Can the Dot Product of Electric and Magnetic Fields be Proven as Invariant?

[Ed Quanta]

If I were to attempt to prove that the dot product of an electric and magnetic field is invariant under the conditions of Einstein's Special Theory of Relativity, how would I do this? Would the proof be very involved and complicated? Or should I just use hypothetical magnetic and electric fields and demonstrate how the dot product is unchanged when dealing with relativistic frames of reference?

 

[turin]

Originally posted by Ed Quanta
If I were to attempt to prove that the dot product of an electric and magnetic field is invariant under the conditions of Einstein's Special Theory of Relativity, how would I do this?

First of all, what makes you think that the dot product is invariant? I don't know for sure that it is not, but I would imagine that it is not. Did you hear or read this somewhere, or is this your own idea?

Originally posted by Ed Quanta
Would the proof be very involved and complicated?

I would personally use the Faraday tensor (in fact, after I'm done typing, I will do this and see for myself if the dot product is invariant). It shouldn't be too complicated. I will just consider some arbitrary Faraday tensor, Lorentz transform it in an arbitrary direction at an arbitrary speed, extract the components for the E & H fields to calculate the dot products, and then compare the results. I have a feeling the dot product will be inconstent from one frame to the next.

Originally posted by Ed Quanta
Or should I just use hypothetical magnetic and electric fields and demonstrate how the dot product is unchanged when dealing with relativistic frames of reference?

Do you have formulas for how the field vectors change from one frame to the other?

 

[lethe]

Originally posted by Ambitwistor
\vec{E}\cdot\vec{B} and E^2-B^2 are the only quadratic invariants you can build out of the electric and magnetic fields.

i didn t know this but it seems believable... basically, to get a lorentz scalar, we would have to contract F^{\mu\nu}F^{\rho\sigma} with something. i think there are only two choices available \epsilon_{\mu\nu\rho\sigma} and \eta_{\mu\rho}\eta_{\nu\sigma}

the first choice will give you E^2-B^2 and the second E\cdot B (i would imagine. i haven t checked it)

i had a question come up the other day... i was trying to remember the formula for the energy density in terms of the electromagnetic 2 form. i know the lagrangian is F\wedge*F=(E^2-B^2)\text{vol}, and since energy density is nearly the same thing, with a plus instead of a minus, i thought i could make a similar formula.

of course i was mistaken, energy density is not a lorentz covariant object, so there will be no way to write down the energy density from the field strength in a covariant way.

but can you write the stress energy tensor? i have seen the Noether formula for the stress energy tensor, but i was hoping for a concise formula from the field strength 2 form. i am suspecting that that doesn t exist either, since the stress tensor is symmetric...

 

[turin]

Originally posted by lethe
i think there are only two choices available \epsilon_{\mu\nu\rho\sigma} and \eta_{\mu\rho}\eta_{\nu\sigma}

the first choice will give you E^2-B^2 and the second E\cdot B (i would imagine. i haven t checked it)

As I started writing to work things out, things started to look like the dot product may be invariant. That does surprise me, given the way the magnetic field arises from the Faraday tensor. From my cursory inspection, though, I would rather have expected contraction with \epsilon_{\mu\nu\rho\sigma} to give the dot product. I have:

E\cdot B = (1/c)(F⁰¹F³² + F⁰²F¹³ + F⁰³F²¹).

(Sorry, I don't know how to make that math looking text.)

Anyway, there are clearly permutations of the indices that suggest nontrivial elements would arise only in the case of a permutation. Are you sure you have the appropriate contractions identified with the corresponding invariants?

Originally posted by lethe
i know the lagrangian is F\wedge*F=(E^2-B^2)\text{vol}, and since energy density is nearly the same thing, with a plus instead of a minus, i thought i could make a similar formula.

That is certainly above my level. I still don't have a satisfactory understanding of what a Lagrangian is.

 

[lethe]

Originally posted by turin
Are you sure you have the appropriate contractions identified with the corresponding invariants?

no, i had them backwards. you are correct... the Levi-Civita contraction gives the dot product, and the metric contraction gives the Lagrangian.

 

[lethe]

Originally posted by lethe
no, i had them backwards. you are correct... the Levi-Civita contraction gives the dot product, and the metric contraction gives the Lagrangian.

actually, i should have guess more cleverly. the lagrangian should be invariant under the full Lorentz group (i.e. parity transformations, as well as Lorentz rotations), whereas the dot product, since if involves a pseudovector, should transform nontrivially under parity transformations.

the Levi-Civita tensor reverses sign under parity transformations, so i should have been able to guess which is which...

silly me.

 

[Ed Quanta]

I was told to prove the following in my electromagnetic theory class:
1) that the dot product of E and B would be relatvistically invariant
2) that the quantity E^2 - c^2B^2 is relativistically invariant.


