Determining 4vector character of a 4tupleby jmcvirgo Tags: 4tuple, 4vector, character, determining 

#1
Feb1810, 06:05 AM

P: 2

Suppose you're given a 4 tuple and told that its scalar product with any 4vector is a lorentz scalar. How do I show that this implies the 4tuple is a 4vector?
Thanks 



#2
Feb1810, 07:16 AM

Emeritus
Sci Advisor
PF Gold
P: 8,987

It can't be done unless you're told that its scalar product with any 4vector is a scalar.
If the given 4tuple is x and the (arbitrary) 4vector is y, [tex]x_\mu y^\mu=x'_\mu y'^\nu=x'_\mu\Lambda^\mu{}_\nu y^\nu[/tex] [tex]x_\mu=\Lambda^\mu{}_\nu x'_\mu[/tex] Now do some raising and lowering of indices and apply a Lorentz transformation to solve for x', and you're done. This post should help with the notation. 



#3
Feb1810, 07:23 AM

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HW Helper
P: 11,863

Just apply the definitions on the <scalar> product. Denoting by F the matrix the 4tuple (index down) uses to transform under a Lorentz group element, you'll end with a matrix equation
[tex] \mathbb{F} \Lambda = \mbox{1}_{4\times 4} [/tex]. Since [itex] \Lambda [/itex] is invertible, the conclusion follows easily. 



#4
Feb1810, 08:49 AM

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Sci Advisor
PF Gold
P: 8,987

Determining 4vector character of a 4tuple
Here's how I would do it in matrix notation:
[tex]x^T\eta y=x'^T\eta y'=x'^T\eta\Lambda y[/tex] [tex]x^T\eta=x'^T\eta\Lambda[/tex] [tex]\eta x=\Lambda^T\eta x'[/tex] [tex]x'=\eta^{1}(\Lambda^T)^{1}\eta x=\eta^{1}(\eta\Lambda\eta^{1})\eta x=\Lambda x[/tex] The fact that [tex](\Lambda^T)^{1}=\eta\Lambda\eta^{1}[/tex] follows from the definition of a Lorentz transformation, [tex]\Lambda^T\eta\Lambda=\eta[/tex]. Just multiply both sides with [tex]\eta^{1}[/tex] from the right. 



#5
Feb1810, 11:41 AM

P: 2




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