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Derivation of VelocityAddition Formula 
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#1
Jul712, 03:42 PM

P: 148

There was an old thread but for some reason it is now closed...
So I'd have to restate this question: Why does any text book doesn't bother to derive the 3vectorvelocities addition formula using the more general 4vector formalism? In detail, the Lorentz Transformation: U[itex]^{1'}[/itex]=γ(v)(U[itex]^{1}[/itex]β(v)U[itex]^{0}[/itex])⇔γ(υ[itex]^{'}_{x}[/itex])υ[itex]^{'}_{x}[/itex]=γ(v)[γ(υ[itex]_{x}[/itex])(υ[itex]_{x}[/itex]cβ(v))] (where υ[itex]_{x}[/itex] the velocity in Σ,υ[itex]^{'}_{x}[/itex] the velocity mesured in Σ', and v the relative velocity between the two systems) solved for υ[itex]^{'}_{x}[/itex] should grant υ[itex]^{'}_{x}[/itex]=(υ[itex]_{x}[/itex]v)/(1vυ[itex]_{x}[/itex]/c[itex]^{2}[/itex]) right? 


#2
Jul812, 01:19 PM

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P: 1,911

However, you can also do it the other way around: You derive the transformation law for velocities using the Lorentz transformations, and then you derive the relativistic form for momentum from that. 


#3
Jul812, 05:58 PM

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PF Gold
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#4
Jul812, 08:28 PM

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P: 1,911

Derivation of VelocityAddition Formula



#5
Jul912, 03:01 PM

P: 148

At first glance the derivation seems straightforward but when put into practice I have some issues with the algebra:
[itex]\frac{u'}{\sqrt{1(u'/c)^{2}}}[/itex]=[itex]\frac{1}{\sqrt{1(v/c)^{2}}}[/itex][itex]\frac{1}{\sqrt{1(u/c)^{2}}}[/itex](u[itex]\frac{v}{c}[/itex]) The above solved for u' would never yield the velocityaddition formula What am I missing? 


#6
Jul912, 07:56 PM

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PF Gold
P: 5,027

[Edit: just worked it through  with the one correction, all the algebra does work out. This is your only mistake.] 


#7
Jul912, 08:50 PM

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Thanks ∞
PF Gold
P: 5,043

Chet 


#8
Jul1012, 12:12 AM

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PF Gold
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u' = (uv)/(1uv/c^2) into the equation and see that it is a solution. To be complete, you then need to provide arguments that it is the only solution. 


#9
Jul1012, 04:42 AM

P: 148

Thanks!
I see it now. cβ(v) is of course v and not v/c... Working SR in 3D problems has always had some troublesome algebra though... 


#10
Jul1012, 10:11 AM

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Thanks ∞
PF Gold
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Chet 


#11
Jul1012, 11:10 AM

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PF Gold
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