What is the significance of the T - V Lagrangian of a system?

[James1238765]

TL;DR Summary

What is the significance of the T - V Lagrangian of a system?

Let E be a fixed immutable quantity. E can be freely exchanged between T and V, as long as $$T + V = E$$

1. What does the quantity $$\int_x T - V $$ signify? What is the importance of this quantity?
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Let E now be the budget of a factory. E can either be spent on T or V in any proportion on any given day.

2. What does the quantity $$\int_x T - V $$ signify for the factory?

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Let E now be the total energy of a system. E can be freely exchanged between the kinetic energy T or the potential energy V as particle moves from point A to point B.

3. What does the quantity $$\int_x T - V $$ signify for the choice of motion of the system?

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Suppose a particle wants to take all possible paths s connecting A to B. Does a weighted quantity of the integral of T - V make sense over infinitely many paths?

4. What does the quantity $$\int_s T - V$$ signify?

 

[Dale]

TL;DR Summary: What is the significance of the T - V Lagrangian of a system?

1. What does the quantity ∫xT−V signify? What is the importance of this quantity?

It is the action. The importance of the action is that actual trajectory of the system is the one that minimizes the action. The energy, while useful, doesn’t distinguish between different trajectories since it is conserved.

TL;DR Summary: What is the significance of the T - V Lagrangian of a system?

What does the quantity ∫xT−V signify for the factory?

Does the factory production obey a Lagrangian principle?

TL;DR Summary: What is the significance of the T - V Lagrangian of a system?

3. What does the quantity ∫xT−V signify for the choice of motion of the system?

See 1 above.

 

[Vanadium 50]

It's not really a quantity.. (Yes, it has a name, action, but in my experience this is more confusing than helpfu). It is a function, a function of coordinates and derivatives. It's value is not in its value (sorry...couldn't resist) but in how it depends on coordinates and their derivatives. For some situations, it's actually zero,

 

[vanhees71]

The importance of the action (which cannot be exaggerated since the action principle is the prime mathematical tool of all of fundamental theoretical physics) is not in that it is an "observable" in any sense.

In the here discussed "Lagrangian form", it's a functional of the possible trajectories of the particle,
$$S[x]=\int_{t_1}^{t_2} \mathrm{d} t L(x,\dot{x}).$$
Where you put in $x(t)$ for $x$ and the derivative $\mathrm{d} x(t)/\mathrm{d} t$ for $\dot{x}$.

Then the trajectory the particle follows is that one which makes the action stationary (usually it makes the action even minimal, and that's why Hamilton's principle is also known as the least-action principle) given the endpoint $x(t_1)=x_1$ and $x(t_2)=x_2$ are fixed.

Variational calculus then leads to the Euler-Lagrange equations for this trajectory:
$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{x}}=\frac{\partial L}{\partial x}.$$
Indeed with
$$L=T-V=\frac{m}{2} \dot{x}^2 -V(x)$$
you get the canonical momentum
$$p=\frac{\partial L}{\partial \dot{x}} = m \dot{x},$$
which for the here used usual Cartesian coordinate is just the usual "kinetic momentum" of the particle. This changes if you use other coordinates or extent the principle to more general equations of motion (e.g., the motion of a charged particle in a magnetic field), but that's not so important here. Then the Euler-Lagrange equation reads
$$\dot{p}=m \ddot{x}=-V'(x).$$
This also generalizes to motion in two or three dimensions, leading to
$$\dot{\vec{p}}=m \ddot{\vec{x}}=-\vec{\nabla} V(\vec{x})=\vec{F}(\vec{x}),$$
i.e., the significance of the action, using the Lagrangian $L=T-V$, is simply that the action principle leads to Newton's equations of motion.

The most obvious advantage is that you can introduce arbitrary "generalized coordinates" to describe the particle trajectories, including also possible "geometric constraints".

