| Topic: |
Science > Physics |
| User: |
"Leonard Pardin" |
| Date: |
31 May 2004 10:30:40 AM |
| Object: |
Why all the fascination with E = mc^2 ?? |
Even though I slept through most of my highschool physics classes, I
still learned that the formula for Kinetic energy is E = 1/2mv^2. I
also remember that Newton found out long ago that for every action
there is an equal and opposite reaction.
So, if any object of mass m is ejected at a velocity v, the kinetic
energy of the object is 1/2 mv^2. The reaction force is exactly the
same, so double the formula and get E = mv^2. That's the total energy
expended by the system.
For example, imagine a spring loaded cannon floating in space. The
spring ejects a cannonball of mass m. We measure the speed of the
cannonball and get v. The energy of the cannonball is 1/2 mv^2.
We don't have to measure the reacton force on the cannon because
we know it is equal to the energy of the cannonball, opposite but
equal. The result: the total energy for the system (cannon and
cannonball) is represented by the formula E = mv^2.
If the object being ejected is an electron, we know the speed is
the speed of light, represented by c. So the formula for the total
energy of the atom is the mass of the ejected electrons times the
speed of light squared: E = mc^2.
The French physicist Poincare explained this long ago. So why is
everyone so fascinated by this simple formula?
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| User: "César Sirvent" |
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| Title: Re: Why all the fascination with E = mc^2 ?? |
31 May 2004 01:19:47 PM |
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"Leonard Pardin" <leoppard@MailAndNews.com> escribió en el mensaje
news:d746a243.0405310730.55a4ca94@posting.google.com...
Even though I slept through most of my highschool physics classes, I
still learned that the formula for Kinetic energy is E = 1/2mv^2. I
also remember that Newton found out long ago that for every action
there is an equal and opposite reaction.
So, if any object of mass m is ejected at a velocity v, the kinetic
energy of the object is 1/2 mv^2. The reaction force is exactly the
same, so double the formula and get E = mv^2. That's the total energy
expended by the system.
No, no. It is E=mc^2 and it is very easy to demonstrate.
We use state-of-the-art dimensional analysis.
E = 1/2 m v^2
dimensionally, we can drop the adimensional 1/2 and take dimensions
[E] = [m][v]^2
now the most important speed in relativity is the speed of light c, so this
v is c and
[E] = [m][c]^2
Now we drop the bracket and we get
E = mc^2
Other elaborated techniques involve iterative proccess of construction:
E = m a^2 -- fist step
E = m b^2 -- second step closer to the solution
E = mc^2 -- third and final step
:-)
Cesar
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| User: "Old Man" |
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| Title: Re: Why all the fascination with E = mc^2 ?? |
31 May 2004 06:38:11 PM |
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"Leonard Pardin" <leoppard@MailAndNews.com> wrote in message
news:d746a243.0405310730.55a4ca94@posting.google.com...
Even though I slept through most of my highschool physics classes, I
still learned that the formula for Kinetic energy is E = 1/2mv^2. I
also remember that Newton found out long ago that for every action
there is an equal and opposite reaction.
So, if any object of mass m is ejected at a velocity v, the kinetic
energy of the object is 1/2 mv^2. The reaction force is exactly the
same, so double the formula and get E = mv^2. That's the total energy
expended by the system.
Idiot. Momentum is conserved. Starting from rest, after decay
the two particles have equal and opposite momentum. The only
way that the two decay products can have equal kinetic energy
is if they have equal masses.
The kinetic energy of the decay products Q = T1 +T2 has
to come from the system's rest mass. Thus,
- Q = delta_M*c2, where M = m1 + m2 + Q / c^2
For example, imagine a spring loaded cannon floating in space. The
spring ejects a cannonball of mass m. We measure the speed of the
cannonball and get v. The energy of the cannonball is 1/2 mv^2.
We don't have to measure the reacton force on the cannon because
we know it is equal to the energy of the cannonball, opposite but
equal. The result: the total energy for the system (cannon and
cannonball) is represented by the formula E = mv^2.
No ! Momentum is conserved. Initial momentum,
p0 = M*v0 = 0 because v0 = 0
and M = m1 + m2 + Q / c^2
After release, p1 = m1*v1 = - p2 = m2*(-v2)
and (1 / 2) [ m1*v1^2 + m2*v2^2 ] = Q
this is assuming v1 << c and v2 << c.
