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Force is equal to the rate of change of the momentum of the particle:

F = dp/dt

The momentum of a particle in relativistic mechanics is calculated by the formula:

p = m v/√(1 - v²/c²)

If you substitute this in the above equation (keeping in mind that the velocity vector has a constant direction during a linear motion, so that we can eliminate the vectors), and differentiate, you get:

F = m/(1 - v²/c²)^3/2 dv/dt


Now, dv/dt is the acceleration wrt the inertial reference frame, and, as is seen from my previous post, it is dependant on velocity by the formula:

a = dv/dt = g (1 - v²/c²)^3/2

Substituting this in the equation for the force, you see that the velocity factors cancel out and all that's left is:

F = m g

That means that the UA exerts a constant  force on the Earth. It also means that you phail. Now die!

Sorry. That's wrong. You understand that mass is dependent on velocity. You understand that velocity constantly increasing, so mass is constantly increasing exponentially, so the force required is constantly increasing exponentially.

Sorry. That's wrong. You understand that mass is dependent on velocity. You understand that
velocity constantly increasing, so mass is constantly increasing exponentially, so the force required is constantly increasing exponentially.

This is a conspiracy too 

Quote from: Gulliver on June 06, 2007, 05:58:24 PM

Sorry. That's wrong. You understand that mass is dependent on velocity. You understand that
velocity constantly increasing, so mass is constantly increasing exponentially, so the force required is constantly increasing exponentially.

This is a conspiracy too 

As I recall, TomB's response is something along the line of the UA is just a natural outcome of (some gibberish) therefore no energy is required. That's better I think than the-ever-growing conspiracy explanation.

Quote from: Bushido on June 06, 2007, 05:12:26 PM

Force is equal to the rate of change of the momentum of the particle:

F = dp/dt

The momentum of a particle in relativistic mechanics is calculated by the formula:

p = m v/√(1 - v²/c²)

If you substitute this in the above equation (keeping in mind that the velocity vector has a constant direction during a linear motion, so that we can eliminate the vectors), and differentiate, you get:

F = m/(1 - v²/c²)^3/2 dv/dt


Now, dv/dt is the acceleration wrt the inertial reference frame, and, as is seen from my previous post, it is dependant on velocity by the formula:

a = dv/dt = g (1 - v²/c²)^3/2

Substituting this in the equation for the force, you see that the velocity factors cancel out and all that's left is:

F = m g

That means that the UA exerts a constant  force on the Earth. It also means that you phail. Now die!

Sorry. That's wrong. You understand that mass is dependent on velocity. You understand that velocity constantly increasing, so mass is constantly increasing exponentially, so the force required is constantly increasing exponentially.

And, of course, you have the calculations to back your claims. Do you even know what exponentially means?

Quote from: Gulliver on June 06, 2007, 05:58:24 PM

Quote from: Bushido on June 06, 2007, 05:12:26 PM

Force is equal to the rate of change of the momentum of the particle:

F = dp/dt

The momentum of a particle in relativistic mechanics is calculated by the formula:

p = m v/√(1 - v²/c²)

If you substitute this in the above equation (keeping in mind that the velocity vector has a constant direction during a linear motion, so that we can eliminate the vectors), and differentiate, you get:

F = m/(1 - v²/c²)^3/2 dv/dt


Now, dv/dt is the acceleration wrt the inertial reference frame, and, as is seen from my previous post, it is dependant on velocity by the formula:

a = dv/dt = g (1 - v²/c²)^3/2

Substituting this in the equation for the force, you see that the velocity factors cancel out and all that's left is:

F = m g

That means that the UA exerts a constant  force on the Earth. It also means that you phail. Now die!

Sorry. That's wrong. You understand that mass is dependent on velocity. You understand that velocity constantly increasing, so mass is constantly increasing exponentially, so the force required is constantly increasing exponentially.

And, of course, you have the calculations to back your claims. Do you even know what exponentially means?

