Newton's Second Law: Acceleration Proportional to Force

[mzntr]

Newton second law: the net force applied to a body produces a proportional acceleration.

Lets fly a rocket. Rocket has some kind of propulsion system that is producing force, so the rocket should accelerate to the infinity. However when the speed is closing to the speed of light, the acceleration slows down, despite the force still beeing applied. Am i right ? What is happening to the Newton formula?
Lets assume that we have reached 99,9999% of c. The force is still beeing aplied, but rocket will not go any faster? When we turn off the propulsion, the rocket will not slow down. So in that moment applying force will not produce any effect ? Am i right ?

 

[Bill_K]

Since you mention a propulsion system, you must mean that the force (and hence the acceleration) is constant in the rest frame of the rocket. That does not hold true in the rest frame of an observer on the ground. In the observer's rest frame the acceleration decreases.

 

[PatrickPowers]

Newton second law: the net force applied to a body produces a proportional acceleration.

Lets fly a rocket. Rocket has some kind of propulsion system that is producing force, so the rocket should accelerate to the infinity. However when the speed is closing to the speed of light, the acceleration slows down, despite the force still beeing applied. Am i right ? What is happening to the Newton formula?
Lets assume that we have reached 99,9999% of c. The force is still beeing aplied, but rocket will not go any faster? When we turn off the propulsion, the rocket will not slow down. So in that moment applying force will not produce any effect ? Am i right ?

Yes and no. It depends on where the measuring is being done.

As far as the rocket is concerned, the more energy you put in the faster it goes, and the faster it goes the sooner you get there. Twice as much energy, half the time, ten times as much energy, one tenth the time. There is no limit to this. So if you have enough energy you can go anywhere in as little time as you like. You can go a light year in two days.

The weirdness comes in because the speed of light is the same in every frame of reference. From the point of view of people at the launch site, you can't go one light year in one day. As far as they are concerned it took you a year and a day. When you turn around and come back it is the same deal. So while the rocket man thinks the whole thing took four days, the Earth people say that it is two years and two days. They are both right.

So how does this look to the rocket man? Assuming unrealistically that he gets this acceleration instantly, then as soon as he is going at this tremendous speed the star IS two light days away. The star is closer by any means of measurement that you use. So as far as the rocket man is concerned he has warped the entire universe, and there is no way to prove him wrong. This concept works for him. On his return the Universe warps in the other direction.

All this violates common sense, but that's just too bad. You have to get used to it.

 

[ghwellsjr]

So as far as the rocket man is concerned he has warped the entire universe, and there is no way to prove him wrong. This concept works for him. On his return the Universe warps in the other direction.

Really? There is a difference in the warping of the universe between the outbound trip and the inbound trip? Please explain.

 

[DaveC426913]

On his return the Universe warps in the other direction.

All this violates common sense, but that's just too bad. You have to get used to it.

Really? There is a difference in the warping of the universe between the outbound trip and the inbound trip? Please explain.

Patrick meant to say that the universe compresses its length along the axis of the spaceship's motion. This is the same regardless of which direction he's going.

While related, not sure it helps answer the OP's question any more than it confuses it.

 

[PatrickPowers]

Really? There is a difference in the warping of the universe between the outbound trip and the inbound trip? Please explain.

The faster you go, the sooner you get there. So if you are going very fast then some star many light years away is one light day away to you. By going so fast you have made it closer. As far as you are concerned the Universe is warped. Other observers do not see it that way. They explain this by saying that your clock is slower than theirs, and with your clock slower there is an illusion that the star is closer.

 

[DaveC426913]

Newton second law: the net force applied to a body produces a proportional acceleration.

Lets fly a rocket. Rocket has some kind of propulsion system that is producing force, so the rocket should accelerate to the infinity. However when the speed is closing to the speed of light, the acceleration slows down, despite the force still beeing applied. Am i right ? What is happening to the Newton formula?
Lets assume that we have reached 99,9999% of c. The force is still beeing aplied, but rocket will not go any faster? When we turn off the propulsion, the rocket will not slow down. So in that moment applying force will not produce any effect ? Am i right ?

You are not taking into account the time dilation.
And you must describe events as observed from chosen frame of reference.

As the observers on Earth observe the ship approach c, they also observe it aging slower. Thus it burns fuel slower. Thus its acceleration plateaus.

Close enough to c, and time aboard the spaceship appears (from Earth's PoV) to have virtually stopped, therefore so has its acceleration.

