View Full Version : Relativistic Invariance
Ed Quanta
Dec5-03, 08:49 PM
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?
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?
Ambitwistor
Dec7-03, 05:20 PM
Originally posted by turin
First of all, what makes you think that the dot product is invariant?
\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.
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...
Ambitwistor
Dec7-03, 05:44 PM
As you say, you can write the electromagnetic stress-energy tensor in terms of the field strength tensor, but not solely by means of operations on differential forms, because the stress-energy tensor isn't a differential form.
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.
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.
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
Dec7-03, 09:08 PM
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.
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.
Ei = 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 = aμαaνβFαβ.
where aμα are the components of the lorentz transformation matrix.
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.
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
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.
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
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)
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.
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
Dec8-03, 11:51 AM
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?
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.
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
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?
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.
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
\begin{multline*}
F^{\mu\nu}F^{\rho\sigma}\epsilon_{\mu\nu\rho\sigma }\\
=4(F^{01}F^{23}+F^{02}F^{13}+F^{03}F^{12})=4\mathb f{E}\cdot\mathbf{B}
\end{multline*}
and that s all i need to do, it is guaranteed to be lorentz invariant.
Ed Quanta
Dec8-03, 05:14 PM
Thank you Turin and everybody. I think I have got it now.
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 signYeah, I'm with you on this one. It just wouldn't have occured 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.
Originally posted by turin
Yeah, I'm with you on this one. It just wouldn't have occured 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\si gma}. 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)
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)
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
Dec9-03, 10:33 AM
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.
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?
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\de lta}...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.
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?
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 gonna 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?
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?
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?
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.
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.
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\de lta}...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 prefered 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
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 cBi = Fjk.
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 Rαμνβ (or Γαμν or gμν)? Is this related to topology?
Originally posted by turin
I'm still not quite getting this. Does this have anything to do with Rαμνβ (or Γαμν or gμν)? 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
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:
D_\mu\lambda^a=\partial_\mu\lambda^a + A^a{}_{b\mu}\lambda^b
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.
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).
Originally posted by turin
I'm still not quite getting this. Does this have anything to do with Rαμνβ (or Γαμν or gμν)? Is this related to topology?
yeah, that Γ 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.
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?
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}?
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
Dec10-03, 01:41 PM
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 [i]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.
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
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 lets 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.
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.
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 lets 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:
\begin{align*}
D_{a\mathbf{v}+b\mathbf{w}}s&=aD_{\mathbf{v}}s+bD_{\mathbf{w}}s\\
D_{\mathbf{v}}(s+t)&=D_{\mathbf{v}}s+D_{\mathbf{v}}t\\
D_{\mathbf{v}}as&=\mathbf{v}(a)s+aD_{\mathbf{v}}s
\end{align*}
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)
Originally posted by selfAdjoint
[i]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. Well, I follow what you're saying to the same extent I've followed it when anyone else has said it. The main thing with which I am uncomfortable is parallel transport. Why would you want to parallel transport a vector when this obviously distorts the vector? It seems like a parallel transported vector isn't really the same vector. Maybe I don't understand what a vector is.
The most popular simplified example that I've seen is the sphere. So parallel transport of a vector is the same as moving an arrow along a great circle while always maintaining the dot product of the arrow with an arrow tangent to the great circle. But this obviously shows that the transported arrow isn't really the same vector as I see it. The arrow obviously changes direction as it moves along a great circle in this manner.
Originally posted by chroot
... 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.Can I not transform any vector into any other vector, proving that every vector is parallel to every other vector?
"Sliding" a vector along some path (a geodesic, for instance) seems to have a different meaning than simply "translating" or "moving" a vector along that same path. It isn't at all clear to me why would want to knowingly change the direction of our vector.
Originally posted by lethe
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.I guess I just got confused by the diagrams. Is not the dual of a vector a set of surfaces, though, or is this flirting too much with the pancake analogy?
Originally posted by lethe
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.Aha. I think things are starting to come together. In the sphere analogy, there is no 3-D space, really? It is useful to visualize, but really we're just talking about a bunch of 2-D tangent spaces, and the 3-D space does not exist? This still makes me a little uncomfortable, do to the readily available 3-D space. I guess there are situations, like space-time, that don't have such a readily available vector space?
Originally posted by lethe
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. Why must it adhere to these axioms? What is the Leibniz law? What are local coordinates (as opposed to "global?" coordinates)?
Ambitwistor
Dec10-03, 08:20 PM
Originally posted by turin
Why would you want to parallel transport a vector when this obviously distorts the vector?
Parallel transport is what you want to do when you don't want to "distort the vector". Parallel transport corresponds to travelling along some path, without letting the vector rotate while you're carrying it.
The most popular simplified example that I've seen is the sphere. So parallel transport of a vector is the same as moving an arrow along a great circle while always maintaining the dot product of the arrow with an arrow tangent to the great circle. But this obviously shows that the transported arrow isn't really the same vector as I see it. The arrow obviously changes direction as it moves along a great circle in this manner.
"Changes direction" relative to what??
Tell me: if I have a vector at the equator pointing north, then which direction should it be pointing at the north pole, in order to "not change direction" from how it was pointing when it was at the equator?
Originally posted by turin
But this obviously shows that the transported arrow isn't really the same vector as I see it.
That's absolutely correct -- and very important! The simple fact is that a vector exists within a vector space. The tangent space at a point P in a manifold is such a vector space. However, another point Q in the manifold has a completely different tangent space, and thus completely different vectors live in it. A vector defined at one point in a manifold doesn't inherently have anything to do with any other vector defined at any other point in the manifold. The vectors are unconnected. There's no way at all to inherently compare a vector at one point P with another at point Q. None at all.
To facilitate the comparison of vectors at P with vectors at Q, you have to introduce some valid mechanism to identify a vector in one tangent space with another (different!) vector in another tangent space. I assume there are lots of ways you can do this -- but the one most often done is to establish the mechanism of parallel transport.
When you parallelly transport a vector from point P to point Q, you are certainly changing the vector -- in fact, you're completely changing the space it lives in. In fact, that's not really possible. What you're doing is destroying the original vector and making an entirely new one in the new space at Q, and just labelling the new one "parallel" to the old one. It's only parellel in the sense that it was found from this operation called parallel transport. Yes, you can see the circularity inherent in the definition.
- Warren
Ambitwistor
Dec10-03, 08:24 PM
Originally posted by turin
Can I not transform any vector into any other vector, proving that every vector is parallel to every other vector?
If you parallel transport a vector along a specific path, then it will end up pointing in one and only one direction, given a connection.
(If you change the connection, then the transported vector can be different: a connection is what defines whether two vectors at different points are parallel, given a path between those points. That's why we say that parallel transport doesn't change the vector: there is no way, other than a connection, to define whether a vector has been changed or not.)
"Sliding" a vector along some path (a geodesic, for instance) seems to have a different meaning than simply "translating" or "moving" a vector along that same path.
Can you define what those words mean? ("Sliding", "translating", and "moving".)
Originally posted by lethe
the notion of a connection is not specific to Riemannian geometry, so lets just say differential geometry.Well, I'll try to remember to respect your wishes on this issue, but neither title means anything to be at all. Apologies for any disrespect to Mr. Riemann or Mr. differential.
Originally posted by lethe
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.I first really understood (I think I understood) what a vector was when I took QM (chapter 1 of Shankar). Are the vectors in QM and the vectors we're talking about now fundamentally different? I suppose the first concrete example of vectors in QM had a coordinatization (the x-basis), but I don't recall ever dealing with parallel transport or anything like that. Is this related to the problem of finding a QM theory of gravity?
Originally posted by lethe
\begin{align*}
D_{a\mathbf{v}+b\mathbf{w}}s&=aD_{\mathbf{v}}s+bD_{\mathbf{w}}s\\
D_{\mathbf{v}}(s+t)&=D_{\mathbf{v}}s+D_{\mathbf{v}}t\\
D_{\mathbf{v}}as&=\mathbf{v}(a)s+aD_{\mathbf{v}}s
\end{align*}
these are just the linearity requirements and the Leibniz rule. this approach is perhaps more abstract, but i much prefer it.I don't quite follow your notation. Apparently, the a and b in the subscript are scalar multiples, the bold v and w are two vectors, the s and t superscripts in the subscript are parameterizations? What is D?
