2
$\begingroup$

I know that tensors are object we use in general relativity to describe phenomenon. They have the property to have the same expression in various coordinates systems.

For example if I take : $\partial_{x_i} A_{j} - \partial_{x_j} A_i$ it is a tensor because if I change the system of coordinates from $\{x\}$ to $\{ y \}$ the quantity will have the same form : $\partial_{x_i} A_j - \partial_{x_j} A_i \rightarrow \partial_{y_i} A'_j - \partial_{y_j} A'_i$.

I have read in various books that it means we describe a quantity that is independant of the system of coordinates. For example I will cite "The geometry of physics" by Frankel : "Tensors are important on manifolds because we are frquently required to construct expressions by using local coordinates, yet we wish our expressions to have intrinsic meaning that all coordinate systems wil agree apon".


But let's imagine I consider $\frac{\partial v^i}{\partial x ^j}$ where $v^i$ is the component of a vector. We know that this quantity is not a tensor.

Why would'nt it describe an "intrinsic" phenomenon ? Okay, it would be the variation of the $i$ component of $v$ according to the direction $e_j$ in the $\{x\}$ coordinate system so it seems to be coordinate dependent. But it could represent a physical phenomenon that "looks like" it in $\{x\}$ and "looks like" differently in another basis but it still would be an intrinsic quantity.

As I understand the philosophy around GR (in fact I didn't really studied it, I am going to, but I know that is is based on tensors), we use tensors because it allows us to write equations the same way in all coordinates systems.

So do we use tensor in GR for thoose "aesthetic" reasons ? Because it would help us to see fundamental symetries easier for example.

If I take the example of newton equation of motion, the differential equation is not the same if I am in polar or cartesian coordinates. And we used Lagrangian mechanics because the equation of motion have the same form in all coordinate systems. And thus we can study easier fundamental properties of motions.

Is it some kind of the same reason that tensor are usefull in GR ?


So, to summarize my questions:

Do we use tensors in GR for aesthetic reasons (we would have the same equation in all coordinates systems). And the advantage of it is to be able to see easier symmetrie or intrinsic properties of the equation : we wouldn't be "biaised" by our specific point of view.

So we use this tool that has the same form in all coordinates system as it will help us to see fundamental properties.

So, what Frankel mean is more that tensors help us to have a point of view independant from the coordinate system on the phenomenon we study rather than tensor have intrinsic meaning. Indeed my example with $\frac{\partial v^i}{\partial x^j}$ has an intrinsic meaning even if it doesn't have the same form in all coordinates system.

Am I right or did I misunderstood things?

$\endgroup$

closed as primarily opinion-based by Kyle Kanos, Jon Custer, sammy gerbil, ZeroTheHero, John Rennie Aug 27 '17 at 10:59

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ @ConfusinglyCuriousTheThird That would make a fine answer; it's not really a great comment. $\endgroup$ – rob Aug 26 '17 at 5:00
  • 2
    $\begingroup$ Newton's laws do not depend on the coordinate system. No physics can ever depend on coordinate system since we are entirely free to use whatever coordinate system we like, and so if laws of physics depended on coordinate system we would be free to change them, which is absurd. $\endgroup$ – tfb Aug 26 '17 at 8:11
  • $\begingroup$ @tfb It depends precisely on what you call "depend on the coordinate system". If your physical predictions will depend of the coordinates system of course it would be absurd. But if the form of the equation depend of the coordinate system it can be valid. In newton law we are doing the derivative of a vector with respect to time. This quantity is not a tensor (it is something like $\frac{\partial v^i}{\partial x^j}$). $\endgroup$ – StarBucK Aug 26 '17 at 12:07
  • $\begingroup$ And this is why I don't understand the argument that says "quantity that don't have the same form in all coordinates system can't be physical". I don't agree with that, it is not because the form of the equation totally changes in different coordinates systems that it can't describe a physical reality. And that's the all point of my question : why do we say that tensor have intrinsic meaning. The example of dv/dt in newton law has intrinsic meaning but it is just if we write it in cartesian and in polar it will not have the same form. $\endgroup$ – StarBucK Aug 26 '17 at 12:10
  • 2
    $\begingroup$ @StarBucK It is a tensor, of course. It is not a tensor in the spacetime of relativity (since newtonian physics does not take place in that spacetime), but it is one in the three-space of newtonian physics. $\endgroup$ – tfb Aug 26 '17 at 13:04
1
$\begingroup$

If you do want to track the change in one component, like the $x$-component, you need to include the one-form that you would use to extract it from the vector explicitly. I.e. to properly talk about the x-component of a vector field $v$ you'd need to take $dx(v)$, which is a properly invariant scalar, and then take the derivative of that. The above formula now correctly carries information about how "measurement of the x-component" works in a coordinate-invariant manner; someone who switches coordinates can pull out the same number you did. An individual vector component is not geometric because its meaning depends on implicit knowledge of the basis that determines the components. If you specify how to pull the desired component out of the vector (through the desired one-form that carries out the operation) this is no longer a problem. If something feels "obviously geometric" but is apparently not a tensor, then the usual problem is that some aspect of what makes it obviously geometric has not been specified by your mathematical object. Usually starting out with a more general tensor and contracting against appropriate basis/dual-basis objects does the trick.


