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

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

Date: 4 Jun 2004 17:14:06 GMT

leoppard@MailAndNews.com (Leonard Pardin) writes:

>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>
 
>> >Let this body send out, in a direction making an angle f with the axis
>> >of x, plane waves of light, of energy 1 / 2L
>>
>> I.e. of energy 1/2 L = L/2.

> Don't you think this is a little too convenient: the body sends
>out radiation on only two sides? Would the result change if the body
>emitted radiation on four sides? Eight sides? Top and bottom?

No to all of the above. All that is necessary is that the body be
stationary in its initial rest frame after radiation. The best way to
ensure this is to pair off all the radiations in such a manner that
in each pair, the same amount is radiated in opposite directions,
relative to the rest frame.

>Correct
>me if I am wrong, but this looks like a set up to get to the desired
>answer.

Only because you are of a suspicious nature. Let us take the case where,
in the rest frame of the body, the body radiates radiation of energy L_1/2
in a direction at an angle of f_1 to the x axis, and simultaneously
radiates radiation of energy L_1/2 in the opposite direction. Meanwhile,
the body simultaneously radiates radiation of energy L_2/2 in a direction
at an angle of f_2 to the x axis, and radiation of energy L_2/2 in the
opposite direction. And so on, ending with the body simultaneously
radiating radiation of energy L_n/2 at an angle of f_n to the x axis, and
radiation of energy L_n/2 in the opposite direction. All this is relative
to the rest frame of the body. Because of the radiation, the energy of
the body reduces by L_1/2 + L_1/2 + ... + L_n/2 + L_n/2 = L_1 + ... + L_n,
relative to the rest frame of the body.

Now, take a second frame that is travelling at speed v in the positive x
direction relative to the rest frame of the body, then in the new frame
of reference, Einstein's transformation law yields that the nett radiated
energy relative to the new frame is:

            L_1/2 (1 - v/c cos f_1)/sqrt(1 - v^2/c^2)

          + L_1/2 (1 + v/c cos f_1)/sqrt(1 - v^2/c^2)

          + L_2/2 (1 - v/c cos f_2)/sqrt(1 - v^2/c^2)

          + L_2/2 (1 + v/c cos f_2)/sqrt(1 - v^2/c^2)

          + ...

          + L_n/2 (1 - v/c cos f_n)/sqrt(1 - v^2/c^2)

          + L_n/2 (1 + v/c cos f_n)/sqrt(1 - v^2/c^2)

        = L_1/sqrt(1 - v^2/c^2) + ... + L_n/sqrt(1 - v^2/c^2)

        = (L_1 + ... + L_n)/sqrt(1 - v^2/c^2).

Using Conservation of Energy in both frames of reference, it follows that
in the second frame, the kinetic energy of the body decreases by
(L_1 + ... + L_n) {1/sqrt(1 - v^2/c^2) - 1}. Since the kinetic energy
of the body decreases, relative to the second frame of reference, due to
radiation, by an amount of

        (L_1 + ... + L_n) {1/sqrt(1 - v^2/c^2) - 1},

but the speed of the body is unchanged (i.e. v), then it follows that the
mass must decrease. For v much smaller than c, the decrease in the
kinetic energy in the second frame of reference is L v^2/(2 c^2) plus
terms too small to worry about, where L = L_1 + ... + L_n. Using the
Newtonian formula for the kinetic energy, the mass decreases by L/c^2.
Recalling that in the rest frame of the body, the body radiates a
quantity of energy equal to L = L_1 + ... + L_n, we still find that a
mass of L/c^2 is converted to an energy of L and is radiated. Note that
this decrease in mass of L/c^2 is observed in ALL frames in which the body
is moving. Since the mass is an invariant, then the mass of the body
decreases by L/c^2 is all inertial frames of reference.

>> >measured relatively to
>> >(x, y, z), and simultaneously an equal quantity of light in the
>> >opposite direction. Meanwhile the body remains at rest with respect to
>> >the system (x, y, z).
>>
>> By Conservation of Momentum.
>>
>> >The principle of energy
>>
>> I.e. Conservation of Energy.
>>
>> >must apply to this
>> >process, and in fact (by the principle of relativity) with respect to
>> >both systems of co-ordinates. If we call the energy of the body after
>> >the emission of light E1 or H1 respectively, measured relatively to
>> >the system (x, y, z) or (x, h, z) respectively, then by employing the
>> >relation given above we obtain E0 = E1 + 1 / 2L + 1 / 2L,
>>
>> I.e. E_0 = E_1 + 1/2 L + 1/2 L = E_1 + L/2 + L/2 = E_1 + L.
>>
>> >H0 = H1 + 1
>> >/ 2L1 - v / c cos f / sqrt(1 - v2/c2) + 1 / 2L1 + v / c cos f / sqrt(1
>> >- v2/c2) = H1 + L / sqrt(1 - v2/c2)
>>
>> I.e. H_0 = H_1 + 1/2 L (1 - v/c cos f)/sqrt(1 - v^2/c^2)
>> + 1/2 L (1 + v/c cos f)/sqrt(1 - v^2/c^2)
>>
>> = H_1 + L/sqrt(1 - v^2/c^2).
>>
>> Here, I have put in the brackets that you should have put in, yourself,
>> if you had chosen to properly follow the rules of precedence, instead
>> of choosing to ignore them. Furthermore, the expressions that I wrote
>> accurately reflect what Einstein wrote. The expressions that you wrote
>> do not.

> Feel free to edit the math as much as you like.