We will assume the two relativistic frames S and S' where x'1=x1, x'2=x2,x'3= j(x3-vt) where j = 1/(square root of 1 - B^2) where B= v/c, and t'=j(t-B/cx3).

When defining the electric and magnetic fields in relativistic frames, current density J and charge density p are not separable nor distinct since a static charge distribution in one frame of reference will be a current density in a moving frame.

In the static frame of reference S we will assume a charge density po in a certain volume, were the change in charge dq=po(dV). Thus in the moving frame of reference S', p=poj. This lorentz contraction and thus change of charge density only occurs in the x3 direction since this is the only dimension which is moving in relativistic frame S'. This is all the information we need to calculate E in the frame of reference S.

Since reference S is static, B=0 because there is no current density. So now we can define the current density J (in frame S' only) as (pu,icp) pu corresponds to the three spatial dimensions, while icp corresponds to the dimension of time. My question is this. How do we calculate E and B in the moving frame of reference S' so I can begin to take the dot products of E and B in both references and see if it is indeed invariant. Sorry if I was confusing or unclear.

 

[turin]

Originally posted by Ed Quanta
When defining the electric and magnetic fields in relativistic frames, current density J and charge density p are not separable nor distinct since a static charge distribution in one frame of reference will be a current density in a moving frame.

In the static frame of reference S we will assume a charge density po in a certain volume, were the change in charge dq=po(dV). Thus in the moving frame of reference S', p=poj.

...

This is all the information we need to calculate E in the frame of reference S.

Since reference S is static, B=0 because there is no current density. So now we can define the current density J (in frame S' only) as (pu,icp) pu corresponds to the three spatial dimensions, while icp corresponds to the dimension of time.

I wouldn't do it this way.

Originally posted by Ed Quanta
This lorentz contraction and thus change of charge density only occurs in the x3 direction since this is the only dimension which is moving in relativistic frame S'.

It doesn't matter what direction; density is density.

Originally posted by Ed Quanta
My question is this. How do we calculate E and B in the moving frame of reference S' so I can begin to take the dot products of E and B in both references and see if it is indeed invariant.

Are you allowed to use the Faraday tensor? Just transform that, take the appropriate components, and compare dot products.

E_i = Fⁱ⁰, cB₁ = F³², cB₂ = F¹³, cB₃ = F²¹.

E'_i = F'ⁱ⁰, cB'₁ = F'³², cB'₂ = F'¹³, cB'₃ = F'²¹.

F'^μν = a^μ_αa^ν_βF^{αβ}.

where a^μ_α are the components of the lorentz transformation matrix.

 

[lethe]

Originally posted by turin

Are you allowed to use the Faraday tensor? Just transform that, take the appropriate components, and compare dot products.

if you are allowed to use relativistic tensors, then it can be even easier.

it is a fact that any tensor expression with all indices contracted must be a (restricted) lorentz scalar. thus, just take the tensor expression above, check that it is equal to E*B, and you re done. you don t even have to check how it transforms, since it is guaranteed to be a scalar.

 

[chroot]

Originally posted by lethe
if you are allowed to use relativistic tensors, then it can be even easier.

it is a fact that any tensor expression with all indices contracted must be a (restricted) lorentz scalar.

This makes sense to me -- if you contract with the metric, you get the Langrangian; if you contract with the Levi-Civita tensor, you get the dot product.

I do have to ask, though -- what exactly is a "Lorentz scalar?" What properties does it have beyond those of the usual scalar?

- Warren

 

[lethe]

Originally posted by chroot

I do have to ask, though -- what exactly is a "Lorentz scalar?" What properties does it have beyond those of the usual scalar?

the (restricted) lorentz scalar is invariant under (restricted) lorentz transformations. and, i could compare this with "the usual scalar", as soon as you tell me which usual scalar you mean.

 

[chroot]

Originally posted by lethe
the (restricted) lorentz scalar is invariant under (restricted) lorentz transformations. and, i could compare this with "the usual scalar", as soon as you tell me which usual scalar you mean.

Hmmm... I guess I always thought that ANY scalar was unchanged by Lorentz transforms.

And I suppose I mean scalar, as per MathWorld: "A one-component quantity which is invariant under rotations of the coordinate system."

- Warren

 

[lethe]

Originally posted by chroot

And I suppose I mean scalar, as per MathWorld: "A one-component quantity which is invariant under rotations of the coordinate system."

yeah, that s about right. so Lorentz scalars are scalars that are invariant under rotations in Minkowski space (the set of all such rotations is called the Lorentz group. hence the name)

 

[lethe]

Originally posted by chroot

And I suppose I mean scalar, as per MathWorld: "A one-component quantity which is invariant under rotations of the coordinate system."

- Warren

note that according to this definition, you might think that energy was a scalar.

and it is, under euclidean rotations. but it is not a Lorentz scalar.

once you choose some coordinate system, i can talk about scalars that are invariant under rotations of that coordinate system.

