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The Simplest Derivation of E = Mc2 The Simplest Derivation of E = mc2 Liu, Jerry Z. PhD Stanford University, California, USA Introduction Mass-energy equivalence, or expressed in the form of E = mc2, is one of the most influential physics principles in Albert Einstein’s theory of relativity. However, after more than 100 years, the theory of relativity is still not well accepted as common knowledge and basic understanding of our universe. The problem may be because we have not done our job to put the knowledge in a way that is more comprehensible to the public. The simple derivations of the mass-energy equation presented here are the efforts we took to make some of the concepts simple, interesting, and easy to be understood for high school students, therefore accessible to the general public. Through the discussions, you will learn where we were wrong in Newtonian physics and appreciate Einstein’s work that led to a deeper understanding of our universe and, therefore, took modern physics onto a different level. The Simplest and Quickest Derivation With Maxwell’s electromagnetic equations, it has been proved that light travels at a constant speed c (~ 3x108 m/s) that has been verified in numerous experiments and observations; and the momentum p of electromagnetic wave, or photons in Albert’s dictionary, is proportional to the energy level E, as described in p = E/c (1) or E = pc. (2) In Newtonian physics, the momentum p of a body is defined as a moving body at speed v with a mass m, as in the following form p = mv. (3) In case of a photon, the definition can be simplified below, since photon travels at the speed of light c, p = mc. (4) Now, put equation (4) into (2), we just derived the famous mass-energy equation: E = mc2. (5) So simple, isn’t it! Wait for a second, a photon does not have mass. Can we use equation (4) above in the derivation? Yes, we can. The momentum definition (3) is one of the few Newtonian definitions that also work in relativistic physics. When people are talking about no mass in a photon, they are referring to the rest/invariant mass, which is usually donated by m0. The mass in the above equations is referring to relativistic mass. In the case of a photon, there is no rest mass but still has relativistic mass. For a stationary body, its relativistic mass has the same value as rest mass, given that there is no other energy difference. It will become more clear when we revisit these concepts later. If you are still not convinced by this short derivation, let’s prove equation (4) for you next. A Derivation of p = mc for Photon Consider a laser pen at rest in an inertial frame, as shown in the figure below, with the x-axis for horizontal space and the vertical axis t for time. At time t1, the laser pen emits a photon to the right. Even though there is no rest mass in photons, light pressure has been proved and observed in experiments, indicating there is momentum and relativistic mass in photons. As a result, the pen and photon move apart as shown at time t2. The pen moved L meters to the left, while the photon moved l meters to the right from their original position o. The mass center of both pen and photon was initially at location C that can be calculated like this Initial Mass Center = (MX1 + mx1) / (M + m) where M and m are the masses of the pen and photon, respectively, and X1 and x1 are the mass centers for the pen and photon. Similarly, the new mass center for both pen and photon at time t2 can be computed as New Mass Center = (MX2 + mx2) / (M + m) = (M(X1 - L)+m(x1 + l)) / (M + m) = (MX1 - ML + mx1 + ml) / (M + m) where X2 and x2 are the mass centers for the pen and photon at time t2. Because there is no external force, the new common mass center should also be at the same location C. Therefore, we have the equation (MX1 + mx1) / (M + m) = (MX1 - ML + mx1 + ml) / (M + m) which can be simplified to ML = ml. By dividing (t2 - t1) from both sides of the equation, it becomes ML / (t2 - t1) = ml / (t2 - t1). Remember the speed for the pen is V = L / (t2 - t1) and the speed for the photon is v = l / (t2 - t1). The equation is reduced to MV = mv. (6) Equation (6) we just proved is the conservation law of momentum. Keep in mind, l is the distance that the photon traveled in time (t2 - t1), so speed v above is actually the speed of the light c, thus the equation can be written as MV = mc. All the momentum discussed in this context, including both photon’s momentum pphoton and the momentum for the pen ppen, are equal in values. Together with definition (3) for momentum, we can connect them in the following equations: pphoton = ppen = MV = mc. (7) This proves equation (4), p = mc, confirming that the definition (3) can be extended from Newtonian to relativistic physics. Therefore, the derivation in the previous section is valid. Next, let’s look at an observation that demonstrates the relativistic mass of photons in action. A Verification of the Existence of Photon Mass It appears to be common sense that light travels in a straight line. Can you imagine that light might be bent? Yes, it has been predicted in Einstein’s theory of general relativity and was observed in the following experiment. Astronomers know the locations of stars with respect to the Sun and Earth in the sky. There is a particular star on the other side of the Sun as shown in the following configuration. If the light of the star were traveling in a straight line, it would have been blocked by the Sun. An observer on the Earth would not be able to see the light from the star. However, the Star was clearly observed in an event of an eclipse, with an illusion as indicated by the dotted line. As we all know, there is energy in light, which is given by E = hf (8) where h is Planck’s constant and f is the frequency of light. With the mass-energy equation (5), we will find relativistic mass m given below, m = hf/c2 (9) for a frequency f of light. It indicates this mass that makes light attracted by the gravity of the Sun, which bends the light and allows the observer on the Earth behind the Sun and Moon to see the light from the star. In general relativity, Einstein explained this phenomenon from a different perspective: gravity curves the space around it. In other words, the light travels in a straight line, but the space is curved. In an extreme case near a black hole, light is curved into the black hole and we cannot see the black hole directly, but observe its effect on nearby objects. An Intuitive Proof for Regular Bodies We just proved the mass-energy equation for photons. Consider from the perspective of the laser pen in the previous derivation, it loses energy El, which is the same amount of energy that the photon takes away from the pen. Thus, the energy loss is 2 El = mc (10) or, equivalent of loss in mass, 2 ml = El/c . (11) This is what Einstein was trying to prove in 1905. However, his logic is hard for many people to follow. Nevertheless, the derivation still has not provided direct proof of the validity of the mass-energy equation for regular bodies with rest mass. Next, we will try to provide an intuitive derivation. Let’s consider a force F applied to a body which will increase its energy and momentum. The energy gain dEg is the work that the force F made over a cause S, given by dEg = FdS. (12) To find Eg, let’s take integral over the cause S on (12). This is one way to derive the mass-energy equation in many textbooks, which we will not repeat here. Instead, let’s take derivative on momentum definition (3) with respect to time, dp/dt = d(mv)/dt. In Newtonian physics, the mass m is assumed to be an invariant. So, it becomes, d(mv)/dt = mdv/dt = ma = F where a is the acceleration of mass m resulted from the applied force F. The right part of the equation is Newton’s second law, which will have problems at high speeds because mass is no longer an invariant. Let’s redefine the second law as the increment of momentum due to the applied force, F = dp/dt. (13) This definition works consistently in both Newtonian and relativistic physics. With definition (3), Newton’s second law would become F = d(mv)/dt = mdv/dt + vdm/dt (14) where the first term is the classical portion for Newton’s law, while the second term indicates the mass appreciation due to the applied force. Replace the force F in (12), the energy gain becomes dEg = d(mv)dS/dt = vd(mv) = vvdm + mvdv or simplified like this 2 2 dEg = v dm + ½mdv . (15) To find the energy gain, we can take the integral over (15). At low speed, there is not much change in mass. So, we can consider dm ~ 0 and the first term can be omitted. The energy gain becomes 2 Eg = Ek = ½mv , (16) which is the kinetic energy in Newtonian physics where all the energy gain is contributed to kinetic energy.
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