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## Problem involving fundamental principle

 Quote by Alcubierre I do see what it's telling me. The second derivative of h is the gravitational pull to the Earth, or $-g$.
Yes, you've got it. Everything up to here is correct.

 Quote by Alcubierre If I integrate it twice, I obtain $\int^{T}_{0}-g dt$ $= -gt = \frac{dh}{dt}$ and $= \int^{T}_{0}-gtdt$ $= -g \frac{t^{2}}{2} = h$.
Yes, although I would recommend including the integration constants when you write these down. So after two integrations, you would have

$$h(t) = - \frac{1}{2} g t^{2} + v_{0} t + h_{0}$$

where $v_{0}$ and $h_{0}$ are the two integration constants. Since h = 0 when t = 0 (the rocket starts out on the ground), you can immediately see that $h_{0} = 0$. Since h = 0 when t = T (1 hour) also, you can plug t = T into the above and solve for $v_{0}$ to get

$$v_{0} = \frac{1}{2} g T$$

 Quote by Alcubierre Then I plug this in the integrand
You can do that if you want to see what the actual value of the maximized proper time is. But you don't have to to know that the function h(t) that is written down above is the one that maximizes the proper time. The Euler-Lagrange equation already guarantees that. So the function h(t) that is written down above *is* the answer to Feynman's question.

The point Feynman was trying to make, of course, as Samshorn pointed out, is that the rocket doesn't have to fire its engines at all, after the initial thrust that gives it its upward velocity of $v_{0}$ at time t = 0. After that it can just free-fall until it returns to the ground. In other words, the path that maximizes proper time is the free-fall path.

 Quote by Alcubierre Is the last part correct, with the integration I mean?
I think you mixed up h and dh/dt when plugging in to the integrand. When I do it, I get this:

$$\tau = \int_{0}^{T} \left( gh - \frac{1}{2} v^{2} \right) dt = \int_{0}^{T} \left[ g \left( - \frac{1}{2} g t^{2} + v_{0} t \right) - \frac{1}{2} \left( - gt + v_{0} \right)^{2} \right] dt$$

which is straightforward, if a bit tedious, to integrate.

 Quote by Alcubierre Moreover, another conceptual question, and perhaps you mentioned it but I missed it. The second derivative of h, that is the acceleration but more explicitly, the velocity and the height that it yields, those are simply the functions, right? So to maximize those functions, I plug the functions in the proper time integral (above, which gave me a large number) or do I have to find a constraint equation so I can maximize the h(t) and v(t) with a Lagrange multiplier?
I think I answered this above, but to repeat: the function h(t) that is obtained above *is* the function that maximizes the integral. You're not trying to maximize the function(s), you're trying to maximize the integral, and once you have the function h(t) as derived above, you're basically done. You can plug the function into the integral and plug in numbers to get an actual numerical value for the maximized integral, as above, but the maximizing is already done once you've got the function h(t) above.
 Although I mixed up the terms, was the integration correct? But what about the point the guy above made, stating that the gravity is not a constant because the trajectory is not parabolic, rather it is cycloidal due to the rocket reaching a radial position more than three times the radius of the Earth in the hour.

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 Quote by Alcubierre The assumption with my calculations was that gravity was a constant thus making the rocket follow a parabolic trajectory. However, I forgot that the rocket will go into space and gravity is weaker there, so gravity was only "constant" until a certain height for a certain amount of time.
Actually, if the problem covers enough range of height that "g" changes (which this one does, I apologize for not pointing that out earlier), then g itself has to be treated as a function. What Samshorn is describing is the most general way of doing that; but he is also using the fact that we already know what trajectory maximizes the proper time--the free-fall trajectory. So what he is really giving you is a way to determine the free-fall trajectory in the general case.

Another way you could try to work the general case would be to recognize that the term $gh$ in the integrand represents the potential energy per unit mass of the rocket, as I mentioned before. To account for g varying, it's sufficient to just substitute a more general expression for potential energy per unit mass:

$$V(h) = \frac{GM}{R} - \frac{GM}{R + h}$$

where G is Newton's gravitational constant and R is the radius of the Earth. This expresses the same thing as the term $gh$ did, the difference in potential energy per unit mass between the ground and the rocket's height, but it does it in a way that is valid for any height. So the new integrand would be:

$$\frac{GM}{R} - \frac{GM}{R + h} - \frac{1}{2} \left( \frac{dh}{dt} \right)^{2}$$

Then you would go through the same process with the Euler-Lagrange equation that we did before.

