[simonh]

d = √(x² + y² + z² - c²t²)

Note that if the space intervals are all zero and the time interval is unit time or 1, the spacetime displacement is equal to C. Thus when at rest physically we progress though the time dimension at the speed of light. Conversely if two points are separated by an interval equal to C, their distance in the time dimension is zero.

The latter result isn't really a surprise, we all know time doesn't pass if you're traveling at light speed, but IMHO it's interesting to see how it arises from the geometry.

[jamesmaniscalco]

> if two points are separated by an interval equal to C

If d = c, I don't think it determines the values of x, y, z, and t. What if x = 2c, y = z = 0, and t = 1? Unless I'm not following your logic correctly. In my recollection, for two given points in spacetime, d is invariant but the values of x, y, z, and t depend on the observer's frame of reference.

[sheerun]

So we don't see the sun as of 8 minutes ago, but as of right now?

[ben_w]

From the point of view of the photons, yes. From the point of view of everything else, no.

One of the bits of relativity that took me the longest to become aware of was that “right now” isn’t even meaningfully and universally defined within it.

[sheerun]

What about my point of view? He said that if my eyes and the sun are separated by X light seconds, then X seconds later sun and my eyes are at time distance of zero

[simonh]

This is for an observer following a trajectory in space time. So for an observer covering a distance at light speed the time interval is zero. For other observers on different trajectories, such as stationary in the space coordinates, that will not be the case.

[platz]

at rest relative to what?

[simonh]

Relative to the point of origin. Not accelerating relative to your inertial frame of reference.

[platz]

This seems odd because there clearly time dilation effects between two observers even when they are not accelerating but merely have a large difference in relative constant speed.

You seem to be saying that there has to be constant acceleration between observers for this effect to take place.

[simonh]

My apologies I really didn’t explain myself very well at all. That was quite confusing. This formula represents a trajectory in spacetime with respect to some frame of reference of an observer. As with any trajectory it has space and time components.

The x, y, z deltas are your displacement through space in those dimensions relative to some frame of reference (of an observer, presumably) and t is the time component. If the space deltas are all zero then you are at rest relative to that inertial frame. You are not accelerating or moving and your motion through the time dimension in unit time, according to this formula, is C. This is odd because C is normally thought of as a motion through space, but in this case your not moving through space (in the reference frame).

[platz]

I see now, thanks!