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Thinking like a theorist

UPDATE: Here is my answer. Enjoy! I'll post another question like this in a day or two.


We've been providing good math and physics problems that we hope everyone enjoys. Above and beyond these problems, which generally require application of known techniques, there is a set of deeper questions about physics and why some math appears but not others. This is the domain of theoretical and mathematical physics. Understanding "why this math?" can lead to a much deeper understanding of the fundamental physical principles of our world.

To further this goal, I'll be asking questions on seemingly simple equations in physics that have deep physical reasons why they are those equations and not others. I hope that our bright users spend some time thinking about them and discover the answer, and, if they prove popular, I'll chime back in after a day or two with a link or post about the answer and pose another question.

Our first question:

Newton's laws (in 1-d) state \(F=ma\), where a is the second time derivative of the position. Why does this equation have two time derivatives and not, say, 1,3,4, etc. time derivatives? In particular, what are the physical reasons behind this?

Note: Answering why this rule exists with another mathematical rule, like "because we work off Lagrangian mechanics and those have single time derivatives" is not okay - it just replaces one rule with another. We want the physical reason...

Note by David Mattingly
4 years, 4 months ago

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The basic physical assumption while formulating laws for the interactions is that when the physical entities are sufficiently away from each other, they don't affect each other's motion. Second assumption that Newton made (which Galileo found experimentally observing spheres' motion on slanted planes) is that if any other body does not affect the motion of the physical object then its velocity doesn't change. When these two assumptions are combined we get concept of inertial frame. So if you observe the particle which is sufficiently isolated from other objects and its velocity is not changing with time then the frame attached to you is an inertial frame. Now any frame moving with constant velocity with respect to an inertial frame itself must be an inertial frame (easy to prove using arguments given above). This construction of inertial frame is the basis of newton's second law F=ma. Now we ask-what happens if the particle, which we are observing from inertial frame, starts interacting with other physical objects? How will it's motion change (if at all)? Now this must be answered experimentally. But to do this experiment we first need to go to an inertial frame. Newton argued that the frame attached to the distant stars is a very good approximation to an inertial frame and this also makes the frame attached to earth almost inertial. Thus we can do experiments on earth and establish the effect of interactions on objects.Thus we can now use for example Galileo's observation about falling objects to establish the rule that interaction causes change in velocity i.e. an acceleration when viewed in an inertial frame. In this case interaction is between falling object and earth. One can do experiments with different objects and establish that the quantity that will characterize the interaction must be acceleration. In this way we find that the law should contain second derivative only and not any other derivatives.

Snehal Shekatkar - 4 years, 4 months ago

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Excellent! The issue of non-interaction between objects is important in this context. We call this concept "locality". Physics is local - what happens at my point at time t doesn't have anything to do with what you do over at your point at time t. Let me pose a specific adaptation of my primary question. I asked, why not higher time derivatives? The extreme case of this is an infinite number of successively higher time derivative, i.e. something like

\(F=m(\sum_n c_n d^n x/dt^n)\)

where n runs from 1 to infinity and the \(c_n\) are coefficients. Can anyone tell me how such a theory would interfere with locality? If we can get rid of such a theory on locality grounds, then we at least have pinned our Newton's laws down to some finite order in derivatives, which is a step. We then have to show that the highest order of allowed derivative is two. And then finally we'd need to argue that having the first derivative is also bad.

David Mattingly Staff - 4 years, 4 months ago

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I also don't understand on what basis you assume this form (sum of all the derivatives) for the law? One may ask for example.. why not to consider square of this expression or even sine of this?

Snehal Shekatkar - 4 years, 4 months ago

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@Snehal Shekatkar You could. This is just an example function. The key point is simply what happens if there are an infinite number of derivatives.

David Mattingly Staff - 4 years, 4 months ago

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Yes.. To know the values of derivatives of all orders up to infinity would require information about function x at all points t and this would violate the assumption of locality. This forces us to restrict ourselves to finite number of derivatives.

Snehal Shekatkar - 4 years, 4 months ago

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@Snehal Shekatkar Bingo, that's neat. If our position function were analytic then it is precisely these derivatives that determine the behavior of our function in the neighborhood of \(t_0\) i.e. you would say force is dependent on knowledge of our function in both the "near" future and the "near" past.

O B - 4 years, 4 months ago

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@Snehal Shekatkar Sorry I am a beginner at physics/calculus so I might be thinking wrongly but cant you find out the nth derivative of any function, for example to find, dn^x/dt^n cos(t) it is merely cyclic -sin(t), -cos(t), sin(t), cos(t) and you just divide n by 4 and find the remainder... and for other functions like a1x^k + a2x^k-1 ... ak-1x + ak if k < n than the derivative will just be zero repeatedly

Satwik Nandala - 4 years, 4 months ago

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@Satwik Nandala No; it requires differentiability. The functions you are looking at are sufficiently 'well-behaved' smooth functions that are infinitely differentiable.

O B - 4 years, 4 months ago

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I have a bit doubt you said that physics is local. what happens at a point in universe doesn't interfere with anything happening elsewhere in that time duration t. But what about this interesting concept of quantum entanglement ? What about bell's experiment ? So please explain to me sir does the concept of locality doesn't work here ?