If the object being ejected is an electron, we know the speed is
the speed of light, represented by c. So the formula for the total
energy of the atom is the mass of the ejected electrons times the
speed of light squared: E = mc^2
Electrons can't travel at c. The total energy of an electron
traveling at speed v1 is
E1 = m1*c^2 / sqrt [ 1 - ( v1/ c )^2 ] ~ m1*v1^2 / 2 if v1 << c
The French physicist Poincare explained this long ago. So why is
everyone so fascinated by this simple formula?
Leonard isn't aware of what he doesn't know. Leonard
operates under delusions of competence. Leonard hasn't
gotten a single thing right here. Even without relativistic
corrections, Leonard has gotten it all wrong. Leonard is
ignorant because he was stupid enough to sleep through
his HS physics classes. Leonard is a physics moron.
[Old Man]
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| User: "David McAnally" |
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| Title: Re: Why all the fascination with E = mc^2 ?? |
31 May 2004 11:18:55 AM |
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(Leonard Pardin) writes:
Even though I slept through most of my highschool physics classes,
As others have pointed out, it is obvious from what you have written below
that you are unfamiliar with even the most basic physics.
I
still learned that the formula for Kinetic energy is E = 1/2mv^2. I
also remember that Newton found out long ago that for every action
there is an equal and opposite reaction.
In Newton's Third Law of Motion, action and reaction are FORCES.
So, if any object of mass m is ejected at a velocity v, the kinetic
energy of the object is 1/2 mv^2. The reaction force is exactly the
same, so double the formula and get E = mv^2.
This is nonsensical. Energy is not force. Try again.
That's the total energy
expended by the system.
No, it isn't.
For example, imagine a spring loaded cannon floating in space. The
spring ejects a cannonball of mass m. We measure the speed of the
cannonball and get v. The energy of the cannonball is 1/2 mv^2.
We don't have to measure the reacton force on the cannon because
we know it is equal to the energy of the cannonball,
Energy and force are not the same thing. Try again.
opposite but
equal. The result: the total energy for the system (cannon and
cannonball) is represented by the formula E = mv^2.
Wrong.
If the object being ejected is an electron, we know the speed is
the speed of light,
Completely false. In non-quantum relativity, the speed of the electron is
never equal to c, since the electron is a material body of nonzero mass.
represented by c. So the formula for the total
energy of the atom is the mass of the ejected electrons times the
speed of light squared: E = mc^2.
Completely invalid derivation.
The French physicist Poincare explained this long ago.
But he didn't do it in the same invalid manner that you did.
So why is
everyone so fascinated by this simple formula?
Perhaps, if you actually learnt the real derivation of the formula, the
derivation that is actually valid, then you might begin to understand.
David
"But I'm always true to you, darlin', in my fashion,
Yes, I'm always true to you, darlin', in my way."
-- Lois Lane
-----
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| User: "John Schoenfeld" |
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| Title: Re: Why all the fascination with E = mc^2 ?? |
01 Jun 2004 11:02:35 AM |
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(Leonard Pardin) wrote in message news:<d746a243.0405310730.55a4ca94@posting.google.com>...
Even though I slept through most of my highschool physics classes, I
still learned that the formula for Kinetic energy is E = 1/2mv^2. I
also remember that Newton found out long ago that for every action
there is an equal and opposite reaction.
So, if any object of mass m is ejected at a velocity v, the kinetic
energy of the object is 1/2 mv^2. The reaction force is exactly the
same, so double the formula and get E = mv^2. That's the total energy
expended by the system.
For example, imagine a spring loaded cannon floating in space. The
spring ejects a cannonball of mass m. We measure the speed of the
cannonball and get v. The energy of the cannonball is 1/2 mv^2.
We don't have to measure the reacton force on the cannon because
we know it is equal to the energy of the cannonball, opposite but
equal. The result: the total energy for the system (cannon and
cannonball) is represented by the formula E = mv^2.
If the object being ejected is an electron, we know the speed is
the speed of light, represented by c. So the formula for the total
energy of the atom is the mass of the ejected electrons times the
speed of light squared: E = mc^2.
The French physicist Poincare explained this long ago. So why is
everyone so fascinated by this simple formula?
Most likely, E != mc^2.
http://en.wikipedia.org/wiki/Dynamic_theory_of_gravity
http://foia.fbi.gov/tesla/tesla1.pdf
http://foia.fbi.gov/tesla/tesla2.pdf
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| User: "Dave Langers" |
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| Title: Re: Why all the fascination with E = mc^2 ?? |
31 May 2004 10:36:28 AM |
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Even though I slept through most of my highschool physics classes
That's quite obvious from the rest of your post.
