direct and inverse Lorentz transformationby bernhard.rothenstein Tags: inverse, lorentz, transformation 

#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: 1,883

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




#3
Sep507, 07:16 AM

Emeritus
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PF Gold
P: 16,101

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 transformationSoft words and hard arguments 



#5
Sep507, 09:17 AM

Mentor
P: 28,808

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

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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

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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

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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

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HW Helper
PF Gold
P: 4,108

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. 


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