Ritvik Choudhary - 4 years, 4 months ago

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@Ritvik Choudhary We started with F=ma so we're squarely in the realm of classical, point particle physics here, not quantum mechanics. However, since you asked :)

If you want to talk properly about locality and causality in quantum physics, then you'd need to use quantum mechanics + special relativity (since SR defines the causal structure of our universe). This is the realm of quantum field theory. Quantum field theory is local, but this statement confuses people because they think of entangled quantum states, non-local wavefunctions, etc. The whole discussion about locality and entanglement is too long here, but the key point is the following. Locality as most people think of it is not enforced on every quantity one could construct in quantum field theory, for example the wavefunctions themselves can have spatial extent, they are manifestly non-local objects. However, observables, i.e. things we can actually measure, do obey what we usually think of as locality, in that if you try to perform two experiments in different places at the same time, the values to get for your measurements will be independent of each other. It's the observables that are the truly important quantities for locality.

An exercise: construct your favorite entanglement scenario and then try to build actually measurable quantities out of the stuff in your scenario. You will find that those observables satisfy locality.

David Mattingly Staff - 4 years, 4 months ago

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@David Mattingly yeah...thanks

Ritvik Choudhary - 4 years, 4 months ago

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The problem with involving an infinite number of derivatives in the expression for force is that we will not be able to ever create a framework for the observation of and recording of the vectors of force.

Let's consider that we are observing the motion of a body from a frame of reference in which we are at rest. The movement of the body is such that the nth order derivative if the object's position (as seen from our frame of reference) with respect to time is the first order derivative to be equal to zero. now, if locality holds, then our own motion should not affect the magnitude of the force acting on the body as seen by us. But we can have many different 'types' of motion...and in every 'type' of motion we shall have some or the other order derivative (let it be the mth order derivative, such that m is less than n) be the highest order derivative that is zero.

For every type of motion we can have, we will observe a different magnitude of force acting on the body. Now, one may argue that our present definition, in which F=ma, if we accelerate with respect to an inertial frame of reference, we'll observe a different magnitude of force. That is true, and the case I am presenting is similar. But, consider if we move such that the third order derivative (pertaining to our motion with respect to an inertial frame of reference) is not zero...then the magnitude of force does not again change...we must only consider our acceleration wrt the object, and not the 'new' aspect of our motion - the third order derivative. In other words, we can create an unchanging framework in which the expression for force works, if we have a finite number of derivatives involved in the expression.

But, if we have an infinite number of derivatives involved in the expression for force, every 'type' of motion we make will involve re-evaluating the magnitude of force according our motion. In other words, the principle of locality will get messed up, because the magnitude of the force on the object will always depend on the 'type' of our motion, and we cannot create any fixed framework in which the expression will invariably hold (like inertial frames of reference).

P.S. - In all the situations I've given above, I'm considering that the motion of the body can always be represented as a polynomial expression of 'x', (the position of the object...as this entire discussion pertains to 1-D motion).

Rish Malviya - 4 years, 4 months ago

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Here is my answer. Enjoy! I'll post another question like this in a day or two.

David Mattingly Staff - 4 years, 4 months ago

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Alright everybody, this is a good discussion and obviously people are interested, so we need a blog post with my answer to this question (which will probably start just as many questions as it answers, but that's okay too). I'll write a post Monday. People are more than welcome to keep going with their questions too, of course, and I'll keep answering as I can.

David Mattingly Staff - 4 years, 4 months ago

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paragraph 1 This must have had some type of experimental reason, for example if I experimentally analyze the change in position of a particle under the action of a non-zero constant force, and plot the points on a position vs time graph, I would obviously get a parabola. Now just for fun, let us assume that the force is equal to a proportionality constant * (the first time derivative of position), If my assumption holds, I would get a straight line (with a non-zero slope) representing force on a force vs time graph, but this quite much contradicts with my assumption that the force is constant. paragraph 2 Now If I again assume (just for fun) that the force = k*(the fourth derivative of time) where k is the proportionality constant. I would get no line representing force on the force vs time graph, which again contradicts with my assumption that the force is non-zero. paragraph 3 This same inductive approach would hold for derivative of any order which can perhaps show that force is equal to a proportionality constant * (the third time derivative of position).

Siddharth Kumar - 4 years, 4 months ago

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It is certainly correct that experimentally this law works so gold star. However there is a purely theoretical reason as well. Think of it this way: if I wrote down a theory that was not f=ma but had higher derivatives i could invalidate that theory either on the grounds of theoretical inconsistency or experimentally. You have given the experimental argument. But what is the theoretical argument? And, it's not technically hard but it requires one to think deeply about what we need in our physical theories. Which is why I asked the question.

David Mattingly Staff - 4 years, 4 months ago

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We look at how force is defined.

We start with a stationary object.

The more mass there is, the more force required to move it in some way.

The greater/faster the movement you want, the greater the force you need.

Hence the unit for force should be \((kg^a) \times ((m/s^b)^c) \times k\), where k is some other unit we may not need to care about, \(a>0\), \(b>0\), \(c>0\)

Step 1: Deriving a. Suppose we have 2 masses we want to move the same way, one m times the mass of the other.

Suppose m is a rational number. We can subdivide the 2 masses into equal parts.

