4-vectors scalar product

[captainjack2000]

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

 

[gabbagabbahey]

=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:

$$\gamma=(1-\beta^2)^{-0.5} \implies \gamma^2=\frac{1}{1-\beta^2}\neq1-\beta^2$$

 

[tiny-tim]

Hi captainjack2000!

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?

 

[captainjack2000]

Oh thanks you..its just me being stupid!

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

thanks