Bernard Schutz defines in his book "A First Course in General Relativity" the components of the stress-energy tensor, in some (arbitrary) frame $O$, in the following way:

$\mathbf T(\tilde{dx}^\alpha, \tilde{dx}^\beta) = T^{\alpha\beta}:= $ { flux of $\alpha$ momentum across a surface of constant $x^\beta$ in frame $O$}

See here (4.14): http://fulviofrisone.com/attachments/article/486/A%20First%20Course%20In%20General%20Relativity%20-%20Bernard%20F.Schutz.pdf

He leaves it to the reader to prove that this really is a tensor. $\{\tilde{dx}^\alpha\}$ is the basis one-form, that is the dual basis of $\{\vec{e}_\beta\}$, defined by $\tilde{dx}^\alpha(\vec{e}_\beta) = \delta^\alpha_\beta$.

Schutz defines a tensor as follows:

An $M \choose N$ tensor is a linear function of $M$ one-forms and $N$ vectors into the real numbers.

Now my question is, why is the object defined as above a tensor?

OK, some more explanations:

In another frame $\overline{O}$ the components are defined accordingly:

$\mathbf T(\tilde{dx}^\overline{\alpha}, \tilde{dx}^\overline{\beta}) = T^{\overline{\alpha}\overline{\beta}}:= $ { flux of $\overline{\alpha}$ momentum across a surface of constant $x^\overline{\beta}$ in frame $\overline{O}$}

The frames $O$ and $\overline{O}$ are connected by a Lorentz transformation $\Lambda$. That means for the basis vectors and the basis one-form ($\tilde{\omega}^\alpha := \tilde{d}x^\alpha$, Schutz notation):

$\vec{e}_\overline{\alpha} = \Lambda^\mu_{\space\space \overline{\alpha}} \space \vec{e}_\mu$

$\vec{e}_\overline{\beta} = \Lambda^\nu_{\space\space \overline{\beta}} \space \vec{e}_\nu$

$\tilde{\omega}^\overline{\alpha} = \Lambda^\overline{\alpha}_{\space\space \mu} \space \tilde{\omega}^\mu$

$\tilde{\omega}^\overline{\beta} = \Lambda^\overline{\beta}_{\space\space \nu} \space \tilde{\omega}^\nu$

Now, IF the function is linear in both arguments, we have:

$T^{\overline{\alpha}\overline{\beta}} = \mathbf T(\tilde{\omega}^\overline{\alpha}, \tilde{\omega}^\overline{\beta}) = \mathbf T(\Lambda^\overline{\alpha}_{\space\space \mu} \space \tilde{\omega}^\mu, \Lambda^\overline{\beta}_{\space\space \nu} \space \tilde{\omega}^\nu) = \Lambda^\overline{\alpha}_{\space\space \mu} \space \Lambda^\overline{\beta}_{\space\space \nu} \space \mathbf T(\tilde{\omega}^\mu, \tilde{\omega}^\nu) = \Lambda^\overline{\alpha}_{\space\space \mu} \space \Lambda^\overline{\beta}_{\space\space \nu} \space T^{\mu\nu}$

This last equation is the usual transformation of a $2 \choose 0$ Tensor. But this only works if we assume that the function is linear.

So the question: WHY is the function (defined at the very top) linear?

EDIT: OK. Put it another way. I reformulate my question. Forget about multilinear maps and one-forms. Simply say a $2 \choose 0$ Tensor is an object with 2 indices, which transforms like this: $T^{\overline{\alpha}\overline{\beta}} = \Lambda^\overline{\alpha}_{\space\space \mu} \space \Lambda^\overline{\beta}_{\space\space \nu} \space T^{\mu\nu}$

Now suppose a physical situation (fluid, dust, electromagnetic fields, whatever) in a frame $O$, and determine the $T^{\alpha\beta}$ as defined above (Flux of $\alpha$ momentum across a surface of constant $x^\beta$). Then you get 16 numbers arranged in a matrix. In principle you can measure these numbers physically by experiment (You can measure energy, momentum, density, etc.). Now observe the same physical situation, at the same point in space and time, from another reference frame $\overline{O}$. Then you get another 16 numbers $T^{\overline{\alpha}\overline{\beta}}$. You also know the Lorentz-Transformation $\Lambda$. So you can in principle test by experiment if the equation $T^{\overline{\alpha}\overline{\beta}} = \Lambda^\overline{\alpha}_{\space\space \mu} \space \Lambda^\overline{\beta}_{\space\space \nu} \space T^{\mu\nu}$ is fulfilled. If it is fulfilled for arbitrary frames, then T is a tensor.

Now my question is: Why is it fulfilled? It is not a question of pure mathematics. It's a physical question. And I think we must take into account that we talk about energy and momentum.

