2013-07-08 20:32:34 8 Comments

Just as background, I should say I am a mathematics grad student who is trying to learn some physics. I've been reading "The Theoretical Minimum" by Susskind and Hrabovsky and on page 134, they introduce *infinitesimal transformations.* Here's the first example they use:

Consider a particle moving in the x,y plane under the influence of a potential, $V$, which depends only on the radius, with Lagrangian:

$L=\frac{m}{2}(\dot{x}^2+\dot{y}^2)-V(x^2+y^2)$

This is clearly invariant under rotations:

$x \rightarrow x\cos \theta + y\sin \theta$

$y \rightarrow -x\sin \theta + y\cos \theta$

All well and good. Now they say "consider what happens ... when the angle $\theta$ is replaced by an infinitesimal angle $\delta$." Already I could say "What the heck is $\delta$ really?", but I'm willing to play along with my intuition. Since $\delta$ is infinitesimal, we work to first order and say

$\cos \delta=1$ and $\sin \delta= \delta$.

Plugging this into our rotation formulas above, we obtain:

$x \rightarrow x+y\delta$

$y \rightarrow y-x\delta$

By differentiating, we see that:

$\dot{x} \rightarrow \dot{x}+\dot{y}\delta$

$\dot{y} \rightarrow \dot{y}-\dot{x}\delta$

Plugging these into the Lagrangian and ignoring terms higher than first order, we see that the Lagrangian is invariant under this transformation.

My main problem with all of this is that I don't understand what the physical nature of an infinitesimal transformation actually is. All I got from the above was that if you do this formal calculation by following rules like "only work to first order in $\delta$," then the Lagrangian is invariant. This is in contrast to the case where we have an actual transformation, like a rotation, where there is no question about what is going on physically.

I would also like to know how all of this relates to rigorous mathematics. In mathematics, I can't recall ever using infinitesimals in an argument or calculation, so it would be useful if there was some way of formulating the above in terms of limits/ derivatives/ differential forms (for example). I sense a connection to Lie Algebras, as the infinitesimal version of the rotation is $(I+A)$ where $I$ is the identity matrix and $A$ is an element of the Lie Algebra of $SO(2)$.

Here are some questions whose answers I believe may be useful to me (feel free to answer some or all):

-What is an infinitesimal quantity like $\delta$ to the physicist?

-Why do physicists argue using infinitesimals rather than "standard" calculus?

-What is the physical meaning of infinitesimal transformation? How does it relate to Lie Algebras?

-Is there a rigorous theoretical apparatus for justifying the computations shown above?

-What is meant by the Lagrangian being invariant under infinitesimal transformations?

If any of the questions seem too vague, please say so. Thanks in advance for your insights!

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## 3 comments

## @Charles Francis 2020-04-11 17:25:31

It is worth noting that the rigorous mathematical development of analysis, due to Weierstrass et al. does not invoke the concept of an infinitesimal, or even admit such a concept. Instead if formalises the notion of something being "really small" in the $\epsilon$-$\delta$ definition of a limit.

As far as physics is concerned $\epsilon$ generally refers to experimental precision, or margin of error. It means that $\delta$ is a small enough number that making it any smaller makes no practical difference to predictions.

## @joshphysics 2013-07-08 21:56:37

When I asked my undergrad analytic mechanics professor "what does it mean for a rotation to be infinitesimal?" after he hand-wavily presented this topic in class, he answered "it means it's really small." At that point, I just walked away. Later that day I emailed my TA who set me straight by pointing me to a book on Lie theory.

Fortunately, I don't intend to write an answer like my professor's.

In general, whenever you see the term "infinitesimal BLANK" in physics, you can be relatively certain that this is merely a placeholder for "first order (aka linear) approximation to BLANK."

Let's look at one of the most important examples.

Infinitesimal transformations.To be more rigorous about this, let's consider the special case of "infinitesimal transformations." If my general terminological prescription above is to be accurate, we have to demonstrate that we can make the concept of a "first order approximation to a transformation" rigorous, and indeed we can.

