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

From: Bjoern Feuerbacher (feuerbac_at_thphys.uni-heidelberg.de)
Date: 06/03/04 

Date: Thu, 03 Jun 2004 17:55:52 +0200

John C. Polasek wrote:
> On Wed, 02 Jun 2004 18:55:01 +0200, Bjoern Feuerbacher
> <feuerbac@thphys.uni-heidelberg.de> wrote:
>
>
>>John C. Polasek wrote:
>>
>>>On Wed, 02 Jun 2004 13:16:41 +0200, Bjoern Feuerbacher
>>><feuerbac@thphys.uni-heidelberg.de> wrote:
>>>
>>>
>>>
>>>>Leonard Pardin wrote:
>>>>
>>>>
>>>>>John C. Polasek <jpolasek@cfl.rr.com> wrote in message news:<35spb0p6cqpd7956bnj7ntudr56paf9ini@4ax.com>...
>>>>>
>
> snip
>
>>>>>>It is a sad commentary that with so many relativistic luminaries
>>>>>>offering their pathetic opinions, that not one of them has responded
>>>>>>to your query, so I thought I might as well bail them out just to
>>>>>>preserve the dignity of this forum.
>>>>>>
>>>>>>You can find the answer on page 65 of my book "Dual Space-the Other
>>>>>>Half of Physics" where I derived it relativistically to wit:
>>>>>>
>
> For Bjoern:
> I have decided to take the time to show you and other students how
> one can deduce important results using first principles.

Unfortunately, you do it wrong.

> Have you ever heard of these?

Of what, specifically?

> It is interesting that your ignorance comes in quanta,
> each seemingly applicable to each new finding as below.

Well, OTOH, your ignorance seems to be macroscopic.

> To clarify the problem, which was never to deduce why E = mc2, but was
> to find out why it's mc2 instead of 1/2 mc2.

Hint: it is neither 1/2 m c^2 nor m c^2 nor 1/2 m v^2, but
gamma m c^2 for the total energy and (gamma - 1) m c^2 for the kinetic
energy.

> The factor 2 is the magic, but that concept may be a bit subtle for you.

There is nothing magical about it.

>>>>>> Int Fdx = 0.5mv^2,
>>>>
>>>>Why should that hold?
>
> First, it holds because it is an unassailable Newtonian identity,

Hint: 0.5 m v^2 is the *classical* expression for the kinetic energy.
It isn't true. The *correct* expression for the kinetic energy
is (gamma - 1) m c^2. This reduces to 0.5 m v^2 for slow velocities,
BTW - so the formula you use is only an *approximation*.

> and
> secondly because it accomplishes an important algebraic task called
> "separation of variables", f(x) on the left, f(v) on the right.

I asked "why should that hold", not "why should one write it this way".
Try to understand the difference.

Hint: I'm well aware of the concept of "separation of variables. Right
now, I'm working with partial wave expansions, which use that very
technique.

> (Copy
> this on your flyleaf: SVAMITB which you will know to translate as
> "separated variables are money in the bank").

*yawn* Try to tell me something new.

>>>>>we can write its differential
>>>>>
>>>>>> Fdx = mvdv
>>>>
>>>>Non sequitur.
>
> Not at all, the differential of (1) establishes a legitimate
> differential equation based on legitimacy of (1). If you still can't
> follow it, email me.

 From int F dx = 0.5 m v^2, it does *not* follow that
F dx = m v dv. Why on earth do you think it does???

Hint: if you claim that E = m c^2, your "m" has to be relativistic mass
(otherwise the equation is either wrong or applies only in the rest
frame of the mass). Hence m = m(v). And therefore
   F dx = m v dv + 0.5 dm v^2.

>>>>>> and since F loses its effect with v (Lorentz)
>>>>>> F' = F*sqrt(1 - v2/c2) = F/gamma
>>>>
>>>>Why do you think so?
>
> One sees that after some time of application of F, that m has acquired
> velocity v, and being relativists, we immediately invoke the Lorentz
> transformation.

One should only "invoke" it if one knows *how* to do this...

For starters, F is a vector. Did it ever occur to you that the Lorentz
transformation is *different* for different components of a vectors?
(simplest example: the position vector)

> In this case we recognize that the local applied force
> F will be reduced in the moving frame carrying m, by the cosine of the
> angle equal to sqrt(1 - v2/c2).

Why should we recognize that? Are you talking about the vector F here,
about its magnitude, about its x component, or what?

And, BTW, where did you get this cosine and angle stuff from?
I know that boosts are analogous to rotations in some senses, but you
seem to do something different here.

> That is one facet of the LT, and the
> other is that clock rates on the moving frame will appear in
> attenuated form in the local frame.

Right, but irrelevant here.

> Forget all that katzenjammer about
> moving frames and lights and clocks, there is a cosine reduction.

Why?

>>>>>> F = gamma*F'
>
> Let me rewrite that:
> F' = F*sqrt(1 - v2/c2)

If you want...

> Therefore since F' is now pushing m we write
> Int F'dx

Err, if you want to look at the term "int F dx" in the primed frame, you
have *also* to consider length contraction, i.e. write
   int F' dx',
where dx' = dx/gamma.

> = Int F*sqrt(1-v2/c2)dx = Int mvdv

Note that on the left hand side, the integral runs from the point
where one started to apply the force to the point where one stopped
applying it. So on the right hand side, the integral has to run
from the velocity at the place where one started to apply the force
to the velocity at the place where one stopped applying the force.
Hence the integral on the right does *not* run from 0 to c, as you
claimed!!! (since one can't accelerate anything to the speed c!)

