector is represented with four components. Component 0 is commonly referred to as the "time" component, and components 1, 2, and 3 are referred to as the "spatial components". An example is a displacement vector between two events $$(ct,x,y,z) \,$$. When referring to the components of a vector as variables, it is labelled using the name of the vector, and a superscript denoting the component. For example if we want to refer to the components of a displacement vector $$A\,$$ in a particular coordinate system we would have $$(A^0,A^1,A^2,A^3)=(ct,x,y,z)\,$$. In equations, a greek letter will be a dummy index to denote the numbers 0 to 3. For example $$A^\alpha \,$$ represents one of the components (or the set of all four depending on the context in discussion) with $$\alpha\,$$ being any value from 0 to 3.

As a geometric object, a vector is independent of its descrption by a coordinate system. If we wish to represent the same vector with a different coordinate system, it will still be labelled with the same name, but a distinguishing mark will be made on the indices to denote which coordinate system the component is from.

Here are several example of components of the vector $$A\,$$:
 * $$A^\alpha,A^\beta,A^0\,$$ are all referring to components of A in the same coordinate system
 * $$A^{\bar\alpha},A^{\bar\beta},A^{\bar{0}}\,$$ are all referring to components of A in the same coordinate system (but a different one than the previous line)

Coordinate transformations of Vector components
A coordinate transformation can be represented as a matrix which we shall denote as $$\Lambda^{\bar{\alpha}}{}_\beta$$. Which coordinate system we are transferring to and from are denoted by the marks on the indices (in this case to the "barred" coordinate system from the "un barred" coordinate system). Here is an example of a Lorentz transformation
 * $$\Lambda^{\bar{\alpha}}{}_\beta = \begin{bmatrix}

\gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$
 * $$x^{\bar\alpha} = \sum_{\beta=0}^{3}\Lambda^{\bar{\alpha}}{}_\beta x^\beta$$

Einstein notation
Because the indices always run from 0 to 3 (unless explicitly stated otherwise), and many sums appear in calculations involving components, Einstein notation simplifies writing equations by removing these sumation signs from the notation completely and there is an implied sum over any repeated index. For example, the coordinate transformation above would be written in Einstein notation as simply:
 * $$x^{\bar\alpha} = \Lambda^{\bar{\alpha}}{}_\beta x^\beta$$

and the sum over $$\beta\,$$ is implied as it is repeated.

Tensors
Tensors are geometric objects and can be represented in components for a given coordinate system. For example the components of a two dimensional tensor would be like a matrix (row,collumn). An "n-dimensional" Tensor would require n indices to label the components. Here are some examples of tensors:
 * $$g_{\alpha\beta} \,$$ this is the metric. It is a two dimensional tensor.
 * $$T^\alpha{}_{\beta\gamma} \,$$ is an example of a three dimensional tensor.

Covariant (subscript) vs Contravariant (superscript)
There is a distinction between $$x^\mu \,$$ which are components of a vector (contravariant vector) and $$x_\mu \,$$ which are components of a covector (covariant vector). Similarly, the components of $$g_{\alpha\beta} \,$$ are not necessarily equal to $$g^{\alpha\beta} \,$$.

The conceptual difference between a vector and a covector is explained later.

Also, you will notice that all implied sums will be between one "top" index and one "bottom" index.

"Raising" or "Lowering" an index
The metric can be used to relate contravariant and covariant components:
 * $$x_\alpha = g_{\alpha\beta}x^\beta \,$$
 * $$x^\alpha = g^{\alpha\beta}x_\beta \,$$

This is often referred to as "raising" or "lowering" indices.

For tensors, this can be applied to each sub or superscript individually. As an example:
 * $$ T^{\alpha}{}_{\beta\gamma} = g_{\beta\nu} T^{\alpha\nu}{}_\gamma \,$$

Coordinate transformations for any object
Covectors transform inverse to vectors.
 * $$x_{\bar\alpha} = \Lambda^\beta{}_{\bar{\alpha}} x_\beta$$

Please note that as "matrices" $$(\Lambda^\beta{}_{\bar{\alpha}}) = (\Lambda^{\bar{\alpha}}{}_\beta)^{-1} $$, for if you transform to a new coordinate system and transform back to the original one, you should return to the same values.

