5
$\begingroup$

Our professor defined a rank $(k,l)$ tensor as something that transforms like a tensor as follows:

$$T^{\mu_1' \mu_2'...\mu_k'}{}_{\nu_1'\nu_2'...\nu_l'} ~=~ \Lambda^{\mu_1'}{}_{\mu_1}...\Lambda^{\mu_k'}{}_{\mu_k}~\Lambda^{\nu_1}{}_{\nu_1'}...\Lambda^{\nu_l}{}_{\nu_l'}~T^{\mu_1\mu_2...\mu_k}{}_{\nu_1\nu_2...\nu_l}$$

Where $\Lambda$ are the Lorentz transformation matrices (translations, rotations, or boosts). I'm not sure if this is only for SR or if also for GR since we've only been talking about SR thus far, though GR is something we'll be covering soon.

He wrote on the whiteboard: if $S_{\mu\nu\rho}=S_{\nu\mu\rho}$ then $S$ is symmetric in $\mu$ and $\nu$.

But let's just talk about a rank (2,0) symmetric contravariant tensor for a second, denoted $S^{\mu\nu}$ and equals $S^{\nu\mu}$. How would we prove this is a tensor? Our book uses $R$ in place of $\Lambda$ in their formulations above, where $R$ might just be rotations. I'm sure General Tensors would have any jacobian and inverse jacobians are matrices rather than just the Lorentz transformations. This is question in Prof. Zee's, "Einstein's Gravity in a Nutshell", Chapter I.4 Exercises 2.

Also if you want to give a student like myself who is new to tensor some advice on learning tensors, and some tensor properties, and how to work with them, be my guest :)

Also, are all transformations homogeneous linear transformations? - These can be read about at: http://www.math.ucla.edu/~baker/149.1.02w/handouts/e_htls.pdf

http://www.lecture-notes.co.uk/susskind/special-relativity/lecture-6/rank-two-tensors/ This lecture gives a nice matrix form of what a symmetric (2,0) tensor looks like. I think this may help, thinking of these tensors as matrices themselves, visually. Basically they're symmetric matrices of the form $A^T=A$. Also we can think of $\Lambda$ if it's a rotation matrix having the property $\Lambda^{-1}=\Lambda$.

Does this picture below do anything for the problem? enter image description here

$\endgroup$
4
$\begingroup$

A tensor is not particularly a concept related to relativity (see e.g. stress tensor), but is a more general concept that describes the linear relationships between objects, independent of the choice of coordinate system.

This coordinate independence results in the transformation law you give where, $\Lambda$, is just the transformation between the coordinates that you are doing. For special relativity this is the Lorentz transform, but in classical physics in may be a simple rotation. The point is that the tensor holds regardless of the change in coordinates.

Therefore to show something is a tensor you just have to show that it obeys the transformation equation and that your transformed answer is still a valid result and can be transformed back to the original by doing the inverse transform.

$\endgroup$
  • $\begingroup$ So if I'm proving that it's a tensor in general, I will need to know all of the properties of $\Lambda$? Otherwise what could I do with an expression such as $\Lambda^{\mu}\Lambda^{\nu}T^{\mu\nu}$? You're saying though that this equals something in which I lost no information & then it's invertible with no loss of information? $\endgroup$ – Adam Staples Sep 13 '15 at 8:48
  • $\begingroup$ I know if the transformation is for rotations then $\Lambda^{\mu}$ is just the orthogonal rotation matrix. If it's for boosts then we have a matrix with hyperbolic functions (I cannot remember if this is orthogonal as well?). But I think when I read the problems I interpret it to asks for the general case of $\Lambda^{\mu}$, but I do not know. $\endgroup$ – Adam Staples Sep 13 '15 at 8:51
  • $\begingroup$ I still do not know how to show that a symmetric tensor is indeed a tensor? I'm completely lost on this. $\endgroup$ – Adam Staples Sep 13 '15 at 10:29
  • $\begingroup$ Have a look at ita.uni-heidelberg.de/~dullemond/lectures/tensor/tensor.pdf particularly section 4. Symmetry is a coordinate transform, so you can show that $t^{uv}=t^{vu}$ $\endgroup$ – nivag Sep 13 '15 at 21:14
3
$\begingroup$

Our professor defined a rank $(k,l)$ tensor as something that transforms like a tensor

Oh dear. All too common, and all too pedagogically faulty.