 

[chroot]

Not necessary, I understand the distinction now. Energy is not a scalar when the group of coordinate transformations in question is the Lorentz group. The best you can say about it that it is the zeroth component of the 4-momentum. Not all "one-component quantities" are scalars -- only those that are invariant under the relevant coordinate transformations.

Thanks as always,

- Warren

 

[Ed Quanta]

I guess my main problem is calculating the 16 components of the electromagnetic field tensor. Once I calculate this, I can just multiply it by the matrix |1 0 0 0 | and |1 0 0 0 |.
|0 1 0 0 | |0 1 0 0 |
|0 0 j iB| |0 0 j-iB|
|0 0-iB j| |0 0 iB j|

This is in accordance to the equation F'= lambda(F)lambda'. Once again I apologize for lack of notation. So in other words, once I can calculate electromagnetic field tensor F, I can calculate the relativistically transformed field tensor F'. And then I should be able to take the dot product of E and B in both relativistic frames,correct?

 

[lethe]

Originally posted by chroot
Not necessary, I understand the distinction now. Energy is not a scalar when the group of coordinate transformations in question is the Lorentz group. The best you can say about it that it is the zeroth component of the 4-momentum. Not all "one-component quantities" are scalars -- only those that are invariant under the relevant coordinate transformations.

sometimes a photon with a time-like polarization is called a scalar photon, not because its a Lorentz scalar, but because its a euclidean scalar.

or, when you do Kaluza-Klein reductions, things that are tensors under SO(4,1) reduce to a tensor, a scalar and a vector under SO(3,1). so one mans scalar is another mans zeroth component. it just depends on which coordinates you feel like working with.

 

[chroot]

Originally posted by lethe
or, when you do Kaluza-Klein reductions, things that are tensors under SO(4,1) reduce to a tensor, a scalar and a vector under SO(3,1). so one mans scalar is another mans zeroth component. it just depends on which coordinates you feel like working with.

Yup.

- Warren

 

[turin]

Originally posted by lethe
thus, just take the tensor expression above, check that it is equal to E*B, and you re done.

Wouldn't this be just as complicated?

Another thing: How do you know that this is going to give you the dot product? I mean, I can see that going throught the calculations would show this, but, how do you know to start with the Levi-Civita contraction in the first place? It seems kind of out-of-the-blue to me. Is there some theorem or something?

 

[turin]

Originally posted by Ed Quanta
... once I can calculate electromagnetic field tensor F, ...

If you can use the Faraday tensor, then you should be able to just say that the corresponding components are the components of the electric and magnetic field, without worrying about the sources. There is no need to calculate these components, you just let them be arbitrary.

Originally posted by Ed Quanta
... I can calculate the relativistically transformed field tensor F'. And then I should be able to take the dot product of E and B in both relativistic frames,correct?

This is basically what I had in mind, but I think lethe has a better idea. I don't quite understand how you can justify starting with the contraction with the Levi-Civita tensor, but apparently it does give the desired result.

 

[lethe]

Originally posted by turin
Wouldn't this be just as complicated?

you be the judge

Another thing: How do you know that this is going to give you the dot product? I mean, I can see that going throught the calculations would show this, but, how do you know to start with the Levi-Civita contraction in the first place? It seems kind of out-of-the-blue to me. Is there some theorem or something?

let s have a look:

the Levi-Civita tensor will select all the terms with even permutations of 1234 with a plus sign, and all the terms with odd permutations with a minus sign


<br /> \begin{multline*}<br /> F^{\mu\nu}F^{\rho\sigma}\epsilon_{\mu\nu\rho\sigma}\\<br /> =4(F^{01}F^{23}+F^{02}F^{13}+F^{03}F^{12})=4\mathbf{E}\cdot\mathbf{B}<br /> \end{multline*}<br />

and that s all i need to do, it is guaranteed to be lorentz invariant.

 

[Ed Quanta]

Thank you Turin and everybody. I think I have got it now.

 

[turin]

Originally posted by lethe
the Levi-Civita tensor will select all the terms with even permutations of 1234 with a plus sign, and all the terms with odd permutations with a minus sign

Yeah, I'm with you on this one. It just wouldn't have occurred to me to use the Levi-Civita until after I had written it out once. I was just wondering if there was some other reason behind it besides seeing that it does in fact work. Like, "given a ..., contraction with the Levi-Civita tensor ..." I don't really know exactly what I want to hear about this. I just don't like proofs to come out of the azz.

