- #26

- 980

- 2

[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.

- Thread starter ehrenfest
- Start date

- #26

- 980

- 2

[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

You have: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.

[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

[tex]A_{\mu} \neq \eta_{\mu 0}A^0[/tex]

- #29

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I see now. Thanks.

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