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Why do we need Newton's First law? And how the First law works?

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Adjoint
#19
Jun27-14, 01:43 PM
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Quote Quote by olivermsun View Post
How would you determine that F = 0? Normally you would measure m and a and infer F, right? So if you're in a rotating frame and you measure a nonzero a for an object which is actually in inertial motion, then what? Is there an F acting on that body?
For me (in a non inertial frame) it would seem so... Well now I am asking myself that how can I identify a frame as non inertial if I am in that frame? Certainly I will feel so; because of my inertia. But really, if I am in a non inertial frame, how can I tell that if an object is moving because a force is acting or because it is in a non-inertial frame?

EDIT: Provide the answer please.
sophiecentaur
#20
Jun27-14, 03:17 PM
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Quote Quote by olivermsun View Post
On a massive wheel, like... the rotating Earth?
No. I meant like Arthur C Clarke's in 2001. i.e. where the rotational forces play a big part in their everyday lives. It took a long time for anyone to measure any rotational effect on Earth.
DrStupid
#21
Jun27-14, 03:59 PM
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Quote Quote by Adjoint View Post
But really, if I am in a non inertial frame, how can I tell that if an object is moving because a force is acting or because it is in a non-inertial frame?
Use the third law.
Adjoint
#22
Jun27-14, 04:05 PM
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I am sorry to say, but it seems like I don't know enough about inertial and non-inertial frames and rotational motion to continue following this discussion properly (though I started this thread )
For example: I cant simply figure out How one person in a non-inertial frame can measure force on an object properly? (I mean avoiding fictitious forces.)

But thank you all for your replies. Meanwhile I should go and do some study!

EDIT: ... thanks DrS! I just had your reply.
To use the third law, I think you mean, I need to find if that accelerating object is giving any reaction force...?
DrStupid
#23
Jun27-14, 04:33 PM
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Quote Quote by Adjoint View Post
To use the third law, I think you mean, I need to find if that accelerating object is giving any reaction force...?
Yes, if there are forces without counter forces than the frame of reference is non-inertial (or you missed interactions).
Adjoint
#24
Jun27-14, 05:07 PM
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Suppose I put a ball on a table and the ball starts moving. It's certainly moving without a counter force. So I decide my frame is non inertial.
But what if there was actually a counter force. Suppose someone with an almost invisible string attached to the ball was moving it around?
So unless I can be sure about all the forces acting on the ball and by the ball, how can I tell if the frame was inertial or not?
And in a certain situation it might not be possible for me to make sure about all the forces acting on and by an object.

As you said too -
Quote Quote by DrStupid View Post
or you missed interactions
It's likely to miss one or two interactions. So in practical case how can we be sure if a frame is inertial or non-inertial?

Also it seems like Newton's first law is not enough to check for an Inertial frame. Because to exclude the fictitious forces We also need the Third law. What's happening?
Jano L.
#25
Jun27-14, 05:27 PM
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Quote Quote by Adjoint View Post
So why do we need to state First law as a separate law?
Apart from the issue of setting up context for the 2nd and 3rd law, the 1st law stands apart of the 2nd law also for another reason: in its original formulation, apart from being consistent with 2nd law, it says also something in addition that cannot be derived from the 2nd law.

How so?

Let the position of the particle at time ##t_0## be ##x_0##. Now imagine the force ##F## is a function of position ##x## only and ##F(x_0) = 0##. What will happen to the particle put at ##x_0##?

Based on the 1st law, we conclude the particle will stay at ##x_0## forever, because there is "no impressed force to compel the particle to change its state of rest".

Based on the 2nd law only, we cannot make such conclusion. This is because in its modern formulation, the 2nd law is this statement: mass times second derivative of position equals net force impressed:

$$
m\frac{d^2 x}{dt^2} = F.
$$

This equation connects force only to the second derivative of the position and disregards any other characteristics of motion. Due to this, in this case it is not sufficient to determine what will happen to the position of the particle: depending on the function ##F(x)##, there may be infinity of solutions distinguished by the time ##t_1## the particle begins to move.
DrStupid
#26
Jun27-14, 05:30 PM
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Quote Quote by Adjoint View Post
So in practical case how can we be sure if a frame is inertial or non-inertial?
There is no way to be sure in practice. You just can make reasonable assumptions. If the sum of all forces is not zero you have to decide weather you describe the situation with unknown interactions or with a non-inertial system (or both).

