| Topic: |
Science > Physics |
| User: |
"Mike" |
| Date: |
20 Jan 2005 10:53:28 AM |
| Object: |
E=mc^2 explained |
There has been a lot of confusion recently in these ng's and many false
statements by many, otherwise considered 'experts on Relativity" about
E=mc^2 and what m stands for in this equation.
Specifically, statements have been made that m stands for rest mass,
not relativistic mass. This statement is half-true and any half-true
statement in physics is also wrong. Here is the correct situation:
- If m is the rest mass in E=mc^2 when E is measured in rest frame then
E is the 'rest' energy as measured in that frame. This is deduced
directly from the mass-energy momentum relation:
E^2 = (mc^2)^2 + (pc)^2
when p = 0 then E = mc^2. But that refers to the mass-energy
relationship in a rest frame.
- If general, E = mc^2, where m is the 'relativistic mass', a term that
has now become outdates but used to differentiate between rest mass and
mass in motion in a FoR.
If we denote as m0 the rest mass, then m = gamma*m0
In this case, E is the energy measured in a moving frame and this
quantity is frame dependent.
Thus, one can understand that there is a notation issue but also some
serious misinterpretation of the equation.
E=mc^2 holds for both rest and relativistic mass, thus for any mass,
whether at rest of in motion.
A better way to clarify this is as follows:
To use E=mc^2 then:
1. in a reference frame at rest use m = rest mass
2. in a FoR where m moves with speed v use instead of m, m*gamma where
m is the m used in case 1 and which is an invariant quantity.
I hope this clarifies this confusion. Save that some (names not
mentioned) have been seriously exposed and lost permanently their
self-proclaimed statues of a 'Relativity expert' [sic] by claiming that
the m in E-mc^2 refers only to rest mass.
Mike
in this case, E is
.
|
|
| User: "Brett Aubrey" |
|
| Title: Re: E=mc^2 explained |
20 Jan 2005 04:03:06 PM |
|
|
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106240008.138590.107850@f14g2000cwb.googlegroups.com...
<snip>
I hope this clarifies this confusion. Save that some (names not
mentioned) have been seriously exposed and lost permanently their
self-proclaimed statues of a 'Relativity expert' [sic] by claiming that
the m in E-mc^2 refers only to rest mass.
Mike
And heck, no one likes to lose their statue! ;-) - Brett.
.
|
|
|
|
| User: "Franz Heymann" |
|
| Title: Re: E=mc^2 explained |
20 Jan 2005 04:43:46 PM |
|
|
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106240008.138590.107850@f14g2000cwb.googlegroups.com...
There has been a lot of confusion recently in these ng's and many
false
statements by many, otherwise considered 'experts on Relativity"
about
E=mc^2 and what m stands for in this equation.
Specifically, statements have been made that m stands for rest mass,
That is correct. The people to whom this matters seriously are the
particle physicists. They use m as the rest mass, or the invariant
mass, in problems involving relativistic kinematics.
not relativistic mass. This statement is half-true and any half-true
statement in physics is also wrong. Here is the correct situation:
- If m is the rest mass in E=mc^2 when E is measured in rest frame
then
E is the 'rest' energy as measured in that frame. This is deduced
directly from the mass-energy momentum relation:
E^2 = (mc^2)^2 + (pc)^2
when p = 0 then E = mc^2. But that refers to the mass-energy
relationship in a rest frame.
- If general, E = mc^2, where m is the 'relativistic mass',
Particle physicists don't usually have to talk about relativistic mass
at all, and in fact they try to avoid using the concept, as it leads
to confusion.
a term that
has now become outdates but used to differentiate between rest mass
and
mass in motion in a FoR.
If we denote as m0 the rest mass,
That is an outdated notation. Particle physicists usually just use m
then m = gamma*m0
Particle physicists prefer to get to the nub of the matter and talk
simply of
E = gamma mc^2
In this case, E is the energy measured in a moving frame and this
quantity is frame dependent.
No. E is the energy measured in *any* frame and it is not an
invariant.
Thus, one can understand that there is a notation issue but also
some
serious misinterpretation of the equation.
Not by particle physicists.
E=mc^2 holds for both rest and relativistic mass, thus for any mass,
whether at rest of in motion.
That is incorrect.
The following two are correct alternatives:
E = gamma mc^2
E^2 = p^2 c^2 +_ m^2 c^4
A better way to clarify this is as follows:
To use E=mc^2 then:
1. in a reference frame at rest use m = rest mass
2. in a FoR where m moves with speed v use instead of m, m*gamma
where
m is the m used in case 1 and which is an invariant quantity.
I hope this clarifies this confusion. Save that some (names not
mentioned) have been seriously exposed and lost permanently their
self-proclaimed statues of a 'Relativity expert' [sic] by claiming
that
the m in E-mc^2 refers only to rest mass.
It is quite correct to state that in terms of the nomenclature used
nowadays by particle physicists,
E = m c^2 refers to the energy of the system measured in the frame in
which its CM is at rest.
Sorry, Mike, but you have simply succeeded in muddying waters which
were perfectly clear before.
Franz
.
|
|
|
| User: "Mike" |
|
| Title: Re: E=mc^2 explained |
20 Jan 2005 05:56:28 PM |
|
|
Franz Heymann wrote:
[snip]
It is quite correct to state that in terms of the nomenclature used
nowadays by particle physicists,
E = m c^2 refers to the energy of the system measured in the frame in
which its CM is at rest.
Sorry, Mike, but you have simply succeeded in muddying waters which
were perfectly clear before.
Franz
Perfectly clear? The issue was wether E=mc^2 applies to any frame.
