Re: e NOT EQUAL TO mc2

From: PD (pdraper_at_yahoo.com)
Date: 02/16/05 

Date: 16 Feb 2005 10:20:57 -0800

vignesh wrote:
> SECTION ONE
>
> Consider the problem of powered space flight. The equation giving the
> kinetic energy obtained by any space vehicle which has been
accelerated
> to a velocity v is of course, K.E. = ½ mv². However observe that to
> double the velocity of any space vehicle (by increasing v to 2v for
> example), four times as much energy would be required each time if
the
> laws of the conservation of energy are to be maintained.

Wrong. Three times as much are required to get from v to 2v, as
required to get from 0 to v.

> This results
> from the relation between velocity and kinetic energy being
non-linear.
> However surely this would go against common sense as the latter would
> tend to suggest that the formula were as follows, K.E. = mv.

Wrong.

>
> Such an equation would simply state that to increase your velocity by
a
> set amount in the frictionless environment of space the same
proportion
> of energy would be required each time no matter what speed you were
> travelling at to begin with.

Consistent with previous wrong statement, and also wrong.

> The laws of conservation of energy would
> suggest that the energy supplied to the craft by its power source
would
> at all times equal the kinetic energy obtained, taking into account
the
> various efficiency ratios involved of course. If the kinetic energy
did
> equal ½ mv² then this would result in increasingly larger amounts
of
> fuel being needed to accelerate the craft which in turn would produce
a
> limiting velocity requiring its own separate equation.

Wrong.

>
> Of course we have all been taught since day one that it is actually
> momentum which equals mv and not kinetic energy. But is it really
that
> straight forward? For example is momentum a force or is it actually a
> form of energy?

Neither.

> Indeed the units it is measured in are rather peculiar,
> for example kilograms per metre per second. But which do these
denote,
> force or energy?

Neither.

>
> Having discovered that there is a possible problem with the current
> equation for kinetic energy and that the exact nature of the physical
> property of momentum is open to question, it is now necessary to
> enquire where the underlying fault would appear to lie. In my opinion
> the best place to start is with the definition of the joule itself
> which we are told equals a force of one Newton applied through a
> distance of 1 metre. An objection which can immediately be raised to
> this apparently straight forward definition is that the work done by
> one Newton through one metre is surely dependant upon the mass it is
> acting upon?

Wrong.

> For example would it not take more work for one Newton to
> move a 2 kg mass through one metre than it would the same force to
move
> 1 kg through the same distance?

No.

> The unit mass is entirely absent from
> the definition.

Right.

>
> To further illustrate my point, using F = ma and s = ½ at², 1
Newton
> acting on a 1 kg mass would take 1.41 seconds to reach one metre (i),
> whereas 1 Newton acting on a 2 kg mass would take 2 seconds to reach
> one metre (ii). Surely it would be the case that if one Newton were
> applied for a longer duration of time as with case (ii) more work
would
> have been done as a consequence?

Wrong.

> However if the equations v = at and
> K.E. = ½ mv² are applied to the accelerations and velocities
obtained
> in (i) and (ii), we do in fact end up with 1 Joule of energy in each
> case as is predicted. In fact this is the very basis of the
definition.
>
>
> But surely it has to be the case that 1 Newton applied for 2 seconds
in
> case (ii) would have carried out an extra 40% more work than in case
> (i) where 1 Newton was only applied for 1.41 seconds?

Wrong.

> This point is
> surely indisputable?

Wrong.

> If it is then a large part of classical physics
> has to be in error and possibly as a consequence many other areas of
> the same science would also be undermined?

Consistent with previous statement, and also wrong.

>
> The above demonstration would clearly suggest therefore, that the
> formula Work Done = Force x Distance, would appear to be not at all
> valid.

Wrong.

> Indeed it is this equation that the definition of the Joule is
> based upon. Instead the correct equation would perhaps be Work Done =
> Force x Time.

Wrong. That's momentum.

> Since the formula for kinetic energy is also derived from
> the former apparently incorrect equation, it too must be unsound.

Consistent with previous statements, and also wrong.

> Indeed it is this formula which is actually responsible for
calculating
> the erroneous 1 Joule of work done for Cases (i) and (ii).
> Interestingly if it is instead assumed that K.E. is proportional to
mv,
> then we obtain the energies of 1.41 Joules and 2 Joules respectively
> for cases (i) and (ii). This is entirely consistent with a 40%
> difference in work done between the two cases.

