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

1. The problem statement, all variables and given/known data

Show that Ui $\frac{dU^i}{d\tau}$ = 0

2. Relevant equations

Raising Indices: Ui = gkiUi = Ui

where gk is a dummy index

3. The attempt at a solution

I'm interpreting this question to mean a scalar multiplying each component of a four vector = 0. Also, since the same index is used twice, once raised and once subscript, this implies summation.

So the sum of vector components multiplied by a scalar is 0? Since when?

I'd imagine I'm misunderstanding something, your help is much appreciated!

Thank you!
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 Recognitions: Homework Help Science Advisor d/dτ doesn't mean multiplication by a scalar. Why would you think that? And ##U## isn't just any vector, it's a four-velocity. What do you know about ##U_i U^i##?
 UiUi would imply a summation. U1*$\frac{dU^1}{d\tau}$+U2*$\frac{dU^2}{d\tau}$... Since Ui is defined as a four-vector in this example, and Ui is present (i once raised once subbed) I imagined that this expression implies a summation. Do I have everything completely wrong here? Thanks for the reply

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

Oh and I'm interpreting Ui to be some scalar since its rank is 1

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 Quote by Ryomega Oh and I'm interpreting Ui to be some scalar since its rank is 1
You aren't really thinking in the right direction. U isn't just any vector. It's the four velocity of something. Look at this http://en.wikipedia.org/wiki/Four-velocity Does any of that look familiar?

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 Quote by Ryomega I'm interpreting this question to mean a scalar multiplying each component of a four vector = 0. Also, since the same index is used twice, once raised and once subscript, this implies summation.
Those two sentences seem contradictory to me. Because...

 So the sum of vector components multiplied by a scalar is 0? Since when?
This sounds like you're saying that
$$a\left(\frac{\partial U^0}{\partial \tau} + \frac{\partial U^1}{\partial \tau} + \frac{\partial U^2}{\partial \tau} + \frac{\partial U^3}{\partial \tau}\right) = 0$$ where a is some number. But then you say
 Quote by Ryomega UiUi would imply a summation. U1*$\frac{dU^1}{d\tau}$+U2*$\frac{dU^2}{d\tau}$
which implies you mean
$$U_0\frac{\partial U^0}{\partial \tau} + U_1\frac{\partial U^1}{\partial \tau} + U_2\frac{\partial U^2}{\partial \tau} + U_3\frac{\partial U^3}{\partial \tau} = 0.$$
 Quote by Ryomega Oh and I'm interpreting Ui to be some scalar since its rank is 1
This suggests an issue with terminology — a scalar is rank 0 — but I'm still confused as to what you mean by your statements above.

http://mathworld.wolfram.com/TensorRank.html
 Yes the contents of the wiki page is what is being covered in class at the moment. I am working to make them very familiar, but as you can see, I have a few kinks I need to work out. Regarding the contradicting sentences, yes, there was a confusion with tensor ranks (thank you for point that out!) but what I meant to say is the latter equation you have posted. I believe the fundamental problem I'm having is not understanding the physical meaning and implication of Ui. My updated understanding of this equation then is: Summation of U_i multiplied by components of vector is 0 I hope I'm thinking in the right direction now. Thank you for replying!
 Recognitions: Homework Help Science Advisor The fact you need about the 4-velocity for this problem is that ##U_i U^i= -c^2##.
 I'm sorry, but I'm afraid I'll need a little bit more than that to go on. I see what you meant with myself not recognising a four-vector earlier. This is what's floating around in my head at the minute: I know that: UiUi = -c2 I also know that: Ui $\frac{dU^i}{d\tau}$ = 0 let since i with index 1,2,3 is 0 at rest frame: U0 $\frac{dU^0}{d\tau}$ = 0 U0 $\frac{dc\gamma\tau}{d\tau}$ = 0 U0 c$\gamma$ = not zero where exactly am I bringing in UiUi = -c2? Thank you!

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 Quote by Ryomega I'm sorry, but I'm afraid I'll need a little bit more than that to go on. I know that: UiUi = -c2
Yes.
 Quote by Ryomega I also know that: Ui $\frac{dU^i}{d\tau}$ = 0
Isn't what you are supposed to show?

What can you do to the equation in the first quote in order to get the equation in the second quote?
 Yes, I do have to prove that the latter equation you have mentioned is = 0. Getting from eq 1 to eq 2 would imply taking the $\frac{d}{d\tau}$ from eq 1. This would give me: $\frac{dU_i}{d\tau}$ $\frac{dU^i}{d\tau}$ = $\frac{dc}{d\tau}$ Letting i = 0, U0 = ct = c$\gamma\tau$ $\frac{dU_0}{d\tau}$ $\frac{c\gamma\tau}{d\tau}$ = $\frac{dc}{d\tau}$ since c = c$\gamma\tau$/t $\frac{dU_0}{d\tau}$ c$\gamma$ = c$\gamma$/t $\frac{dU_0}{d\tau}$ = 1/t and this brings me back to a problem of myself not being aware of what Ui actually means. Thank you for your response!

Mentor
 Quote by Ryomega Getting from eq 1 to eq 2 would imply taking the $\frac{d}{d\tau}$ from eq 1.
Yes.
 Quote by Ryomega This would give me: $\frac{dU_i}{d\tau}$ $\frac{dU^i}{d\tau}$ = $\frac{dc}{d\tau}$
No, on the left side, you haven't used the product rule for calculus correctly.
 After searching, I haven't been able to find product rule for tensor calculus (I did find dot product but has no examples with lower index multiplied by higher index. As previously mentioned I'm unaware of the meaning of U_i. I know that U_i and U^i are related by the metric. Is this the sort of area I am meant to search through? Would you mind pointing me in the right direction? Once again, thank you very much for your response.
 Recognitions: Gold Member What if we told you that what you are doing here is taking the contraction (scalar product) of the 4 velocity with the 4 acceleration? The 4 acceleration has contravariant components of dUi/dτ, and the 4 velocity has covariant components Uj. Let $$U_iU^i=\delta _i^j\delta_k^iU_jU^k=-c^2$$ Take the derivative of $\delta _i^j\delta_k^iU_jU^k$ with respect to proper time τ, recognizing that $\delta _i^j\delta_k^i$ commutes with d/dτ. This should enable you to get to where you want to be.
 oh dear kronicker delta function! I'll have to tackle this tomorrow, it's bed time. Thank you very much for helping me! I'll be back with...something

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 Quote by Chestermiller What if we told you that what you are doing here is taking the contraction (scalar product) of the 4 velocity with the 4 acceleration? The 4 acceleration has contravariant components of dUi/dτ, and the 4 velocity has covariant components Uj. Let $$U_iU^i=\delta _i^j\delta_k^iU_jU^k=-c^2$$ Take the derivative of $\delta _i^j\delta_k^iU_jU^k$ with respect to proper time τ, recognizing that $\delta _i^j\delta_k^i$ commutes with d/dτ. This should enable you to get to where you want to be.