1
$\begingroup$

I’m referring to an instant in time where the velocities of two different particles are identical. One particle experiences a force and the other does not. I think their states of motion are different even if the velocities are identical. Am I correct?

In Figure 1 below, Particle 1 has a velocity to the right and is experiencing a constant force. Particle 2 travels unaccelerated at 10 mps.

In Figure 2 both particles reach points A and B at the same instant in time and the velocities are identical at 10 mps. Particle 1 is experiencing a velocity changing force while Particle 2 experiences no force.

If we try to match the 10 mps velocities, going forward in time, by removing the force on Particle 1 it would have to be at a point past point A which is also a later instant in time. This is because, at the instant in time and location in space of Particle 1 at point A, there is a force on Particle 1. There cannot be both a force and no force on Particle 1 at point A and that instant in time. Thus, if Particle 1 reaches point A and is experiencing the force, the velocities of both particles will never be the same going forward.

enter image description here

When you throw a ball straight up it slows until reaching its apex of travel where for an instant in time there is no velocity. We know it is transitioning from a positive velocity to a negative velocity (relative to earth) but our perception is that there is a pause in the ball’s motion.

If at that instant the ball - with no velocity relative to earth - was experiencing no gravity then it would not move going forward in time. I think that since there is gravitational acceleration at that instant, the ball will move toward earth no matter what at any instant in time later. Math says we can remove the force at that instant and the ball stays put but logic says otherwise.

If we are to believe that the math describes the physics then how could we have, in the two examples, a stationary ball with an applied force and a stationary ball without a force and both having the same results of no velocity at any instant later? This tells me that acceleration is a quantity of motion just as velocity, at any instant in time. That means Particles 1 and 2, though having identical velocities, are not in the same state of motion at points A and B.

$\endgroup$
  • $\begingroup$ Is your question equivalent to "Is there a way of distinguishing an accelerating body from a non-accelerating body with the same velocity at time $t_1$ by means of observations/experiments/ measurements made (only) at $t_1$ ?"? $\endgroup$ – Philip Wood Oct 10 '20 at 15:48
  • $\begingroup$ I'm finding your last paragraph very difficult to follow. $\endgroup$ – Bob D Oct 10 '20 at 15:48
  • $\begingroup$ Starting with its first sentence? $\endgroup$ – Philip Wood Oct 10 '20 at 15:49
  • $\begingroup$ @PhilipWood Yes, starting with the first sentence. Can you make it out? $\endgroup$ – Bob D Oct 10 '20 at 16:00
  • $\begingroup$ No, it seems to be contradicting the first sentence of the previous paragraph. $\endgroup$ – Philip Wood Oct 10 '20 at 16:09
1
$\begingroup$

The key to showing why they behave the same is calculus.

You are correct that a particle cannot simultaneously have a force on it and no forces on it, just due to sheer logic. However, what we can say is that the particle can have a force on it at some time point $t$ (such as when the two objects are at the same velocity), and no force on it at some time $t + dt$, where $dt$ is some infinitesimally small duration of time.

Infinitesimals are a pest. Its really easy to try to make sense of them one way and then find that you generated a contradiction (such as having a force and no force at the same time). They gave us trouble for thousands of years. Newton and Leibnitz's calculus is the tool we use to resolve them. It was the first method of calculating using infinitesimals which didn't generate horrible contradictions.

Using calculus, we can see that the two ball's states are equal after the time t except that one had a force applied to it for the infinitesimal amount of time between $t$ and $t+dt$. When you run math using calculus, you find that the effect of applying a finite force for an infinitesimal amount of time is 0. So thus, if ball $A$ is traveling at $v_a$, ball $B$ is traveling at $v_a + 0$, which of course equals $v_a$. That little infinitesimal difference doesn't matter once you integrate it.

Indeed, the astonishing beauty of calculus is that it can handle these individual, isolated infinitesimal differences and say "they have 0 effect," and yet not write yourself into a corner where you're forced to say "because I can split up any motion into infinitesimally small pieces, and infinitesimally small pieces have no effect, there is no motion." This would be Zeno's most famous paradox. The precise wording of calculus lets us resolve this paradox.

