Hey Brilliant Users!

So I am fairly inexperienced in dealing with tensors (aside from the usual scalars, vectors and linear maps) and am trying to understand the nature of these "objects" a little better. My understanding is as follows, and I'm really not sure if I'm interpreting this correctly so please correct me if I'm wrong:

A 0 order tensor is a scalar. A 1st order tensor is a vector. A second order tensor, by my reasoning, must always be representable as a matrix when one chooses to find such a representation, since every element requires two indices to specify. Now, if we continue this pattern, I reason that a 3rd order tensor must be some kind of geometric object. I imagine this object as a "cube" in which each face is a matrix. This is purely speculation, as I have no proof to back that last claim up but it seems to follow.

As many of you have read in my previous post, I am currently reading Spivak's "Calc on Manifolds" and have recently begun chapter 4, in which I have been formally introduced to tensors for the first time in their full generality. I take back everything I said about the book not being hard, I'll be the first to admit I'm having quite a bit of trouble intuitively understanding what the author is establishing concerning tensors. Any feedback whatsoever is, as usual, greatly appreciated.

I also want to take a brief moment to thank the Brilliant.org community. I've actually learned more from the users on here in the relatively brief time I've been a member of the site than I have from any teacher in any formal class. Great stuff guys!

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TopNewestThe Levi-Cevita tensor e(ijk) is an example of a three tensor, and you are sort of right that it is a 3D matrix. – Willy Penn · 1 year, 9 months ago

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So, you imagine it being something like this? :P Where each entry takes up one block of volume.

3x3 Tensor

Pretty interesting stuff especially in differential geometry. :'-) Also, I enjoyed reading "Introduction to Tensor Analysis and the Calculus of Moving Surfaces" by Pavel Grinfeld. I believe he has some Youtube videos on them as well. Feel free to check it out. ~_~ – Vishnuram Leonardodavinci · 1 year, 9 months ago

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Moment Of Inertia is a tensor quantity......!!!!! can somebody explain......this was the answer i got when i asked my teacher whether it is scalar or vector....now tensor is a new game huh... – Manish Bhargao · 1 year, 9 months ago

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\(I\vec{\alpha} = \sum \vec{T}\)

Since \(\vec{T} = \vec{r} \times \vec{F} \), we must assume everything is in \(\mathbb{R}^3\), since the cross product is not well defined in other dimensions (at least not as it appears here).

Now think about \(I\). This, imprecisely speaking, quantifies a tendency for rotation about a given axis. But we may have many different "lines" about which the object may rotate! We thus need a multilinear function to define \(I\) in its most general form. In elementary physics, scalar \(I\) arises as a result of the moment of inertia tensor being equivalent to (scalar) \(I\) times the identity matrix, in which case we recover the usual rotational analogue of Newton's Law. This usually corresponds to a situation in which rotation is confined to be about a line that may be considered parallel to some basis vector and symmetric density distribution of the rigid body about all basis vectors (i.e. a homogenous sphere centered at the origin). This is the prototypical situation encountered in elementary physics. For a more complicated axis of rotation and mass distribution, the tensor will need to be used.

Another way to think about it is to notice that there is no way to directly isolate \(I\) in the above equation-i.e. there is no mathematically consistent way to divide two vectors.

Finally, another analogue of this is the concept of pressure. "Pressure" is actually also a tensor, commonly called the "stress" tensor, in which the "pressures" or "normal stresses" are the elements along the main diagonal. It turns out we may encounter stresses on an object in many different directions, not just the "normal" direction many people are accustomed to, thus the need for a tensor.

This is my understanding and aspects of it may be incorrect, so any feedback is appreciated. – Ethan Robinett · 1 year, 9 months ago

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sorry for this question but i just want to understand...... – Manish Bhargao · 1 year, 9 months ago

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\(P \vec{A} = \vec{F}\)

Where \(P\) here is the stress tensor, \(\vec{F}\) is the force and \(\vec{A}\) is a vector in \(\mathbb{R}^3\) normal to the surface of the area in consideration. Now suppose we have some surface with a bunch of stresses acting on it in a few different directions. If there exists a stress acting on this surface in a direction that is not precisely normal to the surface, we must find a way of quantifying this, as the direction of \(\vec{F}\) will inevitably be affected by this stress. Note that we cannot represent more than one stress acting on the surface by a single number, at least not in a precise way (we may average the stresses, but again this is not precise). So in order to represent these stresses precisely, we need an "array" of numbers. Now, \(\vec{F} \in \mathbb{R}^3\) and \(\vec{A} \in \mathbb{R}^3\). Then, for dimensional reasons, \(P\) must be a \(3 \times 3\) matrix, which acts on \(\vec{A}\) in what is known as a linear transformation.

Now, just as is the case with the stress tensor, in the case of the inertia tensor, we may be rotating an object about an axis which "presents" little symmetry as far as mass distribution and geometry go, hence the need to consider multiple directions in our calculation of \(I\).

In general, tensors are very abstract objects and there are certain disadvantages to thinking of them in this way. In particular, this method of thinking requires a coordinate system, which we generally wish to avoid. So in order to explain tensors (albeit in a very non-technical way), I will use the following analogy:

Imagine you are observing some property of your environment (maybe temperature, pressure,etc. it doesn't matter). Regardless of whether you precisely measure such a property, it still exists. The precise measurement of the property can be seen as analogous to establishing a coordinate system for the property and the actual measurement is analogous to the coordinates of the property. But the existence of the property should be independent of the coordinate system you choose! Tensors provide a way of quantifying this notion. Each tensor is necessarily equipped with a transformation law telling you how to represent it in a given coordinate system, but independently, outside of the coordinate system, the tensor still exists. There are of course certain properties that change with coordinate systems and those that don't, and with these specifications we begin to encounter the notions of covariant and contravariant tensors, which I will not detail here since I have insufficient knowledge of them to give an explanation. – Ethan Robinett · 1 year, 9 months ago

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One last question....how to identify which is Tensor and which is not...like we say electric current is not a vector since it does not obey Laws of Vector Addition. is there any as such Law.

Thanks again.... :) – Manish Bhargao · 1 year, 9 months ago

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– Manish Bhargao · 1 year, 9 months ago

hey did I guessed it correct or not !!??Log in to reply