8 ms·

Why does F = ma?

153 points by csantini 7y ago

ijiiijji1 7y ago
Remember that F and a are vectors and m is a scalar.
- emsy 7y ago
  My teachers were terrible at explaining that concept. In retrospect they did more damage than good with their explanation. It also doesn't help that vectors are introduced much later in math than mechanics in physics(at least where I live).
- chongli 7y ago
  Yes and that it’s more properly ΣF = ma. A lot of students forget that it’s the sum of the force vectors, not just a single force, when they’re first learning physics.
bobcostas55 7y ago
As Poincaré put it, "Masses are co-efficients which it is found convenient to introduce into calculations."
w0mbat 7y ago
What confuses many people in high school is that F = ma but KE = 1/2 mv^2 (or 1/2 ma).
- bermanoid 7y ago
  Acceleration does not equal velocity squared...
  - w0mbat 7y ago
    That would explain a lot. Our 80’s high school physics teacher taught us that acceleration equals velocity squared, but he was prone to using vast over simplifications to get through the material quickly.
    - deleted 7y ago
      [deleted]
    - elil17 7y ago
      That's not a vast oversimplification, it's completely unrelated to reality
    - anamexis 7y ago
      Perhaps a perversion of their units? E.g. m/s^2 vs (m/s)^2
    - antonvs 7y ago
      If that were true, you would have not been able to reach correct answers on any question involving acceleration, directly or indirectly. Which means much of mechanics. Are you sure you just didn't misremember it?
tel 7y ago
Isn't this derived from conservation of momentum? As I understand it, that arises from a symmetry in spatial configuration: that physics behaves identically along shifts in position. I'm well beyond anything I'm familiar with, but it feels like these more fundamental relations should be where meaning of equations like F=ma arise. If momentum is conserved due to symmetry of position, then forces applied by a field are exactly what disrupt that symmetry and thus are exactly what invoke changes in the conserved quantity? I hope for explanations like that to bear more fruit because they arise from some pretty undeniable facts of reality: there aren't privileged positions in space except for how forces exist in some configuration, there aren't privileged moments in time except for how events occur along some timeline.
dboreham 7y ago
Coincidentally on vacation I'm reading Leonard Suskind's "Special Relativity and Classical Field Theory: The Theoretical Minimum", which (I think, if I'm understanding correctly), goes into this question. Something about Gauge Invariance.. https://g.co/kgs/dygVQN https://g.co/kgs/dygVQN
bluesign 7y ago
I think best way to look at this is m = F/a. Then mass is basically objects resistance to force.
- saagarjha 7y ago
  Inertia is a property of matter…
- Anon84 7y ago
  This is a perfectly fine way of looking at it. But here's an interesting question: - We know that Force is proportional to acceleration. Let's call the proportionality constant, m_1=F/a. - We also know that objects are attracted to the Earth with a proportionality constant: m_2= F r^2/(GM) and we can certainly set: - F == m_1 a == G m_2 M/r^2 to calculate the acceleration of the gravitational force. But why should m_2==m_1? In other words, why is "inertial mass" (m_1) the same as gravitational mass (m_2)?
  - saagarjha 7y ago
    Follow up question: is there an “electrical” inertia? Why does gravity have only one relevant quantity but electromagnetic forces need two?
    - Anon84 7y ago
      Well, it really doesn't. The "problem" is that charge is always associated with mass, with the charge generating electromagnetic fields and the mass being responsible for the gravitational one
  - cygx 7y ago
    Because gravity is a pseudo-force: Free fall is inertial motion, tracing out a straight path through a curved spacetime. In a way, it's more like m·(a - a_grav) = F_nongrav
  - p1mrx 7y ago
    When you drop a hammer, its freefall follows a geodesic, which could be considered a "straight line" through spacetime... at least until the ground gets in the way. In some sense, you could say that the ground is accelerating upward at a constant 9.8 m/s². When we feel weight, that's just regular inertia, pushing back against the ground's acceleration. (I'm not sure to what extent this follows the mainstream interpretation of Einstein's equivalence principle. It at least requires a different definition of "acceleration", given that the Earth's radius is not increasing.)
jules 7y ago
I don't understand the problem the author sees with the conventional explanation. Yes, test particles of different charges behave differently, but this in no way implies that the law should involve the third derivative of position! The author says: > I’ve never worked it out in detail, but wouldn’t be surprised if such an approach can be made to work. Essentially, it’d make acceleration into a free (possibly constrained) parameter of the particle, rather than something completely determined by the distribution of matter and fields. That free parameter would implicitly contain what (in the conventional approach) we think of as the charge information. Indeed, the new equations of motion would have a conserved quantity, corresponding to the charge. But the resulting force laws would be quite a bit uglier. I don't see how this could be true. If it is, I'd indeed like to see it worked out. Particles with different charges still follow F = ma. The force is different for different particles, sure, but the evolution of the position of the particle still follows a second order differential equation and not a third order one. Furthermore, going to a third order equation doesn't even solve the problem that the F is different for particles of different charge, that problem is still there the same as before. I don't think is a problem at all, by the way: the forces always depend on the particular situation we're considering, including on the mass, charge, and so on, and even more so for compound objects. The question "Why does F = ma?" is a good question though. The conventional explanation only explains why classical mechanics is governed by a second order differential equation. It does not explain why F = ma is the right equation. If you interpret F as a general function of the state of the system then indeed F = ma doesn't tell you anything, because by picking the appropriate F you can get any second order differential equation. However, if we already have some prior idea of what force and mass is, which we do, then it's not clear why it should be this differential equation. We do have an intuitive idea of what force is: if you hang 1kg on a rope then the rope pulls with some amount of force, and if you put 2kg then it pulls with double the force. Similarly, if you stretch a spring by 3cm you have some amount of force, and if you stretch two of those springs