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General Relativity: Prove that Four-Vector is 0

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Ryomega
#1
Feb20-13, 11:17 AM
P: 67
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!
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Dick
#2
Feb20-13, 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 four-velocity. What do you know about ##U_i U^i##?
Ryomega
#3
Feb20-13, 01:28 PM
P: 67
UiUi would imply a summation.

U1*[itex]\frac{dU^1}{d\tau}[/itex]+U2*[itex]\frac{dU^2}{d\tau}[/itex]...

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

Ryomega
#4
Feb20-13, 01:29 PM
P: 67
General Relativity: Prove that Four-Vector is 0

Oh and I'm interpreting Ui to be some scalar since its rank is 1
Dick
#5
Feb20-13, 02:34 PM
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Quote Quote by Ryomega View Post
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?
vela
#6
Feb20-13, 02:55 PM
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Quote Quote by Ryomega View Post
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 Quote by Ryomega View Post
UiUi would imply a summation.

U1*[itex]\frac{dU^1}{d\tau}[/itex]+U2*[itex]\frac{dU^2}{d\tau}[/itex]
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 Quote by Ryomega View Post
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
Ryomega
#7
Feb20-13, 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 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!
Dick
#8
Feb20-13, 03:31 PM
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The fact you need about the 4-velocity for this problem is that ##U_i U^i= -c^2##.
Ryomega
#9
Feb20-13, 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 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!
George Jones
#10
Feb20-13, 04:30 PM
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Quote Quote by Ryomega View Post
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 Quote by Ryomega View Post
I also know that:

Ui [itex]\frac{dU^i}{d\tau}[/itex] = 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?
Ryomega
#11
Feb20-13, 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, 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!
George Jones
#12
Feb20-13, 06:16 PM
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Quote Quote by Ryomega View Post
Getting from eq 1 to eq 2 would imply taking the [itex]\frac{d}{d\tau}[/itex] from eq 1.
Yes.
Quote Quote by Ryomega View Post
This would give me:

[itex]\frac{dU_i}{d\tau}[/itex] [itex]\frac{dU^i}{d\tau}[/itex] = [itex]\frac{dc}{d\tau}[/itex]
No, on the left side, you haven't used the product rule for calculus correctly.
Ryomega
#13
Feb20-13, 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.
Chestermiller
#14
Feb20-13, 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 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.
Ryomega
#15
Feb20-13, 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
Dick
#16
Feb20-13, 08:12 PM
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Quote Quote by Chestermiller View Post
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.
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}##.
DimReg
#17
Feb21-13, 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.
Chestermiller
#18
Feb21-13, 01:26 PM
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Quote Quote by Dick View Post
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


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