Originally posted by lethe
define parallel.We could go around and around. Whatever I say, I'm pretty sure you will find some term that I use inadequate. Then, I'll have to define that, and inevitably use another term in that definition that you will find inadequate, and on and on, until I get tired of it. But what the hell, I'll give it a shot:
Two vectors are parallel if they point in the same direction.
I have no idea how to define direction, but for some strange reason, I think I know what it means.
Come to think of it, I really can't think of a good definition for "vector" or "point," either.
Originally posted by Ambitwistor
Tell me: if I have a vector at the equator pointing north, then which direction should it be pointing at the north pole, in order to "not change direction" from how it was pointing when it was at the equator? Towards Polaris.
Originally posted by turin
Come to think of it, I really can't think of a good definition for "vector" or "point," either.
[:D] Well, at least we're making you think! lethe is a big stickler on saying things with mathematic precision, and that's valuable, if sometimes a little grating.
However, it's true that you need to learn to walk (to understand the broad nature of what the mathematics does) before you can fly (to understand the rigorous definitions of each of the words). All of this stuff boils down to the simple concept of vector parallelism you learned about in grade school when the connection coefficients go to zero and the space is flat. As a result, I think it's best to learn this stuff by constantly reminding yourself of how this differential geometry can be boiled down to the stuff you already know.
In the end, it really is just a bunch of circular definitions, so try to keep lethe from leading you around the nose by them! [:D]
- Warren
Originally posted by chroot
A vector defined at one point in a manifold doesn't inherently have anything to do with any other vector defined at any other point in the manifold. The vectors are unconnected. There's no way at all to inherently compare a vector at one point P with another at point Q. None at all.
To facilitate the comparison of vectors at P with vectors at Q, you have to introduce some valid mechanism to identify a vector in one tangent space with another (different!) vector in another tangent space. I assume there are lots of ways you can do this -- but the one most often done is to establish the mechanism of parallel transport.Good supplement to lethe's contribution. Why is it that they hammer a connection between vectors into your head in highschool? Or was that just my experience?
Originally posted by chroot
When you parallelly transport a vector from point P to point Q, ... you're ... destroying the original vector and making an entirely new one in the new space at Q, and just labelling the new one "parallel" to the old one.This is an excellent clarification! Thank you chroot!
Ambitwistor
Dec10-03, 08:38 PM
Originally posted by turin
Towards Polaris.
It can't: there is no 3D embedding space. It has to lie within the surface of the sphere.
Originally posted by turin
Towards Polaris.
You're still stuck in three-space! [:D] If you're an ant living on this two-dimensional surface, you can't look up or down. There's no such thing as Polaris. All you can do with your vectors is to push them around the surface. Imagine starting at some point P and pushing a vector along a great circle. After you've walked all the way around the sphere, the vector ends up being the same at P as it was when you started. That's parallel transport.
If you pushed your vector along another path that was not a great circle, you'll find that, much to your dismay, the vector you end up with at P is not the same as the one you started with at P.
The reason all this mathematical machinery is needed is because we need a way to define curvature without reference to some outside, higher-dimensional space. In other words, we need some way to define what the curvature is at a point on the Earth without any reference at all to any points not also on the surface of the Earth. This is the manifold's intrinsic curvature. Parallel transport allows us to determine this intrinsic curvature, without having to look up!
- Warren
Originally posted by Ambitwistor
If you change the connection, then the transported vector can be different: a connection is what defines whether two vectors at different points are parallel, given a path between those points. That's why we say that parallel transport doesn't change the vector: there is no way, other than a connection, to define whether a vector has been changed or not.I think this is becoming much more clear now. Thanks.
Originally posted by Ambitwistor
Can you define what those words mean? ("Sliding", "translating", and "moving".) No. That's why I made the comment. The term, "sliding" seemed to be used ambiguously. In retrospect, I would say it means "parallelly transporting," or "mapping a vector from the vector space tangent to point P to a vector in the vector space tangent to point Q according to the transformation that corresponds to parallel transport (connection rule)." "Translating" and "moving" I intended to be ambiguous as dramatic contrast.
Originally posted by chroot
[:D] Well, at least we're making you think! lethe is a big stickler on saying things with mathematic precision, and that's valuable, if sometimes a little grating.
In the end, it really is just a bunch of circular definitions, so try to keep lethe from leading you around the nose by them! [:D]Actually I rather enjoy the rigor. I aspire for such in my own treatments. I was somewhat interested to see where the definition of "parallel" winds up. But, judging from the most recent posts, I see that it might not be all that interesting.
Originally posted by Ambitwistor
It can't: there is no 3D embedding space. It has to lie within the surface of the sphere. And now I think I appreciate that a lot more.
Originally posted by Ambitwistor
It can't: there is no 3D embedding space. It has to lie within the surface of the sphere.
To add to this, note that a vector, as we've been discussing, lives in the tangent space at a point in the manifold. In other words, if you take a sphere like the Earth, the tangent space at the north pole is a plane that just touches the Earth at that one point -- like resting a piece of cardboard on the top of a basketball. The tangent space to a 2D manifold is also 2D. In other words, a vector pointing from the north pole towards Polaris is not a member of the tangent space at the north pole. That vector can't exist!
- Warren
Thanks to lethe, chroot and ambitwistor. Y'all are many-fold (pun intended) better at helping me get up to speed than my major professor.
I've got this confusion with the sphere still. I'm trying to understand, really understand, what the metric tensor is, and so I'm trying to understand it on the simplest manifold I can think of that would be non-trivial: the sphere. (if anyone knows of a simpler example, then please share!)
So, the first thing my prof wants to do is define a global coordinate system using the usual polar and azimuthal angles. I didn't like this at the time, but couldn't put my finger on why I didn't like it. It seemed preferential in a way to something. Now, I think I can put my finger on it (almost):
1) How can you have this kind of a coordinate system, (θ,φ), which seems to suggest that you know the surface is a sphere in 3-D space, when we aren't supposed to appeal to this higher D space?
2) Why would you use φ as a coordinate, when taking the partial derivative of position with respect to φ clearly does not demonstrate a geodesic (except the equator)? Does it really make sense to use such a coordinate? Does this issue detract from application of the analogy to GR?
3) At the poles, what is the θ direction? I mentioned that there were singularities at the poles, but he said there weren't.
4) Also, near the poles, it doesn't make sense why you would want to use φ as a coordinate, since going in that direction obviously has a gross difference in ds (close to the poles so sinθ << 1) for a given dφ.
Originally posted by chroot
... a vector pointing from the north pole towards Polaris is not a member of the tangent space at the north pole. That vector can't exist!Right. Like saying you have a vector pointing in the w direction in 3-D space, when the only directions you have are x, y, and z. It just doesn't make sense.
Originally posted by turin
I've got this confusion with the sphere still. I'm trying to understand, really understand, what the metric tensor is, and so I'm trying to understand it on the simplest manifold I can think of that would be non-trivial: the sphere. (if anyone knows of a simpler example, then please share!)
The metric tensor is just a machine into which you can plug two vectors. It enables you to measure the lengths of things in the manifold. It allows you to determine the circumference of a large circle of given radius drawn in a curved space, for example.
1) How can you have this kind of a coordinate system, (θ,φ), which seems to suggest that you know the surface is a sphere in 3-D space, when we aren't supposed to appeal to this higher D space?
You can pick any sort of coordinates you like, actually. People use the spherical polar coordinates simply because they appeal to their knowledge of 2-spheres that are embedded in 3-space. It doesn't actually matter.
2) Why would you use φ as a coordinate, when taking the partial derivative of position with respect to φ clearly does not demonstrate a geodesic (except the equator)? Does it really make sense to use such a coordinate? Does this issue detract from application of the analogy to GR?
This really feeds into your next question...
3) At the poles, what is the θ direction? I mentioned that there were singularities at the poles, but he said there weren't.
There absolutely are singularities in the spherical polar coordinate system on a 2-sphere. In fact, there is no way to cover the entire 2-sphere with just one coordinate system without having singularities.
