Conserved values always derived from Action Integral

[Mike2]

Is every conserved quantity in physics always derived from the invariance of the action integral? Or are some conserved quantities not derived this way? I suppose some physical constants, such as C and h, are not a result of the invariance of the action integral. And I suppose those characteristic of particles/strings that are conserved are derived from the invariance of the action integral with changes of one sort or another, right?

Thanks.

 

[lethe]

most conserved charges are due to some symmetry of the action, but not all.

for example, conservation of baryon number.

 

[Ambitwistor]

There are conservation laws that are not associated with a symmetry:

http://groups.google.com/groups?selm=1994Feb23.085519.11634@cc.usu.edu

It may not be accurate to claim that conservation of baryon number is not associated with a symmetry, depending on what you mean by "symmetry":

http://groups.google.com/groups?selm=mcirvin.762140704@husc8.harvard.edu

Physical constants are not regarded as conserved quantities.

 

[lethe]

Originally posted by Ambitwistor
There are conservation laws that are not associated with a symmetry:

topological charges... of course.

[bIt may not be accurate to claim that conservation of baryon number is not associated with a symmetry, depending on what you mean by "symmetry":[/b]

yeah, i said baryon number because it is one i have heard a lot, but you (or the thread you provided) are definitely right on... baryon number is not conserved by a gauge symmetry, but it is still a symmetry nonetheless.

so i stand corrected.

 

[Mike2]

In reading "Tensors, Differential Forms, and Variational Principles", by Lovelock and Rund, Chapter 6, it would appear that if any path integral of a scalar field is reparameterization invariant and coordinate transformation invariant, then it turns out that the gradient of that field is normal to the path. And this is derived without the use of variational principles. It is interesting that this geometry is implied only by simple symmetries. You'd think that you could specify any path you'd like in any field you'd like and still have the situation not depend on any coordinate system you might arbitrarily impose. But imposing these symmetries on the integral results in the necessity of a normal gradient with respect to the path. Why?

If the path is perpendicular to the gradient, then the value of the field is everywhere the same at every point along the path since you have no change in value with any move along the curve. Do these symmetries also imply no change in value with position on the curve? Having no changes with coordinate transformations is not equivalent to having no changes at all, is it? Or can it be that summing up a function along a line can only be guaranteed intrinsically if the function is a constant? But how can that be when the field an intrinsic object and the path is an intrinsic object?

 

[lethe]

Originally posted by Mike2
In reading "Tensors, Differential Forms, and Variational Principles", by Lovelock and Rund, Chapter 6

i don t have this book, so i m going to just go on what you are saying here.

it would appear that if any path integral of a scalar field

hmm... a path integral of a scalar field? you are only supposed to do path integrals of functionals of paths, which a scalar field is not.

also, this doesn t sound like a quantum field theory book, so i find it a little surprising that it covers path integrals, although i guess that could go under the "Variational principles" heading.

so are you sure that you are talking about path integrals here?

is reparameterization invariant and coordinate transformation invariant,

those two phrases mean the same thing

then it turns out that the gradient of that field is normal to the path.

what path?

And this is derived without the use of variational principles. It is ...

But how can that be when the field an intrinsic object and the path is an intrinsic object?

i didn t follow the rest of this at all...

 

[Ambitwistor]

I think Mike2 is talking about line integrals, which are sometimes called path integrals. (Probably "path integral" was an older usage that fell out of favor when Feynman's path integrals came along.)

 

[lethe]

Originally posted by Ambitwistor
I think Mike2 is talking about line integrals, which are sometimes called path integrals. (Probably "path integral" was an older usage that fell out of favor when Feynman's path integrals came along.)

ahh yes, you re probably right...

i once knew this terminology, but had completely forgotten about it. it didnt even occur to me.

 

[Mike2]

Well, then, so I may be completely controversial, what I'm really trying to get at is a geometric justification of the action integral of the Lagrangian, at least in a classical string theory sense. It would seem that one can derive an Euler-Lagrange vector normal to the world-sheet by nothing more than symmetry/invariance requirements. And when this Euler-Lagrange vector is zero, then the surface is a geodesic and we get dynamical equations of motion.

Isn't it true that the momentum and Hamiltonian can be described in terms of derivatives of the Lagrangian? And derivatives of scalars with respect to coordinates are vectors, right? Can these vectors be described perhaps as components of the Euler-Lagrange vector? If the Euler-Lagrange vector is zero, then wouldn't these vectors live in the tangent space of the surface?

The point being, if a world sheet in a scalar field can be justified for other reasons (as a growing event in "sample space", for example), then perhaps physics can be justified from that.

 

[lethe]

Originally posted by Mike2
It would seem that one can derive an Euler-Lagrange vector

what is an Euler-Lagrange vector?

normal to the world-sheet by nothing more than symmetry/invariance requirements. And when this Euler-Lagrange vector is zero, then the surface is a geodesic and we get dynamical equations of motion.

huh? how can a surface be a geodesic?

