Derive a new equation from $m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$

Question

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

Differentiating the equation,

$$2m \;dm\;c^2-2m\;dm\;v^2-2v\;dv\;m_0^2=0$$

Our book says when we differentiate $m^2c^2-m^2v^2=m_0^2c^2$. We will get the above equation. But, what I understand about Calculus. That I can't derive it anyway. Actually, what we differentiate here? $mass$ or, $velocity$?

Relativistic mass equation :

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

Above question is a constant equation which maybe found from Lorentz Transformation. In the second line I’ve just squared both side. In third line I‘ve just moved "something" right to left. Then, I differentiate. But, I can't understand how they differentiate here.

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After writing the question, I was doing some the sum again.

Relativistic mass changes over time (How fast you travel through space your mass decreases). Here $m$ is relativistic mass and $m_0$ is "normal" mass.

So, I decided to differentiate mass $$(mc)^2-(mv)^2-(m_0c)^2=0$$

We know, $$(f(x))^n=n(f(x))^{n-1} . f' (x)$$

Then :

$$2 (mc) . c - 2 (mv) .v -$$

Then, I can't differentiate anymore. $c$ is speed of light which is constant. That $m_0$ is also constant. So, I stopped there.

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Prove of $E=mc^2$,

$$F=\frac{dp}{dt}$$ $$=\frac{d}{dt} (mv)$$ $$= m \frac{dv}{dt} + v \frac{dm}{dt}$$ ----------------1

$$dW=F .dS$$ $$=>dK=F .dS$$ $$=>dK=[m \frac{dv}{dt} +v \frac{dm}{dt}] .dS$$ $$=m\frac{dS}{dt} .dv + v . \frac{dS}{dt} .dm$$ $$=mvdv+v^2dm$$ -------------------2

$$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ $$\implies m^2=\frac{m_0^2}{1-\frac{v^2}{c^2}}$$ $$\implies m^2c^2-m^2v^2=m_0^2c^2$$ $$2m \;dm\;c^2-2m\;dm\;v^2-2v\;dv\;m_0^2=0$$ $$c^2dm=mvdv+v^2dm$$ ---------------3

$$dK=c^2dm$$ $$\int = \int c^2dm$$ (In LHS Integral starts from 0 and finishes at $k$. In RHS Integral starts from $m_0$ and ends at $m$) $$k=c^2[m-m_0]$$

Hence,

$$Total energy = k + Rest mass energy$$ $$E=c^2[m-m_0]+m_0c^2$$ $$=mc^2-m_0c^2+m_0c^2$$ $$=mc^2$$

$$E=mc^2$$

Answer 1

Say you are given $$ m(t) = \frac{m_0}{\sqrt{1 - \frac{v(t)^2}{c^2}}}, $$ for some $t> 0$, with constants $m_0$ and $c$. Then, differentiating with respect to time, we get that $$ \begin{align} \frac{dm(t)}{dt} = \dot{m}(t) &= m_0 \frac{d}{dt}\left(1 - \frac{v(t)^2}{c^2}\right)^{-\frac{1}{2}} \\ &= \frac{-m_0}{2c^2} \frac{d}{dt}\left(v(t)^2\right) \left(1 - \frac{v(t)^2}{c^2}\right)^{-\frac{3}{2}} \\ &= -\frac{m_0 v(t) \dot{v}(t)}{c^2} \left(1 - \frac{v(t)^2}{c^2}\right)^{-\frac{3}{2}} \tag{1} \end{align} $$ In Equation (1), using the fact that $$ \left(1 - \frac{v(t)^2}{c^2}\right)^{-\frac{3}{2}} = \left(\frac{m(t)}{m_0}\right)^3, $$ we find that $$ \dot{m}(t) = -\frac{m_0v(t)\dot{v}(t)}{c^2}\left(\frac{m(t)}{m_0}\right)^3 $$

Simplifying further, we obtain that

$$ \dot{m}(t) m_0^2 c^2 = \dot{v}(t) v(t) m(t)^3 $$

Instead, in Equation (1), using the fact that $$ \left(1 - \frac{v(t)^2}{c^2}\right)^{-\frac{1}{2}} = \left(\frac{m(t)}{m_0}\right), $$ we can instead obtain that $$ \begin{align} \dot{m}(t) = - \frac{m_0v(t)\dot{v}(t)}{c^2}\left(1 - \frac{v(t)^2}{c^2}\right)^{-1}\left(\frac{m(t)}{m_0}\right) \end{align}. $$ Simplifying this then gives us $$ \dot{m}(t)(c^2 - v(t)^2) = -v(t)\dot{v}(t)m(t) $$ That is, putting everything to one side $$ \dot{m}(t) c^2 + \dot{v}(t) v(t) m(t) - \dot{m}(t) v(t)^2 = 0, $$ which is similar to what you require.

Answer 2

What your question had was a differential equation for energy in terms of two variables, $m$ and $v$. But we could write it in terms of one variable if we knew the Jacobian, which is a fancy word for the scale factor when you convert between $v$ and $m$ variables. This is similar to the manipulation you have seen before in implicit differentiation

$$x^2+y^2=r^2 \implies 2x + 2y\cdot \frac{dy}{dx} = 0$$

but this manipulation is different because it doesn't rely on a parametric variable (one variable inside both - in the above case $x = x(x)$ and $y = y(x)$). It's just like the manipulations you've see for $u$ substitution

$$u = \tan x \implies du = \sec^2x\:dx $$

$$\tan^{-1}u = x \implies \frac{du}{1+u^2} = dx$$

and so on. In physics we can always interpret the Jacobian as the ratio of the two changes i.e.

$$\frac{dx}{du} = \cos^2 x = \frac{1}{1+u^2}$$

What the manipulation does is answer the following question: if we have a little change in each of the variables, how does the little change in one relate to the little change in the other

$$f = f(x,y) \implies df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$$

In this specific case, since we know $f(m,v)$ is a constant we have that

$$df = 0 = \frac{\partial f}{\partial m}dm + \frac{\partial f}{\partial v}dv \implies \frac{dv}{dm} = -\frac{\frac{\partial f}{\partial m}}{\frac{\partial f}{\partial v}}$$

which is actually a famous theorem called the implicit function theorem. With this Jacobian in hand, we are free to replace $dv$, the change in velocity, in the following equation with the related change in mass, $dm$ by

$$dv = \frac{dv}{dm}dm$$

or, the Jacobian (scale factor) times the change in mass.

Again, this doesn't require $m$ nor $v$ to be functions of a parametric time variable. This is important for relativity because time is not suitable to be a universal parametric variable. Both time and spatial coordinates are parametrized in terms of other variables, such as the arclength of a worldline (proper time) or just an arbitrary variable with no physical meaning.