Re: .Re: Why all the fascination with E = mc^2 ??

From: David McAnally (D.McAnally_at_i'm_a_gnu.uq.net.au)
Date: 06/02/04 

Date: 2 Jun 2004 06:18:47 GMT

leoppard@MailAndNews.com (Leonard Pardin) writes:

>D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) wrote in message news:<c9i168$c1o$1@bunyip.cc.uq.edu.au>...
>> D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) writes:
>>
>> >leoppard@MailAndNews.com (Leonard Pardin) writes:
>>
>> >> Don't tell me you are another one of the "relativity" freaks who
>> >>think the formula was derived by some complex mathematics using
>> >>Lorentz transformations. The formula was worked out by Poincare long
>> >>before Einstein's relativity was even published. It's not a
>> >>relativity problem. Why are the relativists so fascinated with the
>> >>formula?
>>
>> >Ah, the arrogance of the ignoramus. You have already admitted ignorance
>> >of classical physics, and yet you presume to lecture those of us who know
>> >classical physics, and what is more, you are demanding that we accept a
>> >statement which directly contradicts Newtonian mechanics, on your
>> >unsupported word.

>> >Anyway, I don't consider E = mc^2 to be one of the fascinating formulae
>> >in relativity. But it is a short memorable equation that catches the
>> >layman's attention. Much easier to understand and remember than the
>> >Lorentz Transformation or Maxwell's Equations.

> Well, I have been accused in this newsgroup of having an agenda.
>In a way I do. As you say, the formula is easy to remember, and the
>general public is led to believe that it has some magical quality. But
>it has been attributed to Einstein who is constantly idolized while
>the brilliant Poincare is virtually unknown.

Poincare is very well known to professionals.

>In almost every case,
>lecturers connect the formula to relativity, suggesting that this
>little formula changed the world as we know it, and it was only made
>possible by the formulation of the theory of relativity.

How do you know what lecturers say?

>But Poincare
>first derived the formula, and it had nothing to do with relativity.
>In my amateurish way, I am trying to encourage the recognition of a
>brilliant scientist who has been buried under an avalanche of public
>relations publicity for the colorful Einstein.

But you did this by writing down statements which were completely
contradictory with Newtonian mechanics. Further, when we stated that you
were wrong in making these contradictory statements, then you claimed that
we disagreed with Newtonian mechanics, when it was Newtonian mechanics
that led us to tell you that you were wrong. You also made the statements
without attempting any sort of proof.

>> My derivation of the formula for energy starts with the Law of
>> Conservation of Momentum, which, in Newtonian mechanics, is a direct
>> consequence of Newton's Third Law of Motion. In relativity, only contact
>> forces are allowed, so that in the relativistic version of Newton's Third
>> Law of Motion, the action and reaction are contact forces, and the Law of
>> Conservation of Momentum becomes the statement that the momentum after a
>> collision is equal to the momentum before the collision. The relativistic
>> momentum of a material body should be given by p = m f(|v|) v, where m is
>> the intrinsic mass of the body, and f is sufficiently nicely behaved
>> function (continuous, differentiable, whatever is required) such that
>> f(0) = 1 (so that when v is much smaller than c, the formula for p
>> approximates to the classical formula p = m v).

> If it approximates the classical formula, why can we not use the
>classical formula? Can't p = mv work without cluttering up the
>derivation with the Lorentz transformation formulae?

The formula for p has nothing to do with the derivation of the Lorentz
transformation. The Lorentz transformation, and its derivation, is
completely kinematic, whereas p is a dynamic quantity. Try reading the
derivation of the Lorentz transformation some time. Obviously, going by
your question above, you have never acquainted yourself with the
derivation.

The reason why p = mv won't work is because if p = mv holds, then
Conservation of Momentum cannot be made to hold in all frames of reference
(specifically, if we adopt the formula p = mv for the momentum, and if
Conservation of Momentum holds in one frame of reference, then there exist
frames of reference in which Conservation of Momentum does not hold), so
the formula p = mv leads to a direct contradiction to Einstein's First
Postulate (the Principle of Relativity). This means that the formula for
momentum must be modified so that Conservation of Momentum holds in all
frames of reference, as required by Einstein's First Postulate.

I would also add that the fact that the formula for p approximates to mv
when v is much smaller than c is important, since Special Relativity is
required to approximate to Newtonian Mechanics when v is much smaller than
c.

>> This means that it is
>> possible to set up an example where conservation of momentum is observed
>> in two different frames. Now comes the first of only two places in the
>> ENTIRE derivation that the Lorentz Transformation is required, and it is
>> only in the guise of the transformation law for velocity that we require
>> it here or at the other place in the derivation. The reason for this is
>> that we have to know the velocities of all objects in both frames of
>> reference.

> Why can't we work with only one frame of reference. If E = mc^2 is
>valid, it is most certainly valid in our own frame of reference.

Because the Law of Conservation of Momentum must hold in all frames, as
required by the Principle of Relativity. This means that we have to
investigate the properties of momentum in different frames. Do you
understand the Principle of Relativity?

