Where Can I Find Resources on Einstein's Field Equations and Their Context?

[professor]

could soemone give me a site, or let me know of a book that gives the field equations in the context which Einstein presented them, along with his discussion of their implications?

 

[masudr]

Wrong forum, mate. I assume you mean Einstein Field Equations by EFE.

Besides, if you have little or no knowledge of tensor calculus, Riemannian (and it's cousin the pseudo-Riemannian) manifolds, curvature tensors etc. then the the "EFE"s will have little meaning to you; and putting them in context will require about 50 pages of text. I recommend you look up general relativity books; someone else on this forum is probably better equipped to give you recommendations.

 

[Physics Monkey]

Depending on your level of sophistication, you may find either "Spacetime Physics" by Wheeler and Taylor, "A First Course in General Relativity" by Schutz, or "Gravitation" by Misner, Wheeler, and Thorne helpful for learning about GR. If you want to read to Einstein's original paper, you may find this Dover book interesting . This collection contains several of his original papers on relativity and its consequences, as well as works from several other important physicists and mathematicians. Be warned however, that Einstein uses some outdated notation in parts of his discussion, notation that you will not generally find in more current treatments. Nevertheless, it is enlightening to here things from the horse's mouth once in your life. Hope that helps.

 

[professor]

i have read gravitation, tanks i will look at the other recomendations (the last looks like what I am looking for)- and ohh drat dident realize what thread i was in, i hope some more poeple find this (qm is atleast related so probably..)



- and ohh yes i began my learning of general calculus from my teachers college book in 7'th grade :) - I am only in tenth now, but i do know a thing or two about tensors.. curvature tensors is what I am looking for

 

[professor]

actually i finally ordered that dover book, the forward by lorentz to go with it grabbed me completely... and only $9

 

[pervect]

i have read gravitation, tanks i will look at the other recomendations (the last looks like what I am looking for)- and ohh drat dident realize what thread i was in, i hope some more poeple find this (qm is atleast related so probably..)
- and ohh yes i began my learning of general calculus from my teachers college book in 7'th grade :) - I am only in tenth now, but i do know a thing or two about tensors.. curvature tensors is what I am looking for

If you've got calculus, the main thing you'll need to learn to learn tensors is linear algebra.

Look for a book on Linear algebra that talks about vector spaces, rather than the engineering approach.

Pay particular attention to vectors and dual vectors. The important theorems are that a map of a vector space to a scalar is also a vector space, which is called the "dual" space, and that the dual of a dual recovers the original vector space.

 

[masudr]

How old is tenth grade?

 

[professor]

well I am 15 now :P 16 in march :)


also- a tensor is a triple vector? - or a three dimensional vector (triple probably not the right word)

 

[masudr]

well I am 15 now :P 16 in march :)
also- a tensor is a triple vector? - or a three dimensional vector (triple probably not the right word)

No. A tensor (I think this is the best way to define them for relativity) is defined by the way it transforms under co-ordinate transformations.

EDIT: a scalar is type (0,0) tensor which is invariant under co-ordinate transformations. In SR, we are interested in Lorentz transformations, which is given as a (1,1) tensor \Lambda^\mu\mbox{}_\nu, which you use to transform tensors from one inertial frame to another.

 

[pervect]

well I am 15 now :P 16 in march :)
also- a tensor is a triple vector? - or a three dimensional vector (triple probably not the right word)

well I am 15 now :P 16 in march :)
also- a tensor is a triple vector? - or a three dimensional vector (triple probably not the right word)

A tensor can have an arbitrarily high rank. A vector is a rank 1 tensor, and a matrix is a specific form of a rnak 2 tensor.

The Riemann tensor is a rank 4 tensor, which is the highest rank tensor used in relativity.

One of the best defintions of a tensor is this:

a rank M+N tensor is a linear map from N vectors and M dual-vectors to a scalar. Linearity is applied to each slot in the tensor individually.

see for instance
http://math.ucr.edu/home/baez/gr/outline2.html

Of course, you have to understand what a dual vector is to make use of or appreciate this simple definition, that's what my other post was about.

