Playing with Gauges
Introduction
A “gauge” is some description of a physical system. Typically, there could be an infinite number of gauges, all of them describing the same system.
Because we could work with any one of them and it would not affect the Physics of a system, we choose the simplest ones and play with them in order to figure out hidden properties of the system that we are working with. This act is called as “fixing a gauge.”
Because of this reason, gauge fixing has become a standard way to do Physics; it can simplify complicated equations and give hidden insights into physical behavior with minimal effort.
Here, we give a very basic example of working with gauges by exploring how a charged particle moves in a magnetic field.
Before we start playing with gauges, we need to have the Lagrangian of a system. So far we have not seen how to construct one; It would be a shame not to do it as it is quite a lot of fun. For this reason, this article is divided into 2 parts:
If you are new to these series of posts then visit this page, to see a list of previous posts, sources, motivation etc.,
This post might feel a little daunting to some of you; it is filled with several beautiful and very clever techniques.
The first time I came across this stuff, it felt a little like black-magic to me. But it is easy enough and can be understood with little effort, after a couple of readings. Think of this as studying computer code, and have fun with this.
The stuff here is pretty standard, one that any average undergraduate student would come across. In case you want to learn more or need help understanding this, you may refer to this book. (People tell me that it is good.) A quick search on Live Journal also yielded this post, by a Physics student. It might be nice to take a look at it, if you want to learn this stuff from a person who has actually taken a formal class on this topic. (as opposed to reading the ramblings of some layman who learns it in his free-time.)
Any doubts or errors, do let me know.
OK, let us begin!
1. Charged particle in a magnetic field
The force acting on a charged particle moving in a magnetic field is given by:
Call this as equation (1)
Here:
We got the above expression for the force experimentally. We want to construct a Lagrangian, in order to give us deeper insight into what is happening. This is what we will do now.
To simplify our calculations, let us assume from now on that:
We will construct the Lagrangian in stages. We first consider the case of a free particle. We then add the effect of the magnetic field to it.
1.1.1 Free particle case
The Lagrangian of a free particle is just its Kinetic Energy.
1.1.2 Magnetic Field
Although, there could be magnetic mono-poles (point-sources of magnetic field), nobody has yet discovered them. Thus we can assume that a magnetic field,B has zero divergence, i.e.:
In such a case, it is always possible to express B in terms of the curl of another field, A, which is called as the “vector potential.”
Call this as equation (2)
We shall be using the vector potential, A instead of the magnetic field, B, while constructing the Lagrangian.
Just as a refresher, for those who are a little rusty in their vector calculus, here are the components of B, expressed in terms of the vector potential A. Please note that these are general equations which apply to all vector products and not just ∇, A and B. We will be using these equations a lot, later on:
1.1.3 Construction
Here is the construction of the Lagrangian. An explanation follows after it.
Equation (6) is the most important one of this lot.
As mentioned earlier, we have constructed the Lagrangian in stages, by considering the case of a free particle (The first integral) and then adding the effect of the magnetic field. (The second integral.)
The first integral is simple enough; Can you recognize the expression for the Kinetic Energy in it?
As for the second integral, what we have done is nothing special. We have simply made use of a standard result from vector calculus: “The effect of a field along some curve is given by the line integral of the field along the curve.” That is it! The rest of the equations are easier to explain.
Equation (7) is got by dividing and multiplying the second integral by dt.
Equation (8) is the expression for the Lagrangian. This simply follows from the definition for the Action which is:
1.2 Testing the Lagrangian
OK, we have constructed a candidate Lagrangian, but how do we know that it is the correct one which describes the physics of the system accurately? The method is as follows:
The above logic works because we have F = ma, from Newton’s Second Law, where m = mass and a = acceleration.
Got it? OK, here is the working for the above logic. Here we will choose the x direction for our purposes. (The other directions have the same operations.)
From the Lagrangian in equation (8), we have:
Equation (11) may seem confusing to you. In order to understand how we got it, you need to see that there are 2 things going on here:
OK, so much for that! Similarly, from the same Lagrangian, we can get:
Call this as equation (12)
From the Euler-Lagrange trick, we can equate equations (11) and (12):
Call this as equation (13)
Re-arranging the terms to make the L.H.S in the familiar F = ma form, we get the following result:
We get equation(14) by canceling the terms involving ẋ on both sides.
The remaining equations all follow from basic results of vector components, which are given in equations (3), (4) and (5).
The result that we get is the one that we were looking for; it is the same as equation (1). We can apply the same procedure to the y direction as well, and we will get the same result. Thus, the Lagrangian that we constructed is the right one!
