And could it look anything like the universe we know?

Don't get hung up on the terminology. In general, you seem to see through it easily, so I don't want any more answers mentioning that. I'll say it's a translation error: My "F" is the quantity that, in the equations of motion discovered by physicists in this universe, occupies the position most similar to that occupied by force in our universe. Things - I guess I can't call them "forces" anymore - like gravity and magnetism confer a velocity rather than an acceleration.

I ask because this way is ancient logic, "common sense", the easier way for humans to understand. But if you look more at how such a world would function, what's impossible? I was thinking about Aristotle's "natural motion" - though don't try to be consistent with Aristotle in your answers. He believed there could be no vacuum because the only thing resisting motion was fluid resistance - that is, inertia was 0 - and any applied force to an object in vacuum would result in infinite speed. I'm taking a different approach, as explained next.

A world with no momentum. If you stop pushing a rolling object, it stops. Objects fall at a constant speed in a vacuum. I may have meant to say "no inertia" here. What I mean is that, where in our universe inertia resists acceleration, in this universe the equivalent to inertia would resist motion.

A recent answer required this clarification: I recognize that a universe entirely filled with highly viscous fluid would create a somewhat similar situation using physics as we know them. Objects would require force to remain in motion. However, I am not looking for that answer.

With real physics, if you are moving and encounter a viscous medium, you will slow down, and you will feel the deceleration. In my world, you feel speed, not acceleration. Driving into a wall won't kill you, but driving too fast will. There's also the minor matter that fluid resistance is, I believe, roughly proportional to speed squared. In my universe, the (force-like quantity) required to move in a vacuum is defined as linearly proportional to speed.

Orbits are impossible. What could the astronomical-scale universe look like?

Are any form of atoms even possible, or would matter have to be continuous?

Are fluids possible? How might they behave? This seems to be the most critical factor in creating a recognizable world.

Assume some form of weak anthropic principle. That's the question: What other laws have to be different to create a universe that could still contain something we could recognize as intelligent life capable of drawing the conclusion (force-like-quantity)=mv?

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    $\begingroup$ The first question: velocity relative to what? $\endgroup$ – Russell Borogove Nov 28 '14 at 19:49
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    $\begingroup$ If I may ask, what put this concept in your mind? $\endgroup$ – Crabgor Nov 28 '14 at 20:04
  • $\begingroup$ If f=mv, I'd assume the world would have to invent a new symbol (like L=ma?) I dunno. These formula's are us describing how the world works, changing the formula just changes how we define it, not how it behaves. Changing the definition of force doesn't impact the universe. If we observed a world where f=mv, then I don't think gravity would work and we couldn't be present in that universe to define it. Curious on an answer to cragor's comment...what are you going for here? $\endgroup$ – Twelfth Nov 28 '14 at 20:09
  • $\begingroup$ How are you going about removing momentum. The way you go about it (mathematically) changes answers. For example, setting the mass of everything to 0 removes momentum, but has dramatic effects all the way at the QM level $\endgroup$ – Cort Ammon Nov 28 '14 at 20:50
  • $\begingroup$ Related: can you define "pushing" in a world with no momentum? That definition may be important for determining whether the world can be consistent or not. $\endgroup$ – Cort Ammon Nov 28 '14 at 21:12

10 Answers 10


Answering this is tricky because $F=ma$ is just 4 symbols. We need to break it apart into its real meanings to start playing with it.

$m$, for mass, isn't changing between the two worlds, so we can get away with not worrying about it here. This is good, since mass is a bugger of a concept to really nail down.

$v$ and $a$ are velocity and acceleration. To haul in calculus terms, velocity is the first derivative of position, and acceleration is the second derivative of position. Because you're using $v$ in your alternate world, I'm going to assume the rules of science still apply.

Now the funny thing about mathematical models like this is that variables are just letters. Force, $F$ is defined to be equal to $ma$. End of story. No can do, folks.

However, the story gets a little more interesting because we do have something which we define to be equal to $mv$, momentum. $p=mv$ is another key equation, often considered more key than $F=ma$ because it handles objects that change mass better, like rockets.

A key difference between force and momentum is that force is an interaction between two objects, and momentum is just an intrinsic trait of an object. This is where the freedom of world-building begins: how can we make a $mv$ interaction?

The cleanest path is probably derivatives. $F=ma$ is actually the derivative of $p=mv$, if you hold $m$ constant. If we want some fictitious force, $F_1=mv$, then it should naturally be the derivative for something momentum-like, giving us $p_1=mx$, where x is the position vector.

