Register to reply 
Direct and inverse Lorentz transformation 
Share this thread: 
#1
Sep507, 12:32 AM

P: 997

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 


#2
Sep507, 02:01 AM

Sci Advisor
P: 1,910

On our knowledge that v'=v. Everything else is just relabeling.



#3
Sep507, 07:16 AM

Emeritus
Sci Advisor
PF Gold
P: 16,091

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.



#4
Sep507, 09:13 AM

P: 997

Direct and inverse Lorentz transformation
Soft words and hard arguments 


#5
Sep507, 09:17 AM

Emeritus
Sci Advisor
PF Gold
P: 29,241

Zz. 


#6
Sep507, 03:17 PM

P: 16

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 :). 


#7
Sep507, 10:16 PM

P: 25

Its all relative, duh



#8
Sep507, 10:19 PM

P: 25

Thought experiment. In Star Trek, a 450 crew starship travelling 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?



#9
Sep607, 11:44 AM

P: 105




#10
Sep607, 02:39 PM

Sci Advisor
P: 8,470

[tex]x' = \gamma (x  vt)[/tex] [tex]y' = y[/tex] [tex]z' = z[/tex] [tex]t' = \gamma (t  vx/c^2)[/tex] with [tex]\gamma = 1/\sqrt{1  v^2/c^2}[/tex] Then you can just solve these equations for x, y, z, and t to get the inverse transformation. 


#11
Sep607, 09:53 PM

P: 997




#12
Sep607, 10:47 PM

Sci Advisor
P: 8,470




#13
Sep707, 02:13 AM

P: 997

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
Sep707, 10:15 AM

Sci Advisor
P: 8,470

[tex]x_{other} = \gamma (x_{mine}  V t_{mine})[/tex] [tex]t_{other} = \gamma (t_{mine}  V x_{mine} /c^2)[/tex] where V represents the velocity of the "other" coordinate system's origin along "my" xaxis. 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
Sep707, 10:23 AM

Sci Advisor
HW Helper
PF Gold
P: 4,139

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 boostwithspatialrotation. 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. 


Register to reply 
Related Discussions  
Direct/Inverse proportion  General Math  3  
How to get inverse Lorentz tranformation from direct Lorentz transformation  Special & General Relativity  13  
Inverse lorentz transformation question  Introductory Physics Homework  5  
Lorentz transformation and lorentzeinstein transformations  Special & General Relativity  1  
Direct, and Inverse Proportion; Invariants.  Precalculus Mathematics Homework  1 