OK, now that we have got our Lagrangian, it is time to do all kinds of cool stuff with it, in order to find out more about the behavior of the system. This is what the remainder of this article will be all about.
2. Magnetic Fields do no work
Let us find out the Hamiltonian (or energy) of the system. It is given by:
We get equation (20), by substituting for the values of L and pi from equations (8) and (10), respectively.
We see that the energy of a charged particle moving around in a magnetic field that is perpendicular to its plane of motion, is the same as that of a free particle! (i.e. it is just equal to the Kinetic Energy.)
Thus, the magnetic field here does no work; any effect it has is to merely changes the direction of motion of the particle.
Now we move on to working with gauges.
3. Gauge Invariance
We mentioned earlier, that a gauge is a redundant description of the physics of a system and that there are an infinite number of them for any given system.
No matter what gauge we end up working with, the physics of a system remains unaffected. This phenomenon is called “gauge invariance.” This section will tell you why this is so.
We have shown the magnetic field, B can be represented in terms of the vector potential A, back in equation (2).
Now let us add a gradient to the vector potential, A. Each component of A is then affected in the following manner:
Call this as equation (23)
Here, λ is some scalar function.
OK, now let us work with this modified vector potential. We will do this by looking at equation(3) and substituting the value for the new value of A. The working is given below:
Equation (28) follows from (27), because it does not matter in what order you take the partial derivatives. (We shall be using this simple but useful “trick” in future articles related to Hamiltonian Mechanics and Poisson Brackets, so it is nice to remember it!)
From the above set of equations we see that the value of Bz remains unchanged no matter what gradient we add, i.e. it remains unaffected no matter what gauge we chose. This illustrates the idea that a scalar field is “curl-free.”
4. Using Gauges
Here, we shall use gauge fixing to find out what is the motion of a charged particle in a magnetic field. This is not readily apparent from equation (1), but we shall see how it is elegantly solved by choosing the appropriate gauges.
In order to simplify our calculations, we had previously made the following assumptions:
To this we will add one more assumption, that the magnetic field is uniform. Again, this is in order to simplify our calculations.
Because of all the above assumptions, we have:
Call this as equation (29)
Here, b is some constant.
Recall, from equation (3), we have:
In order to satisfy equations (29) and (3), we could have the following gauges (of course there are many more):
Gauge 1 Gauge 2 Gauge 3 Ax = 0 Ax = -by Ax = -
y Ay = bx Ay = 0 Ay =
x A z = 0 A z = 0 A z = 0
Among these, the “simplest” ones are Gauge 1 and 2, because they each have only 1 non-zero term in them. Let us work with them.
4.1 Working with Gauge 1
Recall, that for gauge 1, we have:
Ax = 0
Ay = bx
Az = 0
We substitute these values into the Lagrangian in equation (8). This is shown below:
Notice a curious thing about equation (32) - it does not have a “y” term in it. This means that the Lagrangian (and consequently the Action) does not change in the y direction.
Thus, there is a symmetry in the y direction and py is conserved.
Let us find the value of py. It is given by:
Since this is a conserved quantity, its value remains constant.
Let us set the value of this constant to be 0. The idea here is to see if we can derive some properties for the value at 0 and see if we can use the insight that we gain from this, to figure out what the properties could be for any arbitrary constant value, k.
Thus, we have:
We will stop here. Now, let us do the same thing for Gauge 2 and get an expression for ẋ.
4.2 Working with Gauge 2
Recall, that for gauge 2, we have:
Ax = -by
Ay = 0
Az = 0
Again, we substitute these values into the Lagrangian to get:
From the Lagrangian, we notice that px is conserved. Its value is:
Setting the value of px to 0, we get:
OK, let us recap what we have done. Using 2 different gauges, we have found out the equations for x and y. Now we can get an idea of what type of motion is occurring.
4.3 Equations of Motion
Let me tell you before-hand what type of motion is happening here - it is circular motion. It is not hard to see why. The proof is as follows:
Assume a particle moving around the origin, in a circle. In terms of rotational co-ordinates we have:
Let us now assume that the particle moves with some angular frequency ω. Thus we have:
Here, t is the time period.
Thus, the equations for x and y change to:
Taking the time derivative on both sides gives us:
Substituting for x and y values in the above equations we get:
We see that equations (48) and (49) correspond to equations (43) and (36), respectively. This proves that when the momenta are equal to zero, the motion of the charged particle corresponds to that of a circle centered at the origin.