However, you can cover the 2-sphere with two or more coordinate systems -- one that applies to only the northern hemisphere, for example, and a different one that applies only to the southern hemisphere. Where the two coordinate systems overlap, there are functions defined to convert from one to the other.
The resulting "atlas", composed of two different coordinate systems ("charts") stitched together, does not have any singularities. In other words, the singularities that result when you cover the 2-sphere with spherical polar coordinates are coordinate singularities. By picking a better (set) of coordinate system(s) (a better atlas), you can remove them. Some spaces are not so friendly, however. Some spaces contain singularities that cannot be removed by any clever use of charts stitched together.
- Warren
Originally posted by chroot
The metric tensor is just a machine into which you can plug two vectors. It enables you to measure the lengths of things in the manifold. It allows you to determine the circumference of a large circle of given radius drawn in a curved space, for example.From classical mechanics, I gathered that the metric tensor was the machine that calculates dot products. The reason it was introduced, as I understood it, was for use in coordinate systems that were not generally orthonormal, like polar coordinates in 3-D. But this just seemed to be like a using it for different coordinate systems in 3-D (or higher, depending on the number of particles) flat space, and that the coordinate system was curved artificially. In other words, it was allowed to define a vector as an arrow that points from the source to the observation point. Then, the length of that arrow could be found by:
|u|2=gmnumun.
But I have gotten the impression here today that this kind of a vector doesn't make the same kind of sense. It seems like, before you can construct such a vector, you have to define a connection rule, but the connection rule, as I understand it, must be defined in terms of the metric tensor.
Does a particular connection rule have a physical significance? It was said earlier that the connection rule is arbitrary, but, are there certain rules chosen because they have a certain physical significance? Then, do you have to figure out what the metric tensor is from the connection rule. This does seem circular.
Maybe I'll try to reword my confusion with this point in another way:
The metric tensor takes tangent vectors as inputs? These vectors must both be at the same point in the manifold? The vectors don't exist in the manifold, but rather in a space tangent to the manifold at a particular point where all this is goin' down? So then, the metric tensor takes these two objects that do not exist in the manifold, and spits out an object that does exist in the manifold? How can two vectors that are not in the manifold have anything to do with the length of something in the manifold?
Ambitwistor
Dec10-03, 09:49 PM
Originally posted by turin
[some correct statements about the metric]
But I have gotten the impression here today that this kind of a vector doesn't make the same kind of sense.
You can use the metric to take the dot product of vectors and such. It's just that it's not so easy anymore to compare vectors at different points in the manifold, when the manifold is curved (as opposed to curvilinear coordinates in a flat manifold).
It seems like, before you can construct such a vector, you have to define a connection rule, but the connection rule, as I understand it, must be defined in terms of the metric tensor.
You can construct vectors in the tangent space at a point of the manifold, and you can compute their dot products, but you can't compare them directly to vectors in the tangent spaces at other points without a connection.
Connections don't have to be defined in terms of a metric. However, the connection used in general relativity (the Levi-Civita connection) is defined in terms of the spacetime metric.
Does a particular connection rule have a physical significance?
Sure. The connection used in genera relativity has the significance that it preserves the metric (dot product) of vectors under parallel transport: if you parallel-transport two vectors, their lengths and the angle between them remain the same.
Then, do you have to figure out what the metric tensor is from the connection rule.
You can start with the metric, and define a connection from it, or you can go the other way around (assuming the connection you start with can give rise to a metric; not all of them do).
The metric tensor takes tangent vectors as inputs? These vectors must both be at the same point in the manifold? The vectors don't exist in the manifold, but rather in a space tangent to the manifold at a particular point where all this is goin' down?
Yes.
So then, the metric tensor takes these two objects that do not exist in the manifold, and spits out an object that does exist in the manifold?
A metric takes two vectors that exist in a tangent space of the manifold, and spits out a number: their dot product. I wouldn't say that a number is something that "exists in the manifold", although it can characterize properties of the manifold.
How can two vectors that are not in the manifold have anything to do with the length of something in the manifold?
It's safe to think of tangent vectors as lying in the manifold, if you imagine them as infinitesimal.
HallsofIvy
Dec11-03, 07:05 AM
Remember that vectors are derivatives. Two vectors that are not in the manifold can give, for example, a length in the manifold in the same way that, in basic calculus, integrating the tangent vector to a curve can tell you arc length.
Originally posted by turin
Well, I follow what you're saying to the same extent I've followed it when anyone else has said it. The main thing with which I am uncomfortable is parallel transport. Why would you want to parallel transport a vector when this obviously distorts the vector? It seems like a parallel transported vector isn't really the same vector. Maybe I don't understand what a vector is.
its not the same vector. its a parallel transported vector, the rule for parallel transport being whatever i want it to be.
Originally posted by turin
Can I not transform any vector into any other vector, proving that every vector is parallel to every other vector?
you can transform one vector into any other vector you want, but you must single out one of these transformations (per path between points) and call it the connection. then, only those vectors are parallel transported into one another, and others are not.
Originally posted by turin
I guess I just got confused by the diagrams. Is not the dual of a vector a set of surfaces, though, or is this flirting too much with the pancake analogy?
you can call it a "net of surfaces" if you want, as long as i don t hear you use the word perpendicular. there is no canonical notion of perpendicularity
Aha. I think things are starting to come together. In the sphere analogy, there is no 3-D space, really? It is useful to visualize, but really we're just talking about a bunch of 2-D tangent spaces, and the 3-D space does not exist? This still makes me a little uncomfortable, do to the readily available 3-D space. I guess there are situations, like space-time, that don't have such a readily available vector space?
well... there are theorems stating that any n dimensional differentiable manifold can be embedded in R2n. and i think any Riemannian manifold can be embedded isometrically in Rn(n+3)/2 (im not sure if i have that last one correct, but its something like that). that is a theorem proved by russell crowe... err... i mean john nash.
embeddings are hard to find, for some surfaces. they may live in spaces with very high dimension. also, there are many different ways to embed the same manifold.
so defining things in terms of the ambient space can be awkward. also, it makes it hard to tell which properties of your manifold depend only on the manifold, and which properties are particular to the arbitrary way you chose to embed it (intrinsic versus extrinsic properties). for a physicist, this is terrible. for example, physicists are mostly concerned with the manifold that is our spacetime, and it doesn t make physical sense to consider our spacetime as embedded in some higher dimensional space (although sometimes people do do this, see Hawking and Ellis, for example)
so the point is, whether or not there is an ambient space available to our manifold, it is highly desirable to develop all our machinery under the assumption that there is not.
Why must it adhere to these axioms? What is the Leibniz law? What are local coordinates (as opposed to "global?" coordinates)?
well, somewhere along the line, someone thought up the properties he thought a derivative of vectors on a smooth manifold ought to have, and wrote them down. these axioms seem very reasonable to me.
the Leibniz law may be more familiar to you in this form:
\frac{d}{dx}(f(x)g(x))=\frac{df}{dx}g(x)+f(x)\frac {dg}{dx}
the connection must satisfy something like that (if its not obvious how what i wrote above about the connection is the same thing as this rule, that s OK, you may have to study some differential geometry for a while to see that)
and local coordinates are coordinates in some small neighborhood, that don t necessarily extend to the whole manifold (like the polar coordinates on a sphere)
but its not so useful to distinguish local coordinates from global coordinates (except for some trivial examples, no manifolds admit global coordinates)
Originally posted by chroot
manifold. The vectors are unconnected. There's no way at all to inherently compare a vector at one point P with another at point Q. None at all.
none at all? don t forget, the whole point of this discussion is the connection, which is the way to "connect" vectors in different vector spaces. i m sure you knew that, so i don t know why you say "none at all". in fact, there are infinitely many different ways to compare vectors in different tangent spaces.
perhaps you mean "no canonical way" which is certainly true.
Yes, you can see the circularity inherent in the definition.
i can t. and earnestly hope there isn t any.
Originally posted by turin
Well, I'll try to remember to respect your wishes on this issue, but neither title means anything to be at all. Apologies for any disrespect to Mr. Riemann or Mr. differential.
differential geometry is the study of smooth manifolds. Riemannian geometry is the study of smooth manifolds that also have metrics on them.
