# Point-of-View Invariance and Noether’s Theorem

Noether’s theorem is an important theorem that relates invariance of space-time transformations to the laws of conservation: space-translation invariance to the conservation of linear momentum, space-rotation invariance to the conservation of angular momentum, and time-translation invariance to the conservation of energy. The models of physics are point-of-view invariant: physical models cannot depend on any particular position in space or moment in time. A video version of this episode showing the equations is available on YouTube.

Where do the laws of physics come from? This question is the subtitle of Victor Stenger’s 2006 book Comprehensible Cosmos. I think this question is one version of the more general guiding question of my whole intellectual life: why are things the way they are? Stenger has a very interesting response to this question, which is based on what he calls principle of point-of-view invariance “The models of physics cannot depend on any particular point of view.”

The path from this principle to the laws of physics goes through an important theorem known as Noether’s Theorem. This theorem was developed by Emmy Noether in 1918. Put briefly, the theorem says that symmetries in a system generate conserved quantities. Anyone who’s studied (and remembers) physics will know of the conservation of momentum, conservation of angular momentum, and the conservation of energy. These conservation laws are absolutely foundational. And what’s remarkable is that there’s a reason for them. These conservation laws come from symmetries. The conservation of momentum, angular momentum, and energy come from symmetries of translation, rotation, and time.

Stenger puts it this way: “In any space-time model possessing time-translation invariance, energy must be conserved. In any space-time model possessing space-translation invariance, linear momentum must be conserved. In any space-time model possessing space-rotation invariance, angular momentum must be conserved. Thus, the conservation principles follow from point-of-view invariance. If you wish to build a model using space and time as a framework, and you formulate that model so as to be space-time symmetric, then that model will automatically contain what are usually regarded as the three most important ‘laws’ of physics, the three conservation principles.”

To me this is quite remarkable. But maybe I’m just easily impressed. So I went online to see how others view all this. I looked up on Quora responses to the question: “What is the significance of Noether’s theorem?” Here are some of the responses:

“I think it is almost the thing that makes sense of physics. Physics is based on a large number of conservation rules – conservation of energy, momentum etc. Without Noether’s Theorem, all you can say is that they are conserved – they are just givens. With the Theorem, you can say that they arise from the symmetries of the space we live in. [In] a space which did not have these symmetries… these conservations would be so different from the space we know as to be unrecognizable. It derives the otherwise arbitrary conservation rules from intuitively understood symmetries. Brilliant.” (Alec Cawley)

“Most of fundamental physics could be interpreted as positing a symmetry, then handing that symmetry off to Ms. Noether and asking her to tell us what the resulting physics is. In other words, without Noether’s Theorem, there wouldn’t be most of modern physics.” (Brent Follin, PhD in Theoretical Cosmology)

And my favorite.

“It’s a matter of life and death! Being a Physics student, the Noether’s theorem is extremely important with everything I do. If it were falsified, the whole structure of modern physics would crumble!” (Abhijeet Borkar, PhD in Physics (Astrophysics))

So it’s a pretty big deal. Hopefully that sparks some interest. Now let’s dig into it and see how it works.

Invariance and Transformations

First, let’s revisit this idea of point-of-view invariance. One of the first things you do in a physics problem is define your coordinates. If you’re on the surface of the Earth you usually set one axis pointing up from the center of the Earth. This is what we’re used to thinking of as “up”. That’s because in our everyday experience there pragmatically is an obvious coordinate system to use. There’s an up and a down. But that’s because we reference our everyday experience relative to Earth, which we’re living on. But we know, at least since the Copernican revolution, that this coordinate system isn’t absolute. The Earth isn’t the center of the universe, even if it is the center of our lived experience. But it’s not just that. There is no center of the universe at all. There’s no absolute up or down.

That doesn’t mean that we don’t use coordinates. Of course we do. We have to. But it does mean that the coordinate system we use is not absolute. We’ll usually use one that makes things easy for our calculations. But the system we represent in one coordinate system can also be represented in a different coordinate system.

This is easy to see with vectors. Let’s represent a vector on an x-y Cartesian coordinate system. The vector will start from the origin (0,0) and go out to point (4,3). What’s the magnitude of this vector? We calculate that by the equation:

√((x2 – x1)^2 + (y2 – y1)^2)

And plugging in our  values:

√((4 – 0)^2 + (3 – 0)^2) = √((4)^2 + (3)^2) = √(16+ 9) = √(25) = 5

The magnitude of this vector is 5.