A tensor is nothing more or less than a linear map from (possibly multiple copies of) a vector space (and possibly copies of its dual space) into the scalar field.

If I give you components $T_{\mu\nu}$ (16 components in all) in a specific coordinate system/basis, then you can map any two vectors into a scalar with it. Simply consider the vector components in the same coordinate system, and contract indices: $T(\vec{v},\vec{w}) = T_{\mu\nu} v^\mu w^\nu \in \mathbb{R}$. Moreover this map is linear by construction. So it's a tensor. Nothing whatsoever needs to be checked.

What happens, though, is sometimes you simultaneously write down the components for (ostensibly) the same object in two different coordinate systems. For example, you have expressions for all 16 $T_{\mu\nu}$ and another 16 $T_{\mu'\nu'}$. If you did everything right and consistent and you're not being tricked with an intentionally misleading textbook problem, these sets of components should be relatable via whatever transformation law generically takes you from unprimed to primed coordinates. So it might be a good idea to check that this holds. But it's just a sanity check.

Alternatively, someone may have just handed you $T_{\mu\nu}$ and $T_{\mu'\nu'}$ and asked "are these components of the same object, just in different coordinate systems?" Then you can also apply the transformation to check. If it doesn't work, it's not that you don't have a tensor. Rather, you have two distinct tensors.

$\endgroup$
1
$\begingroup$

If I understand correctly, you're asking how to prove that symmetry of a tensor is coordinate independent, but you seem to be having trouble with the definition of a tensor. Well, you're not the first. Let me give you a definition that might help.

First, suppose you have some space (it can be 3-space or spacetime or whatever) and you have a set of coordinates $\{x^i\}$ defined on it. And let's say you have a particle moving in your space, with a trajectory given by $x^i = x^i(t)$. Here $t$ is just a parameter. You can find the components of the velocity in your system of coordinates: $u_x^i(t) = dx^i/dt$. (I am using subscripts to label coordinate systems.) Now here's the thing:

Suppose you calculate the velocity in a different system of coordinates $\{y^i\}$; it would be $u_y^i(t) = dy^i/dt$. But if you know the coordinates $y^i$ as a function of the coordinates $x^j$, you can find out how the two velocities are related:

$$u_y^i(t) = \frac{dy^i(x)}{dt} = \frac{\partial y^i}{\partial x^j} \frac{d x^j}{dt} = \frac{\partial y^i}{\partial x^j} u^j_x(t)$$

I have used the chain rule and the fact that the $y^i$ are functions of the $x^j$. $\partial y^i / \partial x^j$ will have different properties depending on the coordinates. In Euclidean 3-space we tipically use Cartesian coordinates and so $\partial y^i / \partial x^j$ would be a rotation matrix; in Special Relativity it would be a Lorentz transformation, and so on. In General Relativity we use all kinds of coordinates, and the transformations will not in general be linear.

So now we know how the velocity of a particle (or, as the mathematicians would call it, the tangent vector to a curve) transforms when you change coordinates. It is often helpful to regard such a vector as an object $\vec{u}$ that is independent of coordinates. Indeed, this whole business of transformation laws and Einstein convention is a way to make sure that things don't depend on coordinates. The components of a vector (or a tensor) will depend on the coordinates, but if everything transforms the same way, equations made out of tensors will have the same form in different coordinate systems.

Now we can define vectors in general, by asking that they have the same transformation law as velocities:

A vector $\vec{X}$ is a function that assigns a set of numbers (called its components) $X_x^i\ (i = 1\dots n)$ to each coordinate system $\{x^i\}$, such that if $\{x^i\}$ and $\{y^i\}$ are two coordinate systems, the components of $X$ are related by

$$X_y^i = \frac{\partial y^i}{\partial x^j} X_x^j$$

Side note: What I have defined is technically a vector field, not a plain vector. This is not an important distinction here. Also, I am restricting myself to coordinate bases for simplicity.

This is essentially the same as the "set of numbers that transforms like this" definition, but I find it to be a bit clearer and more explicit as to what things are.

