User:Thepigdog/General Relativity - 01 - Shadows of reality

The average person can understand General Relativity. Some of the ideas are actually quite simple, even if the details are complex. The key ideas are,

• Invariance
• Curvature

Invariance is the idea that a quantity, from the real world will have a value that does not depend on how we measure it. This is an abstract idea, but a fairly simple one.

The other idea is curvature. For curvature think of a sphere, like the world we live on. The surface of the sphere is curved. This is easy to picture. The essence of General Relativity is that curvature is the source of gravity, and that mass creates curvature.

This is hard to picture. A couple of points to clarify. The curvature being discussed is intrinsic curvature. It is the distortions that appear when you are actually in the dimension. In the example given with the sphere it is the distortions the ant would see walking on the sphere. Here the surface is 2 dimensional. But for General Relativity the curvature is in a 4 dimensional space (3 dimensions of space and one of time). General Relativity says that this universe we live in is curved by mass, causing gravity.

When we think of curvature we think of the surface of an object being curved. We can see that the surface is curved from outside. But mathematics allows the curvature to be detected from within the space, by measuring distances and angles. This is the curvature described by General Relativity.

This is hard for anyone to understand physically. But is is easy to see that the sphere is curved, and by analogy we can think about the universe being curved.

Curvature will be dealt with later. Lets start with looking at invariance.

Shadows of reality

We perceive the world through our senses. It is always tempting to think that what we sense is actually reality, but it is not. What we think of as reality is an interpretation of the data we receive through our senses.

The way we come to this interpretation is through learning. Learning is finding simpler and simpler models of our world that are consistent with the data provided by our senses.

What we aim to do is to discover the fundamental reality of the world, as facts about the world that are always true.

Consider the sense of vision. It is not a 3 dimensional sense. It is a 2 dimensional grid of pixels received through the camera lens of our eyes. Out of this 2 dimensional world we form our 3 dimensional world model.

My aim is to take you back to this 2 dimensional world and show you how the 3 dimensional world emerges out of it by considering invariance. Then when later I claim that time is simply another dimension which emerges by considering another invariance, it will not seem so shocking.

Our eyesight is complicated by problems of perspective that complicate the matter unnecessarily. Consider instead instead shadows cast by objects from a light source very far away (e.g. the sun). Ignore the complications of the fussy edges of shadows. Lets pretend that the shadows have crisp edges. They are ideal shadows.

The shadow box

There is a shadow box. We do not see directly inside the box. We only see the shadows of the objects inside the box.

The box has two buttons that rotate the objects. One button rotates the objects that go around in circles on the screen. One rotates them so that they appear to tumble over each other.

The shadow of an object whose shadow appears as a right triangle.

By chance (or authors caprice) one object appears as a right angle triangle with a point at the center of rotation. As the button that rotates the object around the center is pushed we see the triangle rotate around the center. Using a ruler the lengths of the sides of the triangle could be measured in each position. The lengths of the sides do not change. They are invariant.

The concept of angle may then be introduced. The angles at the corners of the triangle are defined to be the same if the ratio of the lengths of the sides is the same.

The right triangle rotated, demonstrating Pythagorous.theorem

By considering similar triangles, this leads to Pythagorous theorem. By considering the ratios of sides,

${\frac {A}{C}}{\frac {A}{C}}+{\frac {B}{C}}{\frac {B}{C}}=1$

or,

$A^{2}+B^{2}=C^{2}$

Pythagorous Theorem may then be used to give a measure of the distance between any two points in a co-ordinate system. A co-ordinate system is a way of identifying positions on the screen, by laying out a grid of parallel lines at right angles to the axis. Positions on the axis are then numbered with. A pair of numbers then identifies a position on the screen,

$x=(x_{1},x_{2})$

Here the position named x is identified by a pair of numbers $x_{1}$ and $x_{2}$. The co-ordinates are numbered, 1 and 2. We do not call the co-ordinates x and y, as is sometimes done, because there is going to be more co-ordinates later (eventually 4, for General Relativity). Writing the co-ordinate axis as a subscript will greatly simplify the notation later.

For two points,

$x=(x_{1},x_{2})$

$y=(y_{1},y_{2})$

then the distance measure between two points is,

$|x-y|={\sqrt {(x_{1}-y_{1})^{2}+(x_{1}-y_{1})^{2}}}$

The notation may be simplified by allowing summing over the co-ordinates. For example the the above distance measure formula may be simplified using a sum over co-ordinates as,

$|x-y|={\sqrt {\sum _{\mu }(x_{\mu }-y_{\mu })^{2}}}$

The use of Greek letters is nothing to worry about. Mathematicians use Greek letters to broaden the set of letters they can use for variables. Mathematicians don't usually declare there variables with types, as is done in computer science. But none of this anything to worry about.

