General Relativity: Prove that FourVector is 0by Ryomega Tags: fourvector, general relativity, notation, prove that 

#1
Feb2013, 11:17 AM

P: 67

1. The problem statement, all variables and given/known data
Show that U_{i} [itex]\frac{dU^i}{d\tau}[/itex] = 0 2. Relevant equations Raising Indices: U_{i} = g^{ki}U_{i} = U^{i} where g^{k} 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
Feb2013, 12:01 PM

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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 fourvelocity. What do you know about ##U_i U^i##?




#3
Feb2013, 01:28 PM

P: 67

U_{i}U^{i} would imply a summation.
U_{1}*[itex]\frac{dU^1}{d\tau}[/itex]+U_{2}*[itex]\frac{dU^2}{d\tau}[/itex]... Since U^{i} is defined as a fourvector in this example, and U_{i} 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 



#4
Feb2013, 01:29 PM

P: 67

General Relativity: Prove that FourVector is 0
Oh and I'm interpreting U_{i} to be some scalar since its rank is 1




#5
Feb2013, 02:34 PM

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#6
Feb2013, 02:55 PM

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$$a\left(\frac{\partial U^0}{\partial \tau} + \frac{\partial U^1}{\partial \tau} + \frac{\partial U^2}{\partial \tau} + \frac{\partial U^3}{\partial \tau}\right) = 0$$ where a is some number. But then you say $$U_0\frac{\partial U^0}{\partial \tau} + U_1\frac{\partial U^1}{\partial \tau} + U_2\frac{\partial U^2}{\partial \tau} + U_3\frac{\partial U^3}{\partial \tau} = 0.$$ http://mathworld.wolfram.com/TensorRank.html 



#7
Feb2013, 03:22 PM

P: 67

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 U_{i}. 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! 



#8
Feb2013, 03:31 PM

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The fact you need about the 4velocity for this problem is that ##U_i U^i= c^2##.




#9
Feb2013, 04:10 PM

P: 67

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 fourvector earlier. This is what's floating around in my head at the minute: I know that: U_{i}U^{i} = c^{2} I also know that: U_{i} [itex]\frac{dU^i}{d\tau}[/itex] = 0 let since i with index 1,2,3 is 0 at rest frame: U_{0} [itex]\frac{dU^0}{d\tau}[/itex] = 0 U_{0} [itex]\frac{dc\gamma\tau}{d\tau}[/itex] = 0 U_{0} c[itex]\gamma[/itex] = not zero where exactly am I bringing in U_{i}U^{i} = c^{2}? Thank you! 



#10
Feb2013, 04:30 PM

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What can you do to the equation in the first quote in order to get the equation in the second quote? 



#11
Feb2013, 06:00 PM

P: 67

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, U^{0} = 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 U_{i} actually means. Thank you for your response! 



#12
Feb2013, 06:16 PM

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#13
Feb2013, 06:48 PM

P: 67

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. 



#14
Feb2013, 06:58 PM

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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 dU^{i}/dτ, and the 4 velocity has covariant components U_{j}. 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. 



#15
Feb2013, 07:13 PM

P: 67

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 



#16
Feb2013, 08:12 PM

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#17
Feb2113, 06:49 AM

P: 184

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. 



#18
Feb2113, 01:26 PM

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Chet 


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