For concreteness, let's restrict the discussion to tranformations on normed vector spaces. Let an open interval $I=(a,b)$ containing $0$ be given, and suppose that $T_\epsilon$ is a transformation on some normed vector space $X$ such that $T_0(x)$ is the identity. Let $T_\epsilon$ depend smoothly on $\epsilon$, then we define the infinitesimal version $\widehat T$ of $T_\epsilon$ as follows. For each point $x\in X$, we have $$ \widehat T_\epsilon(x) = x + \epsilon\frac{\partial}{\partial\epsilon}T_{\epsilon}(x)\bigg|_{\epsilon=0} $$ The intuition here is that we can imagine expanding $T_\epsilon(x)$ as a power series in $\epsilon$; $$ T_\epsilon(x) = x + \epsilon T_1(x) + \mathcal O(\epsilon^2) $$ in which case the above expression for the infinitesimal version of $T_\epsilon$ gives $$ \widehat {T}_\epsilon(x) = x+\epsilon T_1(x) $$ so the transformation $\widehat T$ encodes the behavior of the transformation $T_\epsilon$ to first order in $\epsilon$. Physicists often call the transformation $T_1$ the

infinitesimal generatorof $T_\epsilon$.Example.Infinitesimal rotations in 2DConsider the following rotation of the 2D Euclidean plane: $$ T_\epsilon = \begin{pmatrix} \cos\epsilon& -\sin\epsilon\\ \sin\epsilon& \cos\epsilon\\ \end{pmatrix} $$ This transformation has all of the desired properties outlined above, and its infinitesimal version is $$ \widehat T_\epsilon = \begin{pmatrix} 1& 0\\ 0& 1\\ \end{pmatrix} + \begin{pmatrix} 0& -\epsilon\\ \epsilon& 0\\ \end{pmatrix} $$ If we act on a point in 2D with this infinitesimal transformation, then we get a good approximation to what the full rotation does for small values of $\epsilon$ because we have made a linear approximation. But independent of this statement, notice that the infinitesimal version of the transformation is rigorously defined.

Relation to Lie groups and Lie algebras.Consider a Lie group $G$. This is essentially a group $G$ that can also be thought of as a smooth manifold in such a way that the group multiplication and inverse maps are also smooth. Each element of this group can be thought of as a transformation, and we can consider a smooth, one-parameter family of group elements $g_\epsilon$ with the property that $g_0 = \mathrm{id}$, the identity in the group. Then as above, we can define an infinitesimal version of this one-parameter family of transformations; $$ \widehat g_\epsilon = \mathrm{id} + \epsilon v $$ The coefficient $v$ of $\epsilon$ in this first order approximation is basically (this is exactly true for matrix Lie groups) an element of the Lie algebra of this Lie group. In other words, Lie algebra elements are infinitesimal generators of smooth, one-parameter families of Lie group elements that start at the identity of the group. For the rotation example above, the matrix $$ \begin{pmatrix} 0& -1\\ 1& 0\\ \end{pmatrix} $$ is therefore an element of the Lie algebra $\mathfrak{so}(2)$ of the Lie group $\mathrm{SO}(2)$ of rotations of the Euclidean plane. As it turns out, transformations associated with Lie groups are all over the place in physics (particularly in elementary particle physics and field theory), so studying these objects becomes very powerful.

Invariance of a lagrangian.Suppose we have a Lagrangian $L(q,\dot q)$ defined on the space (tangent bundle of the configuration manfiold of a classical system) of generalized positions $q$ and velocities $\dot q$. Suppose further that we have a transformation $T_\epsilon$ defined on this space, then we say that the Lagrangian is

invariantunder this transformation provided $$ L(T_\epsilon(q,\dot q)) = L(q, \dot q) $$ The Lagrangian is said to beinfinitesimally invariantunder $T_\epsilon$ provided $$ L(T_\epsilon(q,\dot q)) = L(q, \dot q) + \mathcal O(\epsilon^2) $$ In other words, it is invariant to first order in $\epsilon$. As you can readily see, infinitesimal invariance is weaker than invariance.Interestingly,

