Inertia is directly proportional to mass but what happens when something travel to speed near to light. Its relativistic mass tends to infinity but that is false mass so I want to know if inertia is applied to relativistic mass. Because if we infinite the mass then the inertia would also be infinite then it will never reach speed of light.


By inertia I assume you mean momentum. The momentum is related to the energy of the object by:

$$ E^2 = p^2c^2 + m^2c^4 $$

and to the velocity by:

$$ p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} $$

The momentum does indeed tend to infinity as $v \rightarrow c$, but note that it will never reach an infinite value because no massive object can travel at the speed of light so $v$ never reaches $c$.

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  • $\begingroup$ "By inertia I assume you mean momentum." Eh.. what? :P As far as I know, inertia of a body is (vaguely!) defined as "the (inherent) quantity that causes a body to resist acceleration". My guess is that it's ill-defined in relativity (speeding up by a certain amount may cost more than stopping altogether, so it should become dependent on $F\cdot v$, and lose its inherent status). $\endgroup$ – Danu Jul 23 '15 at 8:02
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    $\begingroup$ @Danu: inertia is one of those common terms that has no precise meaning (and isn't used by physicists!). But I'd guess it's generally used to mean that proprty of a body that makes it require a force to stop it. Force is of course the rate of change of ... $\endgroup$ – John Rennie Jul 23 '15 at 8:19
  • $\begingroup$ Even if lacking a precise definition, I still think we can be certain it is not momentum (my "definition" above makes this clear, but it appears that you may simply disagree on this definition). It is not important in any case, so let's just leave it at this :) $\endgroup$ – Danu Jul 23 '15 at 9:00
  • $\begingroup$ The books I use define inertia in a qualitative way "the property of resisting acceleration" and momentum and a quantification of a body's inertia between two different states of motion. That said, I believe that you find considerable variation in the precise meaning of "inertia" assigned by different texts. $\endgroup$ – dmckee --- ex-moderator kitten Jul 23 '15 at 16:00

Particles that have no mass, as photons, have momentum based on their energy, more especificaly on the relation $$p=\frac{E}{c}$$

Where $p$ is the momentum, $E$ is the energy and $c$ is the speed of light.

You can think of it intuitively by noticing Einstein's famous equation, $E=mc^2$, and substituting it on the classical momentum formula $p=mv$, using $c$ in $v$'s place.

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A Non-Relativistic Interpretation of Relativistic Results

Million years ago, when self-studying special relativity, an exercise was created in order to understand non-relativistically the relativistic result that a force could not accelerate a particle to speed greater than $\:c\:$. The exercise, the Figure and the solution are already in LaTeX and are copy-paste here. May be is useful.

The solution to the following exercise yields a non-relativistic interpretation of the relativistic result that energy $\:E\:$ has inertia, that is resistance to be accelerated, as material objects. The inertia of material objects is expressed by their inertial mass $\:m\:$ while the inertia of energy $\:E\:$ is expressed by its mass equivalent$\: m_{E}=E/c^{2}\:$. Also, based on this interpretation, an explanation of the impossibility of a force to accelerate an object to velocity greater than that of light ($c$) is given.

enter image description here

A body of mass $\:m_{o}\:$ initially at rest is moving on a surface without friction under the influence of a constant force $\:\mathbf{f}\:$, as in Figure above. As it is moving, it is carrying away material from a straight line above it. The straight line has a constant linear mass density $\:\rho_{\ell}=f/c^{2}\:$, where $\:\rm{f}\:$ the magnitude of the constant force $\:\mathbf{f}\:$ and $\:c\:$ a quantity with dimensions of speed.

The exercise concerns the determination of the following :

(a) the position $\:x(t)\:$, the velocity $\:\upsilon(t)\:$ and the mass $\:m(t)\:$ as functions of time.

(b) any explicit relation, if there exists, between above quantities.