simultaneously you have double the force. You could relate the force in the spring with force by a weight by finding the amount of weight you need to stretch the spring by that 3cm. We can get a definition of force by picking some reference force as being 1 unit of force, and we can similarly come up with an experimental definition of mass. If we accept such a definition then F = ma has physical meaning. For instance, it says that if you double the force then you double the acceleration. That is, if you pull an object with two springs then it accelerates twice as fast as when you pull it with one spring, if you stretch the springs by the same amount in both cases. This is a nontrivial fact about the physical world. The law F = ma also tells you that if you apply a constant force then the velocity increases linearly. That is, if you pull an object with a spring and keep the spring stretched by 3cm, then if the object accelerates to speed v in one second then it will accelerate to 2v in two seconds. This is a nontrivial fact about the physical world too.
KenoFischer 7y ago
I'm not sure the question is particularly well posed, but I think the most coherent answer you'd get from modern physics is symmetry. Restating Newton's law's slightly, you basically get two points: 1. Momentum is conserved 2. All interactions between objects are mediated by forces (i.e. interactions exchange momentum). So you may ask, ok, but why do these two things happen, which is is where symmetry comes in. If you require that your theory is invariant under translations in space and time, your theory must conserve momentum and energy. Then you ask, ok, so what kind of long-range interactions can we have between particles in a theory where energy and momentum are conserved and the answer turns out to be, well those that exchange momentum. For example, in the hypothetical from the blog post where interactions are mediated by exchanges of acceleration (forces are proportional to jerks), you end up with a universe where absent interactions, acceleration is constant. Why is this inconsistent with symmetry? Well, in such a universe, the acceleration of particles would be constant in the absence of interaction, so their kinetic energy would keep growing and growing, violating conservation of energy.
- platz 7y ago
  https://en.wikipedia.org/wiki/Noether%27s_theorem https://en.wikipedia.org/wiki/Noether%27s_theorem
- lordnacho 7y ago
  This seems to lead to another question, which is why is the invariant of the system what it is? Why are momentum (mv) and energy (0.5mv2) conserved?
  - dpq 7y ago
    It directly follows from the assumptions that space is isotropic and homogeneous. I recall that there was a short and elegant proof of this link in Landau and Lifshits vol 1; you may want to look it up.
    - fferen 7y ago
      Indeed, it is at the start of the book. It also relies on the assumption that the Lagrangian is invariant up to a total time derivative under Galilean transformations, which leads to `dL/d(v^2)` being a constant (1/2 m). This is, I believe, the most fundamental way to look at it. Otherwise you could ask why the Lagrangian can't have other terms like v^4.
knzhou 7y ago
> Putting it in somewhat fuzzier terms, and at the risk of repeating myself: F = ma derives its power from the (implicit) assertion that there is a simple unversal force law that lets us figure out F for a particular configuration of matter. And so the configuration of matter completely determines the acceleration of a test particle. There is no a priori reason this ought to be true. It’s an absolutely incredible fact of nature. Yup, this is totally correct. To say it yet another way, we evaluate scientific theories not by looking at the pieces in isolation, but how much explanatory power you get from all the pieces working together (penalized by the total complexity of those pieces). There's nothing mathematically inconsistent about defining "F = mv", it just makes F a less useful quantity. Feynman actually had a remarkably similar discussion in his classic lectures: > For example, if we were to choose to say that an object left to itself keeps its position and does not move, then when we see something drifting, we could say that must be due to a “gorce” — a gorce is the rate of change of position. Now we have a wonderful new law, everything stands still except when a gorce is acting. You see, that would be analogous to the above definition of force, and it would contain no information. > The real content of Newton’s laws is this: that the force is supposed to have some independent properties, in addition to the law F = ma. [...] It implies that if we study the mass times the acceleration and call the product the force, i.e., if we study the characteristics of force as a program of interest, then we shall find that forces have some simplicity; the law is a good program for analyzing nature, it is a suggestion that the forces will be simple.
- analog31 7y ago
  In a lot of problems, there is a "F = mv" as well, just not using the letter m. A velocity dependent force is a "damping" force. Pushing an object through some fluid medium will produce damping. There is also a "F = mx" which is a spring return force. All three of these forces operating together drive the (approximate) equation of motion for a lot of mechanical things, such as the cone of a loudspeaker.
  - semi-extrinsic 7y ago
    Those F's aren't the same as the one in "F = ma" though. The one in "F = ma" means "the (vector) sum of all forces on the object". You can write down, as you say, a bunch of contributions like F_spring and F_damp etc. that can be functions of other variables like x, v etc. "F = ma" then tells you that if you sum up all the forces (F_spring, F_damp etc) you have an expression for the acceleration. This gives you an equation of motion that you can solve to find out how the object will move given some initial state. Basically, if you know the initial position and velocity, you can compute acceleration at that time. Then you can figure out the position and velocity of the object a short time later, and recompute the acceleration using these new values. Then you can step forward again, and again, for as long as you want.
whatshisface 7y ago
F = ma because if you walk outside and push things around that's what you will notice is happening. Any other explanation including a derivation from QFT would just be a different version of that, because whatever deeper underlying theory you were using would have come from the same place as F = ma came from to begin with.
whatshisface 7y ago
>A fun question: how does the universe change if the mass isn’t a scalar, but rather is a matrix, and so a = m-1F is the acceleration? What would this world look like? Is it plausible? This happens in our universe when the coordinate system is squished. In the general curvilinear case, mass is a matrix.
- knzhou 7y ago
