We know that time must shift alongside space because $\frac{dx}{dt} = \frac{dx'}{dt'}$ when $\frac{dx}{dt} = c$. This just means that the speed of light is invariant in all reference frames (light propagates at the same rate of space per time in all inertial reference frames). However it seems harder to state why entities that do not propagate at $c$ (such as events that are separated from the origin by more space than can be traversed by light in the time between them) are shifted in time proportionally to their spatial coordinate relative to the origin, just as their spatial coordinates shift more as more time passes (spatial coordinates shift proportionally to time ($x'=\gamma(x-vt)$) between inertial reference frames. Is there an eloquent way of explaining just why the time coordinate shifts more as spatial coordinates increase (relative to the origin) as shown by the moments transformation equation $t'=\gamma(t-\frac{vx}{c^2})$? (specifically, when they are not along a "light cone" relative to the origin?) As in, why does time shift as a function of space just as space shifts as a function of time? For example, as events that are colocated in one reference frame are not in another because space shifts more over time, is there a good, eloquent way of explaining why events are shifted in time more over space (hence why simultaneous events in one reference frame are not simultaneous in another with relative motion)? It would be especially helpful if this explanation helps build intuition.

Essentially, why does time shift as a function of space just as space shifts as a function of time (Especially for events and entities that are not traveling at C)?

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    $\begingroup$ What would qualify to you as "eloquent"? There are many simple thought experiments that show why you have to have this term; most textbooks use one of these thought experiments to introduce it. However, there's no way to get this term from Galilean intuition, like the $-vt$ term, because it isn't present in Galilean physics at all. $\endgroup$ – knzhou Feb 26 at 2:30
  • $\begingroup$ What I mean by eloquent is relatively simple to understand and good for building intuition. I suppose you are right about your point about not being able to get this from Galilean intuition. I am looking for a simple way to develop and intuition of the (-vt/c^2) term for entities that are not traveling at C away from and observer. My issue is that most thought experiments focus only on events connected by C being shifted (which is valid) but it is more difficult to intuitively understand why, say, an event simultaneous with the origin on one reference frame is shifted by the factor (-vt/c^2). $\endgroup$ – Sciencemaster Feb 27 at 2:15
  • $\begingroup$ Have you seen the train thought experiment? (This is the most common one you find in standard books.) $\endgroup$ – knzhou Feb 27 at 2:16
  • $\begingroup$ This is why I am looking for an explanation for this factor on an intuitive level (hopefully in a way similar to Galilean relativity (although obviously different from it on a fundamental level)) specifically and especially for events that are not connected by C (i.e. events have spacelike separation (separated by more space than can be traversed by light)). $\endgroup$ – Sciencemaster Feb 27 at 2:17
  • $\begingroup$ Yes, I have seen the train thought experiment, but it relies entirely on light signals, which is obviously a large part of special relativity, but it does little to show why this applies to events separated by more distance than C (i.e. $\endgroup$ – Sciencemaster Feb 27 at 2:21

I ll take $(c=1)$ from now on.

I am not sure this is the explanation that you are looking for, but I 'll give it a try.

So for a moment let us forget about the $(x,t)$ coordinate system and focus on the normal cartesian coordinate system (i.e $(x,y)$).

Let us choose a point such that $P(x_p, y_p)$. Let us set another coordinate $(x',y')$ system such that it's rotated with $\theta$ degree with respect to the x-axis. If we wanted to write the point P in terms of $(x',y')$ we need to apply a coordinate transformation, which has a form

$$\begin{pmatrix} x' \\ y' \end{pmatrix}=\begin{pmatrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\\ \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} \tag{1}$$

From here as you can see the length of the line element ($ds$) does not change under this transformation

$$ds^2 = dx^2+dy^2 = dx'^2+dy'^2$$

Now let us think of the SR case. In the SR the line element is defined such that

$$ds^2 = -dt^2+dx^2 = -dt'^2+dx'^2$$ (Which as you know comes from the fact that speed of light must be equal to the every inertial observer)

At this point we can naturally think a way to obtain coordinate transformation from $(x,t)$ to $(x',t')$.

Remember that $\cosh^2w - \sinh^2w =1$. Let me write this coordinate transformation directly

$$\begin{pmatrix} x' \\ t' \end{pmatrix}=\begin{pmatrix} \cosh(w) & -\sinh(w)\\ -\sinh(w) & \cosh(w)\\ \end{pmatrix}\begin{pmatrix} x \\ t \end{pmatrix} \tag{2}$$

This implies that $$x' = x\cosh(w) - t\sinh(w)\\t' = -x\sinh(w) + t\cosh(w)$$

By using (2) you can write that

$$ds^2 = -t^2+x^2 = -t'^2 + x'^2 \tag{3}$$

When you set $x=vt$ for $x'=0$ (and by using $\sinh(w) = \sqrt{\cosh^2(w)-1}$), you will see that $\cosh(w) = \gamma = \frac{1}{\sqrt{1-v^2}}$ and $\sinh(w) = v\gamma$

Note: Here $w$ is not an usual angle. Its an hyperbolic angle and called rapidity

Essentially, why does time shift as a function of space just as space shifts as a function of time?

