# Slightly confused with Spacetime Diagrams

1. Jan 26, 2008

### Biest

Hello,

So I am supposed to draw and find the displacement vector between to points in spacetime, coordinates being (3,2,0,0) and (-2,1,0,0). I somehow feel slightly stupid for asking this, but I was reading the textbook and looking through my lecture notes. What I basically did that draw a standard 2D coordinate system wit ct and x as axis, add the two points draw the vector from the (-2,1,0,0) to (3,2,0,0) and find the distance using the line element formula in Minikowiski space:

$$\Delta S^2 = -(\Delta ct)^2 + \Delta x^2 + \Delta y^2 + \Delta z^2$$

Does that sound about right? I might be reading too much into this as the -2 for the ct component is throwing me off.

Cheers,

Biest

2. Jan 27, 2008

### pam

The sign of a componoent doesn't change anything.

3. Jan 27, 2008

### George Jones

Staff Emeritus
Are you worried about having negative time? t is just a coordinate, so having a negative value for t is no more problematic than having a negative value for the x coordinate, i.e., negative x doesn't mean negative space.

If today is chosen as the origin for the t coordinate than tomorrow has t = +1 day and yesterday has t = -1 day = -24 hours.

4. Jan 27, 2008

### Biest

Thanks, I was just confused cause one half of the textbooks just show positive t and x and the other half show a full coordinate axis.

Thanks