Here's where I make a fool of myself in Physics

[Ben]

So, in this thread I will ask a lot of really dumb questions about physics. When you all get to know me a bit better, you will learn that I am relatively proficient in Math; but here's today's dumb question:

 

Physicists often talk about "the action". What do they mean? The notation sure looks like it is related to a variational principle, but in what context? Is it only in mechanics, or maybe dynamics too?

 

Or do I first need to know what a Lagrangian is? To my limited understanding, this is most definitely used in mechanics, but my understanding of "a Lagrangian" (as a noun) is equally hazy.

 

Try me...

[Haech]

edited: I might be wrong about this. will hold the thought until I get home and check my old text pde and diff eq. books

[CraigD]

Welcome to hypography, Ben! I’m also someone with some education and practical experience in math, but at best a hobbyists understanding of modern physics.

Physicists often talk about "the action". What do they mean?

In the context of discussing or writing about physics, “action” could be a synonym for the more precise modern physics term “interaction”, meaning the exchange of particles, either “charge and mass carrying” fermions, or more usually “force carrying” bosons. When the boson is a gauge boson, such an interaction is considered a fundamental interaction.

Or do I first need to know what a Lagrangian is?

I don’t believe you need to know what a Lagrangian is before knowing what an interaction is. The former can be understood informally, or, more precisely, pseudoclassically, that is, in terms of analogies involving ordinary, macroscopic things like pool balls and string. The latter is the name for something by definition formal.

 

However, when “Lagrangian” and “action” appear near one another, it’s likely “action” refers to a very specific, formal attribute, the change in the wave function of a particle over time. This is about the point where my grasp of the formalism of modern physics fails me, and I must take a spectator’s role in it, accepting work I can’t myself produce.

[Pyrotex]

I am following this thread with intense curiousity.

I, too, want to understand 'action'. I mean, I think I used it in college, but I don't remember.

[freeztar]

I think Craig nailed it with a synonym of interaction. It makes me think of "action at a distance", which could easily be rewritten "spooky interaction at a distance".

 

What are the contexts in which you are finding "action" and "Lagrangian", Ben?

[Erasmus00]

Generally, in modern physics, the action is defined as the integral of the kinetic energy minus the potential energy. S, is generally the character of choice. With v being velocity, and U being potential:

 

[math]\mathcal{S} = \int \frac{1}{2}mv^2 - U dt [/math]

 

The quantity kinetic energy - potential is called the Lagrangian, so you could say the action is the integral of the lagrangian.

 

The reason its talked about so much is that it is very useful. One way of formulating mechanics is to say that particles take the path of least action. So if we compute the action, of various paths, and find the minimum, we know thats the paths particles actually take. If I get some time, I'll try to show how to find the minimum (and that it gives you the standard Newton equations).

 

Also- a question for thought. Paths are non-local things. If a particle traveled from A to B, how does it know which path will be the one with the least action?

[Ben]

Ah right, I believe I know this as "Hamilton's Principle": the motion of a system whose Lagrangian is [math]\mathcal{L} = T - U,\,\,\,(T=[/math] kinetic energy) from time [math]t_a \to t_b[/math] is such as to render the integral [math]\int^{t_b}_{t_a} \mathcal{L}\,dt[/math] an extremum with respect to the coordinate functions [math]x^i(t) [/math].

 

Then, viewing [math]\mathcal{L}(t, x^i, \frac{d x^i}{d t})[/math] as a function with these arguments, the Euler-Lagrange eqn gives, for this extremum,

 

[math] \frac {\partial \mathcal{L}}{\partial x^i} - \frac{d}{dt}(\frac{\partial \mathcal{L}}{\partial \dot{x}^i})=0[/math], where [math]\dot{x} = \frac{dx}{dt}[/math].