This has really disturbing consequences: it suggests there is truly a center of the universe, at the point where momentum approaches 0, and infinite momentum near the edges. There will be a natural tendency for the actions to have more of an effect in one direction than the other (due to the continuous change in momentum). Literally speaking, the laws of physics would be symmetric around this point.

At large $x$, the changes in $x$ for day to day life become smaller, percentage wise. This means we can assume momentum of objects is proportional to mass. Literally speaking, it will be harder to make larger objects move, in any direction.

At this point, you have to make a decision: do all of the forces in the universe adapt to the new interactions, or do we try to just squish things like atoms into the world? Electrostatics governs the shape of the atom. $F_{es}=\frac{1}{4\pi\epsilon_0}\frac{qQ}{r^2}$. If this force was simply applied to the velocity of an electron rather than its acceleration, the effect would be the immediate disintegration of the atom. Electrons would either spring free of the atom, or be sucked into the nucleus. One would have to write a new electrostatic force, such as one without the $r^2$, like $F_{es}=\frac{1}{4\pi\epsilon_0}\frac{qQ}{r}$. This would be a completely different system (without orbits), but it would at least balance better.

We have a large number of $r^2$ based equations in our laws of motion. All of them would need to be rewritten to get anything resembling anything we are familiar with in this world. If you don't rewrite all of them, you can follow a simple rule: "if you can name it, in this world, it cannot exist in the new world."

This is where I have to stop imagining. When there is an opportunity to rewrite every single law of matter, the degrees of freedom are limitless. You can literally create any sort of world you please once you are no longer bound by the laws we believe we are bound by in this world. Want atoms? sure. Want fluids? Sure. Want bird-sized atoms that are sentient and move about on a fluid-like body of electrons? Sure. Literally anything is open if you're rewriting the laws.

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    $\begingroup$ Your use of 'momentum' as a name for p1 is bothering me. If we're building our laws of physics around F=mv, then one of the defining features of this world is surely that objects don't have anything that we would recognise as momentum (For one thing, we've already thrown out or completely redefined conservation of energy...). p1 = mx may still hold, but we need to think more about what p1 means. $\endgroup$ – Toby Y. Nov 28 '14 at 20:43
  • $\begingroup$ True, I sought to keep equations of motions in play because I tried to minimize how much had to be disrupted to change the rules. Given how much my little adjustments did, forcing us to rewrite literally every single law of motion in existence, I think going further might be really really tricky. $\endgroup$ – Cort Ammon Nov 28 '14 at 20:49
  • $\begingroup$ Yeah, it's a challenging one alright. I'll keep thinking about it, and see if I can come up with anything. $\endgroup$ – Toby Y. Nov 28 '14 at 20:53
  • $\begingroup$ I disagree with the part about it indicating that the universe has a center. This effect is merely the artifact of a coordinate system, right? $\endgroup$ – HDE 226868 Nov 28 '14 at 22:58
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    $\begingroup$ @Jan Dvorak: Good point. Let me think on that. Its tricky to handwave enough to rewrite the laws of physics from scratch without making a mistake or two along the way! :-) $\endgroup$ – Cort Ammon Nov 29 '14 at 16:58

tl;dr: Such a world would be quite different from ours. Basically, all modern formulations of classical mechanics fail on it, and you also could not base that world on an underlying quantum theory.

I assume that you want the laws of physics as similar to ours as possible with this restraint. Especially I assume that you want to preserve an underlying quantum theory for which the macroscopic world is the classical limit.

This implies that the world would be governed by a principle of extremal action (the action being essentially the phase of the quantum wave function), and therefore the classical world being described by Lagrangian equations with an appropriate Lagrangian.

Since the intended law of motion does not include acceleration, the Lagrangian must be linear in the velocity, as otherwise the Lagrange equations would generate an acceleration term. So the Lagrangian must have the form $$L(\vec x,\dot{\vec x}) = \vec f(\vec x)\cdot\dot{\vec x} + g(\vec x)$$ Inserting into the Lagrangian equations $$\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot x_k} = \frac{\partial L}{\partial x_k}$$ we get $$\vec\nabla f_k(\vec x)\cdot\dot{\vec x} = \frac{\partial \vec f(\vec x)}{\partial x_k}\cdot \vec x + \vec\nabla g(x)$$ Now the meaning of $g(x)$ is clear: It's just the negative of a "potential function" (quotes, because it is not really a potential function, just like your force is not really a force, as the units show). So let's write $g(x)=-V(x)$. WE have then $\vec F=-\vec\nabla V$ just as in conventional mechanics.