Please provide your answer. As a matter of fact, the m in E=mc^2 stands
for relativistic mass in general and when used as a rest mass refers to
rest energy only.
Mike
.
|
|
|
|
|
| User: "Dirk Van de moortel" |
|
| Title: Re: E=mc^2 explained |
21 Jan 2005 07:16:37 AM |
|
|
"Mike" <eleatis@yahoo.gr> wrote in message news:1106240008.138590.107850@f14g2000cwb.googlegroups.com...
There has been a lot of confusion recently in these ng's and many false
statements by many, otherwise considered 'experts on Relativity" about
E=mc^2 and what m stands for in this equation.
It is very simple.
If m is rest mass then
E = mc^2
is only valid in the rest frame of the particle, i.o.w. when
v =0 and p = 0.
Specifically, statements have been made that m stands for rest mass,
not relativistic mass. This statement is half-true and any half-true
statement in physics is also wrong. Here is the correct situation:
- If m is the rest mass in E=mc^2 when E is measured in rest frame then
E is the 'rest' energy as measured in that frame. This is deduced
directly from the mass-energy momentum relation:
E^2 = (mc^2)^2 + (pc)^2
when p = 0 then E = mc^2. But that refers to the mass-energy
relationship in a rest frame.
- If general, E = mc^2, where m is the 'relativistic mass', a term that
has now become outdates but used to differentiate between rest mass and
mass in motion in a FoR.
If we denote as m0 the rest mass, then m = gamma*m0
In this case, E is the energy measured in a moving frame and this
quantity is frame dependent.
Thus, one can understand that there is a notation issue but also some
serious misinterpretation of the equation.
E=mc^2 holds for both rest and relativistic mass, thus for any mass,
whether at rest of in motion.
A better way to clarify this is as follows:
To use E=mc^2 then:
1. in a reference frame at rest use m = rest mass
2. in a FoR where m moves with speed v use instead of m, m*gamma where
m is the m used in case 1 and which is an invariant quantity.
I hope this clarifies this confusion. Save that some (names not
mentioned) have been seriously exposed and lost permanently their
self-proclaimed statues of a 'Relativity expert' [sic] by claiming that
the m in E-mc^2 refers only to rest mass.
The only person that was confused was *you*.
You lost every shred of credibility you might ever earn with your
stupid and blunt intervention right here:
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
You could have seen from my equation
E = m c^2 / sqrt(1-v^2/c^2)
that in this context m was implicitly *defined* as the rest mass.
That was a giant blunder of yours and you are trying to dodge that
now.
An apology is what you owe some of us (names not mentioned).
Dirk Vdm
.
|
|
|
| User: "Mike" |
|
| Title: Re: E=mc^2 explained |
21 Jan 2005 11:42:42 AM |
|
|
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106240008.138590.107850@f14g2000cwb.googlegroups.com...
There has been a lot of confusion recently in these ng's and many
false
statements by many, otherwise considered 'experts on Relativity"
about
E=mc^2 and what m stands for in this equation.
It is very simple.
If m is rest mass then
E = mc^2
is only valid in the rest frame of the particle, i.o.w. when
v =0 and p = 0.
Specifically, statements have been made that m stands for rest
mass,
not relativistic mass. This statement is half-true and any
half-true
statement in physics is also wrong. Here is the correct situation:
- If m is the rest mass in E=mc^2 when E is measured in rest frame
then
E is the 'rest' energy as measured in that frame. This is deduced
directly from the mass-energy momentum relation:
E^2 = (mc^2)^2 + (pc)^2
when p = 0 then E = mc^2. But that refers to the mass-energy
relationship in a rest frame.
- If general, E = mc^2, where m is the 'relativistic mass', a term
that
has now become outdates but used to differentiate between rest mass
and
mass in motion in a FoR.
If we denote as m0 the rest mass, then m = gamma*m0
In this case, E is the energy measured in a moving frame and this
quantity is frame dependent.
Thus, one can understand that there is a notation issue but also
some
serious misinterpretation of the equation.
E=mc^2 holds for both rest and relativistic mass, thus for any
mass,
whether at rest of in motion.
A better way to clarify this is as follows:
To use E=mc^2 then:
1. in a reference frame at rest use m = rest mass
2. in a FoR where m moves with speed v use instead of m, m*gamma
where
m is the m used in case 1 and which is an invariant quantity.
I hope this clarifies this confusion. Save that some (names not
mentioned) have been seriously exposed and lost permanently their
self-proclaimed statues of a 'Relativity expert' [sic] by claiming
that
the m in E-mc^2 refers only to rest mass.
The only person that was confused was *you*.
You lost every shred of credibility you might ever earn with your
stupid and blunt intervention right here:
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
You could have seen from my equation
E = m c^2 / sqrt(1-v^2/c^2)
that in this context m was implicitly *defined* as the rest mass.
That was a giant blunder of yours and you are trying to dodge that
now.
An apology is what you owe some of us (names not mentioned).
Dirk Vdm
Hey square rooty, you missed the whole point once more. It takes
minimum brain recources to get the point and it seems you lack that.
The point is that some asserted E=mc^2 holds only in the rest frame. My
point is E=mc^2 holds in any frame. In the rest frame m is the rest
mass. In a frame where a body is moving m is the relativistic mass.
Do you have any objections? Do you know how to read and understand?
Do you agree E=mc^2 holds for both rest and relativistic mass?
Do you understand the use of symbols in math and physics?
Obviously you do not know how to post links square rooty.
Mike
.
|
|
|
| User: "Franz Heymann" |
|
| Title: Re: E=mc^2 explained |
26 Jan 2005 10:23:33 AM |
|
|
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106329362.422205.162580@z14g2000cwz.googlegroups.com...