 

[lethe]

Originally posted by turin
Yeah, I'm with you on this one. It just wouldn't have occurred to me to use the Levi-Civita until after I had written it out once. I was just wondering if there was some other reason behind it besides seeing that it does in fact work. Like, "given a ..., contraction with the Levi-Civita tensor ..." I don't really know exactly what I want to hear about this. I just don't like proofs to come out of the azz.

well, this is pretty much how proofs with tensors go. you start with some tensor expression of some rank and you want some other tensor expression of some other rank. there are only a few valid manipulations you can do to these tensors to accomplish this. contracting indices with another tensor is always something you can do. if the space has a metric, then the metric tensor is available for this purpose. if the space has an orientation, then the Levi-Civita tensor is also available. if the space has a connection, then you can also take the covariant derivative.

there are some other stuffs too, but the problem didn t call for any more involved manipulations. maybe you are very familiar with raising and lowering indices with the metric tensor, and not so familiar with raising and lowering indices with the Levi-Civita. most people are. well, let this thread be a lesson to you.

it is worth noting that when i was considering what possible quadratic scalars i can make out of F^{\mu\nu}, i also had to consider F^{\mu\nu}F^{\rho\sigma}\eta_{\mu\nu}\eta_{\rho\sigma}. this is another possible scalar that i could make out of the field strength, but i didn t mention it above because it vanishes (the field tensor is antisymmetric and hence traceless)

 

[lethe]

Originally posted by turin
but, how do you know to start with the Levi-Civita contraction in the first place? It seems kind of out-of-the-blue to me. Is there some theorem or something?

and, as you saw, i didn t know to start with the Levi-Civita in the first place. in fact, i made a mistake and thought the answer would be to start with the contraction with the metric tensor instead.

it wasn t until you pointed out that i was wrong that i realized this.

even so, i had from the beginning recognized the contraction with the Levi-Civita as one possible way of obtaining a scalar.

and in fact, as i argue above, there is evidence why i should have known before hand to start with Levi-Civita (Levi-Civita changes sign when you reflect it in the mirror, and so does E*B)

 

[lethe]

and i ll just say one more reason why i like my proof a lot:

i have never been able to once and for all put to memory exactly what the elements of the Lorentz transformation look like. if i don t know the elements of the Lorentz transformation, i can t figure out how the electric field and magnetic field transform.

so i can t compare the dot product in two frames, without either going to my book to look up some formulas, or else going to first principles and deriving the form of a Lorentz tranformation (\Lambda^Tg\Lambda=g is how it goes, i think?)

this is a very nice feature of the tensor index notation. all you have to do is figure out how to get rid of all the indices, and you are guaranteed to have a Lorentz scalar.

 

[selfAdjoint]

Lethe, this thread is a lesson to us all.

This

actually, i should have guess more cleverly. the lagrangian should be invariant under the full Lorentz group (i.e. parity transformations, as well as Lorentz rotations), whereas the dot product, since if involves a pseudovector, should transform nontrivially under parity transformations.

is so clear, and so handy, and I never would have thought of it by myself. I have now stored it in my brainbox, and I hope I can find it again when I need it.

Thanks.

 

[turin]

Originally posted by selfAdjoint
Lethe, this thread is a lesson to us all.

This ... is so clear, and so handy, ...

Did you ever expect any less from the man himself?

 

[turin]

Originally posted by lethe
there are only a few valid manipulations you can do to these tensors to accomplish this.

That is the first I have heard of this. What about:

\eta_{\alpha\sigma}\eta_{\beta\tau}...\eta_{\kappa\epsilon}\eta_{\mu\eta}F^{\alpha\beta}F^{\gamma\delta}...F^{\kappa\lambda}F^{\mu\nu}

?

Is this not considered distinct from other contractions involving the metric, or is it considered just another version of the same type of contraction. I don't understand how there can only be a few (or even a finite number of) valid manipulations.

(For some reason, It won't let me edit that itex stuff. There are a few typos in it.)

Originally posted by lethe
if the space has an orientation, then the Levi-Civita tensor is also available. if the space has a connection, then you can also take the covariant derivative.

What does it mean for a space to have an orientation or connection?

Originally posted by lethe
maybe you are very familiar with raising and lowering indices with the metric tensor, and not so familiar with raising and lowering indices with the Levi-Civita.

I am not really "familiar" with anything that has to do with tensors. My major prof is so hung up on the facility of the notation that it has been slow going trying to understand the meaning behind any of this stuff. For instance, I thought that raising and lowering (what he calls "index gymnastics") was defined by contraction with the covariant and contravriant components of the metric tensor. This is the first I've heard to the contrary, and now I am utterly confused. I thought that I understood raising and lowering of the indices of a component to be changing that component between contravariant and covariant. I don't understand how the Levi-Civita symbol/tensor can do this, since it doesn't contain any dot products (or does it?).

Originally posted by lethe
well, let this thread be a lesson to you.

As always.

 

[turin]

Originally posted by lethe
(Levi-Civita changes sign when you reflect it in the mirror, and so does E*B)

I got a hold of an undergrad math meth book by Arfken. In the chapter on tensors, it is said that the Levi-Civita symbol isn't really a tensor, but a pseudo-tensor. Any comments on this, and how it relates to the discussion? Does this mean that the dot product in question isn't really a scalar, but a pseudo-scalar? Is this only the case for the Levi-Civita symbol in 3-D, and in 4-D it is a true tensor?

