# What does it mean to transform a tensor?

My electrodynamics lecture only defines a tensor (in $$R^3$$) of rank $$n$$ as a "set of quantities $$T_{i_1i_2...i_n}$$, $$i_1, ... i_n = 1,2,3$$ that transforms under rotation as follows: $$$$T_{i_1...i_n}^{'}=R_{i_1j_1}...R_{i_nj_n}T_{j_1j_n} \ \ \ \ \ \ \ \ \ \ \\ \ \ (1)$$$$". I don't understand what the left side is supposed to mean, nowhere have I found a proper definition for dashed quantities, so this definition makes little sense to me.

From here and here I understand the definition to mean something like $$f(o_1^r, ... o_n^r) = R_{i_1j_1}...R_{i_nj_n}T_{j_1j_n}$$ where $$f(o_1^r, ... o_n^r)$$ is the tensor that has an input of at most $$n$$ (3-dimensional) vectors $$o_i$$ and where $$o_i^r$$ means that they underwent some rotation (although I don't know what that means specifically).

My question is: In equation (\ref{1}) what does the left (and also the right) side mean? How, specifically, can I check for a given tensor that it satisifes the transformation property?

• When you say "given tensor", in what form exactly do you think the tensor is given to you and what there is left to show? The transformation property is part of the definition of the tensor in this approach, so if you have a "given tensor" then there is nothing to show. E.g. I could define a tensor just by writing down the values for the $T_{i_1\dots i_n}$ in one coordinate system and saying "this is a tensor" and the only way for you to compute the l.h.s. in your eq. (1) would be to apply the r.h.s. since that's the definition of how tensors transform. Commented Sep 15, 2023 at 15:37
• By given tensor I mean (1, 2, 3) or $\epsilon_{i,j,k}$ (with the definition of the levi-civita-symbol) or the moment of inertia tensor. I think it remains to show that these quantities are tensors. Or e.g. I would like to prove that the sum of two tensors is a tensor again - but for that I need to show that the left side is equal to the right side, right?
– F L
Commented Sep 15, 2023 at 15:58
• nowhere have I found a proper definition for dashed quantities I think you mean primed quantities. The rotation $R$ relates two Cartesian coordinate systems, $(x,y,z)$ and $(x’,y’,z’)$, that are rotated relative to each other. The primed and unprimed tensor components are the tensor components in the primed and unprimed coordinate systems. Commented Sep 15, 2023 at 17:45
• Yes, thanks, but how are the unprimed components obtained?
– F L
Commented Sep 16, 2023 at 16:57

A tensor on a manifold ($$\mathbb{R}^3$$ in your case) is a coordinate independent object. It is defined as a multilinear map $$T:\mathcal{T}_p\mathcal{M}^n\to\mathbb{R}$$ where $$\mathcal{T}_p\mathcal{M}$$ is the tangent space of the manifold $$\mathcal{M}$$ at a point $$p$$. I would like to answer your question with a short introduction on the concept of coordinate independence in differential geometry.

A manifold can locally be equipped with coordinates $$\{x^\mu\}$$ such that it locally always looks like a subset of $$\mathbb{R}^n$$ (if $$\mathcal{M}\equiv\mathbb{R}^n$$, there are still many different choices of coordinate systems). One can show that the space of directional derivatives fulfills the requirements of a vector space, this space is called the tangent space $$\mathcal{T}_p\mathcal{M}$$. A vector in this space is a directional derivative that acts on functions $$f\in\mathcal{C}^1(\mathcal{M})$$, meaning $$D_v(f)=v^\mu\partial_\mu f$$ Instead of $$D_v$$ one usually just writes $$v$$ ($$D_v$$ might actually denote a different type of derivative, the covariant derivative...that's another story). The coordinate system $$\{x^\mu\}$$ suggests a basis of $$\mathcal{T}_p\mathcal{M}$$, namely $$\{\partial_\mu\equiv\frac{\partial}{\partial x^\mu}\}$$. Therefore $$v=v^\mu\partial_\mu$$. The $$v^\mu$$ are just a set of numbers which the physicist usually calls "vector", though they are really only the components of a vector. They key difference is that $$v$$ is a coordinate independent object; it is a directional derivative and a direction in space is something physical! Its components are coordinate dependent, coordinate systems have no physical reality!

Consider a different coordinate system, $$\{x^{\mu^\prime}\}$$. The tangent space basis $$\{\partial_\mu\}$$ as we know fulfills $$\partial_\mu=\frac{\partial}{\partial x^\mu}=\frac{\partial x^{\mu^\prime}}{\partial x^\mu}\frac{\partial}{\partial x^{\mu^\prime}}=\frac{\partial x^{\mu^\prime}}{\partial x^\mu}\partial_{\mu^\prime}$$ We call $$J^{\mu^\prime}_{\phantom{\mu}\mu}=\frac{\partial x^{\mu^\prime}}{\partial x^\mu}$$ the Jacobian matrix of the coordinate change. If $$v$$ is supposed to be a coordinate independent object, we can now find out how its components must transform: $$v^{\mu^\prime}\partial_{\mu^\prime}=v=v^\mu\partial_\mu=v^\mu J^{\mu^\prime}_{\phantom{\mu}\mu}\partial_{\mu^\prime}\implies v^{\mu^\prime}=J^{\mu^\prime}_{\phantom{\mu}\mu}v^\mu$$ This is what a physicist means by saying "a set of numbers $$v^\mu$$ transforms like a vector"! The physical object $$v$$ is unaffected by this!

