# Einstein Field Equations?

1. Sep 24, 2010

### JDoolin

Einstein Field Equations???

I have not been able to comprehend the Einstein Field Equation, the Stress Energy Tensor, the Ricci Tensor, the Einstein Tensor, and Christoffel Symbols. Though I am reasonably proficient at working with nested loops in programming, and I have a rudimentary knowledge of partial differential equations, there seems to be no source for spanning the gap between diff-eq, and Tensor calculus, and there also seems to be no source that carefully defines the units and terms of the Tensor calculus.

Am I correct in thinking that there is some analogy with the kinematic equations:

v=dx/dt
a=dv/dt ​

Can one make similar equations with 4-displacements, 4-velocities, and 4-accelerations?

What I'd like to do is get from where I am now, to where I can sort of comprehend the articles on wikipeda on these topics.

Like for instance, I know the chain rule in partial differential equations

$$d f(x,y,z)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz$$​

Is this something that could be expressed in tensor notation?

2. Sep 24, 2010

### stevenb

Re: Einstein Field Equations???

I think you can benefit from watching the Leonard Susskind Lectures on General Relativity, available on youtube. He is someone who has the knack for teaching effectively. Although, these lectures aren't the best source for the basic Tensor calculus information, they give the important intuitive explanations that will make it much easier to consult other more formal sources.

I also recommend the lectures by Alex Maloney (audio and notes available). Here you will get into more details and more formality to get everything rigourously correct. However, without the Susskind Lectures, this may be too big a jump for you, given the background you mentioned.

http://www.physics.mcgill.ca/~maloney/514/

3. Sep 24, 2010

### bcrowell

Staff Emeritus
Re: Einstein Field Equations???

http://www.lightandmatter.com/genrel/" [Broken] is my own shot at introducing tensors from scratch. You might also find it helpful to read Exploring Black Holes, by Taylor and Wheeler. It discusses a lot of GR *without* tensors, and if you can understand all of that physics, you should have a very good foundation for understanding the same material using tensors.

Yes, these have direct analogies in terms of 4-vectors: http://www.lightandmatter.com/html_books/genrel/ch04/ch04.html#Section4.2 [Broken] But there is no real analogy with the Einstein field equations. (Don't know if that's what you were asking.)

To understand the physical content of the field equations, try the Feynman lectures and/or Penrose's The Road to Reality.

Yes. For an example, see http://www.lightandmatter.com/html_books/genrel/ch05/ch05.html#Section5.7 [Broken] , subsection 5.7.2.

Last edited by a moderator: May 4, 2017
4. Sep 24, 2010

### Fredrik

Staff Emeritus
Re: Einstein Field Equations???

If you want to really know this stuff, you will at some point have to study differential geometry. I like Lee's books. "Introduction to smooth manifolds" covers the basics (and a lot more). The stuff about connections, covariant derivatives, parallel transport and curvature is in "Riemannian manifolds: an introduction to curvature".

If C:[a,b]→M is a smooth curve, then its tangent vector at C(t) is denoted by $$\dot C(t)$$ and defined by

$$\dot C(t)f=(f\circ C)'(t)$$

for all smooth functions f from an open neighborhood of C(t) into the real numbers. So the tangent vector of a curve, at a specific point on the curve, is an operator that takes functions to numbers. When C is the mathematical representation of the motion of a particle in a 4-dimensional spacetime, its tangent vector field $$\dot C$$ is called its 4-velocity. (In special relativity, these definitions can be simplified).

There's this calculation:

$$df|_p (v)=v(f)=v(x^i)\frac{\partial}{\partial x^i}\bigg|_p f =dx^i|_p(v)\frac{\partial}{\partial x^i}\bigg|_p f$$

(where v is an arbitrary tangent vector at p), which implies

$$df=\frac{\partial f}{\partial x^i} dx^i$$

but to really explain this requires a lot more time than I'm willing to spend tonight. Some of my earlier posts may help, e.g. this one and this one. Note that $$\{dx^i|_p\}$$ is a basis for the cotangent space at p. Specifically, it's the dual basis of $$\bigg\{\frac{\partial}{\partial x^i}\bigg|_p\bigg\}$$

Last edited: Sep 24, 2010
5. Sep 26, 2010

### JDoolin

Re: Einstein Field Equations???

