der.physika said:
What are the equations that describe simultaneity, and hey don't be afraid to throw some math out there. All though I might not be able to grasp things like the Ricci Curvature just yet, I could still derive the equations for Time Dilation & the Lorentz Contraction, so please show me the math, I want the math
Let [itex]\Lambda[/itex] be the Lorentz transformation that does a coordinate change from an inertial frame S to an inertial frame S'.
[tex]\Lambda=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}[/tex]
[tex]\gamma=\frac{1}{\sqrt{1-v^2}}[/tex]
Let [tex]\begin{pmatrix}t\\ x\end{pmatrix}[/tex] be the coordinates in S of an event that is simultaneous in S' with the origin. (Note that this is the x' axis). We have
[tex]\Lambda\begin{pmatrix}t\\ x\end{pmatrix}=\begin{pmatrix}0\\ x'\end{pmatrix}[/tex]
The lower component is irrelevant. The upper one is
[tex]\gamma(t-vx)=0[/tex]
and since [itex]\gamma\neq 0[/itex], we have t=vx. Note that this equation defines a line with slope v in a spacetime diagram (with t increasing in the up direction and x increasing to the right).
This implies that we have the following rule: If the velocity of the spatial origin of S' in S is v, then the t' axis is drawn with slope 1/v, and the x' axis with slope v. And it's not much harder to show
any set of events that are simultaneous in S' is a line with slope v in the diagram. Those lines are called simultaneity lines.
The explicit formula for a Lorentz transformation that I used above follows from a few very naural assumptions about how to make mathematical sense of the ideas in the numbered list in my previous post. Let's start by assuming that a coordinate change between inertial coordinate systems will be a linear function [itex]\Lambda[/itex]. (This is very natural. If these functions don't take straight lines to straight lines, an object that's moving with a constant velocity forever in one inertial coordinate system would be accelerating in another). A linear function can be represented by a matrix, so let's write
[tex]\Lambda=\begin{pmatrix}a & b\\ c & d\end{pmatrix}[/tex]
The lines t=x and t=-x represent light rays. #2 on the numbered list in my previous post suggests that we should assume that any point on the line t=x is taken to a point on the same line, and similarly for t=-x. These assumptions imply that there exist numbers [itex]\alpha[/itex] and [itex]\beta[/itex], such that
[tex]\Lambda\begin{pmatrix}1\\ 1\end{pmatrix}=\alpha\begin{pmatrix}1\\ 1\end{pmatrix}[/tex]
[tex]\Lambda\begin{pmatrix}1\\ -1\end{pmatrix}=\beta\begin{pmatrix}1\\ -1\end{pmatrix}[/tex]
It's easy to show that these equations imply a=d, b=c, so that
[tex]\Lambda=\begin{pmatrix}a & b\\ b & a\end{pmatrix}[/tex]
Now note that the t' axis, expressed in the coordinates of S, is the line x=vt. This combined with the equivalence of different inertial coordinate systems alluded to in #1 on the numbered list, suggests that we should assume that the t axis expressed in the coordinates of S' is the line x'=-vt'. This means that there's a number [itex]\gamma[/itex] such that
[tex]\Lambda\begin{pmatrix}1\\ 0\end{pmatrix}=\gamma\begin{pmatrix}1\\ -v\end{pmatrix}[/tex]
This equation implies [itex]b=-av=-\gamma v[/itex]. We have [itex]a=\gamma[/itex], but this might be a different number for a different v, so I'm redefining [itex]\gamma[/itex] to be a function instead of a number, and I'll write [itex]\gamma(v)[/itex] where I would previously have written [itex]a[/itex] or [itex]\gamma[/itex].
[tex]\Lambda=\gamma(v)\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}[/tex]
#1 also suggests that we should assume that we can obtain the inverse just by changing the sign of the velocity.
[tex]\Lambda^{-1}=\gamma(-v)\begin{pmatrix}1 & v\\ v & 1\end{pmatrix}[/tex]
So
[tex]I=\Lambda\Lambda^{-1}=\gamma(v)\gamma(-v)\begin{pmatrix}1-v^2 & 0\\ 0 & 1-v^2\end{pmatrix}[/tex]
[tex]\gamma(v)\gamma(-v)(1-v^2)=1[/tex]
But #1 also suggests that we should assume that [itex]\gamma[/itex] is an even function. (Without this assumption, statements about time dilation that two inertial observers make about each other won't be symmetrical). And that gives us the final result
[tex]\Lambda=\frac{1}{\sqrt{1-v^2}}\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}[/tex]
If any of this looks difficult because you're unfamiliar with matrices or spacetime diagrams, then I strongly urge you too study those things right away. What I said here can be rephrased without using matrices, but I think it's a terrible idea to go for that option.