If we use E²=m²c⁴+p²c², and we know mass of photon is zero, and they have momentum but why aren't they affected by gravity.
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It's E² on the left, and they are affected by gravity https://en.wikipedia.org/wiki/Gravitational_lens – Conifold Jul 23 '15 at 02:32
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The equation $E = m^2c^2 + p^2 c^2$ is restricted to Special Relativity.
However, in classical physics we have $$ \vec{F} = m \vec{a}, $$ and $$\vec{F} = m \vec{g},$$ whence $$ m \vec{a} = m \vec{g}. $$ This can be written as $$ m \big( \vec{a} - \vec{g}\big) = \vec{0}. $$ From a mathematical point of view we have $$ \big( m = 0 \big) \vee \big( \vec{a} - \vec{g} = \vec{0} \big). $$ But from a physical point of view we could have $$ \vec{a} - \vec{g} = \vec{0}. $$ This means that all bodies satisfy $$ \vec{a} = \vec{g}. $$ As $\vec{g}$ is due to gravity, a photon IS accelerated as well...
johannesvalks
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1As commented in the original question, it's $E^2$, and it should be $m^2c^4$. – Kyle Kanos Jul 23 '15 at 03:16