Hey, I just wanted to say, there's a lot of really good stuff here. It will keep me busy for a while. Thanks to Fredrik, bcrowell, and stevenb

6. Sep 29, 2010

### JDoolin

Re: Einstein Field Equations???

The Leonard Susskind lectures have been really helpful.

Let's see if I've got any of this right:

The covariant tensors (indices downstairs) are like gradients; vector quantities that are functions of position.

The contravariant tensors (indices upstairs) are like differential elements; quantities that represent the positions themselves.

7. Sep 29, 2010

### stevenb

Re: Einstein Field Equations???

Sounds like you're off to a good start.

Just a word of caution about the lectures. They are very streamlined to give you the most critical concepts so that you have a sense of the landscape. From there, it will be much easier to get into the nitty gritty details. As an example, if I remember correctly, he presents the formula for covariant derivative of a covariant tensor with a sign error. There are some other minor mistakes too. Keep in mind that even the best physicists gets signs and factors of two wrong, ... perhaps even more often than the average Joe, for whatever reason.

Consider this lecture series like that "thin book" on a library shelf. One of my favorite professors back in school had a useful piece of advice for his students. He gave it in the form of a question by saying, "If you go to the library to get a book to learn a new subject, which book should you take"? Students would then throw out a bunch of answers until someone finally said, "Pick the thickest book!", to which he would reply, "No, pick the thinnest book."

His point was simply that it's better to start with a brief overview of a subject, hitting the major concepts, before getting into the details. Anyway, the Susskind lectures (even the ones on other topics), are those "thin books" you can read as introductions to new topics. Of course, the book needs to be good as well as thin, but Susskind succeeds there as well, in my opinion.

Last edited: Sep 29, 2010
8. Oct 9, 2010

### JDoolin

Re: Einstein Field Equations???

I haven't had a chance to look at this very much in the last couple of weeks, but maybe something that I was teaching in algebra might apply to rank 0 tensors.

I want to take a function f(x) and stretch it along the y-axis by a factor of V, stretch it along the x-axis by a factor of H, then transpose it along the x and y axis, by x0 and y0.

The end product looks like

$$$g(x)=V f(\frac {1}{H}(x-x_0))+y_0$$$​

Are there contravariant and covariant 0 degree tensors involved here?

9. Oct 9, 2010

### Mentz114

Re: Einstein Field Equations???

What follows may be dumbing down the subject, but it helped me when I was first learning about curved spaces and tensor calculus.

Imagine two axes, representing say, x and y directions. One usually thinks of two straight lines at right angles crossing at the origin x=0, y=0. In this case defining a position vector from the origin to a point x0, y0 is trivial i.e. drop a perpendicular from the point to each axis and read off the value. Now imagine that the axes are not at right angles, and curve smoothly. One can still drop a perpendicular to the axis and read off a value. But there's a nother set of values available - by drawing a line through the point parallel to each axis and reading off the value where this line cuts the other axis.
The two sets of values obtained thus are called the perpendicular projected components, and the parallel projected components.

The motivation for this is that the quantity defined as the inner product of these two 'vectors' is a geometric invariant. It will not change under coordinate transformations.

Writing the perpendicular vector as Xa, and the parallel as Xa, the quantity XaXa=X1X1+X2X2 = L2 is invariant.

This means that if we can write our physical observables in curved space as scalars formed by contracting tensors, they will be covariant under general coordinate transformation.

To summarize : to define an invariant length in curved space, we must allow two ways of expressing positions, the covariant and contravariant vector.

And, the metric of the space, a rank-2 tensor can be used to transform covariant -> contravariant and vice-versa, viz. Xa=gabXb

Last edited: Oct 9, 2010
10. Oct 10, 2010

### JDoolin

Re: Einstein Field Equations???