 Quote by PeterDonis So the new integrand would be: $$\frac{GM}{R} - \frac{GM}{R + h} - \frac{1}{2} \left( \frac{dh}{dt} \right)^{2}$$ Then you would go through the same process with the Euler-Lagrange equation that we did before.

The integrand being the Lagrange, and I'd do the same thing. All right, thank you for your patience. I will let you know if I have any trouble. Just a quick question before I attempt this, the integration of G and R would be something like Gt and Rt or would I factor or pull those out of the integral?

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 Quote by Alcubierre To integrate that, I'd just normally integrate, well separately, and treat v-not and gravity as a number?
Yes, each term can be integrated separately with respect to t.

 Quote by Alcubierre Which means I'd just tag a t term?
I'm not sure what you mean by this, but each term is just a constant times some power of t, so you just use the rule for integrating powers of the integration variable.

 Quote by Alcubierre But what about the point the guy above made, stating that the gravity is not a constant because the trajectory is not parabolic, rather it is cycloidal due to the rocket reaching a radial position more than three times the radius of the Earth in the hour.
See my post just prior to this one.

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 Quote by Alcubierre The integrand being the Lagrange, and I'd do the same thing. All right, thank you for your patience. I will let you know if I have any trouble. Just a quick question before I attempt this, the integration of G and R would be something like Gt and Rt or would I factor or pull those out of the integral?
G, M, and R are all constants, but they won't occur separately; they'll only occur combined in one term, like the first. But what I wrote down isn't what you integrate, not yet; remember that you have to take the partial derivatives with respect to h and dh/dt of what I wrote down. I shouldn't have called it the "integrand"; it's really the "Lagrangian", just like $gh - 1/2 v^{2}$ was before.
 Oh yes, I know what you meant as the integrand, it is the Lagrangian. I was just asking how I'd integrate, or also, derive it, but I'll just treat them as a constant like you mentioned above.
 Peter, this is what I got: The Lagrangian is, $$L = \frac{GM}{R} - \frac{GM}{R + h} - \frac{1}{2}(\frac{dh}{dt})^{2}$$ Plugging that into the Euler-Lagrange yields, $$\frac{\partial L}{\partial h} = \frac{GM}{R} - \frac{GM}{R + h} = (-) \frac{GM}{(R + h)^{2}}$$ And, $$\frac{d}{dx} \frac{\partial L}{\partial v} = - \frac{1}{2}v^{2} = \frac{d}{dx}(-v) = - \frac{d^{2}h}{dt^{2}}$$. Thus, $$(-) \frac{GM}{(R + h)^{2}} + \frac{d^{2}h}{dt^{2}} = 0$$ $$\frac{d^{2}h}{dt^{2}} = (-) \frac{GM}{(R + h)^{2}}$$ I put the negative sign in parenthesis because I am not sure if it is negative or positive, so I will change that once you take a look. When I integrated that, I got $$\frac{dh}{dt} = (-) \frac{GMt}{(R + h)^{2}} + v_{0}$$ and $$h(t) = (-) \frac{GMt^{2}}{2(R + h)^{2}} + v_{0}t + h_{0}$$ Because h0 = 0 when t = 0, it "goes" away. So when I put that in the integral I got, $$\tau = \int^{T}_{0}\left[\left( (-) \frac{GMt^{2}}{2(R + h)^{2}} + v_{0}t \right) - \frac{1}{2} \left( (-) \frac{GM}{(R + h)^{2}}+ v_{0} \right)^{2}\right] dt$$ Again, the negative sign is in parenthesis because it is in question. So what do you think?

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 Quote by Alcubierre I put the negative sign in parenthesis because I am not sure if it is negative or positive
The negative sign should be there in the last equation for d^2h/dt^2. Taking the derivative of the GM/(R+h) term flips the sign, but there's a negative sign in front of that term in the Lagrangian, so the two negatives cancel. Asn they should, because we *want* the negative sign in the equation for d^2h/dt^2, since we want the rocket to be accelerated downward by gravity.