 

[ghwellsjr]

Really? There is a difference in the warping of the universe between the outbound trip and the inbound trip? Please explain.

The faster you go, the sooner you get there. So if you are going very fast then some star many light years away is one light day away to you. By going so fast you have made it closer. As far as you are concerned the Universe is warped. Other observers do not see it that way. They explain this by saying that your clock is slower than theirs, and with your clock slower there is an illusion that the star is closer.

You are explaining the difference in the warping of the universe between the rocket man not traveling and traveling. I'm asking you what you meant by "the Universe warps in the other direction":

So as far as the rocket man is concerned he has warped the entire universe, and there is no way to prove him wrong. This concept works for him. On his return the Universe warps in the other direction.

 

[bobc2]

mzntr, I assume that you are not taking the "universe warping" as a physical reality. The universe doesn't really warp. I think the posters here mean it only in a mathematical sense.

 

[D H]

The above answers, while correct, are a bit too involved for someone starting from this point of view:

Newton second law: the net force applied to a body produces a proportional acceleration.

Newton's second law is not a universal law. It is but an approximation of reality. The approximation that is very close to correct when velocities are small compared to the speed of light. It becomes less correct as relative velocities increase, and is grossly incorrect when relative velocities approach the speed of light.

The problem is that velocity vectors don't add vectorially when relative velocities approach the speed of light. One of the key tenets of relativity is that the velocity of light is the same to all observers. Suppose you are traveling at 99.9999% of c relative to me and both you and I measure the velocity of the light coming from the same source. I will measure that velocity to be c. So will you. This makes absolutely no sense in a Newtonian universe. That's okay though; the universe is not Newtonian.

Lets fly a rocket. Rocket has some kind of propulsion system that is producing force, so the rocket should accelerate to the infinity.

That is exactly what Newtonian mechanics tells us. The problem is that what Newtonian mechanics tells us is wrong.

However when the speed is closing to the speed of light, the acceleration slows down, despite the force still beeing applied. Am i right ? What is happening to the Newton formula?

The short answer is that it has become invalid.

Lets assume that we have reached 99,9999% of c.

Relative to what? You are implicitly assuming a master reference frame here. There is no master reference frame in relativity.

So, let's assume we have reached 0.999999 c relative to some observer. That the rocket continues to fire is not going to change the speed of the rocket by much at all relative to this observer. However, imagine the rocket sends out a non-thrusting satellite to be used as a relay. That satellite will keep moving at the velocity it had when it was ejected from the rocket. That initial observer will see that satellite as moving at 0.999999 c. If the rocket continues firing, the satellite will initially see the rocket accelerate at a=F/m; this will hold until the rocket's speed relative to the satellite becomes a significant fraction of c.

 

[mzntr]

Ha, the only clear thing that comes out of this discussion is that Newton law is wrong, yet it works.

The deeper you get however into the discussion the bigger problems you find.

Something is wrong with this universe.. Is it really so illogical ? :)

Thakns

 

[D H]

Ha, the only clear thing that comes out of this discussion is that Newton law is wrong, yet it works.

Newton's second law works at very low velocities (compared to the speed of light) because (a) momentum is a conserved quantity and (b) the simple vector addition rule you learned in high school or freshman physics is very close to correct at such low velocities.

Conservation of momentum is still a valid principle in relativistic physics, but that simple velocity addition rule is not. Note that relativistic velocity addition is valid even at low speeds.

 

[DrStupid]

Newton second law: the net force applied to a body produces a proportional acceleration.

That's a special case for constant mass. The original second law says: the force applied to a body produces a proportional change of momentum.

However when the speed is closing to the speed of light, the acceleration slows down, despite the force still beeing applied. Am i right ? What is happening to the Newton formula?

It still applies but you have to consider the velocity dependence of mass (as used in Newtons definition of momentum):

[itex]m = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}[/itex]

Then you get

[itex]F = m \cdot a + m \cdot \frac{{v \cdot \left( {v \cdot a} \right)}}{{c^2 - v^2 }}[/itex]

and

[itex]a = \frac{F}{m} - \frac{{v \cdot \left( {v \cdot F} \right)}}{{m \cdot c^2 }}[/itex]

Lets assume that we have reached 99,9999% of c. The force is still beeing aplied, but rocket will not go any faster?

It will go faster but the acceleration is close to zero.

 

[D H]

That's a special case for constant mass. The original second law says: the force applied to a body produces a proportional change of momentum.