Okay. Let's see if I can envision an example. If we've got the idea we can curve space, we could turn an arc-lengths of several concentric circles into a parallel straight lines, and we could take the radial lines rom the center, and also make parallel straight lines from them.

Or, we could take hyperbolas and lines from the origin (see attached) and convert them to parallel straight lines. If we can come up with precise mathematical transformations that do these things, then we could discuss which elements are covariant, contravariant, and invariant, based on how the transformation is accomplished, right?

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11. Oct 10, 2010

### atyy

Re: Einstein Field Equations???

Yes, it's basically partial differential equations. Like Maxwell's equations written with the 4-vector potential - the vector potential is not observable, and there are many vector potentials corresponding to the same physical situation or E and B fields. So in addition to the differential equations, there is a rule saying which potentials correspond to the same physical situation. This identification of different solutions as the same is what we call geometry - the true object that remains unchanged by different descriptions of it. In practical calculations, one can gauge fix.

Last edited: Oct 10, 2010
12. Oct 10, 2010

### Mentz114

Re: Einstein Field Equations???

Sorry, I can't see the relevance of the above to what I wrote. A (crude) diagram is attached to illustrate what I'm saying. The key is that the axes are curved. Notice that if the curved axes deform back to the usual Euclidean axes, the parallel (contravariant) and perpendicular ( covariant) components of the vector coincide. So in flat space you can put tensor indexes high or low.

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13. Oct 22, 2010

### JDoolin

Re: Einstein Field Equations???

It's the concepts of invariant, covariant, contravariant that throw me. From Special Relativity I know that distances, times, simultaneity are all observer-dependent quantities.

But proper time and proper distance between events are invariant. However, when you look at the proper time between events in a gravitational field, this is no longer invariant. If you are looking down,you see things moving in slow-motion. Looking up, you see things moving in fast motion.

If you told me we were working on a problem to find the work, and said $$W= \vec{F}\cdot \vec{d}$$.

I will take a guess and say Work is invariant, Force is covariant, and distance is contravariant. Is that right?

What are some examples of invariant quantities, and do these remain invariant in gravitational fields?

14. Oct 22, 2010

### Mentz114

Re: Einstein Field Equations???

If F and d were contravariant and covariant then W would be invariant under coordinate transformations.

The points you raise are about physics in curved space-time, which is not GR. GR is the theory of how matter curves space-time, and it turns out that pseudo-Riemannian geometry is the correct way to formulate it, and the field equations are the 'final' result.

I realize that just seeing how covariant and contravariant vectors arise in curved spacetime is not enough to understand tensors. I may be overplaying the invariance card. You will have to think abstractly to understand the Riemannian manifold, which is about worldlines and the vectors and vector fields associated with them. The kernel of thing is 'curvature', and you've been looking at that in earlier posts.

Two points about physics in curved spacetime. One obvious adjustment is that the distance2 between points (a,b) and (c,d) is no longer (a-c)^2 + (b-d)^2 ( i.e. dx^2+dy^2) but is now given by integrating ds2=Sum gabdxadxb. The most important is that the usual derivative is replaced by the covariant derivative.

A good book on GR is Stephani's "General Relativity : An Introduction to the Theory of the Gravitational Field" (1986) . It really does start easy with 'Force-free motion ... in Newtonian mechanics'.

 cut manifest falsehoods.

Last edited: Oct 22, 2010
15. Oct 22, 2010

### JDoolin

Re: Einstein Field Equations???

Let me ask some basic questions that I'm getting out of Misner/Thorne/Wheeler; MTW.

If you watch an ant walking around an apple, do you really believe that ant thinks its walking in a straight line? Do you think the apple is flat and geometry is somehow curved, or do you think the apple is actually meaningfully round in some global reference frame?

If you see light bending around the sun in a solar eclipse, do you think that the light traveled in a straight line, but space was somehow bent, or do you think that the light actually meaningfully changed directions in a global refence frame?

