Desk-Lamp
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OK, so I thought I solved the first part, but when I apply the same logic to the second question I don't get the second term, making me think I fluked the first part.

My thoughts were like this:

1. Starting with the hint,  \vec{V}(t) = \frac{d\vec{R}(t)}{dt}

2. Writing the position vector as components and covariant basis vectors:  \vec{V} = \frac{dZ^i\vec{Z_i}}{dt}

3. Expand with the product rule, and write Velocity as components:  V^j\vec{Z_j} = \frac{dZ^i}{dt}\vec{Z_i} + dZ^k\frac{\vec{Z_k}}{dt}

The question is, why does the second term on the right collapse to zero? I was thinking that \frac{\vec{Z_k}}{dt} = 0 because the basis vectors have no explicit time dependence, but maybe I can't do that because the basis depends on Z^k(t)?
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WishingChaff
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What book is this from?

(Original post by Desk-Lamp)
OK, so I thought I solved the first part, but when I apply the same logic to the second question I don't get the second term, making me think I fluked the first part.

My thoughts were like this:

1. Starting with the hint,  \vec{V}(t) = \frac{d\vec{R}(t)}{dt}

2. Writing the position vector as components and covariant basis vectors:  \vec{V} = \frac{dZ^i\vec{Z_i}}{dt}

3. Expand with the product rule, and write Velocity as components:  V^j\vec{Z_j} = \frac{dZ^i}{dt}\vec{Z_i} + dZ^k\frac{\vec{Z_k}}{dt}

The question is, why does the second term on the right collapse to zero? I was thinking that \frac{\vec{Z_k}}{dt} = 0 because the basis vectors have no explicit time dependence, but maybe I can't do that because the basis depends on Z^k(t)?
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Desk-Lamp
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Introduction to tensor analysis and the calculus of moving surfaces, Pavel Grinfeld
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WishingChaff
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For the first part you simply have to utilise the chain rule. Remember that you can write the position vector  \vec{R} as a function of the coordinates  Z^{i} .

Then, using the hint what is  \frac{d \vec{R}(Z)}{dt} ? (Remember the chain rule here)

The second part is a little trickier. You now need to differentiate the expression you should get from above. You may need to relabel some dummy indices here to make it work but it will come out. Hopefully this helps somewhat.



(Original post by Desk-Lamp)
Introduction to tensor analysis and the calculus of moving surfaces, Pavel Grinfeld
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Desk-Lamp
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So for part 1, \vec{V} = \frac{\partial\vec{R}}{\partial Z^i}\frac{\partial Z^i}{\partial t} = \frac{\partial Z^i}{\partial t}\vec{Z_i}

and V^i = \frac{dZ^i}{dt} (since the coordinates are only functions of time)
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WishingChaff
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After reading you initial post again I see you asked why the derivatives of the basis vectors are zero. The answer to this is essentially due to the fact that we are considering general co-ordinate systems. For example, if you compute the velocity vector in cylindrical co-ordinates you will find that the basis vectors are time dependent. This can be seen by thinking about a moving particle constrained to the surface of a cylinder. Anyway, I thought I will post the solution now so you can see how it is done.

For the first question we have:

 \frac{d \vec{R}(Z)}{dt} = \frac{d z^{i}}{dt} \frac{\partial \vec{R}}{\partial z^{i}} = V^{i} \frac{\partial \vec{R}}{\partial z^{i} } = \vec{V}

Comparing coefficients, you find that:

 V^{i} = \frac{ d z^{i} }{dt}

For the second part we have:

 \vec{A} =  A^{i} \frac{\partial \vec{R}}{\partial z^{i}} = \frac{d}{dt} \bigg{(} \frac{d z^{i}}{dt} \frac{\partial \vec{R}}{\partial z^{i}} \bigg{)}

 = \frac{d V^{i}}{dt} \frac{\partial \vec{R}}{\partial z^{i}} + V^{k} \frac{d}{dt} \bigg{(} \frac{\partial \vec{R}}{\partial z^{k}} \bigg{)}

 = \frac{dV^{i}}{dt} \frac{\partial \vec{R}}{\partial z^{i}} + V^{j} V^{k} \Gamma_{jk}^{i} \frac{\partial \vec{R}}{\partial z^{i}}

I have relabelled some indices at the end so things look a bit nicer.