The reaction force is exactly the same, so double the formula
?
We don't have to measure the reacton force on the cannon because
we know it is equal to the energy of the cannonball, opposite but
equal.
?
If the object being ejected is an electron, we know the speed is
the speed of light
?
The French physicist Poincare explained this long ago
Hmmm, but certainly not in the way that you did.
Please go back to your highschool physics books and learn the difference
between force and energy, to start with. Or even better, try doing
politics; there you can probably get away with such gross
misinterpretations and misquotes.
--
M.vr.gr.
Dave
("d-dot-langers-at-wxs-dot-nl")
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| User: "Sam Wormley" |
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| Title: Re: Why all the fascination with E = mc^2 ?? |
31 May 2004 11:30:43 AM |
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Expand your horizons
http://www.amazon.com/exec/obidos/tg/detail/-/0802713521/102-5767734-5434556?v=glance
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| User: "Robert J. Kolker" |
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| Title: Re: Why all the fascination with E = mc^2 ?? |
31 May 2004 11:57:07 AM |
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Sam Wormley wrote:
Expand your horizons
http://www.amazon.com/exec/obidos/tg/detail/-/0802713521/102-5767734-5434556?v=glance
A nifty book. The story of Voltaire and Emile du Chatalet was worth the
price of the book.
Bob Kolker
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| User: "Gregory L. Hansen" |
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| Title: Re: Why all the fascination with E = mc^2 ?? |
31 May 2004 02:33:13 PM |
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In article <d746a243.0405310730.55a4ca94@posting.google.com>,
Leonard Pardin <leoppard@MailAndNews.com> wrote:
Even though I slept through most of my highschool physics classes, I
still learned that the formula for Kinetic energy is E = 1/2mv^2. I
also remember that Newton found out long ago that for every action
there is an equal and opposite reaction.
So, if any object of mass m is ejected at a velocity v, the kinetic
energy of the object is 1/2 mv^2. The reaction
Momentum, not force. The force can vary, e.g. if a cannon fires a shell
and recoils on springs.
force is exactly the
same, so double the formula and get E = mv^2. That's the total energy
expended by the system.
For example, imagine a spring loaded cannon floating in space. The
spring ejects a cannonball of mass m. We measure the speed of the
cannonball and get v. The energy of the cannonball is 1/2 mv^2.
Momenta are equal and opposite. Not kinetic energies.
E = 1/2 m_m v_m^2 + 1/2 m_c v_c^2
m_m v_m = -m_c v_c
where m_m and v_m are the mass and velocity of the marble, m_c and v_c are
the mass and velocity of the cannon. Or, since p_c = -p_m, we could write
it as
E = 1/2 p_m^2 (1/m_c + 1/m_m)
We don't have to measure the reacton force on the cannon because
we know it is equal to the energy of the cannonball, opposite but
equal. The result: the total energy for the system (cannon and
cannonball) is represented by the formula E = mv^2.
If the object being ejected is an electron, we know the speed is
the speed of light, represented by c.
It's never the speed of light. But it can be any speed slower than that.
So the formula for the total
energy of the atom is the mass of the ejected electrons times the
speed of light squared: E = mc^2.
No, it's not.
The French physicist Poincare explained this long ago. So why is
everyone so fascinated by this simple formula?
E = sqrt(m^2 c^4 + p^2 c^2)
with p = m/srt(1-v^2/c^2) for massive objects. (m=0 for photons)
But define a relativistic mass m_r = m/sqrt(1-v^2/c^2) and the above
reduces to
E = m_r c^2
What is now called relativistic mass was derived in the 19th century in
studies of electromagnetism. When you stick it into Newton's 2nd law,
F = dp/dt = d(mv)/dt
it turns out that, if you try to relate it to F=ma, the value of m will
depend on whether the force is applied in the direction of motion or
perpendicular to it; the so-called longitudinal mass and transverse mass.
This reflects the Lorentz-covariance that is present in Maxwell's
equations. And electromagnetism was a great inspiration to Einstein.
All of that came from electrodynamics. The first thing noteworthy about
Einstein's work is he generalized Lorentz covariance to all phenomena, not
just electromagnetic. Second is that he didn't ascribe it to suspicious
behavior of forces and masses, but to the geometry of space and time.
E=mc^2 can be derived from those fundamental assumptions, and the
assumption that mechanics must reduce to Newtonian mechanics when you go
slow.
It's the fundamental change in our view of space and time that makes
Einstein and the theory of relativity so popular. Not to mention the
theory's spectacular success, and Einstein's fashionable hair-do.
