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Homework Help: General Relativity: Prove that Four-Vector is 0

  1. Feb 20, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that Ui [itex]\frac{dU^i}{d\tau}[/itex] = 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!
  2. jcsd
  3. Feb 20, 2013 #2


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    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##?
  4. Feb 20, 2013 #3
    UiUi would imply a summation.


    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
  5. Feb 20, 2013 #4
    Oh and I'm interpreting Ui to be some scalar since its rank is 1
  6. Feb 20, 2013 #5


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    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?
  7. Feb 20, 2013 #6


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    Those two sentences seem contradictory to me. Because...

    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
    which implies you mean
    $$U_0\frac{\partial U^0}{\partial \tau} + U_1\frac{\partial U^1}{\partial \tau} + U_2\frac{\partial U^2}{\partial \tau} + U_3\frac{\partial U^3}{\partial \tau} = 0.$$
    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.

  8. Feb 20, 2013 #7
    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!
    Last edited: Feb 20, 2013
  9. Feb 20, 2013 #8


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    The fact you need about the 4-velocity for this problem is that ##U_i U^i= -c^2##.
  10. Feb 20, 2013 #9
    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 [itex]\frac{dU^i}{d\tau}[/itex] = 0

    let since i with index 1,2,3 is 0 at rest frame:

    U0 [itex]\frac{dU^0}{d\tau}[/itex] = 0

    U0 [itex]\frac{dc\gamma\tau}{d\tau}[/itex] = 0

    U0 c[itex]\gamma[/itex] = not zero

    where exactly am I bringing in UiUi = -c2?

    Thank you!
  11. Feb 20, 2013 #10

    George Jones

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    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?
  12. Feb 20, 2013 #11
    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 [itex]\frac{d}{d\tau}[/itex] from eq 1.

    This would give me:

    [itex]\frac{dU_i}{d\tau}[/itex] [itex]\frac{dU^i}{d\tau}[/itex] = [itex]\frac{dc}{d\tau}[/itex]

    Letting i = 0, U0 = ct = c[itex]\gamma\tau[/itex]

    [itex]\frac{dU_0}{d\tau}[/itex] [itex]\frac{c\gamma\tau}{d\tau}[/itex] = [itex]\frac{dc}{d\tau}[/itex]

    since c = c[itex]\gamma\tau[/itex]/t

    [itex]\frac{dU_0}{d\tau}[/itex] c[itex]\gamma[/itex] = c[itex]\gamma[/itex]/t

    [itex]\frac{dU_0}{d\tau}[/itex] = 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!
  13. Feb 20, 2013 #12

    George Jones

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    No, on the left side, you haven't used the product rule for calculus correctly.
  14. Feb 20, 2013 #13
    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.
  15. Feb 20, 2013 #14
    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

    [tex]U_iU^i=\delta _i^j\delta_k^iU_jU^k=-c^2[/tex]

    Take the derivative of [itex]\delta _i^j\delta_k^iU_jU^k[/itex] with respect to proper time τ, recognizing that [itex]\delta _i^j\delta_k^i[/itex] commutes with d/dτ.

    This should enable you to get to where you want to be.
  16. Feb 20, 2013 #15
    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
  17. Feb 20, 2013 #16


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    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}##.
    Last edited: Feb 20, 2013
  18. Feb 21, 2013 #17
    This is a common problem I see on these forums. Basically, people get confused by notation, and think they don't know how to do things with it, just because it is new to them. You should know how to take derivatives of regular one variable functions, so see if you know a way to write this expression in terms of these one variable functions (hint, you've mentioned it several times in this thread).

    This is a basic tactic of physics (and mathematics) problem solving, see if you can make something unfamiliar into something you understand already. A basic tactic, but one that you basically just need experience to be able to do, so don't feel bad for not knowing to do it. Despite being familiar with tensor calculus already, I ended up doing exactly what I suggest above to check that the product rule makes sense.
  19. Feb 21, 2013 #18
    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.

  20. Feb 21, 2013 #19
    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!
  21. Feb 21, 2013 #20


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    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.
  22. Feb 21, 2013 #21
    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 U1,2,3 would correspond to [x,y,z] or [r,[itex]\theta[/itex], [itex]\varphi[/itex]] or [i,j,k]


    x0 = ct = c[itex]\gamma\tau[/itex]

    If in rest frame:

    UiUi = -c2

    from [eq 1] this becomes:

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

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

    Ui [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*
  23. Feb 21, 2013 #22


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    This isn't correct. There's no implied summation. Equation 1 is a compact way of writing the four equations
    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}##?
  24. Feb 21, 2013 #23
    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


    Ua = gabUb

    where gab and gab are metric and inverse to one another
  25. Feb 21, 2013 #24


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    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.
  26. Mar 14, 2013 #25
    Alright, it took me a while but here's what I got:

    UiUi = 1

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

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

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

    Raising and lowering indices:

    Ui = gkiUk

    Ui = gniUn

    Making these substitutions [EQ1] becomes:

    gkiUk[itex]\frac{d}{d\tau}[/itex](gniUn) +[itex]\frac{d}{d\tau}[/itex] (gkiUk)gniUn = 0

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

    gkigniUk[itex]\frac{d}{d\tau}[/itex](Un) +gkigniUn[itex]\frac{d}{d\tau}[/itex] (Uk) = 0

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

    [itex]\delta[/itex]nk Uk [itex]\frac{d}{d\tau}[/itex](Un) + [itex]\delta[/itex]nk Un[itex]\frac{d}{d\tau}[/itex] (Uk) = 0

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

    (1) Uk [itex]\frac{d}{d\tau}[/itex](Uk) + (1) Uk [itex]\frac{d}{d\tau}[/itex](Uk) = 0

    let k = i

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

    So my question is...what next?

    Sorry for taking so long, midterms and all.
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