Step 1a: We claim that is 2 objects are stationary with respect with each other, then there is no net interaction.

Since there is no net interaction between the divided masses, moving each of them the same way should require the same amount of force.

Thus we can combine the divided parts and prove that force is directly proportional to mass. (Of course, it also requires that the relation between force and mass is continuous, since I only allowed rational numbers into the argument.)

Thus \(a=1\).

Step 2: Showing c=1.

We will use frames of reference. Consider 2 masses, one moving m times faster than the other. Suppose m is rational.

We subdivide the movement into units. By taking the reference frame of the object moved, we can deduce that twice the movement requires twice the force, thrice the movement requires thrice the force and so on.

Thus c=1.

Step 3: Showing b=2. (Which is the main part.)

(This appears to be a circular argument...)

Suppose we exert a positive force on a once stationary mass. (Positive direction.)

Now we remove the force and take the reference frame of the object at that instant and keep the same reference frame (same velocity). At that instant, the object is stationary (in the frame).

When an object is stationary and no force acts on it, the object remains stationary (physically true).

Thus b is not 1.

Supposing b>2 should do something, but I don't think I figured that yet.

Once b is fixed, the rest is dimensional argument.

Clarence Chew - 4 years, 4 months ago

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newtons first law holds for a body moving in a straight line and from there the second law follows but if there be some other derivative of position we will not get a straight line so it is not valid

Superman Son - 4 years, 4 months ago

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@Superman Son Yes, Diganta, it's the same answer which I wrote, but sir is asking not for a mathematical answer or anything mathematical about the law, but sir is actually asking for the basic reason, that why only this mechanism holds in physical situations. Think like a theorist.

Siddharth Kumar - 4 years, 4 months ago

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@Siddharth Kumar yeah i got it

Superman Son - 4 years, 4 months ago

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Well, I'm quite late for this, but I just had an idea, and I'd like to share it.

I think the important thing here is how we define force, in the 'physical' sense. People have pointed out that force arises out of some interaction between two bodies/particles. While that helps elucidate what we 'require' for something to be a 'force', I don't think it quite defines it.

I think the definition which should prove most useful in this discussion is of force as a 'potential (energy) gradient'. This definition is readily derivable mathematically, but if one wants to arrive at this 'definition' more 'physically', one needs to simply consider a simple system such as that of a satellite being launched from Earth, observing the variation in the energy of motion (kinetic energy, and thus the variation in the potential energy) of the satellite. Of course, the 'potential gradient' here must arise from some or the other interaction between two or more bodies (electromagnetic, gravitational, etc.), but this definition allows us to relate force to a much more readily understandable quantity - energy.

Essentially, we can think of force as a means of transferring energy - or for a single body (in an inertial frame) as a means of increasing or decreasing it's energy.

Now, the energy (kinetic only, since F=ma pertains to rigid bodies) of a body can very well be thought of as its capacity to move itself (overcome its intrinsic inertia and change its position or sate of motion). If the position of the body changes linearly with time, its capacity to move (i.e., its energy) remains unchanged. But if it moves more or less than it would have if it continued in the original 'linear' fashion, its 'capacity to move' (i.e., its energy) has clearly changed! Therefore, in order to change the energy of body, the second order derivative of its position with respect to time must change (at least). And an accurate measure of the magnitude of the energy transfer (i.e., the force), would be related to the second order derivative of position, which is the first, most fundamental indication of a change in energy.

Rish Malviya - 4 years, 4 months ago

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Let us start with first derivative of position with respect to time ds/dt i.e. velocity. A particle in an inertial frame can move with constant non-zero velocity for a long time without the influence of any external agent and we can say that a force is an agent with makes change in the state of an body but in above case as we know that no force is applied on the body so ds/dt i.e. the first derivative of position with respect to time times the mass of the body is not the correct way to calculate force on a body.

Now, let is take second derivative of position with respect to time i.e. change in velocity w.r.t. time or acceleration. Again we take a particle in space in an inertial frame moving with a constant non-zero velocity under influence of no external agent. As there is no external force acting on the body it is obvious that the body is not accelerating and if we apply some non-zero force on the body the body starts accelerating (it can be seen experimentally). In this case the state of the body (here velocity) is changing by the application of an external force.Hence velocity is a physical quantity which can change only by the application of external force (condition that the mass should be kept constant). Hence, second derivative of position w.r.t. time times the mass of the body is the appropriate expression to calculate force on a body.

Now one can argue about higher derivatives of position w.r.t. time but that will create a problem. We can see that by the help of an example, for eg. a particle is moving with constant acceleration under a influence of a constant force so the third derivative of position w.r.t. time i.e. rate of change of acceleration will be zero as particle is moving with constant acceleration. In this case if we express force as rate of change of acceleration times the mass of the body it will come out to be zero which is not so as we know that we are applying an external force due to which the particle is accelerating at a constant non-zero rate, similar is for other higher derivatives. This the reason why we do not use higher derivatives of position w.r.t. time and only second derivative.

Rahul Dandwate - 4 years, 4 months ago

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I liked your reasons very much but for higher derivatives you are taking acceleration constant, then, its derivative would be zero, but if the acceleration is not constant, then?

