G(,ab)= metric tensor in relativity
G(,ab)= Lambda(c ,a)lamdba(d,b)G(,cd)

It is given in my lecture notes that Gbar(,ab)=G(,ab)
I'm not too sure what to make of the Gbar, where normally a bared variable denotes a variable as observed in a different intertial frame.
It seems abit odd to me as barring G( a matrix quantity) seems to me like barring numerals like bar1 or bar2.

Anybody can help?

Tensor notation can sometimes be non-standard and confusing, but in this case I assume that your professor simply meant that the numerical value of the components of the metric are the same in both frames. In other words, the metric components always look like -1, 1, 1, 1 (or the opposite depending on your preference) in all frames connected by Lorentz transformations.

I cannot help much with tensor analysis, but I know it would help a moreknowledgable person if you used superscripts & subscripts.
G_ab = L^c_aL^d_bG_cdI think the above is the expression you intended. Perhaps I have the sub's & sup's reversed, and you meant the following.
G^ab = L_c^aL_d^bG^cdNote that there is a pattern to a correct tensor equation, making it obvious when a mistake has been made.

Thanks for the replies, I've slept on the problem and woke up realizing that the definition of G was inertial frame dependent. Ie the Matrix G is the matrix that returns a value G when product with L(the Standard Lorenzt in an inertial frame) as below:
LGL(transpose)=G

As L is the same in all frames, G will be the same in all frames.

Dino,

I tried copying and paste the notation with subscript, but that didnt work. Did not realize that there was a function to quote equations. Thanks for showing it to me though.

If you're into tensors, check out this somewhat old thread:

<A href="http://sciforums.com/showthread.php?t=47366">Tensors</A>

Nameless123: Try X(sub)2(/sub) & X(sup)3(/sup) using less than and greater than characters instead of left & right parentheses. You will get X₂ & X³