0
$\begingroup$

I was studying special relativity, but there is a "proof" than I don't understand:

When we need to prove time dilation, we write the Lorentz equation in form of $\Delta t= \gamma \Delta t'$ because we want to evaluate the time gap in the $S$ system frame, with respect to which the system $S'$ (where the two events happen) is in motion.

But when we want to evaluate $\Delta x$, the proof uses the Lorentz equation for $\Delta x'$. Why? This seems similar to the derivation for time dilation (where we wanted to evaluate $\Delta t$ as a function of $\Delta t'$), so why don't we use the equation for $\Delta x$ directly? But if we did this, we'd see that $\Delta x = \gamma \Delta x'$, which is wrong.

What is the reason for this?

$\endgroup$
1
  • $\begingroup$ Typeset mathematical terms using MathJax. Here's the tutorial. $\endgroup$
    – SarGe
    Aug 28, 2020 at 3:22

2 Answers 2

1
$\begingroup$

A spacetime diagram may help.

enter image description here

Equal time in the Earth frame is determined by horizontal lines, e.g. from the point B. But spacecraft length in the Earth frame is not determined by a vertical line from the point B. Instead it is measured on a horizontal line in the Earth frame. The ends of the space craft at equal time in the Earth frame are not points of equal time in the spacecraft frame.

$\endgroup$
1
$\begingroup$

This is a common source of confusion, and has to do with the definitions of length and time intervals. Let's set the stage: we have two frames $S$ and $S'$, with $S'$ moving with respect to $S$ at a constant velocity $v$. We are looking at how lengths and time intervals that someone observes in $S'$ (i.e. the "proper" length and the "proper" time-interval) appear to an observer in $S$.

Let's also write out the Lorentz Transformations:

\begin{equation} \begin{aligned} &\text{(A)}\quad\Delta x^\prime = \gamma \left(\Delta x - v \Delta t\right)\\ &\text{(B)}\quad \Delta t^\prime = \gamma \left( \Delta t - \frac{v}{c^2}\Delta x\right)\\ \\ &\text{(C)}\quad\Delta x = \gamma \left(\Delta x^\prime + v \Delta t^\prime \right)\\ &\text{(D)}\quad \Delta t = \gamma \left( \Delta t^\prime + \frac{v}{c^2}\Delta x^\prime \right)\\ \end{aligned} \end{equation}

What does it mean to measure length?

Let's say there is a rod in the frame $S'$. To someone sitting in $S$, the rod appears to be moving with respect to them. If the observer in $S$ wants to measure the length of the rod, they need to mark the endpoints of the rod simultaneously. If they wait for a couple of seconds between marking the left hand coordinate and the right hand coordinate of the rod, then the rod would have moved in between these measurements, and one would measure a much larger length than the actual one.

Thus, you want to relate the difference in coordinates observed in $S'$ (which is $\Delta x'$) with the difference in coordinates observed in $S$ (which is $\Delta x$) when the time interval between measurements in $S$ is zero ($\Delta t=0$). And clearly, the equation to do that is $(A)$.

See my answer here for more details on how length intervals are measured: What is wrong in this false derivation of length “dilation”?

What does it mean to measure time intervals?

Let's say there's a clock in $S'$. To someone sitting in $S$, the clock is moving with respect to them. The observer in $S$ wants to measure the time that has elapsed for that clock (which is moving with respect to them) between two events. Now, in the $S'$ frame, the clock appears to be at rest, and therefore its coordinates are fixed between the two measurement events. i.e. $\Delta x'=0$. And we want to relate the time interval elapsed in $S'$ (which is $\Delta t'$), with the one elapsed in $S$ (which is $\Delta t$), and so the equation we need is $(D)$.

See my answer here for more details on how time measurements are related between frames: Why is this assumption made in deriving time dilation?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.