  Not only that, but mass is a matrix for electrons moving in a crystal lattice, even for normal coordinate systems. It's also true if you want F = ma to continue to hold in special relativity, then m is a velocity-dependent matrix.
  - cygx 7y ago
    It's also true if you want F = ma to continue to hold in special relativity, then m is a velocity-dependent matrix. Not if you describe things in terms of four-forces and four-accelerations, which restores the simple proportionality. However, the spatial component of 4-acceleration is not necessarily parallel to 3-acceleration.
roenxi 7y ago
I liked KenoFisher's top level answer but I want to change the perspective because I think momentum is a red herring. The starting point should be energy. Classical physics (and modern physics as far as I recall) determined that literally everything is linked by a single concept of energy. In my view energy is more real than anything else we experience and all our senses are for perceiving energy in different states. Matter is pooled energy, movement is energy, heat is energy (linked to movement), time is defined to some degree in terms of mass so it is linked in there too somehow. And we know energy is conserved because we have observed that everywhere. Once we know energy is conserved the law of momentum makes a lot of sense - there have to be symmetrical laws that don't allow arbitrary creation of energy because that doesn't happen. And then F = ma turns up and it isn't so surprising for the same reason. F = ma is in that sense not a fundamental law; it is a corollary of the conservation of energy. Mathematically there might have been other options, but that is the one that applies in this universe. There was always going to be a mass component in the formula because by observation energy is fundamentally linked to mass.
- klodolph 7y ago
  This only makes sense in a relativistic context, but forces don't exist in relativistic contexts. So you can either have F = ma, or you can have GR, but you don't get to have both at the same time.
  - selimthegrim 7y ago
    Wait, you can’t write the Lorentz force down?
    - cygx 7y ago
      You can: https://en.wikipedia.org/wiki/Maxwell%27s_equations_in_curved_spacetime#Lorentz_force_density https://en.wikipedia.org/wiki/Maxwell%27s_equations_in_curve...
  - pdonis 7y ago
    > forces don't exist in relativistic contexts Yes, they do; the difference in General Relativity is that gravity is not a force, but non-gravitational forces are still there.
ta1234567890 7y ago
> the equation in Newton’s second law isn’t F = ma, but rather the more subtle statement that force is equal to the rate of change of momentum of a body This is probably the most insightful bit and points directly to the potential of fiddling with the equation. Drawing an analogy from electricity, F=ma is in a way the "DC version/expression" of a force, in which m and a are constant and F is a fixed quantity. However we can write an "AC version/expression" of the equation as F(t)=m(t)*a(t), in which the average of F(t)=F. In a way we tend to just be content with dealing with averages instead of looking at the detail and seeing how it varies over time. There's a lot of potential in looking at the variation. Especially as technology enables it through high speed cameras and instruments with higher sampling rates/resolution.
- raverbashing 7y ago
  The "AC" version of that equation is actually F=dp/dt (p = momentum)
  - ta1234567890 7y ago
    Thank you for the formula. Do you know of example advances that have been made or problems that have been solved as a consequence of applying the concept implied in that formula?
    - ianopolous 7y ago
      Rockets
      - ta1234567890 7y ago
        Care to elaborate?
        aerodude 7y ago
        Rockets have an exhaust with constant velocity, and they expel mass in the form of fuel. It's this expulsion of mass that generates the force that propels the rocket. Taking a force balance and assuming no external forces (e.g. gravity), you can integrate to get the rocket equation, which tells us how much fuel mass we need to reach a given velocity (the quantity we care about when talking about stellar distances).
dvt 7y ago
> I find it astounding that a theory like quantum mechanics can have inside it another theory, an approximation, also extremely beautiful mathematically, but radically different. It’s like taking Bach, adding some noise, and getting the best of the Beatles out. I wish I understood better why this can happen. This isn't really the case. Author should've probably mentioned the Ehrenfest theorem[0]. It's not exactly accurate to say that Newtonian mechanics is "embedded" in quantum (or rather Hamiltonian) mechanics. In fact, we need to do some "artificial" re-jigging to make quantum mechanics work at Newtonian scales. [0] https://en.wikipedia.org/wiki/Ehrenfest_theorem https://en.wikipedia.org/wiki/Ehrenfest_theorem
- gus_massa 7y ago
  >> It’s like taking Bach, adding some noise, and getting the best of the Beatles out. It's a very bad analogy. It's difficult to find a good one. Let's try anyway: Imagine that QM is a real piano and CM is how you expect a piano to sound in normal conditions. If you press a C key, you get a C sound. You can modify the sound releasing the key or keeping it pressed, or pressing the pedals. There are some effect where you press another keys so the sound of the original string make the other strings move and modify the sound. You can hit the wood part of the piano to make some drum like sounds. Most of this can be simulate in an electric keyboard and get an accurate piano-like sound. This is like Classic Mechanics. Quantum Mechanics is like the real piano. You can open or close the top. You can fill the piano with a different gas to make the sound slightly different. Or fill it with water or liquid nitrogen. Use drumstick to hit the strings directly. Use a hammer to hit the wood walls. These create a more wide set of sounds that are nor reproducible in a electric keyboard. An electric keyboard is only an approximation of an idealized real piano. (An experiment where the waviness of the particles is the most important part is like dropping a piano from a 8th floor to hear the sound of the crash, or using the piano as firewood to hear the crispy sound of fire in a winter night. I'm not happy with these examples because they are too destructive, but I wanted too examples of sounds that are very far away from the sounds you expect in an electric keyboard.) (For some reason, musicians do more weird destructive acts with guitar than with pianos. Perhaps it would be a better instrument for the analogy. But it is more difficult to make an idealized model of a guitar. A piano is almost discrete.)
andrepd 7y ago
If you're interested, Landau & Lifschitz "Vol 1: Mechanics" has a derivation of classical mechanics from purely physical arguments, from first principles. F=ma is on page 9.
