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liza
When I started to learn special relativity I knew that in a given inertial reference frame I the mass m, the speed u and the momentum p of a given particle are related by
p=mu (1)
Special relativity tought me that in an inertial reference frame I' which moves with speed V relative to I in the positive direction of the overlapped axes OX(O'X') (1) should read
p'=m'u'. (2)
Less or more complicated derivations lead to the following transformation equations for momentum and mass (g(V)=1/sqrt(1-V^2/c^2)
p=g(V)p'(1+V/u') (3)
m=g(V)m'(1+Vu'/c^2) (4)
"Old fashioned" physicists say that (4) relates the "relativistic mass" m and the "relativistic mass m') of the same particle measured by observers from I and I' respectively. If the particle is at rest in I' (u'=0) observers of that frame measure its "rest mass m(0) and (4) leads to
m=g(V)m(0) (5)
Taking into account that c has the same magnitude in all inertial reference frames in relative motion, all transformation equations remain "relativistically correct" if we multiply both their sides by a power of c. Doing so with (4) we obtain nothing interesting because mc and m'c have no physical meaning (no tardyon can move with speed c). Multiplying both sides 0f (4) with c^2 leads to E=mc^2 and E'=m'c^2 respectively which has the physical dimensions of energy (4) becoming
E=g(V)E'(1+Vu'/c^2)=g(V)(E'+Vp') (6)
(3) becoming
p=g(V)(p'+VE'/c^2). (7)
Equation (5) leads to
E=g(V)E(0) (8) E(0) representing the rest energy.
Presenting the transformation equations as (6), (7) and(8) the "new generation" of relativists have nothing to comment.
Is the dispute between the generations solved? Are the frenzied debates motivated?
soft words and hard arguments please

Dallas
QUOTE (liza+Aug 24 2007, 04:55 PM)
When I started to learn special relativity I knew that in a given inertial reference frame I the mass m, the speed u and the momentum p of a given particle are related by
p=mu (1)
Special relativity tought me that in an inertial reference frame I' which moves with speed V relative to I in the positive direction of the overlapped axes OX(O'X') (1) should read
p'=m'u'. (2)
Less or more complicated derivations lead to the following transformation equations for momentum and mass (g(V)=1/sqrt(1-V^2/c^2)
p=g(V)p'(1+V/u') (3)
m=g(V)m'(1+Vu'/c^2) (4)
"Old fashioned" physicists say that (4) relates the "relativistic mass" m and the "relativistic mass m') of the same particle measured by observers from I and I' respectively. If the particle is at rest in I' (u'=0) observers of that frame measure its "rest mass m(0) and (4) leads to
m=g(V)m(0) (5)
Taking into account that c has the same magnitude in all inertial reference frames in relative motion, all transformation equations remain "relativistically correct" if we multiply both their sides by a power of c. Doing so with (4) we obtain nothing interesting because mc and m'c have no physical meaning (no tardyon can move with speed c). Multiplying both sides 0f (4) with c^2 leads to E=mc^2 and E'=m'c^2 respectively which has the physical dimensions of energy (4) becoming
E=g(V)E'(1+Vu'/c^2)=g(V)(E'+Vp') (6)
(3) becoming
p=g(V)(p'+VE'/c^2). (7)
Equation (5) leads to
E=g(V)E(0) (8) E(0) representing the rest energy.
Presenting the transformation equations as (6), (7) and(8) the "new generation" of relativists have nothing to comment.
Is the dispute between the generations solved? Are the frenzied debates motivated?
soft words and hard arguments please