 

[turin]

Originally posted by lethe
i have never been able to once and for all put to memory exactly what the elements of the Lorentz transformation look like. if i don t know the elements of the Lorentz transformation, i can t figure out how the electric field and magnetic field transform.

I do recognize that your preference is your own and that I shouldn't really concern myself, but I just can't resist my curiosity.

Are you saying that you are willing to commit the components of the Faraday tensor to memory before you are willing to commit the elements of the Lorentz transformation? Am I missing something here? It seems like, either way you do this, if you don't have the Faraday tensor memorized, then you're going to have to look it up. And, I don't understand how it would be so much more effort to look up the Lorentz transformation while you're at it.

But, like I said, your preference isn't my business. I'm just curious. Can you fill me in if I'm missing something here?

 

[lethe]

Originally posted by turin
I do recognize that your preference is your own and that I shouldn't really concern myself, but I just can't resist my curiosity.

Are you saying that you are willing to commit the components of the Faraday tensor to memory before you are willing to commit the elements of the Lorentz transformation?

yeah, you bet! but the Faraday tensor is really easy to remember. here it is: any part of the Faraday field tensor with a 0 in it is electric field, and the other nonzero components are magnetic field. that s all i have to know, and its real easy to remember. i ll never have to look that up in a book!

i think there are some plus or minus signs thrown in there for good measure, but who cares about them?

 

[turin]

Originally posted by lethe
any part of the Faraday field tensor with a 0 in it is electric field, and the other nonzero components are magnetic field. that s all i have to know, and its real easy to remember.

Every time I use it, I have to look up which magnetic field component is which Faraday component. What's your trick for remembering this?

 

[lethe]

Originally posted by turin
I got a hold of an undergrad math meth book by Arfken. In the chapter on tensors, it is said that the Levi-Civita symbol isn't really a tensor, but a pseudo-tensor. Any comments on this, and how it relates to the discussion?

this is exactly what i was saying with the above comment about changing sign in a mirror.

the Levi-Civita changes sign under parity transformation in any number of dimensions

Does this mean that the dot product in question isn't really a scalar, but a pseudo-scalar?

you bet! that s exactly what it means!

Is this only the case for the Levi-Civita symbol in 3-D, and in 4-D it is a true tensor?

any number of dimensions.

 

[lethe]

Originally posted by turin
Every time I use it, I have to look up which magnetic field component is which Faraday component. What's your trick for remembering this?

well, the component has two numbers in it, chosen from 1, 2 and 3. thus it doesn t have one of those three numbers. the number it doesn t have is the number for the component of the magnetic field.

for example, the F^{12} doesn t have a 3, so it must be the three component of B.

again, i think there may be some minus signs in there, but i m not sure, and don t really care anyway.

 

[lethe]

Originally posted by turin
That is the first I have heard of this. What about:

\eta_{\alpha\sigma}\eta_{\beta\tau}...\eta_{\kappa\epsilon}\eta_{\mu\eta}F^{\alpha\beta}F^{\gamma\delta}...F^{\kappa\lambda}F^{\mu\nu}

?

Is this not considered distinct from other contractions involving the metric

yeah, that s fine, but for this problem, i was only considering quadratic forms. usually, in any situation, there is some limit like that on the degree.

, or is it considered just another version of the same type of contraction. I don't understand how there can only be a few (or even a finite number of) valid manipulations.

i hope that s clear now. if you are only looking for quadratic forms, there are only 3 choices (one of which vanishes)

(For some reason, It won't let me edit that itex stuff. There are a few typos in it.)

it probably is letting you edit it, but you are just not seeing the edit show up. so after you edit, make sure you reload the page. the same thing happens to me.

What does it mean for a space to have an orientation or connection?

an orientation is a rule for choosing from two classes of coordinates. like, in Rⁿ, there are a bunch of coordinate systems you can choose, and you can change between any two coordinate systems with some matrix, and sometimes the determinant of that matrix will be positive and sometimes it will be negative. all the coordinate systems that are related by a positive determinant are said to have the same orientation, and the others have the opposite orientation.

to say a space has an orientation just means to choose a preferred class from all these coordinate systems. for example, in R³, the choice x,y,z (in that order) is called the right-handed orientation. if i wanted to change to a new coordinate system with y as my x-axis and x as my y-axis (and z the same), then this would be a left-handed orientation.

so the Levi-Civita tensor is sensitive to which orientation you choose (is it clear why?)

a connection is a rule for comparing tensors at different points. it is necessary for taking derivatives

 

[turin]

Originally posted by lethe
for example, the F^{12} doesn t have a 3, so it must be the three component of B.

again, i think there may be some minus signs in there, but i m not sure, and don t really care anyway.