- acqq 7y ago
  And that derivation seems to be known at least since 1788: https://en.wikipedia.org/wiki/Lagrangian_mechanics https://en.wikipedia.org/wiki/Lagrangian_mechanics
  - andrepd 7y ago
    Yes, but Landau derives that in a supremely elegant way, from purely physical arguments.
    - acqq 7y ago
      I personally always prefer first historical approach, only then elegant summaries. I am aware that some like just a distilled material. Still thanks for mentioning Landau and Lifschitz book!
d--b 7y ago
I personally would have gone with: because E=0.5mv^2 And by that i mean that energy is a much more intuitive concept than force. Why is E proportional to v^2? Now THAT’s an interesting question.
- earthicus 7y ago
  Here's a good explanation (it reduces to F=ma, of course!). This explanation starts by answering a different question, then points out that energy is analogous. https://physics.stackexchange.com/questions/67432/intuitive-meaning-of-factor-2-in-formula-of-vertical-throw-max-height-h-v2-2g/67433#67433 https://physics.stackexchange.com/questions/67432/intuitive-...
- XorNot 7y ago
  How about this: because there's a mass involved with a non-zero cross section, so velocity is being added to an area, not a point. Hence v^2 is hiding the complexity of density vs surface area (the law was originally derived by observing steel ball impact craters on a bed of sand vs the height). 0.5 is because only one side hits. EDIT: And what surface area? Well General Relativity does like to mash up the concept of objects affecting space time due to having energy...
ISL 7y ago
We will never know why it does, even though it is a well-motivated prediction of classical and modern theories. It is possible for us to test whether or F=ma, though, and so far, it (with relativistic corrections) checks out. An example of this sort of test from our research group: https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.98.150801 https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.98... (I'd go dig up more citations on the subject, but we have a grant proposal due in two days... :) )
kmm 7y ago
> [...] Two test particles with the same initial position and velocity, but different electric charges, can behave quite differently in the same electric field. > One possible response is to say “oh, maybe our notion of force should really be something like F = mj, where j is the jerk, i.e., the third derivative of position”. > I’ve never worked it out in detail, but wouldn’t be surprised if such an approach can be made to work. It might not be exactly what he's looking for, but in Kaluza-Klein theory, charge is related to the velocity of the particle in an unseen 5th dimension. Through some miracle, if you work out general relativity with one dimension extra, the resulting theory bears a striking resemblance to general relativity in the usual four dimensions plus electromagnetism. If you then assume the 5th dimension is extremely small, which would explain why we can't see it, you get quantization of charge for free. https://en.wikipedia.org/wiki/Kaluza%E2%80%93Klein_theory https://en.wikipedia.org/wiki/Kaluza%E2%80%93Klein_theory
_Microft 7y ago
Hijacking this thread in hope that some fellow physicist could chime in. I'm physicist myself so maybe I should know this or be able to figure it out myself but here's the question: Where is kinetic or rotational energy 'stored'? Is it eventually the transformation of fields that gives rise to kinetic energy? For example an electron whose electric field is not a radially symmetric must be a moving electron and the difference in energy between the field configurations (simple radially symmetric electric field on one hand and the mix of electric and magnetic fields on the other) is exactly the kinetic energy it has?
- selimthegrim 7y ago
  You have hit upon the so called Abraham Minkowski paradox https://en.m.wikipedia.org/wiki/Abraham–Minkowski_controversy https://en.m.wikipedia.org/wiki/Abraham–Minkowski_controvers...
- knzhou 7y ago
  The electromagnetic field configuration does contribute energy/momentum. In general, every field that interacts with the election field contributes: an “electron” stands for a complicated joint excitation of the electron, photon, W/Z boson, quark, gluon, ... quantum fields and what we call the momentum/energy of the electron is just the total of that of all of these fields.
octoboto 7y ago
>A fun question: how does the universe change if the mass isn’t a scalar, but rather is a matrix, and so a = m-1F is the acceleration? What would this world look like? Is it plausible? m is only a scalar for particles though? For a general rigid body mass is a matrix. Unless they specifically mean for translational degrees of freedom, in which case I think that would probably break symmetry
8bitsrule 7y ago
Because Work = Force times Distance = Energy. Acceleration is the result of energy added (by doing work) to a body with inertia (mass). Now if you ask what is inertia: that is the right (IMO unanswered) question!
doggydogs94 7y ago
One time in a physics class we were required to have a “cheat sheet” of formulas for an exam. One student did not have a “cheat sheet” and was told that he was required to have a cheat sheet. He grabbed a sheet of paper and wrote F=ma. He aced the exam.
theothermkn 7y ago
'F = ma' is a special case of 'F = d/dt(mv)'. Expanding the latter via the product rule, we get: F = m dv/dt + v dm/dt, where a=dv/dt and dm/dt is the time rate of change of the mass of the object. This fuller form matters in, for example, aerospace engineering, where the mass of the rocket is changing as fuel is burned, so dm/dt is emphatically not 0. I wonder if the philosophical questions asked in the fine article could have been addressed more satisfyingly from that more general starting point?
millstone 7y ago
One interesting fact that has been missed here is the nature of time reversal. F=ma. Switch on the time reversal machine: t becomes -t. Positions stay the same, but velocities reverse, because dx/d-t -> -dx/dt. BUT accelerations stay the same! dx/d^2t gets a double negative, which is just 1. Gravity attracts in the future and the past. Throw a ball up and it comes back down; reverse the camera and it looks the same. This means that (handwaving) forces are independent of time. And this in turn leads to a time-independent construction of forces - how about as the gradient in space (NOT time) of a potential. This I think is a path to partial insight. A particle moves in a potential, and we can calculate the quantity 'a' from the purely-spatial gradient at the particle's position, without reference to time at all. And we can do this because 'a' is an even derivative of position. This is a crappy argument; at best it argues that it's nicer that f=ma instead of f=mv or f=ma'. Fourth or sixth derivatives are not considered. Still, understanding dynamics as governed by position only is compelling.