Tushar Gopalka - 4 years, 4 months ago

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For eg. if we take acceleration to be increasing such that the rate of change of acceleration is constant and multiply it (rate of change of acceleration) with it's mass then it would be a constant (constant*constant=constant) but we know that for increasing acceleration we need a increasing force (for constant mass) i.e. not constant. But if we see from above if we take rate of change of acceleration (not acceleration) then we get a constant force which is not so. Hence, the problem of taking higher derivatives lies here.

Rahul Dandwate - 4 years, 4 months ago

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@Rahul Dandwate And if the rate change of acceleration is not constant, then?

Tushar Gopalka - 4 years, 4 months ago

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I Think I have got the answer to this question. Imagining we are during Newton's time and we don't know anything about his laws. I think you are asking us to sit and reason that why F=ma, based on day to day observations. Okay, So here's my reason:- We know by observation that a grater force is required to move a more "massive" object that a smaller object, which means that force must somehow depend on mass of the object, the greater the mass, the greater the required force to make the body move. Now, comes the trickier part that is the explanation that why F=ma and not mv. Suppose, we have two balls(both are at rest initially), we apply a force F1 on body 1 to make it move with a velocity "V1" in a time interval t. Now we apply a force F2 on body 2 to make it move with velocity "V2" in the same time t. By pure logic and intuition we can say that if we want V2 to be greater than V1, that is we want the change in velocity in body 2 to be more than body 1 in the SAME TIME INTERVAL, we have to apply a greater force on body 2 The point is that force will not depend on velocity but the rate of change of velocity, it simply dosen't depend on the change in velocity, the time taken for that change does matter which can be stated from simple observation. Hence, force depends on the rate of change of velocity which is acceleration. Hence, F=ma. I think this can be a logical explanation. Can it be, David?

Tushar Gopalka - 4 years, 4 months ago

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It's close, and it gets at getting rid of the possiblity F=mv. One could also ask about higher derivatives too :)

David Mattingly Staff - 4 years, 4 months ago

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higher derivatives..in the sense of what...no offence but can you please explain the concept of higher derivatives elaborately.

Tushar Gopalka - 4 years, 4 months ago

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@Tushar Gopalka Higher derivatives of x with respect to t, so first derivative of x with respect to t is velocity, 2nd derivative is acceleration, etc...

David Mattingly Staff - 4 years, 4 months ago

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This might have been answered already. I haven't looked at these posts in a few days.First off, if we assumed that F=mv where v is velocity, this would mean that any force greater than 0 applied to a stationary object would give an instantaneous discrete change in velocity (from 0 to a value greater than 0, and with no velocities in between), which is intuitively impossible (I'm not sure if that is a sufficient answer, but that's all I've got). For example, it's obvious that something cannot go from 0 m/s to 1 m/s in 0 seconds. Next, I'll try to rule out the possibility of infinite derivatives: picture a graph with position (y) on the y axis and time (t) on the x axis. When a constant force is applied to an object whose position is depicted by the graph (according to F=ma), the function should appear to be in the general form of y=t^2 (there should probably be a coefficient and added constant in there but just ignore that fact). Now assuming that F=m(jerk), this graph would appear as y=t^3. Going on to assume F=md(jerk)/dt (the fourth derivative of position), the graph would appear as y=t^4. You can picture that, with each further time derivative of position, the graph increases more quickly. If we keep going through the entire realm of real numbers for how many derivatives we take and then try infinity, you can imagine that the position will start at 0, but instantaneously shoot up to infinity (in 0 seconds). This would mean that if a pebble hit a rocket ship in space, they would both shoot off with infinite velocity the moment they collided (with the assumed F=m*d^(infinity)x/(dt)^infinity). There are obviously many problems with this: infinite velocity is impossible, velocity over the speed of light is impossible, and a discrete change in velocity over 0 seconds is impossible (as it gives an acceleration of infinity). I realize that answer is a bit mathy, and we are supposed to give a theoretical, physical answer, but there is intuition in that the physical results of the infinite derivative force relation should not be possible, or at least observable. As for all other derivatives in between 2 and infinity, I'm not sure if I can give a sufficient answer.

Casey Shiring - 4 years, 4 months ago

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Today he was supposed to post his answer. Where is that?

Snehal Shekatkar - 4 years, 4 months ago

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I'm working on it :)

David Mattingly Staff - 4 years, 4 months ago

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Okay.. we r waiting for it! :-)

Snehal Shekatkar - 4 years, 4 months ago

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i have one more question. Newton's laws state that Force is directly proportional to the rate of change of momentum. What is the theoretical reason behind this. Also, what exactly do we mean by momentum, what physical reference does it have. Why is the concept of momentum important?

Tushar Gopalka - 4 years, 4 months ago

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David sir would you please comment upon my reply of your question ,i am getting very excited of whether my answer is correct or not.

Rahul Dandwate - 4 years, 4 months ago

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Rahul,

I like your argument for getting rid of the first derivative. As for the higher derivatives, your argument assumes a bit about the existence of Newton's 2nd law in arguing about the higher derivatives, but the existence of the second law is exactly what the question is about.

David Mattingly Staff - 4 years, 4 months ago

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According to the classical machanics the natural state of an object is in rest or constant velocity and the only way to change this natural state is applying a force to it. Thinking on that way, force is associated to this variation on the velocity of the body. That´s why it is related to the acceleration.

