according to the Einstein, no object mass with can reach the speed of light. but in the electron gun the speed of electron can be calculated from $$E_k = \frac{1}{2}mv^2 = eV$$ and because $m$ and $e$ are constants the equation would be

$$V = \frac{\frac{1}{2}mc^2}{e} = 255499.4513$$ $$m = 9.10938188*10^{-31}$$ $$c = 299792458$$ $$e = 1.602176462*10^{-19}$$

does this mean that the electron can travel at the speed of light ?

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    $\begingroup$ Note: $E_k = \frac{1}{2}mv^2$ is a non-relativistic formula. $\endgroup$ – Qmechanic Apr 20 '17 at 11:31
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    $\begingroup$ @Qmechanic That looks like an answer to me. (Meanwhile, ForKoding, maybe have a look at the Wikipedia article for rel. kin. energy. $\endgroup$ – ACuriousMind Apr 20 '17 at 11:45
  • $\begingroup$ i have read this before, i asked because i know that the electronic microscope use this formula to see viruses because $$KE = \frac{1}{2}mv^2=eV$$ and $$λ = \frac{h}{mv}$$ so the higher the v the lower the λ $\endgroup$ – ForKoding Apr 20 '17 at 11:49
  • $\begingroup$ A related question is why only massless particles travel at the speed of light. $\endgroup$ – StephenG Apr 20 '17 at 12:06

There are two types of masses: rest mass and relativistic mass. The rest mass is an intrinsic property of the material and is also known as invariant mass (for reasons you will know soon).

The relativistic mass depends on the velocity whereas the rest mass is independent of the velocity.

The relatvistic mass increases with velocity as follows:

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

where $m_0$ is the rest mass, $\gamma$ is the Lorentz factor, $c$ is the speed of light and $v$ is the speed of the object.

For $v << c$, the $\gamma$ term is approximately equal to one. For such velocities, we assume that the mass of the object remains fixed.

The graph given below represents the variation of the Lorentz factor ($\gamma$) with speed of the object.

enter image description here

In our day to day experiences, we do not deal with such fast moving objects. Therefore, we needn't worry about the relativistic effects. For simple problems in physics, we ignore such effects and take the mass to be equal to rest mass for all velocities.

However, as the speed of the object approaches the speed of light, the relativistic effect cannot be ignored.

As you supply the energy, the velocity keeps increasing. As the velocity increases, the mass increases. As the mass increases, you need to supply more energy than supplied previously to cause the same change in velocity. As the speed of the object approaches the speed of light, the energy that needs to be supplied to increase the object's velocity approaches infinity. Therefore, an object can never reach the speed of light (unless it is massless which allows it to travel at the speed of light).

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    $\begingroup$ Please note that relativistic mass is not an accurate interpretation of physics. It comes from an old style of teaching SR in a way that was meant to make things easier to grasp. However, it caused more problems than it solved and, thus, most physicists try to avoid teaching it and go with the more accurate and pedagogically sound "mass stays the same, it is only the momentum that grows under the Lorentz factor". Relativistic mass is not real and is misleading. Force is $F=\frac{dp}{dt}$. GR shows that moving objects do not generate increased gravity, so mass doesn't increase with speed. $\endgroup$ – Jim Apr 20 '17 at 14:11
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    $\begingroup$ I thought about that and intentionally used the relativistic mass concept because it is easier to understand. I initially wrote an answer which used the $mc^2 - m_0c^2$ definition but that made the answer complicated for a high schooler who seems to have no clue about relativity. So I thought using the relativistic mass concept was easier to get started. Well, most of us are taught that way in the first SR classes. $\endgroup$ – Yashas Apr 20 '17 at 14:15
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    $\begingroup$ Yes, that's what we want to avoid. Most of us are taught that way, so we feel it safe to teach it to others starting out. But, as I said, this has ended up causing more problems than solving. The current trend is to avoid teaching it at all, regardless the temporary ease of understanding it grants. In fact, you'll even find in many modern introductory textbooks a brief section where they mention relativistic mass and explicitly tell the students that it's not proper physics and should be avoided (because they're still likely to see us older guys mention it from time to time) $\endgroup$ – Jim Apr 20 '17 at 14:23
  • $\begingroup$ @jim If KE does not generate gravity, then neither does PE, since their sum is constant. But when a spring is released it loses PE. If this does not affect gravity, then loss of PE cannot affect rest mass, since rest mass does affect gravity. But I think loss of PE does reduce rest mass, since the rest mass of two hydrogen nuclei decreases when they combine to form deuterium. Explain. $\endgroup$ – murray denofsky Apr 21 '17 at 0:47
  • $\begingroup$ @murraydenofsky the total energy of an isolated system is locally conserved. Dropping an object from a height transfers PE to KE, but does not change rest mass. Mass is transformed to binding energy in nuclear fusion reactions, but this does not mean that kinetic energy generates gravity. That is not a sound logical conclusion. If KE generated gravity, then an object travelling close to $c$ would become a black hole. We know this simply does not happen. Therefore, the mass of an object cannot increase like $\gamma m_0$. $\endgroup$ – Jim Apr 21 '17 at 12:07

The correct equation for the kinetic energy of a massive particle is:

$$ E_K = mc^2\left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} - 1\right) $$

And if you set $v=c$ you'll see that you get $\infty$.

  • $\begingroup$ but from wikipedia $$E_{\text{k}}={\tfrac {1}{2}}mv^{2}$$ and from Quantum Theory $$E_{\text{k}} = eV$$ $\endgroup$ – ForKoding Apr 20 '17 at 11:57
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    $\begingroup$ @ForKoding When $v$ is much smaller than $c$, the approximation $\frac{1}{\sqrt{1-x^2}}\approx 1+\frac{x^2}{2}$ implies $E_K\approx mc^2\times\frac{v^2}{2c^2}=\frac{1}{2}mv^2$. $\endgroup$ – J.G. Apr 20 '17 at 12:02
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    $\begingroup$ @ForKoding You'll see that the Wikipedia page has a section on the 'Relativistic kinetic energy of rigid bodies' in which the full equation is given. Quantum mechanics is not relativistic in the basic formulation, which means you can't use it to describe things moving at or near the speed of light. You need quantum field theory for that. $\endgroup$ – gautampk Apr 20 '17 at 12:09

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