Newton only worked with constant mass systems, and for such systems, [itex]F \propto dp/dt[/itex] (F=dp/dt in canonical form) and [itex]F \propto ma[/itex] (F=ma) are identical. They are not identical for systems with time-varying mass. People who work with such systems almost invariably use F=ma.

 

[DrGreg]

mzntr, the arguments that others have given in this thread illustrate that, like other concepts in relativity, acceleration is relative. Different observers may disagree what the acceleration of an object is. In fact[tex]
a_{proper} = \frac{a_{coord}}{\left( 1 - v^2/c^2\right)^{3/2}} \, \mbox{,}
[/tex]where a_proper is acceleration measured by an inertial observer who is momentarily at rest relative to the object being measured, and a_coord is acceleration measured by an inertial observer who is moving at speed v relative to the object being measured (and parallel to the acceleration).

If a_proper remains constant, then as v approaches c, a_coord must approach zero.

Now you can see the problem with Newton's Law: there's no unique definition of what acceleration is. However we can say that F = ma is still true in an inertial frame where the object has zero velocity.

It turns out that, in fact, there is a 4-dimensional vector version of Newton's second law[tex]
F_{\mu} = \frac{DP_{\mu}}{D\tau}
[/tex]but you'll need to understand tensors to make sense of that.

 

[DrStupid]

Newton only worked with constant mass systems

It would be difficult to verify this statement because Newton did not published everything he dealt with. Do you have at least a source verifying that Newton limited his laws of motion to constant mass?

 

[DrStupid]

Now you can see the problem with Newton's Law: there's no unique definition of what acceleration is.

Accerelation is defined by [itex]a: = \frac{{dv}}{{dt}}[/itex].

 

[D H]

It would be difficult to verify this statement because Newton did not published everything he dealt with. Do you have at least a source verifying that Newton limited his laws of motion to constant mass?

Are you truly asking me to prove a negative?

As far as F=ma versus F=dp/dt goes, the reason people who work with dynamic systems prefer F=ma is simple: This choice makes force an invariant quantity. The same force acts on a rocket regardless of the inertial frame of reference used to describe the motion. Choose to use F=dp/dt and force becomes a frame-dependent quantity thanks to the v*dm/dt term.
Getting back to the topic at hand:

Accerelation is defined by [itex]a: = \frac{{dv}}{{dt}}[/itex].

That definition is not invariant in special relativity. It implicitly assumes space is cartesian and that both space and time are absolute (which they are in Newtonian mechanics, but not in relativity).

 

[DrStupid]

Are you truly asking me to prove a negative?

No, I don't even though you made a negative statement and I actually explained why. But I asked you for a source proving the limitation of Newtonian dynamics to constant mass.

As far as F=ma versus F=dp/dt goes, the reason people who work with dynamic systems prefer F=ma is simple: This choice makes force an invariant quantity.

That very well may be. But there is a more fundamental reason for the introduction of F=m·a into classical mechanics. Galilei transformation requires a frame independent inertial mass. Therefore F=m·a applies for all closed systems in classical mechanics. But as we are currently talking about special relativity we have to use Lorentz transformation and with Newton's definition of momentum and Newton's laws of motion it leads to the known relativistic velocity dependence of inertial mass. Of course the resulting force is not invariant but that's not up to the use of a dynamic mass. That's caused by relativity itself.

That definition is not invariant in special relativity.

Who cares?

It implicitly assumes space is cartesian and that both space and time are absolute (which they are in Newtonian mechanics, but not in relativity).

The definition does not depend on a special metric. It works in relativity as well as in Newtonian mechanics. Of course acceleration is frame dependent in special relativity but so is velocity. I do not see why this should be a problem.

 

[chrisbaird]

Ha, the only clear thing that comes out of this discussion is that Newton law is wrong, yet it works.

Actually, Newton's laws are wrong and they don't work. Our clumsy human senses just aren't very good at noticing the error at low speeds. Try to send a rocket to the moon using Newton's laws alone and you will miss it.

 

[D H]

Actually, Newton's laws are wrong and they don't work. Our clumsy human senses just aren't very good at noticing the error at low speeds. Try to send a rocket to the moon using Newton's laws alone and you will miss it.

Uh, no. Maybe Mercury. Not the Moon, not even Mars. There is no reason to go beyond Newtonian mechanics to target the Moon. NASA certainly doesn't.