Suppose you have two vectors - their dot product, A . B, is a rank 2 tensor by defintion. This tensor is usually called the "metric" tensor. It is symmetric because A . B = B . A, i.e. the dot product commutes.

It turns out, by pure logic and the defintion of vectors and dual vectors, that the metric tensor as defined in this way also defines a map from vectors to dual vectors.

I.e. given a map from 2 vectors to a scalar, we can determine a map from
a vector to a (map of a vector to a scalar). This can be done in two ways - if we take A . B, A defines the vector, and the map of (B to a scalar) which is equal to (A . B) defines the dual vector. Or we can let B define the vector, and the map of (A to a scalar) defines the dual. Because of symmetry, the order in which we take the vectors doesn't matter.

This means that the metric tensor can be used to convert vectors into their duals, and vica-versa - or to put it another way, the dot product relation of a vector space which maps two vectors to a scalar defines, by necessity, the mapping of a vector to it's dual.

You may be used to seeing dot products as some vector X = (x,y,z) being

X dot X = x^2 + y^2 + z^2

this works in cartesian coordinates, but not in generalized coordinates (for example, r, theta, phi).

With a non-diagonal metric tensor, the dot product can be expressed in any arbitrary coordinate system. Thus every coordinate system has its own unique defintion of the dot product (metic tensor), and every such defintion of the metric tensor defines a unique relationship between vectors and their duals.

This may seem complicated at first, but it really isn't.

 

[pmb_phy]

could soemone give me a site, or let me know of a book that gives the field equations in the context which Einstein presented them, along with his discussion of their implications?

I wish I knew of a site. Lacking one you may want to see a derivation of them. I whipped this one up at

http://www.geocities.com/physics_world/gr/einsteins_field_equations.htm

Its a cross between Schutz's and Chandrasekhar's derivations. There is an error in it though. Its a small error and I'm probably the only one in the GR world who'd call it such so don't worry about it. It has to do with what the term "relativistic mass" means. The problem does not arise in this page so the point is moot. I just want to point out that its there. To explain would require an entirely new thread and too much work.

Pete

 

[masudr]

I'd seriously recommend Carroll's notes. They're free, and well presented. But I fear you will not understand even the basic concepts as you probably won't have some of the necessary grounding in undergrad physics.

 

[robphy]

These may be helpful with respect to the historical development of the field equations. (I haven't read through them myself yet.)

http://www.mathpages.com/rr/s5-08/5-08.htm
http://www.bun.kyoto-u.ac.jp/~suchii/gen.GR.html

http://philsci-archive.pitt.edu/archive/00002123/
http://www.tc.umn.edu/~janss011/pdf%20files/generalrelativity.pdf
http://www.tc.umn.edu/~janss011/pdf%20files/06-HGR7-031105.pdf
http://www.tc.umn.edu/~janss011/pdf%20files/knot.pdf

 

[professor]

wow you guys are helpfull... and I know of the lorentz transformation, i have used it in thought expirements to calculate time deviation between an observer and the (observee? i guess) , so was i putting this to incorrect use when i tried to plug in some made up velocities and solved for deviation? (such a bad mental block...deviation seems like the wrong word) - actually when i first decided to play around with it, i had got the equation out of something by kip thorne... don't think it was gravitation though


edit: i just looked over the derivation website...thanks for that i had never seen how he (einstien) had come up with that before. I still am having problems with the idea of tensors in the first place... mabye ill post on calc forums for that, or hope my math teacher knows a thing or two about them... he probably will.

 

[pmb_phy]

edit: i just looked over the derivation website...thanks for that i had never seen how he (einstien) had come up with that before. I still am having problems with the idea of tensors in the first place... mabye ill post on calc forums for that, or hope my math teacher knows a thing or two about them... he probably will.

Thanks. Please note that I don't know how Einstein derived his equations. This is how I derived it based on Schutz's derivation as well as Chandrasekhar. I'd imagine that Einstein's wasn't that much different.