For $\vec f(\vec x)$ the situation is a bit more complicated: To get the intended equation of motion, we need $$\vec\nabla f_k(\vec x)-\frac{\partial \vec f(\vec x)}{\partial x_k} = m\vec e_k$$ for all $k$. However, let's write this down for the first component of $k=1$: $$\frac{\partial f_1(\vec x)}{\partial x_1} - \frac{\partial f_1(\vec x)}{\partial x_1} = m$$ Clearly for $m\ne 0$ this equation cannot be fulfilled, since the left hand side is identically $0$. In other words, your law of motion cannot be derived from an action principle, which means that your world cannot be based on quantum mechanics. Given that quantum mechanics in our world is responsible for a lot of the behaviour of materials (including the very fact that there are solid materials to begin with), this means your world must be very different from ours.

Also note that all modern formulations of classical physics (Lagrange, Hamilton, Hamilton-Jacoby) ultimately depend on the stationary action principle. So basically the complete tool set of modern physics could not be applied to your world.

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    $\begingroup$ Also, no Noether's theorem => no explicit reason for conservation laws. Conservation of energy is arbitrary in this world. $\endgroup$ – Irigi Nov 30 '14 at 13:27
  • $\begingroup$ @Irigi: Good point. Unless there's a way to derive Noether's theorem without using the action, that is; whatever principle replaces the action principle in that world might allow its own derivation of Noether's theorem (and the assumed law of motion is certainly not at odds with homogeneity of time; energy in that case would, however, likely look very different from energy in our world) $\endgroup$ – celtschk Nov 30 '14 at 13:37
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    $\begingroup$ Great answer. It might also be worth mentioning that this hurts general relativity, too, because we can't have the Einstein-Hilbert action anymore. $\endgroup$ – HDE 226868 Oct 11 '15 at 23:58
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    $\begingroup$ @HDE226868: Actually already special relativity fails hard because the premise implies that the principle of relativity does not hold: A particle without a force acting on it has velocity $0$ according to that premise, and that's possible only in one frame of reference. $\endgroup$ – celtschk Oct 12 '15 at 6:09
  • $\begingroup$ "I assume that you want the laws of physics as similar to ours as possible with this restraint. Especially I assume that you want to preserve an underlying quantum theory for which the macroscopic world is the classical limit." No, I never thought that. My starting assumption was no relativity, no quantum physics, in fact, nothing discovered in remotely modern times is valid. $\endgroup$ – Tristan Klassen Jul 7 '19 at 22:38

This is what you have in a viscous liquid. The risistence in such liquid is nearly proportional to the force exerted and in absence of foce things stop moving.

There are two possibilities though.

  • In the absence of force things stop instantly

  • In the absence of force things stop gradually but for constant speed you need constant force.

The later is aristotelian world and this law happens in any medium. Even in our universe at large scale: to move constntly at very high speed a body would need constant force to overcome the light pressure of CMB.

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    $\begingroup$ That's getting somewhere. Imagine what laws of mechanics would be determined by a bacterium. They would be surprised to learn rules are different at different scales or away from their familiar habitat. $\endgroup$ – JDługosz Oct 14 '15 at 13:24
  • $\begingroup$ I recognize my concept bears some resemblance to a world with strong fluid resistance everywhere. However, I'm specifically looking for development of your first possibility. $\endgroup$ – Tristan Klassen Oct 14 '15 at 17:22

The definition of "force" is mass times acceleration, so it's not meaningful to describe what the universe would be like it it meant something else. One could, however, consider what it would mean if some quantity other than force were affected by various things much as force is.

Harmonic oscillation of many kinds of objects (e.g. springs), for example relies upon the second-derivative negative feedback relationship between force and position. It doesn't matter what term one uses for the quantities that are interacting, but it's essential that negative feedback affect the second derivative. Without only first-derivative feedback, nothing can oscillate, and higher-order levels of feedback are prone to create systems which, if they oscillate at all, are apt to do so in highly chaotic fashion.

Likewise, the only way planets can have any kind of orbit is if the effect of an object's position upon its gravitational effects has a second-derivative effect. If position had only first-derivative effects, then an object's trajectory would be determined entirely by its position, which would imply that gravity would always pull objects along a straight path toward or away from the center.

In short, in order to define any sort of physics where F=mv work, one would have to define some means by which either an object's position could affect the derivative of force, or the integral of an object's position could affect the force upon it. Such a system would likely be like the current one, but with the word "force" used to describe momentum rather than mass times acceleration.


