So you disagree with the rest of my post there? I always thought that traveling at light-speed would make you see everything at once, so to say. Is this not the case? Because if it is so, that would be the only frame of reference to have them simultaneously, even if not very practical.

In essence, I indeed disagree. You are right though in saying that traveling at light-speed you would see everything at once, in a way. Frames of reference travelling at light-speed are not really practical, since they are degenerate - the x direction becomes equivalent with the t direction and there is no sensible flow of time. But yeah, you can say that time stops and everything happens at once - except that this "everything" is only one 3-dimensional slice, so pretty much like simultaneous events for other observers. The problem is, you never get beyond that one slice, because your time axis is

*in that slice* - so some events

*never* happen.

It's really hard to put into words, so let me use some math:

Let (t, x, y, z) be a coordinate system for some regular, massive observer (like our Mr. Violin).

Our observer O' will be travelling at v=c along the x axis.

Using the traditional Lorentz transformation will give us 0 in the denominator, so that's not really useful. If v<c, then we have:

[math]t' = \frac{t - \frac{v}{c^2}x}{\sqrt{1-\frac{v^2}{c^2}}}[/math][math]x' = \frac{x - vt}{\sqrt{1-\frac{v^2}{c^2}}}[/math][math]y' = y[/math][math]z' = z[/math]

We can safely ignore y and z, since nothing is happening there, and focus on 2D spacetime.

For any observer, simultaneous events are those for which t = const, where t is relative to that observer. This means that for the moving observer simultaneous events will have t'=const:

[math]t' = \frac{t - \frac{v}{c^2}x}{\sqrt{1-\frac{v^2}{c^2}}} = \mathrm{const}[/math]

If v is less than c, the denominator is a nonzero constant and we can write:

[math]ct - \frac{v}{c}x = \mathrm{const}[/math]

So, let's extend that to v=c:

[math]ct - x = \mathrm{const}[/math]

On the other hand, the world line of our observer is x' = 0:

[math]x' = \frac{x - vt}{\sqrt{1-\frac{v^2}{c^2}}} = 0[/math]

That means [math]x - vt = 0[/math]

Once again let's extend that to v=c:

[math]x - ct = 0[/math]

That is a line for which [math]ct - x = 0 = \mathrm{const}[/math] - so all events on the world line of the observer are simultaneous.

But! That means that on the whole world line we have t' = 0 (if we assume it was 0 at some point, which we usually do). On the other hand, the world line should somehow mark the time direction of the observer. This means that for such an observer it really makes no sense to talk about t' not equal to 0.

Now, what about simultaneity? Our observer is always in t' = 0, so we only need to talk about those events, which have t' = 0 - so, [math]ct - x = 0[/math]. Events that have [math]ct - x \neq 0[/math] are not simultaneous to any event on the world line of the observer, and so the observer will never perceive them as simultaneous to his "now".

Actually, t' and x' don't even make a good coordinate system. They are both 0 on [math]ct - x = 0[/math] and undefined anywhere else. It is possible to introduce so-called

*null coordinates*, which are good coordinates and one of them is along the world line of the observer, but then neither of them can be interpreted as time. Time just doesn't flow for an observer travelling at light-speed.

As an interesting fact, if we call such coordinates u and v, the metric (a kind of generalization for the Pythagorean theorem) becomes something like [math]\Delta s^2 = \Delta u \Delta v[/math] (as opposed to the usual [math]\Delta s^2 = c^2 \Delta t^2 - \Delta x^2[/math] in spacetime and [math]\Delta s^2 = \Delta x^2 + \Delta y^2[/math] on a regular plane).

I hope I managed to shed some light on the problem

If something is unclear, feel free to ask, I will try my best to explain.