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106240008.138590.107850@f14g2000cwb.googlegroups.com...
There has been a lot of confusion recently in these ng's and
many
false
statements by many, otherwise considered 'experts on Relativity"
about
E=mc^2 and what m stands for in this equation.
It is very simple.
If m is rest mass then
E = mc^2
is only valid in the rest frame of the particle, i.o.w. when
v =0 and p = 0.
Specifically, statements have been made that m stands for rest
mass,
not relativistic mass. This statement is half-true and any
half-true
statement in physics is also wrong. Here is the correct
situation:
- If m is the rest mass in E=mc^2 when E is measured in rest
frame
then
E is the 'rest' energy as measured in that frame. This is
deduced
directly from the mass-energy momentum relation:
E^2 = (mc^2)^2 + (pc)^2
when p = 0 then E = mc^2. But that refers to the mass-energy
relationship in a rest frame.
- If general, E = mc^2, where m is the 'relativistic mass', a
term
that
has now become outdates but used to differentiate between rest
mass
and
mass in motion in a FoR.
If we denote as m0 the rest mass, then m = gamma*m0
In this case, E is the energy measured in a moving frame and
this
quantity is frame dependent.
Thus, one can understand that there is a notation issue but also
some
serious misinterpretation of the equation.
E=mc^2 holds for both rest and relativistic mass, thus for any
mass,
whether at rest of in motion.
A better way to clarify this is as follows:
To use E=mc^2 then:
1. in a reference frame at rest use m = rest mass
2. in a FoR where m moves with speed v use instead of m, m*gamma
where
m is the m used in case 1 and which is an invariant quantity.
I hope this clarifies this confusion. Save that some (names not
mentioned) have been seriously exposed and lost permanently
their
self-proclaimed statues of a 'Relativity expert' [sic] by
claiming
that
the m in E-mc^2 refers only to rest mass.
The only person that was confused was *you*.
You lost every shred of credibility you might ever earn with your
stupid and blunt intervention right here:
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
You could have seen from my equation
E = m c^2 / sqrt(1-v^2/c^2)
that in this context m was implicitly *defined* as the rest mass.
That was a giant blunder of yours and you are trying to dodge that
now.
An apology is what you owe some of us (names not mentioned).
Dirk Vdm
Hey square rooty, you missed the whole point once more. It takes
minimum brain recources to get the point and it seems you lack that.
The point is that some asserted E=mc^2 holds only in the rest frame.
That is in fact in accordance with the meaning currently associated
with m by the people to whom it matters on a daily basis, namely the
particle physicists.
My
point is E=mc^2 holds in any frame. In the rest frame m is the rest
mass. In a frame where a body is moving m is the relativistic mass.
Your point is not in accord with current nomenclature.
Do you have any objections? Do you know how to read and understand?
Do you agree E=mc^2 holds for both rest and relativistic mass?
No.
Dirk's statement that E = m c^2 / sqrt(1-v^2/c^2)
is in fact the operative relationship between mass and energy.
Do you understand the use of symbols in math and physics?
Regrettably, it is you who are at least a generation out of touch with
present-day nomenclature.
[snip]
Franz
.
|
|
|
| User: "Mike" |
|
| Title: Re: E=mc^2 explained |
29 Jan 2005 06:01:34 AM |
|
|
Franz Heymann wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106329362.422205.162580@z14g2000cwz.googlegroups.com...
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106240008.138590.107850@f14g2000cwb.googlegroups.com...
There has been a lot of confusion recently in these ng's and
many
false
statements by many, otherwise considered 'experts on
Relativity"
about
E=mc^2 and what m stands for in this equation.
It is very simple.
If m is rest mass then
E = mc^2
is only valid in the rest frame of the particle, i.o.w. when
v =0 and p = 0.
Specifically, statements have been made that m stands for rest
mass,
not relativistic mass. This statement is half-true and any
half-true
statement in physics is also wrong. Here is the correct
situation:
- If m is the rest mass in E=mc^2 when E is measured in rest
frame
then
E is the 'rest' energy as measured in that frame. This is
deduced
directly from the mass-energy momentum relation:
E^2 = (mc^2)^2 + (pc)^2
when p = 0 then E = mc^2. But that refers to the mass-energy
relationship in a rest frame.
- If general, E = mc^2, where m is the 'relativistic mass', a
term
that
has now become outdates but used to differentiate between rest
mass
and
mass in motion in a FoR.
If we denote as m0 the rest mass, then m = gamma*m0
In this case, E is the energy measured in a moving frame and
this
quantity is frame dependent.
Thus, one can understand that there is a notation issue but
also
some
serious misinterpretation of the equation.
E=mc^2 holds for both rest and relativistic mass, thus for any
mass,
whether at rest of in motion.
A better way to clarify this is as follows:
To use E=mc^2 then:
1. in a reference frame at rest use m = rest mass
2. in a FoR where m moves with speed v use instead of m,
m*gamma
where
m is the m used in case 1 and which is an invariant quantity.
I hope this clarifies this confusion. Save that some (names not
mentioned) have been seriously exposed and lost permanently
their
self-proclaimed statues of a 'Relativity expert' [sic] by
claiming
that
the m in E-mc^2 refers only to rest mass.
The only person that was confused was *you*.
You lost every shred of credibility you might ever earn with your
stupid and blunt intervention right here:
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
You could have seen from my equation
E = m c^2 / sqrt(1-v^2/c^2)
that in this context m was implicitly *defined* as the rest mass.
That was a giant blunder of yours and you are trying to dodge
that
now.