If you see two satellites travel along geodesic paths and somehow meet at the same point twice, do you think they both traveled straight lines in some warped space-time, or do you think that the satellites actually meaningfully changed directions in a global reference frame?

From the writing, I have this sense that the authors MTW really truly believe there is no global reference frame, and that in each instance, the ant, the light, and the satellites really are moving in straight lines. They manage to convince themselves of this by saying that the light and the satellite feel no forces; thus they must be traveling in straight lines.

These are the arguments they are using to entice me down the rabbit hole, as Einstein said "Why were another seven years required for the construction of the general theory of relativity? The main reason lies in the fact that it is not so easy to free oneself from the idea that coordinates must have an immediate metrical meaning."

I have no such desire to free myself from this idea. To me, any legitimate coordinate system should have immediate metrical meaning. And all coordinate transformations simply change the metrical meanings.

When I think of the distance between two points, I think of the shortest distance between two points. Integrating along the path to find the distance between two points represents the distance traveled along the path between the two points. I don't see how this is a change in space-time. It is a change in what you mean by distance.

16. Oct 22, 2010

### Fredrik

Staff Emeritus
Re: Einstein Field Equations???

Don't take analogies too seriously. You have to imagine a two-dimensional ant and a mathematical apple for this to be really accurate.

It's a "straight line" (technically a geodesic) in spacetime, not space. You need a coordinate system just to define which slice of spacetime to call space, and in the coordinate systems that would be convenient to use in this situation, the path through space is curved.

It sounds like you're asking us to ignore what the theory is saying and tell you what we think actually happens. Since I don't think my intuition is more accurate than general relativity, I can't even tell you what it would mean to do that.

A coordinate system is just a function that assigns a 4-tuple of real numbers to each event.

Yes, and the result is the length of the curve. "Length" is the appropriate word here, not "distance".

In spacetime, the function we're integrating involves a square root of something that can be positive or negative. This means that we can define the "proper length" of a curve only for curves such that the quantity under the square root is positive everywhere on the curve. But we can also define a similar quantity, "proper time", for curves such that the quantity under the square root is negative everywhere on the curve. We just flip the sign of that quantity in the definition, so that we have something positive under the square root.

17. Oct 22, 2010

### JDoolin

Re: Einstein Field Equations???

With some requirements, I hope. You can't just take arbitrary sets of 4 numbers and assign them randomly to the events and call it a coordinate system. The coordinates have to be somehow meaningful.

If you can establish a one-to-one and onto relationship between the coordinate systems then you have a global coordinate system that can be viewed many different and valid ways.

But how do you determine an appropriate way to find the proper length, using an integral of a curve? If you have spacelike separation, there is no valid "path" between the events. In any case, the coordinates of the events are well defined in any given coordinate system. You don't need to do an integral to determine the distance or the time between them.

The space-time interval between events tells you properties only of a hypothethical object, that would have certain properties if it were there. The vast majority of event pairs don't have an object in between them. And the events themselves exist and have locations and times in a global reference frame, describable without doing any integration at all.

18. Oct 22, 2010

### Mentz114

Re: Einstein Field Equations???

If you have a curve y=f(x), the the length along the curve between two points is (as you know)

$$s=\int_{p_0}^{p_1} \sqrt{1+(dy/dx)^2} dx$$

or something similar.

We have to think in terms of curves because worldlines are curves. In Lorentzian space ( t,x) we define the proper length as

$$s=\int_{p_0}^{p_1} \sqrt{1-(dy/dt)^2} dt$$

Most of your remarks are based on flat-space. If the 'axes' of a space are not Euclidean straight lines then there is no canonical distance and you need to define lengths of curve sections.

19. Oct 23, 2010

### JDoolin

Re: Einstein Field Equations???

In your equation you've got events p0, and p1. What if you don't need s? What if you're more interested in what is happening in your own frame of reference, and you're not really concerned about some nonexistent ruler or particle traveling between these events?