only infinitesimal invariance of the lagrangian is required for certain results (most notably Noether's theorem) to hold. This is one reason why infinitesimal transformations, and therefore Lie groups and Lie algebras, are useful in physics.Application: Noether's theorem.Let a Lagrangian $L:\mathscr C\times\mathbb R$ be given where $\mathscr C$ is some sufficiently well-behaved space of paths on configuration space $Q$. Given a one-parameter family of transformations $T_\epsilon:\mathscr C\to\mathscr C$ starting at the identity. The first order change in the Lagrangian under this transformation is $$ \delta L(q,t) = \frac{\partial}{\partial\epsilon}L(T_\epsilon(q),t)\Big |_{\epsilon=0} $$ One (not the strongest) version of Noether's theorem says that if $L$ is local in $c$ and its first derivatives, namely if there is a function $\ell$ such that (in local coordinates on $Q$) $L(q,t) = \ell(q(t), \dot q(t), t)$ and if $$ \delta L(q,t) = 0 $$ for all $c\in\mathscr C$ that satisfy the equation of motion, namely if the Lagrangian exhibits infinitesimal invariance, then the quantity $$ G = \frac{\partial \ell}{\partial \dot q^i}\delta q^i, \qquad \delta q^i(t) = \frac{\partial}{\partial\epsilon}T_\epsilon(q)^i(t)\bigg|_{\epsilon=0} $$ is conserved along solutions to the equations of motion. The proof is a couple lines; just differentiate $G$ evaluated on a solution with respect to time and use the chain rule and the Euler-Lagrange equations to show it's zero.

## @Brian Klatt 2013-07-08 22:55:13

This answer was great. If only the text said "infinitesimals=linear approximation" I would have been fine! As a technical point, where you first define $\hat{T}_{\epsilon}$, the right hand side demands that these transformations occur on a space with an additive structure, so in general manifolds and topological spaces don't seem to support such a concept as stated. Also, the space of transformations where $\epsilon$ takes values would have to have a smooth structure. But none of this has bearing on the particular case I was interested in.

## @joshphysics 2013-07-08 23:43:18

@user26804 What? you mean you don't like excessive verboseness? Thanks for calling me out on being sloppy w.r.t. additive structure. I've edited the answer to consider normed vector spaces. Note, however, that in the case of Lie groups, one does not need such an additive structure to define the Lie algebra which instead is defined rigorously in terms of the tangent space at $\mathrm{id}$ in the manifold. In this context, the corresponding transformation on the group is generated via the exponential map.

## @Brian Klatt 2013-07-09 00:01:34

By text I had meant The Theoretical Minimum, where my issue first cropped up; I greatly enjoyed reading your entire answer. I speak the language of Lie groups, manifolds, tensors, Hilbert spaces, etc. etc. quite well, it's just that physicists can say simple things in a language I'm not totally comfortable with and I get all backwards!

## @Selene Routley 2013-07-09 00:35:33

@joshphysics "Interestingly, only infinitesimal invariance of the lagrangian is required for certain results (most notably Noether's theorem) to hold." Now that is interesting. I'll have to go back and look a proof of Noether's theorem. Can you suggest one for me emphasizing what you've just said? It is relatively easy to prove that $C^1$ assumptions about a Lie group imply $C^\infty$ - probably that's what's going on here. (not to be confused with $C^0\Rightarrow C^\infty$ proven by Montgomery, Gleason, Zippin - Hilbert's fifth problem)

## @joshphysics 2013-07-09 00:42:16

@user26804 Oh haha woops, well I'm glad it wasn't overly verbose. Totally identify with the language barrier; I was a pure math major in undergrad.

## @joshphysics 2013-07-09 01:06:10

@WetSavannaAnimalakaRodVance I added a baby section on Noether's theorem at the end. I'm not sure there's a connection to the theorem on Lie groups to which you refer.