Contravariant vectors can just be thought of as one dimensional tensors, and similarly with covariant vectors. So now that we know how these transform, the components of any object can be transformed into a new coordinate system. Just apply a coordinate transformation for each superscript index, and the inverse of that transformation for each subscript index.

As an example:
 * $$g_{\bar\alpha\bar\beta} = \Lambda^\mu{}_{\bar{\alpha}} \Lambda^\nu{}_{\bar{\beta}} g_{\mu\nu} \,$$

Kronecker delta
The components of the Kronecker delta just look like the identity matrix.
 * $$\delta^\gamma{}_\beta = \begin{bmatrix}

1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Here are somethings that equal the Kronecker delta:
 * $$\Lambda^\mu{}_{\bar{\alpha}} \Lambda^{\bar{\alpha}}{}_\nu = \delta^\mu{}_\nu $$ transform to a different coordinate system and back, you have returned to the same component representation.
 * $$g^{\mu\alpha}g_{\alpha\nu} = \delta^\mu{}_\nu \,$$ lower an index and raise it back, you have returned to the same component representation.

Scalars
A scalar is a geometric quantity. It has no indices, and is just a number.

Since a scalar has no indices, it does not transform when a coordinate system changes. All coordinate systems will agree on this value.

Here are some examples:
 * $$x^\alpha y_\alpha \,$$ after the implied summation this is just a number (referred to as the dot product of x and y)
 * $$x^\alpha g_{\alpha\beta} y^\beta \,$$ same as above, but written with both x and y as vectors
 * $$T^\alpha{}_\alpha \,$$ after the implied summation this is just a number (referred to as the trace of the tensor T)

The Metric
If we can use any coordinate system, what relates these arbitrary labels to anything "real"? If this is the geometric view, what shows us the geometry? It is the metric that gives us a view of the geometry. It tells us the geometric length between a point and points in its close neighbourhood.

This length is $$(ds)^2 = g_{\mu\nu}dx^{\mu}dx^\nu \,$$, and is often referred to as the invariant line element, for all coordinate systems agree on this length. The length of a path is the sum of the line elements along it, and so the length of a given path between two points is independent of coordinate system.

Since we are talking about space-time, this geometric length isn't necessarily a length like a ruler. Depending on your coordinate system choice, the calculation of this geometric length may involve time and spatial coordinates. It is a space-time length.

A 2-D spatial geometry example
Let's first look at some 2-D spatial geometries that we can imagine easily, and then abstract to the space-time of special relativity.

Imagine a flat piece of paper, and use a cartesian coordinate system. We define the length between two points using the usual pythagorean theorem for Euclidean space $$L^2 = x^2 + y^2 \,$$. Now what if we wanted to use polar coordinates? The coordinates of the points change, but we realize that we didn't change the geometry -- the paper is still flat, and we are still talking about the same points.

So how do we find length in polar coordinates? Well, we had a formula giving us length in cartesian coordinates, so if we know how the polar coordinates are related to the cartesian coordinates, we should be able to find it as well. One way to do this is look at the length between a point and points in its close neighbourhood. This is exactly the information that the metric provides.

Since we only have two components here, I'll write out all the components. The metric provides a definition of length:
 * $$(ds)^2=g_{\mu\nu}dx^{\mu}dx^\nu = g_{11}dx^1dx^1 + g_{12}dx^1dx^2 + g_{21}dx^2dx^1 + g_{22}dx^2dx^2 \,$$

For cartesian coordinates our two components are x,y, and from above we see that the invariant line element is just $$(ds)^2=(dx)^2+(dy)^2 \,$$. So for cartesian coordinates, the metric of our 2-d spatial geometry has the same components as the identity matrix.

Since we know the relation between cartesian and polar coordinates is:
 * $$x = r \cos(\theta) \,$$
 * $$y = r \sin(\theta) \,$$

This gives:
 * $$dx = \frac{\partial x}{\partial r}dr + \frac{\partial x}{\partial \theta}d\theta = \cos(\theta)dr - r \sin(\theta) d\theta \,$$
 * $$dy = \frac{\partial y}{\partial r}dr + \frac{\partial y}{\partial \theta}d\theta = \sin(\theta)dr + r \cos(\theta) d\theta \,$$

Putting this back into the line element we find:
 * $$(ds)^2=(dx)^2+(dy)^2 = (\cos(\theta)dr - r \sin(\theta) d\theta)^2 + (\sin(\theta)dr + r \cos(\theta) d\theta)^2 \,$$
 * $$(ds)^2 = (dr)^2 + r^2 (d\theta)^2 \,$$

So we have two options to find length now if we are using polar coordinates. We can transform every point along the path to cartesian coordinates, and add up the lengths. Or we can just calculate the lengths in polar coordinates directly, because we now know what the components of the metric are in polar coordinates.