> But I notice that the two sided energy emission of L has come in
>handy to reach a desired answer.

What on Earth is the matter with the fact that the energy of the radiation
with respect to the second frame of reference is proportional to L? A bit
of careful thought should tell you that the energy has to be proportional
to L. I don't understand what your point is. L can take any value
(within reason), and Einstein needed to calculate the energy of radiation
in the second frame in terms of L, an amount of energy which must be
proportional to L. In other words, you appear to be objecting to
something which is to be expected.

>No wonder some critics claim Einstein
>simply assumed the answer and then manipulated figures to get there.

And you seem to want to join them, without displaying the least idea of
what is actually happening in the argument.

Of course, you would have to learn the basics of relativity first, and
you haven't bothered with that yet.

>> >. By subtraction we obtain from
>> >these equations H0 - E0 - (H1 - E1) = L { 1 / sqrt(1 - v2/c2) - 1 }.
>> >The two differences of the form H - E occurring in this expression
>> >have simple physical significations. H and E are energy values of the
>> >same body referred to two systems of co-ordinates which are in motion
>> >relatively to each other, the body being at rest in one of the two
>> >systems (system (x, y, z)). Thus it is clear that the difference H - E
>> >can differ from the kinetic energy K of the body, with respect to the
>> >other system (x, h, z), only by an additive constant C, which depends
>> >on the choice of the arbitrary additive constants of the energies H
>> >and E. Thus we may place H0 - E0 = K0 + C, H1 - E1 = K1 + C, since C
>> >does not change during the emission of light. So we have K0 - K1 = L {
>> >1 / sqrt(1 - v2/c2) - 1 }. The kinetic energy of the body with respect
>> >to (x, h, z) diminishes as a result of the emission of light, and the
>> >amount of diminution is independent of the properties of the body.
>> >Moreover, the difference K0 - K1, like the kinetic energy of the
>> >electron, depends on the velocity.
>>
>> >Neglecting magnitudes of fourth and higher orders we may place K0 - K1
>> >= 1 / 2 L / c2 v2. From this equation it directly follows that:
>>
>> >If a body gives off the energy L in the form of radiation, its mass
>> >diminishes by L/c2.
>>
>> This because the velocity of the body in the second frame remains
>> unchanged (and we know this because the body remains stationary in the
>> first frame), and so the decrease in kinetic energy can only be accounted
>> for by a corresponding decrease in the mass of the body, by an amount of
>> L/c^2, which has been converted to an amount of energy equal to L, which
>> is radiated from the body.

> There's that convenient L again.

What is that crack supposed to mean???????

L is defined so that it is the quantity of energy radiated relative to
the rest frame of the body. When I want to refer to how much the
energy in the rest frame decreases, of course it is going to be L.
That is what Conservation of Energy demands. Or do you have a problem
with Conservation of Energy? You seem to be making a lot of objections to
results which must logically follow from the argument.

>But the key sentence is, "The
>kinetic energy of the body with respect to (x, h, z) diminishes as a
>result of the emission of light." Thus, only the the observer in the
>second frame perceives any change in the body over there in the
>stationary frame. To the observer in the stationary frame, the body
>appears to be stationary and the kinetic energy is just what it ought
>to be.

Yes, it remains at zero, since the kinetic energy of a stationary body
is zero.

>The measurements obtained in the second frame can have no
>effect on the measurements obtained in the first frame--except by
>Einsteinian Voodoo.

There is the sort of pejorative remark that can only be made by a person
who is too ignorant of the basics of relativity to understand what is
going on. Again, I say to you to learn the basics of relativity before
you embarrass yourself by making any more ignorant remarks.

>> >The fact that the energy withdrawn from the body
>> >becomes energy of radiation evidently makes no difference, so that we
>> >are led to the more general conclusion that: The mass of a body is a
>> >measure of its energy-content; if the energy changes by L, the mass
>> >changes in the same sense by L/(9.10^20), the energy being measured in
>> >ergs, and the mass in grammes.
>>
>> And so there is no loss of energy in either frame. If a stationary body
>> loses energy, then mass from the body is converted into energy. Note that
>> the total energy content is unchanged since the mass is just converted to
>> energy, and no energy disappears.
>>
>> David

> There is no loss of energy in either frame because there is no
>energy transfer to either frame.

An inertial frame is a coordinate system. You can't transfer energy to
or from an inertial frame of reference, since it is nonsensical to
talk about transferring energy to or from a coordinate system.

Again, you show ignorance about the nature of frames of reference.

>Each frame is measuring the same
>thing but from different perspectives.

So you do know more about it than you've been willing to let on. If you
already knew this, then why did you pretend that you thought that I was
writing about two bodies when I was discussing the situation relative to
the two frames?

>The difference isn't
>real--it's merely a difference in perception.

Yes, it is the same physical system, but it is observed differently by
different frames of reference. Furthermore, the situation as observed in
ANY inertial frame of reference must obey the laws of physics. This means
that every single interpretation by every single inertial frame of
reference must obey the laws of physics.

This means that a moving body which radiates energy in such a manner that
it maintains its velocity loses mass. This is because that is what
happens relative to the second frame of reference, and all physical laws
must hold relative to the second frame of reference.

And yet you are telling us that you don't believe that a stationary body
that radiates energy in such a manner that it remains stationary also
loses mass. So tell us what is so special about stationary bodies that
they do not lose mass when they radiate when moving bodies do.

David

        "Men were real men, women were real women, and
         small furry creatures from Alpha Centauri were
         real small furry creatures from Alpha Centauri."

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