4.3.1 Angular Frequency
From equations (48) and (43), we see that the value of ω is:
Call this as equation (50)
This is called as the cyclotron frequency. This is a constant if we assume non-relativistic motion.
This is because m does not change if the charged particle is moving at speeds very much slower than light. Also, q remains constant, and b is constant because we have assumed a uniform magnetic field.
5. Centers and Momenta
Thus far, we have seen that if the canonical momentum was 0, then it corresponds to a circular motion of the particle around the origin. What if the canonical momentum was non-zero? What sort of curve would then result?
We will resolve this question here, albeit in an indirect manner; by moving the center of the circle by an arbitrary amount and seeing how this in turn, affects the momentum. The derivation follows below:
Recall, that for a circle centered at the origin we had the following equations of motion:
If we move the center to some arbitrary point, say (x0,y0), then the above equations change to:
Taking the time derivatives on both sides we get:
Notice that these are the same as equations (46) and (47), respectively.
[Remember, the value of ω does not change when we move the center. This is on account of us having determined ω to be a constant in the previous section.]
Let us now substitute the values of x,y,ẋ and ẏ from above, into the equations for px and py, that we got while playing with Gauges 1 and 2.
Substituting into equation (34) we get:
Note, that we go from equation (57) to (58) by substituting the value for ω from equation (50).
In a similar manner, we get an expression for px by working with equation (41):
Thus, from equations (59) and (64), we see that a non-zero value of the momentum, merely corresponds to a shift in the centers of a circle. This is exactly what happens in real-life! (It is given the name of “Hall Effect.”)
Is this cool or what?!
A “gauge” is some description of a physical system. Typically, there could be an infinite number of gauges, all of them describing the same system.
Because we could work with any one of them and it would not affect the Physics of a system, we choose the simplest ones and play with them in order to figure out hidden properties of the system that we are working with. This act is called as “fixing a gauge.”
Because of this reason, gauge fixing has become a standard way to do Physics; it can simplify complicated equations and give hidden insights into physical behavior with minimal effort.
Here, we give a very basic example of working with gauges by exploring how a charged particle moves in a magnetic field.
Before we start playing with gauges, we need to have the Lagrangian of a system. So far we have not seen how to construct one; It would be a shame not to do it as it is quite a lot of fun. For this reason, this article is divided into 2 parts:
- First, we shall construct the Lagrangian for the above system. This will comprise about a third of length of the article.
- The rest of the article, will show you how to fix a gauge and why it is so useful.
If you are new to these series of posts then visit this page, to see a list of previous posts, sources, motivation etc.,
This post might feel a little daunting to some of you; it is filled with several beautiful and very clever techniques.
The first time I came across this stuff, it felt a little like black-magic to me. But it is easy enough and can be understood with little effort, after a couple of readings. Think of this as studying computer code, and have fun with this.
The stuff here is pretty standard, one that any average undergraduate student would come across. In case you want to learn more or need help understanding this, you may refer to this book. (People tell me that it is good.) A quick search on Live Journal also yielded this post, by a Physics student. It might be nice to take a look at it, if you want to learn this stuff from a person who has actually taken a formal class on this topic. (as opposed to reading the ramblings of some layman who learns it in his free-time.)
Any doubts or errors, do let me know.
OK, let us begin!
1. Charged particle in a magnetic field
The force acting on a charged particle moving in a magnetic field is given by:
Call this as equation (1)
Here:
- q is the charge carried by the particle
is the velocity of the particle
is the magnetic field. We assume that it could have different values at different points in space. But, it does not vary with time.
We got the above expression for the force experimentally. We want to construct a Lagrangian, in order to give us deeper insight into what is happening. This is what we will do now.
To simplify our calculations, let us assume from now on that:
- The charged particle moves in the x∕y plane
- We have a magnetic field acting perpendicular to it, in the z direction.
We will construct the Lagrangian in stages. We first consider the case of a free particle. We then add the effect of the magnetic field to it.
1.1.1 Free particle case
The Lagrangian of a free particle is just its Kinetic Energy.
1.1.2 Magnetic Field
Although, there could be magnetic mono-poles (point-sources of magnetic field), nobody has yet discovered them. Thus we can assume that a magnetic field,B has zero divergence, i.e.:
In such a case, it is always possible to express B in terms of the curl of another field, A, which is called as the “vector potential.”
Call this as equation (2)
We shall be using the vector potential, A instead of the magnetic field, B, while constructing the Lagrangian.