E.g., you can introduce polar coordinates for a pendulum, i.e., a particle fixed at a thread of a given length moving in a plane; then $r=L=\text{const}$, where $L$ is the length of the thread, and the only relevant degree of freedom is the angle wrt. the direction gravitational force of the Earth. Then you have
$$(x_1,x_2)=L (\sin \varphi,\cos \varphi)$$
and
$$T=\frac{m}{2} (\dot{x}_1^2+\dot{x}_2^2) = \frac{m L^2}{2} \dot{\varphi}^2$$
and
$$V=-m g x_2=-m g L \cos \varphi.$$
Thus
$$L=\frac{m L^2}{2} \dot{\varphi}^2 + m g L \cos \varphi.$$
The equation of motion follows from the Euler-Lagrange equation
$$p_{\varphi}=\frac{\partial L}{\partial \dot{\varphi}}= m L^2 \dot{\varphi},$$
i.e., here the canonical momentum is the angular momentum, and
$$\dot{p}_{\varphi}=m L^2 \ddot{\varphi}=\frac{\partial L}{\partial \varphi} = -m g L \sin \varphi,$$
i.e., you get pretty easily directly the well-known equation of motion for the pendulum,
$$\ddot{\varphi}=-\omega^2 \sin \varphi, \quad \omega=\sqrt{\frac{g}{L}},$$
without the need to think about contraint forces (the tension of the thread in this case).

Another advantage is that you can derive Noether's theorem from the action principle, i.e., the relation between symmetries and conservation laws, which is the key element from classical mechanics to understand, how to get to quantum mechanics in a heuristic but pretty convincing and "beautiful" way. For this, however, the Hamilton form of the action principle and the Lie-algebra structure imposed by the Poisson brackets on the phase-space functions, is more appropriate, but this can be derived from the Lagrange form.

 

[Cleonis]

TL;DR Summary: What is the significance of the T - V Lagrangian of a system?

As you point out: we have that over time the sum of potential energy and kinetic energy remains at the same value.
$$ E_{potential} + E_{kinetic} = E_{total} \tag{1} $$
(From here on I will abbreviate as $E_p$ and $E_k$ )

This can be expressed in differential form. As an object is moving along, the derivative with respect to time of the total energy is zero
$$ \frac{d(E_k + E_p)}{dt} = 0 \tag{2} $$
As we know: we can understand that as follows: as potential energy and kinetic energy are being exchanged they both change at the same rate, in opposite direction. That is: (2) can be rearranged to a form that states that the the time derivative of the kinetic energy is equal to the time deriviative of minus the potential energy:
$$ \frac{d(E_k)}{dt} = \frac{d(-E_p)}{dt} \tag{3} $$
You ask about the meaning of the integral with respect to time of the Lagrangian $(E_k - E_p)$.
I assume you are asking about Hamilton's stationary action, since Hamilton's action is defined as the integral with respect to time of the Lagrangian $(E_k - E_p)$.

The variation of a trial trajectory is variation of the position coordinate of the trial trajectory.

The importance of that is this: derivative with respect to variation and derivative with respect to position coordinate are one and the same thing.

The change that is evaluated when applying variation of a trial trajectory is a hypothetical change. In that hypothetical change you are looking for the sweet spot: the spot where rate of change of kinetic energy matches the rate of change of potential energy.

The following expression differentiates with respect to the position coordinate 's', not with respect to the time coordinate 't'.
$$ \frac{d(E_k)}{ds} = \frac{d(E_p)}{ds} \tag{4} $$
The point is: at the point in variation space where the trial trajectory coincides with the true trajectory we have that (3) holds good, and when (3) holds good (4) holds good also.

Notice that compared to (3) the expression above does not have a minus sign. That is because when you apply variation to the trial trajectory the two hypothetical changes are co-changing. When you make the trial trajectory steeper the rate-of-change of the hypothetical kinetic energy and rate-of change of the thypothetical potential energy both become larger; those rates-of-change are co-changing.

(4) can be rearranged as follows:
$$ \frac{d(E_k - E_p)}{ds} = 0 \tag{5} $$
At this point I need to point out a particular property of calculus of variations.

As we know, it was in the wake of the Brachistochrone challenge that Calculus of Variations was developed. Johann Bernoulli had issued the Brachistochrone challenge, and his older brother Jacob Bernoulli was among the mathematicians who was able to solve it. This means Jacob Bernoulli solved the problem without having Calculus of Variations, nor any precursor of it.

Jacob Bernoulli recognized a particular feature of the brachistochrone problem, and he presented that feature in the form of a lemma:

Let ACEDB be the desired curve along which a heavy point falls from A to B in the shortest time, and let C and D be two points on it as close together as we like. Then the segment of arc CED is among all segments of arc with C and D as end points the segment that a heavy point falling from A traverses in the shortest time. Indeed, if another segment of arc CFD were traversed in a shorter time, then the point would move along AGFDB in a shorter time than along ACEDB, which is contrary to our supposition.
(Acta Eruditorum, May 1697, pp. 211-217)

This property that applies in the case of the Brachistochrone problem also applies in the case of Hamilton's stationary action.