An apology is what you owe some of us (names not mentioned).
Dirk Vdm
Hey square rooty, you missed the whole point once more. It takes
minimum brain recources to get the point and it seems you lack
that.
The point is that some asserted E=mc^2 holds only in the rest
frame.
That is in fact in accordance with the meaning currently associated
with m by the people to whom it matters on a daily basis, namely the
particle physicists.
My
point is E=mc^2 holds in any frame. In the rest frame m is the rest
mass. In a frame where a body is moving m is the relativistic mass.
Your point is not in accord with current nomenclature.
Do you have any objections? Do you know how to read and understand?
Do you agree E=mc^2 holds for both rest and relativistic mass?
No.
Dirk's statement that E = m c^2 / sqrt(1-v^2/c^2)
is in fact the operative relationship between mass and energy.
Do you understand the use of symbols in math and physics?
Regrettably, it is you who are at least a generation out of touch
with
present-day nomenclature.
[snip]
Franz
E=mc^2 holds in all frames.
Now, if you do not understand this I'm not suprised. You have proven
you are an idiot already.
Mike
.
|
|
|
| User: "Franz Heymann" |
|
| Title: Re: E=mc^2 explained |
29 Jan 2005 10:26:30 AM |
|
|
"Mike" <eleatis@yahoo.gr> wrote in message
news:1107000094.469430.20860@z14g2000cwz.googlegroups.com...
E=mc^2 holds in all frames.
Now, if you do not understand this I'm not suprised. You have proven
you are an idiot already.
I gave you a detailed explanation of why you are wrong in terms of the
current usage of "m". If you don't accept it, that's your problem.
Franz
.
|
|
|
| User: "Mike" |
|
| Title: Re: E=mc^2 explained |
30 Jan 2005 11:08:49 AM |
|
|
Franz Heymann wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1107000094.469430.20860@z14g2000cwz.googlegroups.com...
E=mc^2 holds in all frames.
Now, if you do not understand this I'm not suprised. You have
proven
you are an idiot already.
I gave you a detailed explanation of why you are wrong in terms of
the
current usage of "m". If you don't accept it, that's your problem.
Franz
When you will learn the difference between what a law asserts and what
symbols assert, your life will become easier and maybe, I say maybe,
you will start understanding physics. In another words, I can say:
buko = muco*futo^2
as long as I make clear that
buko means energy
muco means relativistic mass
futo means speed of light
then I'm ok. You and Dirt confuse symbol definitions with expressions
of law. I repeat for your final convenience:
IN E=mc^2, m is relativistic mass. Now, if you want to redefine the
symbols that's fine as long you define them. The original claim was
that the equation holds only for rest mass and denotes the rest energy.
That's crap. Dirk came along to pound me once more without paying
attention to what I was saying. I repeat once more
IN E=mc^2, m is the relativistic mass. Are you that stupid? I'm
surprised the IW is so low in the ng's.
Mike
.
|
|
|
| User: "Franz Heymann" |
|
| Title: Re: E=mc^2 explained |
30 Jan 2005 03:12:11 PM |
|
|
"Mike" <eleatis@yahoo.gr> wrote in message
news:1107104929.834644.107330@c13g2000cwb.googlegroups.com...
Franz Heymann wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1107000094.469430.20860@z14g2000cwz.googlegroups.com...
E=mc^2 holds in all frames.
Now, if you do not understand this I'm not suprised. You have
proven
you are an idiot already.
I gave you a detailed explanation of why you are wrong in terms of
the
current usage of "m". If you don't accept it, that's your
problem.
Franz
When you will learn the difference between what a law asserts and
what
symbols assert, your life will become easier and maybe, I say maybe,
you will start understanding physics. In another words, I can say:
buko = muco*futo^2
as long as I make clear that
buko means energy
muco means relativistic mass
futo means speed of light
then I'm ok.
You will not only be OK, you will also be talking to yourself only.
Half wits don't usually realise their own shortcomings.
You and Dirt confuse symbol definitions with expressions
of law. I repeat for your final convenience:
IN E=mc^2, m is relativistic mass.
That is not the usual terminology, so if you insist on using it, there
will be a finite probability that you will be misunderstood by someone
oir other.
Now, if you want to redefine the
symbols that's fine as long you define them. The original claim was
that the equation holds only for rest mass and denotes the rest
energy.
That's crap.
No. It is not crap in the current terminology of the folk actually
working in the field.
You are simply making a silly redefinition of the term m.
Dirk came along to pound me once more without paying
attention to what I was saying. I repeat once more
IN E=mc^2, m is the relativistic mass. Are you that stupid? I'm
surprised the IW is so low in the ng's.
On the contrary, the folk who deal in the practicalities of
relativistic kinematics on a daily basis do not nowadays use m as a
symbol signifying relativistic mass. As a matter of fact, the term
"relativistic mass" itself, as I have frequently reminded you, is not
used very often at all, your protestations notwithstanding.
Franz
.
|
|
|
| User: "Y.Porat" |
|
| Title: Re: E=mc^2 explained |
31 Jan 2005 10:43:25 AM |
|
|
did you already realised that there i sjust one kind of mass??
only the rest mass!
mass never swells never iflates
it is th emore energy that is needed to add acceleration to it.
if you didnt got it you will do in the nest 100 years.
Y.Porat
--------------------------
.
|
|
|
|
|
| User: "Dirk Van de moortel" |
|
| Title: Re: E=mc^2 explained |
30 Jan 2005 12:25:02 PM |
|
|
"Mike" <eleatis@yahoo.gr> wrote in message news:1107104929.834644.107330@c13g2000cwb.googlegroups.com...
