How to Prove ##\bar{h}=-h## for the Given Tensor?

[nikhilb1997]

From the tensor, $\bar{h}^{ij}=h^{ij}-1/2\eta^{ij}h$
Where, h=$h^i_i$,
Prove that $\bar{h}=-h$,
Where, $\bar{h}=\bar{h}^i_i$

 

[TSny]

Hello, and welcome to PF!

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Your Latex formating will work if you use ## instead of $.

From the tensor, $\bar{h}^{ij}=h^{ij}-1/2\eta^{ij}h$
Where, $h=h^i_i$ ,
Prove that $\bar{h}=-h$,
Where, $\bar{h}=\bar{h}^i_i$

 

[andrien]

just contract both sides with η_ij.

 

[nikhilb1997]

just contract both sides with η_ij.

I tried it but I guess I made a mistake since I got h(bar) on the left side and on the right side I got h-1/2h. Please help.

 

[andrien]

I tried it but I guess I made a mistake since I got h(bar) on the left side and on the right side I got h-1/2h. Please help.

No,you get on the right side h-(1/2)(4)h=h-2h=-h,can you verify it?

 

[nikhilb1997]

No,you get on the right side h-(1/2)(4)h=h-2h=-h,can you verify it?

I had to find the absolute value of the $n_{ij}$ matrix. Is this is where I went wrong. Thank you very much

 

[andrien]

I had to find the absolute value of the $n_{ij}$ matrix. Is this is where I went wrong. Thank you very much

Not necessarily,after contraction you get
η_ijη^ij=η₀₀η⁰⁰+η₁₁η¹¹
+η₂₂η²²+η₃₃η³³,
all others are zero,right.now this is =(1*1)+(-1*-1)+(-1*-1)+(-1*-1)=4,it does not depend on your signature.you can take as well as(-,+,+,+).

 

[nikhilb1997]

Thanks a lot andrien.