The momentum of a bowling ball in space...

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Motor Daddy

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...that is the question.

Momentum is defined as:

http://en.wikipedia.org/wiki/Momentum.

In classical mechanics, linear momentum or translational momentum (pl. momenta; SI unit kg m/s, or, equivalently, N s) is the product of the mass and velocity of an object. For example, a heavy truck moving fast has a large momentum—it takes a large and prolonged force to get the truck up to this speed, and it takes a large and prolonged force to bring it to a stop afterwards. If the truck were lighter, or moving slower, then it would have less momentum.

Like velocity, linear momentum is a vector quantity, possessing a direction as well as a magnitude:

p=mv



p being the momentum of the object, m being the mass of the object, and v being the velocity of the object.


Hmmmmm......the momentum of the object...the velocity of the object...and the mass of the object.

I have a question about that velocity of the object. There is one object, and it has its own velocity. What is that velocity relative to?
 
...that is the question.

Momentum is defined as:

http://en.wikipedia.org/wiki/Momentum.





p being the momentum of the object, m being the mass of the object, and v being the velocity of the object.


Hmmmmm......the momentum of the object...the velocity of the object...and the mass of the object.

I have a question about that velocity of the object. There is one object, and it has its own velocity. What is that velocity relative to?

It's frame of reference - there is *always* a frame of reference even without other objects.

Besides, there cannot ever be just a single object alone in the universe so the question is totally meaningless.
 
Another object.

So are you saying that the momentum of the object is actually relative to another object because of the dependence on the other object's motion?

So you are saying that v in the momentum equation is for a relative velocity?
 
I have a question about that velocity of the object. There is one object, and it has its own velocity. What is that velocity relative to?
there is not one object, there are 2.
you are forgetting the propelling force, it has to come from somewhere.
so with that in mind the trucks velocity is reckoned from THAT source.
 
For any free particle in special relativity:
$$E \vec{v} = c^2 \vec{p} \\ E^2 - \left( c \vec{p} \right)^2 = \left( m c^2 \right) ^2$$
So $$\vec{v}, \, \vec{p} \, \textrm{and} \, E$$ depend on your choice of coordinate system, while $$c \, \textrm{and} \, m$$ do not.

If $$m > 0$$ then we may, motivated by the trigonometry-like second equation, define $$\rho = \cosh^{\tiny -1} \frac{E}{mc^2}$$ and soon come to the conclusion:
$$ E = m c^2 \, \cosh \rho \\ \vec{p} = m c \, \sinh \rho \; \hat{v} \\ \vec{v} = c \, \tanh \rho \; \hat{v}$$
Since $$ \cosh \, \tanh^{\tiny -1} x = \frac{1}{\sqrt{1 - x^2}} \; \textrm{and} \; \sinh \, \tanh^{\tiny -1} x = \frac{x}{\sqrt{1 - x^2}}$$ these trigonometric identities then allow us to fairly state everything in terms of one free parameter: $$\vec{v}$$
$$ E = \frac{m c^2}{\sqrt{1 - \left(\frac{\vec{v}}{c}\right)^2}} \\ \vec{p} = \frac{m \vec{v}}{\sqrt{1 - \left(\frac{\vec{v}}{c}\right)^2}}$$



If $$m > 0$$ then there exists an inertial coordinate system where $$\vec{v} = 0, \, \vec{p} = 0 , \, \textrm{and} E = mc^2$$ thus we are justified in talking about rest energy, $$E_0 \equiv m c^2$$, and the additional energy associated with movement relative to the chosen inertial coordinate system, or kinetic energy for short, $$\textrm{KE} \equiv E - E_0 = \sqrt{ \left(m c^2 \right) ^2 \, + \, \left( c \vec{p} \right)^2 } - mc^2 = mc^2 \left( \frac{1}{\sqrt{1 - \left(\frac{\vec{v}}{c}\right)^2}} - 1 \right)$$

Therefore, since the velocity of your bowling ball in space depends on a choice of inertial coordinate systems (because only in inertial coordinate systems does a free particle have a constant velocity, by Newton's law of inertia) but its Energy, Kinetic Energy and momentum also depend on this choice in the precise mathematical relationship explained above.

//Edit:
And it is because of precision experiments that we have confidence that both special relativity and the above relationships between velocity, momentum and energy hold to high precision. We see relativity supported over Newtonian predictions in precision experiments with high-speed phenomena as diverse as the speed of light in moving water (Fizeau, 1859) or the relation between energy and velocity for electrons (Bertozzi, 1964; Fan, unpublished). Fan, for example, was a relativity denier and a terrible experimentalist, but even with gross systemic effects his observations agreed in detail with the Relativity that the kinetic energy of a body increases with velocity at a rate higher than Newton predicted.

If you graph velocity squared divided by the speed of light squared on the horizontal axis and kinetic energy divided by the Newtonian prediction of kinetic energy on the vertical axis, then relativity predicts a upward curving line with slope of 3/8 near v=0. Newton predicts zero. Fan, in his paper, predicts -1/8. So the upward slope from Bertozzi (and Fan) got causes us to favor Relativity over the theories of Fan and Newton.

