I What does E(nergy) = m(ass) (times c^2) mean?

  • I
  • Thread starter Thread starter sayetsu
  • Start date Start date
  • Tags Tags
    Mean
AI Thread Summary
Inertia mass (m) is distinct from matter and relates to energy (E) through the principle of mass-energy equivalence, where matter can be converted into energy, as seen in nuclear reactions. When a body with rest mass m is transformed into energy, it can release energy in forms like gamma radiation. Conversely, energy can also contribute to the mass of a system in its rest frame. Finding quality explanations online can be challenging, but there are numerous resources available that cater to different learning styles. Understanding these concepts is crucial for grasping the fundamentals of physics and their applications in nuclear technology.
sayetsu
Messages
46
Reaction score
3
I know m isn't matter; it's inertial mass, but I don't know what that means or how it "equals" energy. I've never found an explanation I understood, so I thought I'd try asking here. Matter can be converted to energy, though, right? That's how nuclear bombs work, or hydrogen bombs...my recollection's a little fuzzy.
 
Physics news on Phys.org
Can you not find a good explanation on line?
 
It means that a body of rest mass ##m## releases energy ##E## if converted to some form without rest mass (e.g. gamma radiation). Or alternatively that energy ##E## contributes ##m## to the mass of some system in its rest frame.
 
  • Like
Likes vanhees71 and berkeman
Hi Sayetsu

There are literally hundreds of videos you can find; the challenge is finding the "good" ones - which will differ for most people depending on background etc

Based on your question I would try these two (in this order)
Special Relativity (E = mc2) Explained A Four minute reminder of what you asked

Why E=mc² is wrong An extra 6 minutes that either reminds you / tells you that there's a bit more to the formula that doesn't often get discussed.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top