Isn't it true that the momentum and Hamiltonian can be described in terms of derivatives of the Lagrangian?

its true that the canonical momentum can be gotten as a derivative of the lagrangian, but this is not true of the Hamiltonian. the Hamiltonian is obtained by a Legendre transformation of the Lagrangian

And derivatives of scalars with respect to coordinates are vectors, right?

more or less...

Can these vectors be described perhaps as components of the Euler-Lagrange vector?

what is the Euler-Lagrange vector?

If the Euler-Lagrange vector is zero, then wouldn't these vectors live in the tangent space of the surface?

you were saying before that this mysterious Euler-Lagrange vector was a normal vector to the surface. now its a tangent vector?

weird...

of course, if the vector in question is zero, then it can be both normal and tangent, since the zero vector is in every vector space, but this is, of course, trivial.

The point being, if a world sheet in a scalar field can be justified for other reasons (as a growing event in "sample space", for example), then perhaps physics can be justified from that.

how can a worldsheet be "in a scalar field"? what is a sample space?

 

[Mike2]

Originally posted by lethe
what is an Euler-Lagrange vector?

$${\bf E}_j ({\rm L})\,\, = \,\,{d \over {dt}}({{\partial {\rm L}} \over {\partial {\rm \dot x}^j }})\,\, - \,\,{{\partial {\rm L}} \over {\partial {\rm x}^j }}$$

where L is the Lagrangian. Lovelock and Rund, page 185 says this is a "generalized gradient" of the Lagrangian. I can see where the second term on the right is the negative of gradient in rectilinear coordinates. But if Ej is a vector, then each term on the right is a vector. I wonder what geometric interpretation each term on the right has. Where are these vectors pointing?

huh? how can a surface be a geodesic?

The minimum surface area from one string to the next.

how can a worldsheet be "in a scalar field"? what is a sample space?

What is a world-sheet? How did it come into existence? If all calculations are done only where the sheet exists, then I can presuppose the existence of a field all I want, as long as I do not do calculation anywhere except on the sheet. You do suppose that whatever is calculated ON the sheet Does change in a continuous manner with respect to time, right? Then that supposes a neighborhood of existence in the direction of time, doesn't it?

 

[lethe]

Originally posted by Mike2
But if Ej is a vector, then each term on the right is a vector.

what on Earth makes you think that this thing is a vector?

The minimum surface area from one string to the next.

this is a very nonstandard definition of geodesic

What is a world-sheet? How did it come into existence? If all calculations are done only where the sheet exists, then I can presuppose the existence of a field all I want, as long as I do not do calculation anywhere except on the sheet. You do suppose that whatever is calculated ON the sheet Does change in a continuous manner with respect to time, right? Then that supposes a neighborhood of existence in the direction of time, doesn't it?

i have no idea what you re talking about here.

 

[Mike2]

Originally posted by lethe
what on Earth makes you think that this thing is a vector?

According to Lovelock and Rund, it transforms as a covariant vector.

 

[lethe]

Originally posted by Mike2
According to Lovelock and Rund, it transforms as a covariant vector.

well i don t have that book.

 

[lethe]

Originally posted by lethe
well i don t have that book.

anyway, if they say it is a covector then it may well be.

but covectors don't "point" anywhere, only tangent vectors point

 

[Mike2]

Originally posted by lethe
anyway, if they say it is a covector then it may well be.

but covectors don't "point" anywhere, only tangent vectors point

right... A covector operates on a vector to produce a scalar. And the book shows that when the Euler-Lagrange covector operates on a velocity vector, the result is zero. This is why they call it normal to the velocity vector, as if the dot product were zero. I suppose this is valid when the metric is the identity matrix. For you can always get a corresponding vector from a covector through multiplying the covector by the metric tensor.

 

[lethe]

Originally posted by Mike2
right... A covector operates on a vector to produce a scalar. And the book shows that when the Euler-Lagrange covector operates on a velocity vector, the result is zero. This is why they call it normal to the velocity vector, as if the dot product were zero. I suppose this is valid when the metric is the identity matrix. For you can always get a corresponding vector from a covector through multiplying the covector by the metric tensor.

there is a natural "inner product" between vectors and covectors, but i think calling that operation an inner product obscures more than it helps. in fact, i downright despise it.

and you maybe see the reason now: covectors and vectors live in different vector spaces! you cannot rightly call a covector perpendicular to a vector just because it returns a value of zero on that vector. that amounts to choosing a rather arbitrary euclidean metric, which i don t like to do

the point of the story is: do not ask which way a covector points. covectors do not point anywhere. vectors point.

 

[Mike2]

Originally posted by lethe
there is a natural "inner product" between vectors and covectors, but i think calling that operation an inner product obscures more than it helps. in fact, i downright despise it.

and you maybe see the reason now: covectors and vectors live in different vector spaces! you cannot rightly call a covector perpendicular to a vector just because it returns a value of zero on that vector. that amounts to choosing a rather arbitrary euclidean metric, which i don t like to do

the point of the story is: do not ask which way a covector points. covectors do not point anywhere. vectors point.

