## direct and inverse Lorentz transformation

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
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 Recognitions: Science Advisor On our knowledge that v'=-v. Everything else is just relabeling.
 Recognitions: Gold Member Science Advisor Staff Emeritus 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.

## direct and inverse Lorentz transformation

 Quote by Hurkyl 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

Mentor
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 Quote by bernhard.rothenstein 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.

 Quote by bernhard.rothenstein 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 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?

 Quote by Peter McKenna 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?
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 :)

Recognitions:
 Quote by bernhard.rothenstein 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.

 Quote by JesseM 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?

Recognitions:
 Quote by bernhard.rothenstein 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.

 Quote by JesseM 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?

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