Time Correction Using Simultaneity Requirement

There is something very confusing about relativity. We have several possibilities for interpreting experiments. First we need to ask about results in the proper frames of both fixed and moving observers. Then we need to ask about how a fixed observer views the results of an experiment done in a moving frame. In a perfectly asymmetric manner we need to ask about how the moving observer views results of an experiment in the fixed frame. The latter question is what I will consider here.

We will use the convention that lower case (x,t) are the parameters used by the observer in the fixed system and upper case (X,T) are the parameters used by the moving system.

First,
let's assume the simplest case scenario where both observers are coincident at
times T=0 and t=0 when both events occur (in the stationary frame) and that the
events are separated in the stationary observer's proper frame by distance D. Also
assume that one of the events in the stationary system is at *x=-D/2* and the other event is at *x=+D/2*.

For the events imagine simultaneous (in the fixed frame)
lightning strikes at *-D/2* and *D/2*.
These will lead to light rays as well as thunder sound which arrives
much later but, if in still air, will result in the same equations as
light. The symbol c will stand for the
speed of sound or speed of light.

Obviously, the stationary observer hears or sees the events at the same time

$$t=\frac{D}{2c}$$ |
(1.1) |

so for him ΔT=0.
The moving observer hears or sees the +*D/2* event sooner at

$${T}_{Front}=\frac{D}{2(c+v)}$$ |
(1.2) |

and the -*D/2* event
later at

$${T}_{Back}=\frac{D}{2(c-v)}$$ |
(1.3) |

The difference between these two receptions is

$$\Delta T=-\frac{Dv}{{c}^{2}-{v}^{2}}$$ |
(1.4) |

We can now generalize the expression for time transformation taking this difference into account

$$\begin{array}{l}T=t-x\text{'}v/({c}^{2}-{v}^{2})\\ X=x-vt\end{array}$$ |

Up to now the equations are okay but we have to take into
account that *x'* in the first of
equations (1.5)
is really x from a different reference frame and must be transformed according
to (see Appendix)

$$x\text{'}=x-vt$$ |
(1.6) |

Collecting terms in t we get:

$$\begin{array}{l}T=t-(x-vt)xv/({c}^{2}-{v}^{2})=[t(1+\frac{{v}^{2}}{{c}^{2}-{v}^{2}})-xv\frac{1}{{c}^{2}-{v}^{2}}]\\ =\frac{{c}^{2}}{{c}^{2}-{v}^{2}}\left(t-xv/{c}^{2}\right)={\gamma}^{2}(t-xv/{c}^{2})\end{array}$$ |
(1.7) |

where

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

Since the expression of X also involves time the same factor
*γ*^{2}
must precede it as well.

$$\begin{array}{l}T(x,t)={\gamma}^{2}\left(t-\frac{vx}{{c}^{2}}\right)\\ X(x,t)={\gamma}^{2}(x-vt)\end{array}$$ |

When we solve for x and t in terms of X and T we need to obtain expressions that look the same as equations (1.8) except for the sign of v. If the expressions looked different then one of the two frames would be "special" and that has been proven not to be valid. We can write the following matrix expression for X and T

$${\gamma}^{2}\left(\begin{array}{cc}1& -v/{c}^{2}\\ -v& 1\end{array}\right)\left(\begin{array}{c}t\\ x\end{array}\right)=\left(\begin{array}{c}T\\ X\end{array}\right)$$ |

To solve for t and x we multiply both sides of (1.9) by the inverse of the matrix on the left and obtain

$${\gamma}^{2}\frac{\left(\begin{array}{cc}1& v/{c}^{2}\\ v& 1\end{array}\right)}{1-\frac{{v}^{2}}{{c}^{2}}}\left(\begin{array}{cc}1& -v/{c}^{2}\\ -v& 1\end{array}\right)\left(\begin{array}{c}t\\ x\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}t\\ x\end{array}\right)=\left(\begin{array}{cc}1& v/{c}^{2}\\ v& 1\end{array}\right)\left(\begin{array}{c}T\\ X\end{array}\right)$$ |
(1.10) |

Which results in the expressions:

$$\begin{array}{l}t=[T+vX/{c}^{2}]\\ x=(X+vT)\end{array}$$ |

Obviously the factor ${\gamma}^{2}=\frac{1}{1-\frac{{v}^{2}}{{c}^{2}}}$ is
missing in front of these expressions and they do ** not** look like equations (1.8). For small v/c this doesn't seem important but
it means that either equations (1.11) or equations (1.8)
are wrong. We can correct both equation pairs by multiplying them by the factor
$\gamma =\frac{1}{\sqrt{1-{v}^{2}/({c}^{2})}}$ at the outset.

