# Conserved values always derived from Action Integral

1. Dec 4, 2003

### 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.

2. Dec 4, 2003

### lethe

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

for example, conservation of baryon number.

3. Dec 4, 2003

### Ambitwistor

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

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":

Physical constants are not regarded as conserved quantities.

4. Dec 4, 2003

### lethe

topological charges... of course.

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.

5. Dec 4, 2003

### 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?

6. Dec 4, 2003

### lethe

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

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?

those two phrases mean the same thing

what path?

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

7. Dec 4, 2003

### 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.)

8. Dec 5, 2003

### lethe

ahh yes, you re probably right...

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

9. Dec 5, 2003

### 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.

10. Dec 5, 2003

### lethe

what is an Euler-Lagrange vector?

huh? how can a surface be a geodesic?

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

more or less...

what is the Euler-Lagrange vector?

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.

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

11. Dec 6, 2003

### Mike2

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

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?

The minimum surface area from one string to the next.

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?

12. Dec 6, 2003

### lethe

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

this is a very nonstandard definition of geodesic

i have no idea what you re talking about here.

13. Dec 6, 2003

### Mike2

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

14. Dec 6, 2003

### lethe

well i don t have that book.

15. Dec 6, 2003

### lethe

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

but covectors dont "point" anywhere, only tangent vectors point

16. Dec 7, 2003

### 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.

17. Dec 8, 2003

### 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.

18. Dec 8, 2003

### 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. 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.

19. Dec 8, 2003

### lethe

what metric?

where did this metric come from all of a sudden?

20. Dec 9, 2003

### Mike2

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?