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There was this video http://www.youtube.com/watch?v=Xo232kyTsO0 where the host explains the "Real meaning of $E=mc^2 $" by saying that even at the most microscopic level even the fundamental particles could be thought of as really a manifestation of binding energy of things that are bound together to create the particle. This only means that there is no such thing as mass and all of physics could be explained via the energy picture only. While I find this somewhat convincing I don't understand how this picture could explain why heavier things are harder to accelerate. What is it about a body that makes it harder to push if it has just more of energy? One possible explanation I found in the comments section say that if you have two bodies of different mass and want them to have the same velocity, obviously the heavier one of them will have more KE and thus you need to input more energy pushing it. Is this correct?

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  • $\begingroup$ Well, what is it about a body that makes it harder to push if it just has more mass? $\endgroup$ – David Z Nov 2 '15 at 15:23
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Well, let's go back to Einstein's paper from September 1905, which demonstrated the famous relation $E=mc^2$. It's quite short, and well worth a read. An English translation is available here. Note that at the time Einstein actually used $L$ where we would now use $E$.

What Einstein does is he considers a stationary body (in some inertial reference frame) which emits an equal amount of light in two opposite directions. Because the light emitted in each direction is the same, the body remains stationary after this emission. However, because the emitted light has energy, the body has necessarily has lost some energy.

Einstein then considers the same body viewed from a different inertial reference frame, moving with velocity $v$ relative to the first frame. By transforming the equations which describe the energy of the light to this new frame, Einstein shows a greater amount of energy is lost in this reference frame.

What can account for the difference? Einstein argues that because the laws of physics hold equally in either frame, that the body's energy before the experiment must be the same in the two reference frames (up to an arbitrary constant) and likewise the energy after the experiment must be the same (up to that same arbitrary constant) - except, in the second frame the body is not at rest! It has an additional form of energy, kinetic energy, both before and after emitting the light.

So, when comparing the change in the body's energy in the first frame to the change in the body's energy in the second frame, the arbitrary constant cancels out, and the difference can only be due to a change in kinetic energy. Specifically, Einstein finds that if the body loses energy $E$ in the rest frame, then it loses additional kinetic energy $E \left( \frac{1}{\sqrt{1-v^2/c^2}} - 1\right)$ in the other frame. In the limit of $v$ much less than $c$ this simplifies to $\frac{1}{2}\frac{E}{c^2}v^2$ - exactly the classical formula for kinetic energy with $m$ replaced with $E/c^2$. (In this case $m$ is really the change in the body's mass, but if we want we can go a step further and imagine the body giving up all its mass to the outgoing radiation.)

So Einstein concluded that when a body loses energy, its mass goes down by a proportional amount, and inferred from this that a body's mass simply is a measure of its energy content.

But suppose we'd never seen the classical formula $E = \frac{1}{2} m v^2$. We might not have then thought to call this energy "mass", but nevertheless Einstein's result shows that more energy can be extracted from a body in motion than from a body at rest, and that this additional energy content in the moving body is proportional to $E/c^2$. If we instead imagine adding energy to a body, the same reasoning holds, and we see that giving a particle an additional rest energy $E$ means that it now takes an additional kinetic energy proportional to $E/c^2$ to get it moving.

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  • $\begingroup$ Thanks but how did you arrive at $\frac{1}{2}\frac{E}{c^2}v^2$ from $E \left( \frac{1}{\sqrt{1-v^2/c^2}} - 1\right)$? $\endgroup$ – Weezy Nov 2 '15 at 19:30
  • $\begingroup$ @Weezy Taylor series expansion at $v/c = 0$, keeping the dominant term. $\endgroup$ – Tim Goodman Nov 2 '15 at 19:43
  • $\begingroup$ That step is only necessary if you want to recover the non-relativistic formula for kinetic energy. In relativity, kinetic energy is $\gamma m c^2 - m c^2$ (where $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$). If Einstein's audience knew that formula, he could have just said "Look, $E$ plays the role of $mc^2$ and skipped the approximation - but he'd only just published his first paper on special relativity 3 months earlier. (Also available online: fourmilab.ch/etexts/einstein/specrel/www ) $\endgroup$ – Tim Goodman Nov 2 '15 at 19:56
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Well, you have Newton's formula $F=\gamma ma$ and Einstein's formula $E=\gamma mc^2$ so you could obtain a relation a little like $F=\frac{E}{c^2}a$ so that $\frac{Fc^2}{E}=a$ you can see from here that the greater the energy, the smaller the resultant acceleration. Even for two identical forces, if you divide it by a greater energy, you'll have a smaller acceleration. That's laws of motion.

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  • $\begingroup$ F=ma doesn't really holds in relativistic regime. So why should we mix the things? $\endgroup$ – Dvij Mankad Nov 2 '15 at 15:51
  • $\begingroup$ Well then if you take into account the relativistic effects, it still does not change much to the example. $\endgroup$ – user97166 Nov 2 '15 at 15:54
  • $\begingroup$ The goal is not to go too much in the details but just to give a little bit of intuition. And no one talked specifically about accelerating something to the speed of light. $\endgroup$ – user97166 Nov 2 '15 at 15:56
  • $\begingroup$ Maybe. But still I think actually what OP wants to ask is why E energy has an inertia of $E/c^2$. $\endgroup$ – Dvij Mankad Nov 2 '15 at 15:56
  • $\begingroup$ The problem changes a bit indeed, in this case. But for the example, it works well. $\endgroup$ – user97166 Nov 2 '15 at 15:56

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