  • 1
    $\begingroup$ Can you be more specific in your question. I mean this, I am assuming you have read Schutz' book from the beginning and this entry is almost 100 pages into the book and you are asking on why this is a tensor. Either, you have failed to grasp the concept of a tensor so far or you are asking something else about the stress energy tensor. $\endgroup$
    – K7PEH
    Commented Jan 6, 2020 at 19:54
  • $\begingroup$ I find the question pretty clear. You have to prove that $\mathbf T$ is a $2 \choose 0$ tensor . I think, you have to prove that this function is linear in both arguments. But why is this so? $\endgroup$
    – Philipp
    Commented Jan 6, 2020 at 20:09
  • 1
    $\begingroup$ I'm with @K7PEH. I don't understand what you are asking. Do you understand that it should be a tensor and need help with the proof requested by the exercise? Or are you unsure why stress-energy should be a tensor in principal? Or something else completely? $\endgroup$
    – Brick
    Commented Jan 6, 2020 at 20:11
  • $\begingroup$ The exercise is to prove that the equation given above defines a tensor. My question is simply how to prove this. (Based on Schutz' definition of a tensor) $\endgroup$
    – Philipp
    Commented Jan 6, 2020 at 20:24
  • $\begingroup$ Mostly in physics we just check whether something transforms as a tensor then by the fact that a tensor is a multilinear map (over smooth/differentiable/continuous functions or over a field) and hence invariant under change of coordinates (in a certain chart) we argue that it is a tensor. Of course you need to check whether this definition is chart dependent $\endgroup$ Commented Jan 7, 2020 at 15:09

3 Answers 3


EDIT: Incidentally, I'm addressing your last comment. You should be able to fill in the details for a $(2,0)$ tensor. I'm using a $(1,1)$ tensor to show how both vectors and co-vectors transform. And a tensor is by definition a multi-linear mapping from one set of algebraic objects to another.

I don't have a copy Schutz's book so your mileage may vary. And I'm using Latin indices instead of Greek to cut down on the number typesetting key strokes. Typically, one doesn't include the sum symbols since it's understood the Einstein summation rule applies.

And the Lorentz transformations are a separate question. However, I would recommend writing them as $ \Lambda^{\beta}_{\nu}$ to remind yourself that Lorentz transformations aren't tensors - it's just a question of style - it's understood it's not a tensor.

In order for a object to be a tensor, it has to transform like a tensor under a coordinate transformation.

I claim $t$ is a $(1,1)$ tensor, $$\begin{equation}\tag{1}t= t^{i}_{\;j} \frac{\partial}{\partial u_{i}} \otimes du^{j}\end{equation}$$

where $\frac{\partial}{\partial u_{i}}$ is a coordinate vector in the tangent plane, and $du^{j}$ is the co-vector in the coordinate co-tangent plane such that $\langle \frac{\partial}{\partial u_{i}}, du^{j}\rangle=\delta^{j}_{i}$.

The vector and co-vector transform according to the following rules:

$$\frac{\partial}{\partial u_{i}}=\frac{\partial w_{k}}{\partial u_{i}} \frac{\partial}{\partial w_{k}}$$

$$du^{j}=\frac{\partial u_{j}}{\partial w_{l}} dw^{l}$$


$$t= t^{i}_{\;j} \frac{\partial w_{k}}{\partial u_{i}} \frac{\partial}{\partial w_{k}}\otimes\frac{\partial u_{j}}{\partial w_{l}} dw^{l}$$

$$t= t^{i}_{\;j} \frac{\partial w_{k}}{\partial u_{i}} \frac{\partial u_{j}}{\partial w_{l}}\frac{\partial}{\partial w_{k}}\otimes dw^{l}$$

$$\begin{equation}\tag{2}t= t^{k}_{\;l} \frac{\partial}{\partial w_{k}}\otimes dw^{l}\;\end{equation}$$ where $$\begin{equation}\tag{3}t^{k}_{\;l} =t^{i}_{\;j} \frac{\partial w_{k}}{\partial u_{i}} \frac{\partial u_{j}}{\partial w_{l}}\end{equation}.$$

And equation $(1)$ is equivalent to equation $(2)$ and transforms like a $(1,1)$ tensor. Typically, one represents a tensor by its tensor coefficient, namely $t^{k}_{\;l}$, and equation $(3)$ shows how the $(1,1)$ tensor $t$ transforms under a coordinate transform.