You want to evaluate a trajectory from a particular start point $t_1$ to an end point $t_2$

Divide the total time interval into sub-intervals. Every single sub-interval is an instance of evaluating a trajectory from a start point to an end point. In order for Hamilton's action to be stationary over the entire interval it is necessary and sufficient that Hamilton's action is stationary over each sub-interval concurrently.

So:
The property of being stationary propagates from the sub-intervals to the entire interval. This propagation from sub-intervals to the entire interval obtains at every scale, down to arbitrarily small sub-intervals.

That means:
The process of ascertaining that the action is stationary for the entire interval is actually a process of ascertaining that the action is stationary along every sub-interval, down to infinitisimally small sub-intervals

There is a more extended discussion of Hamilton's stationary action by me on this forum, I posted that one in december 2021:
https://www.physicsforums.com/threads/stationary-point-of-variation-of-action.1009770/#post-6571395

That extensive discussion is illustrated with diagrams. Those diagrams are animated GIF's; the frames of those GIF's are screenshots of interactive diagrams. The interactive diagrams are on my own website. (Link to my own website is available on my physicsforums profile page.)

Summerizing:

- In the case of Hamilton's stationary action the derivative with respect to variation and derivative with respect to position coordinate are the same thing.

Therefore:
At the point in variation space where the trial trajectory coincides with the true trajectory we have:
$$ \frac{d(E_k - E_p)}{ds} = 0 $$
- In the case of the true trajectory this differential relation holds good on every infinitisimal subsection concurrently. From there the property of being stationary (along the true trajectory) propagates to the integral.

 

[James1238765]

Thank you for the replies. In one dimension, the usual first problem is a ball dropping vertically from a height [here]. The Euler-Lagrange equation leads to a = -g, which once double integrated traces the position path of the ball.

I am still confused if the Lagrangian depends only the final position path taken, or is it dependent on the speed/motion used in traversing this path.

In the 1 dimensional case above, I have calculated the integral L for a ball dropping with a = -g.

I have also calculated alternatively a ball dropping the same height (ie. exact position trajectory) using a different velocity distribution.

I have tried to explain this to no avail. Could you explain why the alternative motion in the second case gets a lower integral L=139 value than the Newton's motion in the first case L=212. Why is this second motion not chosen over the first motion, because the integral value L is lower than the first, according to the least action principle?

 

[Cleonis]

In your message with two images of spreadsheet calculation: it says in the fifth column: 'line integral'

Is that a typo or does your spreadsheet actually evaluate the line integral?

Hamilton's action is the area integral.

 

[James1238765]

In your message with two images of spreadsheet calculation: it says in the fifth column: 'line integral'

Is that a typo or does your spreadsheet actually evaluate the line integral?

Hamilton's action is the area integral.

In the spreadsheet the quantity is total sum of all values above that box. Is this the discrete version of the Lagrangian, or something different? How to define area integral for 1 dimension problem?

 

[Cleonis]

In the spreadsheet the quantity is total sum of all values above that box. Is this the discrete version of the Lagrangian, or something different? How to define area integral for 1 dimension problem?

In a diagram the values $E_k$, $E_p$ and $(E_k - E_p)$ can be plotted as functions of time.

Comparison:
Let's say you want to evaluate the area between a parabola and the x-axis, from x=0 to x=1

For a diagram with 'x' as the horizontal coordinate, and 'y' for the vertical coordinate the corresponding function is: $y = x^2$

The integral: $\int_0^1 x^2 dx$When you have a trajectory as a function of time you can use that to plot the value of the Lagrangian $(E_k - E_p)$ as a function of time. In such a plot you use time as the horizontal coordinate, and the value you want to plot as the vertical coordinate

You can then evaluate the area between the horizontal line at vertical coordinate zero, and the plot of the Lagrangian, from time coordinate $t_1$ to $t_2$

 

[James1238765]

Thank you. I guess my question really should have been this: if we follow the exact same path but use a different speed to get from A to B, will the calculated Lagrangian answer be exactly the same in both cases, or will they be different, because the acceleration/velocity used are different even though the final path is the same?

 

[jbergman]

TL;DR Summary: What is the significance of the T - V Lagrangian of a system?