Now let’s change the coordinate system shifting it 2 to the right and 7 up. Now this same vector starts at (-2,-7) and goes out to (2,-4). What’s the magnitude?

√((-2 – 2)^2 + (-7 – -4)^2) = √((-4)^2 + (3)^2) = √(16+ 9) = √(25) = 5

The magnitude is still 5.

Now let’s go back to the first coordinate system and rotate it 30 degrees counter-clockwise. 30 degrees in radians is π/6 radians. We make this transformation using the rotation matrix

R = [[cos θ,-sin θ], [sin θ, cos θ]]

And multiply R by our vector [[x],[y]].

The result is

Rv = [[x cos θ – y sin θ], [x sin θ + y cos θ]]

Rv = [[3/2 * √(3) – 2], [3/2 * 2 x √(3)]]

Rv = [0.598], [4.964]]

For our transformation θ is π/6 radians. Our new vector coordinates are (0,0) and approximately (0.598,4.964). Now the moment of truth, after all of that. What’s the magnitude? It’s

√((0.598- 0)^2 + (4.964 – 0)^2) = √((0.598)^2 + (4.964)^2) = √(0.358 + 24.642) = √(25) = 5

The magnitude is still 5.

When we look at this visually, it’s actually not surprising. The vector stays the same in all these cases. It’s just the coordinate system that’s moving around. This is the basic idea of invariance. And I think it gives a general sense about how something can remain constant if it doesn’t depend on these coordinate system transformations.

The Lagrangian

Before getting to Noether’s Theorem itself, we need to talk about the Lagrangian because Noether’s Theorem is expressed in terms of it. The Lagrangian is a function that describes the state of a system and is equal to the difference between the total kinetic energy, T, and the total potential energy, V, of a system.

L = T – V

The Lagrangian is used in Lagrangian mechanics and is a different way of looking at systems than Newtonian mechanics. Instead of looking at forces, as in Newtonian mechanics, in Lagrangian mechanics we’re looking at energies. The Lagrangian is a function of spatial coordinates and their derivatives with respect to time. Spatial coordinates could be the familiar Cartesian x,y,z coordinates but it’s customary to generalize these with a single variable. For example, q. For multiple spatial coordinates we can just number them off, q = {q1, q2,…, qn]. The time derivative of q is,. The time derivative of a spatial coordinate is velocity.

So some of the familiar quantities from Newtonian mechanics will be expressed differently in Lagrangian mechanics. Most notably, momentum. In Newtonian mechanics we express momentum as mass times velocity.

p = mv

To express this in terms of a Lagrangian let’s change v to . So,

p = mq̇

Now the Lagrangian is the difference between kinetic energy and potential energy.

L = T – V

Kinetic energy is

T = 1/2 mv^2

Or

T = 1/2 m q̇^2

So we can rewrite the Lagrangian as

L = 1/2 mq̇^2 – V

Now taking the derivative with respect to

δL/δq̇ = mq̇

And mq̇ = p, so

p = δL/δq̇

And that’s the equation for momentum in terms of the Lagrangian.

p = δL/δq̇

So momentum is the derivative of the Lagrangian with respect to velocity. Also the derivative of the kinetic energy with respect to velocity.

The Hamiltonian

Another function I want to go over before moving on to Noether, and that’s the Hamiltonian function. The Hamiltonian is similar to the Lagrangian, except that it’s the sum of kinetic energy and potential instead of the difference between them.

H = T + V

The Hamiltonian is the total energy of the system. And we can express this in terms of the Lagrangian. Since L = T – V we can express the potential energy as

V = T – L

Substituting this into the Hamiltonian

H = T + V

H = T + (T – L)

H = 2T – L

H = 2(1/2 mq̇^2) – L

H = (mq̇)q̇ – L

Since p = mq̇

H = pq̇ – L

And since also p = δL/δq̇

H = (δL/δq̇)q̇ – L

This is the expression for the total energy in terms of the Lagrangian.

H = (δL/δq̇)q̇ – L

The Lagrange-Euler Equation of Motion

One more equation we should introduce before getting into Noether’s theorem is the Lagrange-Euler equation, also called the equation of motion. This has the form

d/dt (δL/δq̇) = δL/δq

What is this equation saying? Let’s translate this out of the Lagrangian form into the more familiar Newtonian quantities. An equivalent form of this equation is:

dp/dt = -δV/δq = F

d(mv)/dt = F

ma = F

This is Newton’s second law. It’s just expressed in a different form with the Lagrangian, which again is:

d/dt (δL/δq̇) = δL/δq

We’ll be plugging this equation into a lot of things in the foregoing so it’s important.

