That's actually from my post. I messed up the quotation though.

You assume that particles with momentum must have rest mass.

momentum does not equal mass times velocity is all cases. Substituting p=mv in the energy equation is incorrect.

p.s. if X = AB, then X^2 is not equal to A^2*B^2.
X^2 = (AB)^2

This is a classic mistake. It's like saying the sqrt(A+B) = sqrt(A) + sqrt(B)

x=ab then
x^2 = (A*B)^2
= (A*B)*(A*B)
= A*A*B*B
= (A*A)*(B*B)
= (A^2)*(B^2)

(
As compared to:
x = a + b
x² = (a + b)²
= (a + b).(a + b)
= a² + ab + ba + b²
= a² + 2ab + b²
)

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Oops! You're both right.

As a substitution, p^2=m^2v^2 does work

I was thinking more along the line of pzkpfw's post.

sorry !

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What´s the physical definition of "particle"?

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If we talk about "particles": what´s momentum, if not "mass times velocity"?

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http://en.wikipedia.org/wiki/Particle_physics

See section title Subatomic Particles, which goes on to say that the word 'particle' is a misnomer.

That´s really new math to me. Which specific math segment are you thinking of?

OMG! I can prove you wrong: A=0, B=0 -> sqrt(0+0)=sqrt(0)+sqrt(0)

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From your wikipedia-link: Strictly speaking, the term particle is a misnomer because the dynamics of particle physics are governed by quantum mechanics. As such, they exhibit wave-particle duality, displaying particle-like behavior under certain experimental conditions and wave-like behavior in others

I don´t see any definition of "particle" there
I´m still thinking of "particle = having mass" (as one of several properties). For me, a photon is not a particle, although it may "display particle-like behavior"

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If a photon is not a particle, what is it?

By definition, the light quanta is the photon. A packet, one thing. A particle.

I have no problem considering many things as particles. A graviton is a particle. Does it have mass? The higgs Boson is a particle, and we say the Higgs mechanism is what gives other particles mass.

Mass is not a pre-req.

technically?

p^2=E^2/c^2+m^2c^4

The case if the photon is the classic example. The photon has 0 rest mass but it does carry momentum. If you measure the momentum of a photon, then you come up with

p=E/c

This may seem like a circular argument, but remember that the photon momentum is measured. It is a known value, and to the limits of observation, the second term in the full momentum equation must be zero, leading to a zero rest mass.

I came up with

p^2 = E^2/c^2 - m^2c^2

edit: can momentum be imaginary?

Do me a huge favor. Please explain why an electron's spin angular momentum is twice as effective at creating magnetic fields as its orbital angular momentum.

We can imagine the electron as having momentum from its orbital velocity and spin, but this is not a correct picture for the electron(a single particle).

Both particle and momentum are misnomers. So, we might want to look at momentum's meaning before giving the definition of particle a hard time.

relativistic momentum: p=mv/sqrt(1-(v^2/c^2))

however photon mass is zero, and its velocity is c. So both numerator and denominator approach 0. To find the photon momentum, simply use the relationship discussed earlier in this thread.

and no, momentum is not imaginary. There is such a law called the conservation of momentum - using this we can build solar sails, etc.

I still don´t know the definition of "particle". Is there any?

That´s interesting. Any literature about this?

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I have not the slightest problem with this. 100% agreed for photons. But what about particles with (rest) mass?

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No idea. Does E²= p²c²+m²c4 include spin angular momenta(momentums?)?

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