Let E be a fixed immutable quantity. E can be freely exchanged between T and V, as long as $$T + V = E$$

1. What does the quantity $$\int_x T - V $$ signify? What is the importance of this quantity?
--------------------

Let E now be the budget of a factory. E can either be spent on T or V in any proportion on any given day.

2. What does the quantity $$\int_x T - V $$ signify for the factory?

--------------------

Let E now be the total energy of a system. E can be freely exchanged between the kinetic energy T or the potential energy V as particle moves from point A to point B.

3. What does the quantity $$\int_x T - V $$ signify for the choice of motion of the system?

--------------------

Suppose a particle wants to take all possible paths s connecting A to B. Does a weighted quantity of the integral of T - V make sense over infinitely many paths?

4. What does the quantity $$\int_s T - V$$ signify?

I think this article is fairly illuminating.

https://sciencemeetsfiction.com/2020/07/04/why-does-the-lagrangian-equal-t-v/

 

[Cleonis]

I think this article is fairly illuminating.

https://sciencemeetsfiction.com/2020/07/04/why-does-the-lagrangian-equal-t-v/

Alex R. Howe, the author of that blog post, utters the following words of desperation:

This is my problem: I’ve never really understood why Lagrangian mechanics works. I get that it works. I kind of get how it works. But I don’t understand why you would subtract the two quantities.

So it is a mystery as to why you would refer to that blog post as 'fairly illuminating'.

It is the opposite: in that blog post Alex R. Howe is espressing his frustration over not understanding the how.

 

[jbergman]

Alex P. Howe, the author of that blog post, utters the following words of desperation:So it is a mystery as to why you would refer to that blog post as 'fairly illuminating'.

It is the opposite: in that blog post Alex P. Howe is espressing his frustration over not understanding the how.

Did you read the thing? That is only the setup to the article. Down below he gives a decent explanation.

 

[Cleonis]

Did you read the thing? That is only the setup to the article. Down below he gives a decent explanation.

Yeah, Alex R. Howe offers the supposition that subtracting $E_p$ from $E_k$ provides the means for comparison of the two, which is indeed the case.

https://www.physicsforums.com/threa...v-lagrangian-of-a-system.1048433/post-6835111

But Alex R. Howe does not follow through. In that blog post Alex R. Howe does not proceed to demonstrate the how.

 

[jbergman]

Yeah, Alex P. Howe offers the supposition that subtracting $E_p$ from $E_k$ provides the means for comparison of the two, which is indeed the case.

https://www.physicsforums.com/threa...v-lagrangian-of-a-system.1048433/post-6835111

But Alex P. Howe does not follow through. In that blog post Alex P. Howe does not proceed to demonstrate the how.

Yes, your linked comment is a more quantitative answer. Nicely done. I still think the blog article is a nice starting point for building some intuition.

 

[Cleonis]

[...] starting point for building some intuition.

I created a set of interactive diagrams for that.

(The diagrams are on my website. Link to my website is available on my physicsforums profile page.)

Animated GIF extracts of the interactive diagrams are available in a post I submitted to physicsforums in december 2021
https://www.physicsforums.com/threads/stationary-point-of-variation-of-action.1009770/#post-6571395

There is a set of three interactive diagrams for exploration of the interconnection between differential calculus and calculus of variations. Those are the ones for which animated GIF versions have been posted to physicsforums.

Additionally there is a set of three interactive diagrams that offers granular controls for modification of the trial trajectory. Specifically: the ascending part of the trial trajectory has 10 individual control points, at equal time intervals. (The descending part mirrors the ascending part.)

I implemented three instances:
- potential increases linear with height
- potential increases with the square of the amplitude
- potential that increases in proportion to the cube of the displacement

Each of those cases is distinct from the other two; distinct enough to merit an instance of its own.

For the case of a potential that increases with the cube of the displacement there is no analytic solution. I homed in on the true trajectory using the diagram itself. (And I sourced independent verification)

The representation in the interactive diagrams is mathematically correct; the backend implementation evaluates the actual integrals.

 

[Orodruin]

if we follow the exact same path but use a different speed to get from A to B, will the calculated Lagrangian answer be exactly the same in both cases, or will they be different

They will be different.

 

[James1238765]

@Orodruin Thank you. Does the least action principle state that, the observed velocity of motion must also be the least integral sum of T - V compared to all possible ways of traversing the same path by using different accelerations/velocities? Or is acceleration/velocity irrelevant, and only the position path trajectory is predicted by least action principle?