Franz Heymann wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1107000094.469430.20860@z14g2000cwz.googlegroups.com...
E=mc^2 holds in all frames.
Now, if you do not understand this I'm not suprised. You have
proven you are an idiot already.
I gave you a detailed explanation of why you are wrong in terms of
the current usage of "m". If you don't accept it, that's your problem.
Franz
When you will learn the difference between what a law asserts and what
symbols assert, your life will become easier and maybe, I say maybe,
you will start understanding physics. In another words, I can say:
buko = muco*futo^2
as long as I make clear that
buko means energy
muco means relativistic mass
futo means speed of light
then I'm ok. You and Dirt confuse symbol definitions with expressions
of law. I repeat for your final convenience:
IN E=mc^2, m is relativistic mass. Now, if you want to redefine the
symbols that's fine as long you define them. The original claim was
that the equation holds only for rest mass and denotes the rest energy.
That's crap. Dirk came along to pound me once more without paying
attention to what I was saying. I repeat once more
IN E=mc^2, m is the relativistic mass. Are you that stupid? I'm
surprised the IW is so low in the ng's.
In the context of the equation
E = m c^2 / sqrt(1-v^2/c^2)
that was already put in place *before* you decided to jump
in and fart your silly misplaced remark
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com ,
the "m" was already defined as rest mass.
You can continue squealing like a throated pig, but there is
nothing you can do about it.
Dig yourself in a bit deeper, pig, and squeal ;-)
Dirk Vdm
.
|
|
|
|
|
|
|
| User: "tj Frazir" |
|
| Title: Re: E=mc^2 energy equal mass it displaces |
28 Jan 2005 05:44:20 PM |
|
|
Mass is equal the energy it displaces.
.
|
|
|
|
|
| User: "Dirk Van de moortel" |
|
| Title: Re: E=mc^2 explained |
21 Jan 2005 01:21:58 PM |
|
|
"Mike" <eleatis@yahoo.gr> wrote in message news:1106329362.422205.162580@z14g2000cwz.googlegroups.com...
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106240008.138590.107850@f14g2000cwb.googlegroups.com...
[snip]
I hope this clarifies this confusion. Save that some (names not
mentioned) have been seriously exposed and lost permanently their
self-proclaimed statues of a 'Relativity expert' [sic] by claiming
that the m in E-mc^2 refers only to rest mass.
The only person that was confused was *you*.
You lost every shred of credibility you might ever earn with your
stupid and blunt intervention right here:
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
You could have seen from my equation
E = m c^2 / sqrt(1-v^2/c^2)
that in this context m was implicitly *defined* as the rest mass.
That was a giant blunder of yours and you are trying to dodge that
now.
An apology is what you owe some of us (names not mentioned).
Dirk Vdm
Hey square rooty, you missed the whole point once more.
The point that you are trying to make, is to dodge the documented
*fact* that you were rude and wrong:
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
You can scream and squeal like the disgusting Undeniable Bill Smith
pig that we all know you are, and there is *nothing* you can do
about it :-)
Dirk Vdm
.
|
|
|
| User: "Mike" |
|
| Title: Re: E=mc^2 explained |
21 Jan 2005 02:32:51 PM |
|
|
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106329362.422205.162580@z14g2000cwz.googlegroups.com...
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106240008.138590.107850@f14g2000cwb.googlegroups.com...
[snip]
I hope this clarifies this confusion. Save that some (names not
mentioned) have been seriously exposed and lost permanently
their
self-proclaimed statues of a 'Relativity expert' [sic] by
claiming
that the m in E-mc^2 refers only to rest mass.
The only person that was confused was *you*.
You lost every shred of credibility you might ever earn with your
stupid and blunt intervention right here:
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
You could have seen from my equation
E = m c^2 / sqrt(1-v^2/c^2)
that in this context m was implicitly *defined* as the rest mass.
That was a giant blunder of yours and you are trying to dodge
that
now.
An apology is what you owe some of us (names not mentioned).
Dirk Vdm
Hey square rooty, you missed the whole point once more.
The point that you are trying to make, is to dodge the documented
*fact* that you were rude and wrong:
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
You can scream and squeal like the disgusting Undeniable Bill Smith
pig that we all know you are, and there is *nothing* you can do
about it :-)
Dirk Vdm
So go ahead idiot square rooty and tell me what's wrong with my
statement that:
" m stands for rest mass in your energy-momentum equation.
In E = mc^2, m is the relativistic mass."
Do you know any physics besides being an average psycho?
"...The term m in the equation E=mc^2 does not represent rest-mass; it
represents relativistic mass, which is the inertial mass of a body when
it is in a state of motion relative to an inertial frame... "
quote taken from:
http://plato.stanford.edu/entries/equivME/
Idiot quare rooty.
Mike
.
|
|
|
| User: "Dirk Van de moortel" |
|
| Title: Re: E=mc^2 explained |
21 Jan 2005 05:34:27 PM |
|
|
"Mike" <eleatis@yahoo.gr> wrote in message news:1106339571.024046.320570@c13g2000cwb.googlegroups.com...
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106329362.422205.162580@z14g2000cwz.googlegroups.com...
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106240008.138590.107850@f14g2000cwb.googlegroups.com...
[snip]
I hope this clarifies this confusion. Save that some (names not
mentioned) have been seriously exposed and lost permanently
their self-proclaimed statues of a 'Relativity expert' [sic] by
claiming that the m in E-mc^2 refers only to rest mass.
The only person that was confused was *you*.