Thanks, I think I figured it out:

You want ijk to be a cyclic permutation in the scheme cB_i = F^jk.

 

[turin]

Originally posted by lethe
i hope that s clear now. if you are only looking for quadratic forms, there are only 3 choices (one of which vanishes)

Very clear. Quadratic, I'm assuming, refers to the Faraday tensor appearing twice in the product.

Originally posted by lethe
... you can change between any two coordinate systems with some matrix, and sometimes the determinant of that matrix will be positive and sometimes it will be negative. all the coordinate systems that are related by a positive determinant are said to have the same orientation, and the others have the opposite orientation.

Very clear.

Originally posted by lethe
so the Levi-Civita tensor is sensitive to which orientation you choose (is it clear why?)

I'm assuming because the permutations have a sense, so, taking them in the other sense shows up as a minus sign, just like your example with flipping one of the axes.

Originally posted by lethe
a connection is a rule for comparing tensors at different points. it is necessary for taking derivatives

I'm still not quite getting this. Does this have anything to do with R^&alpha;_{&mu;&nu;&beta;} (or &Gamma;^&alpha;_&mu;&nu; or g_&mu;&nu;)? Is this related to topology?

 

[chroot]

Originally posted by turin
I'm still not quite getting this. Does this have anything to do with R^&alpha;_{&mu;&nu;&beta;} (or &Gamma;^&alpha;_&mu;&nu; or g_&mu;&nu;)? Is this related to topology?

The \Gamma^a_{bc} are called "connection coefficients." As lethe said, they provide a means to compare vectors at two different points in a manifold. In other words, they "connect" the tanget space of one point in the manifold with the tangent spaces of neighoring points. This allows the identification of "parallel" vectors at two different points in the manifold.

The connection coefficients appear in the construction of the covariant derivative, which is basically a generalization of the partial derivative that transforms like a tensor (the partial derivative does not). In order for differentiation to obey tensor character, we have to use component differences at the same point in the manifold -- something not done by normal partial differentiation. The connection coefficients allow us to identify a vector at one point in the manifold with a parallel one at a neighboring point -- they define the operation of parallel transport. Thus the covariant derivative uses the component differences between a vector at one point and a vector at a neighboring point by first using the connection coefficients to parallely transport the neighboring vector so that both are defined at the same point.

The covariant derivative of a vector \lambda^a is thus defined as

\lambda^a_{;c} \equiv \frac{\partial \lambda^a}{\partial x^c} + \Gamma^{a}_{bc} \lambda^a

where x^c are the coordinates of the point at which this covariant derivative is taken.

- Warren

 

[lethe]

Originally posted by chroot
The \Gamma^a_{bc} are called "connection coefficients." As lethe said, they provide a means to compare vectors at two different points in a manifold. In other words, they "connect" the tanget space of one point in the manifold with the tangent spaces of neighoring points.

not just the tangent space, you can have a connection for any vector bundle (including the cotangent bundle, and higher products of bundles)

The covariant derivative of a vector \lambda^a is thus defined as

\lambda^a_{;c} \equiv \frac{\partial \lambda^a}{\partial x^c} + \Gamma^{a}_{bc} \lambda^a

where x^c are the coordinates of the point at which this covariant derivative is taken.

this formula, as it stands, is a little confusing, since when we are doing gauge theory, we usually try to use different symbols for the gauge components of the connection than for the spacetime components.

so how about this formula instead:

<br /> D_\mu\lambda^a=\partial_\mu\lambda^a + A^a{}_{b\mu}\lambda^b<br />
when the connection is the metric connection from GR, then the gauge indices really ought to be the same as spacetime indices, (and we use a capital gamma, and call them Christoffel symbols), but when we are talking about electromagnetic gauge theories, the indices are "internal symmetry" indices, not the same as spacetime indices, and we use the symbol A and call these components the vector potential.

but its all really a matter of semantics or notation or whatever.

 

[lethe]

Originally posted by turin
For instance, I thought that raising and lowering (what he calls "index gymnastics") was defined by contraction with the covariant and contravriant components of the metric tensor. This is the first I've heard to the contrary, and now I am utterly confused.

um... yes, don t get confused here, you raise and lower indices with the metric, by definition.

but you can contract indices between any tensors you want.

I thought that I understood raising and lowering of the indices of a component to be changing that component between contravariant and covariant. I don't understand how the Levi-Civita symbol/tensor can do this, since it doesn't contain any dot products (or does it?).

no, the Levi-Civita doesn t contain the metric, only the orientation. so you can t quite raise and lower indices with the Levi-Civita. but like i said, you can contract. for example, F^{\mu\nu}\epsilon_{\mu\nu\rho\sigma} is not the same thing as F_{\rho\sigma}. it is a new tensor (called the dual tensor to the electromagnetic field tensor).