 

[Orodruin]

The result of solving the Euler—Lagrange equation is position as a function of time. You cannot reparametrise it and still obtain a stationary function of the action.

 

[vanhees71]

Thank you for the replies. In one dimension, the usual first problem is a ball dropping vertically from a height [here]. The Euler-Lagrange equation leads to a = -g, which once double integrated traces the position path of the ball.

I am still confused if the Lagrangian depends only the final position path taken, or is it dependent on the speed/motion used in traversing this path.

In the 1 dimensional case above, I have calculated the integral L for a ball dropping with a = -g.

View attachment 319288

I have also calculated alternatively a ball dropping the same height (ie. exact position trajectory) using a different velocity distribution.

View attachment 319290

I have tried to explain this to no avail. Could you explain why the alternative motion in the second case gets a lower integral L=139 value than the Newton's motion in the first case L=212. Why is this second motion not chosen over the first motion, because the integral value L is lower than the first, according to the least action principle?

That's usually a source of confusion when you start learning "analytical mechanics", at least it was for me, and it's a bit due to the somewhat sloppy (but very effective as soon as one gets used to) notation of the physicists.

On one hand the Lagrangian is a function of the generalized coordinates $q_k$ and the generalized velocities (and sometimes also explicitly on time): $L=L(q_1,\ldots,q_f;v_1,\ldots,v_f,t)$. The physicists usually write $L(q,\dot{q},t)$, i.e., they use $\dot{q}$ for the $f$ independent velocities $v$.

On the other hand you define the action as a functional of trajectories in configuration space, i.e., the $q_k$ become functions of time and the velocities are the time derivatives of these funktions, $v_k(t)=\dot{q}_k(t)$. Then the action functional is defined as
$$S[q]=\int_{t_1}^{t_2} \mathrm{d} t L[q(t),\dot{q}(t),t].$$
When taking partial derivatives with respect to the $q_k$ the $v_k=\dot{q}_k$ you have to consider these $2f$ variables as independent variables of the function you take derivatives of. The partial time derivative only acts on the explicit time dependence of such a function. E.g., taking the total time derivative of the Lagrangian for an arbitrary given trajectory $q(t)$ reads
$$\frac{\mathrm{d}}{\mathrm{d} t} L=\sum_{k=1}^f \left [ \dot{q}_k \frac{\partial L}{\partial q_k} + \ddot{q}_k \frac{\partial L}{\partial{\dot{q}_k}} \right] + \frac{\partial L}{\partial t}.$$
Usually also the physicists don't write the sum symbol explicitly but just use Einstein's summation convention, according to which any term, which includes a pair of equal indices, has to be summed over this index pair.

 

[James1238765]

Thank you. Just to confirm this, when we say that the Lagrangian allows us to derive the "equations of motion" of the system, is what we really mean only that the Lagrangian allows us to derive the position trajectory to be followed by the system but that the velocity and acceleration used to traverse this trajectory are not actually forced by the extrema of the Lagrangian?

 

[PeroK]

Thank you. Just to confirm this, when we say that the Lagrangian allows us to derive the "equations of motion" of the system, is what we really mean only that the Lagrangian allows us to derive the position trajectory to be followed by the system but that the velocity and acceleration used to traverse this trajectory are not actually forced by the extrema of the Lagrangian?

The path of an object falling vertically under gravity is a straight line, however you look at it. Newton's laws and the equivalent Euler-Lagrange equations tell you the trajectory as a function of time, from which the velocity and acceleration profiles are implied.

Or, vice versa, they tell you the acceleration profile, from which the trajectory as a function of time is implied.

 

[Orodruin]

Thank you. Just to confirm this, when we say that the Lagrangian allows us to derive the "equations of motion" of the system, is what we really mean only that the Lagrangian allows us to derive the position trajectory to be followed by the system but that the velocity and acceleration used to traverse this trajectory are not actually forced by the extrema of the Lagrangian?

No. Very much no. As I said explicitly in #19, the equations of motion give you position as a function of time. This means you can definitely find both velocity and acceleration as a function of time. You do not only get the spatial shape of the path. That would make things much less useful.

 

[Cleonis]

Thank you. Just to confirm this, when we say that the Lagrangian allows us to derive the "equations of motion" of the system, is what we really mean only that the Lagrangian allows us to derive the position trajectory to be followed by the system but that the velocity and acceleration used to traverse this trajectory are not actually forced by the extrema of the Lagrangian?

