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## General Relativity: Prove that Four-Vector is 0

 Quote by Dick I don't think that's very useful. The point is that (assuming this is special relativity where the metric tensor, ##g_{ab}## is constant), that ##U^i U_i=g^{ab} U_a U_b=-c^2##. Take the ##\frac{d}{d \tau}## of both sides and use the product rule and the symmetry of ##g^{ab}##.
Yes. After further consideration, I agree. In my defense, I was assuming that this proof had to be done under the assumption that it is not SR. That makes the problem much more complicated, and requires you to prove that the covariant derivative of the metric tensor is zero. But, in any event, the approach I had recommended would not have worked. For SR, I too would have recommended the same approach that you described. Sorry for any confusion I might have caused.

Chet

 I'd like to take a moment while I'm struggling through this problem to thank you guys for helping and guiding me. Thank you so much!
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus Since the notation is apparently confusing you, start by expanding ##U_iU^i = -c^2## to get $$-U_0 U_0 + U_1 U_1 + U_2 U_2 + U_3 U_3 = -c^2.$$ Differentiate that, and then see how you could get the same result using the summation convention and without having to expand everything out.
 Ok, I've resolved some issues with my understanding of index notations: Ui = $\frac{dx^i}{d\tau}$ [eq 1] Such that writing out component form: Ui = $\frac{dx^0}{d\tau}$ + $\frac{dx^1}{d\tau}$ + $\frac{dx^2}{d\tau}$ + $\frac{dx^3}{d\tau}$ Where U1,2,3 would correspond to [x,y,z] or [r,$\theta$, $\varphi$] or [i,j,k] and x0 = ct = c$\gamma\tau$ If in rest frame: UiUi = -c2 from [eq 1] this becomes: Ui $\frac{dx^i}{d\tau}$ = -c2 taking $\frac{d}{d\tau}$ Ui $\frac{d^2x^i}{d\tau^2}$ = -$\frac{dc^2}{d\tau}$ In rest frame, it follows that Ui with index i =1,2,3 are 0 so: U0 $\frac{d^2c\gamma\tau}{d\tau^2}$ = 0 U0 0 = 0 since U0 is just an index: 0 = 0 I really hope I got that right. If not, *sigh*

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 Quote by Ryomega Ok, I've resolved some issues with my understanding of index notations: Ui = $\frac{dx^i}{d\tau}$ [eq 1] Such that writing out component form: Ui = $\frac{dx^0}{d\tau}$ + $\frac{dx^1}{d\tau}$ + $\frac{dx^2}{d\tau}$ + $\frac{dx^3}{d\tau}$
This isn't correct. There's no implied summation. Equation 1 is a compact way of writing the four equations
\begin{align*}
U^0 &= \frac{dx^0}{d\tau} \\
U^1 &= \frac{dx^1}{d\tau} \\
U^2 &= \frac{dx^2}{d\tau} \\
U^3 &= \frac{dx^3}{d\tau} \\
\end{align*} On the other hand, in the expression ##U_a U^a = -c^2##, the index a is repeated, so there is a summation so that ##U_a U^a = U_0 U^0 + U_1 U^1 + U_2 U^2 + U_3 U^3##.

Can you tell us how to lower an index? That is, how do you get ##U_a## from ##U^b## and the metric tensor ##\eta_{ab}##?

 Ah yes, I see where I went wrong on [eq 1]. I was coming from the convention that: Vector U = Uaea where ea can represent [i,j,k] [x,y,z] and in case of tangents, it can represent partial derivatives of correspondent components. I forgot to include that, now a sum is implied and [eq 1] would be correct. Yes, I do know how to lower and raise indices (just covered in class) Ua = gabUb or Ua = gabUb where gab and gab are metric and inverse to one another
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus OK, so you have ##U^aU^a = g_{ab}U^bU^a##. What do you get when you differentiate ##g_{ab}U^bU^a## with respect to ##\tau##? Just ignore the fact that it's being summed for now and apply the product rule.
 Alright, it took me a while but here's what I got: UiUi = 1 $\frac{d}{d\tau}$ (UiUi) = 0 chain rule:$\frac{d}{dx}$ f(x)g(x) = f(x)g'(x)+ f'(x) g(x) Ui$\frac{dU^i}{d\tau}$ + $\frac{dU_i}{d\tau}$ Ui = 0 [EQ1] Raising and lowering indices: Ui = gkiUk Ui = gniUn Making these substitutions [EQ1] becomes: gkiUk$\frac{d}{d\tau}$(gniUn) +$\frac{d}{d\tau}$ (gkiUk)gniUn = 0 since g has no dependence on tau, it can be taken out: gkigniUk$\frac{d}{d\tau}$(Un) +gkigniUn$\frac{d}{d\tau}$ (Uk) = 0 from definition product of gki gni is the kroecker delta function: $\delta$nk so that: $\delta$nk Uk $\frac{d}{d\tau}$(Un) + $\delta$nk Un$\frac{d}{d\tau}$ (Uk) = 0 from definition of kroenicker delta function. Let n=k so that $\delta$kk = 1 (1) Uk $\frac{d}{d\tau}$(Uk) + (1) Uk $\frac{d}{d\tau}$(Uk) = 0 let k = i Ui $\frac{d}{d\tau}$(Ui) + Ui $\frac{d}{d\tau}$(Ui) = 0 So my question is...what next? Sorry for taking so long, midterms and all.
 Recognitions: Homework Help Science Advisor Don't raise and lower both of them. Just do one, so you get ##U_i \frac{d U^i}{d \tau} + U_i \frac{d U^i}{d \tau} = 0##. Now what?
 That would give me: 2Ui$\frac{dU^i}{d\tau}$ = 0 And that would be the proof. The identity Ui$\frac{dU^i}{ds}$ = 0 Similarly, identity Ui$\frac{dU^i}{d\tau}$ would be zero

 Tags four-vector, general relativity, notation, prove that