Now as shapes are rotated by pushing the first button (the one that appears to rotate the objects around the center of the screens) the lengths of sides remains unchanged. In other words the length is invariant. If we measure the same sides after rotations which we will name $\alpha$ and $\beta$

$\sum _{\mu }(x_{\mu }^{\alpha }-y_{\mu }^{\alpha })^{2}=\sum _{\mu }(x_{\mu }^{\beta }-y_{\mu }^{\beta })^{2}$

These rotations may also be considered co-ordinate transforms. Actually a co-ordinate system is defined by us in examining the world, which is somewhat hard to think about. It is easier, at least to start with, to think about the world being rotated around our fixed co-ordinate system. But later we must get used to the idea of different co-ordinate system.

Because co-ordinate systems are compared so often in relativity, instead of using alpha and beta we will often refer to the primed and un-primed co-ordinate system.

$\sum _{\mu }(x_{\mu }-y_{\mu })^{2}=\sum _{\mu }(x'_{\mu }-y'_{\mu })^{2}$

In understanding mathematical notation, we will often see that the goal of the mathematician is the simplest most concise notation possible (even if it is somewhat un-intuitive). Mathematics arose long before computers, and everything had to be written by hand. We will need to get use to this mathematical notation. But however it is dressed up mathematical notation usually means something that is essentially simple.

The dot product

The dot product is needed.

$x\cdot y=x_{1}y_{1}+x_{2}y_{2}$

Or more generally,

$x\cdot y=\sum x_{\mu }y_{\mu }$

The dot product may be used to express the measure of distance,

$(x-y)\cdot (x-y)=|x-y|^{2}$

The distance from the will be written,

$x\cdot x=|x|^{2}$

The Euclidian metric

A metric is a matrix of data required to measure distance. It is represented as g in,

$|x|^{2}=\sum _{\mu ,\nu }g_{\mu \nu }x_{\mu }x_{\nu }$

Comparing this with,

$|x|^{2}=\sum _{\mu }x_{\mu }^{2}$

Then $g=\delta$ where $\delta$ is called the Kronecker delta, defined as,

$\mu \neq \nu \implies \delta _{\mu \nu }=0$

$\delta _{\mu \nu }=1$

Then $\delta =g$ gives,

$\sum _{\mu ,\nu }\delta _{\mu \nu }x_{\mu }x_{\nu }=\sum _{\mu }x_{\mu }x_{\mu }=\sum _{\mu }x_{\mu }^{2}=|x|^{2}$

The metric will be important later when we consider curved space. Then $g\neq \delta$ and may vary with position.

Rotation in the third dimension (the second button)

When the second button is pushed the shapes change, and appear to tumble over each other. Small rotations allow us to track objects as they change shape, but distances change.

Everything we had when pushing the first button disappears. Distances are no longer invariant. This would be quite a shock if previously we had only known about 2 dimensions.

A simple generalization of the distance measure resolves the problem,

$|x-y|^{2}=(x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2}+(x_{3}-y_{3})^{2}=\sum _{\mu }(x_{\mu }-y_{\mu })^{2}$

Then by assuming this distance is invariant it is possible to assign distances to objects that are consistent with the apparent changing of shape of the objects. What this shows is that the shadows we see are not the reality. The reality is that the shadows are images of an object in 3 dimensions displayed in 2 dimensions.

The special relativity shadow box

Around the 1900s physics, which had been doing very well in explaining the world, started to have a few problems. The behavior of fast moving electrons did not seemed to make sense. They behaved as if there length had been contracted in the direction of movement.

Physics had also met problems in another area. Maxwell's equations of electo-magnetism had unified the electric and the magnetic fields and even explained light. But they also describe a wave, and as we all know, waves need a medium to travel through. The physicists named this medium was named the ether.

The earth is travelling fairly fast through the universe, so if the ether was stationary, it should be possible to detect the motion of the earth through the universe. What you would expect is that the speed of light would be different if measured in different directions. Michelson and Morley set up an experiment to test this result using an incredibly accurate device called an interferometer.

But, to there surprise, there results were negative. Whichever way they looked the speed of light was stubbornly the same. This seemed to defy the laws of logic because how can the speed of something be the same, no matter how fast you are going relative to it.