I first really understood (I think I understood) what a vector was when I took QM (chapter 1 of Shankar). Are the vectors in QM and the vectors we're talking about now fundamentally different? I suppose the first concrete example of vectors in QM had a coordinatization (the x-basis), but I don't recall ever dealing with parallel transport or anything like that. Is this related to the problem of finding a QM theory of gravity?
not really related, except in the sense that both concepts are vector spaces.
but in both cases, there is the issue of coordinate dependence.
I don't quite follow your notation. Apparently, the a and b in the subscript are scalar multiples, the bold v and w are two vectors, the s and t superscripts in the subscript are parameterizations? What is D?
the s and t are vectors (not necessarily tangent vectors, but you may assume that they are tangent vectors if you like).
D is the connection. a connection is any linear operator satisfying the rules i listed. these are the natural rules that characterise the notion of comparing vectors in neighboring vector spaces on a smooth manifold.
Come to think of it, I really can't think of a good definition for "vector" or "point," either.
actually, the definition of vector is a little tricky on a smooth manifold. there are many equivalent choices, my favorite of which is "derivation on the algebra of smooth functions on the manifold". but it will probably take quite some time to see why this definition has anything to do with what you normally think of a vector.
a point is just an element of the underlying set of the manifold.
Originally posted by chroot
[:D] Well, at least we're making you think! lethe is a big stickler on saying things with mathematic precision, and that's valuable, if sometimes a little grating.
think im grating eh?
However, it's true that you need to learn to walk (to understand the broad nature of what the mathematics does) before you can fly (to understand the rigorous definitions of each of the words).
perhaps this is true for you. it is not true for everyone
In the end, it really is just a bunch of circular definitions, so try to keep lethe from leading you around the nose by them! [:D]
time for my mathematical sticklerism to show up: why do you keep saying the definition is circular? a circular definition is actually not a definition at all, and i claim the connection can be consistently defined in terms of the mathematical objects on the set (a linear derivation on sections and functions)
Originally posted by turin
I've got this confusion with the sphere still. I'm trying to understand, really understand, what the metric tensor is, and so I'm trying to understand it on the simplest manifold I can think of that would be non-trivial: the sphere. (if anyone knows of a simpler example, then please share!)
you remember the dot product from vector algebra? well, the metric tensor is just this same old dot product, defined pointwise, in a smooth way, on the manifold for vectors in the tangent space at each point.
this allows you do have concepts on your manifold like perpendicular, area, length, angle, parallel. it also allows you to have an isomorphism between the tangent vectors and their dual vectors. with a notion of length on hand, you can also have a notion of shortest length. i.e. the geodesic.
1) How can you have this kind of a coordinate system, (θ,φ), which seems to suggest that you know the surface is a sphere in 3-D space, when we aren't supposed to appeal to this higher D space?
"aren t supposed to" is not the same thing as "not allowed to"
but um... i don t think he used the 3D space. if he specifies a point by two points (θ and φ), then he is parametrizing the sphere with 2D space.
2) Why would you use φ as a coordinate, when taking the partial derivative of position with respect to φ clearly does not demonstrate a geodesic (except the equator)? Does it really make sense to use such a coordinate? Does this issue detract from application of the analogy to GR?
who cares if its not a geodesic? or, alternatively, maybe i can produce a funny metric on the sphere so that this vector does follow a geodesic.
3) At the poles, what is the θ direction? I mentioned that there were singularities at the poles, but he said there weren't.
there are no intrinsic singularities on the sphere. it is smooth anywhere. you just made a bad choice of coordinates (but remember, there is no such thing as global coordinates, so you can hardly be blamed for your bad choice)
here is an example of the importance of distinguishing properties that are intrinsic from properties that depend on your choice of coordinates or your embedding.
Originally posted by turin
From classical mechanics, I gathered that the metric tensor was the machine that calculates dot products. The reason it was introduced, as I understood it, was for use in coordinate systems that were not generally orthonormal, like polar coordinates in 3-D. But this just seemed to be like a using it for different coordinate systems in 3-D (or higher, depending on the number of particles) flat space, and that the coordinate system was curved artificially.
you can t really tell whether the space is curved until you calculate the curvature. even in spherical coordinates, flat space has zero curvature.
however, you are right, the metric tensor allows you to deal with non orthonormal bases on flat space. it also allows you to deal with nonflat space.
In other words, it was allowed to define a vector as an arrow that points from the source to the observation point. Then, the length of that arrow could be found by:
|u|2=gmnumun.
yeah.
But I have gotten the impression here today that this kind of a vector doesn't make the same kind of sense. It seems like, before you can construct such a vector, you have to define a connection rule, but the connection rule, as I understand it, must be defined in terms of the metric tensor.
you don t need a connection to define the length of a vector. the length of a vector is given by the inner product of that vector with itself.
the connection allows you to compare neighboring vectors. i don t see any relationship between the two.
Does a particular connection rule have a physical significance? It was said earlier that the connection rule is arbitrary, but, are there certain rules chosen because they have a certain physical significance?
no. choice of coordinates is completely arbitrary
Then, do you have to figure out what the metric tensor is from the connection rule. This does seem circular.
there is nothing circular going on here, depsite chroots warnings.
the connection and the metric are completely independent, although if you want the connection s concept of parallel transport to agree with the metrics idea of parallel (inner product = 0), along with preserve length, then there is some consistency relationship they must satisfy. a connection that satisfies this consistency relationship is called a Levi-Civita connection.
Maybe I'll try to reword my confusion with this point in another way:
The metric tensor takes tangent vectors as inputs? These vectors must both be at the same point in the manifold?
yep.
The vectors don't exist in the manifold, but rather in a space tangent to the manifold at a particular point where all this is goin' down?
yep
So then, the metric tensor takes these two objects that do not exist in the manifold, and spits out an object that does exist in the manifold?
no, the metric spits out a number in the real line. this is not in the manifold
How can two vectors that are not in the manifold have anything to do with the length of something in the manifold?
you get vectors in the tangent space by taking the derivative. you can get distances between points on the manifold by reversing that process, i.e. integrating the inner product of vectors.
I have just hit a major learning curve.
Going back to the metric in classical mechanics. So, we only dealt with flat space, so the position vector could make sense as the "arrow" that starts at the origin and terminates at the point at that position. Generally, in flat space, vectors can make sense as starting at some point and terminating at another different point in the flat space, because a flat space is itself a vector space.
That's why I emphasised the coordinate system being curved, while, in actuality, the space is flat. If the space is curved, then the arrow that starts at some point will literally disappear from the manifold before it ever makes it out the door - it doesn't seem to make sense, from what I have gathered, to even have a position vector, or any vector to define the distance between two points in the manifold, no matter how small. No matter how close you choose two points to be, it just seems like an arrow starting at one of them cannot possibly point to the other, because the other isn't even in the same vector space.
Is there some notion that two tangent spaces, given that the points in the manifold to which they are tangent are sufficiently close together, can be said to be sufficiently the same tangent space?
Originally posted by turin
I have just hit a major learning curve.
Hopefully you mean you've reached the summit!
it just seems like an arrow starting at one of them cannot possibly point to the other, because the other isn't even in the same vector space.
Bingo! That's exactly right. Vectors don't exist between points in manifolds -- they exist only at one point. To connect a vector at one point with another vector at another point, you need a set of connection coefficients.
Is there some notion that two tangent spaces, given that the points in the manifold to which they are tangent are sufficiently close together, can be said to be sufficiently the same tangent space?
Yes. In many manifolds, such as the one used in general relativity, the space can be assumed locally flat. For two points close together, the space between them is approximately flat, and the effect of curvature is negligible. This means the metric of the spacetime can be expressed as
g_{\mu\nu} \equiv \eta_{\mu\nu} + h_{\mu\nu}
where the h_{\mu\nu} are small.
How exactly you define "small" depends upon the problem at hand!
- Warren
Ambitwistor
Dec11-03, 02:06 PM
Originally posted by turin
That's why I emphasised the coordinate system being curved, while, in actuality, the space is flat.
There isn't quite a notion of "curvature of a coordinate system" like there is "curvature of a connection", although if the coordinate system is non-Cartesian, people say that it is "curvilinear".
