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 \Lambda be the Lorentz transformation that does a coordinate change from an inertial frame S to an inertial frame S'.
\Lambda=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}
\gamma=\frac{1}{\sqrt{1-v^2}}
Let \begin{pmatrix}t\\ x\end{pmatrix} 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
\Lambda\begin{pmatrix}t\\ x\end{pmatrix}=\begin{pmatrix}0\\ x'\end{pmatrix}
The lower component is irrelevant. The upper one is
\gamma(t-vx)=0
and since \gamma\neq 0, 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 \Lambda. (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
\Lambda=\begin{pmatrix}a & b\\ c & d\end{pmatrix}
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 \alpha and \beta, such that
\Lambda\begin{pmatrix}1\\ 1\end{pmatrix}=\alpha\begin{pmatrix}1\\ 1\end{pmatrix}
\Lambda\begin{pmatrix}1\\ -1\end{pmatrix}=\beta\begin{pmatrix}1\\ -1\end{pmatrix}
It's easy to show that these equations imply a=d, b=c, so that
\Lambda=\begin{pmatrix}a & b\\ b & a\end{pmatrix}
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 \gamma such that
\Lambda\begin{pmatrix}1\\ 0\end{pmatrix}=\gamma\begin{pmatrix}1\\ -v\end{pmatrix}
This equation implies b=-av=-\gamma v. We have a=\gamma, but this might be a different number for a different v, so I'm redefining \gamma to be a function instead of a number, and I'll write \gamma(v) where I would previously have written a or \gamma.
\Lambda=\gamma(v)\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}
#1 also suggests that we should assume that we can obtain the inverse just by changing the sign of the velocity.
\Lambda^{-1}=\gamma(-v)\begin{pmatrix}1 & v\\ v & 1\end{pmatrix}
So
I=\Lambda\Lambda^{-1}=\gamma(v)\gamma(-v)\begin{pmatrix}1-v^2 & 0\\ 0 & 1-v^2\end{pmatrix}
\gamma(v)\gamma(-v)(1-v^2)=1
But #1 also suggests that we should assume that \gamma 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
\Lambda=\frac{1}{\sqrt{1-v^2}}\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}
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.
