Can Spacelike and Timelike Points Undergo the Same Transformations?

[LCSphysicist]

Homework Statement

"When (x−y)2<0 we can perform a Lorentz transformation taking (x−y)→−(x−y). Note that if (x−y)2>0 there is no continuous Lorentz transformation that takes (x−y)→−(x−y)"

Relevant Equations

The metric is majority -

I want to understand bettew what this statement says. Maybe later we could try to put it mathematically, but for while i want to know if my interpretation is right.

When we lie outside the light cone, the physics regarding the limit of the velocity is break, and technically we could go faster than light. So, if we want, we could perorm a transformation in x and y such that they got alterned sign instantaneally in a reference frame. So we can have a transformation that leads (x−y)→−(x−y) at the same time if the points are spacelike.

But, if they are timelike, we couldn't go so fast (it need to be avaliated at the same time), and we can't find a transformation that change the signs.

Is this the right interpretation? Let me know any error

 

[stevendaryl]

There is a discussion of this here:

https://physics.stackexchange.com/q...tum-field-theory-from-improper-lorentz-transf

The idea is that you want to perform rotations and Lorentz transformations to turn the vector

$(A^x,A^y,A^z,A^t)$

into

$(- A^x, - A^y, -A^z, -A^t)$
So the way you can do this is through a sequence of transformations:

1. Rotate about the z-axis to make $A^y \rightarrow 0$.
2. Rotate about the y-axis to make $A^z \rightarrow 0$.
3. Perform a boost (a regular Lorentz transformation) to make $A^t \rightarrow 0$.
4. Perform a rotation about the z-axis to make $A^x \rightarrow - A^x$
5. Undo the boost in 3 to turn $A^t$ into the negative of its original value.
6. Undo the rotation in 2 to turn $A^z$ into the negative of its original value.
7. Undo the rotation in 1 to turn $A^y$ into the negative of its original value.

It's sort of like solving the Rubiks cube. You perform a bunch of moves, then you switch two pieces, then you undo those moves to get back to where you started, but with two pieces switched.

The only step that is impossible for timeline vectors is step 3. If $(A^x)^2 + (A^y)^2 + (A^z)^2 \lt (A^t)^2$, then there is no Lorentz transformation that can make $A^t \rightarrow 0$.