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4-vectors scalar product

  • #1

Homework Statement


If a and b are 4-vectors give the definition of the scalar product a.b and demonstrate its Lorentz invariance


Homework Equations





The Attempt at a Solution


So (with 4-vectors double underlined!)
a.b = a0b0-a1b1-a2b2-a3b3

a' = (a0*gamma - beta*gamma a1 , -ao*beta*gamma + a1* gamma, a2 , a3)
b' = (bo*gamma - beta*gamma b1, -bo*beta*gamma + b1* gamma, b2 , b3)

a'.b' = two brackets above multiplied together
=(aobo*gamma^2+a1b1*beta^2*gamma^2 -aob1*beta*gamma^2 - a1bo*gamma^2*beta)-(aobo*beta^2*gamma^2 - aob1*beta*gamma^2 - a1bo*beta*gamma^2 +a1b1*gamma^2) -a2b2-a3b3

=a0b0(gamma^2-beta^2*gamma^2) + a1b1(beta^2*gamma^2-gamma^2) - a^2b^2-a^3b^3
substituting gamma=(1-beta^2)^-0.5 gives
=aobo(1-2beta^2+beta^4)+a1b1(-beta+2beta^2-beta^4)-a2b2-a3b3

which doesn't equal a.b.!!!
Could someone please tell where I am going wrong. I am really confused
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
6
=a0b0(gamma^2-beta^2*gamma^2) + a1b1(beta^2*gamma^2-gamma^2) - a^2b^2-a^3b^3
substituting gamma=(1-beta^2)^-0.5 gives
=aobo(1-2beta^2+beta^4)+a1b1(-beta+2beta^2-beta^4)-a2b2-a3b3
Hi CaptainJack,

Your error seems to be in making the above substitution... remember to take care of the negative sign in the exponent:

[tex]\gamma=(1-\beta^2)^{-0.5} \implies \gamma^2=\frac{1}{1-\beta^2}\neq1-\beta^2[/tex]

:wink:
 
  • #3
tiny-tim
Science Advisor
Homework Helper
25,832
249
Hi captainjack2000! :smile:

This is your second post that I have found virtually unreadable …

can you please try to use β and γ, and the X2 and X2 tags just above the Reply box? :smile:
 
  • #4
Oh thanks you..its just me being stupid!

Really sorry about that didn't notice them..will definitely use them next time!

thanks
 

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