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I am learning mass-energy equivalence, but I have been having a hard time understanding what the $E$ in $E=mc^2$ represents. Does this represent that if an object has mass, it inherently has energy associated with that mass, sort of like how if an object has velocity is inherently has kinetic energy associated with that velocity, or if gas particles are at a certain temperature, they inherently have internal energy associated with that temperature? In other words, is the energy in an object stored as mass unrelated to all the other forms of energy (potential, kinetic, etc), or is it related to them? If I increased an object's potential energy by lifting it up, would its mass change?

Edit: I've also just read something about rest frames, and how $E=mc^2$ applies for an object in its rest frame. I'm a bit new to relativity in general, so does this rest frame mean that the object has no kinetic energy since everything else is moving, not it? Does the rest frame also imply a specific potential energy, or does it vary depending on the object's position still? Would that mean that potential energy is related to the energy stored as mass, but kinetic energy is not?

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(a) "Does this represent that if an object has mass, it inherently has energy associated with that mass[?]" Yes.

(b) "[I]s the energy in an object stored as mass unrelated to all the other forms of energy (potential, kinetic, etc), or is it related to them?" No. The mass of the object includes all the energies stored in the object, as measured in the frame in which the object's centre of mass is at rest. So, for example a sample of gas whose kinetic energy of random molecular motion amounts to 25 J will have a contribution to its mass of $(25\ \text J)/c^2$.

(c) "If I increased an object's potential energy by lifting it up, would its mass change?" You must remember that potential energy cannot be ascribed to a single body but to the system of bodies between which forces act (such as the Earth and the body that you lift up). The system's energy will indeed increase – if you are not including the lifter in your system.

(d) "[D]oes this rest frame mean that the object has no kinetic energy[?]" Yes, that's what we'd usually say. We wouldn't usually count, for example, the random motion of molecules relative to a body's centre of mass frame as part of an object's kinetic energy.

(e) "Does the rest frame also imply a specific potential energy, or does it vary depending on the object's position still?" See (c) above.

(f) In Special Relativity a body's KE can still be expressed in terms of its mass and speed: $$KE = mc^2(\gamma-1)\ \ \ \ \text{in which} \ \ \ \ \gamma ={\left(1-v^2/c^2\right)}^{-1/2}$$ Regarding the KE as a function of mass and speed, since $m$ includes the body's energy relative to its centre of mass, its kinetic energy does depend on its energy in its centre of mass frame frame, and that includes the potential energy of interactions between its particles.

Note My $m$ is the invariant mass (independent of a body's speed) or simply the mass. It used to be called 'rest mass'. Anna v (see her answer) denotes it by $m_0$. I do not use the notion of relativistic mass.

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  • $\begingroup$ If you were looking for the total energy of the body outside of its rest frame, is that where the (pc)^2 term comes in in E^2=(mc^2)^2 + (pc)^2? Can you say that momentum is always 0 in a rest frame by definition? $\endgroup$
    – Akash
    Jan 24, 2021 at 18:41
  • $\begingroup$ Yes and yes! The beauty of it is that $E$ is the time component of a 4-vector and $pc$ is the combined magnitude of the three spatial components, so $mc^2$ is the magnitude of the 4-vector – which is independent of your frame of reference. Recommended reading: Spacetime Physics by Taylor and Wheeler. $\endgroup$ Jan 24, 2021 at 19:06
  • $\begingroup$ If I had an atom of U-238 decay into Th-234 plus an alpha particle, would the decrease in mass (which I assume has something to do with the arrangement of nucleons becoming more stable and therefore having less energy) be accounted for by the non-zero momentum of the individual daughter nuclei (even though the net momentum is still 0)? Or would you have to consider energy leaving the system? In general, if you are using the energy equation with momentum, do you use the net momentum or the momentum of individual particles? $\endgroup$
    – Akash
    Jan 24, 2021 at 19:59
  • $\begingroup$ I'm sorry, but I find this difficult to follow. Could you give some equations? $\endgroup$ Jan 25, 2021 at 8:42
  • $\begingroup$ Here are the equations I'd use to solve your problem. Using no suffix for the U nucleus and suffices 1 and 2 for the products of the split, and $E$ for the total energy (rest energy + kinetic) of a particle: $$E_1+E_2=mc^2,$$ $$p_1+p_2=0,$$ $$E_1^2-c^2p_1^2=m_1^2c^4,$$ $$E_2^2-c^2p_2^2=m_2^2c^4.$$ Hope this helps. $\endgroup$ Jan 25, 2021 at 17:03
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All types of energy which are independent of the reference frame contribute to an objects rest mass. If you add potential energy to a spring by compressing it, its mass will increase. If you heat an object up, its rest mass increases. A system of two photons flying in opposite directions with momenta $p$ and $-p$ does have a rest mass, even though individual photons are massless.

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  • $\begingroup$ That's a lovely example. $\endgroup$ Jan 24, 2021 at 17:56
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The $E$ in the $E=mc^2$ refers to an object's rest mass energy, which is the intrinsic energy the object has due to its mass. The more general result is given by $E^2 = {\bf{p}}^2c^2 + m^2c^4$, which gives you the energy for massless particles as well. All of this can be derived by solving the Lagrangian for a free particle in $\Bbb{R}^4$.

