# Why does action have to be invariant?

1. Jul 30, 2011

### mmmboh

In classical mechanics, isn't kinetic energy not a Galilean scalar? So the action isn't invariant under Galilean transformations, but we can still use it with Galilean transformations. So why must it be a scalar in special relativity?
I think I'm missing something...

2. Jul 30, 2011

### bcrowell

Staff Emeritus
To be more concrete, we should probably talk about what system we have in mind.

For a system consisting of a single particle that isn't interacting with anything, the Lagrangian in SR is just the particle's proper time, which is a Lorentz scalar. In the nonrelativistic limit, this Lagrangian becomes simply the universal time coordinate that everyone agrees on. It is invariant under Galilean transformations.

One thing to keep in mind is that the way of expressing the Lagrangian isn't unique. You can have different Lagrangians that make the same predictions. This may be helpful: http://www.eftaylor.com/pub/variational1D.html See section V, where they go from the "abbreviated action" to the "just plain action."

When you want to make particles interact, relativistically it doesn't make sense to do L=T-V, because a universal potential like V would represent instantaneous interaction at a distance, which SR doesn't allow.

3. Jul 30, 2011

### mmmboh

I'm confused. If you launch a ball in a train, and you are in the train, the kinetic energy of the ball will be less than if you were standing outside the train, so the calculated actions will be different, won't they? Which would mean the action isn't invariant. So why would we assume the action is invariant under Lorentz transformations?

4. Jul 30, 2011

### atyy

In Zwiebach's string theory text, he says the Galilean example shows it may not be necessary, and might be too strong a constraint, and just luckily it turns out ok in the cases he discusses.

5. Jul 30, 2011

### Dickfore

The action is not an invariant, but it's differential is determined up to a total differential of an arbitrary function of coordinates:
$$dS' = dS + df(q, t)$$
This does not affect the extremal properties of the action since it simply gives an additive constant.

6. Jul 30, 2011

### ghwellsjr

If you perform an experiment at rest in one Frame of Reference, recording the coordinates of all the significant events and then you transform all those events into a new FoR moving with respect to the first one, you will not see the experiment behaving the same way because it will not be at rest in this new FoR. What you have to do is repeat the experiment while at rest in the new FoR and then the experiment will behave the same way as before.

Last edited: Jul 30, 2011
7. Jul 30, 2011

### bcrowell

Staff Emeritus
Now you're talking about a theory in which a ball interacts with a planet through instantaneous action at a distance. For the reasons I gave in #2, you can't just model that with L=T-V and expect it to work relativistically.

8. Jul 30, 2011

### bcrowell

Staff Emeritus
I know the OP was asking in the context of SR, but, e.g., #3 involves gravity which brings us into GR. In the context of GR, it clearly wouldn't make sense to formulate an action that wasn't a Lorentz scalar. If you tried to do that, it would be coordinate-dependent, and GR doesn't even come equipped with a default set of coordinates.

9. Jul 30, 2011

### mmmboh

Ok so you can't use L=T-V in SR, but why does that imply the action is invariant? The action being invariant is the only justification I've read for why the SR Lagrangian is what it is. But in classical mechanics, Galilean transformations are analogous to Lorentz transformations...so if the action doesn't have to be Galilean invariant in classical mechanics, why is it assumed it is invariant in SR?

I've studied non-relativistic Lagrangian mechanics, but not relativistic. I know the justification for why L=T-V in CM (class. mech) and why nature minimizes the action, but from what I've read in SR, this seems really contrived. They just assume that there is an integral that is invariant under Lorentz transformations, and that nature minimizes it.

Does it have something to do with allowable coordinate transformations? I know in classical mechanics the allowable coordinate transformations are rotations and translations...although I don't see why it's just those two...

Thanks for all the responses so far.

Last edited: Jul 30, 2011
10. Jul 30, 2011

### Dickfore

Could you please enlighten us why Nature chooses to do so?

11. Jul 30, 2011

### mmmboh

I didn't mean it like that, I mean I've seen the derivation of the Lagrange equations without using the action, and that minimizing the action gives the same results.

12. Jul 31, 2011

### Dickfore

Oh, ok. So, I guess what you meant was that the principle of least action is equivalent to Newton's laws or D'Alambert's principle. This is only true for purely mechanical systems. Nevertheless, the principle of least action is valid for a far wider range of phenomena. For example, as was pointed out above, due to the finite speed of propagation of interactions in Nature, the concept of action at a distance, which is essential in the notion of a potential energy, loses its meaning. One needs to ascribe a field as a continuum with its own degrees of freedom and its own action. The interaction between the particle and the field is then described by another term in the action.

On another note, in the first part of Landau Lifgarbagez, they derive the Lagrangian for a free particle on principle of homogeneity of space and time and isotropy of space and invariance (up to a total differential as I pointed out in #5) of the action differential w.r.t. to infinitesimal Galilean transformations.

Let us follow this route in deriving the Lagrangian but w.r.t. to Lorentz invariance. Again, the Lagrangian has to be a function of the square of the velocity of the particle $L = L(v^{2})$.