Solution :

For the mass we have \begin{equation} dm=\rho_{\ell} dx=\dfrac{f}{c^{2}}dx \Longrightarrow \nonumber \end{equation} \begin{equation} m\left(x\right)=m_{o}+\dfrac{f}{c^{2}}x \tag{01} \end{equation} Now, \begin{equation} f=\dfrac{dp}{dt}=\dfrac{d\left(m\upsilon\right)}{dt}=m\dfrac{d\upsilon}{dt}+\upsilon\dfrac{dm}{dt} \tag{02} \end{equation} that is \begin{align*} fdx & =\left( m\dfrac{d\upsilon}{dt}+\upsilon\dfrac{dm}{dt}\right)\upsilon dt=m\upsilon d\upsilon +\upsilon^{2}dm =\dfrac{1}{2}md\upsilon^{2}+\upsilon^{2}\dfrac{f}{c^{2}}dx\\ & =\dfrac{1}{2}\left(m_{o}+\dfrac{f}{c^{2}}x\right) d\upsilon^{2}+\upsilon^{2}\dfrac{f}{c^{2}}dx \end{align*} so \begin{equation*} \left(1-\dfrac{\upsilon^{2}}{c^{2}}\right)fdx = \dfrac{1}{2}\left(m_{o}+\dfrac{f}{c^{2}}x\right)d\upsilon^{2} \end{equation*} Separating the variables $\:x\:$ and $\:\upsilon \:$ \begin{equation*} \dfrac{fdx}{\left(m_{o}+\dfrac{f}{c^{2}}x\right)} = \dfrac{1}{2}\dfrac{d\upsilon^{2}}{\left(1-\dfrac{\upsilon^{2}}{c^{2}}\right)}\:\Longrightarrow \: \dfrac{fdx}{\left(m_{o}+\dfrac{f}{c^{2}}x\right)} = \dfrac{1}{2}\dfrac{d\upsilon^{2}}{\left(1-\dfrac{\upsilon^{2}}{c^{2}}\right)} \end{equation*} or \begin{equation} \dfrac{d\left(m_{o}+\dfrac{f}{c^{2}}x\right)}{\left(m_{o}+\dfrac{f}{c^{2}}x\right)} =\;-\; \dfrac{1}{2}\dfrac{d\left(1-\dfrac{\upsilon^{2}}{c^{2}}\right)}{\left(1-\dfrac{\upsilon^{2}}{c^{2}}\right)} \tag{03} \end{equation}

Note: We don't know a priori that the expression $\:\left(1-\dfrac{\upsilon^{2}}{c^{2}}\right)\:$ will not take negative values as the body moves in time. Since at $\:t=0\:$ the body is at rest ($\:\upsilon = 0 \:$), the value of this expression is positive ($\:=1\:$) at the beginning . Our analysis will prove then that this expression remains always positive. Anyway, to overcome this difficulty we could do something like this :

Since \begin{equation} \dfrac{dx}{x}=\dfrac{1}{2}\dfrac{dx^{2}}{x^{2}} \nonumber \end{equation} equation (03) can be written as \begin{equation} \dfrac{d\left(m_{o}+\dfrac{f}{c^{2}}x\right)}{\left(m_{o}+\dfrac{f}{c^{2}}x\right)} =\;-\; \dfrac{1}{4}\dfrac{d\left(1-\dfrac{\upsilon^{2}}{c^{2}}\right)^{2}}{\left(1-\dfrac{\upsilon^{2}}{c^{2}}\right)^{2}} \nonumber \end{equation} so that in the equations that follow to replace the expression $\:\sqrt{1-\dfrac{\upsilon^{2}}{c^{2}}}\:$ by $\:\sqrt{\left\vert 1-\dfrac{\upsilon^{2}}{c^{2}}\right\vert}\:$.

Integration yields \begin{equation} \ln\left(m_{o}+\dfrac{f}{c^{2}}x\right) =\;-\; \dfrac{1}{2}\ln\left(1-\dfrac{\upsilon^{2}}{c^{2}}\right)+C\:\Longrightarrow \: \left(m_{o}+\dfrac{f}{c^{2}}x\right)=C\cdot\left(1-\dfrac{\upsilon^{2}}{c^{2}}\right)^{-\frac{1}{2}} \nonumber \end{equation}

At $\:x=0\:$ we have $\:\upsilon=0\:$, so $\:C=m_{o}\:$ and above equation yields \begin{equation} x\left(\upsilon\right)=\dfrac{m_{o}c^{2}}{f}\left( \dfrac{1}{\sqrt{1-\dfrac{\upsilon^{2}}{c^{2}}}}-1\right) \tag{04} \end{equation} Replacing $\:x\:$ by this expression in equation (01) yields \begin{equation} \bbox[#FFFF88,5px,border:1px solid black]{ m\left(\upsilon\right)=\dfrac{m_{o}}{\sqrt{1-\dfrac{\upsilon^{2}}{c^{2}}}}} \tag{05} \end{equation}