For instance, what if p0 and p1 are two supernova explosions, that happen in two different parts of the sky several thousand years apart. Certainly we could calculate the space-time interval between those two events, but what possible use would it have?

It would be an answer to this question: (What is the distance between those events in a reference fram where the two events appear to be simultaneous?) but that question is completely irrelevant unless you happen to be traveling in just the right velocity where the two events were simultaneous.

With the Euclidian distance $$r^2=x^2+y^2+z^2[/itex], there are all kinds of things you can do with it: Work = Force dot distance, Torque =Force Cross Distance, velocity = distance over time, etc, etc. But with this s distance $s^2=t^2-x^2-y^2-z^2$, what good physics comes from this? I think none. I think that t, x, y, z are contravariants. s is invariant. But if you want to talk about velocity, momentum, these are covariant quantities, based on properties of matter within the space. Perhaps it is here where general relativity can shine. But it is not because somehow you warped the space; up down, left, right, forward, backward, future and past, are still defined in a global inertial reference frame. You KNOW you're spinning, so what was once left is now forward. But that's a global change! EVERYTHING that was once to your left is now forward. You can't make the claim that the universe is somehow only "locally Euclidian," when everyday experience tells you that turning your head allows you to see what's over to your left or right--no matter how far away it is. Extreme distances do not free events from the laws of Euclidian rotation. Why should extreme distance free events from the laws of Lorentz Transformation? Yet that's what the argument seems to be in Chapter 1 of Misner Thorne Wheeler. Well, technically, they argue that "the geometry of a sufficiently limited region of spacetime in the real physical world is Lorentzian" and "the geometry of a tiny thumbprint on the apple is Euclidian." (I may be making thehttp://www.jimloy.com/logic/converse.htm" [Broken]). However, with Chapter 7 entitled "Incompatibilty of Gravity and Special Relativity" I suspect they also meant the inverse: that Lorentzian spacetime can only exist in sufficiently limited regions. Last edited by a moderator: May 5, 2017 20. Oct 23, 2010 ### JDoolin Re: Einstein Field Equations??? MTW Chapter 7 Section 3: "... Schhild's [argument uses] a global Lorentz frame tied to the earth's center. It makes no demand that free particles initially at rest remain at rest in this global Lorentz frame" Now I see at least part of the problem. You can't tie a global inertial frame to anything unless it never accelerates. The title of the chapter is "Incompatibility of Gravity and Special Relativity" and it is "Track-2" material, meaning that except for the obvious mistake of trying to tie a global inertial reference frame to an accelerating object, it mostly uses notation that I don't know. 21. Oct 23, 2010 ### Mentz114 Re: Einstein Field Equations??? I don't think you can mean this. Probably none, but only because you've chosen a bad example. How would you calculate how long it takes light to get from the source to an observer ? How would you calculate what time is showing on the observers clocks ? To answer these questions you need to know the worldline of the source, and of the observers, and the metric of the spacetime. The local clock times are integrals of the proper length ( from which no good physics can arise !) of the observers worldlines, and getting the light travel times mean working with null geodesics. Of course, if a galaxy deflects the light things