Everyone is fascinated with E=mc^2 because it's associated with all of
that, and because it's a formula that the average person can easily
recognize.
--
"A good plan executed right now is far better than a perfect plan
executed next week."
-Gen. George S. Patton
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| User: "" |
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| Title: Re: Why all the fascination with E = mc^2 ?? |
31 May 2004 05:33:19 PM |
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In article <d746a243.0405310730.55a4ca94@posting.google.com>, (Leonard Pardin) writes:
Even though I slept through most of my highschool physics classes,
That's pretty apparent from what follows.
I still learned that the formula for Kinetic energy is E = 1/2mv^2. I
also remember that Newton found out long ago that for every action
there is an equal and opposite reaction.
So, if any object of mass m is ejected at a velocity v, the kinetic
energy of the object is 1/2 mv^2. The reaction force is exactly the
same, so double the formula and get E = mv^2. That's the total energy
expended by the system.
Confusion of energy and force. Different things.
For example, imagine a spring loaded cannon floating in space. The
spring ejects a cannonball of mass m. We measure the speed of the
cannonball and get v. The energy of the cannonball is 1/2 mv^2.
We don't have to measure the reacton force on the cannon because
we know it is equal to the energy of the cannonball,
Bad, really bad.
opposite but
equal. The result: the total energy for the system (cannon and
cannonball) is represented by the formula E = mv^2.
Not at all.
If the object being ejected is an electron, we know the speed is
the speed of light, represented by c.
No, we most certainly know that it isn't c.
So the formula for the total
energy of the atom is the mass of the ejected electrons times the
speed of light squared: E = mc^2.
The French physicist Poincare explained this long ago. So why is
everyone so fascinated by this simple formula?
Well, if the above is supposed to be a joke, then fine. If it is in
earnest, too bad.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "The Ghost In The Machine" |
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| Title: Re: Why all the fascination with E = mc^2 ?? |
31 May 2004 07:00:02 PM |
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In sci.physics, Leonard Pardin
<leoppard@MailAndNews.com>
wrote
on 31 May 2004 08:30:40 -0700
<d746a243.0405310730.55a4ca94@posting.google.com>:
Even though I slept through most of my highschool physics classes, I
still learned that the formula for Kinetic energy is E = 1/2mv^2.
For Newtonian physics, yes.
You might try the following.
In E = mc^2, m is the *relativistic mass* = m_0 / sqrt(1 - v^2/c^2).
If one expands sqrt(1 - v^2/c^2) using an infinite binomial, one
gets
E = m_0 c^2 / sqrt(1-v^2/c^2)
= m_0*c^2 + 1/2*m_0*v^2 + 3*m_0/(8*c^2)*v^4 + 5*m_0/(16*c^4)*v^6 + ...
Note in particular the second term.
I also remember that Newton found out long ago that for every action
there is an equal and opposite reaction.
Yes, that's the Third Law.
So, if any object of mass m is ejected at a velocity v, the kinetic
energy of the object is 1/2 mv^2. The reaction force is exactly the
same, so double the formula and get E = mv^2. That's the total energy
expended by the system.
Reaction force? Wrong units.
Look up "rocket equation", if you want to go farther along these
lines. It's a pretty form when expressed differentially;
if one ejects a mass dM_r from the rocket with velocity v_e,
conservation of momentum requires that
dM_r/dt * v_e = -M_r * dv_r/dt
(This is using Newtonian concepts, so it's not quite right.
There is a relativistic variant.)
If one separates, one gets
dM_r / M_r = -dv_r / v_e
Solving it gets the more conventional form:
v_r = v_0 + v_e * log(M_i / M_f)
assuming constant ejecta velocity and fuel flow.
For example, imagine a spring loaded cannon floating in space. The
spring ejects a cannonball of mass m. We measure the speed of the
cannonball and get v. The energy of the cannonball is 1/2 mv^2.
We don't have to measure the reacton force on the cannon because
we know it is equal to the energy of the cannonball, opposite but
equal. The result: the total energy for the system (cannon and
cannonball) is represented by the formula E = mv^2.
Doesn't quite work that way. *Momentum* is conserved, yes
(and energy too, just not in that fashion).
If one has a 1 kg cannonball fired from a 100 kg cannon, and
the 1 kg is moving at 100 m/s, the cannon will be moving in
the opposing direction at 1 m/s.
cannonball energy: 1/2 * 1 * (100)^2 = 5000 J
cannon energy: 1/2 * 100 * (1)^2 = 50 J
The total amount of energy consumed would be 5050 J.