If the space is curved, then the arrow that starts at some point will literally disappear from the manifold before it ever makes it out the door
The way you should visualize a tangent vector is to consider a curve in the manifold between two different points, and then shrink the curve down towards infinitesimal by bringing its endpoints closer together, in the limit that they're brought to the same point. At all times, you are considering something in the manifold itself.
It's just like calculus: the slope of a curve at a point is not defined for any finite separation between two points, but exists only as a limit (of "infinitesimal separation", although you don't have to speak in terms of infinitesimals to make the concept of limit rigorous).
it doesn't seem to make sense, from what I have gathered, to even have a position vector, or any vector to define the distance between two points in the manifold, no matter how small.
You can rigorously make sense of an "infinitesimal" displacement vector in the manifold (i.e., a tangent vector); there are several ways to define it rigorously, including as a differential operator, or as an equivalence class of curves passing through a point without intersecting each other elsewhere.
Originally posted by chroot
Hopefully you mean you've reached the summit!Huh. I wish. I feel like I fell over the cliff. Y'all are a tremendous help, though!
Originally posted by chroot
In many manifolds, such as the one used in general relativity, the space can be assumed locally flat. For two points close together, the space between them is approximately flat, and the effect of curvature is negligible.This is my source of confusion. In light of this, I wonder how the metric tensor could ever be fundamentally different from that of flat space. For some curve, at everypoint on that curve, there is a tangent space where the vectors live. If, at every one of these points on the curve, the space can be assumed locally flat, (and apparently approximated as the tangent space?), then I don't understand why one even talks about the metric.
For instance, I'm sitting here on the surface of the earth. It certainly does look flat to me. I can go forward/backward or right/left. I have two degrees of freedom. If I move to the right 1 meter, then I move along the surface of the earth 1 meter. If I move forward 1 meter, then I move along the surface of the earth 1 meter. The metric seems to be the Kronecker-Delta. I can go to the other side of the world, but the situation seems the same to me. I will still have access to forward/backward and right/left, with the same consequence to my displacement along the surface of the earth. In mathematical notation, I can have a tangent vector of the general form:
dxeright + dyeforward,
where the basis is:
eright,eforward.
The metric is:
(ds)2 = (dx)2 + (dy)2.
From this, I infer that the metric tensor is δmn.
I don't see where the metric tensor for a sphere differs from the metric tensor for a flat plane.
turin,
Sure -- at every point on the earth, the space around you is approximately flat, and it's metric is approximately the Kronecker delta. That's what you're talking about -- you're walking around and measuring the local space around each point, and it's always locally flat.
Now, instead of considering the space at just individual points, try considering it over larger areas. For example, try drawing some circles on the surface of the Earth. To make the idea concrete, stand on the north pole as you draw these circles.
If you draw a small circle, the relationship between its circumference and diameter is approximately your good friend \pi. If you draw a much larger circle around yourself -- say you draw a huge one, the equator -- you'll notice that the ratio of circumference to radius is no longer \pi! This deviation is caused by the metric not being the Kronecker delta. The larger the circle you draw, the more the curvature matters.
- Warren
Ambitwistor
Dec11-03, 03:34 PM
Originally posted by turin
[...]
From this, I infer that the metric tensor is δmn.
I'm not sure that I followed your argument, but it looks to me like you demonstrated that you can find an orthonormal basis at every point in the manifold (a frame field). That's true, but you can't find a coordinate system on a curved space in which the metric components are the Kronecker delta everywhere.
Originally posted by turin
I have just hit a major learning curve.
Going back to the metric in classical mechanics. So, we only dealt with flat space, so the position vector could make sense as the "arrow" that starts at the origin and terminates at the point at that position. Generally, in flat space, vectors can make sense as starting at some point and terminating at another different point in the flat space, because a flat space is itself a vector space.
you don t have to do classical mechanics in R3, or in a linear space at all. for example, you may like to do classical mechanics on the sphere. in which case, you cannot call the position a vector.
it just seems like an arrow starting at one of them cannot possibly point to the other, because the other isn't even in the same vector space.
that s right.
Is there some notion that two tangent spaces, given that the points in the manifold to which they are tangent are sufficiently close together, can be said to be sufficiently the same tangent space?
only in the limit that the two points are actually the same point.
in other words, no.
Originally posted by chroot
Originally posted by turin
Is there some notion that two tangent spaces, given that the points in the manifold to which they are tangent are sufficiently close together, can be said to be sufficiently the same tangent space?
Yes. In many manifolds, such as the one used in general relativity, the space can be assumed locally flat. For two points close together, the space between them is approximately flat, and the effect of curvature is negligible. This means the metric of the spacetime can be expressed as
g_{\mu\nu} \equiv \eta_{\mu\nu} + h_{\mu\nu}
where the h_{\mu\nu} are small.
How exactly you define "small" depends upon the problem at hand!
what on earth does a nearly flat metric have to do with the tangent spaces nearly coinciding (which is what turin asked about)?
Originally posted by Ambitwistor
I'm not sure that I followed your argument, but it looks to me like you demonstrated that you can find an orthonormal basis at every point in the manifold (a frame field). That's true, but you can't find a coordinate system on a curved space in which the metric components are the Kronecker delta everywhere.
i think i already knew this, i mean, its clear, that if the space has nonzero curvature, then there are no coordinates in which the metric is flat.
but... i think i must not understand this idea as well as i should. why can t you just compute the integral curves of the basis vectors of the vielbein and use those to parametrize the space? perhaps these integral curves don t exist, but why not?
Originally posted by lethe
what on earth does a nearly flat metric have to do with the tangent spaces nearly coinciding (which is what turin asked about)?
Oh, give me a break -- you understood what he was asking and you understood my answer, too.
But let's clarify it a bit, since I was speaking without much precision. What I want to know is this: in flat space, you can slide a vector around all you want without changing its direction. In other words, the connection coefficients go to zero. In a flat space, can the tangent spaces at points P and Q be said to be "the same" in some way? Or does each point maintain its own independent tangent space even in flat space?
- Warren
Originally posted by chroot
Oh, give me a break -- you understood what he was asking and you understood my answer, too.
i certainly understood what he was asking, and i believe i also understood your answer, but i fail completely to see what your answer has to do with his question.
In a flat space, can the tangent spaces at points P and Q be said to be "the same" in some way? Or does each point maintain its own independent tangent space even in flat space?
- Warren
here is what you are saying: in flat space, since the transport of the vectors is independent of path, there is a canonical isomorphism between any two tangent spaces, and thus they can be regarded as the same.
fine.
in flat space, everything is dandy, if we were only dealing with flat space, then the intuition we learned about R3 would be sufficient.
but turin was asking a question about smooth manifolds. not manifolds with connections... he wants to know if you can consider neighboring tangent spaces the same, and the answer is simply no! in general not!
insofar as i can approximate the space by a trivial space, i can treat the tangent spaces as approximately equal.
Originally posted by lethe
insofar as i can approximate the space by a trivial space, i can treat the tangent spaces as approximately equal.
Isn't that exactly what I said? lol
- Warren
Originally posted by chroot
Now, instead of considering the space at just individual points, try considering it over larger areas. For example, try drawing some circles on the surface of the Earth. To make the idea concrete, stand on the north pole as you draw these circles.
If you draw a small circle, the relationship between its circumference and diameter is approximately your good friend \pi. If you draw a much larger circle around yourself -- say you draw a huge one, the equator -- you'll notice that the ratio of circumference to radius is no longer \pi! This deviation is caused by the metric not being the Kronecker delta. The larger the circle you draw, the more the curvature matters.I don't interpret what you say to mean I should consider literally drawing circles, but that I should consider mathematical circles in R3 that coincide with the sphere S2 which is imbedded in R3. Otherwise, I can't make sense of doing all of this from the north pole.
1) Isn't suggesting that the equator is a circle to be suggesting that a straight line is a circle? I'm uncomfortable with that.
2) I suppose we can define a circle as the collection of evey point that is the same geodesic (straight line) proper distance away from some given point (i.e. the north pole). In order to "draw" (construct) such a thing, I would have to walk along all possible geodesics that pass through the north pole for a certain distance, r, and stick a flag in the ground or something. I don't understand how I can do that while standing at the north pole.