Why does F = ma?

153 points by csantini 7y ago

ijiiijji1 7y ago

Remember that F and a are vectors and m is a scalar.

emsy 7y ago
My teachers were terrible at explaining that concept. In retrospect they did more damage than good with their explanation. It also doesn't help that vectors are introduced much later in math than mechanics in physics(at least where I live).
chongli 7y ago
Yes and that it’s more properly ΣF = ma. A lot of students forget that it’s the sum of the force vectors, not just a single force, when they’re first learning physics.

bobcostas55 7y ago

As Poincaré put it, "Masses are co-efficients which it is found convenient to introduce into calculations."

w0mbat 7y ago

What confuses many people in high school is that F = ma but KE = 1/2 mv^2 (or 1/2 ma).

bermanoid 7y ago
Acceleration does not equal velocity squared...
- w0mbat 7y ago
  That would explain a lot. Our 80’s high school physics teacher taught us that acceleration equals velocity squared, but he was prone to using vast over simplifications to get through the material quickly.
  - deleted 7y ago
    [deleted]
  - elil17 7y ago
    That's not a vast oversimplification, it's completely unrelated to reality
  - anamexis 7y ago
    Perhaps a perversion of their units? E.g. m/s^2 vs (m/s)^2
  - antonvs 7y ago
    If that were true, you would have not been able to reach correct answers on any question involving acceleration, directly or indirectly. Which means much of mechanics. Are you sure you just didn't misremember it?

tel 7y ago

Isn't this derived from conservation of momentum? As I understand it, that arises from a symmetry in spatial configuration: that physics behaves identically along shifts in position. I'm well beyond anything I'm familiar with, but it feels like these more fundamental relations should be where meaning of equations like F=ma arise. If momentum is conserved due to symmetry of position, then forces applied by a field are exactly what disrupt that symmetry and thus are exactly what invoke changes in the conserved quantity? I hope for explanations like that to bear more fruit because they arise from some pretty undeniable facts of reality: there aren't privileged positions in space except for how forces exist in some configuration, there aren't privileged moments in time except for how events occur along some timeline.

dboreham 7y ago

Coincidentally on vacation I'm reading Leonard Suskind's "Special Relativity and Classical Field Theory: The Theoretical Minimum", which (I think, if I'm understanding correctly), goes into this question. Something about Gauge Invariance.. https://g.co/kgs/dygVQN https://g.co/kgs/dygVQN

bluesign 7y ago

I think best way to look at this is m = F/a. Then mass is basically objects resistance to force.

saagarjha 7y ago
Inertia is a property of matter…
Anon84 7y ago
This is a perfectly fine way of looking at it. But here's an interesting question: - We know that Force is proportional to acceleration. Let's call the proportionality constant, m_1=F/a. - We also know that objects are attracted to the Earth with a proportionality constant: m_2= F r^2/(GM) and we can certainly set: - F == m_1 a == G m_2 M/r^2 to calculate the acceleration of the gravitational force. But why should m_2==m_1? In other words, why is "inertial mass" (m_1) the same as gravitational mass (m_2)?
- saagarjha 7y ago
  Follow up question: is there an “electrical” inertia? Why does gravity have only one relevant quantity but electromagnetic forces need two?
  - Anon84 7y ago
    Well, it really doesn't. The "problem" is that charge is always associated with mass, with the charge generating electromagnetic fields and the mass being responsible for the gravitational one
- cygx 7y ago
  Because gravity is a pseudo-force: Free fall is inertial motion, tracing out a straight path through a curved spacetime. In a way, it's more like m·(a - a_grav) = F_nongrav
- p1mrx 7y ago
  When you drop a hammer, its freefall follows a geodesic, which could be considered a "straight line" through spacetime... at least until the ground gets in the way. In some sense, you could say that the ground is accelerating upward at a constant 9.8 m/s². When we feel weight, that's just regular inertia, pushing back against the ground's acceleration. (I'm not sure to what extent this follows the mainstream interpretation of Einstein's equivalence principle. It at least requires a different definition of "acceleration", given that the Earth's radius is not increasing.)

jules 7y ago

I don't understand the problem the author sees with the conventional explanation. Yes, test particles of different charges behave differently, but this in no way implies that the law should involve the third derivative of position! The author says: > I’ve never worked it out in detail, but wouldn’t be surprised if such an approach can be made to work. Essentially, it’d make acceleration into a free (possibly constrained) parameter of the particle, rather than something completely determined by the distribution of matter and fields. That free parameter would implicitly contain what (in the conventional approach) we think of as the charge information. Indeed, the new equations of motion would have a conserved quantity, corresponding to the charge. But the resulting force laws would be quite a bit uglier. I don't see how this could be true. If it is, I'd indeed like to see it worked out. Particles with different charges still follow F = ma. The force is different for different particles, sure, but the evolution of the position of the particle still follows a second order differential equation and not a third order one. Furthermore, going to a third order equation doesn't even solve the problem that the F is different for particles of different charge, that problem is still there the same as before. I don't think is a problem at all, by the way: the forces always depend on the particular situation we're considering, including on the mass, charge, and so on, and even more so for compound objects. The question "Why does F = ma?" is a good question though. The conventional explanation only explains why classical mechanics is governed by a second order differential equation. It does not explain why F = ma is the right equation. If you interpret F as a general function of the state of the system then indeed F = ma doesn't tell you anything, because by picking the appropriate F you can get any second order differential equation. However, if we already have some prior idea of what force and mass is, which we do, then it's not clear why it should be this differential equation. We do have an intuitive idea of what force is: if you hang 1kg on a rope then the rope pulls with some amount of force, and if you put 2kg then it pulls with double the force. Similarly, if you stretch a spring by 3cm you have some amount of force, and if you stretch two of those springs simultaneously you have double the force. You could relate the force in the spring with force by a weight by finding the amount of weight you need to stretch the spring by that 3cm. We can get a definition of force by picking some reference force as being 1 unit of force, and we can similarly come up with an experimental definition of mass. If we accept such a definition then F = ma has physical meaning. For instance, it says that if you double the force then you double the acceleration. That is, if you pull an object with two springs then it accelerates twice as fast as when you pull it with one spring, if you stretch the springs by the same amount in both cases. This is a nontrivial fact about the physical world. The law F = ma also tells you that if you apply a constant force then the velocity increases linearly. That is, if you pull an object with a spring and keep the spring stretched by 3cm, then if the object accelerates to speed v in one second then it will accelerate to 2v in two seconds. This is a nontrivial fact about the physical world too.

