Re: .Re: Why all the fascination with E = mc^2 ??
From: Bjoern Feuerbacher (feuerbac_at_thphys.uni-heidelberg.de)
Date: 06/04/04
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Date: Fri, 04 Jun 2004 13:44:59 +0200
Leonard Pardin wrote:
> D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) wrote in message news:<c9muid$qk4$1@bunyip.cc.uq.edu.au>...
>
>>leoppard@MailAndNews.com (Leonard Pardin) writes:
>>
>><snip>
I notice that you *still* don't supply a reference to Poincare's work,
and even simply snipped the request for that.
You are beginning to look more and more dishonest, do you know that?
>>> Einstein seems to be saying just what you said: mass/energy in one
>>>frame appears to be reduced in another frame.
>>
>>No. Einstein took the classical Conservation of Energy is both frames.
>>He
>>had already calculated how the energy of radiation transforms between
>>frames, so he could use Conservation of Energy in both frames, and
>>calculate how the energy of the body in question changes as a consequence
>>of the radiation.
>>
>>Einstein starts off with a body of a certain mass (M, say).
>
>
> Where are you getting that from? Einstein doesn't even mention
> mass until the conclusion.
So what? That doesn't change the fact that the mass was there all the time.
Do you think something only appears as soon as one mentions it?
>>He uses the
>>frame in which the body is stationary. He then allows, in the frame in
>>which the body is stationary, the body to radiate radiation of energy L/2
>>in one direction, and radiation of energy L/2 in the opposite direction.
>>This allows the body to remain stationary after the radiation (the
>>momentum of one lot of radiation cancels the momentum from the other lot
>>of radiation, so that the momentum of the body remains unchanged, i.e. at
>>zero).
>
>
> Okay. Total mass of the stationary body in the STATIONARY SYSTEM
> is L (1/2 L one way, and 1/2 L the other). Fine so far.
>
>
>
>>Einstein then investigates the same set-up in another frame which
>>is moving relative to the original frame (at velocity -v). In this frame,
>>the body has mass M and is moving at velocity v, and then it radiates a
>>certain amount of energy in two directions
>
>
>
> HOLD IT!! Are we reading the same Einstein? That passage I
> posted is a quote from Einstein's 1905b paper. He says nothing about
> a second body in another frame.
Err, David didn't mention a second body, so what are you talking
about??? It is still the same body, only seen from another frame!
> Instead, Einstein says that the energy radiated by the body in the
> STATIONARY SYSTEM is measured from the MOVING SYSTEM with different
> results. There is only one body, and it is located in the STATIONARY
> SYSTEM. So we are going to get two measurements: one set as measured
> from the STATIONARY SYSTEM and one set as measured from the MOVING
> SYSTEM.
That's *exactly* what David said!
>>(with the energies being
>>calculated using the formula that Einstein had already derived), and then,
>>after radiation, the body continues to move with velocity v.
>
>
>
> WAIT A MINUTE! The body doesn't move at all in the STATIONARY
> SYSTEM.
Err, David was taling about the *other* frame here. He said that clearly
above!
> It just sits there radiating light. The motion of the body as
> measured from the MOVING SYSTEM is due only to the motion of the
> MOVING SYSTEM.
It's unclear if it makes any sense to say that the motion is "due to"
that. In the second system, the body is moving. If you want to
distinguish between "stationary" and "moving system" and imply that the
body in the second system only moves because the system is moving, then
you are implying that absolute speed exists.
>
>
>>Later in
>>Einstein's working, he additionally assumes that |v| is much smaller than
>>c (this is so that he can use Newton's approximation for the formula for
>>the kinetic energy).
>
>
>
> Lucky for Einstein Newton provided a classical formula for kinetic
> energy.
AFAIK, the formula for kinetic energy was *not* provided by Newton, but
was derived only much later.
And even if it had not been provided by anyone before - it is quite easy
to derive that formula. Einstein didn't need to know that formula before
- he simply could have derived it on his own.
>>In the second frame of reference, the kinetic energy of the body before
>>radiation is 1/2 M v^2 (plus a series of terms too small to worry about).