 

In terms of your example, this yields (though I tell you frankly I am a little unsure about this)

 

[math]-\frac{\partial U}{\partial x^i}- \frac{d}{dt}(m \dot {x^i}) =0[/math] This is, of course

 

[math]-\frac{\partial U}{\partial x^i}-m\ddot{x^i}=0[/math] or

 

[math]m\ddot{x^i} =-\frac{\partial U}{\partial x^i}[/math].

 

Now the RHS is the vector sum of forces acting on a particle in the [math]x^i[/math] direction, so write [math]\frac{\partial}{\partial x^i} = \nabla[/math] and obviously [math]\ddot{x^i}[/math] is acceleration, hence

 

[math]ma = - \nabla U = \mathbf{F}[/math], something Newton dreamed up.

 

As to your "question to ponder". Let our particle be such that, to it, we may associate a wave function. Let us assume that there is a shortest path [math]A \to B[/math]. Then all other paths are longer and thus, if they are to "arrive" at [math]B[/math] at the same time, must have different wave functions which interfere destructively, leaving the shortest path as the only remainder. In other words, our particle sets out on all possible paths, but all save one kill each other.

 

How about that for a piece of science?

[Qfwfq]

Well Ben you're not making that much of a fool of yourself, really!

 

Is it only in mechanics, or maybe dynamics too?

OK here's one basic correction: dynamics is part of mechanics, not vice versa.

 

Yes, it is a variational principle and yes, it's dynamics. Lagrangian and Hamiltonian formulation are the ones used, the two being connected by the Legendre transformation. If you're a mathematician that wants to delve into the topic (especially classical) you might especially like "Lectures in Analytical Mechanics" by F. Gantmacher.

 

These formulations are fundamental to quantum mechanics, but for the detail that things appear as operators. The Schrödinger equation is actually based on the system's Hamiltonian. For a relativistic treatment the Lagrangian is handier, basically because energy isn't a Lorentz-scalar.

[Boof-head]

So you're saying a photon is a dynamical operator (that would be: "on electrons and charged particles w/mass")?

 

P.S. see this thread: http://hypography.com/forums/physics-and-mathematics/19329-physics-question-about-a-particle.html

[Ben]

Qfwfq: I thank you for that, though I need to brush up on the Legendre transformation. However....

For a relativistic treatment the Lagrangian is handier, basically because energy isn't a Lorentz-scalar.

....what is a Lorentz-scalar?

 

Let me try this, then we'll see who the fool is. It is an elementary fact from linear algebra that, given the transformation [math]A: V \to V'[/math], where the [math]V,\,\,V'[/math] are arbitrary vector spaces, then this operator will be a linear operator iff [math]A(\alpha v) = \alpha A(v) = \alpha v'[/math], where [math]\alpha[/math] is scalar. ( edited [math]T \to A[/math])

 

In this circumstance I will say that the scalar [math]\alpha[/math] is invariant under the action of the transformation [math]A[/math], that is [math]\alpha = \alpha'[/math]. By an abuse of language I may say that [math]\alpha[/math] is an [math]"A-scalar"[/math].

 

Clearly energy written in the form [math]\tfrac{1}{2}mv^2[/math] or even [math]mc^2[/math] depends on the vector quantities [math]v,\,\,c[/math], and therefore is not invariant under the action of any transformation, Lorentz or otherwise.

 

Anywhere near close?

[Qfwfq]

Anywhere near close?

Pretty close except that you seem unfamiliar with the specific case of the Lorentz group (of space-time coordinate transformations). Energy is one of the components of the energy-momentum vector, corresponding to the time axis while momentum is the spatial part. It might strike you odd but in special relativity c is a scalar and [imath]mc^2[/imath] isn't the total energy, it's like the modulus of the energy-momentum.

 

Actually, that last remark of mine wasn't really essential unless you want to actually get into the relativistic case.

 

Moreover I'd say that an important result in analytical mechanics is Noether's theorem, a link between dynamic symmetries and conserved quantities. It is actually a fundamental reason for Lagrangian and Hamiltonian formulation being so all-important.