Just as a refresher, for those who are a little rusty in their vector calculus, here are the components of B, expressed in terms of the vector potential A. Please note that these are general equations which apply to all vector products and not just ∇, A and B. We will be using these equations a lot, later on:
1.1.3 Construction
Here is the construction of the Lagrangian. An explanation follows after it.
Equation (6) is the most important one of this lot.
As mentioned earlier, we have constructed the Lagrangian in stages, by considering the case of a free particle (The first integral) and then adding the effect of the magnetic field. (The second integral.)
The first integral is simple enough; Can you recognize the expression for the Kinetic Energy in it?
As for the second integral, what we have done is nothing special. We have simply made use of a standard result from vector calculus: “The effect of a field along some curve is given by the line integral of the field along the curve.” That is it! The rest of the equations are easier to explain.
Equation (7) is got by dividing and multiplying the second integral by dt.
Equation (8) is the expression for the Lagrangian. This simply follows from the definition for the Action which is:
1.2 Testing the Lagrangian
OK, we have constructed a candidate Lagrangian, but how do we know that it is the correct one which describes the physics of the system accurately? The method is as follows:
- We set up the Euler-Lagrange equation, from this Lagrangian.
- We then, we re-arrange the terms so that we have “ma” on the L.H.S.
- If the R.H.S can be reduced to the expression of the force, given in equation (1), then we know that our Lagrangian is correct.
The above logic works because we have F = ma, from Newton’s Second Law, where m = mass and a = acceleration.
Got it? OK, here is the working for the above logic. Here we will choose the x direction for our purposes. (The other directions have the same operations.)
From the Lagrangian in equation (8), we have:
Equation (11) may seem confusing to you. In order to understand how we got it, you need to see that there are 2 things going on here:
- We know that the magnetic field, B is a function of only the positions. It follows from this that A is also a function of the positions. How can we then take the time derivative here?
This is because, even if A does not explicitly depend on time, (i.e. there is no time term in its arguments) there is an implicit time dependence here. That is because we are considering a particle moving in the field, A. Thus, the effect of A on the particle will vary with time.
- Another thing to remember, (I kept forgetting this point and hence got stuck one to many a time while deriving this stuff :-)) is that A is a function of the positions, means that each Ai is a function of the positions.
In other words, you have Ax as a function of x,y,z. A similar logic holds for Ay and Az. Thus, it is legal to have expressions like
here.
OK, so much for that! Similarly, from the same Lagrangian, we can get:
Call this as equation (12)
From the Euler-Lagrange trick, we can equate equations (11) and (12):
Call this as equation (13)
Re-arranging the terms to make the L.H.S in the familiar F = ma form, we get the following result:
We get equation(14) by canceling the terms involving ẋ on both sides.
The remaining equations all follow from basic results of vector components, which are given in equations (3), (4) and (5).
The result that we get is the one that we were looking for; it is the same as equation (1). We can apply the same procedure to the y direction as well, and we will get the same result. Thus, the Lagrangian that we constructed is the right one!
OK, now that we have got our Lagrangian, it is time to do all kinds of cool stuff with it, in order to find out more about the behavior of the system. This is what the remainder of this article will be all about.
2. Magnetic Fields do no work
Let us find out the Hamiltonian (or energy) of the system. It is given by:
We get equation (20), by substituting for the values of L and pi from equations (8) and (10), respectively.
We see that the energy of a charged particle moving around in a magnetic field that is perpendicular to its plane of motion, is the same as that of a free particle! (i.e. it is just equal to the Kinetic Energy.)
Thus, the magnetic field here does no work; any effect it has is to merely changes the direction of motion of the particle.
Now we move on to working with gauges.
3. Gauge Invariance
We mentioned earlier, that a gauge is a redundant description of the physics of a system and that there are an infinite number of them for any given system.
No matter what gauge we end up working with, the physics of a system remains unaffected. This phenomenon is called “gauge invariance.” This section will tell you why this is so.
We have shown the magnetic field, B can be represented in terms of the vector potential A, back in equation (2).
Now let us add a gradient to the vector potential, A. Each component of A is then affected in the following manner:
Call this as equation (23)
Here, λ is some scalar function.
OK, now let us work with this modified vector potential. We will do this by looking at equation(3) and substituting the value for the new value of A. The working is given below:
Equation (28) follows from (27), because it does not matter in what order you take the partial derivatives. (We shall be using this simple but useful “trick” in future articles related to Hamiltonian Mechanics and Poisson Brackets, so it is nice to remember it!)
From the above set of equations we see that the value of Bz remains unchanged no matter what gradient we add, i.e. it remains unaffected no matter what gauge we chose. This illustrates the idea that a scalar field is “curl-free.”