A tensor can be defined as something that transforms as products of vectors: If we take two vectors $\vec{u}$ and $\vec{v}$ and define the (coordinate-dependent) quantity $T_x^{ij} = u^i_x v^j_x$, then in two different coordinate systems we find (defining $\Lambda^i_{\ j} = \frac{\partial y^i}{\partial x^j}$):

$$T^{ij}_y = \Lambda^i_{\ k} \Lambda^j_{\ l} T^{kl}_x$$

Following the definition of a vector, we can define a $(2,0)$ tensor $T$ (not necessarily a product of vectors as above) as a function that assigns a set of numbers $T^{ij}_x$ to each coordinate system, such that the components in two different systems follow the above transformation law.

Now let's get to your question. You ask how to prove that a symmetric tensor is a tensor, but this is a tautological question, because a symmetric tensor obviously is a tensor! I suspect that the actual question is as follows. You defined a symmetric tensor as one that has the property $T^{ij} = T^{ji}$. This is a valid definition, but it is a priori coordinate dependent. We would like to prove that if the above identity is true in one coordinate system, it is true in all of them.

So let's suppose in some coordinates $\{x^i\}$ it happens that $T_x^{ij} = T_x^{ji}$ for all $i,j$. Let $\{y^i\}$ be an arbitrary coordinate system. Then

$$T_y^{ij} = \Lambda^i_{\ k} \Lambda^j_{\ l} T^{kl}_x = \Lambda^i_{\ k} \Lambda^j_{\ l} T^{lk}_x = \Lambda^j_{\ l} \Lambda^i_{\ k} T^{lk}_x = T_y^{ji}$$

To get the second equality I used that $T_x^{kl} = T_x^{lk}$, to get the third equality I moved the $\Lambda$s around, and in the first and last equalities I used the transformation law for a tensor. So we have found out that if a tensor is symmetric in some coordinate system, it is symmetric in any coordinate system. Therefore, it makes sense to say that symmetry is a property of the tensor instead of its representation in a particular coordinate system.

One final remark: As you said, a tensor with two indices can be represented as a matrix. The derivatives of the transformation $\partial y^i / \partial x^j$ can also be represented as a matrix. These matrices have different meanings! A tensor is a coordinate independent object, and its matrix will change if you change coordinates. A transformation is defined only between a specific pair of coordinate systems. If you have a matrix $\Lambda^i_{\ j}$ relating coordinates $x$ and $y$ as above, it makes no sense to ask what $\Lambda$ looks like in coordiantes $z$. So even though a symmetric tensor has a symmetric matrix ($A^T = A$) and a rotation matrix is orthogonal ($A^{-1} = A^T$), these properties are unrelated to each other.

$\endgroup$
  • $\begingroup$ To get the third equality, you moved the lamdas around. But that would be possible only if lorentz transformation is commutative. What am I missing? $\endgroup$ – MycrofD Jan 30 '18 at 14:10
  • $\begingroup$ @MycrofD when using component notation, everything is numbers, so it doesn't matter what order you write the factors in. Remember that there's an implied summation sign. $\endgroup$ – Javier Jan 30 '18 at 14:11
  • $\begingroup$ I get what you are saying, somewhat. Can you please show some reference? I have been searching the net, but could find no reliable source. I even found a question saying that in general they are non-commutative. physicsforums.com/threads/… $\endgroup$ – MycrofD Jan 30 '18 at 14:18
  • 1
    $\begingroup$ @MycrofD say we have matrices $A$, $B$ with $C = AB$. Then $C$ is definitely not the same as $BA$. But in component notation, $C_{ij} = \sum_k A_{ik} B_{kj}$, and this is the same as $\sum_k B_{kj} A_{ik}$, because the components are just numbers; the sum is a sum of products of numbers. As an exercise, write $BA$ in component notation and see the difference. $\endgroup$ – Javier Jan 30 '18 at 14:25
  • $\begingroup$ Right! Yes, component notation is just numbers and not matrices anymore. Thanks for clarifying. I even got the implied summation you meant, which was initially not obvious to me. $T$ on the left is a single number equal to the sum of all $\Lambda\Lambda T$ terms on the right (for certain i,j,y). $\endgroup$ – MycrofD Jan 30 '18 at 14:35

Not the answer you're looking for? Browse other questions tagged or ask your own question.