# To person A standing on a railway platform, person B on the train travelling past would seems to be aging slower (if such a thing were perceptible) and to person B it would appear that person A was aging faster. But in the absence of a train, or a platform, or even a planet, the two people would appear to be moving apart. Without a frame of reference it would be difficult to say which is moving and which is not. However, they can't both be appearing to age faster than the other. So which one would be aging faster and which slower? Or rather, how can one tell which one is moving?

Amina Richards 2022-10-17 Answered
To person A standing on a railway platform, person B on the train travelling past would seems to be aging slower (if such a thing were perceptible) and to person B it would appear that person A was aging faster.
But in the absence of a train, or a platform, or even a planet, the two people would appear to be moving apart. Without a frame of reference it would be difficult to say which is moving and which is not. However, they can't both be appearing to age faster than the other.
So which one would be aging faster and which slower? Or rather, how can one tell which one is moving?
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Ramiro Sosa
First: ditch $A$ and $B$. Their names are now Unprimed and Primed.
So Unprime is on the platform in frame $S$, and Prime is on the in-bound train in frame ${S}^{\prime }$, moving at $\beta =\sqrt{3}/2$ in the $x$ direction.
"Unprime" has two events: starting a stop watch at:
${E}_{0}=\left({t}_{0}=0,{x}_{0}=0\right)$
and stopping it $T=32$ seconds latter at:
${E}_{3}=\left({t}_{3}=T,{x}_{3}=0\right)$
The time difference is, not surprisingly:
$\mathrm{\Delta }t={t}_{3}-{t}_{0}=\left(T-0\right)=32s$
In Prime's frame of reference:
${E}_{0}=\left({t}_{0}^{\prime }=0,{x}_{0}^{\prime }=0\right)$
${E}_{3}=\left(\gamma \left[{t}_{3}-\beta {x}_{3}\right],\gamma \left[{x}_{3}-\beta {t}_{3}\right]\right)$
$=\left(\gamma T,-\gamma \beta T\right)$
Thus, Prime says the time between events is:
$\mathrm{\Delta }{t}^{\prime }={t}_{3}^{\prime }-{t}_{0}^{\prime }=\gamma T=64s$
Hence: Prime says that Unprime's clock is running slower.
Now suppose Prime also had a running stop-watch for $T=32$ seconds. The start is the same event as Unprime's start:
${E}_{0}=\left({t}_{0}^{\prime }=0,{x}_{0}^{\prime }=0\right)$
while the stopping event occurs at:
${E}_{2}=\left({t}_{2}^{\prime }=T,{x}_{2}^{\prime }=0\right)$
Of course, the time difference is:
$\mathrm{\Delta }{t}^{\prime }={t}_{2}^{\prime }-{t}_{0}^{\prime }=T=32s$
How does this look in Unprimes's frame? We know ${E}_{0}$, but we need to use the inverse of Lorentz transform we used before to transform ${E}_{2}$ to $S$:
${E}_{2}=\left(\gamma \left[{t}_{2}^{\prime }+\beta {x}_{2}^{\prime }\right],\gamma \left[{x}_{2}^{\prime }+\beta {t}_{2}^{\prime }\right]\right)$
$=\left(\gamma T,\gamma \beta T\right)$
so that:
$\mathrm{\Delta }t=\gamma \left({t}_{2}^{\prime }-{t}_{0}^{\prime }\right)=\gamma T=64s$
and now Unprime says that Prime's clock in running slower.
That seems paradoxical at first, but it is resolved by the fact that Unprimes's stopwatch stops at $x=0$, while Primes's stopwatch stops at ${x}^{\prime }=0$, or $x=+55$ light-seconds ($ls$), and there is no way to define "simultaneous" across $55$ light-seconds.
We can assign an event to the moment Unprime says Prime has stopped his watch, call it ${E}_{1}$. We have in $S$ that:
${E}_{2}=\left(\gamma T,\gamma \beta T\right)=\left(64\phantom{\rule{thinmathspace}{0ex}}s,55\phantom{\rule{thinmathspace}{0ex}}ls\right)$
The event of Unprimed saying Prime's watch has stopped occurs simultaneously at Unprime's origin:
${E}_{1}=\left(\gamma T,0\right)$
If we transform this to ${S}^{\prime }$, that event occurs at ${t}^{\prime }={\gamma }^{2}T$, or $128$ s. So Prime stops his watch at $32$s, which Unprime says happened after $64$ seconds, while Prime says Unprimed said he stopped his watch after $128$ seconds. Likewise in the other direction, Unprimed says Prime thought Unprimed watch was stopped after $8$ seconds.
It is the relativity of simultaneity that allows each frame to see the other's clock moving more slowly.
Now the idea that you can't move relative to space is a good one:
1. There is no absolute rest frame.
2. The speed of light is always $c$ in all directions.
So no matter how fast you move, space looks the same. There is a global electromagnetic field with $\stackrel{\to }{E}=0$ and $\stackrel{\to }{B}=0$, and any disturbance in that field propagates at:
$c=\frac{1}{\sqrt{{ϵ}_{0}{\mu }_{0}}}$
(and of course, charges and currents look "the same" in the lab through ${ϵ}_{0}$ and ${\mu }_{0}$).
Moreover, we could add a third frame, Doubleprime, on the outbound train, moving at ${\beta }^{″}=-\frac{\sqrt{3}}{2}$, and discover that both the other clocks move slower, while Unprime says it moves at the same speed as Prime.
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Danika Mckay
Your statement about aging is incorrect. Instead, time dilates relative to stationary, so they each see the other moving more slowly through time, and therefore aging more slowly.
Also, you say several times "with no frames of reference". There are frames of reference, it's just that there's no absolute frame of reference.