Derivation of the Noether current - Lorentz Transformation

[Higgsono]

We make an infinitesimal Lorentz transformation of the Lagrangian and require it to be invariant. We then arrive at the following expression.

$$\epsilon^{\mu\nu}j_{\mu\nu} = P_{\mu}\epsilon^{\mu\nu}X_{\nu}$$ which can be written as

$$\epsilon^{\mu\nu}j_{\mu\nu} = -\frac{1}{2}\epsilon^{\mu\nu}(X_{\mu}P_{\nu}-P_{\nu}X_{\mu})$$

The left hand side is anti symmetric since $\epsilon^{\mu\nu}$ is anti symmetric matrix. Next in the derivation the book says that we can cancel $\epsilon^{\mu\nu}$ on both sides of the equation. And we get

$$j_{\mu\nu} = -\frac{1}{2}(X_{\mu}P_{\nu}-P_{\nu}X_{\mu})$$

But I don't understand why. How can we cancel $\epsilon^{\mu\nu}$ when we are summing over it's indices on both sides?

 

[haushofer]

You can't. Your j, as it stands there contracted with the epsilon symbol, is only determined up to a symmetric 2-tensor. Since I don't know which book you use, I can't comment on its details or intentions.

 

[Higgsono]

You can't. Your j, as it stands there contracted with the epsilon symbol, is only determined up to a symmetric 2-tensor. Since I don't know which book you use, I can't comment on its details or intentions.

It's from "A first course in string theory by Barton Zwieback"

 

[stevendaryl]

The issue of "cancelling off" $\epsilon^{\mu \nu}$ is this:

If you have two antisymmetric tensors $A_{\mu \nu}$ and $B_{\mu \nu}$, and for every possibly tensor $\epsilon^{\mu \nu}$,

$\epsilon^{\mu \nu} A_{\mu \nu} = \epsilon^{\mu \nu} B_{\mu \nu}$

then it must be that

$A_{\mu \nu} = B_{\mu \nu}$

This is easy enough to prove. Pick two indices, $\alpha$ and $\beta$, and let $\epsilon^{\mu \nu}$ be defined as follows:

• If $\mu = \alpha$ and $\nu = \beta$, then $\epsilon^{\mu \nu} = 1$
• If $\mu = \beta$ and $\nu = \alpha$, then $\epsilon^{\mu \nu} = -1$
• For all other values of $\mu$ and $\nu$, $\epsilon^{\mu \nu} = 0$

Then in that case,

$\epsilon^{\mu \nu} A_{\mu \nu} = A_{\alpha \beta} - A_{\beta \alpha} = 2 A_{\alpha \beta}$
$\epsilon^{\mu \nu} B_{\mu \nu} = 2 B_{\alpha \beta}$

So $\epsilon^{\mu \nu} A_{\mu \nu} = \epsilon^{\mu \nu} B_{\mu \nu}$ implies that

$A_{\alpha \beta} = B_{\alpha \beta}$

 

[Higgsono]

The issue of "cancelling off" $\epsilon^{\mu \nu}$ is this:

If you have two antisymmetric tensors $A_{\mu \nu}$ and $B_{\mu \nu}$, and for every possibly tensor $\epsilon^{\mu \nu}$,

$\epsilon^{\mu \nu} A_{\mu \nu} = \epsilon^{\mu \nu} B_{\mu \nu}$

then it must be that

$A_{\mu \nu} = B_{\mu \nu}$

This is easy enough to prove. Pick two indices, $\alpha$ and $\beta$, and let $\epsilon^{\mu \nu}$ be defined as follows:

• If $\mu = \alpha$ and $\nu = \beta$, then $\epsilon^{\mu \nu} = 1$
• If $\mu = \beta$ and $\nu = \alpha$, then $\epsilon^{\mu \nu} = -1$
• For all other values of $\mu$ and $\nu$, $\epsilon^{\mu \nu} = 0$

Then in that case,

$\epsilon^{\mu \nu} A_{\mu \nu} = A_{\alpha \beta} - A_{\beta \alpha} = 2 A_{\alpha \beta}$
$\epsilon^{\mu \nu} B_{\mu \nu} = 2 B_{\alpha \beta}$

So $\epsilon^{\mu \nu} A_{\mu \nu} = \epsilon^{\mu \nu} B_{\mu \nu}$ implies that

$A_{\alpha \beta} = B_{\alpha \beta}$

I don't understand. It is simply not true that $\epsilon^{\mu \nu} = +-1$ in general. Why do you make that assumption? $\epsilon^{\mu \nu}$ have 6 independent parameters.

 

[Higgsono]

oh wait, is it because $\epsilon^{\mu \nu}$ for fixed indices define an independent parameter so that we can wary each individual parameter by itself and equality should still hold? Like what you do when you equate the coordinates between two vectors if you know that A=B. But in this case, we have a sum of scalars on both sides, but it is a sum in terms of independent parameters, so maybe we must equate terms that have the same parameter?

 

[stevendaryl]

I don't understand. It is simply not true that $\epsilon^{\mu \nu} = +-1$ in general. Why do you make that assumption? $\epsilon^{\mu \nu}$ have 6 independent parameters.

If the equation $\epsilon^{\mu \nu} A_{\mu \nu} = \epsilon^{\mu \nu} B_{\mu \nu}$ is true for EVERY value of $\epsilon^{\mu \nu}$, then it must be true for the special case that I gave.

 

[stevendaryl]

Maybe it would be more obvious if instead of dealing with an abstract index $\mu \nu$, we picked specific indices.

Noether's theorem tells us that if $\mathcal L$ is left unchanged by the variation $x^\mu \rightarrow x^\mu + \delta x^\mu$, then the quantity

$\frac{\partial \mathcal L}{\partial \dot{x}^\mu} \delta x^\mu$

is a constant of the motion. Which can be rewritten as:

$P_\mu \delta x^\mu$

So let's look at the specific case where:

$\delta x^0 = \epsilon x_1$
$\delta x^1 = - \epsilon x_0$

and $\delta x^\mu = 0$ for $\mu = 2$ or $3$.

then the conserved quantity is:
$P_\mu \delta x^\mu$
$= P_0 \delta x^0+ P_1 \delta x^1$
$= P_0 \epsilon x_1 - P_1 \epsilon x_0$

So we come to the conclusion that $\epsilon (P_0 x_1 - P_1 x_0)$ is conserved. Dividing through by $\epsilon$, we come to the conclusion that:

$P_0 x_1 - P_1 x_0$

is a conserved quantity.

If instead, we had chosen a variation with $\delta x^0 = \epsilon x_2$ and $\delta x^2 = -\epsilon x_0$, then we would come to the conclusion that

$P_0 x_2 - P_2 x_0$

We can reason, for all 6 possibilities for $\mu$ and $\nu$ that

$P_\mu x_\nu - P_\nu x_\mu$

is a conserved quantity.