Theory:
attachment.php
Fan's Conjecture (green) versus Imprecise Reflections from Reality (Fan's experimental data, blue):
attachment.php


All other precision observation also supports special relativity in its domain of applicability (where gravity effects can be ignored).
 
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...that is the question.

Momentum is defined as:

http://en.wikipedia.org/wiki/Momentum.





p being the momentum of the object, m being the mass of the object, and v being the velocity of the object.


Hmmmmm......the momentum of the object...the velocity of the object...and the mass of the object.

I have a question about that velocity of the object. There is one object, and it has its own velocity. What is that velocity relative to?
Momentum isn't intrinsic to the bowling ball. It can be measured from various frames, and therefore changes depending on who is doing the measuring. For example, the bowling ball itself will claim that its momentum is zero.
 
For example, the bowling ball itself will claim that its momentum is zero.
No it won't, unless the ball performs the measurement of its own velocity and finds its velocity to be zero. Light travel time defines distance. There are numerous tests that the bowling ball can perform as an observer to measure its own velocity, and therefore know its own momentum as long as it knows its own mass.
 
there is not one object, there are 2.
you are forgetting the propelling force, it has to come from somewhere.
so with that in mind the trucks velocity is reckoned from THAT source.

Acceleration is not in question here, velocity is in question.

If a bowling ball is accelerated by a means of propulsion, and then the acceleration stops (shut off propulsion), the bowling ball has a velocity. The means at which the ball was propelled at a prior time is irrelevant.
 
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No it won't, unless the ball performs the measurement of its own velocity and finds its velocity to be zero. Light travel time defines distance. There are numerous tests that the bowling ball can perform as an observer to measure its own velocity, and therefore know its own momentum as long as it knows its own mass.

The bowling ball cannot measure its own velocity, without knowing the velocity of some object of reference.

If it had an accelerometer it could know how much its velocity had changed during any given time frame. But without knowing its initial velocity, it could have no diffinitive knowledge of its own velocity.

Everything it looked at would seem to be moving relative to it. There is no way, without some outside frame of reference, which is not available to mere mortals, let alone bowling balls, for the bowling ball to know its own velocity.

This all assumes we are talking about a sentient bowling ball, which can think, to begin with.
 
No it won't, unless the ball performs the measurement of its own velocity and finds its velocity to be zero. Light travel time defines distance. There are numerous tests that the bowling ball can perform as an observer to measure its own velocity, and therefore know its own momentum as long as it knows its own mass.
I thought you were asking the question in terms of Physics, not MotorDaddyWorld. In Physics an inertial body can always consider itself to be at rest (and therefore its momentum to be zero). Using any tricks involving light travel time would simply lead the bowling ball to conclude that the objects it is using as reference points are themselves moving.
 
I thought you were asking the question in terms of Physics, not MotorDaddyWorld. In Physics an inertial body can always consider itself to be at rest (and therefore its momentum to be zero). Using any tricks involving light travel time would simply lead the bowling ball to conclude that the objects it is using as reference points are themselves moving.

Light travel time defines distance, it is not a trick.
 
No it won't, unless the ball performs the measurement of its own velocity and finds its velocity to be zero.

It cannot measure its own velocity without a reference. Just as you cannot accurately measure your own speed while sitting in a seat in an airliner.

Light travel time defines distance. There are numerous tests that the bowling ball can perform as an observer to measure its own velocity

Incorrect. An observer always observes light traveling at the speed of light.
 
In Physics an inertial body can always consider itself to be at rest (and therefore its momentum to be zero).

In physics, how do you differentiate the velocity between the two objects traveling away from a central point at different constant rates, when they both are entitled to assume their own velocities to be zero? If two objects are traveling away from each other at different rates, the objects will agree on the closing speed. In Einstein's world they are also entitled to assume their own velocity to be zero and the other object has the entire velocity. We know that is not always true so how is it possible that one is correct to assume their own velocity to be zero? Why not assume their velocity to be any other velocity other than zero? There is only 1 chance in an infinite amount of possibilities as to guessing one's velocity correctly. The odds are the same to pick zero as they are to pick .23654c, .103c, 9 m/s, .01843m/s.
 
It cannot measure its own velocity without a reference. Just as you cannot accurately measure your own speed while sitting in a seat in an airliner.



Incorrect. An observer always observes light traveling at the speed of light.

Incorrect, observers define their standard of distance by light travel time. His concept of 2 meters could be entirely different from another observer's concept of 2 meters.
 
In Einstein's world they are also entitled to assume their own velocity to be zero and the other object has the entire velocity.

Hate to break it to you but the same is true in Galilei/Newton world. I am perfectly aware that you'll never get that , so carry on.
 
Incorrect, observers define their standard of distance by light travel time.

Yes, they do. And since light travel time is invariant, they all see the same speed; zero.

His concept of 2 meters could be entirely different from another observer's concept of 2 meters.

He could think that 2 meters really means 2 miles. Doesn't matter. He always sees light take the same time to travel any distance he chooses.
 
Hate to break it to you but the same is true in Galilei/Newton world. I am perfectly aware that you'll never get that , so carry on.

I have issues with both. Neither of those two individuals knew a method of determining one's own velocity in space.
 
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