In the case of Galilean Relativity, then the Law of Conservation of
Momentum holds in all inertial frames with the classical formula p = mv,
provided the Law of Conservation of Mass also holds (without Conservation
of Mass, Conservation of Momentum cannot be made to hold in all frames).

In the case of Special Relativity, then the Law of Conservation of
Momentum holds in all inertial frames with the relativistic formula
p = mv/sqrt(1-v^2/c^2), provided the Law of Conservation of Energy also
holds (with the formula E = mc^2/sqrt(1-v^2/c^2)).

>> It is provable from this point that the only differentiable
>> function of v which works is f(|v|) = 1/sqrt(1 - v^2/c^2), so that the
>> momentum is p = m v/sqrt(1 - v^2/c^2). It is also provable that there is
>> an additional law that if m/sqrt(1 - v^2/c^2) is evaluated for every
>> object in a collision, then the incoming sum of these quantities and the
>> outgoing sum of these quantities must be equal, i.e. there must be a law
>> of conservation of the sum of quantities of the form m/sqrt(1 - v^2/c^2)
>> over all bodies involved in the collision (this is the second place where
>> the Lorentz Transformation is required).

> If the sums of the ingoing and outgoing quantities must be equal,
>then why not simply use m rather than m adjusted by the Lorentz
>numbers? wouldn't the law of conservation of the sum of quantities
>still be preserved?

No, they would not. It is quite simple to come up with a collision for
which the formula p = mv will not lead to Conservation of Momentum in all
frames of reference in the Theory of Relativity.

>> Since the energy of a body is the integral of velocity with respect to
>> momentum then E = integral v.dp = m c^2/sqrt(1 - v^2/c^2) + K, where K
>> is a constant, after a bit of calculation, or for the difference in
>> energy for two velocities v_1 and v_2,
>>
>> E_1 - E_2 = m c^2/sqrt(1 - v_1^2/c^2) - m c^2/sqrt(1 - v_2^2/c^2),
>>
>> and the kinetic energy of a body is given by
>>
>> K.E. = m c^2/sqrt(1 - v^2/c^2) - m c^2,
>>
>> which approximates to the classical formula of K.E. = 1/2 m v^2 when v is
>> much smaller than c.

> Okay, so why not just use the classical formula? K.E. = 1/2 mv^2

Because the energy is equal to mc^2/sqrt(1-v^2/c^2) + K for some constant
K. The formula K.E. = 1/2 m v^2 is not consistent with this formula for
the energy. The formula for the energy was derived from the formula for
the momentum, and the formula for the momentum was derived from the fact
that the Law of Conservation of Momentum holds in all inertial frames of
reference.

The approximation of K.E. to 1/2 m v^2 for v much smaller than c is
important since Special Relativity must approximate to Newtonian Mechanics
when v is much smaller than c.

>> This only requires the first application of the
>> transformation law for velocity above, and does not require the second.
>>
>> From the previous paragraph, we require that the sum of quantities of
>> the form m c^2/sqrt(1 - v^2/c^2) for all incoming bodies for a collision
>> is equal to the sum of quantities of the same form for all outgoing
>> bodies.

> So you require the sums of the quantities be the same for all
>incoming and all outgoing.

No. If we evaluate m v/sqrt(1 - v^2/c^2) for all incoming bodies and
evaluate the sum of all these values over all incoming bodies, and if we
evaluate m v/sqrt(1 - v^2/c^2) for all outgoing bodies and evaluate the
sum of all these values over all outgoing bodies, then the two sums are
equal. That is the Law of Conservation of Momentum.

If we evaluate m c^2/sqrt(1 - v^2/c^2) for all incoming bodies and
evaluate the sum of all these values over all incoming bodies, and if we
evaluate m c^2/sqrt(1 - v^2/c^2) for all outgoing bodies and evaluate the
sum of all these values over all outgoing bodies, then the two sums are
equal. That is the Law of Conservation of Energy.

There is no requirement that the quantities be the same for all bodies,
only that the sum before the collision and the sum after the collision
be equal.

>Isn't that the same as if I had insisted
>in my original erroneous model that the mass of the cannon and the
>cannonball be the same?

But cannons are much more massive than cannonballs, otherwise they
wouldn't work properly.

Before the cannonball is fired, there is energy stored in the gunpowder,
etc. The nett energy before the firing is the sum of the kinetic energies
(i.e. zero and zero) and the energy stored in the gunpowder, etc. The
nett energy after the firing is the sum of the two kinetic energies. In
other words, the energy stored in the gunpowder, etc. is distributed to
the cannon and the cannonball as kinetic energy. There is no reason why
the energy should be partitioned in such a manner that equal kinetic
energies should be distributed to the cannon and the cannonball. All that
the Law of Conservation of Energy tells us is that the energy initially
stored up in the gunpowder, etc. is partitioned into kinetic energies of
the cannon and cannonball without imposing any other quantitative
restrictions on the kinetic energies, i.e. the Law only tells what the sum
of the two kinetic energies must be.

