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This question already has an answer here:

I was watching a video on Youtube which deduce Einstein's relation $E=mc^2$ and the process of deduction used the relation between relativistic mass and rest mass, which is

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

So I look for a nice deduction of this relation.

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marked as duplicate by Ben Crowell, Emilio Pisanty, Qmechanic Sep 2 '13 at 11:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ A classic question! $\endgroup$ – Ali Jul 21 '13 at 15:07
  • $\begingroup$ @Ali , and What is the classic answer?! $\endgroup$ – Fawzy Hegab Jul 21 '13 at 15:33
  • $\begingroup$ From Lorentz transformations and 4-D interval (proper length between two events) in Minkowski space. Here $\gamma=\frac1{\sqrt{1-\frac{v^2}{c^2}}}$ is called the Lorentz factor And there is its derivation. $\endgroup$ – Tirsky Igor Jul 21 '13 at 15:49
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    $\begingroup$ Be aware that while there is nothing actually wrong about this relationship and "relativistic mass", many (but not all) physicists discourage its use. It's not necessary and mostly serves to preserve the functional form of the relationship $p = mv$ from Newtonian mechanics and can result in some confusion when miss applied. $\endgroup$ – dmckee Jul 21 '13 at 18:39
  • $\begingroup$ Check my anwser. Ask if you need further explaination. $\endgroup$ – 71GA Jul 22 '13 at 11:23
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The question -- once you understand that $E=m$ (see [a] for a correct proof) -- is equivalent to asking "How does energy transform from the rest frame of the body to a moving frame?"

Consider the standard set-up in which $E=m$ is proven -- a body at rest gives out two flashes of light in opposite directions, then you analyse the frame from a relatively moving reference frame. In this frame, one of the beams' energy transforms as $\sqrt {\frac{{1 + v}}{{1 - v}}} \frac{E}{2}$ and the other transforms as $\sqrt {\frac{{1 - v}}{{1 + v}}} \frac{E}{2}$. By energy conservation, the energy lost -- and thus the mass reduction -- must equal the total of these quantities, which is $\gamma E$, as required.

Archive

My original answer -- which you can see the comments are relevant to -- considered a bizarre approach where you write $m=m_0+\frac12m_0v^2$, then keep replacing the $m_0$ with the expansion for $m$ i.e. $m=m_0+\frac12mv^2$, which leads to $m=m_0/(1-v^2/2)$, which is wrong. But the approach is arbitrary, it is pretending that kinetic energy transforms as $\mathrm{KE}\to\gamma\,\mathrm{KE}$ (it doesn't, this is wrong), for no good reason.

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    $\begingroup$ How are $K_1$ and $K_2$ different if the observers are both moving at $v$ wrt the object? It would be clearer if you could label the observers 1 and 2. $\endgroup$ – Larry Harson Jul 27 '13 at 20:44
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    $\begingroup$ If $K_1$ and $K_2$ are the same, then you should label them the same. I don't see the point in having two observers moving at the same velocity. $\endgroup$ – Larry Harson Jul 28 '13 at 10:37
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    $\begingroup$ but you say the observer moves at velocity $v$ wrt object, and another observer moves at velocity $v$ wrt the object, so why aren't the kinetic energies the same since they're moving at the same velocity wrt the object? $\endgroup$ – Larry Harson Jul 28 '13 at 12:14
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    $\begingroup$ No, this is not right--- you don't get a square root. The result of this infinite series is (m-.5mv^2/(1-v^2)), it isn't right because the rational coefficients of the expansion of the square root is 1/2, -1/4, 3/8,15/16, and so on, involving successive products of odd numbers on top, while your thing has 1/2,1/4,1/8,1/16 and so on. It's not right, but it was very clever. There is no reason to assume that the mass shift can be used recursively like this--- the correct formula is found geometrically, from requiring relativity. $\endgroup$ – Ron Maimon Aug 22 '13 at 22:43
  • $\begingroup$ @DImension10AbhimanyuPS: Ok, you know E= m + .5mv^2 to order v^2, why should the v^2 term suddenly change it's coefficient? If you went to order v^4, E = m + .5mv^2 - 1.5 mv^4, you could substitude the fouth-order expression for the mass in front of v^2, or in front of v^4 (maybe both). The issue is that the actual v^2 dependence is only a lowest order thing in v, it is intrinsically modified, not just the coefficient. It is hard to say why something isn't true. It just isn't. You should think about why you think it should be true, and figure out the mistake, because it gives the wrong answer. $\endgroup$ – Ron Maimon Aug 23 '13 at 6:00
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$$ \int\vec{\rm F}\cdot{\rm d}\vec{r} = c^{2}\Delta m $$

Just replace $$ \vec{F} = {{\rm d} \over {\rm d}t}\left\lbrack{m\vec{v} \over \sqrt{1 -v^{2}/c^{2}}}\right\rbrack \quad\mbox{and}\quad {{\rm d}\vec{r} \over {\rm d}t} = \vec{v} $$

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    $\begingroup$ This is the same as 71GA's answer and is circular. $\endgroup$ – Abhimanyu Pallavi Sudhir Sep 1 '13 at 7:03
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The easiest way for me was to write down the relativistic momentum equation $p=mv\gamma(v)$. Then from this you can write down the relativistic second Newton's law:

\begin{align} \substack{\text{I start with a clasic}\\\text{second Newton's law}}\longrightarrow \boxed{Fdt = dp} \Longrightarrow F &= \frac{dp}{dt} \longleftarrow\substack{\text{here we insert formula for}\\\text{a relativistic momentum $p=mv\gamma(v)$}}\\ F &= \frac{d [mv\gamma(v)]}{dt}\\ F &= \frac{d}{dt}[ m v \gamma(v)] \longleftarrow \substack{\text{here we name the product $m\gamma(v)$}\\\text{a "relativistic mass" $\tilde{m}$}}\\ F &= \frac{d}{dt}[ \tilde{m} v] \end{align}

So we can clearly see the relativistic mass $\tilde{m}$ is the product of a rest mass $m$ and Lorentz factor (for speed) $\gamma(v)$:

$$\tilde{m} = m\gamma(v) = \frac{m}{\sqrt{1 - v^2/c^2}}$$

The only time I used the relativistic mass was when dealing with a photon in a gravitational field. I don't know any other ways to deal with that topic. If anyone knows it, please do tell =)

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    $\begingroup$ This doesn't really answer the question. It's trivially true that obtaining the gamma in $p=mv\gamma$ is equivalent to replacing $m$ with $m\gamma$. $\endgroup$ – Ben Crowell Jul 27 '13 at 14:22

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