Direct and inverse Lorentz transformation

Click For Summary

Discussion Overview

The discussion revolves around the direct and inverse Lorentz transformations, focusing on the mathematical relationships and conceptual underpinnings of these transformations in the context of special relativity. Participants explore the implications of changing physical quantities and the sign of relative velocity, as well as the broader principles of reciprocity and symmetry in inertial frames.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the inverse Lorentz transformation can be derived by changing primed quantities to unprimed ones and altering the sign of the relative velocity, based on the equivalence of all inertial frames.
  • Others argue that the relationship v' = -v is a key aspect of understanding the transformations, emphasizing that the process is largely a matter of relabeling.
  • A participant suggests a method to verify that Lorentz boosts with velocities v and -v are inverse operations by applying both transformations and checking for the identity transformation.
  • Concerns are raised about the use of the term "mindless" in the context of verifying transformations, with some participants expressing that it may not be appropriate in a collaborative knowledge-sharing environment.
  • Some participants discuss the algebraic derivation of the inverse transformation, indicating that it can be achieved by manipulating the Lorentz equations directly.
  • A later reply emphasizes the importance of understanding reciprocity and symmetry in the context of the transformations, noting that both observers in different frames should arrive at the same transformation equations despite using opposite signs for velocity.
  • Strictly speaking, it is noted that the discussion pertains to pure boosts, and that spatial rotations complicate the relationship between direct and inverse transformations.

Areas of Agreement / Disagreement

Participants express a range of views on the derivation and interpretation of the Lorentz transformations, with no clear consensus on the best approach or terminology. Disagreements arise particularly around the appropriateness of certain language and the implications of the transformations in broader contexts.

Contextual Notes

Some participants highlight that the discussion is limited to pure boosts and does not fully address cases involving spatial rotations, where simply flipping the sign of the relative velocity may not suffice to obtain the inverse transformation.

bernhard.rothenstein
Messages
988
Reaction score
1
When we state that:
Knowing the direct Lorentz transformation we obtain the inverse one by changing the corresponding primed physical quantities with unpriomed ones and changing the sign of the relative velocity, we are based on what?
Thanks
 
Physics news on Phys.org
On our knowledge that v'=-v. Everything else is just relabeling.
 
If you want to verify that a Lorentz boost with velocity v and a Lorentz boost with velocity -v are inverse operations, there is always the mindless approach: apply both and see if you get the identity transformation.
 
mindless?

Hurkyl said:
If you want to verify that a Lorentz boost with velocity v and a Lorentz boost with velocity -v are inverse operations, there is always the mindless approach: apply both and see if you get the identity transformation.
English is not my first language, but I think, after looking for the meaning of "mindless" , that people who are ready to share knowledge will never use such a word!
Soft words and hard arguments
 
bernhard.rothenstein said:
English is not my first language, but I think, after looking for the meaning of "mindless" , that people who are ready to share knowledge will never use such a word!
Soft words and hard arguments

You need to consider that you have misinterpreted the context of the usage of that word in what Hurkyl's wrote. Nothing in there was disparaging.

Zz.
 
bernhard.rothenstein said:
When we state that:
Knowing the direct Lorentz transformation we obtain the inverse one by changing the corresponding primed physical quantities with unpriomed ones and changing the sign of the relative velocity, we are based on what?
Thanks

In my opinion it's as Ich said. It simply comes from the fact that all inertial frames are equivalent. Hence, it doesn't matter which one we call S and which S', the Lorentz transformation should look the same. Changing all primed quantities with unprimed ones (or viceversa) is valid simply because we're free to label each frame however we want.

About the sign of the velocity: you technically don't need to change it if you just stick to the rule of "changing all primed quantities with unprimed ones". It's just that it turns out that v' = -v, so you might as well not change the velocity and flip its sign, it's all the same.

I hope it made sense :).
 
Its all relative, duh
 
Thought experiment. In Star Trek, a 450 crew starship traveling at the speed of light or greater has to stop on a dime so Captain Kirk can see his girlfriend. How far in advance would they have to make that decision?
 
Peter McKenna said:
Thought experiment. In Star Trek, a 450 crew starship traveling at the speed of light or greater has to stop on a dime so Captain Kirk can see his girlfriend. How far in advance would they have to make that decision?

