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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.

2  
A classic question! –  Ali Jul 21 '13 at 15:07
    
@Ali , and What is the classic answer?! –  A string Jul 21 '13 at 15:33
    
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. –  Tirsky Igor Jul 21 '13 at 15:49
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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. –  dmckee Jul 21 '13 at 18:39
    
Check my anwser. Ask if you need further explaination. –  71GA Jul 22 '13 at 11:23

3 Answers 3

The most intuitive method (since it involves peturbation (not to be confused with peturbation theory (quantum mechanics) ))

An observer and an object are stationary wrt to each other.

enter image description here

Now, let us say that the object suddenly emits a flash of light in all directions (isn't this one of the most standard things you hear in SR?.

enter image description here

Then, the observer observes the object lose an energy of $E$.

Now, the observer starts moving at a speed of $v$ (this is implicitly a Lorentz transform.) . Then, from the observer’s reference frame, the object starts moving at $v$ therefore gaining a Kinetic Energy of $K_1$.

Another observer is moving at a velocity of $v$ with respect to the same exact object, and therefore observes it gain a Kinetic Energy of $K_2 $.

enter image description here

Now, if the object emits a flash of light, the observer observes the object losing an Energy of $E\left(1+\frac{v^2}{2c_0^2}\right) $ .

So, the first observer observes a net energy change in the object of $K_1-E$ whereas the second observer observes a net energy change in the object of $K_2-E\left(1+\frac{v^2}{2c_0^2}\right)$.

Equating the two, and applying the Kinetic Energy forrmula: :

$$E=m_0c_0^2 $$

That's mass-energy-equivalence.

This is only the energy of the rest mass of the object (since we are assuming the velocity is the same in both cases). The total energy of a moving object would be then what?

Zeroth - order approximation:

$$E=m_0c_0^2$$

First - order approximation:

$$E=m_0c_0^2+\frac{1}{2}m_0 v^2 $$

Second - order approximation:

$$E=m_0c_0^2+\frac{1}{2}\left( m_0 + \frac{\frac{1}{2}m_0v^2}{c_0^2 } \right) v^2 $$

Wait, what did we do? We should have the relativistic mass instead of $m$, right? But we don't know what the relativistic mass is, so we put in the first-order approximation for the relativistic mass instead of the relativistic mass.

Third - order approximation:

$$E=m_0c_0^2+\frac{1}{2}\left( m_0 + \frac{\frac{1}{2}\left( m_0 + \frac{\frac{1}{2}m_0 v^2 }{c_0^2} \right)v^2}{c_0^2 } \right) v^2 $$

Repeated the process.

Fourth - order approximation:

$$E=m_0c_0^2+\frac{1}{2}\left( m_0 + \frac{\frac{1}{2}\left( m_0 + \frac{\frac{1}{2}\left(m_0 + \frac{\frac{1}{2}m_0 v^2 }{c_0^2} \right) v^2 }{c_0^2} \right)v^2}{c_0^2 } \right) v^2 $$

Same.

Infinite - order approximation:

$$E=m_0c_0^2+\frac{1}{2}\left( m_0 + \frac{\frac{1}{2}\left( m_0 + \frac{\frac{1}{2}\left(m_0 + \frac{\frac{1}{2}(m_0+... ) v^2 }{c_0^2} \right) v^2 }{c_0^2} \right)v^2}{c_0^2 } \right) v^2 $$

Now, how do you solve this? Let:

$$\varkappa = \frac{1}{2}\left( m_0 + \frac{\frac{1}{2}\left( m_0 + \frac{\frac{1}{2}\left(m_0 + \frac{\frac{1}{2}(m_0+... ) v^2 }{c_0^2} \right) v^2 }{c_0^2} \right)v^2}{c_0^2 } \right) v^2$$

Then,

$$E=m_0c_0^2+\varkappa = m_0c_0^2+ \frac{1}{2} \left(m_0+\frac{\varkappa}{c_0^2 }\right)v^2 $$

$$\varkappa=\frac{1}{ 2}\left(m_0+\frac{\varkappa}{c_0^2 }\right) v^2 $$

Solving, we obtain:

$$\varkappa=\frac{\frac{1}{2}mv^2}{1-\frac{v^2}{c_0^2} }$$

Substituting in,

$$E=\frac{m_0 c_0^2 }{\sqrt{1 - \frac{v^2}{c_0^2} }}$$

Defining the "relativistic mass" $M$, as $$E=Mc_0^2 $$, we get that, :

$$M =\frac{m_0 }{\sqrt{1 - \frac{v^2}{c_0^2} }}$$

Q.E.D.

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1  
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. –  Larry Harson Jul 27 '13 at 20:44
    
@LarryHarson: ? It's the order of the flash of the light beam and the moving which is different. –  Dimensio1n0 Jul 28 '13 at 1:59
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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. –  Larry Harson Jul 28 '13 at 10:37
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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? –  Larry Harson Jul 28 '13 at 12:14
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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. –  Ron Maimon Aug 22 '13 at 22:43

$$ \int\vec{\rm F}\cdot{\rm d}\vec{r} = c^{2}\Delta m $$

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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1  
This is the same as 71GA's answer and is circular. –  Dimensio1n0 Sep 1 '13 at 7:03

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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You should be using GR for photon in Gravitational fields... –  Dimensio1n0 Jul 21 '13 at 18:54
    
I haven't come so far yet with my study but thank you on the comment. –  71GA Jul 21 '13 at 19:19
    
nice method $\large{\color{green}{+1}}$! –  Ali Jul 21 '13 at 19:52
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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 $m\gamma$. –  Ben Crowell Jul 27 '13 at 14:22
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@Ali: Ben Crowell is right, the answer just "uses" the Lorentz factor. This is unfortunately the approach that most textbooks take. –  Dimensio1n0 Jul 28 '13 at 2:01

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