 

[lethe]

Originally posted by turin

I'm still not quite getting this. Does this have anything to do with R^&alpha;_{&mu;&nu;&beta;} (or &Gamma;^&alpha;_&mu;&nu; or g_&mu;&nu;)? Is this related to topology?

yeah, that &Gamma; thingy is a component of the connection (as you have no doubt read by now from other posts). this has to do with geometry, not topology, except inasmuch as topology and geometry are connected.

 

[turin]

Originally posted by lethe
not just the tangent space, you can have a connection for any vector bundle (including the cotangent bundle, and higher products of bundles)

I'm lost with this. We used that classical dynamics book by Jose & Salen this semester in my classical mechanics, and I think this stuff was in there. But I also think that my prof was a mathe-phobe, and he avoided appealing to the notation and termonolgy at every turn. So, this is about all I undertand:

The tangent bundle (called TQ in the book) of say a 1-D problem is the 1-D manifold of q (some curve), with a flat line tangent to the curve at every point to represent q_dot.

The cotangent bundle (calle T*Q in the book) of the same would be the same 1-D manifold but with flat lines attached perpendicular to the curve to represent p.

Do I have anything incorrect here? What is a higher product of a bundle?

Originally posted by lethe
... since when we are doing gauge theory, we usually try to use different symbols for the gauge components of the connection than for the spacetime components.

... when the connection is the metric connection from GR, then the gauge indices really ought to be the same as spacetime indices, ... but when we are talking about electromagnetic gauge theories, the indices are "internal symmetry" indices, not the same as spacetime indices, ...

I'm a little confused. It still looks like your taking the derivatives with respect to the spacetime coordinates. What is the capital D on the L.H.S.? Are you saying that this is a derivative with respect to the components of the potential?

 

[turin]

Originally posted by lethe
for example, F^{\mu\nu}\epsilon_{\mu\nu\rho\sigma} is not the same thing as F_{\rho\sigma}. it is a new tensor (called the dual tensor to the electromagnetic field tensor).

OK, I think I see what you're saying. The indices can literally be lowered by the Levi-Civita, but you don't get F_{\rho\sigma}. Instead you get *F_{\rho\sigma}?

 

[turin]

Originally posted by chroot
The \Gamma^a_{bc} are called "connection coefficients." As lethe said, they provide a means to compare vectors at two different points in a manifold. In other words, they "connect" the tanget space of one point in the manifold with the tangent spaces of neighoring points. This allows the identification of "parallel" vectors at two different points in the manifold.

This is something that confuses the hell out of me in this curved space/Riemannian geometry business. Lately, from classical mechanics, and my discussions with my major prof about GR, it has been hammered into my head that vectors are objects that exist independent of the coordinate system. But then these Christoffel symbols show up and yank the vectors around, stretching and twisting and bending and redirecting. It seems so artificial. If two vectors are parallel, then they are parallel, correct? Why should the components matter? Is it just that we need a way to find these components? In GR, are the vectors really more than 4-D and they only project onto 4-D?

 

[selfAdjoint]

A vector, as a rank 1 tensor shares the fact that tensorial equations are invariant under coordinate changes ("covariant" in a different sense of the overused word). So if a vector is zero in one coordinate system it is zero in every coordinate system. But obviously when you change coordinates the componentsof the vector will change, and there are specific formulas for doing that, one for covariant vectors and one for covariant vectors.

Connection components, aka Christoffel symbols in this case (but take notice of the generalizations mentioned above) are not tensors: they don't change right under new coodinates. So everything you do with them assumes you are for the moment in a fixed frame of reference with a specified coordinate frame. So within that frame{/i] you move the vector parallel to itself in a small (differentially small) closed curve and its direction changes. From the amount of turning you can deduce the connection components and derive the covariant derivative, in that coordinate system. Then you have to prove that unlike the connection coefficients, the covariant derivative is, uh, covariant.

 

[chroot]

Originally posted by turin
If two vectors are parallel, then they are parallel, correct?

Indubitably, but how do you determine whether or not they are parallel?

Think about a sphere, like the Earth -- a positively-curved two-dimensional Riemannian manifold.

Put a vector on the equator somewhere -- say in Africa -- that points due east. Now, put another on the opposite side of earth, also pointing due east. Are these two vectors parallel, or not? They both point in the same direction, east, in one sense -- yet, as viewed from above the north pole, they seem to point in exactly opposite directions!

The bottom line is that the notion of parallelism is non-trivial in curved space. You can't simply look at the vectors and declare them parallel or non-parallel like you can in flat space. You have to have some rigorous way of defining parallelism, and the connection coefficients provide the mechanism. For example, if you slide the vector in Africa around the equator, you can transform it into the vector on the other side of the Earth -- proving they are, in fact, parallel. This "sliding" action, called parallel transport, depends upon some numbers specific to the curvature of the space at each point along the path used for that sliding. These numbers are called connection coefficients. In flat space, the connection coefficients vanish -- they're all zero.

