# Thread: Vector as a tensor.

1. ## Vector as a tensor.

Sorry if this is a stupid question but I am a little confused and would like to clear something up. It has been a while since I was formally educated.

A scalar is a tensor rank of 0.
A vector is a tensor rank 0f 1. This can be represented as a one dimensional array. That is the part I am confused about...

Why is it represented as a one dimensional array?
Lets take force for example:
Does the one dimensional array mean that for all directions there is a net force in that directions? Or why is it a one dimensional array?  Reply With Quote

2. Originally Posted by tommac Why is [a vector] represented as a one dimensional array?
A vector, has a starting point in n-dimensional space and an ending point. It can be mapped so that the starting point is at the origin and the ending point is (p1, p2, p3, ... pn).

Voila. A one-dimensional array.

Wikipedia: Vector (spatial)

Representation of a vector
[...]
In three dimensional Euclidean space (or R3), vectors are identified with triples of numbers corresponding to the Cartesian coordinates of the endpoint (a,b,c). These numbers are often arranged into a column vector or row vector, particularly when dealing with matrices [...]
(Note there the use of "column vector". The relationship of vectors to one-dimensional arrays is so pervasive that a one-dimensional array is often called a "vector".)  Reply With Quote

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For comparison, a 2-dimensional array would be a matrix, that is something like this:

a11 a12 ... a1n
a21 a22 ... a2n
...
am1 am2 ... amn  Reply With Quote

4. Originally Posted by 01101001 A vector, has a starting point in n-dimensional space and an ending point. It can be mapped so that the starting point is at the origin and the ending point is (p1, p2, p3, ... pn).

Voila. A one-dimensional array.

Wikipedia: Vector (spatial)
Oh .. OK ... so ending point meaning

starting point + the vector = end point?

Sorry still dont get it. Why so many end points? One for each dimension?  Reply With Quote

5. Originally Posted by tommac Why so many end points?
It's an ordered collection of n Cartesian coordinate values indicating a single point in n-space.

Like (5, 20, -10) indicates a single point in 3-space.  Reply With Quote

6. Originally Posted by tommac Oh .. OK ... so ending point meaning

starting point + the vector = end point?

Sorry still dont get it. Why so many end points? One for each dimension?
The end point establishes a one-dimensional length. Thus, this single line segment offers both direction and length. The length represents the magnitude of whatever it is you wish to represent, e.g. force. This math method is often superior to other math methods.  Reply With Quote

7. We gotta be careful here about mixing up the terminology. This tensor wiki page points out that a tensor of rank 3 might have different dimensions
For example, a rank 3 tensor might have dimensions 2, 5, and 7. Here, the indices range from «1, 1, 1» through «2, 5, 7»; thus the tensor would have one value at «1, 1, 1», another at «1, 1, 2», and so on for a total of 70 values.
In that terminology, a vector would have rank 1, but n dimensions. Saying "a vector is a one-dimensional array" is the same thing as saying "a tensor (vector) has rank 1". I can see the possibility for confusion.  Reply With Quote

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## Tensor example

I'm afraid I'm also mixing up vectors , matrices and tensors I've been looking for a while for good and easy to understand examples of tensors . In fact till now I've always done my work using matrices .
Someone can provide an example or a link ?  Reply With Quote

9. Originally Posted by frankuitaalst Someone can provide an example or a link ? Originally Posted by hhEb09'1 I knew that was going to happen--my message was too short!  Reply With Quote

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## Tensor example

Thanks for the link to wikipedia !
Reading this I realised I have been using tensors without even knowing ...  Reply With Quote

11. I am reading this now:
http://www.lerc.nasa.gov/WWW/K-12/Nu...2002211716.pdf

it is pretty good Originally Posted by frankuitaalst I'm afraid I'm also mixing up vectors , matrices and tensors I've been looking for a while for good and easy to understand examples of tensors . In fact till now I've always done my work using matrices .
Someone can provide an example or a link ?  Reply With Quote

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Didn't there used to be a poster named "Tensor"?