The components of the velocity transform during an infinitesimal Lorentz transformation (with $\beta = \epsilon \rightarrow 0$):
$$v'_{x} = \frac{v_{x} - c \, \epsilon}{1 - \frac{v_{x} \, \epsilon}{c}} = (v_{x} - c \, \epsilon) \, \left(1 + \frac{v_{x} \, \epsilon}{c} + o(\epsilon)\right) = v_{x} - c \, \left(1 - \frac{v^{2}_{x}}{c^{2}} \right) \, \epsilon + o(\epsilon)$$
$$v'_{y} = \frac{v_{y} \, \sqrt{1 - \epsilon^{2}}}{1 - \frac{v_{x} \, \epsilon}{c}} = (v_{y} + o(\epsilon)) \left(1 + \frac{v_{x} \, \epsilon}{c} + o(\epsilon)\right) = v_{y} + \frac{v_{x} \, v_{y}}{c} \, \epsilon + o(\epsilon)$$
$$v'_{z} = v_{z} + \frac{v_{x} \, v_{z}}{c} \, \epsilon + o(\epsilon)$$
The square of the speed then expands as:
$$v'^{2} = v'^{2}_{x} + v'^{2}_{y} + v'^{2}_{z} = v^{2}_{x} - 2 \, c \, v_{x} \, \left(1 - \frac{v^{2}_{x}}{c^{2}}\right) \, \epsilon + v^{2}_{y} + 2 \, \frac{v_{x} \, v^{2}_{y}}{c} \, \epsilon + v^{2}_{z} + 2 \, \frac{v_{x} \, v^{2}_{z}}{c} \, \epsilon + o(\epsilon) = v^{2} - 2 c \, v_{x} \, \left(1 - \frac{v^{2}}{c^{2}}\right) \, \epsilon + o(\epsilon)$$
Thus, the Lagrange's function has an expansion:
$$L' \equiv L(v'^{2}) = L(v^{2}) - 2 c \, \epsilon \, v_{x} \, \frac{\partial L}{\partial v^{2}} \, \left(1 - \frac{v^{2}}{c^{2}}\right) + o(\epsilon)$$

But, we are not done yet. We also have to transform the time differential up to linear powers in $\epsilon$:
$$dt' = \frac{dt - \epsilon \, \frac{dx}{c}}{\sqrt{1 - \epsilon^{2}}} = dt - \frac{dx}{c} \, \epsilon + o(\epsilon)$$

Then, for the action differential, we get:
$$L' \, dt' = L \, dt - \epsilon \, \left[\frac{L}{c} \, dx + 2 c \, \left(1 - \frac{v^{2}}{c^{2}}\right) \, \frac{\partial L}{\partial v^{2}} \, v_{x} \, dt\right] + o(\epsilon)$$
The term in the square bracket is a total differential of a function of coordinates and time (remember that $v_{x} \, dt = dx$ along the trajectory of the particle) if and only if:
$$\frac{L}{c} + 2 c \, \left(1 - \frac{v^{2}}{c^{2}}\right) \, \frac{\partial L}{\partial v^{2}} = K$$
where K is a constant independent on the speed of the particle. This can be rearranged as an inhomogeneous first order linear ordinary differential equation (we make the substitution $u = v^{2}/c^{2}$ for the argument, which gives $\frac{d L}{d u} = c^{2} \, \frac{\partial L}{\partial v^{2}}$):
$$2 (1 - u) \, \frac{d L}{d u} + L = c \, K$$
$$\frac{d L}{d u} + \frac{1}{2(1 - u)} \, L = \frac{c \, K}{2 (1 - u)}$$
The integrating factor is:
$$\mu(u) = \exp{\left(\int{\frac{du}{2(1- u)}}\right)} = \exp{\left[-\frac{1}{2} \, \mathrm{Log}{(1 - u)}\right]} = (1 - u)^{-1/2}$$
Then, the equation becomes:
$$\frac{d}{d u}\left((1 - u)^{-1/2} \, L\right) = c \, K \, (1 - u)^{-3/2}$$
Integrating once:
$$(1 - u)^{-1/2} \, L = C_{1} - c \, K \, \frac{(1 - u)^{-1/2}}{(-1/2)}$$
$$L = C_{1} \, (1 - u)^{1/2} + 2 \, c \, K$$
The additive constants are unimportant in the definition of the Lagrangian. Therefore we may omit the second constant term.
$$L = C_{1} \, \sqrt{1 - \frac{v^{2}}{c^{2}}}$$
Expanding for small speeds ($v/c \ll 1$), we get:
$$L = C_{1} \, \left[1 - \frac{v^{2}}{2 c^{2}} + o\left(\frac{v^{2}}{c^{2}}\right)\right]$$
Again, we may ignore any constant terms that arise. Comparing this with the expression for the Lagrangian of a free particle (kinetic energy) in classical mechanics:
$$L_{\mathrm{cl}} = \frac{m v^{2}}{2}$$
we conclude that:
$$C_{1} = -m \, c^{2}$$
and we finally get the well known result.

13. Jul 31, 2011

### mmmboh

Wow that's a very interesting derivation, thanks! By the way, has it ever been the case that someone successfully found the laws of a theory based on minimizing the action, without knowing the laws first? Like just by guessing a Lagrangian? I mean when the least action principle came along, the equations of classical mechanics were already known, and the equations for SR were already known before someone used the Lagrangian method as well.

14. Jul 31, 2011

### Dickfore

Yes. The weak forces and the strong forces were discovered in this way, although one does not minimize the action in Quantum Mechanics, but it still plays an important role.