Pedro Soares - 4 years, 4 months ago

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According to newtons first law a body moving with constant velocity will continue to move till all its forces are balanced and it will try to remain in this state only due to inertia. Now all forces are balanced so force is not dependent on velocity as net force acting on it is zero. So its not the first derivative on which force depends.

Karan Kewalramani - 4 years, 4 months ago

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Are you asking why F=ma?. I am in 11th grade and have just studied newton's laws, this is the question which I have asked to every of my physics tutor and they say it's a law, I asked them what was the logical reason behind this saw and they kept on saying that it's a law. That's why now I am thinking why is F=ma and I am happy that this post would answer my question.....

Tushar Gopalka - 4 years, 4 months ago

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Yep, I will be answering this question Monday, so check back!

David Mattingly Staff - 4 years, 4 months ago

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And glad we can help out with a question that's been bugging you.

David Mattingly Staff - 4 years, 4 months ago

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I have a question rather than an answer and it is that if we apply force on a object(push or pull it) and the acceleration is zero then the force acc. to Newtons law is 0. So, if we push a wall or a large rock for a whole day trying to move it and the acceleration is zero then probably I applied no force.. So isn't it a flaw in the theory ?? or am I wrong somewhere??

Pranjal Rs - 4 years, 4 months ago

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No, you applied a force but as there was no displacement, you did no work, according to physics. The amount of Energy you provided was not enough to move the wall, The mass of the wall is too high, so magnitude of acceleration was almost zero, negligible.

Tushar Gopalka - 4 years, 4 months ago

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The first time derivative of position is velocity, the second is the rate of velocity change with respect to time (Acceleration). Firstly we can easily know it by experiment, and that's how Newton probably did it, but I guess you're looking for a more complex answer. Mass can be thought of as the amount of inertia a body has. A body is at rest or moving in constant velocity if the net forces acting on it is zero. Thus it cannot be one time derivative (velocity), or else we'll be saying that a body moving with constant velocity will require a force to keep it moving, which is not the case. However in order to change the motion of a body, to change its velocity, requires a force. This change in velocity is the acceleration we use in the equation. Now why would a theorist think of further derivatives if he has already figured the variables out? Anyway, to make further derivatives of time will be out of inertial frame of reference, since time will be cubed and so on, I can possibly have an infinite amount of inertial frames of references at one point, which is not the case either.

Mohab Sameh - 4 years, 4 months ago

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Let me preface my post by saying that I'm not in any way adept at physics. My background is in pure math, and I barely scraped by in my college physics courses. So I'm not going to claim I know what I'm talking about.

That said, I think that the underlying concept here is that old thought experiment where an observer sees an object moving at a constant velocity, and wonders whether it is the object which is moving and the observer is at rest, or if it might be equally said that the object is at rest and it is merely the observer who is in motion.

For if a person is traveling in, say, a car or a train at a fixed speed, we do not perceive (aside from the occasional bump or turn) that we are moving at all. The Earth is rotating about its axis and orbiting the Sun, yet for all this movement, we perceive nothing, since we are moving along with it.

So, it seems perfectly reasonable to characterize the laws of mechanics as being invariant for any inertial reference frame (we'll forgive that the Earth's rotational and orbital movements are, strictly speaking, not in such a frame). In that sense, Newton's second law pertains to "force" as the notion that an object experiences a change in velocity with respect to some inertial reference frame.

Hero P. - 4 years, 4 months ago

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You are on the right track. Can you use this argument or a variation to remove the possibility that Newton's laws could be \(F=mv\)?

David Mattingly Staff - 4 years, 4 months ago

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In thinking about this question, we should recall that Forces are a result of particle interactions. That is, a particle experiences a force when it interacts with another particle, say a graviton, or a gluon.

As to why it does not involve only 1, or even 0 time derivatives, we should recall the rules of relativity. A common example used to understand relativity is that given you were in an opaque, non-accelerating box, it would be physically impossible to tell if the box was moving. We could tell if it had an acceleration, but not if it only had a velocity. More generally, the laws of physics behave identically in all inertial, non-accelerating reference frames.

If our force interaction is going to be significant, it makes little sense for it to involve something that is relative, that is velocity, or position, the 0th and 1st time derivatives. Then, the simplest change that this interaction should probably have is the second derivative, because of its relation to non-inertial frames. I then just go off on a limb and say that the force probably acts in the simplest manner.

Mark Brown - 4 years, 4 months ago

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I hope I'm not being too basic about this:

We all know that speed or velocity is just the change in position, and acceleration is the change in velocity, or change in change in position. Let's say, for example, that you are driving in a frictionless car. You are travelling at a nice constant (non-zero) speed. You're moving, but not accelerating. So let's say you come up to another frictionless car that's moving slower than you. You come up to the car and bump into it, causing you to obviously slow down. In this instance you are putting a force on the other car when you hit it. This is the kind of force that we would call in every day life. But at the same time, you are reducing your speed, which is a change in velocity, which is acceleration (albeit, negative). So, there should be some relationship between the force that the car created and the acceleration the car underwent. And the relationship, intuitively turns out to be proportion to the mass of the object. Think: if you had a bigger car, you would hit the other car harder. If you drove a small car, you wouldn't effect the other vehicle as much. So \(F=ma\) makes sense in our every day life.

