How to prove that relative speed between a pair of Lorentz transformations is < c ?

1. Jun 11, 2012

swampwiz

[ T'' ] = [ T ] [ T' ]

[ T'' ] =

[ ( γ'' ) , ( - γ'' β'' ) ]

[ ( - γ'' β'' ) , ( γ'' ) ]

γ'' = γ γ' ( 1 + β β' ) → no problem, proven to be b/w [ 1 , ∞ )

β'' = ( β + β' ) / ( 1 + β β' ) → can't figure out how to prove it's b/w [ 0 , 1 )

2. Jun 11, 2012

Muphrid

Re: How to prove that relative speed between a pair of Lorentz transformations is < c

Each of the individual betas ought to be less than 1, right?

3. Jun 11, 2012

swampwiz

Re: How to prove that relative speed between a pair of Lorentz transformations is < c

Yes, for each β:

0 ≤ β < 1

4. Jun 11, 2012

Muphrid

Re: How to prove that relative speed between a pair of Lorentz transformations is < c

So in the limiting case of each beta going to 1, what do you get for the final result?

5. Jun 12, 2012

Staff: Mentor

Re: How to prove that relative speed between a pair of Lorentz transformations is < c

Hint: Assuming both β and β' are in [0,1), what happens to β'' if β (or β') increases?

Can you combine both to a formal proof?

6. Jun 12, 2012

Whovian

Re: How to prove that relative speed between a pair of Lorentz transformations is < c

Perhaps yet another way to go from here is to note that that looks awfully like the tangent addition formula ... and then making a substitution.

7. Jun 12, 2012

swampwiz

Re: How to prove that relative speed between a pair of Lorentz transformations is < c

I've figured it out.

( β + β' ) / ( 1 + β β' ) - 1 + 1

[ ( β + β' ) - ( 1 + β β' ) ] / ( 1 + β β' ) + 1

[ - ( 1 - β ) ( 1 - β' ) ] / ( 1 + β β' ) + 1

1 ≤ ( 1 + β β' ) < 2

0 < ( 1 - β ) ≤ 1

0 < ( 1 - β ) / ( 1 + β β' ) ≤ 1

0 < ( 1 - β ) ( 1 - β' ) / ( 1 + β β' ) ≤ 1

-1 ≤ - ( 1 - β ) ( 1 - β' ) / ( 1 + β β' ) < 0

0 ≤ - ( 1 - β ) ( 1 - β' ) / ( 1 + β β' ) + 1 < 1