Noether’s Theorem

Now, let’s move to Noether’s theorem. We’ll look at Noether’s theorem for the conservation of momentum, the conservation of angular momentum, and for the conservation of energy.

We start with the Lagrangian as a function of position, q, and velocity, .

L(q, q̇)

What we’re going to do is apply the following transformation on q and .

q q(s)

q̇(s)

If our Lagrangian has symmetry it should not change under this transformation to s. Expressed mathematically this means

d/ds L(q(s), q̇(s)) = 0

Let’s propose that under this transformation that there is a conserved quantity, C, of the following form:

C = (δL/δq̇)(δq/δs)

And since it is a conserved quantity it does not change over time. That is

dC/dt = 0

And here’s the proof for that. Take the proposed conserved quantity C and take the time derivative of it.

C = (δL/δq̇)(δq/δs)

dC/dt = d/dt ((δL/δq̇)(δq/δs))

Since we have two variables, q and , we need to apply the product rule:

dC/dt = d/dt (δL/δq̇) * (δq/δs) + (δL/δq̇) * (δq̇/δs)

Now, recall the Euler-Lagrange equation of motion.

d/dt (δL/δq̇) = δL/δq

We’re going to plug that in here to get.

dC/dt = (δL/δq)(δq/δs) + (δL/δq̇)(δq̇/δs)

What do we have here? The right hand side of this equation is what we get when we apply the chain rule to the derivative of the Lagrangian with respect to s.

d/ds L(q(s), q̇(s)) = (δL/δq)(δq/δs) + (δL/δq̇)(δq̇/δs)

And this is equal to 0. So

dC/dt = (δL/δq)(δq/δs) + (δL/δq̇)(δq̇/δs) = d/ds L(q(s), q̇(s)) = 0

And

dC/dt = 0

So what’s been proved here is that if the Lagrangian, L, does not change with respect to transformation, s, than the conserved quantity, C, doesn’t either.

That’s Noether’s Theorem. Now let’s look at some applications, examples of conserved quantities that result from different symmetries.

Conservation of Linear Momentum

To get the conservation of linear momentum we’re going to say that the Lagrangian is symmetric under continuous translations in space. Our spatial coordinates are

q = {q1, q2,…, qn].

And we’ll apply the transformation

q q(s)

where

q(s) = q + s

So we’re just sliding our coordinate system over by an interval, s.

The conserved quantity C is

C = (δL/δq̇)(δq/δs)

Taking the derivative of q with respect to s

δq/δs = δ/δs (q + s) = 1

So C becomes

C = (δL/δq̇) = p

Which is momentum. So when we apply the spatial transformation

q à q(s)

The conserved quantity, C, is momentum, p. In other words, the conservation of momentum results from symmetry in space. To give some interpretation, this means that the system has no dependence on where it is in space. It’s not being acted upon by any external forces. If there were an external force then it would depend on it’s location in space.

Recall that force is equal to

F = ma

F = m(dv/dt)

F = d/dt (mv)

F = dp/dt

Force is equal to the rate of change in momentum with respect to time. So clearly if there is a non-zero external force acting on the system momentum is not constant.

If there is an applied force external to the system, like with a spring, then momentum is obviously not conserved. And with such forces location makes a difference. With a spring it matters how much the spring is stretched. So momentum is not conserved in such cases where there’s not symmetry in space for that system. But in systems that do have symmetry in space, momentum is conserved.

Conservation of Angular Momentum

To get the conservation of angular momentum we’re going to say that the Lagrangian is symmetric under continuous rotations in space.