For an introduction to tensors please see

http://www.geocities.com/physics_world/ma/intro_tensor.htm

Pete

 

[professor]

ha! I see a reiman tensor explains 4 dimensional space... and is why it is used in the field equations.. I also understand the meaning of when you earlier stated that a rank 3 tensor was a vector as was a rank 1 tensor. and the metric tensor being a dot product... my only question is... is a rank one tensor a scalar or a vector? Pervect and the site seem to be saying different things (though at this point, and in most equations it won't make a huge difference i don't think)

 

[selfAdjoint]

A rank oone tensor is a vector. A scalar is a rank 0 tensor. For low ranks you can display the components, given a basis; rank 2 as a square array that looks like a matrix, rank one as a list that really is a vector, and rank 0 as a single number.

A contravariant tensor of rank k is a multilinear function that takes k tangent vectors into a single vector. A covariant tensor of rank k is a multilinear function that takes k tangent vectors into a coefficient number.

 

[pmb_phy]

A contravariant tensor of rank k is a multilinear function that takes k tangent vectors into a single vector.

Huh!

A contravariant tensor of rank k is a multilinear function that maps k vectors into scalars. The vectors need not be tangent vectors. They could be displacement vectors such as 4-position in special relativity and flat spacetime. This is an important 4-vector in SR. Such a vector is an example of what is called a "Lorentz 4-vector".

A covariant tensor of rank k is a multilinear function that takes k tangent vectors into a coefficient number.

A covariant tensor of rank k is a multilinear function that takes k 1-forms and maps them into scalars.

Pete

 

[pervect]

A contravariant tensor of rank k is a multilinear function that takes k tangent vectors into a single vector. A covariant tensor of rank k is a multilinear function that takes k tangent vectors into a coefficient number.

I don't see how to make this defintion match up with Baez's at

http://math.ucr.edu/home/baez/gr/outline2.html

 

[professor]

thanks... and if anyone isent tired of answering ym wuestions yet, could someone confirm that a lorentz-4 four vector as mentioned above can also be called a reiman curvature tensor (I'm more familiar witht his term)

 

[selfAdjoint]

No, a Lorentz 4-vector is not a Riemannian curvature tensor. The curvature tensor is of rank 4, R^{\alpha}_{\mu\nu\rho} and the 4-vector is only of rank 1, A^{\mu}.

 

[masudr]

What do you think the Riemann curvature tensor is? Due to your lack of knowledge of tensor calculus, I highly doubt that you are familiar with it in any way.

Don't get me wrong, I'm not trying to put you down here at all (but can see that this is how this post can come across), but I'm interested to see how someone can understand what the Riemann curvature tensor is without really knowing what a tensor is, or a "Lorentz 4-vector" (whatever that means).

 

[pervect]

There's a short defintion of the Riemann curvature tensor on Baez's gr tutorial webpage with the general geometric view in mind that uses the notion of parallel transport.

http://math.ucr.edu/home/baez/gr/outline2.html

Parallel transport is not something we've discussed in this thread, but it can be somewhat intuitive.

A more typical use of the Riemann in GR is to calculate tidal forces. You can think of the Riemann as taking in three vectors and putting out a fourth. In the case of tidal forces, two of the three input vectors are the time vector of the observer who is experiencing the tidal force. This vector has to be duplicated because the tensor is a linear machine, and the behavior of tidal forces with time is quadratic and not linear. The third vector is the direction, and the output vector is the force.

It may seem odd to say that time is a vector, but this concept comes out of special relativity.

 

[HallsofIvy]

Be careful about terminology: A "four-vector" is a vector (i.e. rank one tensor in a 4 dimensional space) and has 4 components. The Riemann curvature tensor is a rank 4 tensor. Strictly speaking it can be in any dimensional space but in General Relativity it is still over 4 dimensions (as Einstein said "a 4 dimensional space-time continuum") and so has 4⁴= 256 components.