If we model the gunpowder as a spring with this energy,
and the acceleration of cannon and ball occurs within 1.01
m and the spring is sitting exactly at the mass-center of
the two (presumably deep in the barrel), the spring will
still be sitting there after it releases its energy.
The cannonball will accelerate through 1 m with an
acceleration a = 2d/t^2 (the exact value of t depends on
the spring's stiffness) while the cannon takes the other
1 cm.
At the end of the time the spring has 5050J less energy; it's
all been transferred to the cannon and cannonball.
If the object being ejected is an electron, we know the speed is
the speed of light, represented by c. So the formula for the total
energy of the atom is the mass of the ejected electrons times the
speed of light squared: E = mc^2.
Not quite that simple either. Electrons simply cannot be
ejected at the speed of light in vacuum (although they can
equal or exceed the speed of light in a fluid, generating
Cherenkov radiation). Only objects with zero rest mass
(such as photons) can travel at lightspeed.
It's possible (and routine) to give an electron more energy
than 1/2 * m_e * c^2 = 1/2 * 9.10938188*10^-31 * (2.99792458 * 10^8)^2
= 4.09355207 * 10^-14 J = 0.2555 MeV.
The French physicist Poincare explained this long ago. So why is
everyone so fascinated by this simple formula?
Well, for starters, because it works. :-)
--
#191,
It's still legal to go .sigless.
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| User: "Uncle Al" |
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| Title: Re: Why all the fascination with E = mc^2 ?? |
31 May 2004 11:02:23 AM |
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Leonard Pardin wrote:
Even though I slept through most of my highschool physics classes,
Idiot alert.
I
still learned that the formula for Kinetic energy is E = 1/2mv^2. I
also remember that Newton found out long ago that for every action
there is an equal and opposite reaction.
So, if any object of mass m is ejected at a velocity v, the kinetic
energy of the object is 1/2 mv^2. The reaction force is exactly the
same, so double the formula and get E = mv^2. That's the total energy
expended by the system.
1) Look up the derivation of E=mc^2 from four-momentum.
2) Look up the derivation of E=mc^2 from four-momentum.
3) Look up the derivation of E=mc^2 from four-momentum.
4) Look up the derivation of E=mc^2 from four-momentum.
5) Look up the derivation of E=mc^2 from four-momentum.
...
100) Look up the derivation of E=mc^2 from four-momentum.
[snip crap]
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
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| User: "John Anderson" |
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| Title: Re: Why all the fascination with E = mc^2 ?? |
01 Jun 2004 09:00:32 PM |
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Leonard Pardin wrote:
Even though I slept through most of my highschool physics classes, I
still learned that the formula for Kinetic energy is E = 1/2mv^2.
That's the Newtonian expression that's only valid if v << c.
I
also remember that Newton found out long ago that for every action
there is an equal and opposite reaction.
So, if any object of mass m is ejected at a velocity v, the kinetic
energy of the object is 1/2 mv^2. The reaction force is exactly the
same, so double the formula and get E = mv^2. That's the total energy
expended by the system.
That just reveals that you don't understand Newtonian physics let
alonerelativity.
For example, imagine a spring loaded cannon floating in space. The
spring ejects a cannonball of mass m. We measure the speed of the
cannonball and get v. The energy of the cannonball is 1/2 mv^2.
We don't have to measure the reacton force on the cannon because
we know it is equal to the energy of the cannonball, opposite but
equal. The result: the total energy for the system (cannon and
cannonball) is represented by the formula E = mv^2.
Again, that just shows that you don't understand Newtonian physics.
If the object being ejected is an electron, we know the speed is
the speed of light, represented by c
Electrons don't travel at the speed of light. Yet more physics that
youdon't understand.
So the formula for the total
energy of the atom is the mass of the ejected electrons times the
speed of light squared: E = mc^2.
The French physicist Poincare explained this long ago. So why is
everyone so fascinated by this simple formula?
He didn't explain what you just "did". Go learn something before
youintellectually drool in this group again. It would be a lot less
embarassing for you.
John Anderson
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| User: "Robert J. Kolker" |
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| Title: Re: Why all the fascination with E = mc^2 ?? |
31 May 2004 10:49:45 AM |
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Leonard Pardin wrote:
The French physicist Poincare explained this long ago. So why is
everyone so fascinated by this simple formula?
It is easier to memorize than Maxwell's Equations.
Bob Kolker
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