Now there's two caveats to this which are probably useful going forward.

  • Practically speaking, you can't stop applying a force instantaneously. No practical device is capable of going from a finite force to zero force instantaneously. So when your intuition says there's something different between $A$ and $B$, you're right. This perfect equality only exists in the mathematically exact corner case that you are investigating.
  • When you get further into physics, you'll learn about Lagrangian formulations of motion. In this, there is one formula for a system, called the Lagrangian, which never changes for the system, and all particles are defined using just their position and velocity. The fact that we were able to define the physics of all systems using only position and velocity, bundling up any forces into that one unchanging equation, is probably one of the more fascinating things I've come across in physics. We can define Lagrangians with higher order terms, like accelerations, but it turns out that, in our universe, we don't actually need them.
$\endgroup$
  • $\begingroup$ Thanks @Cort Ammon. To determine the velocity of either particle we need to measure the distance traveled over time. If we do this again we find Particle 1 velocity is changing. As you noted; by position and velocity we define the physics of all systems. The same goes for the change (acceleration) of Particle 1. I think that acceleration is a quantity of the particle’s “state of motion” just as velocity is but I just don't see it considered that way. Note that I added another example to the body of the question. $\endgroup$ – Nectac Oct 11 '20 at 16:28
  • $\begingroup$ you find that the effect of applying a finite force for an infinitesimal amount of time is 0 Shouldn't the effect be (m)dv which isn't equal to zero? $\endgroup$ – Deschele Schilder Oct 16 '20 at 22:35
  • $\begingroup$ @DescheleSchilder My phrasing was informal to avoid getting mired in technicalities. I was going after the idea that $\int f(x) dx = \int f(x) + g(x) dx$ where g(x) is finite for x=0 and g(x)=0 everywhere else. Of course, that involved evaluating an integral. Your answer is more precise, not requiring an integral, but doesn't do anything to get rid of the confusing infinitesimals. The answer I was seeking pointed out that the time histories of the position and velocity will be identical (since we do integrate the accelerations), so we don't need to worry... $\endgroup$ – Cort Ammon Oct 16 '20 at 22:48
  • $\begingroup$ .. whether one operation occurs instantnaiously before or after the other. $\endgroup$ – Cort Ammon Oct 16 '20 at 22:49
0
$\begingroup$

I’m referring to an instant in time where the velocities of two different particles are identical. One particle experiences a force and the other does not. I think their states of motion are different even if the velocities are identical. Am I correct?

The state of motion of an object is defined by its velocity. See the following:

https://www.physicsclassroom.com/class/newtlaws/Lesson-1/State-of-Motion#:~:text=The%20state%20of%20motion%20of,resist%20changes%20in%20its%20velocity.&text=Such%20an%20object%20will%20not,upon%20by%20an%20unbalanced%20force.

Based on that definition, if the velocities of the two particles at an instant in time are the same, their state of motion is the same. A change in velocity is therefore a change in state of motion. If you are only told the instantaneous velocity of two particles is the same you have no way of knowing what, if any, forces are acting on the particles.

I have no problem with your statements for Figures 1 and 2. But I'm having trouble following your last paragraph. For example, the first sentence of the last paragraph seems to contradict the first sentence of the prior paragraph.

I believe you need to further clarify the last paragraph to enable a complete answer to your post.

In any case, to the extent possible, hope this helps.

$\endgroup$
  • $\begingroup$ Thanks @Bob D Note the edit that might fix the problem. $\endgroup$ – Nectac Oct 11 '20 at 16:39
0
$\begingroup$

in the moment, both particles have the same velocity it is of cause the same, for example if both were cars of the same mass they would cause the same damage in a collision. What is different is not the velocity, but the acceleration. So if you compare at the same moment and in in this moment the acceleration stops both continue with 10mps.But for many people the idea of a velocity in a moment is hard to grasp, if the velocity is changing continuously, since you could not measure it, to me measure a velocity, you have to have an interval of time.

$\endgroup$
  • $\begingroup$ Thanks @trula. Good point on the interval of time. $\endgroup$ – Nectac Oct 11 '20 at 16:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.