1) No orbits:

1a) That means no atoms, no chemistry.

1b) That means no planets to live on.

1c) No orbits. Galaxies would collapse into a supermassive black hole.

1d) No orbits. Galactic clusters would collapse into a supermassive black hole.

2) Temperature is motion. Motion dies away--everything very quickly cools to absolute zero.

3) Everything at absolute zero means ordinary stars can't exist. Fusion could still occur once the pressure gets high enough. That means degenerate matter (if electron degeneracy means anything in a world without orbits) and the normal thermal regulation doesn't work. You would have dark bodies that simply sat there until they reached 1.41 solar masses and then detonate in a bang that would outshine a supernova.

However, the mass thrown off by the detonation would soon stop and then fall back due to gravity. The light of the supernova would fall off much faster than normal and the star would soon reform, albeit slightly lighter. It would detonate again when enough material had fallen on it, each subsequent detonation would be a lot weaker than the previous ones as the material went up the periodic table. They would eventually cease when the star had been replaced with a great iron ball.

Note that no mass escapes, the interstellar medium remains as it did after the big bang. Planets can't exist because there's nothing to make them out of.

  • $\begingroup$ Can you justify “no orbits”? $\endgroup$ – JDługosz Apr 3 '17 at 9:32
  • $\begingroup$ @JDługosz I was showing a bunch of implications of no orbits. I do see how to improve the answer, though... $\endgroup$ – Loren Pechtel Apr 3 '17 at 17:35
  • $\begingroup$ But you can't assume nuclear energy exists, since this is clearly a non-relativistic universe as pointed out in the comments to worldbuilding.stackexchange.com/a/4758/2318 $\endgroup$ – Tristan Klassen Apr 12 '17 at 15:08
  • $\begingroup$ Also, in this world, if there's anything they call "temperature", it's not the same thing we call temperature. One of the things I was trying to ask was, what are possible laws of physics that would allow a non-collapsing universe given that momentum doesn't exist? $\endgroup$ – Tristan Klassen Jul 7 '19 at 22:35

(Force) (is proportional to) (speed difference) is a formula that describes "dynamic friction". It is a significant part of many real-world physics (aka engineering) problems. Natural and artificial bearings (including hips) are designed based on this force.

(Force) (is proportional to) (fluid density) * (area) * (speed) * (speed) is a formula that describes "air resistance". It is a major part of many engineering designs, including the shapes of animals, plants, cars, and planes.

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    $\begingroup$ Yes, but in all those cases, the force is still $F = ma$. And we can actually model it: $F = -kv = ma \Rightarrow v(t) = A\exp(-t k/m)$. $\endgroup$ – Physicist137 Nov 29 '14 at 21:46

This obviously throws all of physics out of the window.

For starters, as @Russell points out, the velocity has to be relative to something. There therefore has to be a fixed notion of position, and being static. This overthrows one of the fundamental assumptions of the Theory of Relativity, and so the Big Bang is out of the question. You're going to have to think of another way to get any sort of astronomical scale universe.

  • $\begingroup$ I don't see why you need to even consider the Big Bang or any alternative to have a universe. You could simply have a universe. Just like the sofa in "Dirk Gently's Holistic Detective Agency" (if you've read it), it could simply exist without any possible steps by which it could have reached its current state. $\endgroup$ – Wildcard Jul 18 '17 at 3:41

Basically, you're proposing a universe in which constant forces are required to maintain constant velocity, instead of to change it as our own universe allows. In addition, an increase in force is an instantaneous increase in velocity, or infinite acceleration.

In such a universe, the energy required to exert force would be depleted almost instantly, and F=ma=mv=0.

  • $\begingroup$ What makes you think you could instantly change the force? $\endgroup$ – celtschk Oct 14 '15 at 21:10
  • $\begingroup$ Because this equation doesn't consider time at all. F=ma, when you consider a as dv/dt, rearranges to the impulse/momentum equation, Ft=mv. Remove t from the equation and we implicitly assume it's infinitesimal (not zero but so close it doesn't matter), which when we rearrange again produces arbitrarily large accelerations for even small variations in force. $\endgroup$ – KeithS Oct 14 '15 at 21:17
  • $\begingroup$ Sorry, but I can't follow you here. Note that in our world you also cannot instantly change the acceleration (i.e. have an infinite $da/dt$) despite the time derivative of the acceleration not appearing in F=ma. $\endgroup$ – celtschk Oct 14 '15 at 21:22
  • $\begingroup$ Elastic collisions, like those of a Jacob's cradle, are ideally instantaneous. In reality they simply occur on such a short timescale, like nanoseconds, that any attempt to plot instantaneous velocity at a reasonable resolution will have a jump discontinuity, in turn causing the derivative of this graph (acceleration over time) to spike off the charts at the same point. $\endgroup$ – KeithS Oct 14 '15 at 21:39
  • $\begingroup$ Nanoseconds are not instantaneous. And large is not infinite. Indeed, it's infinitely far away from it. $\endgroup$ – celtschk Oct 14 '15 at 21:47