Still nothing of any interest since you can find all these formulas (derived much cleanly) in any textbook. So, why are you posting this stuff? This is the forum for {b]new[/b] theories , not about your attepts to understand [/b]established[/b] theories. If you are trying to learn how momentum and energy transform in relativity by writing about it, then fine but there is nothing new in what you write. Plus, your derivations are not very good.
Sicewa
Oh really? According to the name its "Relativity, Quantum Mechanics, New Theories" not "New Theories". Besides, most "new theories" on this part of the forum are crackpot theories.
Dallas
QUOTE (liza+Aug 24 2007, 04:55 PM)
When I started to learn special relativity I knew that in a given inertial reference frame I the mass m, the speed u and the momentum p of a given particle are related by
p=mu (1)
Special relativity tought me that in an inertial reference frame I' which moves with speed V relative to I in the positive direction of the overlapped axes OX(O'X') (1) should read
p'=m'u'. (2)
Less or more complicated derivations lead to the following transformation equations for momentum and mass (g(V)=1/sqrt(1-V^2/c^2)
p=g(V)p'(1+V/u') (3)
m=g(V)m'(1+Vu'/c^2) (4)
"Old fashioned" physicists say that (4) relates the "relativistic mass" m and the "relativistic mass m') of the same particle measured by observers from I and I' respectively. If the particle is at rest in I' (u'=0) observers of that frame measure its "rest mass m(0) and (4) leads to
m=g(V)m(0) (5)
Taking into account that c has the same magnitude in all inertial reference frames in relative motion, all transformation equations remain "relativistically correct" if we multiply both their sides by a power of c. Doing so with (4) we obtain nothing interesting because mc and m'c have no physical meaning (no tardyon can move with speed c). Multiplying both sides 0f (4) with c^2 leads to E=mc^2 and E'=m'c^2 respectively which has the physical dimensions of energy (4) becoming
E=g(V)E'(1+Vu'/c^2)=g(V)(E'+Vp') (6)
(3) becoming
p=g(V)(p'+VE'/c^2). (7)
Equation (5) leads to
E=g(V)E(0) (8) E(0) representing the rest energy.
Presenting the transformation equations as (6), (7) and(8) the "new generation" of relativists have nothing to comment.
Is the dispute between the generations solved? Are the frenzied debates motivated?
soft words and hard arguments please

This is very simple, in a frame K the relativistic energy-momentum is :

E=gama(u).m_0 .c^2
p=gamma(u).m_0.u

(for dervation see Spacetime Physics by Taylor and Wheeler, for example)

m_0=rest mass, u=speed of the massive particle in frame K, E=energy,p=momentum, gamma(u)=1/sqrt(1-(u/c)^2)

In a different frame K':

E'=gama(u').m_0 .c^2
p'=gamma(u').m_0.u'

If frames K and K' are in relative motion with speed v, then, the above formulas result into the tranformation of the energy-momentum according to :

E=gamma(v)(E'+beta*p'c)
pc=gamma(v)(beta*E'+p'c)

where beta=v/c

No use of any "relativistic mass". There are no "frenzied debates". Period.
Farsight
The dispute is settled liza. Now we know what mass is. It's simply a measure of the amount of energy that is not moving with respect to you. Check out page 105 paragraph one of The Trouble with Physics for backup. But do note it takes a while for this sort of thing to sink in, and meanwhile all sorts of meaningless arguments bubble on.

QUOTE (Farsight+)

In pair production, a gamma photon of slightly more than 1022KeV is effectively broken over a nucleus to create an electron and a positron of 511KeV apiece. They’re like two half-wavelength “eddies” spinning off in opposite directions. Apart from a little wastage on the motion of the particles, most of the energy/momentum is stopped down from c and re-presented as inertia. We converted travelling kinetic energy or "relativistic mass" into non-travelling energy or "rest mass". If we simplify matters by discarding the positron and considering the electron to be at rest, we can look at those equations again and say:

E = hc/λ → mc˛   therefore   m ≡ h/λc

That seems to be saying the photon has mass. That sounds wrong, because nowadays we define mass to be rest mass. But we know that both matter and energy cause gravity. Einstein told us that with his mass/energy stress tensor. Energy has what’s called “active” gravitational mass. And since a photon has energy, it has gravitational mass too. A 511KeV photon contributes the same amount of gravitational attraction as an electron. What’s important is that energy causes gravity, not mass. Whilst a mass does cause gravity, that’s because of its energy content.