You lost every shred of credibility you might ever earn with your
stupid and blunt intervention right here:
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
You could have seen from my equation
E = m c^2 / sqrt(1-v^2/c^2)
that in this context m was implicitly *defined* as the rest mass.
That was a giant blunder of yours and you are trying to dodge
that now.
An apology is what you owe some of us (names not mentioned).
Dirk Vdm
Hey square rooty, you missed the whole point once more.
The point that you are trying to make, is to dodge the documented
*fact* that you were rude and wrong:
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
You can scream and squeal like the disgusting Undeniable Bill Smith
pig that we all know you are, and there is *nothing* you can do
about it :-)
Dirk Vdm
So go ahead idiot square rooty and tell me what's wrong with my
statement that:
" m stands for rest mass in your energy-momentum equation.
There is nothing wrong with that, as I already told you.
There is something severely wrong with your attitude
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
Don't project your own confusion and frustration upon others.
It shows.
Dirk Vdm
.
|
|
|
| User: "Mike" |
|
| Title: Re: E=mc^2 explained |
22 Jan 2005 05:14:41 AM |
|
|
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106339571.024046.320570@c13g2000cwb.googlegroups.com...
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106329362.422205.162580@z14g2000cwz.googlegroups.com...
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106240008.138590.107850@f14g2000cwb.googlegroups.com...
[snip]
I hope this clarifies this confusion. Save that some (names
not
mentioned) have been seriously exposed and lost permanently
their self-proclaimed statues of a 'Relativity expert'
[sic] by
claiming that the m in E-mc^2 refers only to rest mass.
The only person that was confused was *you*.
You lost every shred of credibility you might ever earn with
your
stupid and blunt intervention right here:
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
You could have seen from my equation
E = m c^2 / sqrt(1-v^2/c^2)
that in this context m was implicitly *defined* as the rest
mass.
That was a giant blunder of yours and you are trying to dodge
that now.
An apology is what you owe some of us (names not mentioned).
Dirk Vdm
Hey square rooty, you missed the whole point once more.
The point that you are trying to make, is to dodge the documented
*fact* that you were rude and wrong:
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
You can scream and squeal like the disgusting Undeniable Bill
Smith
pig that we all know you are, and there is *nothing* you can do
about it :-)
Dirk Vdm
So go ahead idiot square rooty and tell me what's wrong with my
statement that:
" m stands for rest mass in your energy-momentum equation.
There is nothing wrong with that, as I already told you.
There is something severely wrong with your attitude
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
Don't project your own confusion and frustration upon others.
It shows.
Dirk Vdm
Look who is talking about attitude! The Dirt of sci.physics who
maintains a site to thraw dirt on people.
You are seriously psychotic quare rooty. All your friends who insisted
m in E=mc^2 refers to rest mass only have disappeared and you are left
alone with your only weapon you have, that is throwing dirt around.
Get lost crank. You and some others here have one and only one goal: to
preserve your ill-concept of physics.
Time after time all of your wrong conceptions about Relativity and
physics in general have been rebuted by several people who at this
point don't pay attention to you any longer and laugh at you every time
you posy something. This is exactly what I am going to do from now on.
Every time I see you posting something I will laugh at the smallest
square-root brain I have ever come across in these ng's.
Mike
.
|
|
|
| User: "Dirk Van de moortel" |
|
| Title: Re: E=mc^2 explained |
22 Jan 2005 05:49:04 AM |
|
|
"Mike" <eleatis@yahoo.gr> wrote in message news:1106392481.624299.245590@z14g2000cwz.googlegroups.com...
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106339571.024046.320570@c13g2000cwb.googlegroups.com...
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106329362.422205.162580@z14g2000cwz.googlegroups.com...
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1106240008.138590.107850@f14g2000cwb.googlegroups.com...
[snip]
I hope this clarifies this confusion. Save that some (names not
mentioned) have been seriously exposed and lost permanently
their self-proclaimed statues of a 'Relativity expert' [sic] by
claiming that the m in E-mc^2 refers only to rest mass.
The only person that was confused was *you*.
You lost every shred of credibility you might ever earn with
your stupid and blunt intervention right here:
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
You could have seen from my equation
E = m c^2 / sqrt(1-v^2/c^2)
that in this context m was implicitly *defined* as the rest mass.
That was a giant blunder of yours and you are trying to dodge
that now.
An apology is what you owe some of us (names not mentioned).
Dirk Vdm
Hey square rooty, you missed the whole point once more.
The point that you are trying to make, is to dodge the documented
*fact* that you were rude and wrong:
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
You can scream and squeal like the disgusting Undeniable Bill
Smith pig that we all know you are, and there is *nothing* you
can do about it :-)
Dirk Vdm
So go ahead idiot square rooty and tell me what's wrong with my
statement that:
" m stands for rest mass in your energy-momentum equation.
There is nothing wrong with that, as I already told you.
There is something severely wrong with your attitude
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
Don't project your own confusion and frustration upon others.
It shows.
Dirk Vdm
Look who is talking about attitude! The Dirt of sci.physics who
maintains a site to thraw dirt on people.
The site collects the dirt people throw at themselves without
being aware of it. The only thing I do, is provide a title and
some colour.
You are seriously psychotic quare rooty. All your friends who insisted
m in E=mc^2 refers to rest mass only have disappeared and you are left
alone with your only weapon you have, that is throwing dirt around.
No, I'm rubbing your nose in your own dirt:
http://groups.google.co.uk/groups?&threadm=1105783543.600180.173130@z14g2000cwz.googlegroups.com
Most puppies and kitties don't like it, but you seem to like
the taste and the smell of it :-)
Dirk Vdm
.
|
|
|
|
| User: "Sam Wormley" |
|
| Title: Re: E=mc^2 explained |
22 Jan 2005 07:42:54 AM |
|
|
Mike wrote:
Look who is talking about attitude! The Dirt of sci.physics who
maintains a site to thaw[throw] dirt on people.