Minkowski metric

What if the 3 dimensional space we see is a shadow box, of a 4 dimensional structure? The Michelson, Morley experiment takes us straight to the fundamental result. It says that the speed of light is invariant, no matter how you measure it.

Then in a 3 dimensional co-ordinate system, with time, if a beam of light is omitted from the origin and received at a point x some time t later, then.

$\sum _{\mu }x_{\mu }^{2}=c^{2}t^{2}$

Or,

$\sum _{\mu }x_{\mu }^{2}-c^{2}t^{2}=0$

and this will be true even in any co-ordinate system no matter co-ordinate system moving at uniform speed relative to each other. Then,

$\sum _{\mu }x_{\mu }^{2}-c^{2}t^{2}$

may be thought of as a generalized distance in space time. The time t is then identified as a dimension in x and is called $x_{0}$. The metric is then given by,

$g_{0,0}=-c^{2}$

$\mu \neq 0\implies g_{\mu ,\mu }=1$

$\mu \neq \nu \implies g_{\mu ,\nu }=0$

Hopefully the Michelson Morley result is not so shocking in light of the shadow box experiment. The invariance it represents is a deeper order in the universe, showing the interrelation of space-time as a single four dimensional co-ordinate system, with its own metric.

Lorentz transformation

???? work in progress - lot of tidying up to do - not going very smoothly ????

We say a frame of reference is inertial if it is moving with a constant velocity to another. How do positions in one frame of reference transform into another frame of reference?

The assumptions are,

• The two axis of the reference frames are parallel.
• The transformation between frames depends only the velocity.
  • The transformation from frame to the primed frame uses a velcoity v.
  • The transformation from primed frame to the frame uses a velcoity -v.
• The speed of light as measured in the two frames is constant.
  • The difference in position divided by the difference in time of a light particle recorded at two positions is always the constant c.
• In the transformation, each component in the reference frame is a linear combination of the components in the primed frame.

The last condition is expressed as,

$x_{\mu }=\sum _{\nu }m_{\mu ,\nu }x'_{\nu }$

where $m_{\mu ,\nu }$ are the components of the transformation matrix.

One approach would be two put this transformation into the Minkowski metric,

$(\sum _{\mu }x_{\mu }^{2})-c^{2}t^{2}=(\sum _{\mu }(x'_{\mu }-v_{\mu }t')^{2})-c^{2}t'^{2}$

$(\sum _{\mu }(x_{\mu }+v_{\mu }t)^{2})-c^{2}t^{2}=(\sum _{\mu }x'^{2})-c^{2}t'^{2}$

This is mathematically difficult. See Hyperbolic geometry.

Derivation

The usual approach is to assume that the frames move relative to each other along one co-ordinate axis. This approach is described at Derivations of the Lorentz transformations. These derivations are simpler in some ways, but also more confusing in other ways. They rely on understanding the physical situation, where as I want a derivation that puts all the facts in to the formula, and arrives at the result. Each step should be a simple no-brainer, even the derivation is a bit longer.

To make the derivation more understandable take $t=x_{0}$ and let the indexes go from 1 to 3 (missing out 0). Then,

$x_{\mu }=m_{\mu ,0}t'+\sum _{\nu }m_{\mu ,\nu }x'_{\nu }$ (Equation 1)

$t=m_{0,0}t'+\sum _{\nu }m_{0,\nu }x'_{\nu }$ (Equation 2)

Distance and the moving frames

The two reference systems are moving with a relative constant velocity v in a direction e (given by a unit vector). So at, $\mathbf {x} =(0,0,0)$, $\mathbf {x} '=vt'\mathbf {e}$. This equation may be written,

$x'_{\nu }=vt'e_{\nu }$

Note: There is an assumption here that $\mathbf {e}$ and $\mathbf {e} '$ are the same. This assumes that the co-ordinate axis are parallel.

(Equation 1) which applies for any $x'_{\mu }$. The general formula may be applied to describe the origin of the co-ordinate systems. Substituting,

