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While reading some papers on Einstein's theory of relativity, seeing how the flow of time is not the same for everyone, a doubt occurred to me:

Let us imagine a photon moving in a well-defined space at the speed of light. While it will not "feel" the passage of time, according to our point of view the photon will come to us after a finite amount of time. This is because its speed is finite. But how can it be that for us the photon takes time to travel if it does not interact with space-time? Then wouldn't this imply that the spatial distance is the same for everyone?

Stream Sphere
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2 Answers2

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Distance isn't the same for everyone even in classical physics.

You stand by a road. A car drives by. At $t_0$ it is at $x_0$. At $t_1$, it is at $x_1$. The distance it traveled in time $t_1-t_0$ is $x_1-x_0$.

The driver sits in the car. He says the distance the car moved is $0$. At $_0$, the seat is right under him. At $t_1$, the seat is still right under him. At $t_0$, $x_0$ is at a certain place. At $t_1$, $x_1$ is in the same place.

Both he and you are right. You are just using two different inertial frames of reference.

mmesser314
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There are two event, Transmission ($T$) and Receive ($R$). They have coordinates (in a well chosen frame, $S$, $c=1$):

$$ T = (t=0, x=0) $$ $$ R = (T=L, x=L) $$

So it travels a distance $L$ in time $L$.

Now Lorentz transform to any other frame, $S'$, moving at $\beta \in (-1, 1)$:

$$ T = (t'=0, x'=0) $$ $$ R = \big(t'=\gamma(L-\beta L), x'=\gamma(L-\beta L) \big) = (f_DL, f_DL) $$

with:

$$ f_D = \gamma(1-\beta) = \sqrt{ \frac{1-\beta}{1+\beta} } $$ so it travels a distance $L'$ in time $L'$ with

$$ L'= f_DL \in (0, \infty) $$

and that is an open interval.

Oh, and $f_D$ is the relativistic Doppler shift, which is has to be, since phase:

$$ \phi(x, t) = kx-\omega t = k^{\mu}x_{\mu} $$

is a Lorentz scalar. All observers see the same number of oscillations between $T$ and $R$.

JEB
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