get even more interesting. Because the coordinate speed of light can vary in curved spacetimes, SR is no longer applicable globally. What we have to work with is local frames which are transported along worldlines. We can define these as inertial frames if they are geodesics, provided we don't make the volume they cover too large ( I see you are getting this from MTW). The key to all this is intrinsic curvature. In another thread you started analysing curvature but it's not as straightforward as you think, judging by what you've written. It is not difficult to define the curvature of a line. For a 2D surface embedded in 3D it can also be done, but you obviously need 2 numbers at least to describe the 2D curvature. To define curvature in 4D we have to consider 3D hyperslices, each of which can be sliced into 3 orthogonal embedded 2D sufaces - my point being that it takes a whole bunch of numbers to fully describe intrinsic curvature. Up to 20 in fact. In order for curvature to be covariant it must be expressed as tensors. Keep reading. Although what you've quoted from MTW sounds to me like a waffly attempt to justify what needs no justification. But I've never like fat books that try to explain everything. Good luck with it, it will probably get to the point eventually. Last edited: Oct 23, 2010 22. Oct 23, 2010 ### Fredrik Staff Emeritus Re: Einstein Field Equations??? It has to be a continuous bijection from an open subset of the manifold into an open subset of $\mathbb R^n$, with a continuous inverse. I don't know how you would define "meaningful", but I wouldn't say that coordinates have to be meaningful. Two coordinate systems $x:U\rightarrow\mathbb R^n$ and $y:V\rightarrow\mathbb R^n$ are said to be smoothly compatible if $U\cap V=\emptyset$ or $x\circ y^{-1}$ is smooth. (That means that its partial derivatives exist up to arbitrary order). A set of smoothly compatible coordinate systems is called an atlas. An atlas is said to be maximal if it's not a proper subset of another atlas. The set of coordinate systems on a smooth manifold is required to be a maximal atlas. There are no other requirements than that (unless of course I forgot some minor technicality). Many manifolds don't have a global coordinate system. The simplest example is a sphere. I don't understand the question. The proper length of a curve is defined as an integral. Sure there is. Right, but the coordinate distance and the coordinate time difference between the events depend on the coordinate system. Proper length and proper time depend only on the curve. 23. Nov 5, 2010 ### JDoolin Re: Einstein Field Equations??? Susskind Lecture 5 First ten minutes Struggling my way through this topic. We want to discuss the properties of geometries of various kinds. We can talk about properties of the earth without introducing coordinates. But it is convenient to be able to talk about components of vectors, tensors and so forth are the way to describe properties, in general. We want relationships which are independent of the coordinates. The fact that the earth is almost a sphere is independent of coordinates The fact that Mt. Everest is a lump on the earth is idependent of coordinates Properties are coordinate independent. But what we want to do is express those facts in such a way that the numerical left side and right side do depend on the coordinates, but the fact that they are equal does not depend on the coordinates. If we can write our laws of physics in such a way that when properties are equal in one coordinate system then they are necessarily equal in all coordinate systems, then the physics does not depend on the coordinate system. [tex]W_{nm}(x)=V_{nm}(x) \Leftrightarrow W_{nm}(y)=V_{nm}(y)$$