Whappen to him/her?

Pete

Uhhh.. nvrmnd...he just posted like 5 minutes ago :blush
Last edited by peter eldergill; 2008-May-13 at 04:09 AM. Reason: I'm a complete idiot :)  Reply With Quote

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6,208 Originally Posted by peter eldergill Didn't there used to be a poster named "Tensor"?

Whappen to him/her?

Pete

Uhhh.. nvrmnd...he just posted like 5 minutes ago :blush
Yeah, I'm still here. At the time I joined, I was deep in the self study of GR. Since GR, absolutely depends on tensors, that was the name I took for the board. I haven't been quite as active, but that is due to health concerns, more than anything else. I was going to post a smart alec comment in this thread, but kinda let it go.  Reply With Quote

14. Hmm. If you want to be very math-pedantic - vectors and tensors are described or parameterized with n-D arrays of scalar values, but they're actually mathematical objects in their own right. Given different frames, sets of basis vectors, or changing frames and basis vectors, there are sets of transforms that you use to get from one to the other.

Depending on what frame you are in, d/dt {velocity vector} isn't always {dvx/dt,dvy/dt,dvz/dt} like you would expect for a pure array.  Reply With Quote

15. Originally Posted by ASEI Depending on what frame you are in, d/dt {velocity vector} isn't always {dvx/dt,dvy/dt,dvz/dt} like you would expect for a pure array.
I'm not sure what to expect there, what does dvx/dt mean, in that context?  Reply With Quote

16. Originally Posted by hhEb09'1 I'm not sure what to expect there, what does dvx/dt mean, in that context?

acceleration in the x direction ?  Reply With Quote

17. Originally Posted by tommac acceleration in the x direction ?
Maybe. I could see that, but why would we expect it to be the "velocity vector"?  Reply With Quote

18. Originally Posted by frankuitaalst Thanks for the link to wikipedia !
Reading this I realised I have been using tensors without even knowing ...
I am also quite familiar with matrices, but that wiki link didn't exactly allow me to transform my understanding of matrices into some understanding of tensors. Isn't there some simplification here, assuming a background in matrix (and abstract) algebra?
Last edited by Cougar; 2008-May-13 at 03:01 PM. Reason: abstract  Reply With Quote

19. try the link I posted above. It is a 20 page doc of easy reading. Originally Posted by Cougar I am also quite familiar with matrices, but that wiki link didn't exactly allow me to transform my understanding of matrices into some understanding of tensors. Isn't there some simplification here, assuming a background in matrix (and abstract) algebra?  Reply With Quote

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Vector: both magnitude (340 mph) and direction (240 deg, horizontal)

Tensor: varying (linear, geometric, ln, or other various functions) throughout a 3 dimentional space. One example would be the varying bending moment along a beam with a load equally distributed along the length of the beam when said beam is anchored on one end only. The moment = 0 at the free end and is at a maximum at the anchored end. The moment varies linearly along the beam in this simplest of cases.  Reply With Quote

21. I dont think it is limited to 3 dimensional space right? Originally Posted by mugaliens Vector: both magnitude (340 mph) and direction (240 deg, horizontal)

Tensor: varying (linear, geometric, ln, or other various functions) throughout a 3 dimentional space. One example would be the varying bending moment along a beam with a load equally distributed along the length of the beam when said beam is anchored on one end only. The moment = 0 at the free end and is at a maximum at the anchored end. The moment varies linearly along the beam in this simplest of cases.  Reply With Quote

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14,315 Originally Posted by tommac I dont think it is limited to 3 dimensional space right?
Correct, and it can even include the element of time, such as how heat spreads throughout a steel plate. When a source of heat is applied along one edge.  Reply With Quote

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234 Originally Posted by tommac A scalar is a tensor rank of 0.
A vector is a tensor rank 0f 1. This can be represented as a one dimensional array. That is the part I am confused about...