3) Once I've stuck all my flags in the ground, I could walk around this path of flags, pull them out of the ground as I go, and keep track of how far I walked. When I was done pulling out all the flags, I would compare the distance I walked to r, and find that it is less than 2πr. (The first thing I would think to myself is that I miscounted something)
4. I still don't see how this helps me understand the metric tensor, or, more importantly for my purposes, how to determine its components. No matter what I do, I take the metric tensor with me. Every step I take, I see Kronecker Delta componentes of my metric tensor. Drawing circles seems to instead be demonstrating an Rαμνβ
Originally posted by Ambitwistor
... you can't find a coordinate system on a curved space in which the metric components are the Kronecker delta everywhere. Maybe that's my problem. I don't think I understand a coordinate system on a curved space, at least not without viewing it from the higher D Euclidean space. I am thinking of the coordinate system in terms of the tangent space, because that is what seems the most physical to me. I don't see signs on the highway that say, "Houston, 5 degrees lattidude;" I see signs that say, "Houston, 300 miles." These 300 miles I experience in my own personal tangent plane that I take with me.
Originally posted by lethe
what on earth does a nearly flat metric have to do with the tangent spaces nearly coinciding (which is what turin asked about)? I thought that you could make the metric exactlty flat at a point. I thought that that might mean there is some neighborhood of that point that exists in the tangent space at that point.
Originally posted by lethe
why can t you just compute the integral curves of the basis vectors of the vielbein and use those to parametrize the space?Can you explain to us laymen what you are asking? This looks suspiciously like how I'm trying to picture it except for that "vielbein" thing.
Originally posted by lethe
insofar as i can approximate the space by a trivial space, i can treat the tangent spaces as approximately equal. What does it mean for a space to be trivial?
Originally posted by turin
I thought that you could make the metric exactlty flat at a point.
the word flat doesn t apply to a single point. what you can do is choose an orthonormal basis for the tangent space which makes the metric look like a kronecker delta at a point. but you cannot in general do this in any neighborhood.
I thought that that might mean there is some neighborhood of that point that exists in the tangent space at that point.
the points in the manifold are never in the tangent space. ever.
Originally posted by chroot
Isn't that exactly what I said? lol
make careful note of my usage of the phrase "insofar as". this means that in some cases, such an approximation might apply, only when we have added additional structure to our manifold
in general, the answer is a definite "no", which is a stark contrast to your answer, which was "yes"
you add things to the problem so that you can answer a different question than that which was asked, which makes it irrelevant. why do you introduce a metric when turin asks about tangent spaces?
Originally posted by lethe
the points in the manifold are never in the tangent space. ever. I think a light bulb just came on, though not the one I wanted to.
Points in a manifold are not vectors, period, so it doesn't make sense to say that they would be in any vector space, whether it be a tangent space to the manifold or otherwise? The point of the manifold to which some vector space is tangent is not even in the vector space?
Ambitwistor
Dec11-03, 07:05 PM
Originally posted by turin
Maybe that's my problem. I don't think I understand a coordinate system on a curved space, at least not without viewing it from the higher D Euclidean space.
What's so hard about visualizing lines of latitude and longitude? Why do you need a higher dimensional space?
I am thinking of the coordinate system in terms of the tangent space, because that is what seems the most physical to me.
What do tangent spaces have to do with coordinate systems?
I don't see signs on the highway that say, "Houston, 5 degrees lattidude;" I see signs that say, "Houston, 300 miles."
A sign that says "Houston, 300 miles" doesn't specify the location of Houston, relative to you or anything else. So what does this have to do with coordinate systems?
These 300 miles I experience in my own personal tangent plane that I take with me.
You don't "take a tangent plane with you". A tangent plane is associated with a point in space; it doesn't go anywhere.
Originally posted by turin
Can you explain to us laymen what you are asking? This looks suspiciously like how I'm trying to picture it except for that "vielbein" thing.
i think it is the same thing that you are thinking about. a vielbein, or frame or vierbein or tetrad in 4d, is an orthonormal basis for the tangent space.
you can use the gram-schmidt process to choose such an orthonormal basis at every point in the manifold. Ambitwistor says that this does not imply that you can choose coordinates that will make the space flat, which is a perfectly reasonable statement, since flatness is a coordinate independent property.
but my question is, if i can choose a basis for the tangent bundle that is orthonormal, cannot i just integrate the vectors in this basis to obtain some curves on the manifold, whose tangent vectors are this basis, and therefore coordinates for which the metric is flat?
clearly there is something wrong with my reasoning, since i know that flatness is coordinate independent, but what?
Originally posted by turin
What does it mean for a space to be trivial?
this means different things in different contexts. in this context, i mean euclidean or flat.
Originally posted by turin
Points in a manifold are not vectors, period, so it doesn't make sense to say that they would be in any vector space, whether it be a tangent space to the manifold or otherwise?
that is correct.
The point of the manifold to which some vector space is tangent is not even in the vector space?
correct.
points in a manifold are simply not vectors. you cannot add them or multiply them by numbers, which are the properties that characterize vectors.
just try to imagine a way to add or multiply the points in a sphere. it cannot be done, let alone in the linear fashion that a vector space would require.
so there is associated to each point in the manifold a vector space, but the point is in on sense in the vector space.
Originally posted by Ambitwistor
What's so hard about visualizing lines of latitude and longitude? Why do you need a higher dimensional space?They are not hard for me to visualize, even on a "chart" (I think that's the term I'm looking for, a piece of the manifold that is presented as a flat map). The issue I have with lat. & long. is that, if I look at a map of Scandanavia, for instance, I see this lat. & lon. coordinate system dramatically curved, but I see no good reason for it in the context of the chart. If I were to extend myself into 3-D and look at Scandanavia on the globe, then I would definitely have no problem understanding why these lines were curved, but that entails considering Scandanavia as residing on a globe in R3, which is a higher D space than what I should be considering.
Originally posted by Ambitwistor
What do tangent spaces have to do with coordinate systems?I thought that a coordinate system had a set of basis vectors. I must be very confused about what a coordinate system is.
Originally posted by Ambitwistor
A sign that says "Houston, 300 miles" doesn't specify the location of Houston, relative to you or anything else. So what does this have to do with coordinate systems?It specifies to me that Houston is 300 miles directly in front of me. How much more specific can one be about location? Specifying the distance in degrees of lattitude I find less specific, in the sense that I have a standard for the mile that will tell me how many times my tires will have to rotate before I get there, whereas I don't have a standard for a degree of lattitude that will tell me anything at all about the duration of my trip in terms of tire rotation or anything else. I would have to know what the radius of the Earth was, and that, again, seems to suggest required knowledge about the 3-D situation.
Originally posted by Ambitwistor
You don't "take a tangent plane with you". A tangent plane is associated with a point in space; it doesn't go anywhere. So do I jump from one to another? Where am I in between tangent planes?
Originally posted by lethe
but my question is, if i can choose a basis for the tangent bundle that is orthonormal, ...Does "orthonormal" mean something as opposed to "perpendicular?"
Originally posted by lethe
... cannot i just integrate the vectors in this basis to obtain some curves on the manifold, ...What does it mean to integrate a vector?
Originally posted by lethe
... whose tangent vectors are this basis, and therefore coordinates for which the metric is flat?How did you go go from vectors to coordinates? I tried to suggest going from vectors to coordinates, but apparently I was doing it incorrectly or at least wording it inappropriately.
selfAdjoint
Dec11-03, 07:43 PM
orthonormal means "mutually perpendicular and of magnitude one". Like a set of unit vectors spanning the space. v_i v_j = \delta_j^i
Originally posted by selfAdjoint
orthonormal means "mutually perpendicular and of magnitude one". Like a set of unit vectors spanning the space. v_i v_j = \delta_j^i I'm assuming this is supposed to be a dot product? Don't you need to raise one of the indices to do that? I don't mean to be picky, but I was under the impression that you must contract covariant components with contravariant components, but that lethe found the notion of "perpendicular" distasteful. I just wanted to hear how "orthonormal" does not require a notion of "perpendicular."
Originally posted by turin
I'm assuming this is supposed to be a dot product? Don't you need to raise one of the indices to do that?
yes, if he wants to use the einstein summation notation, one of those indices should be raised.