It can get a little confusing because there are lots of different ways of talking about mass. Whilst the accepted definition is rest mass, this is also  called “invariant mass” or “intrinsic mass” or “proper mass”. The term “relativistic mass” is really a measure of energy, which is why it applies to a massless photon. When you apply it to a cannonball travelling at 1000m/s, it’s a measure that combines the rest mass and the kinetic energy into total energy. There’s also “inertial mass”, which is a measure of how much force you need to apply to accelerate an object according to the equation F=ma. If you think of decelerating the cannonball using sheets of cardboard, it’s clear that this is the same thing as relativistic mass. There’s also “passive gravitational mass”, which is a a measure of how much an object is attracted by gravity. But it’s best not to get hung up on all these terms, because what’s important is this:

A photon has no rest mass, because rest mass is just rest energy, and the photon is never at rest. Because when it is at rest, it’s not a photon any more.
Dallas
QUOTE (Farsight+Aug 25 2007, 04:38 PM)
The dispute is settled liza. Now we know what mass is. It's simply a measure of the amount of energy that is not moving with respect to you. Check out page 105 paragraph one of The Trouble with Physics for backup. But do note it takes a while for this sort of thing to sink in, and meanwhile all sorts of meaningless arguments bubble on.

This is not what she asked. Of course, you couldn't read a line of her math , so you felt compelled to make up an irrelevant answer.
liza
QUOTE (Dallas+Aug 25 2007, 03:21 PM)
This is very simple, in a frame K the relativistic energy-momentum is :

E=gama(u).m_0 .c^2
p=gamma(u).m_0.u

(for dervation see Spacetime Physics by Taylor and Wheeler, for example)

m_0=rest mass, u=speed of the massive particle in frame K, E=energy,p=momentum, gamma(u)=1/sqrt(1-(u/c)^2)

In a different frame K':

E'=gama(u').m_0 .c^2
p'=gamma(u').m_0.u'

If frames K and K' are in relative motion with speed v, then, the above formulas result into the tranformation of the energy-momentum according to :

E=gamma(v)(E'+beta*p'c)
pc=gamma(v)(beta*E'+p'c)

where beta=v/c

No use of any "relativistic mass". There are no "frenzied debates". Period.

I think that we start teaching SR with p=mu and p'=m'u' trying to find out the relativistic extensions of m(m') and u(u').
In order to see how "fierce" the debates are have a look at Okun's papers.
Dallas
QUOTE (liza+Aug 26 2007, 06:43 AM)
I think that we start teaching SR with p=mu and p'=m'u'

"We"? Who is "we"? Taylor and Wheeler don't do that . You should check the book I recommended.

QUOTE

trying to find out the relativistic extensions of m(m') and u(u').

Hmm, I don't think so. "Relativistic mass" is not a useful concept, time to give up on it.

QUOTE (->
 QUOTE trying to find out the relativistic extensions of m(m') and u(u').

Hmm, I don't think so. "Relativistic mass" is not a useful concept, time to give up on it.

In order to see how "fierce" the debates are have a look at Okun's papers.

I know his papers, the debate is over. There is no pedagogical use for relativistic mass. Quite the opposite, it confuses things. Time to upgrade , "liza"
liza
thanks. that is your oppinion! there are probably others who think in an other way. As you know "De gustibus nihil disputandum". If you read with patience my thread you will see that I offer a solution for the pros and for the contras of the concept of relativistic mass and that is its essence.
Please use in our discussion a less superiority style. We all are participants on the Forum with equal opportunity.
Dallas
QUOTE (liza+Aug 26 2007, 05:29 PM)
thanks. that is your oppinion! there are probably others who think in an other way. As you know "De gustibus nihil disputandum". If you read with patience my thread you will see that I offer a solution for the pros and for the contras of the concept of relativistic mass and that is its essence.
Please use in our discussion a less superiority style. We all are participants on the Forum with equal opportunity.

Try to upgrade, Bernhard
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