I am very new to physics and have recently come across the term stress-energy tensor. I am completely clueless as to what this is, and the Wikipedia page seems to confuse me even more. Can someone explain to me just exactly what this might be? Please be patient with me, as this is an entirely new concept to my understanding.

  • 2
    $\begingroup$ have a look here physics.stackexchange.com/q/184042 $\endgroup$ – anna v Sep 10 '18 at 4:23
  • 1
    $\begingroup$ Understanding advanced scientific concepts requires a certain minimum background in physics and mathematics. What specific qualifications do you have to believe that you can be successful in understanding this particular concept? Could it be that you didn't understand the Wiki page simply because you didn't have the necessary background? In this case, what makes you believe that a different explanation would be more productive than studying the prerequisites? $\endgroup$ – safesphere Sep 11 '18 at 7:17
  • $\begingroup$ safe, I am 13. I am just trying to find simplicity in the complicated fields of physics, trying to gain a fundamental understanding of my questions for physics to not be completely stumped in high school. I do not have a background, and that is why you will see in my account posts why I ask for simple answers. I’m sorry if I come across as too advancing, I would just like to know everything. $\endgroup$ – Curious Fish Sep 12 '18 at 0:01

So just to give a first example, when we want to describe how charge flows through a continuum, we invent a charge density field $\vec J$. This at every point specifies the direction that the charge is actually flowing, but also the magnitude of its flow. The idea is that if you took a tiny little surface perpendicular to that flow of tiny little area $dA$ then $|J|~dA$ would be the electric current through that surface. If we instead orient the area to some direction in space, we can use the dot product, $dI= \vec J\cdot d\vec A$.

So a scalar like current gets, for its flow description, a vector flow, plus a unit shift from amps to amps per square meter, $\text{A}\mapsto\text{A}/\text m^2.$

Another thing that can flow is momentum. It is conserved, just like charge is. Well, not just like: it has three directional components that are all independently conserved, and which combine together into a vector. But otherwise, it's conserved locally. We generally think of changes in momentum per unit time as forces, so the units of our flow entity are going to be force per unit area—in other words, pressure.

Engineers also have a context where they have to use units of force per unit area, and it is when you are looking at, say, a steel cable that is under tension. If you think about a cable holding a mass that is near its breaking point, and you were to double its length, the resulting cable would also be near its breaking point, but it would be stretched out from rest about twice as much as the original cable was. By contrast if you were to put two cables in parallel, they would stretch out less, and not be near their breaking point.

So if you wanted them to be microscopically similar to the original, you would want to really double the force that's on them: and that comes from the idea that if you were to look at it a certain way, you would have just two cables next to each other with the original weight on them, side by side, and then you masslessly glue the loads together. If you did this, you can see in your mind's eye that the stretch length from rest would be the same. The engineers' conclusion: to get closer to what is happening microscopically, you first off need to measure stretch as a percentage, not the literal extra length that the cable has stretched, but that amount relative to the rest length of the cable: this is called “strain.” Secondly, you have to divide the load or force by the cross-sectional area that it is pulling on, so that you normalize for how many cables there are in parallel: this is called “stress.”

So we steal that word to describe momentum flows.

Only three facts need to be added. The first is that we are trying to extract a momentum flow, a vector quantity, through a tiny oriented area $d\vec A$: another vector area. So the stress object as a geometrical entity needs to be a rank-2 tensor. That should not be surprising if you see that a scalar needs a vector flow: a vector would stand to reason to have a matrix describing it's flow. The second thing is more surprising, though you can find the derivation in many textbooks: it is that this tensor is symmetric, so rather than having nine independent components, it only has six: three on the diagonal that represent conventional stresses, three off that represent shear stresses.

Finally, we want to insert relativity into this picture. Relativity takes us from a three dimensional Euclidean space to a four dimensional Minkowski space. In the process, it tells us that the time component of our momentum vectors is actually an energy $E/c$. So we're talking about energy flows, too. So if you think about a car traveling down the highway, we would say that it is carrying a lot of momentum down that Highway. But actually, in the car is frame of reference that you are in when you are in the driver's seat, you don't appear to be moving any momentum anywhere: you appear to be just at rest. Relativity says that you actually are moving a sort of momentum, that momentum is your rest energy $E=mc^2$, and it is oriented in the time direction for you. so the car is still moving through spacetime, it's just not moving through the space parts of spacetime.

Similarly if you form the relativistic generalization of the stress tensor we just found, it must be a four-by-four symmetric matrix with 4 diagonal components and 6 off-diagonal components. The three new off diagonal components have to do with momentum flow through time, and just like the car kind of effortlessly moves through time and we just never thought about it, so to momentum density must effortlessly move through time and we just never thought about it, so the momentum density at any given point fills up those 3 components. And that leaves, for the remaining diagonal component, the energy density divided by $c$.

Now it turns out that the flow of momentum in spacetime is what determines the curvature of spacetime. So this little symmetric tensor suddenly becomes hugely important to understanding the cosmos. It should also not be lost on you that for a real system, often the curvature of space-time would be needed in order to understand how the things are moving, to decide how the momentum is flowing. That sort of self-consistency is why practical applications of this theory often require lots of mathematics and big computers.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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