Is spatial distance objective?

Question

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?

In the discussion on p.48 of his 2021 pop-sci book titled "The God Equation", the physicist Michio Kaku describes the fact that Special and General Relativity can each have effects opposite the other's (in the GPS System, for instance), and similar considerations have been brought out by the well-known team of Lineweaver and Davis in their numerous papers whose titles contain the phrase "Expanding Confusion", so I'm attempting to add a Special Relativity tag. The notion that photons "feel" anything seems (to me) to allude to whatever faint role they might have in biological systems. — Edouard, Jan 05 '24 at 16:54
Biological cognizance develops at a rate vastly slower than the speed of light. — Edouard, Jan 05 '24 at 17:04

score 1 · Answer 1 · answered Jan 05 '24 at 15:06

1

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.

answered Jan 05 '24 at 15:06

mmesser314

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You may additionally write that the car driver saw the OP travel the same distance as the OP saw him. That might help the OP too!! – Proscionexium Jan 05 '24 at 15:11
I understood the analolgy, but does the same apply to photons? – Stream Sphere Jan 05 '24 at 15:38
This might help - A photon travels in a vacuum from A to B to C. From the point of view of the photon, are A, B, and C at the same location in space and time? – mmesser314 Jan 05 '24 at 16:05

score 1 · Accepted Answer · answered Jan 05 '24 at 17:05

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 $R$.

Is spatial distance objective?

2 Answers2