You are seriously psychotic [s]quare rotoy. All your friends who insisted
m in E=mc^2 refers to rest mass only have disappeared and you are left
alone with your only weapon you have, that is throwing dirt around.
Get lost crank. You and some others here have one and only one goal: to
preserve your ill-concept of physics.
Time after time all of your wrong conceptions about Relativity and
physics in general have been rebut[t]ed by several people who at this
point don't pay attention to you any longer and laugh at you every time
you posy something. This is exactly what I am going to do from now on.
Every time I see you posting something I will laugh at the smallest
square-root brain I have ever come across in these ng's.
Mike
Why is it that the untutored (like Mike) are the one that get pissed
at the rest for pointing out correct physics? One would think they
should be more humble and grateful.
Dirk's recording of immortal fumbles is educational and enlightening.
.
|
|
|
| User: "Mike" |
|
| Title: Re: E=mc^2 explained |
29 Jan 2005 05:57:09 AM |
|
|
Sam Wormley wrote:
Mike wrote:
Look who is talking about attitude! The Dirt of sci.physics who
maintains a site to thaw[throw] dirt on people.
You are seriously psychotic [s]quare rotoy. All your friends who
insisted
m in E=mc^2 refers to rest mass only have disappeared and you are
left
alone with your only weapon you have, that is throwing dirt around.
Get lost crank. You and some others here have one and only one
goal: to
preserve your ill-concept of physics.
Time after time all of your wrong conceptions about Relativity and
physics in general have been rebut[t]ed by several people who at
this
point don't pay attention to you any longer and laugh at you every
time
you posy something. This is exactly what I am going to do from now
on.
Every time I see you posting something I will laugh at the smallest
square-root brain I have ever come across in these ng's.
Mike
Why is it that the untutored (like Mike) are the one that get
pissed
at the rest for pointing out correct physics? One would think they
should be more humble and grateful.
Dirk's recording of immortal fumbles is educational and
enlightening.
shame on you copy and paste impecile
Mike
.
|
|
|
|
| User: "" |
|
| Title: Re: E=mc^2 explained |
29 Jan 2005 08:07:09 AM |
|
|
Sam Wormly wrote:
Why is it that the untutored (like Mike) are the one that=AD get
pissed
at the rest for pointing out correct physics? One would t=ADhink they
should be more humble and grateful.
Dirk's recording of immortal fumbles is educational and
e=ADnlightening.
Perhaps because your "correct physics" is subject to fashion instead of
science.
More enlightening might be: how many of his own fumbles did Dirk
record?
Harald
.
|
|
|
|
|
| User: "" |
|
| Title: Re: E=mc^2 explained |
29 Jan 2005 07:56:47 AM |
|
|
Mike wrote:
Get lost crank. You and some others here have one and only o=ADne goal:
to
preserve your ill-concept of physics.
Time after time all of your wrong conceptions about Relativi=ADty and
physics in general have been rebuted by several people who a=ADt this
point don't pay attention to you any longer and laugh at you=AD every
time
you posy something. This is exactly what I am going to do fr=ADom now
on.
Every time I see you posting something I will laugh at the s=ADmallest
square-root brain I have ever come across in these ng's.
Right (except for the exaggeration)! That seems to pretty well describe
Dirk's agenda. He starts to sound like Androcles - except for the math
of course.
Harald
.
|
|
|
| User: "Franz Heymann" |
|
| Title: Re: E=mc^2 explained |
29 Jan 2005 10:23:19 AM |
|
|
<harald.vanlintel@epfl.ch> wrote in message
news:1107007007.840650.144410@z14g2000cwz.googlegroups.com...
Mike wrote:
Get lost crank. You and some others here have one and only one goal:
to
preserve your ill-concept of physics.
Time after time all of your wrong conceptions about Relativity and
physics in general have been rebuted by several people who at this
point don't pay attention to you any longer and laugh at you every
time
you posy something. This is exactly what I am going to do from now
on.
Every time I see you posting something I will laugh at the smallest
square-root brain I have ever come across in these ng's.
Right (except for the exaggeration)! That seems to pretty well
describe
Dirk's agenda. He starts to sound like Androcles - except for the math
of course.
Where are the headers and ths arrtibution marks?
I see you have buggered the thread so much that even my own current
attribution marks don't appear.
Fix your newsreader before contaminating every thresd in the ng.
Franz
Harald
.
|
|
|
| User: "" |
|
| Title: Re: E=mc^2 explained |
29 Jan 2005 05:50:33 PM |
|
|
Comment:
At home the old Google groups program can't find the links so that
users are condemned to use the new Google Beta crap which requires
improvisation as it doesn't even _have_ any quoted text in the reply
window...
Harald
.
|
|
|
|
|
| User: "" |
|
| Title: Re: E=mc^2 explained |
29 Jan 2005 05:55:12 PM |
|
|
Hmm, I see that the new Google Beta Crap even messes up existing text
and adds spaces like a madman...
Harald
.
|
|
|
|
|
|
|
|
|
| User: "" |
|
| Title: Re: E=mc^2 explained |
22 Jan 2005 03:36:51 PM |
|
|
But Mike, Dirl never asserted that m was rest mass in E=mc^2. He said
that m is rest mass in E = m c^2 / sqrt(1-v^2/c^2).
The extra stuff is Lorentz' gamma - the correction factor for fast
objects.