Note also, the components of the metric in polar coordinates are not independent of position (unlike in cartesian coordinates). So if you are familiar with polar coordinates, you already know an example of a non-linear transformation! So you already know many of the concepts of non-linear transformations in SR (which is involved in considering coordinate systems where accelerating observers are at rest)!

Let's look at the metric a bit further. Some might complain that if we can choose any coordinate system, can't we purposely construct one to make the components of the metric always look like whatever we want? The answer is no.

Consider the surface of the globe. This clearly isn't a 2-D Euclidean geometry. Can we choose a coordinate system to make the metric appear as if it was? No, this is not possible. If one chooses a point and took all the points a constant distance away from this, it would make a circle in both the globe and flat paper examples. However if we measured the length of this circle, on the flat paper it would give $$2\pi r$$ while for the globe it would give a value < $$2\pi r$$. Since the length is invariant of the coordinate system, the infinite set of possible metrics of the flat paper are not possible in the globe surface, and vice versa.

For the picky:
 * Note however that at a single point, the components of the metric for the globe could be made to look like that at a single point of the paper. (This is why locally on the globe, we can use cartesian coordinates to draw a map without much distortion.) But mathematically, this is only exact in the limit that we are considering a single point.  You still cannot make the metric on a "small patch" of the flat paper look like the metric on a "small patch" of the globe no matter what coordinate system you choose.

Minkowski metric
In an inertial coordinate system, the metric is:
 * $$g_{\alpha\beta} = \begin{bmatrix}

-1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

This shows why the inertial coordinate system is such a convenient coordinate system. Notice that the components of the metric are independent of position (space-time is described homogenously - the same everywhere). Also, the spatial components describe space isotropically (it looks the same in all directions). Many consider this the defining characteristics of inertial coordinate systems: space and time are described homogenously and isotropicly.

Because the components of the metric are independent of position for inertial coordinate systems, we can easily go from neighbourhoods of points to large distances:
 * $$(ds)^2 = -(c \ dt)^2 + (dx)^2 + (dy)^2 + (dz)^2$$
 * $$(s)^2 = -(c \ t)^2 + (x)^2 + (y)^2 + (z)^2$$

Please remember though that this is just a special feature of inertial coordinate systems. In general we will have to perform a line integral to obtain these distances.

Also note that this metric requires that the speed of light is the same in all inertial coordinate systems. This does not make special relativity's speed of light postulate into a tautology (true by definition), for as mentioned in the previous section, we cannot make the metric look like whatever we what just by an appropriate choice of coordinate system. Experiment has shown that we can indeed describe spacetime in this manner.

Also note that any linear transformation of coordinates will transform a linear path through space-time according to inertial coordinates to a linear path through space-time according to the new coordinates. As not all linear transformations are Lorentz transformations, this means there are an infinite number of coordinate systems in which Newton's first law holds, but the speed of light is not constant. So this is a reminder that "inertial motion of objects" is not a sufficient condition to specify an inertial coordinate system.

Geometric distances in SR
Due to the negative sign in the time component of the metric, lengths in SR can be both negative and positive.

Using the fact that massive objects in SR are limited to travelling slower than the speed of light, we can make the following comments about the geometric distance:
 * $$(ds)^2 = 0$$ corresponds to space-time points that could lay on the path of a light beam.
 * $$(ds)^2 < 0$$ corresponds to space-time points that could lay on the path of a massive object (so these are called time-like separated points). Events at one of these points can influence events at the other (they are not causally separated).
 * $$(ds)^2 > 0$$ corresponds to space-time points that could not lay on the path of a massive object (so these are called space-like separated points). Events on these points cannot influence events on the other (they are causally separated).

Proper time
If we use the metric to calculate what the geometric length of an object's space-time path (worldline) is, what does this correspond to?