  • $\begingroup$ Nice answer, but one comment. Lorentz transformations are basis transformations in Minkowksi space and can be thought of as linear maps between $\mathbb{R}^4$ as a vector space over the reals (with Minkowski inner product). Hence you can regard this as (part of) $\mathrm{End}(\mathbb{R}^4)$. In that sense you can actually regard a basis transformation on a finite-dimensional vector space as a $(1,1)$-tensor since for such a vector space $V$, we have that $V \otimes V^\star \cong \mathrm{End}(V)$ $\endgroup$ Commented Jan 7, 2020 at 15:12
  • $\begingroup$ Thanks for this answer. But I suppose this does not answer my question, because your Ansatz already assumes that 𝐓 is a tensor, a thing I want to have proven. For my example this means: $\mathbf T = T^{\mu\nu} \vec{e}_\mu \otimes \vec{e}_\nu = T^{\mu\nu} \Lambda^{\overline{\alpha}}_{\space\space\mu} \Lambda^{\overline{\beta}}_{\space\space\nu} \vec{e}_{\overline{\alpha}} \otimes \vec{e}_{\overline{\beta}} = T^{\overline{\alpha}\overline{\beta}} \vec{e}_{\overline{\alpha}} \otimes \vec{e}_{\overline{\beta}}$.It follows the transformation for $T^{\overline{\alpha}\overline{\beta}}$ given above $\endgroup$
    – Philipp
    Commented Jan 7, 2020 at 17:24
  • $\begingroup$ But that's a Lorentz transformation - which is a boost and or a rotation - the frames are moving at constant velocities. In a tensor transformation, it's a coordinate transform where both the tensor coefficient the basis vectors change so the tensor remains the same. $\endgroup$ Commented Jan 7, 2020 at 19:40
  • $\begingroup$ And the indices on the Lorentz transformation are for different coordinate systems - where the indices on a tensor refer to the same coordinate system. $\endgroup$ Commented Jan 7, 2020 at 20:29
  • $\begingroup$ @Mathphysmeister: Is it even possible to construct a $(1,1)$ tensor $V\otimes V^{*}$ when the $V$ is at point $p$ and $V^{*}$ is at point $q$ where $p\neq q$? Why do we use a separate symbol $\Lambda(v)^{\mu}_{\nu}$ for each pair of indices? Why not use the tensor product $\Lambda(v)^{\mu \alpha}_{\;\;\;\nu \beta}$? $\endgroup$ Commented Jan 8, 2020 at 21:34

The physical answer is that the stress-energy tensor represents the physical energy and momentum is a particular place. And it does that by doing a linear sum of the stress energy tensor of all the components contributing there. What else could be but such a sum?

Each component has a stress energy tensor that fundamentally comes from its energy-momentum 4-vector. Again, what else could it be?

And those underlying 4-vectors transform like, well, 4-vectors.

If we live in a Lorentz-transforming space-time universe, the stress energy tensor has to have its conventional transformations.

Of course, whether we do live in that a universe is an experimental question.


"Physical" answer inspired by your final sentence. Consider that we have two frames, $S$ and $\tilde{S}$, related to each other via the coordinate transform $\tilde{x}^{\mu} = (\Lambda)^{\mu}_{\nu} x^{\nu}$.

Now consider a 3-element $\Delta$, given by $dS_{\mu}$ in the frame $S$. In a frame, if $dS_{\mu}$ has only a time component, $\Delta$ represents a spatial volume element in that frame. $P^{\nu} = dS_{\mu} T^{\mu \nu}$ is the total Energy-Momentum contained in the $\Delta$.

(I) This quantity must transform covariantly under the above transformation, ie, $\tilde{P}^{\mu} = (\Lambda)^{\mu}_{\nu} P^{\nu}$. This is a result of the way "Energy" and "Momentum" are defined : as generators of spacetime translations. If we allow a coordinate transform that mixes time and components of space in a certain way, "Energy" and components of "Momentum" must mix accordingly (eg, let us say what we call the x axis in $S$ is called the y axis in $\tilde{S}$, accordingly, what we call $P^{x}$ in $S$ must be called $\tilde{P}^{y}$ in $\tilde{S}$ etc.)

(II) Because the 3-element is constructed out of $dx^{\mu}$, its transformation under the above coordinate transform is also understood : $\tilde{dS}_{\mu} = (\Lambda^{-1})_{\mu}^{\nu} dS_{\nu}$

Now, we can conclude the law of transformation for $T^{\mu \nu}$. Point (I) fixes the transformation law for the second index $\nu$, whereas Point (II) fixes the transformation law for the first index $\mu$. Note that two indices in $T^{\mu \nu}$ are not formally the same, after all, this quantity represents the "$\mu$th component of flux of $\nu$th component of Energy-momentum", so it should not be surprising that we had to invoke different arguments for motivating tensorial properties of the two indices.

Anyway. Pulling together everything we have said above, we obtain: $\tilde{T}^{\mu \nu} = \Lambda^{\mu}_{\alpha} \Lambda^{\nu}_{\beta} T^{\alpha \beta}$, which is the required transformation property. As you said, this transformation law is not a mathematical triviality. We do need to think about what we physically mean by "Energy" and "Momentum", and how these transform under the coordinate transformation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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