- #26

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Ahh, this is what I was doing wrong.Please also remember that the accelerating object isnotan inertial observer

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- #26

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Ahh, this is what I was doing wrong.Please also remember that the accelerating object isnotan inertial observer

- #27

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In relativity theory gamma = 1/sqrt(1-v^2/c^2), and mass and impulse are increasing with this factor and Kinetic energy = (m-m_r)*c^2.

- #28

HallsofIvy

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Where did you get the idea that "force" decreases with distance?Since force obviously decreases with distance could you explain what you mean when you say you are applying aconstantforce to a mass that is rapidly moving away from you?

- #29

HallsofIvy

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- #30

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so that is fast! in the first term I miss the mass and in the second dv/dtdp/dt= (1- v^{2}/c^{2})^{1/2}(dv/dt)- (v^{2}/c^{2})(1- v^{2}/c^{2})^{-3/2}= F. Solve that for dv/dt.

- #31

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F= dp/dt = gamma^3*m_0*dv/dt

So with increasing velocity gamma is increasing and with constant force dv/dt must decrease!

- #32

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I agree with the gravitational force but the electro-magnetic force, the weak nuclear force and the strong nuclear force are all forces. Hint: why do you think they are called forces in the fist place?Where did you get the idea that "force" decreases with distance?

Gravitionalforce between two masses decreases with distance andelectricalforce between two charges decreases with distance but those are not all forces.

I think you are confused by Newtonian dynamics. In fact Newtonian dynamics by itself does not really deal with forces as macro objects "collide" due to EM forces not real collisions. The difference is obviously irrelevant at the macro level but at the micro level two objects coming very close together do not need to collide to feel a force. And fact is, that force depends on distance.

If I use one magnet to accelerate another magnet the force on the pushed magnet does decrease with distance.

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- #33

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I'am not so familliar with forces on the microlevel, so how is this dependence on distance there?The difference is obviously irrelevant at the macro level but at the micro level two objects coming very close together do not need to collide to feel a force. And fact is, that force depends on distance.

If I use one magnet to accelerate another magnet the force on the pushed magnet does decrease with distance.

secondly

If the magnet repulsion remains constant then Mejennifer let us take a very long tube; place a magnet at the one end and put in a magnetic canon-ball.

- #34

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The point is that it does not remain constant over distance.I'am not so familliar with forces on the microlevel, so how is this dependence on distance there?

secondly

If the magnet repulsion remains constant then Mejennifer let us take a very long tube; place a magnet at the one end and put in a magnetic canon-ball.

- #35

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You are answering none of my two questions and discussing the starter of this thread.

Suppose F constant: WHAT IF.

This question has resulted in the explaining of Impulse as a function of velocity and acceleration at high velocities. Surely a micro-physics knowledger as you must know that it is an important question whether velocity has an upper limit.

- #36

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Isnt it true that as v => c

Mass => infinite

Processes/time =>0

Force/energy required for acceleration => infinte ???

This would seem to mean that in this universe both proper accceleration and coordinate acceleration would slow to a halt. Otherwise you have Zenos Rocket perpetually accelerationg off down the rabbithole.

Or am I just taking this discussion to literally?

- #37

JesseM

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Imagine you are driving an object forward by aiming a constant stream of pellets at its backside, and imagine taking the limit as the spacing between the pellets (and their individual mass) goes to zero so the stream is applying a continuous force rather than a discrete series of impulses. Using this method it would certainly be possible in principle to apply a force to accelerate the object which is constant in our own rest frame, although it would not appear constant in the instantaneous inertial rest frame of the object from one moment to another.The point is that it does not remain constant over distance.

- #38

HallsofIvy

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- #39

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I assume it is clear that I would have to accelerate as well to achieve that. But think of the impossibility of Born rigidy in this context, since what you are describing is nothing else than the acceleration of a rod from the back.Imagine you are driving an object forward by aiming a constant stream of pellets at its backside, and imagine taking the limit as the spacing between the pellets (and their individual mass) goes to zero so the stream is applying a continuous force rather than a discrete series of impulses. Using this method it would certainly be possible in principle to apply a force to accelerate the object which is constant in our own rest frame, although it would not appear constant in the instantaneous inertial rest frame of the object from one moment to another.

- #40

JesseM

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Why would you have to accelerate? The gun firing the pellets can have a single fixed rest frame (though it might have to steadily increase the velocity which with it fires successive pellets in order to achieve a constant force on the object, if that's what you mean).I assume it is clear that I would have to accelerate as well to achieve that.

If we're just discussing the fact that the force needed to accelerate an object to c in finite time would be infinite, I don't see why it's necessary to impose the condition of Born rigidity at all.MeJennifer said:But think of the impossibility of Born rigidy in this context, since what you are describing is nothing else than the acceleration of a rod from the back.