Bob Krueger - 4 years, 4 months ago

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In statics, you can still have a force without acceleration so F is independent of a. F is the cause of the change in the position of an object initially at rest in some frame. To give it physical meaning, you have to define how it's to be measured and one way would be to define 1 unit of F causing one unit of compression in some standard spring.

Now if F causes a body at rest to change its position, then over a time dt the postion has changed by dx. Your job as a physicist is to construct an equation relating F to the change in velocity of the body.

So with all this in mind, what would happen if F=m∗d^3x/dt^3 ?

It would mean that even though F is the cause behind the change in velocity of a body, there are some changes in the velocity possible where F=0 such as for a=const. You would end up with particles accelerating in arbitary directions for F=0.

Harsa Mitra - 4 years, 4 months ago

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This is closer....what is intrinsically wrong with particles accelerating without a force? Is it mathematically forbidden? Is it physically forbidden? If so, why?

David Mattingly Staff - 4 years, 4 months ago

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After reading some Books and doing a bit research on the internet i found out:- We tend to not use higher derivative theories. It turns out that there is a very good reason for this, but that reason is rarely discussed in textbooks. We will take, for concreteness
<math xmlns="http://www.w3.org/1998/Math/MathML"> <mi>L</mi> <mrow> <mo>(</mo> <mi>q</mi> <mo>,</mo> <mrow class="MJX-TeXAtom-ORD"> <mover> <mi>q</mi> <mo>&#x02D9;<!-- ˙ --></mo> </mover> </mrow> <mo>,</mo> <mrow class="MJX-TeXAtom-ORD"> <mover> <mi>q</mi> <mo>&#x00A8;<!-- ¨ --></mo> </mover> </mrow> <mo>)</mo> </mrow> </math> a Lagrangian which depends on the 2nd derivative in an essential manner. Inessential dependencies are terms such as qq¨ which may be partially integrated to give q˙2. Mathematically, this is expressed through the necessity of being able to invert the expression <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"> <mfrac> <mrow> <mi mathvariant="normal">&#x2202;<!-- ∂ --></mi> <mi>L</mi> </mrow> <mrow> <mi mathvariant="normal">&#x2202;<!-- ∂ --></mi> <mi>q</mi> </mrow> </mfrac> <mo>&#x2212;<!-- − --></mo> <mfrac> <mi>d</mi> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mfrac> <mfrac> <mrow> <mi mathvariant="normal">&#x2202;<!-- ∂ --></mi> <mi>L</mi> </mrow> <mrow> <mi mathvariant="normal">&#x2202;<!-- ∂ --></mi> <mrow class="MJX-TeXAtom-ORD"> <mover> <mi>q</mi> <mo>&#x02D9;<!-- ˙ --></mo> </mover> </mrow> </mrow> </mfrac> <mo>+</mo> <mfrac> <msup> <mi>d</mi> <mn>2</mn> </msup> <mrow> <mi>d</mi> <msup> <mi>t</mi> <mn>2</mn> </msup> </mrow> </mfrac> <mfrac> <mrow> <mi mathvariant="normal">&#x2202;<!-- ∂ --></mi> <mi>L</mi> </mrow> <mrow> <mi mathvariant="normal">&#x2202;<!-- ∂ --></mi> <mrow class="MJX-TeXAtom-ORD"> <mover> <mi>q</mi> <mo>&#x00A8;<!-- ¨ --></mo> </mover> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>0.</mn> </math> and get a closed form for q¨(q,q˙,P2). Note that usually we also require a similar statement for q˙(q,p), and failure in this respect is a sign of having a constrained system, possibly with gauge degrees of freedom. In any case, the non-degeneracy leads to the Euler-Lagrange equations in the usual manner: <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"> <mtable columnalign="right left right left right left right left right left right left" rowspacing="3pt" columnspacing="0.278em 2em 0.278em 2em 0.278em 2em 0.278em 2em 0.278em 2em 0.278em"> <mtr> <mtd> <msub> <mi>P</mi> <mn>1</mn> </msub> </mtd> <mtd> <mo>=</mo> <mfrac> <mrow> <mi mathvariant="normal">&#x2202;<!-- ∂ --></mi> <mi>L</mi> </mrow> <mrow> <mi mathvariant="normal">&#x2202;<!-- ∂ --></mi> <mrow class="MJX-TeXAtom-ORD"> <mover> <mi>q</mi> <mo>&#x02D9;<!-- ˙ --></mo> </mover> </mrow> </mrow> </mfrac> <mo>&#x2212;<!-- − --></mo> <mfrac> <mi>d</mi> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mfrac> <mfrac> <mrow> <mi mathvariant="normal">&#x2202;<!-- ∂ --></mi> <mi>L</mi> </mrow> <mrow> <mi mathvariant="normal">&#x2202;<!-- ∂ --></mi> <mrow class="MJX-TeXAtom-ORD"> <mover> <mi>q</mi> <mo>&#x00A8;<!-- ¨ --></mo> </mover> </mrow> </mrow> </mfrac> <mo>,</mo> <mtext>&#xA0;</mtext> <msub> <mi>P</mi> <mn>2</mn> </msub> </mtd> <mtd> <mo>=</mo> <mfrac> <mrow> <mi mathvariant="normal">&#x2202;<!-- ∂ --></mi> <mi>L</mi> </mrow> <mrow> <mi mathvariant="normal">&#x2202;<!-- ∂ --></mi> <mrow class="MJX-TeXAtom-ORD"> <mover> <mi>q</mi> <mo>&#x00A8;<!-- ¨ --></mo> </mover> </mrow> </mrow> </mfrac> <mo>.</mo> </mtd> </mtr> </mtable> </math> This is then fourth order in t, and so require four initial conditions, such as q, q˙, q¨, q(3). This is twice as many as usual, and so we can get a new pair of conjugate variables when we move into a Hamiltonian formalism. We follow the steps of Ostrogradski, and choose our canonical variables as Q1=q, Q2=q˙, which leads to \begin{align} P1 &= \frac{\partial L}{\partial \dot q} - \frac{d}{dt}\frac{\partial L}{\partial \ddot q}, \ P2 &= \frac{\partial L}{\partial \ddot q}. \end{align} Note that the non-degeneracy allows q¨ to be expressed in terms of Q1, Q2 and P2 through the second equation, and the first one is only necessary to define q(3).

