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dekoi
Dec10-04, 05:23 PM
edit 3:
How does one go about deriving the formula:mr =\frac{m0}{\sqrt{1 - \frac{v^2}{c^2}}} ?

jcsd
Dec10-04, 05:30 PM
Mass in your equation does not change with velocity.

Some people call \gamma m the relatvistic mass and this cleraly does increase with velocity, but this defintion is not widely used and it's not a very good defintion anyway.

dekoi
Dec10-04, 05:47 PM
Mass in your equation does not change with velocity.

Some people call \gamma m the relatvistic mass and this cleraly does increase with velocity, but this defintion is not widely used and it's not a very good defintion anyway.
But \gamma m isn't mass itself; it is the multiple of mass by a factor of \gamma. Or is this the "wrong" way of looking at the situation?

dekoi
Dec10-04, 05:56 PM
[ec^2][\sqrt{1-\frac{v^2}{c^2}}]=m

Therefore, as \sqrt{1-\beta^2 increases due to the increase in velocity, m increases as well.

I am not sure whether this is right or not.
Mass in your equation does not change with velocity.
You're right... According to my equation, mass will decrease. If velocity is say, 0.7c, \sqrt{1-\beta^2 will equal 0.714 (as compared to the a value of 1 which it would be if velocity was much smaller than c). Therefore, mass would decrease.

I don't understand how this is so.

jcsd
Dec10-04, 06:06 PM
Almost universally m is regarded as the mass aka the rest mass, but a few people regard \gamma m as the mass aka the relatvistic or transverse mass.

Defining m as the mass is useful as it doesn't change from frame to frame. Defining \gamma m as the mass means that it does change from frame to frame and you find for conssitency you have to define a second kind of mass \gamma^3 m called the longitudinal of mass.

The reason for the two definitions is due to how the concepts of mass and force are related, if you think that Netwon's second law is \vec{F} = m\vec{a} then you might argue that the concept of relatvistic mass is just sticking with this concpet, but most physcists like to think of Netwon's second law as \vec{F} =\frac{d\vec{p}}{dt} anyway.

dekoi
Dec10-04, 07:00 PM
Ah. I have confused several concepts in SR. I now understand the fact that:

mr =\frac{m0}{\sqrt{1 - \frac{v^2}{c^2}}}

Thanks.

dekoi
Dec10-04, 10:10 PM
edit 3: Topic changed.