4. Using Gauges
Here, we shall use gauge fixing to find out what is the motion of a charged particle in a magnetic field. This is not readily apparent from equation (1), but we shall see how it is elegantly solved by choosing the appropriate gauges.
In order to simplify our calculations, we had previously made the following assumptions:
- The charged particle moves in the x∕y plane
- We have a magnetic field acting perpendicular to it, in the z direction.
To this we will add one more assumption, that the magnetic field is uniform. Again, this is in order to simplify our calculations.
Because of all the above assumptions, we have:
Call this as equation (29)
Here, b is some constant.
Recall, from equation (3), we have:
In order to satisfy equations (29) and (3), we could have the following gauges (of course there are many more):
Gauge 1 Gauge 2 Gauge 3 Ax = 0 Ax = -by Ax = -
y Ay = bx Ay = 0 Ay =
x A z = 0 A z = 0 A z = 0
Among these, the “simplest” ones are Gauge 1 and 2, because they each have only 1 non-zero term in them. Let us work with them.
4.1 Working with Gauge 1
Recall, that for gauge 1, we have:
Ax = 0
Ay = bx
Az = 0
We substitute these values into the Lagrangian in equation (8). This is shown below:
Notice a curious thing about equation (32) - it does not have a “y” term in it. This means that the Lagrangian (and consequently the Action) does not change in the y direction.
Thus, there is a symmetry in the y direction and py is conserved.
Let us find the value of py. It is given by:
Since this is a conserved quantity, its value remains constant.
Let us set the value of this constant to be 0. The idea here is to see if we can derive some properties for the value at 0 and see if we can use the insight that we gain from this, to figure out what the properties could be for any arbitrary constant value, k.
Thus, we have:
We will stop here. Now, let us do the same thing for Gauge 2 and get an expression for ẋ.
4.2 Working with Gauge 2
Recall, that for gauge 2, we have:
Ax = -by
Ay = 0
Az = 0
Again, we substitute these values into the Lagrangian to get:
From the Lagrangian, we notice that px is conserved. Its value is:
Setting the value of px to 0, we get:
OK, let us recap what we have done. Using 2 different gauges, we have found out the equations for x and y. Now we can get an idea of what type of motion is occurring.
4.3 Equations of Motion
Let me tell you before-hand what type of motion is happening here - it is circular motion. It is not hard to see why. The proof is as follows:
Assume a particle moving around the origin, in a circle. In terms of rotational co-ordinates we have:
Let us now assume that the particle moves with some angular frequency ω. Thus we have:
Here, t is the time period.
Thus, the equations for x and y change to:
Taking the time derivative on both sides gives us:
Substituting for x and y values in the above equations we get:
We see that equations (48) and (49) correspond to equations (43) and (36), respectively. This proves that when the momenta are equal to zero, the motion of the charged particle corresponds to that of a circle centered at the origin.
4.3.1 Angular Frequency
From equations (48) and (43), we see that the value of ω is:
Call this as equation (50)
This is called as the cyclotron frequency. This is a constant if we assume non-relativistic motion.
This is because m does not change if the charged particle is moving at speeds very much slower than light. Also, q remains constant, and b is constant because we have assumed a uniform magnetic field.
5. Centers and Momenta
Thus far, we have seen that if the canonical momentum was 0, then it corresponds to a circular motion of the particle around the origin. What if the canonical momentum was non-zero? What sort of curve would then result?
We will resolve this question here, albeit in an indirect manner; by moving the center of the circle by an arbitrary amount and seeing how this in turn, affects the momentum. The derivation follows below:
Recall, that for a circle centered at the origin we had the following equations of motion:
If we move the center to some arbitrary point, say (x0,y0), then the above equations change to:
Taking the time derivatives on both sides we get:
Notice that these are the same as equations (46) and (47), respectively.
[Remember, the value of ω does not change when we move the center. This is on account of us having determined ω to be a constant in the previous section.]
Let us now substitute the values of x,y,ẋ and ẏ from above, into the equations for px and py, that we got while playing with Gauges 1 and 2.
Substituting into equation (34) we get:
Note, that we go from equation (57) to (58) by substituting the value for ω from equation (50).
In a similar manner, we get an expression for px by working with equation (41):
Thus, from equations (59) and (64), we see that a non-zero value of the momentum, merely corresponds to a shift in the centers of a circle. This is exactly what happens in real-life! (It is given the name of “Hall Effect.”)
Is this cool or what?!