Understanding Negative t Coordinates in Lorentz Transformations

[dpa]

In lorentz transformation,
why has t -sign

 

[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.

 

[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.

 

[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.

 

[bobc2]

In lorentz transformation,
why has t -sign

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.

 

[dpa]

thanks
matterwave and bobc2