KenoFischer 7y ago

I'm not sure the question is particularly well posed, but I think the most coherent answer you'd get from modern physics is symmetry. Restating Newton's law's slightly, you basically get two points: 1. Momentum is conserved 2. All interactions between objects are mediated by forces (i.e. interactions exchange momentum). So you may ask, ok, but why do these two things happen, which is is where symmetry comes in. If you require that your theory is invariant under translations in space and time, your theory must conserve momentum and energy. Then you ask, ok, so what kind of long-range interactions can we have between particles in a theory where energy and momentum are conserved and the answer turns out to be, well those that exchange momentum. For example, in the hypothetical from the blog post where interactions are mediated by exchanges of acceleration (forces are proportional to jerks), you end up with a universe where absent interactions, acceleration is constant. Why is this inconsistent with symmetry? Well, in such a universe, the acceleration of particles would be constant in the absence of interaction, so their kinetic energy would keep growing and growing, violating conservation of energy.

platz 7y ago
https://en.wikipedia.org/wiki/Noether%27s_theorem https://en.wikipedia.org/wiki/Noether%27s_theorem
lordnacho 7y ago
This seems to lead to another question, which is why is the invariant of the system what it is? Why are momentum (mv) and energy (0.5mv2) conserved?
- dpq 7y ago
  It directly follows from the assumptions that space is isotropic and homogeneous. I recall that there was a short and elegant proof of this link in Landau and Lifshits vol 1; you may want to look it up.
  - fferen 7y ago
    Indeed, it is at the start of the book. It also relies on the assumption that the Lagrangian is invariant up to a total time derivative under Galilean transformations, which leads to `dL/d(v^2)` being a constant (1/2 m). This is, I believe, the most fundamental way to look at it. Otherwise you could ask why the Lagrangian can't have other terms like v^4.

knzhou 7y ago

> Putting it in somewhat fuzzier terms, and at the risk of repeating myself: F = ma derives its power from the (implicit) assertion that there is a simple unversal force law that lets us figure out F for a particular configuration of matter. And so the configuration of matter completely determines the acceleration of a test particle. There is no a priori reason this ought to be true. It’s an absolutely incredible fact of nature. Yup, this is totally correct. To say it yet another way, we evaluate scientific theories not by looking at the pieces in isolation, but how much explanatory power you get from all the pieces working together (penalized by the total complexity of those pieces). There's nothing mathematically inconsistent about defining "F = mv", it just makes F a less useful quantity. Feynman actually had a remarkably similar discussion in his classic lectures: > For example, if we were to choose to say that an object left to itself keeps its position and does not move, then when we see something drifting, we could say that must be due to a “gorce” — a gorce is the rate of change of position. Now we have a wonderful new law, everything stands still except when a gorce is acting. You see, that would be analogous to the above definition of force, and it would contain no information. > The real content of Newton’s laws is this: that the force is supposed to have some independent properties, in addition to the law F = ma. [...] It implies that if we study the mass times the acceleration and call the product the force, i.e., if we study the characteristics of force as a program of interest, then we shall find that forces have some simplicity; the law is a good program for analyzing nature, it is a suggestion that the forces will be simple.

analog31 7y ago
In a lot of problems, there is a "F = mv" as well, just not using the letter m. A velocity dependent force is a "damping" force. Pushing an object through some fluid medium will produce damping. There is also a "F = mx" which is a spring return force. All three of these forces operating together drive the (approximate) equation of motion for a lot of mechanical things, such as the cone of a loudspeaker.
- semi-extrinsic 7y ago
  Those F's aren't the same as the one in "F = ma" though. The one in "F = ma" means "the (vector) sum of all forces on the object". You can write down, as you say, a bunch of contributions like F_spring and F_damp etc. that can be functions of other variables like x, v etc. "F = ma" then tells you that if you sum up all the forces (F_spring, F_damp etc) you have an expression for the acceleration. This gives you an equation of motion that you can solve to find out how the object will move given some initial state. Basically, if you know the initial position and velocity, you can compute acceleration at that time. Then you can figure out the position and velocity of the object a short time later, and recompute the acceleration using these new values. Then you can step forward again, and again, for as long as you want.

whatshisface 7y ago

F = ma because if you walk outside and push things around that's what you will notice is happening. Any other explanation including a derivation from QFT would just be a different version of that, because whatever deeper underlying theory you were using would have come from the same place as F = ma came from to begin with.

whatshisface 7y ago

>A fun question: how does the universe change if the mass isn’t a scalar, but rather is a matrix, and so a = m-1F is the acceleration? What would this world look like? Is it plausible? This happens in our universe when the coordinate system is squished. In the general curvilinear case, mass is a matrix.

knzhou 7y ago
Not only that, but mass is a matrix for electrons moving in a crystal lattice, even for normal coordinate systems. It's also true if you want F = ma to continue to hold in special relativity, then m is a velocity-dependent matrix.
- cygx 7y ago
  It's also true if you want F = ma to continue to hold in special relativity, then m is a velocity-dependent matrix. Not if you describe things in terms of four-forces and four-accelerations, which restores the simple proportionality. However, the spatial component of 4-acceleration is not necessarily parallel to 3-acceleration.