OK. But where does the contravariant version of the covariant Euler-Lagrange covector point to, since we can convert the covector to a vector/contravarian vector by use of the metric tensor. Where does it point? I don't think you can say that this new contravariant vector is now perpendicular to the velocity since not you've multiplied it by the metric tensor. But since it is then a vector, can we say where it is pointing, or what relation it has to the velocity vector ?

Thanks.

 

[lethe]

Originally posted by Mike2
OK. But where does the contravariant version of the covariant Euler-Lagrange covector point to, since we can convert the covector to a vector/contravarian vector by use of the metric tensor.

what metric?

Where does it point? I don't think you can say that this new contravariant vector is now perpendicular to the velocity since not you've multiplied it by the metric tensor.

where did this metric come from all of a sudden?

 

[Mike2]

Originally posted by lethe
what metric?

where did this metric come from all of a sudden?

Well obviously, if there is a velocity equal to some distance per time, then there is a metric to specify the distance.

So the question remains: where does the contravariant version of the covariant Euler-Lagrange covector point?

Now, I seem to remember this vector is used in the variational principles to point in the direction of the greatest increase in the Action integral if a differential line segment at that point should move in that direction. So if this vector is zero everywhere on the line, then the line gives a minimum for the action integral, or in other words, a geodesic. Is this right?

 

[lethe]

Originally posted by Mike2
Well obviously, if there is a velocity equal to some distance per time, then there is a metric to specify the distance.

not only is this not obvious, it is not correct.

So the question remains: where does the contravariant version of the covariant Euler-Lagrange covector point?

my objection remains: a covariant vector does not point anywhere

Now, I seem to remember this vector is used in the variational principles to point in the direction of the greatest increase in the Action integral if a differential line segment at that point should move in that direction. So if this vector is zero everywhere on the line, then the line gives a minimum for the action integral, or in other words, a geodesic. Is this right?

i don t know what this means, so i cannot comment on whether it is correct.

 

[Mike2]

Originally posted by lethe
i don t know what this means, so i cannot comment on whether it is correct.

Perhaps this is an easier question:

What does it mean that the action integral is a minimum for a given line? Does that mean that the integral is "conserved" for small changes in the path? Or is something being conserved along the line?

Is there such a thing as a differential action, the action from one point to an infinitesimal distance to the next?

 

[lethe]

Originally posted by Mike2
Perhaps this is an easier question:

What does it mean that the action integral is a minimum for a given line? Does that mean that the integral is "conserved" for small changes in the path? Or is something being conserved along the line?

no, the integral is not "conserved". rather, it is stationary under small changes in the path. not the same thing.

something is conserved along any specific changes in configuration that leave the action invariant, whether or not it is stationary: the conserved current.

Is there such a thing as a differential action, the action from one point to an infinitesimal distance to the next?

sure...

 

[Mike2]

Originally posted by Ambitwistor
There are conservation laws that are not associated with a symmetry:

http://groups.google.com/groups?selm=1994Feb23.085519.11634@cc.usu.edu

Can a distinction be made if one is asking whether the conserved quatity is one associated with propagation as opposed to some other type of conserved value? Or does conservation always have to do with some value being constant along a path? I'm thinking maybe there are values that are conserved and do not change with a change in the path used to get from one point to the next, etc. And perhaps this is distinguished from conserved values at each point on the path.

Thanks.

 

[Mike2]

Originally posted by Ambitwistor
There are conservation laws that are not associated with a symmetry:

http://groups.google.com/groups?selm=1994Feb23.085519.11634@cc.usu.edu

Can a distinction be made if one is asking whether the conserved quatity is one associated with propagation as opposed to some other type of conserved value? Or does conservation always have to do with some value being constant along a path? I'm thinking maybe there are values that are conserved and do not change with a change in the path used to get from one point to the next, etc. And perhaps this is distinguished from conserved values at each point on the path.

Thanks.

 

[Mike2]

quote:
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Is there such a thing as a differential action, the action from one point to an infinitesimal distance to the next?
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Originally posted by lethe
sure...

Can you think of an example? Would this differential action be the Lagrangian itself?

 

[Mike2]

Originally posted by lethe
my objection remains: a covariant vector does not point anywhere

Lovelock and Rund use this Euler-Lagrange vector quite liberally. They use it in the proof of Nother's theorem, multiplying it at will with contravariant vectors without any mention of a metric. This has me a little concerned.

Plus, what is a gradient, a vector or covector?

 

[Mike2]

quote:
--------------------------------------------------------------------------------
Originally posted by Mike2
OK. But where does the contravariant version of the covariant Euler-Lagrange covector point to, since we can convert the covector to a vector/contravarian vector by use of the metric tensor.
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Originally posted by lethe
where did this metric come from all of a sudden?

How can you do Action integrals or Feynman path integrals if there is no metric to define distance?