Einstein notation

  • Thread starter ehrenfest
  • Start date
  • #26
980
2
Hmm. I see you really have yet to understand index mechanics.

[tex]x_\mu = \eta_{\mu\nu} x^\nu[/tex] does NOT mean that [tex]x_0 = \eta_{00}x^0[/tex].

Therefore, you've got twice the correct answer.
 
  • #27
918
16
Agreed. Applying your formula, I get

[tex] \eta_{00} \frac{d x ^ {0}}{d \tau}
+\eta_{01} \frac{d x ^ {1}}{d \tau}
[/tex]

[tex] = \frac{d x _ {0}}{d \tau}
+ \frac{d x _ {0}}{d \tau}
[/tex]

which is two times what you posted.
You have:
[tex]\eta_{\mu \nu}A^{\nu} = A_{\mu} + A_{\mu}[/tex]
This is not the same as what I had in my last post:
[tex]\eta_{\mu \nu}A^{\nu} = A_{\mu}[/tex]

To be more specific:
[tex]A_{\mu} = \eta_{\mu \nu}A^{\nu} = \Sigma_{\nu = 0}^{3}\eta_{\mu \nu}A^{\nu} = \eta_{\mu 0}A^{0} + \eta_{\mu 1}A^{1} + \eta_{\mu 2}A^{2} + \eta_{\mu 3}A^{3}[/tex]
 
Last edited:
  • #28
918
16
I think genneth has hit upon your problem. You should sum over greek letters, not over numbers:
[tex]A_{\mu} \neq \eta_{\mu 0}A^0[/tex]
 
  • #29
2,012
1
I see now. Thanks.
 

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