We can then proceed in the usual fashion, and find the Hamiltonian through a Legendre transform: \begin{align} H &= \sumi Pi \dot{Q}i - L \ &= P1 Q2 + P2 \ddot{q}\left(Q1, Q2, P2\right) - L\left(Q1, Q_2,\ddot{q}\right). \end{align} Again, as usual, we can take time derivative of the Hamiltonian to find that it is time independent if the Lagrangian does not depend on time explicitly, and thus can be identified as the energy of the system.

However, we now have a problem: H has only a linear dependence on P1, and so can be arbitrarily negative. In an interacting system this means that we can excite positive energy modes by transferring energy from the negative energy modes, and in doing so we would increase the entropy — there would simply be more particles, and so a need to put them somewhere. Thus such a system could never reach equilibrium, exploding instantly in an orgy of particle creation. This problem is in fact completely general, and applies to even higher derivatives in a similar fashion.

Harsa Mitra - 4 years, 4 months ago

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@Harsa Mitra This is correct. There are ill-behaved solutions to our physics if we had too many derivatives. Can you boil this down to a simple point particle or wave example, find extra non-energy conserving solutions (for example a wave that is exponentially growing or decaying) without invoking a Hamiltonian analysis? Many people here, who don't know Hamiltonian's or Lagragians, would be appreciative.

David Mattingly Staff - 4 years, 4 months ago

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@Harsa Mitra First paragraph is very difficult to understand. Can you format those equations using latex perhaps?

Snehal Shekatkar - 4 years, 4 months ago

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@Snehal Shekatkar MathXL language wasn't working properly,apologies.

Harsa Mitra - 4 years, 4 months ago

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force without acceleration and acceleration without force has a simple error ...that is in our observation ...we are not using a correct frame of reference ..i.e. inertial frame of reference so i can say in that way ...that here physics is not correct (using only newtonian physics)

Chitres Guria - 4 years, 4 months ago

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I think it shall be physically forbidden because everything has inertia (I won't go deep into it's cause just like Higgs Bosons and stuff like that). But still if a particle is accelerating without a force, this would mean that its kinetic energy shall be increasing, but wait a minute where is the source of kinetic energy for this particle. So, it looks like the energy is coming out from nowhere, that must be theoretically inconsistent (it breaks the law of conservation of energy).

Siddharth Kumar - 4 years, 4 months ago

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@Siddharth Kumar Very good! The issue of having a well defined lowest energy and conservation of energy is definitely part of this. Particles just speeding up out of nowhere would violate those fundamental principles.

David Mattingly Staff - 4 years, 4 months ago

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@David Mattingly Honestly speaking I don't think so. Its always possible to have universe where such laws exist which allow particles to speed up. There cannot be a purely theoretical base for this question. It has to be experimental. In our universe it is the second derivative which exists. Other things I have already explained in my post above. By the way, even the concept of energy is something that we have created for our convenience and you just can't say that there is no source of energy or something like that to "prove" Newton's laws as Siddharth says above.

Snehal Shekatkar - 4 years, 4 months ago

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@Snehal Shekatkar I agree that it's mathematically possible. And I agree that conservation of energy and the existence of a lowest energy state is an experimental fact of our universe, which we promote to a "physical principle" of our universe. That's almost precisely the point of the initial question above, though. There are physical principles that we hold to be true in our universe, and those principles give significant restrictions on the allowed physical laws we have. This is incredibly useful when you are trying to "think like a theorist" and design new physical theories.

In other words, from an experimentalist's standpoint it is enough to say, we have F=ma because I observe a cart to accelerate uniformly when I apply a constant force. End of story. A more theoretical perspective is, we notice that if we have F=ma we have this set of physical principles. i.e. Newton's laws are sufficient to show them. We also notice that if we want these physical principles to hold, then all the other possibilities for Newton's laws don't work for one reason or another, i.e. we run the argument in reverse and discover that Newton's laws are not only sufficient, but necessary. This is a much stronger statement, and the difference is incredibly important. When one designs new theories, one keeps a set of physical principles and discards others, and those principles constrain the mathematical form of the new theory. It is exactly the connection between broad physical principles (rather than a specific cart experiment) and the mathematical implementations that we use that I want to get across to our users, as it's a different way of looking at physics than many are used to. Make sense?