$x_{\nu }$ for $0$

$x'_{\nu }$ for $vt'e_{\nu }$

in (equation 1) gives,

$0=m_{\mu ,0}t'+\sum _{\nu }m_{\mu ,\nu }vt'e_{\nu }$

which simplifies to,

$0=m_{\mu ,0}+v\sum _{\nu }m_{\mu ,\nu }e_{\nu }$

and then,

$-{\frac {m_{\mu ,0}}{v}}=\sum _{\nu }m_{\mu ,\nu }e_{\nu }$

This suggests a more natural constant for $(\mu ,0)$. Define, $n_{\mu ,0}$ as,

$n_{\mu ,0}=-{\frac {m_{\mu ,0}}{v}}$ (definition 1)

then,

$n_{\mu ,0}=\sum _{\nu }m_{\mu ,\nu }e_{\nu }$

Then define $n_{\mu ,\nu }$ as,

$m_{\mu ,\nu }=n_{\mu ,0}n_{\mu ,\nu }$ (definition 2)

then,

$n_{\mu ,0}=\sum _{\nu }n_{\mu ,0}n_{\mu ,\nu }e_{\nu }=n_{\mu ,0}\sum _{\nu }n_{\mu ,\nu }e_{\nu }$

gives,

$1=\sum _{\nu }n_{\mu ,\nu }e_{\nu }$ (equation 3)

Substituting the (definition 1) and (definition2) into (equation 1) gives,

$x_{\mu }=n_{\mu ,0}(-vt'+\sum _{\nu }n_{\mu ,\nu }x'_{\nu })$ (equation 4a)

$x'_{\mu }=n_{\mu ,0}(vt+\sum _{\nu }n_{\mu ,\nu }x_{\nu })$ (equation 4a)

Distance and the constant speed of light

Two reference frames are moving relative to each other in a direction e with a velocity v.

A light particle moving from the origin to a point x in the direction of e will always be measured to have a velocity c,

$\mathbf {x} =ct\mathbf {e}$

The same light particle will appear to travel from the origin to x',

$\mathbf {x} '=ct'\mathbf {e}$

These equations may be written as,

$x_{\mu }=cte\mu$

$x'_{\mu }=ct'e\mu$

These equations represent another specific situation which the general equations must satisfy. Substituting,

$x_{\mu }$ by $cte\mu$

$x'_{\mu }$ by $ct'e\mu$

in (equation 4a) and (equation 4b) gives,

$cte_{\mu }=n_{\mu ,0}(-vt'+\sum _{\nu }n_{\mu ,\nu }ct'e_{\nu })$

$ct'e_{\mu }=n_{\mu ,0}(+vt+\sum _{\nu }n_{\mu ,\nu }cte_{\nu })$

which may be simplified as,

$cte_{\mu }=t'n_{\mu ,0}(-v+c\sum _{\nu }n_{\mu ,\nu }e_{\nu })$

$ct'e_{\mu }=tn_{\mu ,0}(+v+c\sum _{\nu }n_{\mu ,\nu }e_{\nu })$

But (equation 3) is,

$\sum _{\nu }n_{\mu ,\nu }e_{\nu }=1$

so,

$cte_{\mu }=t'n_{\mu ,0}(c+v)$

$ct'e_{\mu }=tn_{\mu ,0}(c-v)$

multiplying each side of each equation gives,

$c^{2}e_{\mu }^{2}=n_{\mu ,0}^{2}(c+v)(c-v)$

$n_{\mu ,0}^{2}={\frac {e_{\mu }^{2}}{1-{\frac {v^{2}}{c^{2}}}}}$

Define $\alpha$ as,

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

then,

$n_{\mu ,0}=e_{\mu }\alpha$

and,

$x_{\mu }=e_{\mu }\alpha (-vt'+\sum _{\nu }n_{\mu ,\nu }x'_{\nu })$ (equation 5a)

$x'_{\mu }=e_{\mu }\alpha (vt+\sum _{\nu }n_{\mu ,\nu }x_{\nu })$ (equation 5a)

Time and moving reference frames

As before the two reference systems are moving with a constant velocity v in a direction e (given by a unit vector). So at, $x=0$, $\mathbf {x} '=vt'\mathbf {e}$.

From (equation 2) setting $x'_{\nu }=ve_{\nu }t'$ gives,

$t=m_{0,0}t'+\sum _{\nu }m_{0,\nu }ve_{\nu }t'$

which simplifies to,

$t=t'(m_{0,0}+v\sum _{\nu }m_{0,\nu }e_{\nu })$

From (equation 2) by the principle of relaticity

$t'=m_{0,0}t-\sum _{\nu }m_{0,\nu }x_{\nu }$

Setting $\mathbf {x} =(0,0,0)$ gives,

$t'=tm_{0,0}$

Solving both equations for $t/t'$ gives,

${\frac {1}{m_{0,0}}}=t/t'=m_{0,0}+v\sum _{\nu }m_{0,\nu }e_{\nu }$

which simplifies to,

${\frac {1}{m_{0,0}}}-m_{0,0}=v\sum _{\nu }m_{0,\nu }e_{\nu }$

and then,

$m_{0,0}({\frac {1}{m_{0,0}}}^{2}-1)=v\sum _{\nu }m_{0,\nu }e_{\nu }$

Let,

$m_{0,\nu }=m_{0,0}n_{0,\nu }$

then,

$({\frac {1}{m_{0,0}}}^{2}-1)=v\sum _{\nu }n_{0,\nu }e_{\nu }$ (equation 7)

and,

$t=m_{0,0}(t'+\sum _{\nu }n_{0,\nu }x'_{\nu })$ (equation 8)