(Original post by WishingChaff)
For the first part you simply have to utilise the chain rule. Remember that you can write the position vector  \vec{R} as a function of the coordinates  Z^{i} .

Then, using the hint what is  \frac{d \vec{R}(Z)}{dt} ? (Remember the chain rule here)

The second part is a little trickier. You now need to differentiate the expression you should get from above. You may need to relabel some dummy indices here to make it work but it will come out. Hopefully this helps somewhat.
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Desk-Lamp
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This is how I work it, following the same process as the first part:

\vec{A} = \frac{d\vec{V}}{dt}               =  \frac{\partial\vec{V}}{\partial Z^i} \frac{\partial Z^i}{\partial t} =  \frac{\partial V^j \vec{Z_j}}{\partial Z^i} V^i  = \frac{\partial V^j  }{\partial Z^i} \vec{Z_j} V^i + \frac{\partial \vec{Z_k}}{\partial Z^m}  V^m V^k

 = \frac{\partial V^j }{\partial Z^i} \frac{\partial Z^i}{\partial t} \vec{Z_j} + \Gamma_{mk}^l \vec{Z_l} V^k V^m

 = \frac{\partial V^j }{\partial t} \vec{Z_j} + \Gamma_{mk}^j V^k V^m \vec{Z_j}
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Desk-Lamp
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Oh, you beat me to it
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Desk-Lamp
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Just out of curiousity, was my original method wrong or is it possible to reduce Z^i \frac{d \vec{Z_i}}{dt} to zero?
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WishingChaff
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Nice answer. One of the more interesting examples of this is when we consider non-euclidean spaces. The use of the Christoffel symbols (or connection coefficients) is really useful in general relativity.

Anyway, good question and the book looks rather good as well. Cheers.

(Original post by Desk-Lamp)
This is how I work it, following the same process as the first part:

\vec{A} = \frac{d\vec{V}}{dt}               =  \frac{\partial\vec{V}}{\partial Z^i} \frac{\partial Z^i}{\partial t} =  \frac{\partial V^j \vec{Z_j}}{\partial Z^i} V^i  = \frac{\partial V^j  }{\partial Z^i} \vec{Z_j} V^i + \frac{\partial \vec{Z_k}}{\partial Z^m}  V^m V^k

 = \frac{\partial V^j }{\partial Z^i} \frac{\partial Z^i}{\partial t} \vec{Z_j} + \Gamma_{mk}^l \vec{Z_l} V^k V^m

 = \frac{\partial V^j }{\partial t} \vec{Z_j} + \Gamma_{mk}^j V^k V^m \vec{Z_j}
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WishingChaff
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I think it is as you cannot assume that the basis vectors are time independent. Pick cylindrical co-ordinates for example. These are indeed time dependent. That is actually a good exercise to see.

Take the position vector   \vec{R} = r \hat{r}

Then what is the velocity vector in a cylindrical co-ordinate basis?

to make it slightly simpler, consider a 2-dimensional plane with:

 x = r \cos(\theta)

and

 y = r \sin(\theta)

This will hopefully enlighten you as to why you cannot assume the basis vectors are time independent.

(Original post by Desk-Lamp)
Just out of curiousity, was my original method wrong or is it possible to reduce Z^i \frac{d \vec{Z_i}}{dt} to zero?
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Desk-Lamp
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Yeah, i can see that \frac{d \vec{Z_i}}{dt} isn't zero now, that was wishful thinking in the first place.

But following my original calculation, \frac{d \vec{Z_i}}{dt}Z^i has to be zero which doesn't seem right at all.
I just can't see where my mistake is.
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WishingChaff
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I think the problem arises as we need to pick a particular basis in which to expand the corresponding vectors. The question is ambiguous in this respect. This makes sense physically as the components of the velocity are different with respect to different basis vectors.
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Desk-Lamp
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Oh, I did something stupid .
When expanding the position vector R(t) the coefficients are NOT the same as the coordinates of the point, in general.
So my calculation was just completely wrong and I somehow fudged the right answer.

Anyway, thanks for the help, much appreciated.
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