For instance, we could have
$$V_{nm}(x)= \left( \begin{matrix} V_{xx}(x,y,z) & V_{xy}(x,y,z) & V_{xz}(x,y,z) \\ V_{yx}(x,y,z) & V_{yy}(x,y,z) & V_{yz}(x,y,z) \\ V_{zx}(x,y,z) & V_{zy}(x,y,z) & V_{zz}(x,y,z) \end{matrix} \right)$$and $$V_{nm}(y) = \begin{pmatrix} V_{rr}(r,\theta,\varphi) & V_{r\theta}(r,\theta,\varphi) & V_{r\varphi}(r,\theta,\varphi) \\ V_{\theta r}(r,\theta,\varphi) & V_{\theta \theta}(r,\theta,\varphi) & V_{\theta \varphi}(r,\theta,\varphi) \\ V_{\varphi r}(r,\theta,\varphi) & V_{\varphi \theta}(r,\theta,\varphi) & V_{\varphi \varphi}(r,\theta,\varphi) \end{pmatrix}$$

Obviously $V_{nm}(x) \neq V_{nm}(y)$ but the fact that V=W will be true in both coordinate systems.

The objects that have this property are called tensors.

We'll simplify this question a little bit, just considering a vector instead of a 3X3 tensor.
$$V_n(y) = \frac{\partial x^r}{\partial y^n}V_r(x)$$

Expanded out, this equation looks something like this:

$$\begin{pmatrix} V_r(r,\theta,\varphi) \\ V_\theta(r,\theta,\varphi) \\ V_\varphi(r,\theta,\varphi) \\ \end{pmatrix} =\begin{pmatrix} \frac{\partial x}{\partial r} V_x + \frac{\partial y}{\partial r} V_y+\frac{\partial z}{\partial r} V_z \\ \frac{\partial x}{\partial \theta} V_x + \frac{\partial y}{\partial \theta} V_y+\frac{\partial z}{\partial \theta} V_z \\ \frac{\partial x}{\partial \varphi} V_x + \frac{\partial y}{\partial \varphi} V_y+\frac{\partial z}{\partial \varphi} V_z \end{pmatrix}$$

Another way of saying that $V_{nm}(x)= W_{nm}(x)$ is to say that $V_{nm}(x) - W_{nm}(x)=0$. So I guess, imagine that we're talking about the difference of two Vector Fields that comes out to be zero.

If all of the components of V(x) are zero, then all of the components of V(y) are zero. With transformations like this, (or more complicated ones) then all of the components of V are zero in the y-coordinates. If you know they're all zero in one coordinate, and you know they are tensors, then you know they will be zero in all coordinates.

Not all objects are tensors. We need to have a concept of a derivative of a tensor. Why is it problematic?

To be continued.

Last edited by a moderator: Sep 25, 2014
24. Nov 5, 2010

### JDoolin

Re: Einstein Field Equations???

http://www.youtube.com/watch?v=WtPtxz3ef8U" notes, continued (t=10 minutes to t~25 minutes)

Scalar fields:

We know that a scalar field should be the independent of the coordinate system. i.e.

$$\Phi(y) = \Phi(x)$$
(expanded)
$$\Phi(r,\theta,\varphi) = \Phi(x,y,z)$$
Assuming that x and y correspond to the same point.

We're interested in how the derivative of the scalar field transforms.
$$\frac{\partial \Phi} {dy^n}=\frac{\partial x^n}{\partial y^n} \frac{\partial \Phi}{\partial X^m}$$

(expanded)

$$\begin{pmatrix} \frac{\partial \Phi}{\partial r} \\ \frac{\partial \Phi}{\partial \theta} \\ \frac{\partial \Phi}{\partial \varphi} \\ \end{pmatrix} =\begin{pmatrix} \frac{\partial x}{\partial r} \frac{\partial \Phi}{\partial_x} + \frac{\partial y}{\partial r} \frac{\partial \Phi}{\partial_y}+\frac{\partial z}{\partial r} \frac{\partial \Phi}{\partial_z} \\ \frac{\partial x}{\partial \theta} \frac{\partial \Phi}{\partial_x} + \frac{\partial y}{\partial \theta} \frac{\partial \Phi}{\partial_y}+\frac{\partial z}{\partial \theta} \frac{\partial \Phi}{\partial_z} \\ \frac{\partial x}{\partial \varphi} \frac{\partial \Phi}{\partial_x} + \frac{\partial y}{\partial \varphi} \frac{\partial \Phi}{\partial_y}+\frac{\partial z}{\partial \varphi} \frac{\partial \Phi}{\partial_z} \end{pmatrix}$$

Whatever coordinates I use, if the temperature is constant in this room, then it will be constant in every set of coordinates. So if the derivative of temp in thermal equilibrium is zero in cartesian coordinates, it will also be zero in polar, spherical, or shmoolivitz coordinates.

Let's take a vector field, though, that is constant in space. It has the same direction everywheres and has the same length everywheres. (The obvious meaning of saying that a vector field is constant) Now, what can you say about the components of this vector as you move from point to point in this same coordinates, x^1 and x^2. For example, let's take the covariant components (the projections onto the axes.) It looks like what we are saying is that the covariant components of the vector are zero.

Now let's take exactly the same situation, but take it in curvilinear coordinates. We have a constant vector field, but the covariant components (projection onto the axes) are not constant. Because the axes are moving and bending around. The same vector; exactly the same vector, pointing in exactly the same direction, but the components will be different.

Incidentally, if we changed from one cartesian coordinate system to another, the components will remain constant. But if the coordinates are flopping around as you move around then the components will also change.