>> Now take a stationary body of mass m, which then explodes into miniscule
>> particles each travelling at nearly the speed of light, then the sum of
>> the kinetic energies of the bodies after the explosion is very nearly
>> equal to m c^2, and can be made as close below m c^2 as one would like.
>> This means that a body of mass m has the potential to supply a nett
>> kinetic energy of m c^2 after an explosion.

> You are assuming a quantity of internal energy that will cause an
>explosion with particles travelling outward at nearly the speed of
>light.

Not at all. Many subatomic particles decay naturally.

>Your formula requires that the sum of the mass of those
>particles be a certain quantity.

No, it doesn't. Let the initial particle be a stationary particle of mass
m. Suppose that it decays into N particles with masses m_i and velocities
v_i for i = 1, 2, ..., N. Then the Law of Conservation of Momentum tells
us that

        0 = sum_i m_i v_i/sqrt(1 - v_i^2/c^2),

and the supplementary requirement for the Law of Conservation of Momentum
(i.e. the supplementary requirement which must be satisfied in order that
the Law of Conservation of Momentum be satisfied in all inertial frames
of reference) tells us that

        m = sum_i m_i/sqrt(1 - v_i^2/c^2).

Since the kinetic energy of particle i is

        K.E._i = m_i c^2/sqrt(1 - v_i^2/c^2) - m_i c^2,

then the nett kinetic energy is given by m c^2 - sum_i m_i c^2.

Let us take the case where the particle decays into two particles of equal
mass (so N = 2 and m_1 = m_2), then v_1 = -v_2 by Conservation of
Momentum, Let U be the common speed of the particles, so that U = |v_1|
= |v_2|. Then the supplementary requirement tells us that

        m = 2 m_1/sqrt(1 - U^2/c^2),

so that 2 m_1 = m sqrt(1 - U^2/c^2), and so the nett kinetic energy after
decay is m c^2 (1 - sqrt(1 - U^2/c^2)). If m_1 = m_2 is very small, then
U is very close to c, and so the kinetic energy after decay is very close
to m c^2. This can be made as close to m c^2 as you like by making m_1
as small as you like. This means that a stationary body of mass m has the
potential to deliver a nett kinetic energy of m c^2 after decay.

>Isn't that begging the question?

No, because the reasoning behind the conclusion was not based on
Conservation of Energy or on a definition of energy. The reasoning was
based on the fact that there is an additional requirement which must be
satisfied in order that the Law of Conservation of Momentum holds in all
inertial frames of reference.

>This means that there is
>> already a quantity of energy m c^2 present in a stationary body of mass
>> m, and such a body moving with velocity v has energy m c^2/sqrt(1 - v^2/c^2).
>> Once this identification of the energy of a mass is established, the law
>> which necessarily accompanies the Law of Conservation of Momentum in the
>> Theory of Relativity becomes the Law of Conservation of Energy, i.e.
>> Conservation of Momentum requires Conservation of Energy (and, in fact,
>> Conservation of Energy requires Conservation of Momentum), and the two
>> laws unite in the language of 4-vectors into the Law of Conservation of
>> 4-Momentum (the 3-Momentum being the spatial component of the 4-Momentum,
>> and the Energy being the temporal component).
>>
>> David

> But couldn't the same result have been reached without the Lorentz
>transformations being employed?

The Lorentz transformations were required because we needed to transform
momenta between different inertial frames of reference. We needed to
do this in order to satisfy the Principle of Relativity as far as the
Law of Conservation of Momentum was concerned.

>Shouldn't the formula work in only
>one frame of reference?

No. The point is that we require a formula that works in ALL inertial
reference frames, otherwise there would be a special frame of reference,
contradicting the Principle of Relativity. In asking this question, you
have demonstrated that you have missed the entire point of the Principle
of Relativity.

>And doesn't your derivation assume the
>presence of a quantity of energy in order to prove that quantity of
>energy?

No. My derivation pointed out that there was an additional quantity Q
associated with each mass such that the sum of the values of Q for all the
incoming masses is equal to the sum of the values of Q for all the
outgoing masses. That the sum of the incoming values of Q and the sum of
the outgoing values of Q must be equal follows from the Law of
Conservation of Momentum and the Principle of Relativity. For Galilean
Relativity, Q = m (i.e. Q is the mass), and for Special Relativity,
Q = m/sqrt(1 - v^2/c^2). This has nothing to with a definition of energy.

> I went back to check on Poincare's derivation. I mistakenly
>remembered the use of electrons. In fact, Poincare postulated a
>mathematical relationship between mass and "momentum of radiation." He
>proposed that mass was equal to the "flux of radiation" divided by the
>speed of light squared. Mass, then, was equal to radiation energy
>divided by the speed of light squared. Poincare was doing nothing less
>than redefining mass. That translated into Energy equal to mass times
>the speed of light squared. No relativity mathematics was involved in
>the derivation.

I have asked you to supply a reference. Again, I ask you to supply a
reference.

David

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