Ok, I've seen the solution on Physics GRE.net, but I don't get it.

If I draw a spacetime diagram, the slope of a lightlike line is 1, and x=ct, right?

So to have a spacelike line, I need my slope to be bigger than 1; I need (ct)/x > 1. But this implies that (x/t)< c, which would be answer B, which is not correct!!!

What am I not getting here??

## 0177 #34

### Re: 0177 #34

u need slope < 1 for space like : ct < x

### Re: 0177 #34

Oh, I think I get it. If I just draw a spacetime diagram, the only way two events can occur simultaneously is when they are lying along the spatial axis, right? So I had it backwards. One doesn't even really need to compute anything to figure this out.

And the spacelike vectors are OUTSIDE the cone!!! That explains SOOO much. I can't believe I forgot all this.

And the spacelike vectors are OUTSIDE the cone!!! That explains SOOO much. I can't believe I forgot all this.