I don't mean to be picky, but I was under the impression that you must contract covariant components with contravariant components, but that lethe found the notion of "perpendicular" distasteful. I just wanted to hear how "orthonormal" does not require a notion of "perpendicular."
when you have a metric present, the notion of perpendicular is perfectly well acceptable. necessary even. two vectors are perpendicular if their dot product is zero.
but note that you have dual vectors independently of the metric, and that the metric does not allow you to take inner products with a vector and a dual vector, but rather only two vectors (or also two dual vectors).
so it makes no sense to talk about a vector and a dual vector being perpendicular, and that is what i was objecting to before.
Ambitwistor
Dec11-03, 10:15 PM
Originally posted by turin
They are not hard for me to visualize, even on a "chart" (I think that's the term I'm looking for, a piece of the manifold that is presented as a flat map). The issue I have with lat. & long. is that, if I look at a map of Scandanavia, for instance, I see this lat. & lon. coordinate system dramatically curved, but I see no good reason for it in the context of the chart.
Whether they look curved on a chart depends on the chart. If you choose a something like a Mercator projection, where latitude and longitude lines get mapped to straight lines in the Euclidean plane, then they'll look straight. If you don't, they won't.
But what does whether they look straight or curved have to do with your ability to visualize them?
I thought that a coordinate system had a set of basis vectors. I must be very confused about what a coordinate system is.
A coordinate system is a collection of functions on the manifold, each of which assigns a number (a coordinate) to a given point.
You don't need a coordinate system in order to define a tangent space, and you don't need a tangent space in order to define a coordinate system.
It specifies to me that Houston is 300 miles directly in front of me.
I was taking your statement rather literally: most road signs do not indicate that something is directly ahead of you, they just indicate how far away it is (and maybe a rough sense of direction).
How much more specific can one be about location? Specifying the distance in degrees of lattitude I find less specific, in the sense that I have a standard for the mile that will tell me how many times my tires will have to rotate before I get there, whereas I don't have a standard for a degree of lattitude that will tell me anything at all about the duration of my trip in terms of tire rotation or anything else.
Coordinates themselves don't tell you anything about distances. The metric tells you about distances.
So do I jump from one to another? Where am I in between tangent planes?
You're never "between tangent planes". You are always in some tangent plane, because you are always at some point, and every point has a tangent plane associated with it.
Ambitwistor
Dec11-03, 10:19 PM
Originally posted by lethe
why can t you just compute the integral curves of the basis vectors of the vielbein and use those to parametrize the space?
Does your question have anything to do with anything I said?
Originally posted by Ambitwistor
Does your question have anything to do with anything I said?
i think so... you said
Originally posted by Ambitwistor
but it looks to me like you demonstrated that you can find an orthonormal basis at every point in the manifold (a frame field).
ok, we have a bunch of vectors in the tangent space at each point that are orthonormal. this is a vielbein, or frame field.
That's true, but you can't find a coordinate system on a curved space in which the metric components are the Kronecker delta everywhere.
right. a nonflat space shouldn t have a flat metric. i agree, but then i thought of a question...
my question is this: if i have this orthonormal basis at each point, can i not integrate them to find the flow? can i not parametrize along these flows? will this parametrization not give me back the Kronecker delta as the components of the metric in these coordinates?
that would be a contradiction if i were able to do that. so clearly it must not be possible. but on the other hand, i am under the impression that one can always integrate a vector field to find its flow.
so this is my question. i think it has something to do with what you said, but i guess if you don t think so, well.... i must be awfully confused about a lot of things.
Ambitwistor
Dec11-03, 10:52 PM
Originally posted by lethe
my question is this: if i have this orthonormal basis at each point, can i not integrate them to find the flow? can i not parametrize along these flows? will this parametrization not give me back the Kronecker delta as the components of the metric in these coordinates?
Take polar coordinates. The integral curves of the \vec{r} and \vec{\theta} vector fields are lines and circles, respectively. Now replace those fields with their normalized versions \hat{r} and \hat{\theta}. The \hat{r} vector at a point is still tangent to the same line, and the \hat{\theta} vector is still tangent to the same circle, so you could say that those curves are integral curves of the frame field. But if you try to build coordinates out of those curves by constructing their tangent vectors (derivative operators), you don't get the frame field back, you get back \vec{r} and \vec{\theta}, which are coordinate vector fields. The frame field isn't a coordinate vector field; the vectors don't commute with each other. And thus when you try to find the components of the metric in that coordinate system, you're not going to get the Kronecker delta.
Originally posted by Ambitwistor
Take polar coordinates. The integral curves of the \vec{r} and \vec{\theta} vector fields are lines and circles, respectively. Now replace those fields with their normalized versions \hat{r} and \hat{\theta}. The \hat{r} vector at a point is still tangent to the same line, and the \hat{\theta} vector is still tangent to the same circle, so you could say that those curves are integral curves of the frame field.
would you? i suspect two circles would be parametrized differently, and therefore have different magnitudes of the tangent vectors.
But if you try to build coordinates out of those curves by constructing their tangent vectors (derivative operators), you don't get the frame field back, you get back \vec{r} and \vec{\theta}, which are coordinate vector fields.
by the very definition of integral curve of a vector, i must get that vector back as the tangent vector of the integral curve. this is the definition of integral curve, so if this isn t satisfied, this ain t an integral curve.
should i infer that these integral curves don t exist? can you explain to me how come this integral curve does not have the vector it is an integral curve of as tangent?
The frame field isn't a coordinate vector field; the vectors don't commute with each other.
sure, i understand this, but i don t see how this implies your other statements. are you going to say something like the tangent vectors of integral curves of noncommuting vectors fields are commuting vector fields? does noncommutativity somehow make the tangent of a flow of a vector be a different vector?
And thus when you try to find the components of the metric in that coordinate system, you're not going to get the Kronecker delta.
i am going to try a calculation, perhaps.
Originally posted by lethe
... dual vectors independently of the metric, and that the metric does not allow you to take inner products with a vector and a dual vector, but rather only two vectors (or also two dual vectors).OK, now we've uncovered more of my confusion/inconsistency in the way I'm trying to view this stuff.
From QM, I remember that the inner product was a bra multiplying a ket (at least, that's how it was done in Shankar). I got the impression in QM that the bra and ket are literally different objects, even if it is the bra version and ket version of the same state. There was never a metric idea introduced, but orthogonality was still an important issue.
From classical mechanics, I remember that any vector has two different kinds of components and two different kinds of bases, contravariant and covariant. The contravariant basis is like arrows, and the covariant basis is like a set of planes (stack of pancakes). Taking a dot product (which I'm still trying to decide if this is really the same as an "inner product") of two vectors is summing the products of the contravariant components of one of them with the covariant components of the other. If the dot product is zero, I interpretted it as the arrow being parallel to the planes in the set of planes. The dot product is maximum when the arrow is perpendicular to the set of planes (so that it passes through the maximum number of planes).
Are you saying that my interpretations from classical mechanics are inappropriate? Should I think about it more in terms of the way I think about the inner product in QM?
Originally posted by Ambitwistor
A coordinate system is a collection of functions on the manifold, each of which assigns a number (a coordinate) to a given point.
You don't need a coordinate system in order to define a tangent space, and you don't need a tangent space in order to define a coordinate system.After letting this soak overnight, I think I realize what y'all are saying. I don't know if I quite follow your set of functions description though. But now it does make sense to me that the points in the manifold exist, period. The coordinate system is a way to keep track of them, like assigning an index to the elements of a set? Since the points are not discrete elements, we can't use an index (at least not like i = 1,2,3,...), so we use a continuous variable and call it a coordinate. Just like I could index the elements of a set differently, I can use a different coordinate.
But now I am confused about why one needs two coordinates instead of one. Is there no way to use just one coordinate to indicate points. I am having trouble shifting my mind from elements of a set needing only one index and points in a manifold needing more than one coordinate. (I am at this point going to assume that my analogy was correct that points:manifold::elements:set) In a vector space, I've got the idea of linear independence dictating how many independent vectors I will need in my basis to span the vector space, so the dimension is clear. I don't see the same indicator in a manifold.
Can you explain the collection of functions a bit?