Your formulas do the job quite nicely. Yes. Consider that the denominator contains the term: √(1 - v²/c²)

That's not an exponential function, it's an irrational algebraic expression. Also, please be exact which formula you are referring to.

That's not an exponential function, it's an irrational algebraic expression. Also, please be exact which formula you are referring to.

F= m₀/(√(1-v²/c²)*a where m₀ is the rest mass.

As far of the exponentially increasing behavior, I hope the following table will assist your understanding:

v   relativistic term
1   1
10   1.000000001
100   1.000000056
1000   1.000005556
10000   1.000556019
100000   1.060660172
200000   1.341640786
290000   3.905667329
299000   12.2576677
299900   38.73306135
299990   122.4755078
299999   387.2986574
299999.9   1224.744974
299999.99   3872.983376
299999.999   12247.44872
299999.9999   38729.83186

your formula is incorrect.

your formula is incorrect.

I'd love to know what's wrong with it. You, yourself, said F=ma. m=m₀/(√(1-v²/c²) is explained here: physlink.com. Simply algebraic substitution yields the formula that you say is incorrect.

Quote from: Bushido on June 07, 2007, 02:24:16 AM

your formula is incorrect.

I'd love to know what's wrong with it. You, yourself, said F=ma. m=m₀/(√(1-v²/c²) is explained here: physlink.com. Simply algebraic substitution yields the formula that you say is incorrect.

Where did I say that?

Quote from: Gulliver on June 07, 2007, 02:31:35 AM

Quote from: Bushido on June 07, 2007, 02:24:16 AM

your formula is incorrect.

I'd love to know what's wrong with it. You, yourself, said F=ma. m=m₀/(√(1-v²/c²) is explained here: physlink.com. Simply algebraic substitution yields the formula that you say is incorrect.

Where did I say that?

Please note my emphasis below:

Force is equal to the rate of change of the momentum of the particle:

F = dp/dt

The momentum of a particle in relativistic mechanics is calculated by the formula:

p = m v/√(1 - v²/c²)

If you substitute this in the above equation (keeping in mind that the velocity vector has a constant direction during a linear motion, so that we can eliminate the vectors), and differentiate, you get:

F = m/(1 - v²/c²)^3/2 dv/dt  <---- where dv/dt is a


Now, dv/dt is the acceleration wrt the inertial reference frame, and, as is seen from my previous post, it is dependant on velocity by the formula:

a = dv/dt = g (1 - v²/c²)^3/2

Substituting this in the equation for the force, you see that the velocity factors cancel out and all that's left is:

F = m g   <------- where g is the acceleration, a.

That means that the UA exerts a constant  force on the Earth. It also means that you phail. Now die!

Yes, but that is the final derived formula for the force according to the inertial reference frame expressed with the rest mass (although, it's an obsolete term) of the Earth m (or m₀, according to you) and the acceleration of free fall as measured by the observer in the reference frame co-moving with the Earth. As you can see, the force is constant, regardless of the velocity (according to the inertial reference frame) of the Earh.

Yes, but that is the final derived formula for the force according to the inertial reference frame expressed with the rest mass (although, it's an obsolete term) of the Earth m (or m₀, according to you) and the acceleration of free fall as measured by the observer in the reference frame co-moving with the Earth. As you can see, the force is constant, regardless of the velocity (according to the inertial reference frame) of the Earh.

Well, I'm glad we agreed on my formula.

Now for your next error: Since the FE is accelerating, it does not have an inertial frame of reference of itself. We know that it must be going at least as fast as any object accelerated at 1g over the lifetime of the Earth.

ROFLMAO.

Stop posting, dude. Seriously.

ROFLMAO.

Stop posting, dude. Seriously.

Seriously, any acceleration, linear, rotational, or revolving causes the lost of the inertial frame of reference. See for example, Einstein, 1924, Relativity: The Special and General Theory, page 10.

If you know you're moving with respect to the Universe, then you're not in an inertial frame of reference.