Consistent, and also wrong.

>
> Indeed it would even appear possible to prove by integration that
> kinetic energy equals mv if it is assumed that Work Done = Force x
> Time, as follows:
>
> K.E. = ∫ F.dt = ∫ m dv/dt . dt = ∫ m dv = mv.

Wrong | <--------- right ----------------->

>
> or alternatively :
>
> K.E. = Work Done = Force x Time = F x t = m a x t (and because v =
at)
> = m v

Second equality is wrong. Rest is right.

>
> Further evidence that would suggest that kinetic energy were equal to
> mv is the observation that when two objects of known masses collide,
it
> is the conservation of momentum mv which would apply and not the
> conservation of kinetic energy according to the currently accepted
> equation which covers it, ½mv².

Half-right. Conservation of momentum does apply. If the collision is
elastic, conservation of kinetic energy also applies.

> Indeed the two cases produce quite
> different results : Consider two bowls each with a mass of 1.5 kg.
The
> first bowl has a velocity of 2 m/sec and impacts at a slight angle
with
> the second which is stationary. If the velocity of the first bowl
after
> impact is 0.5 m/sec,
> the velocity of the second bowl based upon the
> conservation of momentum would be 1.5 m/sec (i.e. 2 – 0.5 = 1.5
> m/sec).

Wrong. Not in an impact at a slight angle.
Moreover, if it's a head-on impact, and the collision is elastic, the
second bowl would never have a velocity of 0.5 m/s after impact.
Verifiable experimentally.

> But if it were assumed that the conservation of kinetic energy
> applied instead, the velocity of the second bowl would be as follows
:
>
> V = √(2² - 0.5²) = 1.94 m/sec - A different value altogether.

Wrong. Same reason as before.

>
> Since conservation would only occur experimentally where mv is used,
> perhaps it can be assumed therefore that the kinetic energy also
equals
> mv?

Wrong.

>
>
>
>
> SECTION TWO
>
> According to the Special Theory of Relativity the equation for
kinetic
> energy is as follows:
>
> K.E. = mc² / √(1 - v²/c²) - mc²
>
> It is apparent that this equation is entirely based upon E = mc².
> However in this case m happens to be the mass increase

Wrong.

> which is equal
> to the relativistic mass minus the rest mass or to formulate it
> algebraically, m / √(1 - v²/c²) – m. According to the Special
> Theory the relativistic mass is obtained by dividing the rest mass by
> the Lorentz factor √(1 - v²/c²) which in turn is obtained from
the
> Lorentz transformation.

Right.

> The mathematical properties of this factor are
> as follows: when v is small compared with c, the speed of light,
√(1
> - v²/c²) = 1 and when v approaches c, √(1 - v²/c²) = 0 making
the
> relativistic mass tend towards infinity.

Right.

> Multiplying the mass
> differential quoted above, m / √(1 - v²/c²) – m, by c² gives
the
> stated equation for relativistic kinetic energy. An alternative way
of
> looking at it is that the mass-energy

Wrong.

> of the relativistic particle
> minus the rest energy of the particle equals the relativistic kinetic
> energy of the particle as a result of v, the velocity.
>
> If you put all the relevant numbers into the equation for
relativistic
> kinetic energy it does actually produce the same results as ½ mv²
but
> only when v is small compared with c, the speed of light, of course.

Right.

> You can try this yourself at home with an appropriate software
package
> that can handle large numbers of decimal places. Most spreadsheets
can
> actually do this and you can divide the equation into a number of
parts
> which are then calculated in adjacent cells (or alternatively you can
> do it all in the same cell if you are clever enough!). Knowing the
> speed of light to be 3 x 10e8 m/sec (such that c² then equals 9 x
> 10e16) and that v, the velocity is also calculated in m/sec, the
maths
> is quite straightforward. The mass m can be left out of the equation
> because it cancels on both sides or simply assume it to have a value
of
> 1 kg which adds up to much the same thing.
>
> Having proved in the first section that the kinetic energy could
never
> equal ½ mv² and that the classical equation is therefore wrong,

Consistent, and also wrong.

> this
> brings into question the relativistic equation for kinetic energy
since
> it produces the equivalent results as ½ mv² at low velocities.
> However altering the relativistic equation for kinetic energy
> completely undermines the logic supporting the conservation of
> mass-energy and hence also the equation of mass-energy equivalence
> itself, E = mc².

Consistent, and also wrong.

PD

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