With the correct sign as above, everything to this point looks correct.

 Quote by Alcubierre When I integrated that, I got $$\frac{dh}{dt} = (-) \frac{GMt}{(R + h)^{2}} + v_{0}$$ and $$h(t) = (-) \frac{GMt^{2}}{2(R + h)^{2}} + v_{0}t + h_{0}$$
Here's where it gets tricky. There is an h in the denominator of the integrand, so you can't just treat it as a constant like "g" was before. That makes the integral a lot harder if you want to try to do it directly. I don't have time to go into that now; the only comment I can make is that for this version of the problem, it's better to "cheat", in a way, by using the knowledge we got from the simpler version we've already done, which showed us that the trajectory that maximized the proper time was the free-fall trajectory. There are lots of different ways to figure out what the free-fall trajectory is, besides maximizing the Lagrangian integral. Samshorn talked about some of them. You can see some of the equations that result on these pages:

http://en.wikipedia.org/wiki/Equatio...a_falling_body

 Quote by PeterDonis ...the only comment I can make is that for this version of the problem, it's better to "cheat", in a way, by using the knowledge we got from the simpler version we've already done, which showed us that the trajectory that maximized the proper time was the free-fall trajectory. There are lots of different ways to figure out what the free-fall trajectory is, besides maximizing the Lagrangian integral. Samshorn talked about some of them.
That's exactly backwards. I explained how to derive the trajectory purely and explicitly from the condition that it maximizes the proper time, as requested. The approach you attempted did not relate to the proper time at all. You simply asserted - incorrectly - that the incremental proper time equals the integral of the classical Lagrangian of a particle in free-fall. But it doesn't, so your whole approach is false from the start. At best (after correcting your erroneous assumption of constant g), all you are doing is struggling to derive the classical trajectory of a particle in free fall. You're not giving a derivation from maximizing proper time, which is what the OP requested. So you have things exactly backwards.

Also, please note that the "simpler derivation" you attempted did not "show us that the trajectory that maximized the proper time was the free-fall trajectory". First, your derivation didn't have anything to do with proper time (aside from an erroneous assertion). Second, and more importantly, free-fall trajectories (i.e., timelike geodesics) are DEFINED as the paths that maximize proper time, so there was nothing to "show". That was the whole point of Feynman's story - that many people (who have learned things by some kind of rote) will go off and struggle with the problem without even realizing that what they are doing is deriving the geodesic equations. Maybe the OP was just trolling to illustrate Feynman's point?

 Quote by Samshorn That was the whole point of Feynman's story - that many people (who have learned things by some kind of rote) will go off and struggle with the problem without even realizing that what they are doing is deriving the geodesic equations.
And it's a really good point Feynman makes, that can be made extensive to science in general wrt people just preaching what they learned by rote, not really understanding it.

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 Quote by Samshorn That's exactly backwards. I explained how to derive the trajectory purely and explicitly from the condition that it maximizes the proper time, as requested.
Hm, yes, you're right, you were talking about doing it using the geodesic equation. Sorry for the mixup on my part.

 Quote by Samshorn The approach you attempted did not relate to the proper time at all. You simply asserted - incorrectly - that the incremental proper time equals the integral of the classical Lagrangian of a particle in free-fall.
Please go back and re-read all my posts. You are correct that the expression $gh - 1/2 v^{2}$ is not the classical Lagrangian, strictly speaking. It's minus the classical Lagrangian divided by the particle's mass. I said that explicitly in at least one previous post, and it seems like the OP understands the distinction.

But minus the classical Lagrangian divided by the particle's mass also happens to be the equation for the elapsed proper time of the particle following the trajectory h(t), provided that g can be considered constant over the range of height covered. That can be shown directly from the metric.

You pointed out, correctly, that g can't be considered constant for the problem as specified, which I hadn't spotted because I hadn't run the actual numbers. If g can't be considered constant, then the integral of minus the Lagrangian divided by the mass gets more complicated, as I noted in my last post.

 Quote by Samshorn At best (after correcting your erroneous assumption of constant g), all you are doing is struggling to derive the classical trajectory of a particle in free fall. You're not giving a derivation from maximizing proper time, which is what the OP requested.
The Euler-Lagrange equation is a general method for finding the maximum of any integral of the given general form. So given an integral that yields elapsed proper time, which the one I gave does for the case where g can be considered constant, applying the Euler-Lagrange equation gives you the function h(t) which maximizes the proper time.