Anyway, Born rigidity is an idealization, but I would guess that approximate Born rigidity is achieved "naturally" as an equilibrium of all the forces between atoms in a solid object when you push the object from the back (since each atom is only directly being accelerated by the force from nearby atoms, when you push an object's back you aren't directly applying a force to atoms far from the point you're pushing, only indirectly since the atoms next to the point being pushed then apply a force to farther atoms, which apply their own force to farther atoms, etc.--so while the initial push will at first cause a compression wave which distorts the shape of the object, if the external force is constant eventually an equilibrium should be reached where each atom is being accelerated at a constant rate by nearby atoms.) My reasoning is that the equivalence principle says that a small object sitting on a platform on the Earth should behave like the same object sitting on a platform which is accelerating at 1G in empty space, with gravitational time dilation between clocks at the top and bottom of the object being equivalent to the time dilation of clocks at front and back of an accelerating object that satisfies the Born rigidity condition (see this post). If the object being pushed from the back at 1G did not "naturally" achieve an equilibrium distribution of forces such that the front was being accelerated at a lesser rate, then you wouldn't have the same time dilation between clocks at "top" and "bottom" of the object as you do with clocks at the top and bottom of an object resting on Earth, and an observer standing on the ground next to the object could determine whether he was on a gravitating body or an accelerating platform in deep space, which would seem to violate the equivalence principle.

- #41

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That is not a fact at all, you are mistaken. The Einstein velocity addition has no limit. I suggest you read up on limits and hyperbolic functions.If we're just discussing the fact that the force needed to accelerate an object to c in finite time would be infinite, I don't see why it's necessary to impose the condition of Born rigidity at all.

- #42

JesseM

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I'm not sure what "the Einstein velocity addition has no limit" means in this context. No limit to what equation as what variable approaches what value? And what would this have to do with my statement "the force needed to accelerate an object to c in finite time would be infinite"? It's certainly true that if we imagine applying a constant proper acceleration to an object for some finite time (time as measured in the inertial frame where the object is at rest initially), then in the limit as the acceleration goes to infinity, the object's final speed at the end of the time interval approaches c. You can also draw an "impossible" worldline which reaches c in some finite time, like v(t) = (c/1 year)*t (where t and v are defined in terms of a particular inertial frame, so the coordinate velocity reaches c at t=1 year), and then calculate the proper acceleration at any point on this worldline--the thing that makes it impossible is that in the limit as t approaches 1 year, the proper acceleration approaches infinity.That is not a fact at all, you are mistaken. The Einstein velocity addition has no limit.

- #43

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Indeed itIt's certainly true that if we imagine applying a constant proper acceleration to an object for some finite time (time as measured in the inertial frame where the object is at rest initially), then in the limit as the acceleration goes to infinity, the object's final speed at the end of the time interval approaches c.

- #44

JesseM

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It is standard when talking about limits to say that the function "approaches" a certain value in the limit as some variable "approaches" another value (if a mathematician wants to verbally describe the symbols [tex]\lim_{x\rightarrow a} f(x)[/tex], they'll use words like 'the limit of f(x) as x approaches a'). c satisfies the formal definition of a limit in this case:Indeed itapproachesc but never reaches c and that is what you claimed. The limit is not c as the limit does not exist.

If in this case we replace z with t so the function f(t) is the instantaneous velocity of the object accelerating with velocity given by f(t) = (c/1 year)*t, then it's true that for any velocity you choose which is less than c by some tiny amount epsilon, it's possible to find a delta such that at the time (1 year - delta) the velocity given by f(t) becomes larger than (c - epsilon). Just name any epsilon, I'll give you a delta! For example, say you chose epsilon = 0.0001c, so that c - epsilon was equal to 0.9999c. Then I could pick any delta smaller than 0.0001 years, say delta=0.00009 years, and f(t) would be closer to c than 0.9999c at time f(1 year - delta). This is all that's required for f(t) to have the limit c as t approaches 1 year, that the function f(t) getsA function f(z) is said to have a limit [tex]\lim_{z\rightarrow a} f(z)[/tex] = c if, for all [tex]\epsilon[/tex] > 0, there exists a [tex]\delta[/tex] > 0 such that | f(z) - c | < [tex]\epsilon[/tex] whenever 0 < | z - a | < [tex]\delta[/tex].

- #45

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Apparently you do not want to learn that some functions do not have limits.

- #46

JesseM

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I don't believe it is physically possible, since it would require infinite energy and infinite proper acceleration. But as a purely mathematical matter it is certainly true that the function v(t) = (c/1 year)*t approaches c in the limit as t approaches 1 year, and it is also true that theJesseM, it is clear to me that your teacup is full, just keep believing that massive objects can reach c in the limit.

I am not sure what bizarre misreading of my posts could lead you to think that I am denying the statement "some functions do not have limits". Of course some functions don't have limits. The velocity function v(t) = (c/1 year)*t does have a well-defined limit of c in the limit as t approaches 1 year, however (but the different function that describes proper acceleration as a function of time for a worldline with that velocity function wouldMeJennifer said:Apparently you do not want to learn that some functions do not have limits.

It would really help if you would give some mathematically precise statement of what you are arguing instead of speaking in nebulous generalities. What

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- #47

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As I claimed earlier HallsofIvy: the mass is asumed to have relation to velocity, so dm/dt has a dv/dt in it and it is not fair to isolate this dv/dt this way!

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