Quote Quote by Adjoint View Post
Also it seems like Newton's first law is not enough to check for an Inertial frame.
The first law is neither necessary nor sufficient for the identification of inertial systems. The second and the third law can do the job if you know all interactions.
DrStupid
#27
Jun27-14, 05:38 PM
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Quote Quote by Jano L. View Post
Let the position of the particle at time ##t_0## be ##x_0##. Now imagine the force ##F## is a function of position ##x## only and ##F(x_0) = 0##. What will happen to the particle put at ##x_0##?

Based on the 1st law, we conclude the particle will stay at ##x_0## forever, because there is "no impressed force to compel the particle to change its state of rest".
Only for ##v_0 = 0##.

Quote Quote by Jano L. View Post
Based on the 2nd law only, we cannot make such conclusion.
With the same starting conditions the second law leads to the same conclusion.
Jano L.
#28
Jun27-14, 05:57 PM
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Quote Quote by DrStupid View Post
Only for ##v_0 = 0##.
Yes, that is the idea.

With the same starting conditions the second law leads to the same conclusion.
Not always. If ##F(x_0) = 0## then the equation of motion may not have unique solution.
DrStupid
#29
Jun27-14, 06:13 PM
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Quote Quote by Jano L. View Post
If ##F(x_0) = 0## then the equation of motion may not have unique solution.
Can you give me an example that violates the first law?

To my knowledge the first law and the original wording of the second law differ in a single special case only: In contrast to the first law

[itex]F = \dot p = m \cdot \dot v + v \cdot \dot m[/itex]

allows accelerations without force for

[itex]\dot v = - v \cdot \frac{{\dot m}}{m} \ne 0[/itex]

I don't know if this was intended by Newton but this means that the use of forces for open systems might be problematic.
bcrowell
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Jun27-14, 07:25 PM
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Newton's first law, in its original formulation, is a special case of the second law. Newton didn't intend his laws to be logically independent, and it's not possible to read them as being independent. He probably just wrote the first law separately in order to emphasize that he was making a break with Aristotelianism.

Ernst Mach wrote a book, The Science Of Mechanics, 1919, http://archive.org/details/scienceofmechani005860mbp , which critiqued the logical basis of Newton's laws. Most people hear about Mach only through Mach's principle, which was not formulated precisely until long after Mach's death, so they get the impression that Mach was some kind of fuzzy-headed philosopher. But his critique of Newton's laws was very logically sound, and was influential. Probably influenced by Mach, some modern textbook authors started presenting Newton's laws in a rewritten form, with the first law rewritten as a statement that inertial frames exist. (It's not just a definition, it's an existence claim.) It's only in this rewritten form of Newton's laws that the first law is not a special case of the second.
Jano L.
#31
Jun28-14, 02:58 AM
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Quote Quote by DrStupid View Post
Can you give me an example that violates the first law?
When the particle begins to change its rectilinear motion at some instant ##t_1## while the acceleration at this instant vanishes.

For example, according to the 2nd law, particle put at rest on top of a sphere at instant ##t_0## may begin to change its velocity at any subsequent time ##t_1##. The 2nd law does not determine ##t_1##.

According to the 1st law, the particle will stay put.

To my knowledge the first law and the original wording of the second law differ in a single special case only: In contrast to the first law

[itex]F = \dot p = m \cdot \dot v + v \cdot \dot m[/itex]

allows accelerations without force for

[itex]\dot v = - v \cdot \frac{{\dot m}}{m} \ne 0[/itex]

I don't know if this was intended by Newton but this means that the use of forces for open systems might be problematic.
The second law is not
$$
F = \dot{m}v + m\dot v.
$$
It is
$$
F = m\dot{v}.
$$
In the last form, the 2nd law applies also to systems with variable mass.

The form ##F=\frac{dp}{dt}## is correct only if in addition, ##m## is assumed constant.
DrStupid
#32
Jun28-14, 06:32 AM
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Quote Quote by Jano L. View Post
For example, according to the 2nd law, particle put at rest on top of a sphere at instant ##t_0## may begin to change its velocity at any subsequent time ##t_1##.
Show me your calculation.