You have asked the same question three times now, in the postings #10 #18, and #21 of this thread.

I wonder whether possibly you are conflating two different action concepts, the one introduced by Maupertuis, and the one introduced by William Rowan Hamilton.

I have never looked at Maupertuis' action; I quote the following statement from the wikipedia article about Maupertuis' action:

Hamilton's principle determines the trajectory q(t) as a function of time, whereas Maupertuis's principle determines only the shape of the trajectory in the generalized coordinates. For example, Maupertuis's principle determines the shape of the ellipse on which a particle moves under the influence of an inverse-square central force such as gravity but does not describe per se how the particle moves along that trajectory. (However, this time parameterization may be determined from the trajectory itself in subsequent calculations using the conservation of energy.) By contrast, Hamilton's principle directly specifies the motion along the ellipse as a function of time.

We have the three interconnected quantities of theory of motion: position, velocity, and acceleration. If you have an expression for a trajectory as a function of time then you have the three of them in one go, since velocity is the time derivative of position and acceleration is the time derivative of velocity. $v=\tfrac{ds}{dt}, a=\tfrac{dv}{dt}$

By contrast, you can have the case where an orbital trajectory is given as radial distance as a function of angular position. Then you do have the shape of the trajectory, but you don't have the motion of the orbiting object as a function of time.

 

[vanhees71]

Maupertuis's principle is a special case of Hamilton's principle, if $L$ doesn't depend explicitly on time and thus energy is conserved. For this case it's in a way a "Legendre transformed Hamilton principle", i.e., you hold the energy rather than the space-time endpoints but only the spatial enpoints of the trajectory fixed.

 

[James1238765]

@Cleonis I'm scratching my head too. My mental block seems to be that I have obtained a lower L value (in #4) by using a constant velocity $v = \sqrt {mgh_{mid} - m} / 2$ motion on the same path, compared to the Newtonian accelerated motion. Because L depends on v, thus v can be toyed with to minimize the integral further. Could you reproduce the lower L value, or help show the calculation error please?

 

[PeroK]

@Cleonis I'm scratching my head too. My mental block seems to be that I have obtained a lower L value (in #4) by using a constant velocity $v = \sqrt {mgh_{mid} - m} / 2$ motion on the same path, compared to the Newtonian accelerated motion. Because L depends on v, thus v can be toyed with to minimize the integral further. Could you reproduce the lower L value, or help show the calculation error please?

You haven't understood the idea of minimising the action. The end points in space and time are fixed and you minimise the action with those constraints.

If you take the case of motion with no external force, then you are looking at motion from point $A$, which we can take to be the origin at $t = 0$, to point $B$ at some time $t = T$. We can take $B$ to be the point $(X, 0, 0)$.

We then minimise the action under these constraints and find the straight line path at constant velocity, namely $\frac X T$.

 

[James1238765]

@PeroK thank you. I think i am starting to understand, the time taken to complete the motion must be the same regardless.

 

[Orodruin]

I have obtained a lower L value (in #4) by using a constant velocity $v = \sqrt {mgh_{mid} - m} / 2$ motion on the same path, compared to the Newtonian accelerated motion.

No, you haven’t. Your expression does not even make dimensional sense. You cannot subtract a mass from an energy.

 

[PeroK]

@PeroK thank you. I think i am starting to understand, the time taken to complete the motion must be the same regardless.

In general, if you allow the time to vary and consider only constant speed motion between A and B, then we have:$$\int_A^B \frac 1 2 mv^2 \ dt = \frac 1 2 mv^2 (\frac X v) = \frac 1 2 mvX$$This has no extrema, as the integral tends to $0$ as $v \to 0$ and increases without bound as $v \to \infty$.

Clearly, that's not what's meant. Instead, the start end end points must have fixed times. This then gives you a whole class of geodesic solutions, one for each time interval. In practice, the velocity is then determined by the initial velocity and you recover Newton's first law.

 

[PeroK]

PS in this case, you can see that we have a common average velocity between all potential paths. And, any attempt to vary the velocity produces a greater increase in the action when $v > \frac X T$ than the reduction when $v < \frac X T$. In other words, the $v^2$ term ensures that uniform velocity minimises the integral. Although, it's a bit more effort to prove that formally.