## @Qmechanic 2013-07-09 22:16:14

Suggestion to answer (v10): Replace

"[...]in such a way that the group multiplication is also smooth"with"[...]in such a way that the group multiplication and inversion maps are also smooth".## @Selene Routley 2013-07-10 01:28:36

@joshphysics Thanks for the proof. It is deceptively simple and intuitive. There likely isn't a direct connexion as you say, but I suspect the definition $\mathscr C$ ("sufficiently well behaved") is what is nailing things down to make first order behaviour suffice.

## @joshphysics 2013-07-10 02:34:45

@WetSavannaAnimalakaRodVance Sure thing.

## @Jahan Claes 2016-05-30 17:27:17

"As you can readily see, infinitesimal invariance is weaker than invariance." I don't think that's true. If you're infinitensimally invariant under an infinitesimal transformation, that means you are invariant under an "integrated" version of the transformation. In other words, if you're infinitesimally invariant under an infinitesimal version of a continuous transformation, you're invariant under that continuous, non-infinitesimal transformation.

## @joshphysics 2016-05-30 18:55:31

@JahanClaes What about $T_\epsilon(q,\dot q) = (\epsilon q, \dot q)$ and $L(q, \dot q) = m\dot q^2/2 + kq^2/2$? Notice that $L$ does not change to first order in $\epsilon$, but it is not invariant under $T_\epsilon$.

## @Jahan Claes 2016-05-30 19:12:54

@joshphysics Maybe I'm reading your notation wrong, but is $L$ really invariant under that transformation? Before the transformation, $L$ has a potential energy term $kq^2/2$, and after the transformation it has no potential energy term to first order, since the potential energy is quadratic in $\epsilon$.

## @joshphysics 2016-05-30 19:15:31

@JahanClaes no youre right, that was a lapse in brain function. Ill attempt to think of a legitimate counterexample.

## @Jahan Claes 2016-05-30 19:19:42

@joshphysics Thinking about it a little more, I think my first statement was correct, my second wasn't. In other words, if you integrate an infinitesimal transformation you get a continuous transformation which preserves $L$, but there exist continuous tranformations that don't preserve $L$, yet do preserve $L$ infinitesimally.

## @Jahan Claes 2016-05-30 19:22:48

For example, in a rotationally invariant $L$, rotations preserve $L$ but the transformation $S_a(x,y)=(x+ay,y-ax)$ does not preserve $L$, even though the infinitesimal versions of both transformations are the same. However, if you integrate the infinitesimal transformation, you get a rotation, not $S_a$.

## @joshphysics 2016-05-30 19:41:46

@JahanClaes I'm inclined to agree, although I'd need to think about that first statement more. In any event, your $S_a$ is a good counterexample.

## @rschwieb 2018-08-17 20:51:37

This is a very useful answer for mathematicians like me struggling to understand physicists. Thanks!

## @user4552 2013-07-18 21:02:29

To most physicists, it means the same thing it meant to Newton, Leibniz, and Euler. It means something that's small enough that we can apply a certain informally defined body of techniques to it and get correct answers.

To physicists who know more about math after 1960, it means the same thing, except that they are aware that the body of techniques was eventually formally defined and proved to be consistent. There are in fact multiple ways of doing this, and for a physicist's purposes, it never matters which formalization is used. Some examples of formalizations are non-standard analysis and smooth infinitesimal analysis.

The important thing to understand here is that the results obtained by people like Euler were

right. There is nothing wrong with the informal versions of the techniques.Infinitesimals

werethe standard calculus for hundreds of years. The reason the subject was originally developed in terms of infinitesimals is because it's the most natural and comfortable way of reasoning about the subject. Often one finds that when a particular argument can be expressed either using infinitesimals or using epsilon-delta methods, the depth of quantifiers is lower by one in the former case.In the physical example you gave, it means what it says: an infinitesimal rotation. The Lie group of rotations is a continuous group connected to the identity. You can use infinitesimals as generators.

Yes, in fact there's more than one, as explained above.

## @Qmechanic 2014-07-20 16:37:50

It's a keeper:

Often one finds that when a particular argument can be expressed either using infinitesimals or using epsilon-delta methods, the depth of quantifiers is lower by one in the former case.