David Mattingly Staff - 4 years, 4 months ago

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@David Mattingly I do agree that while constructing a new theory we need to preserve few principles and all this philosophy you already explained nicely in this post. My only point is that question about particular form for the physical law cannot be explained purely using theoretical arguments.

Snehal Shekatkar - 4 years, 4 months ago

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Conservation of energy and momentum would be broken, and those follow from Noethers first theorem by asserting that physics is symmetric in time and space. This symmetry is not obviously true, though. I believe a universe where this is not true is concieveable. As a side note, does Noethers theorem work in converse? That is, if I find a conserved size, is there a corresponding symmetry?

Arthur Mårtensson - 4 years, 4 months ago

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@Arthur Mårtensson I believe the converse works. Let's see...let there be a conserved quantity Q for all configurations (modulo perhaps some mathematical requirements that I don't know cause I haven't thought deeply about this). I can use Q to generate motion in the phase space. That motion should be a symmetry of the action, i.e. you get the corresponding symmetry from simple knowledge of Q.

David Mattingly Staff - 4 years, 4 months ago

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I think it can well be proved that the equation holds on the grounds of dimensional consistency.

Nishant Sharma - 4 years, 4 months ago

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@Nishant Sharma if initially the newton's law would have been established in some other form(involving higher derivatives) then force might have some other dimensions .

Brilliant Member - 4 years, 4 months ago

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we clearly know or proved that F=ma now i think that you ask why is "a" only second derivative why not 3rd or fourth or something else ? isn't it ?. so i think that from newton's second law we clearly can make out F=ma Now what is "a" it's nothing but change in velocity per unit time that is dv/dt and then what's velocity it's dx/dt so hence from these two we can clearly see that a = d^2x/dt^2. Hence only second derivative is there for acceleration. (My calculus is a bit weak so i hope what i wrote made sense...)

Now one other way is that thinking F=m * d^3/dt^3 now here even though it's relating mass with velocity but it's actually giving a relation of mass with jerk !( it's relating force with change in pressure i think) so if acceleration is an derivative of 1,3,4 or any other order there can be some instances when there is force being applied but our acceleration derivative is constant showing that a object is accelerating even though the force is zero or it can relate something to something else toppling the laws of nature. we can then also see a clear violation of Newton's First law and it's clearly physically forbidden but mathematically i'm not sure (at my level i'm not into deep mathematics) Hope i'm not too basic !

Ritvik Choudhary - 4 years, 4 months ago

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@Ritvik Choudhary Dear Ritvik, you did not understand the question. Everyone knows that acceleration is second derivative of displacement wrt time and why that is so. Question is why should the law contain acceleration only and not velocity or rate of change of acceleration etc. :)

Snehal Shekatkar - 4 years, 4 months ago

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Dear Harsa, I think first sentence in your post is wrong. F is never independent of a. In statics, net force on an object is always zero so there is no acceleration.

Snehal Shekatkar - 4 years, 4 months ago

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by one way we can look at this problem is that force is directly proportional to the change in velocity with time and then if there is a change in velocity at a larger time interval thn we have to look at the taylor series expansion of velocity and thn the change in velocity will be the sum of n time derivatives of the velocity!

Sabyasachi Tewari - 4 years, 3 months ago

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the basic aim to find about the behaviour of the system is to find the how the system moves in the space wrt to time,or what all will be the coordinates of the system in the due course of the time.So if we have the position of an object and velocity of that object at particular instants of time,we are able to calculate the next position and velocity.So the minimum derievative that one should posess should be the second derievative of an equation.So to describe a physical system only a second derievetive is neccessary

Anupam Ah - 4 years, 4 months ago

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According to the Newton mechanics, any non-zero force will cause acceleration. Acceleration can be measured only as second time derivative.

Roman Podoynitsyn - 4 years, 4 months ago

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Correct, but the question here is whether Newton's laws could be different, or whether there's a mathematical or physical obstruction?

David Mattingly Staff - 4 years, 4 months ago

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There are two types ofequations. One that are formed by observing a quantity's variation with respect to other and striking a relationship between them. For eg. if v=1,4,9 for x=1,2,3 it is evident that v=x^2. But there is another type of equation for which we don't have datas as in the previous case. We define the quantity by the equation. F=ma is an example for this kind. Force is defined to be this way. I've no idea why but this is how they have defined.

Abhishiekh Ramesh - 4 years, 4 months ago

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I think the idea of using second derivative comes from "gravity", when physicist started studying about motions of bodies they must have found out that the "rate of change of velocity with respect to time of an object under free fall" is constant and so they must have started using second derivative of displacement w.r.t time as an important physical quantity.

Brilliant Member - 4 years, 4 months ago

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i think like a theorist.

Jess J - 4 years, 4 months ago

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Great! You sometimes have to be careful acting like one though, they can tend to twitch :)

David Mattingly Staff - 4 years, 4 months ago

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Comment deleted Jun 24, 2013

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We are discussing something else here. You can go to discussion home and start new discussion with your question! :)

Snehal Shekatkar - 4 years, 4 months ago

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