We apply the transformation.

q q(s)

In which case s is some angle of rotation. This is a two-dimensional case where q is represented by the matrix

[[q1],[q2]]

We make this transformation using the rotation matrix

R = [[cos s,-sin s], [sin s, cos s]]

And multiply R by our matrix [[q1],[q2]]

The result is

Rq = [[cos s,-sin s], [sin s, cos s]] * [[q1],[q2]]

For very small values of s near 0

sin(s) ≈ s

cos(s) ≈ 1

That’s from Taylor’s series expansion to the first order. This makes the rotation matrix is equal to

[[1, -s], [s, 1]]

So the transformation is

[[1, -s], [s, 1]] * [[q1],[q2]]

The result of this transformation is that

q1 → q1 – s * q2

q2 → q2 + s * q1

For reasons that will be clear shortly, let’s differentiate these.

dq1/ds = -q2

dq2/ds = q1

Now let’s bring in our conserved quantity, C

C = (δL/δq̇)(δq/δs)

And since

q = {q1,q1}

C = (δL/δq̇1)(δq1/δs) + (δL/δq̇2)(δq2/δs)

Or in terms of momentum, p

C = p1 * (δq1/δs) + p2 * (δq2/δs)

The derivatives in this equation are equal to the derivatives we just calculated for q1(s) and q2(s). So, plugging those in:

C = q1 * p2 – q2 * p1

And this is equal to the cross product

C = q x p

Which is angular momentum L. Angular momentum is equal to the cross product of linear momentum and the position vector. So

C = L

The conserved quantity, C, is angular momentum, L. In other words angular momentum results from symmetry of rotation. To give some interpretation again, this is the condition in which the system has no external rotational forces, i.e. torque. To use the example of a spring again, if this were a system where we’re winding up a torsion spring then angular position very much matters. The tighter we wind it up the higher the torque. In that kind of system angular momentum is not conserved. But in the absence of that kind of torque, angular position and rotation don’t matter. So angular momentum is conserved.

Conservation of Energy

To get the conservation of energy we’re going to say that the Lagrangian is symmetric in time. So we have our Lagrangian

L(q, q̇)

And we’re going to say that it doesn’t change with time

dL/dt = 0

Let’s see what follows from this. First let’s to the derivative of the Lagrangian with respect to time. To do this we apply the chain rule.

dL/dt = (δL/δq)(δq/δt) + (δL/δq̇)(δq̇/ δt) + δL/δt

We already set δL/δt to 0 so that goes away. And Let’s simplify δq/δt to q̇ and δq̇/ δt to q̈.

dL/dt = (δL/δq) * q̇ + (δL/δq̇)* q̈

Recall from the Euler Lagrange equation that

δL/δq = d/dt (δL/δq̇)

And we can plug this in to get

dL/dt = d/dt (δL/δq̇) * q̇ + (δL/δq̇)* q̈

This is actually a result of the following application of the product rule:

d/dt (q̇ * (δL/δq̇)) = d/dt (δL/δq̇) * q̇ + (δL/δq̇)* q̈

So we can plug that in to get this more compact result:

dL/dt = d/dt (q̇ * (δL/δq̇))

Rearranging we get:

0 = d/dt (q̇ * (δL/δq̇) – L)

Maybe this looks familiar. Recall that the Hamiltonian, which is equal to the sum of kinetic and potential energy has the following form, expressed in terms of the Lagrangian.

H = (δL/δq̇)q̇ – L

So we can plug this into our equation to get

d/dt (H) = 0

Let’s go ahead express this in terms of kinetic energy, T, and potential energy, V.

H = T + V

d/dt (T + V) = 0

So from our starting condition

dL/dt = 0

We get

d/dt (T + V) = 0

If we set the condition where the Lagrangian doesn’t change with time then the total energy is conserved. This is the Noether symmetry-conservation relation.

What would it be like if things weren’t this way? Under time symmetry things like the gravitational constant and the masses of fundamental particles are constant across time. What if they weren’t? An object elevated above the Earth’s surface has potential energy

V = mgh

Where m is mass, g is acceleration due to gravity, and h is height. Acceleration due to gravity is a function of the gravitational constant G.

g = – GM/r^2

Where M is the mass of the gravitational field source, like the Earth, and r is the distance from the center of the Earth. For the elevated object in our example, none of these values is changing. But what if we could change the gravitational constant G? Say we increase it. Now acceleration due to gravity, g, is higher and potential energy, V, is higher. We’ve created energy from nowhere.

Or another example. At one moment in time you throw a ball up into the air with a certain velocity. So it starts off with a kinetic energy that gets converted to potential energy as it goes up into the sky. But then right as it reaches its highest point you turn the gravitational constant, G, way up and the ball slams to the ground at a much faster velocity than you started with. Again, we’ve created energy from nowhere.

But that doesn’t happen because the laws of physics don’t change over time.