Quote Quote by Jano L. View Post
The second law is not
$$
F = \dot{m}v + m\dot v.
$$
According to the original wording of the second law

http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/46

and the definition of momentum

http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/26

it is.
Jano L.
#33
Jun28-14, 07:28 AM
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Quote Quote by DrStupid View Post
Show me your calculation.
I made a mistake above, the particle has to be put on a dome with special shape,not sphere. Check out John Norton's calculation:

http://www.pitt.edu/~jdnorton/Goodies/Dome/

Norton argues that actually the first law is valid even here, but he uses different "1st law" than the one Newton wrote. I only say that Newton's first law predicts something that Newton's 2nd law does not: that the particle will stay at the top. But it shows that Newton's first law is different kind of law: it is not a mathematical equation, but causality statement. Unfortunately, it is impossible to check this distinction experimentally.

According to the original wording of the second law

http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/46

and the definition of momentum

http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/26

it is.
Yes, if you are making logical conclusions based on these two statements only.

But Newton wrote a whole book on this. In the section on definitions you cited, he deals with "body", which most of the time means that mass is assumed constant.

I did not read whole of Newton's book, but I am sure he did not meant actually to include the term ##\dot{m}v## from the derivative of momentum as a part of the equation of motion. That would lead him to incorrect results and I'm sure he would check before publishing them.

His 2nd law is thus correct only if in addition mass of the body is assumed to be constant, something he forgot/did not care to say or he said it elsewhere.
DrStupid
#34
Jun28-14, 10:16 AM
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Quote Quote by Jano L. View Post
Check out John Norton's calculation:

http://www.pitt.edu/~jdnorton/Goodies/Dome/
It seems he missed the initial condition ##r_0=0##. I get the additional solution

[itex]r = \frac{{t{}^4}}{{144}}[/itex]

Edit: Now I got it. He mixed the time-dependent solution for the initial condition ##r(T) = 0## with the trivial solution ##r(t) = 0## and spliced them at ##t = T##. The resulting hybrid is still a solution of the differential equation and meets the initial condition ##r(0) = 0##.

There is either no force and no acceleration (for ##r=0##) or both force and acceleration (for ##r \ne 0##). Whether this is a violation of the first law or not seems to be a matter of interpretation.

Quote Quote by Jano L. View Post
Newton's first law is different kind of law: it is not a mathematical equation, but causality statement.
Yes, the first law says that force is the only cause of changes in motion. This is the qualitative definition of force. As bcrowell already mentioned this was important to distinguish Newton's force from Aristoteles' force. According to Aristoteles force was required to keep the state of motion. According to Newton force is required to change the state of motion. That's what the first law is about.

The second law does not care about causality but gives a quantitative definition of force.

Quote Quote by Jano L. View Post
But Newton wrote a whole book on this. In the section on definitions you cited, he deals with "body", which most of the time means that mass is assumed constant.
The second law is not limited to bodies. Even if the quantity of matter (we better do not use the term "mass" here in order to avoid possible confusions with rest mass) of a body is constant, a system consisting of a variable number of bodies may have a variable quantity of matter. As the second law is not limited to bodies it is not forbidden to apply it to such systems. But as already mentioned it might be problematic to use the second law this way. It only works if you do not mix forces for constant and variable quantity of matter.

Quote Quote by Jano L. View Post
I did not read whole of Newton's book, but I am sure he did not meant actually to include the term ##\dot{m}v## from the derivative of momentum as a part of the equation of motion. That would lead him to incorrect results and I'm sure he would check before publishing them.
To my knowledge Newton didn't refer to this topic. I guess he started from conservation of momentum and stopped with the conclusion that the sum of all alterations of momentum need to be zero (that's what the second and third law say). In that sense his laws of motion are universal and also work for open systems or even for different transformations. Thus is wasn't necessary to go into further details at this point. Everything else can be derived from his definitions for particular conditions.

Quote Quote by Jano L. View Post
His 2nd law is thus correct only if in addition mass of the body is assumed to be constant, something he forgot/did not care to say or he said it elsewhere.
That's the result for closed systems and Galilean transformation. That's the usual situation in classical mechanics (that's why it is so popular). But other condition may return other results.


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