In this case it is best to replace the terms with new terms so that we can use different units. We could use the terms SRF for second rate force, srm for second rate mass, srp for second rate momentum, and SRE for second rate energy. For force we have the equation F=p/t so for second rate force we have the equation SRF=srp/t and if we use the equation and so if we use the equation SRF=(srm)*v for the second rate force then that means we need to use srp=(srm)*x for the second rate momentum. Just as momentum in our universe is direction dependent the second rate momentum in your universe would also be direction dependent. For the second rate kinetic energy we can use the equation SRKE=(srm)*x^2 so that the second rate kinetic energy does not depend on direction. This would mean that the units for the second rate mass would be srkg for second rate kilograms, and the units of second rate momentum would be (srkg)*m, the units for second rate force would be (srkg)*m/s, and the units for second rate energy would be (srkg)*m^2.

In the universe you are describing the second rate momentum and the second rate energy would be conserved as that way objects would be unable to move without interacting with each other as that would violate conservation of second rate energy, and second rate momentum. This would also ensure that two objects that are applying second rate forces to each other will experience equal and opposite second rate forces.

In this case there would not be relativity in terms of speeds as anything that would be moving would experience a second rate force meaning it would be possible to tell which objects would be moving and which are standing still. There would however be relativity in terms of position values however as physics would not depend on an objects position value and so if there were two object it would not be possible to tell which object is at position zero and which has a none zero position. This would also mean that the second rate momentum and second rate energy of an object would depend on the position reference frame that their second rate energy and second rate momentum is being measured from.

In our universe in addition to force being a type of interaction there is another type of interaction known as elastic collision in which the energy before and after the collision is the same. In this case we might imagine that in the universe you describe there would be an interaction known as second rate inelastic collisions in which the second rate kinetic energy would be the same before and after the collision meaning that there would be no conversion of second rate kinetic energy into second rate potential energy. If we assume that second rate elastic collisions work by first converting second rate kinetic energy into second rate potential energy and then back to second rate potential energy then the second rate elastic collisions could produce harmonic motion as two objects keep having elastic collisions with each other repeatedly. This also means that stable structures could form that would be held together by the equivalent of elastic collisions instead of by the equivalent of forces. Objects that are made of particles that experience constant second rate elastic collisions would behave like a cross between a liquid and a solid as they would not flow and would tend to retain their shape but particles could move through them freely through the equivalent of elastic collisions. This is the reverse of our universe as in our universe if something only has elastic collisions in it then it will disperse over time while attractive forces can cause stability. In the universe you describe any unbalanced force would prevent the universe from being stable but the equivalent of elastic collisions would lead to stable motion and potentially complex structures.


Changing the formula doesn't change the world.

The easiest, and most obvious answer to your question is: "Yes, of course".

Just switch the meaning of 'v' and 'a'.

So in your world, the word 'velocity' has been replaced by 'arrow-speed' and the word 'acceleration' has been replaced by 'voom-crease'.

So the great Physicist 'Oldton' derived F=mv.

It's important to get that there are (at least) two schools of mathematics - Platonists, and Formalists.

Platonists believe that formula actually correspond to the universe somehow. A few physicists are Platonists.

Formalists see formula as expressions that can be juggled according to a set of rules.

In Physics, almost every physicist accept that the formula we have for describing the world that we measure models, simulations, representations. If we didn't accept that, we would have to start looking for ideal objects in nature, and we don't find them.

So, another way of looking at this is regarding the progress of science. Pre-Newtonian physics used all sorts of different answers, such as the theory of impetus. Post Newtonian physics replaces f=ma with relativistic version of f=ma

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    $\begingroup$ But the question explicitly states that it's not about changing the letters in formulas, it's about a world with fundamentally different physics. The inertia there doesn't resist acceleration, it resists the velocity itself. $\endgroup$ – avek Apr 2 '17 at 18:38
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    $\begingroup$ You might try using MathJax markup instead of a png file image. $\endgroup$ – JDługosz Apr 3 '17 at 9:34

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