Relativistic mass, how did Einstein thought of it?by misogynisticfeminist Tags: einstein, mass, relativistic 

#1
Jun505, 12:03 AM

P: 387

I've a question here, is there a way to derive [tex] m'= \gamma m [/tex]? So far, i have not seen any derivation to this. I have heard that relativistic mass is a definition, but how did Einstein thought of this?




#2
Jun505, 07:11 AM

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In order to LT momentum, it has to be made part of a fourvector.
The fourth (usually the 0th component) is energy (as identified by the NR reduction). The invariant length of the momentum fourvector is given by m^2=E^2p^2 (letting c=1), with m being the (Invariant) mass of the object. Using v=p/e, this can be rewritten as E=m\gamma. Einstein thought up "relativisitic mass" by identifying the increase in E with v as an increase in "relativisitic mass". But the energy increase is no longer described this way. (I am pretty sure even Einstein stopped using "relativisitic mass".) Mass is considered a relativistic invariant, with E increasing because of the \gamma factor. 



#3
Jun505, 09:08 AM

P: 2,955

Note: Word of caution. E = mc^2 holds only in special cases. Pete 



#4
Jun605, 06:10 AM

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Relativistic mass, how did Einstein thought of it?
The history is interesting, but m=p/v or light having mass are not part of current SR theory. There is an energymomentum tensor, but mass is a scalar invariant.




#5
Jun605, 06:41 AM

P: 2,955

Consider this example: Take a rod with equal and opposite charges on each end. Let the proper mass of this "dumbell" be m_{0}. Let there be a uniform electric field in S where the rod is at rest and laying on the xaxis. The the Efield be parallel to the xaxis. Now transform to S' which is moving in the +x direction. The relation [tex]E^2  (pc)^2 = m_0^2 c^4[/tex] Does not hold for this dumbell. There is also an interesting problem in Ohanian's new SR text. He states the energy density of a magnetic field and then asks you to find the mass density. The answer in the back of the text is incorrect. He uses energy density = "rest mass density" *c^2 which is an invalid relationship in this situation. Pete 



#6
Jun605, 02:12 PM

P: 445





#7
Jun605, 04:31 PM

P: 2,955

I created a derivation on my web site. This one is pretty straight forward.
http://www.geocities.com/physics_wor...rtial_mass.htm You'll notice that it is Weyl's definition of mass that is used when one derives [itex]m = \gamma m_0[/itex]. Notice also that it is assumed from the start that the particles are tardyons (particles which move at v < c) and not luxons (particles which move at v = c). Pete 



#8
Jun705, 10:53 AM

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pmbphy: Of course m^2=E^2p^2 has to be changed if there are potentials present.
Look at any good text to see how. Ohanion is a good popularizer, but should not be quoted as an authority. 



#9
Jun705, 11:10 AM

P: 2,955

However if you're speaking about a single charged 'point' particle in an electric field for which the potential energy of position was nonzero then "m^2=E^2p^2" is still valid. Pete 


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