Philosophical reflections

If you were to create a universe how would you do it? I don’t know how to create a universe but if I did my inclination would be to make it as self-designing as possible. Set a few basic rules and let things develop from there. This seems to be the most efficient and elegant way to configure things. I think what makes Noether’s Theorem so marvelous is that we get a great deal of purchase from a rather simple principle: symmetry.

This reminds me a little of what Immanuel Kant tried to do in his moral philosophy. In his 1785 Groundwork of the Metaphysics of Morals he proposed that all moral principles could be derived from one master principle, called the categorical imperative, which was the following:

“Act only according to that maxim whereby you can, at the same time, will that it should become a universal law.”

This is also known as the principle of universalizability. This reminds me of Noether’s Theorem in two ways. First, it’s a simple principle from which others can be derived. Second, it’s a principle of universalizability. We could say that Kant is making his ethics point-of-view invariant. I should act only according to a maxim that could be a universal law, that is not only applicable to me, but to anyone. That’s what it means for it to be universalizable.

In Comprehensible Cosmos Victor Stenger also proposed a principle of universalizability, but for physics. “The models of physics cannot depend on any particular point of view.” That’s the principle of point-of-view invariance. Stenger says of this principle:

“Physics is formulated in such a way to assure, as best as possible, that it not depend on any particular point of view or reference frame. This helps make possible, but does not guarantee, that physical models faithfully describe an objective reality, whatever that may be… When we insist that our models be the same for all points of view, then the most important laws of physics, as we know them, appear naturally. The great conservation principles of energy and momentum (linear and angular) are required in any model that is based on space and time, formulated to be independent of the specific coordinate system used to represent a given set of data. Other conservation principles arise when we introduce additional, more abstract dimensions. The dynamical forces that account for the interactions between bodies will be seen as theoretical constructs introduced into the theory to preserve that theory’s independence of point of view.”

Sort of like Kant’s principle of universalizability, point-of-view invariance keeps us honest. Repeatability of experiments by multiple observers, holding constant only those factors relevant to the experiment, is what ought to finally convince others of the validity of our observations. It won’t do much good if I have a singular experience that only I observe that, in other words, is not universalizable, not point-of-view invariant, but rather strictly tied to me and my point of view. That’s not to say that we don’t have private, subjective experiences that are real. They’re just phenomena of a different nature. Here’s more from Stenger on this point:

“So, where does point-of-view invariance come from? It comes simply from the apparent existence of an objective reality—independent of its detailed structure. Indeed, the success of point-of-view invariance can be said to provide evidence for the existence of an objective reality. Our dreams are not point-of-view invariant. If the Universe were all in our heads, our models would not be point-of-view invariant. Point-of-view invariance generally is used to predict what an observer in a second reference frame will measure given the measurements made in the first reference frame.”

I think that’s well put. And that line that “Our dreams are not point-of-view invariant” is one I think about a lot.

Noether’s Theorem is absolutely foundational. It’s been said that Noether’s theorem is second only to the Pythagorean theorem in its importance for modern physics. It’s remarkable that just one, compact principle can produce so much of what we observe in the world.

Reference Material

Baez, J. (2020b, February 17). Noether’s Theorem in a Nutshell. University of California, Riverside. Retrieved March 25, 2022, from https://math.ucr.edu/home/baez/noether.html

Branson, J. (2012, October 21). Recalling Lagrangian Mechanics. University of California San Diego. Retrieved March 25, 2022, from https://hepweb.ucsd.edu/ph110b/110b_notes/node86.html

Greene, B. (2020, May 11). Your Daily Equation #25: Noether’s Amazing Theorem: Symmetry and Conservation. YouTube. Retrieved March 25, 2022, from https://www.youtube.com/watch?v=w7Q5mQA_74o&t=428s

Khan, G. J. H. What Is Noether’s Theorem? Ohio State University. Retrieved March 25, 2022, from https://math.osu.edu/sites/math.osu.edu/files/Noether_Theorem.pdf

Stenger, V. J. (2006). The comprehensible cosmos: Where do the laws of physics come from? Prometheus Books.

Washburn, B. (2018, March 13). Introduction to Noether’s Theorem and Conservation Principles. YouTube. Retrieved March 25, 2022, from https://www.youtube.com/watch?v=XxxUEHD8OZM&t=827s

## One thought on “Point-of-View Invariance and Noether’s Theorem”

1. Brent Meeker says:

Vic was a friend and I’m pleased that his ideas are being appreciated.

Like