For the Lagrangian coordinates $\{x^\mu(\tau)\}$ with the action given by $$S=\cfrac{1}{2}m\int_\Gamma g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu d\tau$$ we take the conjugate momentum, to get $$p_{\mu}=\cfrac{\partial L}{\partial\dot{x}^\mu}=mg_{\mu\nu}\dot{x}^\nu=m{\dot{x}}^\mu$$ Now by using the index rule we have that $p^\mu=g^{\mu\nu}p_\nu$ to get conservation of mass: $$p^\mu p_\mu=-mc^2$$ Now we switch to the Cartesian coordinates by using $g_{\mu\nu}=\eta_{\mu\nu}$ to get $$\cfrac{d p_{\mu}}{d\tau}=0$$ Since $p^\mu p_\mu=\eta_{\mu\nu}p^\mu p^\nu$, we can use the conservation of mass to normalise the 4-velocity and get $$\eta_{\mu\nu}\dot{x}^\mu\dot{x}^\nu=-c^2\dot{t}^2+\dot{x}^2+\dot{y}^2+\dot{z}^2=-c^2$$ We can now rearrange this to end up with the following $$\dot{t}^2=\cfrac{c}{\sqrt{c^2-{\bf{v}}^2}}=\gamma$$ where ${\bf{v}}$ is the velocity 3-vector. Now by applying the chain rule, we can get the components for the 4-velocity. As an example $$\dot{x}=\cfrac{d x}{d\tau}=\cfrac{dx}{dt}\cfrac{dt}{d\tau}=\gamma v_x$$ Similarly we have $\dot{y}=\gamma v_y$ and $\dot{z}=\gamma v_z$. As for the first component, we simply use $x^0=ct$. Now that we have all of our 4-velocity components, we can write $$\dot{x}^\mu=(\gamma c, \gamma{\bf{v}})$$ We can now write down the momentum 4-vector, but before we do, you should know that this result was derived from considering the Lagrangian for a free particle in $\Bbb{R}^3$, you should try it yourself. The full momentum 4-vector is given by $$p^\mu=\left(\cfrac{E}{c}, {\bf{p}}\right)$$ Similarly, we also have $$p_{\mu}=\left(-\cfrac{E}{c},{\bf{p}}\right)$$ Finally, by using the equations above and the conservation of mass, we get to the result $$E^2=m^2c^4+{\bf{p}}^2c^2$$ For ${\bf{p}}=0$, you get to the desired $E=mc^2$. Now bare in mind that the relativistic effects such as the particle gaining mass from being accelerated only start to take affect when sufficiently high speeds are reached, i.e. when $$1-\cfrac{{\bf{v}}^2}{c^2} << 1$$

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The $m$ in $E=mc^2$ is called "the relativistic mass" and is part of the algebra of special relativity.

$$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma m_0$$ $m_0$ = "rest mass"

Note the dependence on velocity, and note that velocity so this mass is variable, it is the inertial mass of an object , i.e. the resistance to becoming accelerated in classical terms. It is no longer used in particle physics because of this variability. What characterizes particles is the rest mass, or invariant mass. This, in the four vector algebra, is the length of the given four vector:

$$\sqrt{P\cdot P}=\sqrt{E^2-(pc)^2}=m_0c^2$$

The length of this 4-vector is the rest energy of the particle. The invariance is associated with the fact that the rest mass is the same in any inertial frame of reference.

It is the $m_0c^2$ that is the inherent available energy of a particle due to the mass. When the particle is at rest, i.e momentum is equal to zero, all the energy is in the rest mass as the formula shows. When a particle meets an antiparticle twice this mass-energy is available in the annihilation.

Also in nuclear physics , the rests mass of the nuclei shows if it is possible to fission or fusion into other nuclei.

In the last formula, the total energy of a particle minus the kinetic energy, give the invariant mass.The concept of potential energy does not enter when using four vector algebra.

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  • $\begingroup$ Even though Einstein initially used the expressions "longitudinal" and "transverse" mass in two papers (see previous section), in his first paper on {\displaystyle E=mc^{2}}E = mc^2 (1905) he treated m as what would now be called the rest mass.[2] Einstein never derived an equation for "relativistic mass", and in later years he expressed his dislike of the idea:[27] $\endgroup$
    – Gert
    Jan 24, 2021 at 16:35
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    $\begingroup$ It is not good to introduce the concept of the mass ${\displaystyle M=m/{\sqrt {1-v^{2}/c^{2}}}}M = m/\sqrt{1 - v^2/c^2}$ of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the ’rest mass’ m. Instead of introducing M it is better to mention the expression for the momentum and energy of a body in motion. — Albert Einstein in letter to Lincoln Barnett, 19 June 1948 (quote from L.B. Okun (1989), p. 42[5]). Both comments from Wikipedia: en.wikipedia.org/wiki/… $\endgroup$
    – Gert
    Jan 24, 2021 at 16:35
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    $\begingroup$ @Azzinoth the equation is correct, it is just not useful, it will become useful when/if space travel between stars ever happens and velocities close to the velocity of light are attained, in the calculations for the fuel needed for the trip.. the formula is the one I have quoted. $\endgroup$
    – anna v
    Jan 24, 2021 at 17:15
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    $\begingroup$ @Gert I outlined with italics that it is not a useful formula $\endgroup$
    – anna v
    Jan 24, 2021 at 17:16
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    $\begingroup$ I see exactly where you are coming from in the first line of your answer. $E$ usually denotes total energy, both internal and kinetic. So with this interpretation we have $E=c^2\gamma \text{(invariant mass)}$. The trouble is that beginning students are usually taught to use $E=(\text{invariant mass})c^2$ to calculate the energy 'released', say when a Uranium nucleus splits. What they are really doing is calculating the change in internal energy. But the '$E$' notation is wrong for this. $\Delta U$ would be better or $\Delta E_0$. $\endgroup$ Jan 24, 2021 at 19:49

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