Zwiebach Page 197: Checking Translations & Commutators

[ehrenfest]

Homework Statement

In equation 11.60, Zwiebach is supposed to be checking whether p^mu(tau) generates translations.

Firstly, I am not sure why he checks i epsilon_rho p^rho(tau) instead.

Second, I am not sure why taking the commutator with x^mu(tau) shows that this generates translations. He claims that taking commutators is the way to generate symmetry transformations above, but I do not see where he justifies that statement?

Homework Equations

The Attempt at a Solution

 

[Jimmy Snyder]

This is an example of one of Zwiebach's habits that drives me up a wall. He is not the only author of physics books to do it. In school, I studied mathematics, not physics and in all of the textbooks I read, I never saw this practice. Here I go up on my soapbox.

Zwiebach writes:
\delta x^{\mu}(\tau) = [i\epsilon_{\rho}{p^{\rho}(\tau),x^{\mu}(\tau)] = i\epsilon_{\rho}(-i\eta^{\rho\mu}) = \epsilon^{\mu}.

If this were changed to
[i\epsilon_{\rho}{p^{\rho}(\tau),x^{\mu}(\tau)] = i\epsilon_{\rho}(-i\eta^{\rho\mu}) = \epsilon^{\mu} = \delta x^{\mu}(\tau).

Then there would be 2 major improvements. First, each of the = signs in the equation could be justified by the previous text. The way Zwiebach wrote it, there is no justification for the first equal sign. Second, you could read the meat of the equation by removing the interior of the equation. In other words, if the author is trying to show that A = B, then A = B = C = D has B buried in the interior of the line where the reader has to dig it out. while A = C = D = B has the intended equation on the ends of the line where they are readable.

In the present case, Zwiebach is trying to show that

[i\epsilon_{\rho}{p^{\rho}(\tau),x^{\mu}(\tau)] = \delta x^{\mu}(\tau).
This is what he means by the expression i\epsilon_{\rho}p^{\rho}(\tau) generates the translation (11.57)

 

[ehrenfest]

I see. Thanks.