 Quote by Samshorn Also, please note that the "simpler derivation" you attempted did not "show us that the trajectory that maximized the proper time was the free-fall trajectory".
Yes, it did; it showed that the function h(t) which maximizes the proper time integral has the property $d^{2}h / dt^{2} = - g$, for the case where g can be considered constant. That's equivalent to saying that the particle falls freely. If you use the more complicated expression that's valid when g can't be considered constant, you still get the same expression for $d^{2}h / dt^{2}$, just with the "g" dependent on h on the RHS.

 Quote by Samshorn free-fall trajectories (i.e., timelike geodesics) are DEFINED as the paths that maximize proper time, so there was nothing to "show". That was the whole point of Feynman's story - that many people (who have learned things by some kind of rote) will go off and struggle with the problem without even realizing that what they are doing is deriving the geodesic equations. Maybe the OP was just trolling to illustrate Feynman's point?
Even if that's the definition, it's still a worthwhile exercise, IMO, to see how the answer can arise by different routes. But I'll let the OP comment on whether the exercise was worthwhile and what he was trying to illustrate.

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 Quote by PeterDonis But minus the classical Lagrangian divided by the particle's mass also happens to be the equation for the elapsed proper time of the particle following the trajectory h(t), provided that g can be considered constant over the range of height covered. That can be shown directly from the metric.
I should clarify one point here. Strictly speaking, integrating minus the classical Lagrangian divided by the mass gives the *difference* between the elapsed proper time along the trajectory h(t) and the elapsed time at some constant height. The formula $gh - 1 / 2 v^{2}$, or the more complicated formula in h for the case where g is not constant, gives the difference relative to h = 0, i.e., relative to the clock on the ground. For the purpose of answering Feynman's question, finding the function h(t) that maximizes the difference is sufficient; but to actually find the elapsed time, the time on the clock on the ground needs to be added back in. Apologies for not making that clear earlier.

Edit: also, to calculate the actual elapsed time, the formula in h needs to be divided by $c^{2}$, so the units are correct.

 Quote by PeterDonis But minus the classical Lagrangian divided by the particle's mass also happens to be the equation for the elapsed proper time of the particle following the trajectory h(t)...
No it isn't. You'd see this if you actually tried to go from the metric line element to that classical Lagrangian. Please note that changing the sign and dividing by the particle's mass are not the issue. Those are trivial. The issue is that the proper time is the square root of the quadratic line element, whereas the Lagrangian is essentially proportional to the quadratic line element itself - not to the square root. So they are not the same things, nor are they even proportional to each other. So it's incorrect to claim that the classical Lagrangian represents d tau.

 Quote by PeterDonis That can be shown directly from the metric.
No it can't, because it's false. What CAN be shown is that extremizing the classical Lagrangian yields (approximately) the same paths as maximizing the elapsed proper time, but this does not imply that the classical Lagrangian is proportional to the elapsed proper time (which it isn't). And of course, showing that these two things yield the same trajectories is precisely what is needed here, and it's precisely what we do when we derive the geodesic equations from the metric. By leaving out this derivation, and simply taking for granted that the paths yielded by the classical Lagrangian are the ones that maximize proper time, you are simply assuming the very thing you've been asked to demonstrate.

 Quote by PeterDonis Please go back and re-read all my posts. You are correct that the expression $gh - 1/2 v^{2}$ is not the classical Lagrangian, strictly speaking. It's minus the classical Lagrangian divided by the particle's mass. I said that explicitly in at least one previous post, and it seems like the OP understands the distinction.
Again, that's not the issue. The difference between the metric expression for d tau and the classical Lagrangian for a free-falling particle is greater than you realize. You would see this if you actually try to go from one to the other. Furthermore, by assuming the classical Lagrangian of a free particle in a gravitational field with acceleration g, you have already assumed the solution. I mentioned this in my first post. The definition of "g", the acceleration of gravity, is d^2r/dt^2, so this already gives the equation of the trajectory.