Originally posted by Ambitwistor
You're never "between tangent planes". You are always in some tangent plane, because you are always at some point, and every point has a tangent plane associated with it. But I thought that the tangent planes for any two points, no matter how close they are, have absolutely nothing to do with each other. Unless I'm allowed to, in some sense, take the tangent plane with me, then I have to go from one to another. But, if adjacent tangent planes have nothing to do with each other, then this process seems disjoint, abrupt, or something. There just seems to be a discontinuity here, but I can't put my finger on it. I'll have to let this idea soak in my mind, too.
Originally posted by Ambitwistor
The integral curves of the \vec{r} and \vec{\theta} vector fields are lines and circles, ...What vector fields? I only know of vector fields like the electric field and such. Can you explain these vector fields?
Originally posted by Ambitwistor
... if you try to build coordinates out of those curves by constructing their tangent vectors Doesn't the fact that you're trying to build coordinates by using vectors automatically tell you that it doesn't make sense? I thought that was the lesson I was learning on the sphere.
Originally posted by Ambitwistor
The frame field isn't a coordinate vector field; the vectors don't commute with each other.I don't understand commuting/noncommuting vectors. The only experience I have with nontrivial commutation is in QM, but I only understand the concept with operators, not vectors. Can you explain?
Originally posted by turin
From QM, I remember that the inner product was a bra multiplying a ket [...]
From classical mechanics, I remember that any vector has two different kinds of components and two different kinds of bases, contravariant and covariant. The contravariant basis is like arrows, and the covariant basis is like a set of planes (stack of pancakes). Taking a dot product (which I'm still trying to decide if this is really the same as an "inner product")
yes, dot product and inner product are synonymous.
of two vectors is summing the products of the contravariant components of one of them with the covariant components of the other. If the dot product is zero, I interpretted it as the arrow being parallel to the planes in the set of planes. The dot product is maximum when the arrow is perpendicular to the set of planes (so that it passes through the maximum number of planes).
Are you saying that my interpretations from classical mechanics are inappropriate? Should I think about it more in terms of the way I think about the inner product in QM?
in both cases, you are making heavy use of a certain fact without realizing it. when a vector space has an inner product, then there is an isomorphism between the vector space and the dual space. in the case of quantum mechanics, where your vector space is actually an infinite dimensional L2 Hilbert space, this is actually heavy theorem (Riesz), whereas in the finite dimensional case its rather trivial, but either way, its true.
so one way to think about an inner product is this: take two vectors in the vector space, feed them to the inner product machine, and get a number out. this is the better way to think of it, in my opinion.
another way to think of it, is, take two vectors, then use the isomorphism determined by the inner product to find the dual of one of the vectors, and then feed the other vector to that dual, getting a number out.
this latter is what you do in both descriptions you gave above. so in some sense, if you have an inner product, you can just not bother to distinguish between vectors and their dual. this is how you are thinking about vectors in classical: the only difference between a vector and a covector is whether it has raised or lowered indices. in QM, the only difference between a vector and a covector is whether it is a bra or a ket.
so, im not going to tell you that this way of thinking about vector spaces is wrong, but i have learned that it is very useful to, at the very least, remain aware of when you are using your inner product, and for what purposes. you never know when you may need to change your inner product, and then this isomorphism will change, and all the dual vectors you had previously written down will be something different. better still, don t use the isomorphism at all, and only write down a vector if you actually mean a vector, and a covector if you mean a covector.
Originally posted by turin
But now it does make sense to me that the points in the manifold exist, period. The coordinate system is a way to keep track of them, like assigning an index to the elements of a set?
yeah, this sounds good. the points in the manifold are the points, and there may be many different coordinate systems that you can use to label the point, but we think of the point existing in some sense independently of which coordinate system we use.
Since the points are not discrete elements, we can't use an index (at least not like i = 1,2,3,...), so we use a continuous variable and call it a coordinate. Just like I could index the elements of a set differently, I can use a different coordinate.
yep
But now I am confused about why one needs two coordinates instead of one.
you need n coordinates, where n is the dimension of the manifold.
Is there no way to use just one coordinate to indicate points.
only if the manifold is 1 dimensional (a curve)
I am having trouble shifting my mind from elements of a set needing only one index and points in a manifold needing more than one coordinate.
if you label the elements of the set A with an index i, then you would use two indices (i,j) to label the elements of the set AxA. in the same sense, you need one real number to specify a point on the line (=R), 2 real numbers to specify a point on the plane (=RxR), etc.
i guess in a set theoretic sense, you don t actually need two indices. the cartesian product of two sets has the same cardinality, so you could count the elements of the new set with the same index, if you had a mind to, but this would muck up the other nice properties of the set, so it is much preferred to just stick to the double index concept.
(I am at this point going to assume that my analogy was correct that points:manifold::elements:set)
yes, its a good analogy
In a vector space, I've got the idea of linear independence dictating how many independent vectors I will need in my basis to span the vector space, so the dimension is clear. I don't see the same indicator in a manifold.
perhaps it would be useful to know the definition of a manifold. leaving out some technical details, the definition is this: an n-dimensional manifold is a space X such that for any point x in X, there is a neigborhood of x U and a continous invertible map with continuous inverse (homeomorphism) between U and Rn
in other words, if you look at a manifold up closely enough, it looks like Rn. these local homeomorphisms are the charts of the manifold, they are the coordinates.
Can you explain the collection of functions a bit?
perhaps it would be useful to see an example. a sphere can be written as the map (u,v)\mapsto(\sin u\cos v,\sin u\sin v,cos v). here u and v are the coordinates. there are two of them because a sphere is a 2 dimensional manifold
But I thought that the tangent planes for any two points, no matter how close they are, have absolutely nothing to do with each other.
thats right. nothing at all (unless you like chroots approximations above. i don t.)
Unless I'm allowed to, in some sense, take the tangent plane with me, then I have to go from one to another. But, if adjacent tangent planes have nothing to do with each other, then this process seems disjoint, abrupt, or something. There just seems to be a discontinuity here, but I can't put my finger on it. I'll have to let this idea soak in my mind, too.
i m not sure what exactly you re saying here. i can assure you that, for example, if you have a smooth curve on the manifold, then this curve will have a tangent vector which is in the tangent space to the manifold at each point the curve passes through. furthermore, this tangent vector to the curve will vary from tangent space to tangent space, as the curve goes along, in a smooth way.
there is no discontinuity. perhaps it would be useful to know the following fact: a smooth manifold, together with the tangent space at each point, considered as one larger set, is also a smooth manifold. this manifold is called the tangent bundle.
Originally posted by turin
What vector fields? I only know of vector fields like the electric field and such. Can you explain these vector fields?
a vector field is any mapping that assigns a vector to each point of the space. electric field is an example of a vector field on minkowski space, sure. you know how every vector space has a basis? a nice basis for cartesian space is i, j, and k. these are the unit vectors pointing along each coordinate. since there is a tangent space to every point in space, these form a basis at every point in space. hence they are vector fields (not just vectors). the vectors Ambitwistor are the same thing for polar coordinates.
this should be apparent to you. if you know that electric field is a vector field, and you know that electric field can be written \mathbf{E}=E_x\hat{\imath}+E_y\hat{\jmath}+E_z\hat {k}, so those i, j, and k are also vector fields. you can choose to work in polar coordinates, and thats all we are doing here.
Doesn't the fact that you're trying to build coordinates by using vectors automatically tell you that it doesn't make sense? I thought that was the lesson I was learning on the sphere.
points in the sphere are not vectors. never the less, given points in the sphere, you can calculate the tangent vectors, and given tangent vectors, you can calculate coordinates for points in the sphere.
I don't understand commuting/noncommuting vectors. The only experience I have with nontrivial commutation is in QM, but I only understand the concept with operators, not vectors. Can you explain?
vectors are differential operators. if you would like to understand why, well, i explained this concept at length once, when i started writing a tutorial for differential forms, here (http://www.physicsforums.com/showthread.php?s=&threadid=2953).
in short, there are various things you can do to define tangent vectors in an intrinsic way. a nice starting way is to define a tangent vector as an equivalence class of curves all of which have the same derivative at a point. with a little work, you see that this is canonically isomorphic to the space of first order differential operators on functions on the manifold. therefore i can simply say they are the same thing, and call a tangent vector on a manifold a differential operator.
this is tremendously useful, though unfamiliar.
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