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Direct and inverse Lorentz transformation

by bernhard.rothenstein
Tags: inverse, lorentz, transformation
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bernhard.rothenstein
#1
Sep5-07, 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
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Ich
#2
Sep5-07, 02:01 AM
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On our knowledge that v'=-v. Everything else is just relabeling.
Hurkyl
#3
Sep5-07, 07:16 AM
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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.

bernhard.rothenstein
#4
Sep5-07, 09:13 AM
P: 997
Direct and inverse Lorentz transformation

Quote Quote by Hurkyl View Post
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
ZapperZ
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Sep5-07, 09:17 AM
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Quote Quote by bernhard.rothenstein View Post
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.
Meithan
#6
Sep5-07, 03:17 PM
P: 16
Quote Quote by bernhard.rothenstein View Post
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 :).
Peter McKenna
#7
Sep5-07, 10:16 PM
P: 25
Its all relative, duh
Peter McKenna
#8
Sep5-07, 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?
RetardedBastard
#9
Sep6-07, 11:44 AM
P: 105
Quote Quote by Peter McKenna View Post
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 :)
JesseM
#10
Sep6-07, 02:39 PM
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Quote Quote by bernhard.rothenstein View Post
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:

[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.
bernhard.rothenstein
#11
Sep6-07, 09:53 PM
P: 997
Quote Quote by JesseM View Post
Just plain algebra. If you write the Lorentz equations like this:

[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.
Thanks. I.e. combining the first and the last equation I obtain the inverse one?
JesseM
#12
Sep6-07, 10:47 PM
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Quote Quote by bernhard.rothenstein View Post
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.
bernhard.rothenstein
#13
Sep7-07, 02:13 AM
P: 997
Quote Quote by JesseM View Post
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?
JesseM
#14
Sep7-07, 10:15 AM
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Quote Quote by bernhard.rothenstein View Post
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:

[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" 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.
robphy
#15
Sep7-07, 10:23 AM
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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.


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