planck42 said:
I've read contrasting sources concerning the concept of an object increasing in mass at relativistic velocities. Some of my older calculus texts mention this as being accepted by physicists, while a newer(by comparison) textbook called Principles of Physics: A Calculus-Based Text by Serway and Jewett claims that relativistic mass is outdated. Normally, the newer book would be correct, but I don't think there is any other reasonable explanation of the momentum of photons. Can the good people at PF please shed some light on this matter?
I think it's never been outdated and will never be. Relativistic mass is calculated so:
for example, kinetic enerji can be calculated with KE=mc^2-m_0c^2
it's also KE=mc^2-\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}c^2=mc^2-\frac{m_0c}{\sqrt{c^2-v^2}} and if we add momentum to this equ., it would be E^2=m_0^2c^4+p^2c^2=E=\sqrt{m_0^2c^4+p^2c^2} where momentum isn't relativistic. Remeber that relativistic p is p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}
after we make momentum relativistic in KE equ., we'll find out the relativistic value of energy, which includes p. It's so: E=\frac{m_0^2c^3}{\sqrt{c^2v^2}}+\frac{p^2cm_0}{\sqrt{c^2-v^2}}=\frac{m_0c(m_0c^2+p^2}{\sqrt{c^2-v^2}}
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we can also say for pc when E=\sqrt{m_0^2c^4+p^2c^2} pc is pc={\sqrt{p_0^2-m_0^2c^4}} and so, we can say p^2c^2=\frac{m_0^2v^2c^2}{1-\frac{v^2}{c^2}}=\frac{m_0^2\frac{v^2}{c^2}c^4}{1-\frac{v^2}{c^2}}
after a quick calculation you'll find p^2c^2=\frac{m_0^2c^4[\frac{v^2}{c^2}-1]}{1-\frac{v^2}{c^2}}+\frac{m_0^2c^4}{1-\frac{v^2}{c^2}}=-m_0^2c^4+m^2c^4=(mc^2)^2\Rightarrow E=pc
You can find the basic ideas of this calculations (made by me :) ) at the notes part of the book "Relativity: The Special and General Theory" by Albert Einstein.
If you see the equations as text, you can see them as pic.s by copy-pasting them onto
http://www.sitmo.com/latex/