 Quote by PeterDonis You pointed out, correctly, that g can't be considered constant for the problem as specified, which I hadn't spotted because I hadn't run the actual numbers. If g can't be considered constant, then the integral of minus the Lagrangian divided by the mass gets more complicated, as I noted in my last post.
Sure, but we still have (by definition) d^2r/dt^2 = g, recognizing that g = -m/r. This is nothing but the geodesic equation (in the classical approximation). Solving this gives the cycloid trajectory. My point is that you aren't showing how this emerges from maximizing the proper time in a spherically symmetrical metric field, you are simply assuming it, and then disguising the assumption by submerging it in some irrelevant manipulations of the classical Lagrangian.

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 Quote by Samshorn The issue is that the proper time is the square root of the quadratic line element
Agreed.

 Quote by Samshorn whereas the Lagrangian is essentially proportional to the quadratic line element itself - not to the square root.
Can you clarify the relationship? The OP may not understand what that means.

 Quote by Samshorn By leaving out this derivation, and simply taking for granted that the paths yielded by the classical Lagrangian are the ones that maximize proper time, you are simply assuming the very thing you've been asked to demonstrate.
Actually, what I did has no necessary connection to the classical Lagrangian. I derived the expression I wrote down for dtau by taking the square root of the line element, as you say. I probably should have made it clearer in my original post that that was what I was doing.

 Quote by Samshorn Furthermore, by assuming the classical Lagrangian of a free particle in a gravitational field with acceleration g, you have already assumed the solution.
I didn't assume that at all. As I said, I derived an expression for the proper time by taking the square root of the line element. I did note that the formula I came up with was equal to minus the classical Lagrangian divided by the mass, but I'll agree that the connection between the two is not as simple as I made it seem. I did not mean to imply that I was taking the classical Lagrangian as a starting point; I wasn't. I made no use of it at all in my actual derivation.

 Quote by Samshorn I mentioned this in my first post. The definition of "g", the acceleration of gravity, is d^2r/dt^2, so this already gives the equation of the trajectory.
That's not the definition of g, at least not if you do the derivation by taking the square root of the line element; "g" is then $GM / R^{2}$, where R is the radius of the Earth.

 Quote by Samshorn My point is that you aren't showing how this emerges from maximizing the proper time in a spherically symmetrical metric field, you are simply assuming it
No, I'm not. I've explained what I did above, but just to recap quickly:

(1) I took the square root of the line element for a purely radial trajectory;

(2) I expanded the square root and retained only leading order terms;

(3) I subtracted out the constant potential at the surface of the Earth;

(4) I wrote r = R + h, where R is the radius of the Earth, and made the approximation h << R (which, as you noted, is not actually valid for the problem as stated, but would be valid for a short enough trajectory);

(5) I wrote "g" for $GM / R^{2}$;

(6) I left out the constant factor $1 / c^{2}$ in the formulas. This changes the units to "energy per unit mass", or "action per unit mass" when integrated with respect to time, but doesn't change anything else;

(7) I wrote down the integral of the resulting expression with respect to time, and applied the Euler-Lagrange equation to it, which finds the function h(t) that maximizes the integral. This will be the function h(t) that maximizes the proper time; and to find the actual proper time, the units will need to be corrected and the (constant) time elapsed on the ground clock (which I effectively subtracted off by subtracting the constant potential at the Earth's surface) will need to be added back.

Nowhere in the above did I use the classical Lagrangian; I merely commented on the similarities of the formula I obtained to the classical Lagrangian. I just did a derivation from the line element. If you want to say that's a roundabout way of deriving the geodesic equation, that's fine. But it's certainly not the same as *assuming* the geodesic equation or the classical Lagrangian; the only thing I *assumed* was the metric.

For ease of reference, here's what you wrote in post #4, where you introduced your derivation:

 Quote by PeterDonis The full equation for proper time in a gravitational field is rather complicated, but the following approximate formula, which is derived from the exact equation (which comes from General Relativity), should work fine for this problem: τ = ∫dτ = ∫ [gh − 1/2(dh/dt)^2]dt where h is your height above the ground (as a function of the time t elapsed on the ground clock), g is the acceleration due to gravity, and T is the time elapsed on the ground clock when you are supposed to return (1 hour)--you're assumed to start at time 0.
So you clearly defined, or at least stated, that g is the acceleration of gravity. Also, there's no indication of how you go from the expression for proper time to this expression (which I think is what the OP was asking for), you just asserted it. Obviously once you get to that expression, which is the classical Lagrangian for a free-falling particle, and have identified g as the acceleration of gravity, i.e., d^2h/dt^2 = -g, you're done. That's the geodesic equation for the trajectory.

