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Trying to understand E=mc2 here !!!!!!!

Kal'thzar

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ok so to an observer it would appear to have infinite mass as per equation:
m = m0/(square root of) (1 - v^2/c^2) (m0 is stationary mass)
(as mass over zero goes infinate times) (note i don't think this equation is completely accurate either is it?)

You know i think i should just drop this as it IS impossible (all the above would require infinite force etc to do), and any speculation on the matter is pointless. (the saving grace comment ;) ) And secondly, trawling through a whole bunch of threads is...tiring to say the least....
 
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Tenacious-D

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But m in your equation is not mass. It is a combination of mass and kinetic energy. The relativistic mass is the time component of the mass - energy four-vector. This is not, I repeat not the true mass which is just m0, the rest mass, which does not change.

If you expand the equation out for small v you will see that it reduces to the standard kinetic energy term 1/2 m v^2.
 
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Kal'thzar

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its the mass an observer would observe. I think. The mass itself of the particle doesn't change it only appears to change, to the observer. (i.e. I'm not saying that m is the true mass)

I was interested in the fact that if you had an infinite force you would be able to apply that to a mass it would be able to reach the speed of light, and it would appear to the observer to have infinite mass. (i was speculating what the observer would see.
I just realised that unfortunatley there would be no observer as the observer would be used up creating that infinate force. (as would the rest of the universe....well the other thing being that you cannot get 'infinite force')

please though, drop?
 
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Aeschylus

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Kal'thzar said:
its the mass an observer would observe. I think. The mass itself of the particle doesn't change it only appears to change, to the observer. (i.e. I'm not saying that m is the true mass)
Your m is the 'transverse mass' γm[sub]0[/sub] (where m[sub]0[/sub] is the rest mass and γ = (1 - v[sup]2[/sup]/c[sup]2[/sup])[sup]-1/2[/sup]). The reason it is thought of by some as a kind of mass is that, in a given reference frame, when the 3-velocity vector is perpendicular to the 3-acceleration vector the 3-force vector f and the 3-acceleration vector a are related by (note: a 3-vector is a normal 3-dimensional vector so 3-velocity is just the velocity vector, 3-accelartion is just the acceleration vector, etc.):

f = γm[sub]0[/sub]a

But this 'F = ma' relationship doesn't hold in the other cases as when the 3-velocity and 3-acceleration vector are parallel then 3-force and 3-acceleration are related by:

f = γ[sup]3[/sup]m[sub]0[/sub]a

where γ[sup]3[/sup]m[sub]0[/sub] is what is know as the longitudinal mass.

When referring to mass it is more proper to call m[sub]0[/sub] the mass. When we talk about the situation in terms of 4-vectors then the 4-force F and the 4-acceleration A are related by:

F = m[sub]0[/sub]A


I was interested in the fact that if you had an infinite force you would be able to apply that to a mass it would be able to reach the speed of light, and it would appear to the observer to have infinite mass. (i was speculating what the observer would see.
I just realised that unfortunatley there would be no observer as the observer would be used up creating that infinate force. (as would the rest of the universe....well the other thing being that you cannot get 'infinite force')
There is no finite force that will acclerate any massive object to c, that's probably the best way to state it as infinite forces are ruled out a priori.
 
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