# Translational Invariance

1. Apr 18, 2016

### spaghetti3451

1. The problem statement, all variables and given/known data

Consider a system of objects labeled by the index $I$, each object located at position $x_{I}$. (For simplicity, we can consider one spatial dimension, or just ignore an index labeling the different directions.) Because of translational invariance

$x'_{I}=x_{I}+\delta x$

where $\delta x$ is a constant independent of $I$, we are led to define new variables

$x_{IJ} \equiv x_{I}-x_{J}$

invariant under the above symmetry. But these are not independent, satisfying

$x_{IJ}=-x_{JI}, \qquad x_{IJ} + x_{JK} + x_{KI} = 0$

for all $I,J,K$. Start with $x_{IJ}$ as fundamental instead, and show that the solution of these constraints is always in terms of some derived variables $x_{I}$ as in our original definition. (Hint: What happens if we define $x_{1}=0$?) The appearance of a new invariance upon solving constraints in terms of new variables is common in physics: e.g., the gauge invariance of the potential upon solving the source-free half of Maxwell’s equations.

2. Relevant equations

3. The attempt at a solution

If $x_{1}=0$, then $x'_{1}=\delta x$.

Not sure where to go from here.

2. Apr 18, 2016

### Samy_A

If you set $x_1=0$, how could you define $x_I$ such that the constraints $x_{I1}=-x_{1I}$ are verified?

3. Apr 18, 2016

### spaghetti3451

Aren't the constraints $x_{I1} = - x_{1I}$ always satisfied regardless of whether $x_{1} = 0$ or $x_{1} \neq 0$, simply because $x_{IJ} = x_{I} - x_{J}$?

4. Apr 18, 2016

### Samy_A

What the exercise asks you to do is the following:
Assume that you have the numbers $x_{IJ}$, and that they satisfy the constraints $x_{IJ}=-x_{JI}, \qquad x_{IJ} + x_{JK} + x_{KI} = 0$.
Now show that there exists numbers $x_I$, such that $x_{IJ}=x_I-x_J$.

In other words, the $x_I$ are not given, you have to define them to suit the given $x_{IJ}$.

The hint is to set one value, $x_1=0$.
Now for an arbitrary index $I$, how could you define $x_I$ so that they suit the given $x_{IJ}$ and satisfy the constraint $x_{I1}=-x_{1I}$.
(I use the index 1 here because you start by setting $x_1=0$. Of course you will have to show that the $x_I$ you construct satisfy the constraints $x_{IJ}=-x_{JI}, \ x_{IJ} + x_{JK} + x_{KI} = 0$).

5. Apr 18, 2016

### spaghetti3451

I need to find the solutions $x_{IJ}$ (in terms of some variables $x_{I}$ and $x_{J}$) which satisfy the constraints $x_{IJ}=-x_{JI}$ and $x_{IJ}+x_{JK}+x_{KI}=0$.

The funny part is that we already know that the solution is $x_{IJ} = x_{I} - x_{J}$, but we have to assume that we don't know the solution $x_{IJ}$ and set about finding it.

For $x_{1}=0$ and $x_{I1} = - x_{1I}$, one possible choice of $x_{I1}$ and $x_{1I}$ are $x_{I}$ and $-x_{I}$ respectively.

$x_{IJ}$ is antisymmetric, and $x_{I} - x_{J}$ is also antisymmetric, therefore $x_{IJ} = x_{I} - x_{J}$ is a possible solution of the constraints.

To prove that the solution $x_{IJ} = x_{I} - x_{J}$ satisfies the constraints, we need to plug in the solution into the constraints.

Am I on the right track?

My idea is to conjecture a possible form of $x_{IJ}$ and then to check if the conjectured solution satisfies the constraints.

6. Apr 18, 2016

### Samy_A

You misunderstand the exercise.
First they show you how the numbers $(x_I)_I$ allow to define the numbers $(x_{IJ})_{IJ}$ by setting $x_{IJ}=x_I-x_J$.
It then appears that these numbers $(x_{IJ})_{IJ}$ satisfy the constraints $x_{IJ}=-x_{JI}, \ x_{IJ} + x_{JK} + x_{KI} = 0$.

Now they want you to reverse this process.
Assume you have numbers $(x_{IJ})_{IJ}$ that satisfy the constraints $x_{IJ}=-x_{JI}, \ x_{IJ} + x_{JK} + x_{KI} = 0$. No $x_I$ are given or assumed.
What you are asked to do is: show that you can define numbers $(x_I)_I$ such that $x_{IJ}=x_I-x_J$.

7. Apr 18, 2016

### spaghetti3451

I still don't see how my understanding of the exercise differs from yours.

Perhaps if you provide a few initial lines of the solution, that will clarify things.

8. Apr 18, 2016

### Samy_A

Set $x_1=0$. Now, you need $x_I$ to satisfy $x_{1I}=x_1 - x_I$. That should give you a clue about how to define $x_I$.
Once that is done, you will have to show that $x_{IJ}=x_I-x_J$ holds for all indices $I,J$.

9. Apr 18, 2016

### spaghetti3451

Let's define an origin such that $x_{1} = 0$ and let's define a coordinate $x_{I}$ for particle I such that $x_{I} = x_{I1}$.

Now, the first constraint $x_{IJ} = -x_{JI} \implies x_{I1} = -x_{1I} \implies x_{1I} = - x_{I}$.

Next, the second constraint $x_{1I} + x_{IJ} + x_{J1} = 0 \implies - x_{I} + x_{IJ} + x_{J} = 0 \implies x_{IJ} = x_{I} - x_{J}$.

Am I correct?

10. Apr 18, 2016

### Samy_A

As the $x_{IJ}$ are given, and you have to define the $x_I$, it would make more sense to write the last equation as $x_I=-x_{1J}$. But given that, that is a good definition for $x_I$.
Yes.

11. Apr 18, 2016

### spaghetti3451

I modified my post a little (before I was alerted about your new post). Would you check if my modified post is a better answer?

12. Apr 18, 2016

### Samy_A

Yes, totally correct.