Well, assuming the length of a dime is 2.0136296 10 ^ -18 lightyears, and the ship has instantaneous acceleration (which it doesn't but for the sake of simplicity), they don't have plan in advance at all :)
 
Last edited:
  • #10
bernhard.rothenstein said:
When we state that:
Knowing the direct Lorentz transformation we obtain the inverse one by changing the corresponding primed physical quantities with unpriomed ones and changing the sign of the relative velocity, we are based on what?
Thanks
Just plain algebra. If you write the Lorentz equations like this:

x' = \gamma (x - vt)
y' = y
z' = z
t' = \gamma (t - vx/c^2)
with \gamma = 1/\sqrt{1 - v^2/c^2}

Then you can just solve these equations for x, y, z, and t to get the inverse transformation.
 
  • #11
direct and inverse LT

JesseM said:
Just plain algebra. If you write the Lorentz equations like this:

x' = \gamma (x - vt)
y' = y
z' = z
t' = \gamma (t - vx/c^2)
with \gamma = 1/\sqrt{1 - v^2/c^2}

Then you can just solve these equations for x, y, z, and t to get the inverse transformation.
Thanks. I.e. combining the first and the last equation I obtain the inverse one?
 
  • #12
bernhard.rothenstein said:
Thanks. I.e. combining the first and the last equation I obtain the inverse one?
Right, combining the x' equation and the t' equation will give you the equation for x and t.
 
  • #13
inverse and direct

JesseM said:
Right, combining the x' equation and the t' equation will give you the equation for x and t.
Tanks again. My problem is which are the roots of that fact: reciprocity, symmetry...
When I want to save the derivation of the inverse transformation stating the rule in my thread what should I say more to the learner?
 
  • #14
bernhard.rothenstein said:
Tanks again. My problem is which are the roots of that fact: reciprocity, symmetry...
When I want to save the derivation of the inverse transformation stating the rule in my thread what should I say more to the learner?
Well, the fact that the equations come out looking identical except for the v being replaced by -v can be explained in terms of reciprocity, the laws of physics should be the same in every inertial frame so if two observers have a set of physical rulers and clocks synchronized by the Einstein synchronization convention which they use to define their respective coordinate systems, they'd better get the same set of equations for transforming their coordinates to the other coordinate system. And don't be confused by the fact that one uses v and the other uses -v, in fact they are both using exactly the same general transformation, it's just that the velocity of the (x',t') system as seen in the (x,t) system is opposite to the velocity of the (x,t) system as seen in the (x',t') system...you can think of both of them using this general transformation:

x_{other} = \gamma (x_{mine} - V t_{mine})
t_{other} = \gamma (t_{mine} - V x_{mine} /c^2)

where V represents the velocity of the "other" coordinate system's origin along "my" x-axis. If "my" coordinate system is the (x,t) system and the "other" is (x',t') then V=v, while if "my" coordinate system is the (x',t') system and the "other" is (x,t) then V=-v.
 
  • #15
Strictly speaking, as Hurkyl mentioned, this discussion only applies to a pure boost.
A spatial rotation is technically a Lorentz Transformation, and so is a boost-with-spatial-rotation.
So, in these last cases, flipping the sign of the relative velocity "V" is not sufficient to obtain the inverse.

As I often suggest for questions like the OP's question, one could study the Euclidean analogue of the question.
 

Similar threads

Undergrad Follow-up on Index notation for inverse Lorentz transform
  • · Replies 1 ·
Replies
1
Views
1K
Graduate Show Lagrangian is invariant under a Lorentz transformation without using generators
  • · Replies 3 ·
Replies
3
Views
1K
Undergrad Lorentz transformation on mode functions
  • · Replies 7 ·
Replies
7
Views
2K
Undergrad Assumption in the derivation of the Lorentz transformation
  • · Replies 33 ·
2
Replies
33
Views
3K
Undergrad How can Lorentz transforms be inverted?
  • · Replies 13 ·
Replies
13
Views
2K
Undergrad Lorentz Transformation: Premises + Derivation
  • · Replies 5 ·
Replies
5
Views
2K
High School Confusion in notation of Lorentz Transformations
  • · Replies 6 ·
Replies
6
Views
2K
Undergrad Coord Transform in de Sitter Space: Phys Significance &Linearity?
  • · Replies 5 ·
Replies
5
Views
2K
Undergrad Gravitational Field Transformations Under Boosted Velocity
  • · Replies 14 ·
Replies
14
Views
3K
Undergrad Derive the Lorentz transformation in Minkowski four-dimensional spacetime spacetime
  • · Replies 8 ·
Replies
8
Views
1K