# Why is t coordinate negative

1. Feb 14, 2012

### dpa

In lorentz transformation,
why has t -sign

2. Feb 14, 2012

### Matterwave

The sign doesn't go on the t but on the v. Depending on direction of your boost you can have either +vt or -vt.

3. Feb 14, 2012

### dpa

i mean for the metric,
s2=-t2+x2+y2+z2.

I beg your pardon if i did not get you. Would you mind to clarify.

4. Feb 14, 2012

### Matterwave

Oh, in the metric the sign is negative because you want a "length" which is preserved under the natural transformations of the space (Lorentz transformations in this case).

ds^2=dt^2+dx^2+dy^2+dz^2 would not be preserved under a Lorentz transformation. The version with the minus sign is. In other words, ds with the minus sign definition is something that all observers would agree on, but ds with the plus sign definition, different observers would measure different ds's.

5. Feb 14, 2012

### bobc2

Because it is the leg of a triangle that is being computed, not the hypotenuse. Google "special relativity space-time diagram" to study how to interpret the sketch below. We have red and blue guys moving in opposite directions at the same relativistic speed relative to the black coordinates. The blue guy uses the equation to compute the length of some object in the red guy's coordinates. Then, it's the red leg of the triangle that's being computed.

Given the way the coordinates are oriented for red and blue in special relativity, you can identify a right triangle--then just use Pythagorean theorem and solve for the red length. dX1'^2 +dX4^2 = dX1^2. Or, use vector addition as shown.

6. Feb 14, 2012

### dpa

thanks
matterwave and bobc2