If we consider coordinate systems in which the object is at rest, the only coordinate left to describe the path is the time coordinate. So even without doing the transformations we can see that $$(ds)^2 = -(c\ dt)^2$$ where t is the time measured by a clock following that path.

In the cases where a path is not specified, the proper time between two time-like separated events is defined as the extremal time of the family of paths connecting those two events.

Proper length
If an observer defined the path from one end of an object to the other such that every point along the path occurred at the same time, this is what we would normally refer to as the length of the object according to that observer. So if the geometric length of a path is positive, we could construct a coordinate system where this would correspond to a spatial length measurement.

Unlike proper time, given an object, this path is not well defined as it depends on how we choose to define simultaneous. However given two events that are space-like separated, proper length can be defined as the extremal length of the family of paths connecting those two events.

Contravariant and covariant
The metric can also be used to switch between the dual vector spaces -- contravariant and covariant vectors. To help people see the idea behind a dual vector space, a common example is row vectors and collumn vectors in linear algebra. You can multiply them by any matrix you want and they are still distinct, and there is no defined way of how to add them.

A vector with a subscript is referred to as a "covariant" vector (covector) because it varies "like the basis". A vector with a superscript is referred to as a "contravariant" vector becuase it varies "opposite of the basis".

To see this better, imagine for instance x,y cartesian axes, and a displacement vector with components (1,1). We rotate the axes 90 degrees clockwise (so x axis -> where -y axis used to be, y axis -> where x axis was). The new components of the vector are (-1,1), it seems to have rotated instead COUNTER-clockwise.

So what we usually refer to as vectors in physics (displacements, velocity, etc.) are contravariant vectors. $$x^\alpha \,$$

As mentioned previously, the metric can be used to easily transform between the two.
 * $$x_\alpha=g_{\alpha\beta}x^\beta  \,$$

In inertial coordinate systems, this means the only difference between the components of a vector and the components of a covector are that the time component has a different sign (go back and look at the metric in inertial coordinate systems if you need convincing of this). And thus, in an inertial coordinate system:
 * $$x_\alpha = (-x^0,x^1,x^2,x^3) \,$$
 * $$\begin{matrix}

x^\alpha x_\alpha &=& x^0x_0 +x^1x_1 +x^2x_2 +x^3x_3 \\ \ &=& -(x^0)^2 + (x^1)^2 + (x^2)^2 + (x^3)^2 \end{matrix}$$

Coordinate systems for an accelerating observer
In special relativity, a uniformly accelerating point particle follows a hyperbola instead of a parabola. (If it followed a parabola, it would eventually travel faster than the speed of light, so clearly that is not the path it will take. As for the hyperbola, I will not be deriving that here, but taking it as given.)

There are of course many ways to define a coordinate system for an accelerating observer such that he is at rest. But if we want the time coordinate to refer to the time measured by a clock that is stationary in these coordinates (in this case time is labelled by the clock at the spatial origin), and we want the spatial coordinates labelled so that in some sense they stay the same distance appart (maybe the round trip time a light pulse takes to go between two spatial coordinates is the same regardless of when the experiment is performed ... or maybe something similar like throwing a ball with a specific speed).

You can verify for yourself that the following coordinates fit those requirements:
 * $$x^0 = (\frac{c^2}{a}+y^1)\sinh(a\ y^0/c^2) \,$$
 * $$x^1 = (\frac{c^2}{a}+y^1)\cosh(a\ y^0/c^2) \,$$
 * $$x^2 = y^2 \,$$
 * $$x^3 = y^3 \,$$

where x coordinates are inertial coordinates and y coordinates are for the accelerating coordinate system, which is accelerating in the $$x^1$$ or $$y^1$$ direction.

Going though the math like with the polar coordinates example, we find the line element to be:
 * $$(ds)^2 = -(1+ ay^1 / c^2)^2(dy^0)^2 + (dy^1)^2 + (dy^2)^2 + (dy^3)^2 \,$$

For a clock at rest, only the $$g_{00} \,$$ component of the metric is needed to calculate the time it measures along its path for only the time coordinate changes on its path. Looking at the result, this means that clock synchronization by a light pulse will not work in this coordinate system. For the rate at which a clock runs depends on its position in the direction of acceleration. One would see the clocks ahead of us running fast and those behind us running slow.