How would one describe to a non-mathematician (an undergraduate physicist actually- so please do use mathematics-i am not even sure if they can be described without mathematics!) what do tensors represent geometrically?
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asked Apr 22 '16 at 11:53
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$\begingroup$ Depends what you mean by tensors. Are we talking in the sense of "stress tensors" and general relativity, or in the sense of QM (i.e. the tensor product of Hilbert spaces) $\endgroup$ – Omnomnomnom Apr 22 '16 at 12:00
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$\begingroup$ Wow, i didn't think about that as i am yet to encounter them. I would say, all three kinds of tensors. $\endgroup$ – TheQuantumMan Apr 22 '16 at 12:19
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$\begingroup$ Have you taken linear algebra yet? $\endgroup$ – Omnomnomnom Apr 22 '16 at 15:47
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$\begingroup$ Yes. Also, i know a bit of multivariable calculus, differential equations and calculus of variations and complex analysis(i don't know if the last 3 are relevant). $\endgroup$ – TheQuantumMan Apr 22 '16 at 16:16
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$\begingroup$ ...@QuanticMan, linear algebra that includes vector-duality? $\endgroup$ – janmarqz Apr 22 '16 at 16:27
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Here's the quick way to describe what's going on:
In linear algebra, you learned primarily about linear transformations. In particular, $T:V \to W$ is a function that takes a single vector, and produces another vector in a linear way. In particular $T$ satisfies $T(ax + by) = aT(x) + bT(y)$. It turns out that linear transformations can naturally be thought of as matrices, so that's what you end up working with most of the time.
A tensor, by contrast, is a map $T:V_1 \times V_2 \times \cdots \times V_k \to W$. That is, it takes several vectors as its input and produces an output-vector in a multi-linear way. That is, $T$ will satisfy $$ T(v_1,v_2,\dots v_{k-1},ax + by,v_{k+1},\dots,v_n) =\\ aT(v_1,v_2,\dots v_{k-1}, x,v_{k+1},\dots,v_n) + bT(v_1,v_2,\dots v_{k-1}, y,v_{k+1},\dots,v_n) $$ Rather than thinking of tensors as matrices, we think of them as multidimensional arrays.
Another example of a Tensor that can be thought of as a matrix is a "bilinear form" $T:V_1 \times V_2 \to \Bbb R$. In particular, given a matrix $A$, the function $$ T(v_1,v_2) = v_1^T A v_2 $$ is exactly this kind of tensor. One commonly used tensor that works this way is the stress tensor. The question it answers is "what is the component of pressure on the plane perpendicular to $v_1$ in the direction of $v_2$?" Two others are the dot-product and, more generally, metric tensors.
answered Apr 22 '16 at 16:40
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We can add something to the already excellent answer given by Omnomnomnom.
The wonderful Gravitation by Misner,Thorpe and Wheeler (affectionately called MTW) on page 75 describes the most general (m,n) tensor as:" a linear machine with n input slots for n 1-forms and m input slots for m vectors, given the requested input, it puts out a real number..." (1-forms are simply the linear operators that take a vector and give out a number).
Tensors are useful in physics because (roughly): each observer is described by a different 4D basis in spacetime and each measurement is described by the value of a specific component of a specific tensor in the observer's basis. Tensor components in one basis are (linearly) related to the components in another basis, so if we know what one observer is measuring, we can find out what another observer will measure.
In short: measurements made by different observers are related in the same way that certain tensor components are related when passing from one basis to another basis
