Common images suggest that Lorentz boosts can be interpreted as the coordinate transformation between two observers that have chosen the same basis and origin. However, I have read that one must be careful when talking about parallel axes: As far as I understand, the point is that given a Lorentz boost, the coordinate axes of one observer are not viewed as orthogonal from another observer. So does it even make sense to say that two observers have chosen the same basis?
Asked
Active
Viewed 174 times
1
-
1Lorentz transformations are linear transformations. So, [straight] lines are mapped to [straight] lines. (Maybe "straight" is the wrong word for the concept you are thinking about?) – robphy Aug 19 '22 at 20:17
-
Can you formulate the Euclidean analogue of your question for a rotation of 3D-space between two choices of rays from the origin? – robphy Aug 19 '22 at 21:26
-
1@filippo: all posts on SE sites are version controlled, as you can see the revision history of your post here. There is never any need for marking it as "edited" – Kyle Kanos Aug 20 '22 at 10:54
-
@robphy "Maybe "straight" is the wrong word for the concept you are thinking about?" - You were absolutely right. Thank you for pointing that out. As far as I understand, the point is that the axes of one observer are not viewed as orthgonal from another observer. – Filippo Aug 20 '22 at 11:12
-
@robphy I was unsure whether I should change the title and ask the question again (without the mistake) or correct the mistake/edit the question. I chose the second option, I hope thats okay for you. – Filippo Aug 20 '22 at 11:14
-
Whatever you do, I think it's important that changes are small enough that the answers already given to your original question still make sense. – robphy Aug 20 '22 at 11:18
1 Answers
1
I think the concern about "parallel axes" (like $x$ and $x'$) is that:
- while these "spatial [3-]vectors" are both "colloquially parallel [in space]" in the sense of pointing in the "directions of increasing x and x' (keeping y,y' and z,z' unchanged)",
- the corresponding spatial 4-vectors---call them $\tilde X$ and $\tilde X'$---(orthogonal to the inertial-observer 4-velocities $\tilde T$ and $\tilde T'$) are not-"parallel" in spacetime (just like, in an ordinary rotation of the $xy$-plane, the $y$- and $y'$-axes are not parallel)
I think the better spacetime-viewpoint way to describe the typical situation in a relativity boost problem (in (1+1)-spacetime) is to say that
- "the relative motion is coplanar with the inertial observer worldlines (with their y- and z-axes orthogonal to this plane of motion)"
- and then the
"projection of $\tilde X'$ onto $\tilde X$ in the $TX$-plane"
is equal to the "projection of $\tilde X$ onto $\tilde X'$ in the $T'X'$-plane"
robphy
- 11,748
-
Thank you for your answer. Unfortunately, I don't understand yet: It is not clear to me how $\tilde X$, $\tilde T$ and $TX$ are defined. – Filippo Aug 20 '22 at 12:05
-
In case that this is common knowledge in SR, could you please provide a reference? – Filippo Aug 20 '22 at 12:16
-
@Filippo T and T’ are 4-velocities, which are unit vectors on a Spacetime diagram along their time axes with tips on the unit hyperbola, corresponding to a unit of time (say 1 sec). X and X’ are their corresponding unit vectors (corresponding to their x-axes) which are perpendicular-in-spacetime to their time axes. X is parallel to the tangent to the hyperbola at the tip of T, and similarly for X’ and T’. See https://www.desmos.com/calculator/kv8szi3ic8 – robphy Aug 20 '22 at 12:51
-
TBH, I have no experience with spacetime diagrams, so I dont understand what is going on, but I guess that this is not the right place to discuss this. Anyways, I appreciate your efforts, so +1. – Filippo Aug 20 '22 at 13:11
-
@Filippo look at Spacetime physics at https://www.eftaylor.com/download.html . A Spacetime diagram is a real-world position-vs-time diagram. The PHY 101 position-vs-time diagram is an approximation. – robphy Aug 20 '22 at 13:18
-