 Quote by PeterDonis I derived the expression I wrote down for dtau by taking the square root of the line element, as you say. I probably should have made it clearer in my original post that that was what I was doing... If you do the derivation by taking the square root of the line element; "g" is then $GM / R^{2}$, where R is the radius of the Earth.
Well, you stated in your original post that your "g" represented the acceleration of gravity (see above), so it's understandable that people would think that's what you meant.

By the way, isn't there something odd about your use of "g" (=m/r^2) here? You supposedly derived the expression from the square root of the metric, but that gives the factor (1 - 2m/r)^(1/2), and hence to the first approximation 1 - m/r, and the variable part is just -m/r. At no point does the quantity m/r^2 appear in this derivation. In other words, the metric gives the gravitational potential directly as -m/r, and yet you wrote the gravitational potential not as -m/r, but rather as g*r, where you defined g = -m/r^2. It's hard to see why anyone would write it like that if they were deriving it from the metric. Do you see what I mean?

And when you go on to call "g" the acceleration of gravity, well... Also, the use of h instead of r is strange, as is the fact that you omitted any discussion of deriving it from the square root of the metric to maximize proper time until just now, even though that's precisely what the OP requested...

 Quote by PeterDonis I've explained what I did above, but just to recap quickly:
I don't disagree with what you outlined there, other than the use of the past tense. What you're describing now is not what you presented previously. (See the quote above.) Essentially what you're describing now is what I outlined for deriving the geodesic equation from the metric line element, although you stop short of actually deriving the cycloid solution.

 Quote by PeterDonis If you want to say that's a roundabout way of deriving the geodesic equation, that's fine.
Yes, what you're outlining now is a roundabout way of deriving the geodesic equation from the condition of maximizing the proper time. What you presented previously wasn't.

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 Quote by Samshorn So you clearly defined, or at least stated, that g is the acceleration of gravity.
Which is equal to GM/R^2 at the surface of the Earth. I already agreed that I should have been clearer about the initial presentation. See further comments below.

 Quote by Samshorn Obviously once you get to that expression, which is the classical Lagrangian for a free-falling particle
Only it isn't; it's minus the classical Lagrangian divided by the mass of the particle. And you yourself have already pointed out that the relationship of this expression with the classical Lagrangian is not as simple as I made it seem.

 Quote by Samshorn By the way, isn't there something odd about your use of "g" (=m/r^2) here? You supposedly derived the expression from the square root of the metric, but that gives the factor (1 - 2m/r)^(1/2), and hence to the first approximation 1 - m/r, and the variable part is just -m/r.
Yes, but r is not height above the Earth's surface; it's distance from the Earth's center. The equation I wrote down has h, height above the Earth's surface. As you noticed:

 Quote by Samshorn Also, the use of h instead of r is strange
Not if you're trying to answer the question Feynman actually asked: how to maximize the proper time on the rocket clock, given that an hour elapses *on the ground clock*. The integral I wrote down has "t" as the time of the ground clock, *not* Schwarzschild coordinate time. That means you need to adjust the "zero" of the potential to the radius of the Earth, which is the radius of the ground clock; and it also means you want to change variables from r to h, so that your distance variable is distance from the ground. That's what steps 3 and 4 in what I posted last time are for. When you do all that, you end up with a term GM/R^2 times h; so the "g" pops out, and you might as well recognize that. That's what step 5 in what I posted last time is for.

 Quote by Samshorn as is the fact that you omitted any discussion of deriving it from the square root of the metric to maximize proper time until just now, even though that's precisely what the OP requested...
I thought the OP was asking about the maximization part, not the part about how you derive the initial expression that then gives you the integral to maximize. I would be glad to post more details about how the equation I wrote down gets derived from the metric if the OP requests it.

 Quote by Samshorn Yes, what you're outlining now is a roundabout way of deriving the geodesic equation from the condition of maximizing the proper time. What you presented previously wasn't.
What I outlined in my last post *is* what I presented previously, just with a clearer explanation of the steps. See above.