Then equations (1.8) and equations (1.11) become:

$$\begin{array}{l}T=\gamma [t-xv/{c}^{2}]\\ X=\gamma (x-vt)\end{array}$$ |

and

$$\begin{array}{l}t=\gamma [T+vX/{c}^{2}]\\ x=\gamma (X+vT)\end{array}$$ |
(1.13) |

We might think of the factor γ' as an expansion factor since it is always greater than 1. It is important to understand that γ' arose only because simultaneity required that we add a spatial dependence to time measurements.

When applied to the time equation, γ' is called the time dilation factor.

There is something tricky about how to apply these Lorentz equations to the measurements of the lengths of moving objects. Since distance is linked with time, we must be sure that we measure the two ends of the object simultaneously. Think of the two measurements as events. We can send light pulses from the center toward the ends and observe the elapsed time required to arrive. If the length is L, then in a frame where the object is stationary, the times will both be t=L/2c so that the apparent length is ct=L. In the frame viewing the moving object, the times will be from equations (1.12)

$$\begin{array}{l}{T}_{Front}=\gamma \left[\frac{L}{2c}-\frac{Lv}{2{c}^{2}}\right]=\frac{L(c-v)}{2{c}^{2}}\\ {T}_{Back}=\gamma \left[\frac{L}{2c}+\frac{Lv}{2{c}^{2}}\right]=\frac{L(c+v)}{2{c}^{2}}\end{array}$$ |

These times are not the same so we must correct the apparent distance measurement for this difference which is presently:

$$l=c({T}_{Back}-{T}_{Front})=\frac{\gamma Lv}{c}$$ |
(1.15) |

To make the correction, multiply the first equation in(1.14) by (1+v/c) and the second equation by (1-v/c):

$$\begin{array}{l}T{\text{'}}_{Front}=\gamma \frac{L}{2{c}^{2}}(c-v)(1+v/c)=\gamma L/c(1-{v}^{2}/{c}^{2})=L/(2c\gamma )\\ T{\text{'}}_{Back}=\gamma \frac{L(c+v)(1-v/c)}{2{c}^{2}}=\gamma L/c(1-{v}^{2}/{c}^{2})=L/(2c\gamma )\end{array}$$ |
(1.16) |

The length $L\text{'}=c(T{\text{'}}_{Front}+T{\text{'}}_{Back})/\gamma =L/\gamma $ in this frame is the length in the stationary frame divided by γ and is therefore longer since γ is always greater than 1 when v is not zero.

This transformation is used when converting the lack of simultaneity of light pulse receptions
for an event separation *D* into a *x*-dependent clock correction. Essentially in our expression for time
difference D becomes x'=x-vt which results in the relativistic expression for time:

$$T(x,t)=\alpha {\gamma}^{2}(t-xv/{c}^{2})$$ |
(1.17) |

where α is to be determined by requiring that t(X,T)
look the same as T(x,t) except for the sign of the term *xv/c ^{2}*.

The first comment we can make is that it postulates that x' is zero when x and t are zero. It also stipulates, assuming v>0, that x' becomes more negative when t becomes more positive while x remains constant. For an observer starting at x'=0 to move at time t=0 at positive velocity v in the x direction looking at a particular position x in a stationary frame, x'=x-vt would be the x' coordinate that he would observe. Note that this value is fixed in the moving observer's frame of reference and therefore requires no relativistic correction.

I feel more comfortable computing the observations of a
fixed observer who sees a moving frame moving to the right at speed v. It turns out that the x dependence of time is
almost the same if we use x'=x+vt except for the sign of the vx/c^{2}
term .

$$T\text{'}(x,t)=\alpha {\gamma}^{2}(t+xv/{c}^{2})$$ |
(1.18) |