roenxi 7y ago

I liked KenoFisher's top level answer but I want to change the perspective because I think momentum is a red herring. The starting point should be energy. Classical physics (and modern physics as far as I recall) determined that literally everything is linked by a single concept of energy. In my view energy is more real than anything else we experience and all our senses are for perceiving energy in different states. Matter is pooled energy, movement is energy, heat is energy (linked to movement), time is defined to some degree in terms of mass so it is linked in there too somehow. And we know energy is conserved because we have observed that everywhere. Once we know energy is conserved the law of momentum makes a lot of sense - there have to be symmetrical laws that don't allow arbitrary creation of energy because that doesn't happen. And then F = ma turns up and it isn't so surprising for the same reason. F = ma is in that sense not a fundamental law; it is a corollary of the conservation of energy. Mathematically there might have been other options, but that is the one that applies in this universe. There was always going to be a mass component in the formula because by observation energy is fundamentally linked to mass.

klodolph 7y ago
This only makes sense in a relativistic context, but forces don't exist in relativistic contexts. So you can either have F = ma, or you can have GR, but you don't get to have both at the same time.
- selimthegrim 7y ago
  Wait, you can’t write the Lorentz force down?
  - cygx 7y ago
    You can: https://en.wikipedia.org/wiki/Maxwell%27s_equations_in_curved_spacetime#Lorentz_force_density https://en.wikipedia.org/wiki/Maxwell%27s_equations_in_curve...
- pdonis 7y ago
  > forces don't exist in relativistic contexts Yes, they do; the difference in General Relativity is that gravity is not a force, but non-gravitational forces are still there.

ta1234567890 7y ago

> the equation in Newton’s second law isn’t F = ma, but rather the more subtle statement that force is equal to the rate of change of momentum of a body This is probably the most insightful bit and points directly to the potential of fiddling with the equation. Drawing an analogy from electricity, F=ma is in a way the "DC version/expression" of a force, in which m and a are constant and F is a fixed quantity. However we can write an "AC version/expression" of the equation as F(t)=m(t)*a(t), in which the average of F(t)=F. In a way we tend to just be content with dealing with averages instead of looking at the detail and seeing how it varies over time. There's a lot of potential in looking at the variation. Especially as technology enables it through high speed cameras and instruments with higher sampling rates/resolution.

raverbashing 7y ago
The "AC" version of that equation is actually F=dp/dt (p = momentum)
- ta1234567890 7y ago
  Thank you for the formula. Do you know of example advances that have been made or problems that have been solved as a consequence of applying the concept implied in that formula?
  - ianopolous 7y ago
    Rockets
    - ta1234567890 7y ago
      Care to elaborate?
      - aerodude 7y ago
        Rockets have an exhaust with constant velocity, and they expel mass in the form of fuel. It's this expulsion of mass that generates the force that propels the rocket. Taking a force balance and assuming no external forces (e.g. gravity), you can integrate to get the rocket equation, which tells us how much fuel mass we need to reach a given velocity (the quantity we care about when talking about stellar distances).

dvt 7y ago

> I find it astounding that a theory like quantum mechanics can have inside it another theory, an approximation, also extremely beautiful mathematically, but radically different. It’s like taking Bach, adding some noise, and getting the best of the Beatles out. I wish I understood better why this can happen. This isn't really the case. Author should've probably mentioned the Ehrenfest theorem[0]. It's not exactly accurate to say that Newtonian mechanics is "embedded" in quantum (or rather Hamiltonian) mechanics. In fact, we need to do some "artificial" re-jigging to make quantum mechanics work at Newtonian scales. [0] https://en.wikipedia.org/wiki/Ehrenfest_theorem https://en.wikipedia.org/wiki/Ehrenfest_theorem

gus_massa 7y ago
>> It’s like taking Bach, adding some noise, and getting the best of the Beatles out. It's a very bad analogy. It's difficult to find a good one. Let's try anyway: Imagine that QM is a real piano and CM is how you expect a piano to sound in normal conditions. If you press a C key, you get a C sound. You can modify the sound releasing the key or keeping it pressed, or pressing the pedals. There are some effect where you press another keys so the sound of the original string make the other strings move and modify the sound. You can hit the wood part of the piano to make some drum like sounds. Most of this can be simulate in an electric keyboard and get an accurate piano-like sound. This is like Classic Mechanics. Quantum Mechanics is like the real piano. You can open or close the top. You can fill the piano with a different gas to make the sound slightly different. Or fill it with water or liquid nitrogen. Use drumstick to hit the strings directly. Use a hammer to hit the wood walls. These create a more wide set of sounds that are nor reproducible in a electric keyboard. An electric keyboard is only an approximation of an idealized real piano. (An experiment where the waviness of the particles is the most important part is like dropping a piano from a 8th floor to hear the sound of the crash, or using the piano as firewood to hear the crispy sound of fire in a winter night. I'm not happy with these examples because they are too destructive, but I wanted too examples of sounds that are very far away from the sounds you expect in an electric keyboard.) (For some reason, musicians do more weird destructive acts with guitar than with pianos. Perhaps it would be a better instrument for the analogy. But it is more difficult to make an idealized model of a guitar. A piano is almost discrete.)

andrepd 7y ago

If you're interested, Landau & Lifschitz "Vol 1: Mechanics" has a derivation of classical mechanics from purely physical arguments, from first principles. F=ma is on page 9.

