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screenshot of text

First, I believe there is a trivial error. The second equation should have another $\Delta t$ multiplying everything on the right. It is divided out later when the equation I set equal to 0.

  1. Given that $L$ is a function of $ x_7, x_8, x_9$ how can he justify evaluating it at single points?

  2. Also, why are points 8 and 9 used rather than 7 and 8?

For context, this is the last equation on page 112 and the first equation on page 113 of Susskind's The Theoretical Minimum.

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Yes, I agree it looks like the second equation is missing an overall factor of $\Delta t$ on the right.

The action is a function of all the $x$s for the whole trajectory, but the Lagrangian is not. It is only a function of a position and a velocity. So it makes perfect sense to evaluate it using the position and velocity for a single time.

Points 8 and 9 are used because (I guess) the action is defined as a sum of terms involving $x_n$ and $x_{n-1}$, so the terms that involve $x_8$ are the 8th and 9th terms. If he had defined the action using $x_n$ and $x_{n+1}$ then he would have had to take the 7th and 8th terms instead. It's just a choice of convention.

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  • $\begingroup$ I don't understand how he can have an equation with two variables and then treat it as though it is composed of one in the next step. Also, the $x_7$ and $x_8$ terms are included in the series also. See the second line of the first equation. $\endgroup$
    – MadScientist
    Commented Dec 11, 2014 at 8:12
  • $\begingroup$ @MadScientist No, the $n$th term depends on $x_n$ and $x_{n-1}$. So the $n = 8$ term depends on $x_8$ and $x_7$, while the $n = 9$ term depends on $x_9$ and $x_8$. Those are the only terms that depend on $x_8$. So to get the total dependence on $x_8$ he only needs to evaluate things at $n = 8$ and $n = 9$. $\endgroup$
    – Nathan Reed
    Commented Dec 11, 2014 at 8:15
  • $\begingroup$ I'm not seeing why he can't choose points 7 and 8. $\endgroup$
    – MadScientist
    Commented Dec 11, 2014 at 8:18
  • $\begingroup$ @MadScientist The $n = 7$ term only depends on $x_7$ and $x_6$, so it's irrelevant to $x_8$. And if you leave out the $n = 9$ term you don't capture all the dependence of the action on $x_8$. $\endgroup$
    – Nathan Reed
    Commented Dec 11, 2014 at 8:19
  • $\begingroup$ Thanks. I see it now. Everything makes sense. I was confused. $\endgroup$
    – MadScientist
    Commented Dec 11, 2014 at 8:25
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In order to add more context to anyone else that stumbles upon this, in the book directly above the screenshot posted, the author defines the action as follows: $$A = \sum_{n} L\left(\frac{{x_{n+1}} - x_n}{\Delta t},\frac{{x_n} + x_{n+1}}{2}\right)\Delta t.$$

I agree with the accepted answer that a factor of $\Delta t$ is missing on the right.

According to the convention chosen by the author, I believe that the 7th and 8th points should have been used for evaluation.

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Here are my thoughts (I may be wrong). The first equation states the action which is the integral of the Lagrangian with respect to time. Integrating is the equivalent of finding the areas of strips, which is why the first equation has delta t multiplied by the Lagrangian evaluated at two points. Differentiating the Action removes the delta t terms.

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I still think it might be wrong. the 8th term is dependant on x8 and x9 and the 7th term is dependant on x7 and x8. Therefore L should be evaluated at n=8 and n= 7.

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