Time and the constant speed of light

The constancy of the speed of light allows us to instantiate,

$x=ct\mathbf {e}$

$x'=ct'\mathbf {e}$

In (equation 2) giveing,

$t=m_{0,0}(t'+\sum _{\nu }n_{0,\nu }ct'e_{\nu })$

which simplifies to,

$t=t'm_{0,0}(1+c\sum _{\nu }m_{0,\nu }e_{\nu })$

and from the principle of relativity,

$t'=tm_{0,0}(1-c\sum _{\nu }m_{0,\nu }e_{\nu })$

Multiplying both sides of the two equations gives,

$1=m_{0,0}^{2}(1-c^{2}(\sum _{\nu }n_{0,\nu }e_{\nu }))^{2}$

and by dividing both sides by $m_{0,0}^{2}$, and subtracting 1,

${\frac {1}{m_{0,0}}}^{2}-1=-c^{2}(\sum _{\nu }n_{0,\nu }e_{\nu }))^{2}$

(Equation 7) gives another expression for ${\frac {1}{m_{0,0}}}^{2}-1$,

${\frac {1}{m_{0,0}}}^{2}-1=v\sum _{\nu }n_{0,\nu }e_{\nu }$

So,

$-c^{2}(\sum _{\nu }n_{0,\nu }e_{\nu }))^{2}={\frac {1}{m_{0,0}}}^{2}-1=v\sum _{\nu }n_{0,\nu }e_{\nu }$

Which simplifies to,

$\sum _{\nu }n_{0,\nu }e_{\nu }=-{\frac {v}{c^{2}}}$ (equation 9)

This expression may be substituted in (equation 5) giving,

${\frac {1}{m_{0,0}}}^{2}-1=-v{\frac {v}{c^{2}}}=-{\frac {v^{2}}{c^{2}}}$

This equation may be re-arranged as,

$m_{0,0}^{2}={\frac {1}{1-{\frac {v^{2}}{c^{2}}}}}=\alpha ^{2}$

so

$t=\alpha (t'-\sum _{\nu }n_{0,\nu }x'_{\nu })$ (equation 10)

A simple expression that satisfies (equation 9) is,

$n_{0,\nu }={\frac {v}{c^{2}}}e_{\nu }$ (equation 11)

???? This result is not derived here. ????

substituting in,

$\sum _{\nu }n_{0,\nu }e_{\nu }=-{\frac {v}{c^{2}}}\sum _{\nu }e_{\nu }^{2}$

But e is defined as a unit vector so,

$\sum _{\nu }e_{\nu }^{2}=1$

Giving,

$\sum _{\nu }n_{0,\nu }e_{\nu }=-{\frac {v}{c^{2}}}$

Which gives,

$t=\alpha (t'-{\frac {v}{c^{2}}}\sum _{\nu }e_{\nu }x'_{\nu })$ (equation 12)

Lorentz transform

$x_{\mu }=\alpha (x'_{\mu }-vt'e_{\mu })$

$t=\alpha (t'-{\frac {v}{c^{2}}}\sum _{\nu }e_{\nu }x'_{\nu })$

But this is what I am expecting ???? how ????

$x_{\mu }=\alpha e_{\mu }(x'_{\mu }-vt')+(1-e_{\mu })x'_{\mu }$

Using $t=x_{0}$

$x_{\mu }=\alpha (x'_{\mu }-vx'_{0}e_{\mu })$

$x_{0}=\alpha (x'_{0}-{\frac {v}{c^{2}}}\sum _{\nu }e_{\nu }x'_{\nu })$

Equivalence of mass and energy

$E=mc^{2}$

The inevitability of curvature

Einstein had no choice but to consider a curved universe because, curvature is an inevitable consequence of general relativity.

Light gaining energy, and so changing frequency

$E=fv$

Photon gains energy as it falls. So frequency increases. Time speeds up.

Rotating planet

Different parts moving at different speeds.

Gravity and mass

The impossibility of distinguishing between gravity and acceleration .