acqq 7y ago
And that derivation seems to be known at least since 1788: https://en.wikipedia.org/wiki/Lagrangian_mechanics https://en.wikipedia.org/wiki/Lagrangian_mechanics
- andrepd 7y ago
  Yes, but Landau derives that in a supremely elegant way, from purely physical arguments.
  - acqq 7y ago
    I personally always prefer first historical approach, only then elegant summaries. I am aware that some like just a distilled material. Still thanks for mentioning Landau and Lifschitz book!

d--b 7y ago

I personally would have gone with: because E=0.5mv^2 And by that i mean that energy is a much more intuitive concept than force. Why is E proportional to v^2? Now THAT’s an interesting question.

earthicus 7y ago
Here's a good explanation (it reduces to F=ma, of course!). This explanation starts by answering a different question, then points out that energy is analogous. https://physics.stackexchange.com/questions/67432/intuitive-meaning-of-factor-2-in-formula-of-vertical-throw-max-height-h-v2-2g/67433#67433 https://physics.stackexchange.com/questions/67432/intuitive-...
XorNot 7y ago
How about this: because there's a mass involved with a non-zero cross section, so velocity is being added to an area, not a point. Hence v^2 is hiding the complexity of density vs surface area (the law was originally derived by observing steel ball impact craters on a bed of sand vs the height). 0.5 is because only one side hits. EDIT: And what surface area? Well General Relativity does like to mash up the concept of objects affecting space time due to having energy...

ISL 7y ago

We will never know why it does, even though it is a well-motivated prediction of classical and modern theories. It is possible for us to test whether or F=ma, though, and so far, it (with relativistic corrections) checks out. An example of this sort of test from our research group: https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.98.150801 https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.98... (I'd go dig up more citations on the subject, but we have a grant proposal due in two days... :) )

kmm 7y ago

> [...] Two test particles with the same initial position and velocity, but different electric charges, can behave quite differently in the same electric field. > One possible response is to say “oh, maybe our notion of force should really be something like F = mj, where j is the jerk, i.e., the third derivative of position”. > I’ve never worked it out in detail, but wouldn’t be surprised if such an approach can be made to work. It might not be exactly what he's looking for, but in Kaluza-Klein theory, charge is related to the velocity of the particle in an unseen 5th dimension. Through some miracle, if you work out general relativity with one dimension extra, the resulting theory bears a striking resemblance to general relativity in the usual four dimensions plus electromagnetism. If you then assume the 5th dimension is extremely small, which would explain why we can't see it, you get quantization of charge for free. https://en.wikipedia.org/wiki/Kaluza%E2%80%93Klein_theory https://en.wikipedia.org/wiki/Kaluza%E2%80%93Klein_theory

_Microft 7y ago

Hijacking this thread in hope that some fellow physicist could chime in. I'm physicist myself so maybe I should know this or be able to figure it out myself but here's the question: Where is kinetic or rotational energy 'stored'? Is it eventually the transformation of fields that gives rise to kinetic energy? For example an electron whose electric field is not a radially symmetric must be a moving electron and the difference in energy between the field configurations (simple radially symmetric electric field on one hand and the mix of electric and magnetic fields on the other) is exactly the kinetic energy it has?

selimthegrim 7y ago
You have hit upon the so called Abraham Minkowski paradox https://en.m.wikipedia.org/wiki/Abraham–Minkowski_controversy https://en.m.wikipedia.org/wiki/Abraham–Minkowski_controvers...
knzhou 7y ago
The electromagnetic field configuration does contribute energy/momentum. In general, every field that interacts with the election field contributes: an “electron” stands for a complicated joint excitation of the electron, photon, W/Z boson, quark, gluon, ... quantum fields and what we call the momentum/energy of the electron is just the total of that of all of these fields.

octoboto 7y ago

>A fun question: how does the universe change if the mass isn’t a scalar, but rather is a matrix, and so a = m-1F is the acceleration? What would this world look like? Is it plausible? m is only a scalar for particles though? For a general rigid body mass is a matrix. Unless they specifically mean for translational degrees of freedom, in which case I think that would probably break symmetry

8bitsrule 7y ago

Because Work = Force times Distance = Energy. Acceleration is the result of energy added (by doing work) to a body with inertia (mass). Now if you ask what is inertia: that is the right (IMO unanswered) question!

doggydogs94 7y ago

One time in a physics class we were required to have a “cheat sheet” of formulas for an exam. One student did not have a “cheat sheet” and was told that he was required to have a cheat sheet. He grabbed a sheet of paper and wrote F=ma. He aced the exam.

theothermkn 7y ago

'F = ma' is a special case of 'F = d/dt(mv)'. Expanding the latter via the product rule, we get: F = m dv/dt + v dm/dt, where a=dv/dt and dm/dt is the time rate of change of the mass of the object. This fuller form matters in, for example, aerospace engineering, where the mass of the rocket is changing as fuel is burned, so dm/dt is emphatically not 0. I wonder if the philosophical questions asked in the fine article could have been addressed more satisfyingly from that more general starting point?

millstone 7y ago

One interesting fact that has been missed here is the nature of time reversal. F=ma. Switch on the time reversal machine: t becomes -t. Positions stay the same, but velocities reverse, because dx/d-t -> -dx/dt. BUT accelerations stay the same! dx/d^2t gets a double negative, which is just 1. Gravity attracts in the future and the past. Throw a ball up and it comes back down; reverse the camera and it looks the same. This means that (handwaving) forces are independent of time. And this in turn leads to a time-independent construction of forces - how about as the gradient in space (NOT time) of a potential. This I think is a path to partial insight. A particle moves in a potential, and we can calculate the quantity 'a' from the purely-spatial gradient at the particle's position, without reference to time at all. And we can do this because 'a' is an even derivative of position. This is a crappy argument; at best it argues that it's nicer that f=ma instead of f=mv or f=ma'. Fourth or sixth derivatives are not considered. Still, understanding dynamics as governed by position only is compelling.