Solving (04) with respect to $\:\upsilon\:$ we have \begin{equation} \left(\dfrac{\upsilon}{c}\right)^{2} = 1-\dfrac{1}{\left(1+\dfrac{f}{m_{o}c^{2}}x\right)^{2}} \tag{06} \end{equation} that is \begin{equation} \upsilon\left(x\right) = c \sqrt{1-\dfrac{1}{\left(1+\dfrac{f}{m_{o}c^{2}}x\right)^{2}}} \tag{07} \end{equation} From this last equation we'll find the function $\:x\left(t\right)\:$ :

\begin{equation} \dfrac{dx}{dt} = \upsilon = c \sqrt{1-\dfrac{1}{\left(1+\dfrac{f}{m_{o}c^{2}}x\right)^{2}}} \tag{08} \end{equation} Changing the variable $\:x\:$ to $\:y\:$ where \begin{equation} y=1+\dfrac{f}{m_{o}c^{2}}x \tag{09} \end{equation} and since from this \begin{equation*} dx =\dfrac{m_{o}c^{2}}{f}dy \quad \text{and} \quad y>1 \end{equation*} equation (08) yields

\begin{equation*} \dfrac{y}{\sqrt{y^2-1}}dy =\dfrac{f}{m_{o}c}dt \end{equation*} or \begin{equation} d\left(\sqrt{y^2-1}\right) =\dfrac{f}{m_{o}c}dt \:\Longrightarrow \: y^2 =\left[ \left(\dfrac{f}{m_{o}c}\right)t + \text{constant}\right]^{2} +1 \tag{10} \end{equation} Since $\: t=o \Rightarrow x=0 \Rightarrow y=1 \Rightarrow \text{constant}=0 \:$, returning to variable $\:x\:$

\begin{equation} x\left(t\right)=c \left[\sqrt{t^2+\left(\dfrac{m_{o}c}{f}\right)^{2}}-\left(\dfrac{m_{o}c}{f}\right)\right] \tag{11} \end{equation}

Inserting this expression of $\:x\:$ on one hand in (07) yields

\begin{equation} \upsilon\left(t\right)=c \sqrt{1-\dfrac{1}{\left(\dfrac{f}{m_{o}c}t\right)^2+1}} \tag{12} \end{equation}

and on the other hand in (01) gives

\begin{equation} m\left(t\right)=\sqrt{m_{o}^{2}+\left(\dfrac{f t}{c}\right)^{2}} \tag{13} \end{equation}

From above equations we have :

\begin{equation} \lim_{t\rightarrow +\infty}x\left(t\right) = +\infty \tag{14} \end{equation}

\begin{equation} \bbox[#FFFF88,5px,border:1px solid black]{ \lim_{t\rightarrow +\infty}\upsilon\left(t\right) = c \;,\quad \upsilon\left(t\right) < c} \tag{15} \end{equation}

\begin{equation} \lim_{t\rightarrow +\infty}m\left(t\right) = +\infty \tag{16} \end{equation}

The force could not accelerate the particle to speed greater than $\:c\:$ because its work is feeding continuously back as mass (inertia) to this particle.

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    $\begingroup$ The problem with taking $\gamma m v$ as "inertia" in relativity is that it is not correct for forces applies transverse to the current direction of motion. See for instance mathpages.com/home/kmath674/kmath674.htm . To go this route you end up with two different relativistic inertias for the same particle (one in the direction of boost and one perpendicular to boost). People have gone that road in the past, and some defined it as clear and useful, but I never mastered it to my own satisfaction. $\endgroup$ – dmckee --- ex-moderator kitten Jul 23 '15 at 16:05

Thanks. Looks good. I trust there are no errors. Saved me a bunch or work. Let me share the reasons for the request.

Say you have an interstellar craft which, though having only a modest thrust, can just keeps the foot on the gas without stop, if so desired. So, given that one will want to stop, eventually, when is it good to turn off the thrust? The advantage to keeping the gas on is that hibernating passengers will age less, and that, if you are looking for an exoplanet, how much ground can you cover? (If that sounds like a book, well ...) And yes, if you are only turning the craft about and using the same thrust to stop, it will take the same time to stop as you spent accelerating.

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