Why is it represented as a one dimensional array?
Lets take force for example:
Does the one dimensional array mean that for all directions there is a net force in that directions?
No. There is a net force only in directions for which component of the vector is not zero. Which points to one important feature of a vector, its compnents.

Or why is it a one dimensional array?
Because the components of a vector may be written as a one dimensional array. Eg. [3,0,-4] and if this is some vector of some force there is no force in second directon. The brackets mean that it is a real vector or so called "column" vector and the array should be actually written vertically (as a column, but the square brackets take care of that).

It points to another thing, that not every one dimensional array is a vector (many people call any one dimensional array a "vector" but when you learn physics it's better to learn from the beginning what physicists call "vector": an arrow that has a lenght and a direction). Example of a one dimensional array that is not a vector is (3,0,-4) where "round" parenthises mean that it is 1-form, a.k.a. covariant vector.

There are a few other features of a vector and one rather imporant is its squared magnitude which for 3 dimensional vector (so called 3-vector) is its squared length. If vector v = [3,0,-4] its squared magnitue is 9+0+16 = 25. An interesting feature of magnitude of a vector is that when the vector is rotated by any angle in any direction its magnitude stays the same. E.g. when you turn a vector so that it points exactly backwards to the directon that it pointed to, its all components change from [3,0,-4] to [-3,0,4] but its magnitude stays the same 5.

The same thing happens with 4-vectors used in relativistic physics. Those 4-vectors have the first component in direction of time and their squared magnitude is calculated not as a sum of its components squared (as for 3-vectors) but as the difference of the first component squared and the sum of all other components squared. 4-vector with components [4,1,2,3] will have squared magnitude equal 16-(1+4+9) = 2. If this is a momentum 4-vector, the first component represents total energy of the particle and others are the components of 3-momentum of the particle. In regular relativistic physics both energy and momentum are conserved separately and therefore so is the 4-momentum. In the Big Bang cosmology though the energy is thought not to conserve separately from momentum (since otherwise the universe wouldn't be seen as expanding as astronomers insist it is). The Big Bang theory would collapse. The Big Bang theorists maintain that only 4-momentum is conserved (so called "momenergy") and not the energy and the momentum separately as it is in relativistic physics. As we can see to achieve that there must be possible the conversion of momentum into the non conserved energy to keep the magnitude of 4-momentum conserved.

I asked how astronomers explain such a conversion of momentum into energy that is required by the Big Bang Theory. Unfortunately the thread with this question was closed by the moderator on the assumption that I try to gain ponts for Einstein's static universe, by suggesting that the Big Bang cosmology is using a creationist physics. Apparently the moderator thinks that this creationist side has a legitimate explanation just it shouldn't be discussed in a forum where people should rather believe what they are told by experts than to compare it with their knowledge of relativistic physics which might be wrong despite millions of experiments confirming its validity. And that's probably why Feynman didn't want to discuss the Big Bang with the cosmologists.

I'd like to discuss the Big Bang, but they either don't respond or just close my thread. Apparently to keep the Big Bang alive they prefer the new creative assumptions of their math to observationally confirmed predictions just following the old relativistic physics. Ignoring totally Fenman's warning: "Let me also say something that people who worry about mathematical proofs and inconsistencies seem not to know. There is no way of showing mathematically that a physical conclusion is wrong or inconsistent. All that can be shown is that the mathematical assumptions are wrong. If we find that certain mathematical assumptions lead to a logically inconsistent description of Nature, we change the assumptions, not nature."
Last edited by JimJast; 2008-May-13 at 10:17 PM.  Reply With Quote

24. JimJast, ATM discussions are not appropriate in the Q&A forum. You've been cautioned about this before. I'm not quite ready to suspend you, but that's what's going to happen if you do this again.

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