> But we need to get back to Int Fdx because Fdx is our measurable
> effort,

Why is int F' dx (or int F' dx') *not* measurable?

> and F' is an unhappy and unknown admixture of v and x:

Huh?

> E = Int Fdx = Int (mvdv)/sqrt(1-v2/c2)

How on earth does that follow logically from anything you wrote before?

And don't you notice that you are contradicting your own earlier
equation here??? You can't claim simultaneously that
   Int F dx = 0.5 m v^2
and
   Int F dx = Int (mvdv/sqrt(1-v^2/c^2)!

Int F dx can't be two different things at once!!! Isn't that totally
obvious?!?!?

> (Please note that it is easy at this point to come up with the term
> relativistic mass but such does not exist).

If you write E = m c^2, then relativistic mass *has* to exist - since
energy increases with velocity. Since c is constant, this can only be
explained by using a mass which depends on velocity.

>>>>>> Int Fdx = m*Int(vdv/(1-v2/c2)) from 0 to c
>>>>
>>>>What has the right side to do with F'?
>
> Have you seen me abandon the equal sign since eq. (1)?

No. But you wrote the equation for F' directly above, so it seemed
to me has if you somehow inserted the gamma factor for F' here on the
right hand side. And I didn't understand why you did do that.

>>>>And why on earth should one integrate over v on the right side, and why
>>>>from 0 to c???
>
> The left side whose variable of integration is x cannot be evaluated
> except to equate it always to E.

To be more precise, the left side is the *work* which is required for
acceleration - hence not the *energy* of the mass, but the *change* in
the energy of the mass when you accelerate it.

> Neither the function F(x) nor the extent x is of any consequence.
> The variable of integration on the right side is velocity and we are
> clearly constrained to make the only possible choices, 0 and c.

Err, no. The *clear* constraint is that the lower limit of integration
is the velocity *before* the force started, and the upper limit is
the velocity *after* the force ceased. That is *math*. You can't choose
the limits willy-nilly!

Your "choices" are not the "only possible" ones - they are simply nonsense!

> And so we find the result of our effort in this local frame, working
> through Lorentz to deliver force to the other frame (God knows how),

Forces can't be delivered from one frame to another. You aren't making
any sense.

> results in the exertion of twice what would be expected, namely mc2.

Why on earth *should* anyone expect 1/2 m c^2 ??????????

The formula E = m c^2 gives either the energy of a mass *at rest* (then
one would expect classically E = 0), or, if m is the relativistic mass,
the energy of a moving mass. In contrast, classically, E = 1/2 m v^2
gives the energy of a mass moving with velocity v. Why on earth should
have anyone inserted v=c in this formula in order to get the energy of
a mass *at rest*???

> snip
>
>>>>>>Dual Space has a solid logical reason for E = mc2. This is the
>>>>>>relativistic version-no model-just math.
>
>
> In Dual Space, mc2 is clearly shown to be a 2 step process, the first
> being 1/2 mc2

That first step makes no sense at all.

> and similarly the second step, giving a total of mc2.
> Dual Space makes everything right in physics including explanation of
> the Pioneer 10 anomaly.

*yawn* As if you were the only one who claims to be able to explain
that. As if standard physics had no explanations for that. (hint: that
you don't like these explanations doesn't change the fact that they exist)

> Dual Space is superior to both relativities, (while giving the same
> answers),

Oh? Please show me how one gets the precession of Mercury's perihelion
from "Dual Space".

> so the derivation from first principles is just a sop to the
> relativists, who I note have not put up a similarly simple derivation.

Einstein derived the formula in 1905. I gave a standard derivation
in another post in this thread.

Your so-called derivation above is full of absolutely trivial
mathematical errors, nonsensical statements and illogical conclusions.

>>>>Hint: Einstein derived this already in 1905. With a calculation which
>>>>made sense, in contrast to yours.
>
> He did it in 1905, so maybe you can derive it in 2005, as you haven't
> in 2004.

See another post in this thread.

>>>>>>Mr. Dual Space
>>>>>>(If you have something to say, write an equation.
>>>>>>If you have nothing to say, write an essay).
>>>>
>>>>Well, despite having written some equations, you had nothing sensible
>>>>to say.
>
> Look again, and pardon my telegraphic mode in the previous note.

Your elaborations in this post showed only more that you don't
understand maths.

> snip
>
>>>>Don't you think that it is fascinating that energy and mass are equivalent?
>
> Except that it is impossible to extract the equivalent energy,

Huh? Ever heard of electron-positron annihilation?

> and that energy is not actually affected by gravity.

Huh? Why do you think so?

> snip
> snip
>
>>>If you don't know why a force in
>>>THIS frame has a reduced effect in THAT frame, then despite everything
>>>you may have memorized, you still don't understand.
>>
>>True, a force in another frame *has* a reduced effect - but the formula
>>you used for that reduction was simply wrong. The right formulas for a
>>relativistic force for velocity vector parallel to the x-axis are
>>
>> F_x = m_o gamma^3 a_x
>> F_y = m_o gamma a_y
>> F_z = m_o gamma a_z
>
> Huh?

Thanks for showing that you *really* have no clue what you yourself
pretended to talk about.

What on earth is your problem with these equations???

Bye,
Bjoern

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