[Pyrotex]

Qfwfq: I thank you for that, though I need to brush up on the Legendre transformation. However........what is a Lorentz-scalar?...Clearly energy written in the form [math]\tfrac{1}{2}mv^2[/math] or even [math]mc^2[/math] depends on the vector quantities [math]v,\,\,c[/math], and therefore is not invariant under the action of any transformation, Lorentz or otherwise...

Ben, your understanding of physics is obviously less rusty than mine, but back in grad school, Energy was always a scalar. :)

 

As to invariance, hmmm. I have this tiny hunch that the energy of the object isn't "increased" by the Lorentz factor. This may just be because I do not remember ever seeing an equation in my grad physics classes that expressed the energy of a particle (object) as a function containing the Lorentz factor. Mass, yes. Longitudinal length, yes. But energy? :confused: I don't think so. :shrug: But it has been a loooooong time since grad school. :)

[lawcat]

In other words, our particle sets out on all possible paths, but all save one kill each other.

 

How about that for a piece of science?

 

I am not sure about that, and here is why. In electrical circuits, the current will take as many paths as are provided, and the voltage will support. Thus, first, the path must be real, and second, the force must be able to overcome the resitance of the path. For example, if there are 4 resistors in a circuit--4, 10, 1000, 1,000,000 ohms--current will travel through all four resistors. but, current will not travel through air, because air or insulation on wires is not real conductor. Thus, current, energy, heat can be reconstituted at the final destination after taking many different paths piecemeal, but objects hardly can. It appear then that nature of the action depends on the nature of the object , available paths, and nature of the paths.

 

As far as action, since Newton, I thought action is change in state of object due to a force resulting from interaction.

[Qfwfq]

Mass, yes.

That's because, historically, Minkowski's much neater formulation came later, was initially snubbed by Albert (till he realized how essential it is, by the time he completed GR) and unfortunately did not so rapidly take hold on most physicists. Indeed the so-called relativistic mass, when multiplied by [imath]c^2[/imath], is actually the energy (kinetic + rest energy).

[Ben]

Yeah, that was my understanding too. Anyway look, I didn't come here to discuss relativity, but as long as we are, let me say this;

 

I was told here that the Special Theory threats [math]c[/math] as a (Lorentz) scalar. Now I am well aware that this theory is predicated on the postulate that [math]c[/math] is invariant under the Lorentz transformation, which suggests the following definition:

 

a mathematical object will be called a "scalar" iff it is invariant under an arbitrary coordinate transformation.

 

Corollary: a scalar is a rank zero tensor - well we knew that already, right?

 

But it illuminates (but doesn't wholly excuse!) the annoying (to me) habit of physicists to define tensors in terms of how they transform. I prefer to define them in terms of what they actually are, but I can now see where you guys are coming from.

 

Now it is easy to show that energy is not a Lorentz scalar, as follows:

 

Suppose there is a body A, and let it emit a steady stream of radiation for some fixed time [math]t_A[/math] as measured by its own clock. Then obviously, there will be energy withdrawn from A, so its final "energy content" will be, say, [math]E_{initial} - E_{final} = \Delta E_A[/math], and this, let us suppose, will depend only on [math]t_A[/math] and nothing else.

 

Now suppose there is an identical body in uniform motion relative to A, call it B. If the radiation emitted by B is timed by a clock attached to B, time dilation (by the Lorentz transformation) tells us that [math]t_A \ne t_B[/math] and accordingly [math]\Delta E_A \ne \Delta E_B[/math], thus energy is not a Lorentz scalar.

 

I have more questions, many more, none of them explicitly relativistic, but they will have to wait.

[Pyrotex]

I did an hour of research in Wiki today on "action", "principal of least action", Hamiltonians and Lagrangians. Fascinating. I did study this in grad school, but I remember so little. Shame. Must study harder.

[arkain101]

Now suppose there is an identical body in uniform motion relative to A

 

What do you mean exactly by "in uniform motion"?