That's exctly the same as your article in stanford said here:
http://plato.stanford.edu/entries/equivME/e=m.jpg
It immeditly followed the part you quoted,
Now, what do REAL scientists always say when they're wrong?
"I apologize"?
No! That's optional.
They say "Oops, I was wrong".
Nobody needs to apologize for being wrong if they change their mind
when they're proven wrong.
Whether you apologize for your aggressive insults to Dirk is up to you.
=[ d
.
|
|
|
| User: "Dirk Van de moortel" |
|
| Title: Re: E=mc^2 explained |
23 Jan 2005 05:58:51 AM |
|
|
<dt041054@yahoo.com> wrote in message news:1106429811.159492.236680@f14g2000cwb.googlegroups.com...
But Mike, Dirl never asserted that m was rest mass in E=mc^2. He said
that m is rest mass in E = m c^2 / sqrt(1-v^2/c^2).
In which case E = mc^2 is valid when v = 0, i.o.w, when E is
measured in the rest frame of the particle. So you can safely
assume that I do assert that m is rest mass in E = mc^2,
provided you take the context into account.
The extra stuff is Lorentz' gamma - the correction factor for fast
objects.
That's exctly the same as your article in stanford said here:
http://plato.stanford.edu/entries/equivME/e=m.jpg
It immeditly followed the part you quoted,
It does not matter anyway, since "relativistic mass" is an obsolete
and virtually useless concept
http://hermes.physics.adelaide.edu.au/~dkoks/Faq/Relativity/SR/mass.html
And, as can be seen, it tends to confuse some people ;-)
Now, what do REAL scientists always say when they're wrong?
"Bill Smith" aka "Undeniable" aka "Eleatis" aka "Mike"... a scientist?
Good one :-)))
"I apologize"?
No! That's optional.
They say "Oops, I was wrong".
Nobody needs to apologize for being wrong if they change their mind
when they're proven wrong.
Whether you apologize for your aggressive insults to Dirk is up to you.
=[ d
Since the only people in the world who could really insult me are the
members of my closest family, he should not (and technically even
could not) apologize to me :-)
Anyway, these people don't even understand the concept to begin
with...
Cheers,
Dirk Vdm
.
|
|
|
| User: "Ken S. Tucker" |
|
| Title: Re: E=mc^2 explained |
26 Jan 2005 03:23:32 PM |
|
|
Dirk Van de moortel wrote:
....
Since the only people in the world who could really insult me are the
members of my closest family, he should not (and technically even
could not) apologize to me :-)
Anyway, these people don't even understand the concept to begin
with...
Probably too busy swinging in trees and having banana
fights, to explain anything to you.
Cheers,
Dirk Vdm
.
|
|
|
|
| User: "Mike" |
|
| Title: Re: E=mc^2 explained |
29 Jan 2005 05:58:53 AM |
|
|
idiot dirt fan de matel
Mike
.
|
|
|
|
|
|
|
| User: "Y.Porat" |
|
| Title: Re: E=mc^2 explained |
21 Jan 2005 10:07:04 AM |
|
|
Mr (pompous) Van Der Shmatte
do you know how 'relativistic mass' is created??
--------------
Y.Porat
------------------------
.
|
|
|
|
|
| User: "" |
|
| Title: Re: E=mc^2 explained |
28 Jan 2005 04:06:12 PM |
|
|
Mike wrote:
- If general, E = mc^2, where m is the 'relativistic mass', a
term that has now become outdate[d] but used to differentiate
between rest mass and mass in motion in a FoR.
Not quite. See, for instance, the discussion in s.p.r. regarding
the general classification in the Poincare/Galilei/Euclidean group.
The generator E that appears in the Lie algebra should really be
interpreted as the KINETIC energy, rather than the total energy. This
is what provides a unified 'hyperbolic'/'parabolic'/'elliptical'
description of Poincare/Galilei/Euclidean and helps show where the
transition takes place between them.
Repeating what's discussed there, in general the Lie algebra has
generators (L_a,K_a,P_a: a=1,2,3), T (we'll use T instead of E for
clarification), with the commutators:
[L_a,L_b] = e^c_{ab} L_c
[L_a,K_b] = e^c_{ab} K_c
[L_a,P_b] = e^c_{ab} P_c
[L_a,T] = 0
[K_a,K_b] = -A e^c_{ab} L_c
[K_a,P_b] = delta_{ab} M
[K_a,T] = P_a
[P_a,P_b] = 0
[P_a,T] = 0
where M = m + AT is the total mass, m the (sole) parameter of the
central extension of the Lie algebra; e^1_{23}=e^2_{31}=e^3_{12}=1;
e^c_{ab}=-e^c_{ba}, e^c_{ab} = 0 else; delta is the Kronecker delta.
For Galilei, A = 0; for Poincare' A > 0 (= 1/c^2), for Euclid A < 0 (=
-1/r^2 for a suitable value of r).
The invariants (besides M - AT) are then
P^2 - 2MT + AT^2; |ML + KxP|^2 - A(L.P)^2.
In all these considerations, it's the TOTAL MASS and KINETIC ENERGY
that enter naturally into play. The total energy E = M/A is only there
for non-zero A and does not transition over in any meaningful way to
all 3 cases.
Second (and closely related to the generalized Wigner classification),
the relativistic mass enters naturally into the Hamiltonian and
Lagrangian which have the relations
mH = ML; H = pv - L; p = Mv
where M = m + H/c^2. Solving these for the cases c -> infinity, m = 0
and the general cases gives you a large part of the Wigner
classification (including mass 0 particles EVEN IN the Galilean case!)
.
|
|
|
|
|
Related Articles |
|
|