So $$\frac {\partial V_m(x)}{\partial x^n} = 0 \xrightarrow[]{does not}\frac {\partial V_m(y)}{\partial y^n}=0$$

21:00
The derivatives of scalars make perfectly good vectors, but the derivatives of the components of vectors do not transform as the components of tensors. You have to do something a little bit trickier to define derivatives of tensors and make them tensors. The process is called covariant differentiation. (The covariant is not because it has to do with a lower index, it is a completely different use of the term covariant.) What it means is that it's a form of differentiation you can apply to tensors and get back tensors; you can apply to vectors and get back vectors.

Let's see if we can figure out what kind of thing a covariant derivative might be. Let's assume something is a tensor and then prove that it's not.

22:45
Supposing there is a tensor in x-coordinates (rectangular cartesian coordinates) that is equal to the derivative of some vector with respect to x in the x-coordinate system.
What must it be in the y-coordinate system? Does it or does it not become the derivative

$$T_{mn}(x)=\frac{\partial V_m(x)}{\partial x}\overset{?}{\rightarrow}\frac{\partial V_m(y)}{\partial x}$$

What must the components be in the y-coordinate system? If it really is a tensor, then the components must be:

$$T_{mn}(y)=\frac{\partial x^r}{\partial y^m}\frac{\partial x^s}{\partial y^n}T_{rs}(x)$$

(expanded with two dimensions, that's)
$$\begin{pmatrix} T_{rr}&T_{r \theta} \\ T_{\theta r} & T_{\theta \theta} \end{pmatrix}= \begin{pmatrix} (\frac{\partial x}{\partial r}\frac{\partial x}{\partial r}T_{xx}+ \frac{\partial x}{\partial r}\frac{\partial y}{\partial r}T_{xy}+ \frac{\partial y}{\partial r}\frac{\partial x}{\partial r}T_{yx}+ \frac{\partial y}{\partial r}\frac{\partial y}{\partial r}T_{yy}) & (\frac{\partial x}{\partial r}\frac{\partial x}{\partial \theta}T_{xx}+ \frac{\partial x}{\partial r}\frac{\partial y}{\partial \theta}T_{xy}+ \frac{\partial y}{\partial r}\frac{\partial x}{\partial \theta}T_{yx}+ \frac{\partial y}{\partial r}\frac{\partial y}{\partial \theta}T_{yy}) \\ (\frac{\partial x}{\partial \theta}\frac{\partial x}{\partial r}T_{xx}+ \frac{\partial x}{\partial \theta}\frac{\partial y}{\partial r}T_{xy}+ \frac{\partial y}{\partial \theta}\frac{\partial x}{\partial r}T_{yx}+ \frac{\partial y}{\partial \theta}\frac{\partial y}{\partial r}T_{yy}) & (\frac{\partial x}{\partial \theta}\frac{\partial x}{\partial \theta}T_{xx}+ \frac{\partial x}{\partial \theta}\frac{\partial y}{\partial \theta}T_{xy}+ \frac{\partial y}{\partial \theta}\frac{\partial x}{\partial \theta}T_{yx}+ \frac{\partial y}{\partial \theta}\frac{\partial y}{\partial \theta}T_{yy}) \end{pmatrix}$$

(I think in three dimensions it would be a 3x3 matrix with nine terms in each one. Getting pretty ugly.)

...to be continued

Last edited by a moderator: Apr 25, 2017
25. Nov 6, 2010

### Blackforest

Re: Einstein Field Equations???

Sorry for intervening in your discussion. But I get myself a fundamental difficulty with that apparently trivial assertion... Since the positioning of the 4-tuple (what we call in fact the geometry) is changing within a GTR approach... is there really a difference between the concept of coordinate system and the concept of event ? I don't know if I am clear enougth. With other words: In a "changing geometry as GTR makes it possible" is not the set of all coordinates an event itself? And is not the deformation of this set also another kind of event?

It was just to trying to develop what JDoolin was perhaps trying to say (?)