General Relativity: Prove that Four-Vector is 0

Chestermiller

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

Ryomega

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!

vela

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.

Ryomega

Ok, I've resolved some issues with my understanding of index notations:

Uⁱ = [itex]\frac{dx^i}{d\tau}[/itex] [eq 1]

Such that writing out component form:

Uⁱ = [itex]\frac{dx^0}{d\tau}[/itex] + [itex]\frac{dx^1}{d\tau}[/itex] + [itex]\frac{dx^2}{d\tau}[/itex] + [itex]\frac{dx^3}{d\tau}[/itex]

Where U_1,2,3 would correspond to [x,y,z] or [r,[itex]\theta[/itex], [itex]\varphi[/itex]] or [i,j,k]

and

x⁰ = ct = c[itex]\gamma\tau[/itex]

If in rest frame:


U_iUⁱ = -c²


from [eq 1] this becomes:


U_i [itex]\frac{dx^i}{d\tau}[/itex] = -c²


taking [itex]\frac{d}{d\tau}[/itex]


U_i [itex]\frac{d^2x^i}{d\tau^2}[/itex] = -[itex]\frac{dc^2}{d\tau}[/itex]


In rest frame, it follows that Uⁱ with index i =1,2,3 are 0 so:


U₀ [itex]\frac{d^2c\gamma\tau}{d\tau^2}[/itex] = 0

U₀ 0 = 0


since U₀ is just an index:


0 = 0


I really hope I got that right. If not, *sigh*

vela

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}##?

Ryomega

Ah yes, I see where I went wrong on [eq 1]. I was coming from the convention that:

Vector U = U^ae_a

where e_a 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)


U^a = g^abU_b

or

U_a = g_abU^b


where g^ab and g_ab are metric and inverse to one another

vela

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.

Ryomega

Alright, it took me a while but here's what I got:

U_iUⁱ = 1

[itex]\frac{d}{d\tau}[/itex] (U_iUⁱ) = 0

chain rule:[itex]\frac{d}{dx}[/itex] f(x)g(x) = f(x)g'(x)+ f'(x) g(x)

U_i[itex]\frac{dU^i}{d\tau}[/itex] + [itex]\frac{dU_i}{d\tau}[/itex] Uⁱ = 0 [EQ1]

Raising and lowering indices:

U_i = g_kiU^k

Uⁱ = gⁿⁱU_n

Making these substitutions [EQ1] becomes:

g_kiU^k[itex]\frac{d}{d\tau}[/itex](gⁿⁱU_n) +[itex]\frac{d}{d\tau}[/itex] (g_kiU^k)gⁿⁱU_n = 0

since g has no dependence on tau, it can be taken out:

g_kigⁿⁱU^k[itex]\frac{d}{d\tau}[/itex](U_n) +g_kigⁿⁱU_n[itex]\frac{d}{d\tau}[/itex] (U^k) = 0

from definition product of g_ki gⁿⁱ is the kroecker delta function: [itex]\delta[/itex]ⁿ_k so that:

[itex]\delta[/itex]ⁿ_k U^k [itex]\frac{d}{d\tau}[/itex](U_n) + [itex]\delta[/itex]ⁿ_k U_n[itex]\frac{d}{d\tau}[/itex] (U^k) = 0

from definition of kroenicker delta function. Let n=k so that [itex]\delta[/itex]^k_k = 1

(1) U^k [itex]\frac{d}{d\tau}[/itex](U_k) + (1) U_k [itex]\frac{d}{d\tau}[/itex](U^k) = 0

let k = i

Uⁱ [itex]\frac{d}{d\tau}[/itex](U_i) + U_i [itex]\frac{d}{d\tau}[/itex](Uⁱ) = 0

So my question is...what next?

Sorry for taking so long, midterms and all.

Dick

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?

Ryomega

That would give me:

2U_i[itex]\frac{dU^i}{d\tau}[/itex] = 0

And that would be the proof. The identity U_i[itex]\frac{dU^i}{ds}[/itex] = 0
Similarly, identity U_i[itex]\frac{dU^i}{d\tau}[/itex] would be zero