E=mc² is Incomplete
Why Physics Stack Exchange recommends it
Physics Stack Exchange users link it when someone is confused about whether E=mc² or the energy-momentum relation is correct, especially for photons and moving objects.
Recommended by 1 people across 1 different months, 2013 to 2022.
What it is
A two-minute MinutePhysics animation by Henry Reich explaining that E=mc² is only the special case for an object at rest. The full energy-momentum relation, E² = (mc²)² + (pc)², covers moving objects and massless particles like light.
What people said
Recommends Mike · 2013-02 · Physics Stack Exchange · on “Confusion regarding photons?”
$E = mc^2$ is a popular formula but only valid in a special case when "the total additive momentum is zero for the system under consideration" $^1$ The more general "energy-momentum relation" is: $E^2 = (mc^2)^2 + (pc)^2$ (Also, here's a neat little video for your classmates ;) -> [this video] ) You can read more about this and the correct four-vector notation under: $^1$ [LINK] Even today in the time of quantum optics, there are still quite a lot of papers, that try to grasp what a "photon" is, e.g.: What is a photon - David Finkelstein Light reconsidered - Arthur Zajonc The concept of the photon - revisited - Ashok Muthukrishan, Marlan O. Scully, M.Suhail Zubairy and many more all together in t
Cites THE · 2013-10 · Physics Stack Exchange · on “Is $E=mc^2$ false?”
Recently, I found a [this video] saying that $$E=mc^2$$ is false. It says that the real one is $$ E^2=m^2 c^4 + p^2 c^2 $$ where $E$: energy, $m$: mass, $c$: speed of light and $p$: momentum. Is there anything true in it? It says that if it was $E=mc^2$ then as light had only momentum it would not have any energy.
Cites user34331 · 2014-10 · Physics Stack Exchange · on “Is $E^2=(mc^2)^2+(pc)^2$ or is $E=mc^2$ the correct one?”
I have been having trouble distinguishing these two equations and figuring out which one is correct. I have watched a video [this video] that says that $E^2=(mc^2)^2+(pc)^2$ is correct, but I do not know why. It says that $E=mc^2$ is the equation for objects that are not moving and that $E^2=(mc^2)^2+(pc)^2$ is for objects that are moving.
Critical Tyma Gaidash · 2022-10 · Physics Stack Exchange · on “Is $E^2=(mc^2)^2+(pc)^2$ or is $E=mc^2$ the correct one?”
@Prahar It is clear now, so when there is $0$ momentum, $m$ turns into $m_0$. Unfortunately, this video [this video] both does not use $m$ vs $m_0$ and does not say give an interpretation as to why a right triangle with legs $m_0 c^2$ and $pc$ with hypotenuse $E$ is drawable. It is clear that the interpretation of the $m_0 c^2$, $pc$, $E$ right triangle comes not from geometry, but rather algebra from other Physics equations transformed into $E^2=(pc)^2+(m_0 c^2)^2$. Is this why the $m_0 c^2$,$pc$, $E$ right triangle exists? Thanks