- Warren

 

[lethe]

Originally posted by turin
The tangent bundle (called TQ in the book) of say a 1-D problem is the 1-D manifold of q (some curve), with a flat line tangent to the curve at every point to represent q_dot.

to represent the vector space spanned by q-dot, yes.

The cotangent bundle (calle T*Q in the book) of the same would be the same 1-D manifold but with flat lines attached perpendicular to the curve to represent p.

no, i don t agree with this picture at all. in general, there isn t even any such notion as perpendicularity with things that are not in the tangent space, and sometimes even with things that are int the tangent space.

some GR books like to use a stack of pancakes to represent cotangent vectors, but i never really liked that. so i just say a cotangent vector is a dual vector to a tangent vector, and i don t need to draw a picture at all.

Do I have anything incorrect here? What is a higher product out of a bundle?

like, the tensor product, symmetrized tensor product, or antisymmetrized tensor product of any number of copies of the tangent bundle and the cotangent bundle.

also, you can just have an arbitrary vector bundle that is not built out of the tangent or cotangent bundle at all. just some arbitrary vector space bundle over some manifold. all these bundles admit connections.

I'm a little confused. It still looks like your taking the derivatives with respect to the spacetime coordinates.

I am

What is the capital D on the L.H.S.?

the (covariant) derivative

Are you saying that this is a derivative with respect to the components of the potential?

remember how a derivative works: you have to subtract the value of the thing you are taking the derivative of at two neighbouring points, and then take the limit as these points approach each other.

only thing is, there is no way to subtract one vector from another if they do not live in the same vector space. so a covariant derivative is just a rule that let's you do this. what the rule looks like is completely arbitrary, you can define comparison between neighboring vectors any way you want, as long as you adhere to a few simple axioms: linearity and the Leibniz law for products.

such a linear derivation is called a connection.

no matter what it looks like, i must be able to express it in local coordinates, and that is just what i have done on the right hand side of that equation.

 

[lethe]

Originally posted by turin
OK, I think I see what you're saying. The indices can literally be lowered by the Levi-Civita, but you don't get F_{\rho\sigma}. Instead you get *F_{\rho\sigma}?

that s right. i can take any tensor A^{\mu\nu} and get a new tensor with lowered indices by contracting like this: B_{\mu\nu}=A^{\rho\sigma}t_{\rho\sigma\mu\nu} . if the tensor i use to lower those indices is the metric tensor, then i pretend that A and B are really the same tensor. if the tensor that i used to lower was the Levi-Civita, then i would say that A and B are dual to one another. in the general case, i wouldn t really say they have any important relationship.

in the problem at hand, there were only two natural tensors available in the problem, the metric tensor and the Levi-Civita tensor.

but if i were doing, say, symplectic geometry, then i would have another tensor available, the symplectic form, and i could have used that to construct a scalar as well.

 

[lethe]

Originally posted by turin
This is something that confuses the hell out of me in this curved space/Riemannian geometry business.

the notion of a connection is not specific to Riemannian geometry, so let's just say differential geometry.

Lately, from classical mechanics, and my discussions with my major prof about GR, it has been hammered into my head that vectors are objects that exist independent of the coordinate system.

good! that s a good lesson to learn!

But then these Christoffel symbols show up and yank the vectors around, stretching and twisting and bending and redirecting.

the way vectors are often introduced, one writes their components, and decrees how they must transform under coordinate transformations.

this totally obscures the fact that they are geometric objects completely independent of what their components may be in some basis! i really dislike this definition of a vector.

for the exact same reason, i really dislike defining a connection to be a set of Christoffel symbols. that is how i saw them the first time, and it was really confusing to me!

so now i much prefer to just think of a connection as a linear rule for comparing neighboring vectors in a derivative like way. in other words, forget the component formula you saw above, with the Christoffel symbols, and remember the following rules:

<br /> \begin{align*}<br /> D_{a\mathbf{v}+b\mathbf{w}}s&amp;=aD_{\mathbf{v}}s+bD_{\mathbf{w}}s\\<br /> D_{\mathbf{v}}(s+t)&amp;=D_{\mathbf{v}}s+D_{\mathbf{v}}t\\<br /> D_{\mathbf{v}}as&amp;=\mathbf{v}(a)s+aD_{\mathbf{v}}s<br /> \end{align*}<br />
these are just the linearity requirements and the Leibniz rule. this approach is perhaps more abstract, but i much prefer it.

It seems so artificial. If two vectors are parallel, then they are parallel, correct?

define parallel..

Why should the components matter?

the components don t matter.

Is it just that we need a way to find these components? In GR, are the vectors really more than 4-D and they only project onto 4-D?

in GR, tangent vectors live in a 4D vector space (if you are doing 4D GR)