The simple answer is to preserve the $ds$ term invariant under the coordinate transformation (i.e equation 3) which is a crucial point in SR. This coordinate transformation (2) is similar to the one which I did for equation (1). So the $vx$ term comes naturally, without it the transformations do not make sense.

Edit :

So as I said we need a transformation from $(x',t')$ to $(x,t)$ or vice versa. So assume a linear form of transformation such that

$$x' = ax + bt$$ $$t' = cx + dt$$

When you impose certain conditions such that

(i) $x = vt$ for $x' = 0$

(ii) $x' = -vt'$ for $x = 0$

(iii) $x = ct$ and $x'=ct'$ (Invariance of the speed of light)

You ll find that $a = \gamma$, $b = -v\gamma$ etc.

I am not sure how the $\cosh(w)$ and $\sinh(w)$ terms are derived but It's not hard to guess.

Again take the form of equations,

$$x' = ax + bt$$ $$t' = cx + dt$$

Square them

$$x'^2 = a^2x^2 + 2axbt + b^2t^2$$ $$t'^2 = c^2x^2 + 2cxdt + d^2t^2$$

So we know that,

$x'^2 - t'^2 = x^2-t^2$

This implies

$$[a^2x^2 + 2axbt + b^2t^2] - [c^2x^2 + 2cxdt + d^2t^2] = x^2 - t^2$$

so we have $a^2-c^2 = 1$ , $ab = cd$ and $b^2 - d^2 = 1$

We also know that $\cosh^2(w) - \sinh^2(w) = 1$ as you can see there's a similarity between these equations so we can set

$a = b = \cosh(w)$ and $c = d = \sinh(w)$ to satisfy the equations.

It may not seem certain, which is not, but you can see that the form must be somehow related to the $\cosh(w)$ and $\sinh(w)$.

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    $\begingroup$ Yes, this is the type of answer I was looking for. I was looking for an answer as to why the -(vx)/c^w factor exists and why it applies to entities that do not move at C from the origin in a well-written (i.e. somewhat easy to understand) way. Your answer accomplishes this. Thank you! $\endgroup$ – Sciencemaster Mar 19 at 21:16
  • $\begingroup$ How does the invariance of C imply that the sine in the bottom right corner of the rotation matrix becomes negative? Why is this all that it implies in terms of this matrix? Essentially, would you please elaborate on how using spacetime as a coordinate system and how using the invariance of C as an axiom changes the rotation matrix/function to create the 'new' matrix shown in figure 2 above that describes spacetime? $\endgroup$ – Sciencemaster Mar 29 at 20:13
  • $\begingroup$ @Sciencemaster Its not a rotation matrix as what we generally do in cartesian coordinate. Invarience of C determines the metric of the spacetime. I recently answered a question about it. Here is the link. physics.stackexchange.com/questions/540138/… $\endgroup$ – Layla Mar 30 at 9:03
  • $\begingroup$ @Sciencemaster I edited my post a bit. Hope that helps $\endgroup$ – Layla Mar 30 at 9:32
  • $\begingroup$ Thank you for writing the edits. At first glance they seem to be intuitive. but there are some things that I noticed and would like some clarification on. First of all, in the original post you showed how we can derive the cosh and sinh using x^2-t^2 and so on and so forth. However, as t^2-x^2 and so forth is also a valid equation that instead applies to the time coordinate of events (rather than applying to the space coordinate of events like your first equation). Is your sinh and cosh derivation still valid for t^2-x^2 and so forth? $\endgroup$ – Sciencemaster Apr 5 at 3:02

Let's say an alarm clock going off is an event.

That event shifts in space when the alarm clock is moved in space.

If the moving of the clock does not cause time dilation of the clock, then the event of the clock going off does not shift in time, when the clock is shifted in space.

Those observers that observe the clock to become more time dilated, or less time dilated, when it is being moved, agree that doubling the time that the clock is being moved doubles the amount that the event gets shifted in time because of it being shifted in space.

And doubling the time that the clock is being moved also doubles the amount that the clock/event is shifted in space.

I think I should explain what clock being moved means. Well, think about an alarm clock at the rear of a long spaceship. An astronaut slowly carries the clock to the front of the spaceship. That is what clock being moved means here. A slow change of x-coordinate in the coordinate system attached to the spaceship. An astronaut carrying the clock very slowly causes a very small amount of time dilation in the spaceship frame, but has large effect on the clock in some other frames.

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