# 4-vectors scalar product

## Homework Statement

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

## 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

gabbagabbahey
Homework Helper
Gold Member
=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
Science Advisor
Homework Helper
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? Oh thanks you..its just me being stupid!

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

thanks