We are missing one final step to get the final answer to your question: There exists a way of turning vectors into scalars. The dual space $$\mathcal{T}^*_p\mathcal{M}$$ is defined to be the set of linear maps $$\omega:\mathcal{T}_p\mathcal{M}\to\mathbb{R}$$ so if $$v$$ is a vector and $$\omega$$ is a dual vector, then $$\omega(v)$$ is a real number. It can again be expressed in coordinates, $$\omega(v)=\omega(v^\mu\partial_\mu)\overset{\text{linearity of }\omega}{=}v^\mu\omega(\partial_\mu)\equiv v^\mu\omega_\mu$$ Again, we call $$\omega_\mu$$ the (coordinate dependent) components of the dual vector (also called $$1$$-form) $$\omega$$. And again, even though we expressed it in coordinates, $$\omega$$ is a coordinate independent object, meaning that $$\omega(v)$$ should give the same real number no matter what coordinate system we used. In formula: $$v^{\mu}\omega_{\mu}=v^{\mu^\prime}\omega_{\mu^\prime}\overset{\text{transformation rule for }v^{\mu}}{=}v^\mu J^{\mu^\prime}_{\phantom{\mu}\mu}\omega_{\mu^\prime}\implies \omega_\mu=J^{\mu^\prime}_{\phantom{\mu}\mu}\omega_{\mu^\prime}$$ Notice how the transformation rule for $$\omega$$ is in some sense inverse to the transformation rule of $$v$$. We can also write $$\omega_{\mu^\prime}=(J^{-1})^\mu_{\phantom{\mu}\mu^\prime}\omega_\mu$$

Finally, tensors are objects that do not only take 1 vector (as $$\omega$$ did), but many vectors (say $$n$$), and turn them into a scalar. We would call them a multilinear map (whereas $$\omega$$ before was just a linear map, it had only 1 argument). Therefore they do not only have 1 index labelling the components (like the $$\mu$$ in $$\omega_\mu$$), but many indices (for example $$T_{\mu_1\mu_2\dots\mu_n}$$). However, by exactly the same arguments as before, in order to be coordinate independent objects on the manifold, the tensor components must transform like $$T_{\mu_1^\prime\dots\mu_n^\prime}=(J^{-1})^{\mu_1}_{\phantom{\mu}\mu_1^\prime}\dots(J^{-1})^{\mu_n}_{\phantom{\mu}\mu_n^\prime}T_{\mu_1\dots\mu_n}$$

A formal definition of this can be obtained from wikipedia:

A tensor is an assignment of a multidimensional array $$T^{}_{}[f]$$ to each basis $$f = (e_1, ... , e_n)$$ of an n-dimensional vector space such that, if we apply a change of basis $$f \rightarrow f \cdot R = (e_iR_1^i, ... , e_iR^i_n)$$ then the multidimensional array obeys the transformation law: $$T^{i'_1...i'_p}_{j'_1...j'_p}[f\cdot R] = (R^{-1})^{i'_1}_{i_1}...(R^{-1})^{i'_p}_{i_p}T^{i_1,...,i_p}_{j_1,...,j_q}[f]R^{j_1}_{j'_1}...R^{j_q}_{j'_q}$$

In physics, tensors are sometimes said to be defined by how they transform. In this case, the left hand side is the transformed tensor, and the right hand side is the transformation acting on the initial tensor. If you want to prove that your entity is a tensor using this definition, you need to calculate the change of basis elements (in your case this seems to be a rotation) and show how the tensor components can be related by the transformation according to this definition.

Addendum: It seems you are working specifically with the Kronecker delta, which makes use of a trick. We want to prove $$\delta^{i'}_{j'} = \delta^k_l\frac{\partial x_{i'}}{\partial x_k}\frac{\partial x_l}{\partial x_{j'}}$$

Beginning with the RHS, we can write $$\delta^k_l\frac{\partial x_{i'}}{\partial x_k}\frac{\partial x_l}{\partial x_{j'}} = \delta^k_k\frac{\partial x_{i'}}{\partial x_k}\frac{\partial x_k}{\partial x_{j'}}$$

by definition of the Kronecker delta. Then, we can see that the numerator of the second fraction and the denominator of the first fraction reduce to one. This leaves us with $$\delta^k_l\frac{\partial x_{i'}}{\partial x_k}\frac{\partial x_l}{\partial x_{j'}} = \delta^k_k\frac{\partial x_{i'}}{\partial x_k}\frac{\partial x_k}{\partial x_{j'}} = \frac{\partial x_{i'}}{\partial x_{j'}}$$

Now, since these are coordinates, we can note that this term will be 1 when $$i' = j'$$ and that it will be zero when $$i' \neq j'$$. This is the definition of $$\delta^{i'}_{j'}$$, so we have shown that $$\delta^k_l\frac{\partial x_{i'}}{\partial x_k}\frac{\partial x_l}{\partial x_{j'}} = \delta^{i'}_{j'}$$ as we have set out to show.

• Thanks. I am still unclear on how to find the transformed tensor. E.g. if I want to prove that the Kronecker symbol $T=d_{i,j}$ is a tensor of rank 2, the right side is $R_{ik}R_{jl}T_{kl}$. To find T' I have to transform $d_{i,j}$ in another coordinate system but I do not understand what that means
– F L
Commented Sep 16, 2023 at 17:05
• Please let me know if my edit helped @FL Commented Sep 16, 2023 at 22:15
• So, since the Kronecker Delta does not depend on coordinates $\delta'_{ij}=\delta_{i,j}$?
– F L
Commented Sep 17, 2023 at 9:11
• @FL I have cleaned up the notation a bit, which might clarify what you are asking about, but when we show that $\delta^{i'}_{j'}$ transforms as a tensor, we are showing that it doesnt depend on coordinates because this is what a tensor is. You can read more about this here physics.stackexchange.com/questions/357797/…. Commented Sep 17, 2023 at 20:51