[edna314]

For those who wonder what the bottom line in the article is why time isn’t just another dimension: There is a minus sign and the speed of light in front of time in the space-time metric. It is argued that the minus sign comes about because time is imaginary (in complex number terms). I don’t find this convincing. One could still argue that time is just an ordinary dimension and the structure of the metric is just a property of how we measure distances.

[msla]

> There is a minus sign and the speed of light in front of time in the space-time metric.

Yes, that's called the Minkowski metric, and it's absolutely nothing new.

You can do the exact same physics with the opposite sign convention (called the signature of the metric), where time has a positive sign and all three spatial dimensions have negative signs; the only rule is that time has to be the odd one out, so rotations in a space-time plane obey hyperbolic geometry as opposed to Euclidean. You can get rid of the factor of c by moving to a different system of units, commonly called the natural units, where the speed of light is 1.

All of this is covered in any real introduction to Special Relativity, which, in turn, is at the beginning of any course on Modern Physics, as opposed to the Newtonian Physics.

Welcome to Spacetime:

https://www.av8n.com/physics/spacetime-welcome.htm

THE GEOMETRY OF SPECIAL RELATIVITY with the quote:

Lorentz transformations are just hyperbolic rotations.

http://sites.science.oregonstate.edu/~tevian/physics/paradig...

[sriku]

You're right in that the article doesn't quite "explain" it .. and it cannot be explained relative to normal newtonian "intuition". Working with Maxwell's equations to understand how electric fields can transform to magnetic fields and vice versa, working out the "wave equation" and seeing the speed of electromagnetic waves emerge from the equation independent of the reference frame was what nailed it for me.

[antman]

Let’s not forget that Maxwell equations are shaped as such due to the respective algebra. In clifford algebric spaces it is simply one equation. One idea is that not only it is important to conceptualize a complex phenomenon but to also whether we can formalize it in a simple way.

[sriku]

"simply one equation" that expands to the same set of vector equations which expands to the coordinate-system-specific equations.

∂F=J (barring constants) is a beautiful restatement, but I think it puts the cart before the horse to focus on it because that formulation is possible because of the invariances that hold and that comes from the raw Maxwell's equations .. at least historically.

The abstract formalisation is even harder to convey (at least for me, and so far) since it takes away the familiar "electricity" and "magnetism" and you need to think about the more complex F that combines both. One way perhaps is to start with circuits - which are discrete and circuit laws can be expressed with the same equation and then argue for the continuous case .. but speed of light invariance would still be a long way from that compared to the raw Maxwell's equations route.

Or maybe I misunderstood what you're suggesting.

[simiones]

> One could still argue that time is just an ordinary dimension and the structure of the metric is just a property of how we measure distances.

The math doesn't work unless time is different from the 3 spatial dimensions. In particular, distances in space-time can be negative, unlike distances in space.

[edna314]

Well, I'm just arguing that it isn't clear whether this prefactor (that shows up in the metric in front of time) is a property of time or a property of the metric. It is pretty clear that the math wouldn't work if the prefactor was different, though.

[simiones]

Given that there is a physical difference between events separated by positive and negative distances in spacetime, I think it's pretty clear that the prefactor is a property of spacetime.

[edna314]

Yes, it is mostly semantics but the question really is what you think the space-time metric is. In order to attribute the prefactor to time you would need to argue that the space-time metric really is Euclidean, but because time is different from space there is this negative prefactor showing up in the metric. But you can also claim that time and space are the same and the space time metric is hyperbolic. Since there is there is no good argument why metrics necessarily need to be Euclidean, I find the second option more convincing.

[dfilppi]

Implying that there is space separate from time denies spacetime

[simiones]

Not at all - spacetime is real, but it is not a 4-dimensional Cartesian coordinate system where all 4 dimensions are of the same kind. For example, for any two events that are time-like separated, any two observers will agree on the order in which they happen, whereas for two events that are space-like separated, there will be observers that see these two events happen in either order. Time is different from space, though they are related.

[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!