For the question of “Does the pendulum analogy hold water?” the answer is “Yes”, but it relates only to one small aspect of the whole Higgs and mass story. I’ll try to add some context around it and also plant some seeds for other parts of your questions. I don’t know your physics background, so let me know if I need to back up in parts.
Potential energy
A ball high up on a hillside is said to have “potential energy”, meaning that the ball can spontaneously pick up substantial kinetic energy by rolling down. That newfound kinetic energy had to come from somewhere, and it came from the original potential energy.
With effort, you can bring the ball back up the hill. You have to work hard to do it, spending some of your own energy to lift it back up there. In contrast, you don’t have to work hard to get the ball down the hill. It just goes.
So, physical systems will spontaneously evolve to lower potential energy arrangements. A ball resting in the valley at the bottom will never jump back up to the top, but a ball up the slope will happily head down to the valley below all on its own.
To tighten things up, let’s look at a specific shape for a valley. Here’s a picture:
The red shape represents “ground level”, so there’s a nice deep valley right in the middle.
If we place a ball in this picture somewhere, it will settle into the bottom of the valley. When it’s there, the horizontal position of the ball, call it X, will be at 0. The vertical position of the ball at that point, call it V, is at roughly 1 in the graph. So, X=0 and V=1.
Important for later: the potential energy associated with a given vertical height is not specified here. We know that the bottom of the valley, at X=0 and V=1, has the lowest potential energy, but what that actual potential energy value is there is partly arbitrary. Maybe the height of V=5 corresponds to 100 joules of potential energy and V=1 corresponds to 30 joules, or instead maybe these are 10 joules and -60 joules. All that matters to the physics is the difference in potential energy between the two heights, and in this numerical example, the lower point is 70 joules “better” than the higher point. The actual potential energy at the individual points is not important. In practice, one sets the “zero point” of potential energy to wherever is most convenient for the work at hand.
Symmetry breaking
Consider now this blue version, where there is a hill in the middle with valleys on the sides:
If we place a ball right in the middle, it’s not stable. There are better, lower-potential-energy arrangements possible. So, the ball will roll down into one or the other valley. Let’s say it happens to roll to the right. The stable equilibrium arrangement for this system would become X=1 and V=1.
The blue curve in the picture is 100% symmetric around the middle, i.e., around X=0. But the actual physical arrangement that happens naturally has X=1 (say). How did a symmetric “universe” lead to a non-symmetric outcome? It’s because the shape of the blue hills and valleys make it so that the obvious symmetric answer (X=0) isn’t the stable one.
This idea generally is called “spontaneous symmetry breaking”. The physical laws are symmetric, but the outcome has to be a random choice that hides that symmetry.
Carroll’s pendulum analogy
You can maybe see the connection between the “balls in valleys” situation to the pendulum analogy you cited. In the latter, the upside-down rigid-stick pendulum can’t stay in the upright position. It has to fall one way or the other, just like a ball that tries to stay in the middle of the blue hill.
Removing “height” from the story
If we push the ball to a different X value, it changes potential energy because we are forced by the ground’s shape to either fight gravity (if climbing) or benefit from gravity (if falling). But the “height” part can be abstracted away. In fact, it’s entirely redundant with X, since we know the vertical position V perfectly if we know X. So, we can skip the middle man and just say that different positions X have different potential energies, with no more reference to height. This is important: all that matters in general is the connection between the system’s arrangement (however specified) and the potential energy of that arrangement.
Connecting to the Higgs field
The value of the Higgs field throughout spacetime corresponds to X in the above picture. It’s the “free variable”. The potential energy for different choices of X looks like the blue example above, with “Higgs field value” on the x axis and “potential energy” on the y axis. For different values of the Higgs field, there are correspondingly different potential energies.
The Higgs field spontaneously settles into a value with minimum potential energy, which is away from X=0. This is the “vacuum expectation value” of the Higgs, and it’s 246 GeV. This is not the potential energy value itself; that’s the vertical axis. 246 GeV is the field value (horizontal axis). The actual value of the potential energy at X=246~\rm{GeV} is arbitrary, as noted previously. Nothing in the Standard Model dictates where the “zero” point should be set.
Dark energy?
You asked if this could relate to dark energy. There is a non-zero value to the Higgs field everywhere, but the field itself doesn’t represent any energy unless there is an excitation of the field (an actual particle) or if we can specify some definite potential energy for it. But we can’t – the Standard Model is mute on the actual value of any potential energy for the Higgs field even when it has a non-zero value. Particle physics doesn’t care about the arbitrary offset. But! General relativity does care. So, what to do?
A naive thing to try is to just set potential energy for the Higgs field to zero when the Higgs field value is itself zero. That is, choose X=0 to be the point where V=0 by sliding the whole blue-curve picture up or down suitably. To be clear: this is an arbitrary choice. But if you do that, the vacuum value for the Higgs field (X=246~\rm{GeV}) corresponds to a massively negative potential energy density that both has the wrong sign (negative instead of positive) and is ludicrously off in magnitude to explain dark energy.
You could instead choose the potential energy to be zero when X=246~\rm{GeV}, the stable vacuum situation. That is extra arbitrary since the “vertical” offset for V has no reason to be tied to the dynamic equilibrium value that X settles out to. But, if you did this, then by definition the Higgs field would provide zero dark energy.
You could also choose to set the potential’s zero point in the Standard Model to whatever is needed to exactly explain dark energy. But that doesn’t “explain” anything since it would just mean setting an arbitrary shift of the zero point to get whatever you needed back out. Further, the extra shift you need would require a perfect tuning of the shift out to dozens and dozens of decimal places to nearly perfectly cancel the ludicrous negative value noted in the first version above. This (and some related points) is at the heart of the cosmological constant problem. It remains a big open question.
Not addressed here
- Why should the Higgs potential behave this way?
- How does this relate to the Higgs mass itself, which is a different value (125 GeV)?
- How does any of this lead to particle masses in general?
That last item (how to get particle masses out) connects to the other analogy in the other thread, namely the idea that the Higgs field is like some drag force on particles. That’s the analogy I greatly dislike, finding it more harmful than helpful. I can perhaps talk about that piece of the story as well.