Transform Bases for 4-Vectors in Ref. Frames

[Silviu]

Hello! Why do we need to impose a change on the basis vector, when going from a reference frame to another. I understand that the components of the vector and the basis change using inverse matrices (the components use a matrix and the vector basis the inverse). But the transformation condition we impose on the components of the 4-vector, already impoose that the vector has the same length in any frame. So why do we need to make sure the vector looks the same in all frames (i.e. why we bother to apply any transformation to the basis, if we are able to keep the length constant anyway)?

 

[stevendaryl]

Hello! Why do we need to impose a change on the basis vector, when going from a reference frame to another. I understand that the components of the vector and the basis change using inverse matrices (the components use a matrix and the vector basis the inverse). But the transformation condition we impose on the components of the 4-vector, already impoose that the vector has the same length in any frame. So why do we need to make sure the vector looks the same in all frames (i.e. why we bother to apply any transformation to the basis, if we are able to keep the length constant anyway)?

Well, do you understand how it works in the case of two-dimensional plane geometry?

Take a piece of paper, and draw two points on it, and call them A and B. There are many ways to get from A to B. In the picture below, we show two of them:

• the red path: Go a distance 60 units in direction $\hat{x}$ and then go a distance 25 units in direction $\hat{y}$
• the blue path: Go a distance 39 units in direction $\hat{x'}$ and then go a distance 52 units in direction $\hat{y'}$

So the vector from A to B can be described by the coordinates $(60, 25)$ in the coordinate system $(x,y)$ or it can be described by the coordinates $(39, 52)$ in the coordinate system $(x', y')$. The lengths of the vector is the same in either coordinate system: 65 units. But the basis vectors have changed: from $\hat{x}, \hat{y}$ to $\hat{x'}, \hat{y'}$.