The other problem here is that the correct vector field derivative needs to account for changes in basis as well. A vector is not only given by its components, but its components and the basis they're expressed in. To talk about a geometrically meaningful derivative of vector fields we need to incorporate all this information. We need to take the whole vector field $v^ie_i$ (using the Einstein summation convention) and differentiate it togther:

$$\frac{\partial}{\partial x^j}(v^ie_i)=\frac{\partial v^i}{\partial x^j}e_i+v^i\frac{\partial e_i}{\partial x^j}$$

where I used the product rule to split up the "coefficient derivative" and the yet-to-be-defined "basis derivative". The second term is supposed to represent the rate of change of the ith basis vector in the jth direction. This will just be some other vector, which we can expand in the given basis: $$\frac{\partial e_i}{\partial x^j}=\Gamma^l_{ij}e_l$$ for some coefficients $\Gamma^l_{ij}$. The specification of these will tell us which vector fields count as constant, which may be specified explicitly or computed from some other data, as in the Levi-Civita connection associated to a metric. The sum of these together is

$$\frac{\partial v^i}{\partial x^j}e_i+v^i\Gamma^l_{ij}e_l=(\frac{\partial v^i}{\partial x^j}+v^l\Gamma^i_{lj})e_i$$

where the second equality swaps the l and i labels on the last term. And this is just the covariant derivative of $v$.

$\endgroup$
  • $\begingroup$ I am not very familiar to levi civita connection (I haven't started to read it). But as far as I know it is not a tensor. But is it something that has the same form in all coordinates system ? (but the coefficient of basis change would not be the same as tensor). I need to know this before I can ask my real question :D $\endgroup$ – StarBucK Aug 26 '17 at 12:15
0
$\begingroup$

A vector can be introduced in two ways, either as a quantity that has magnitude and direction, or as the coordinates of a position in space after we have imposed a coordinate system to measure our coordinates.

The advantage of the former is that we do not need to specify a coordinate system, whereas in the latter, we do; further, when we rotate our coordinate system the coordinates of this point changes, but of course the position of this point manifestly has not changed. This means, that in this description we need to supplement out description with a law of transformation which tells us how these coordinates change when we change our coordinate system; typically, in physics tensors are introduced by a generalisation of this.

However, given that vectors have an intrinsic description we can ask is there an intrinsic description of tensors. There is, but it's typically not used in physics textbooks as it requires a lengthy introduction and perhaps, more to the point, it's not traditional.

In brief, first of all there is the tensor product of vector spaces $U$ and $V$ written as $U \otimes V$. Now any manifold $M$ has a tangent bundle $TM$ which is a vector bundle, and it turns out that the usual operations that works for vector spaces such as dualising and tensoring extend in a straight-forward manner to this setting, we simply do the operations fibre-wise, though the technical details can get, well, technical; so, we can also dualise this space to get $T^*M$. Now, writing $T^p_qM= TM \otimes ... TM \otimes T^*M \otimes ... T^*M$ where the first set of factors has $TM$ repeated $p$ times, and the second set has $T^*M$ repeated $q$ times.

This is the tensor bundle, and it's sections over an open set (over which the bundle is trivialisable) can be shown to be exactly those introduced in the usual physics textbooks, essentially by taking two different coordinate systems and showing that we replicate the description of tensors as well as their transformation law.

What this shows, is that like vectors, tensors do have an intrinsic description.

$\endgroup$
  • $\begingroup$ Well, I am not very familiar to bundle, but as far as I understood you explained why a tensor can live without having any coordinate system. My question is thus : am I right if I say that tensor can describe physical reality because they exist independantly of a coordinate system. So the important thing in physics when we use them is not that their expression is the same in all coordinate system but rather that it is a quantity that exist independantly of a coordinate system. So it has "intrinsic" meaning. $\endgroup$ – StarBucK Aug 26 '17 at 12:25
  • $\begingroup$ Indeed, if I take again my example of newton equation of motion, the dv/dt is not a tensor (if we expand it in polar or cartesian the equation will be totally different) but it represents an intrinsic reality. Thus the important thing is to have quantity that lives independantly of coordinates system rather than having the same form in all coordinates system. Thus tensors are just a particular case of "intrinsic" object : they have the same form in all coordinates system but we could not ask this in general $\endgroup$ – StarBucK Aug 26 '17 at 12:29
  • $\begingroup$ @starBuck: "I am not very familiar with to bundles", are you familiar with grammar? It should be "I am not very familiar with bundles". $\endgroup$ – Mozibur Ullah Aug 26 '17 at 13:15
  • $\begingroup$ I wanted to say with the bundles and i miswrote it. $\endgroup$ – StarBucK Aug 26 '17 at 13:41
  • $\begingroup$ @Starbuck: Yes, thats why I don't like to see language being 'dis-respected'. $\endgroup$ – Mozibur Ullah Aug 26 '17 at 13:46

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