>>The total energy content of the body in the second frame of reference is
>>equal to the total energy content of the body in the first frame of
>>reference plus the kinetic energy.
>
>
>
> Einstein mentions no body in the "second frame" (the MOVING
> SYSTEM); there is only one body, and it's in the STATIONARY SYSTEM.
> Are you getting your facts from a differenc source?
Err, no. The body is in *both* frame. That it is *resting* in only one
frame does not mean that it does not exist in the other!
David is simply talking about observing the same body in two different
frames. Exactly as Einstein did.
>>We know how the total energy content
>>of the body changes in the first frame of reference: the total energy
>>content reduces by L. Einstein used his transformation formulae for
>>radiated energy to show that the total energy content of the body reduces
>>by L + L v^2/(2 c^2) plus terms too small to worry about.
>
>
>
> The measurement in the STATIONARY SYSTEM shows that the energy
> content of the body is reduced by L. Nothing unusual happens in the
> STATIONARY SYSTEM. Only the MOVING SYSTEM perceives a measurement that
> is less than the measurement taken in the STATIONARY SYSTEM.
What's your point?
>>Since the total
>>energy content of the body in the second frame after radiation is the
>>total energy content of the body in the first frame after radiation plus
>>the kinetic energy of the body in the second frame after radiation, then
>>the kinetic energy of the body after radiation drops by Lv^2/(2c^2) plus
>>terms too small to worry about.
>
>
>
> There is no second body in the second frame;
David didn't ever mention a second body, so why do you keep attacking
that straw man?
> the MOVING SYSTEM contains no body.
Nonsense. It contains the same body as the stationary frame!!!
> Energy measurements by the MOVING SYSTEM are
> converted to Kinetic energy for the body in the STATIONARY SYSTEM.
The body is in *both* frames.
> Kinetic energy as measured from the MOVING SYSTEM is less than the
> kinetic energy as measured from the STATIONARY SYSTEM containing the
> body. Measurements in the STATONARY SYSTEM where the body is located
> would show no loss of kinetic energy.
What's your point?
>>The kinetic energy in the second frame
>>after the radiation is 1/2 M v^2 - 1/2 L/c^2 v^2 plus terms too small to
>>worry about, so that the mass of the body after radiation is M - L/c^2,
>>and the body has lost a nett mass of L/c^2.
>
>
>
> Only as measured from the MOVING SYSTEM. The kinetic energy
> measurements taken by the MOVING SYSTEM are smaller than the same
> kinetic energy measurements taken by the STATIONARY SYSTEM.
What's your point?
> There are only three variables in the formula for kinetic energy
> (KE = 1/2 mv^2).
That formula is only an *approximation*. It is valid only for
*small* speeds. The *right* formula is KE = m c^2 (gamma - 1).
> The speed of light must remain constant, so v
> becomes c and must stay the same.
This sentence made no sense at all, sorry. What on earth was it supposed
to mean???
> Therefore, as measured from the
> MOVING SYSTEM, if KE is less, then m must be reduced--BUT ONLY FOR THE
> MOVING SYSTEM.
Err, no. For both systems. In the "stationary" system, the mass
reduction is different - but nevertheless it happens.
> There is an only an apparent reduction of mass and
> only as viewed from the MOVING SYSTEM.
Completely wrong.
>>The decrease of the mass of the body takes place in BOTH frames.
>
>
>
> What a jump! The MOVING SYSTEM sees a reduction in mass in a body
> located in the STATIONARY SYSTEM.
Yes.
> Those in the STATIONARY SYSTEM
> can't see it, can't measure it.
Wrong. They see a mass reduction, too. A different one, but not a zero one.
> It is happening right under their eyes
> but they can't measure it.
They can.
> To know what is happening, they have to
> contact the MOVING SYSTEM and get the measurements.
Wrong.
> But it must be
> happening because Einstein says that the reduced measurements taken in
> a different frame must apply to all frames.
Care to prove the opposite?
If you think that E=mc^2 does not hold, then how do you explain
electron-positron annihilation and production?
> This is the famous Einstein derivation of E = mc^2 ?? You are
> joking, aren't you?
No, you are simply not understanding it.
Bye,
Bjoern
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