What is the time dilation due to acceleration?

Question

what is the formula to calculate time dilation due to acceleration. I have tried to search in google to find out the answer but I couldn't find any equation to calculate the time dilation due to acceleration

Answer 1

There is no time dilation due to acceleration. Kinematic time dilation is due to velocity in an inertial frame and gravitational time dilation is due to gravitational potential.

Answer 2

There are two answers here, depending on your viewpoint.

Inertial observer

An inertial observer sees a rocket whose occupants experience constant acceleration, what motion does the inertial observer see, and what speed do they ascribe to the rocket?

Well, in relativity the additive measure of speed is called rapidity and is a dimensionless angle $\phi$ such that the actual velocity is $v=c~\tanh\phi$, where tanh is the hyperbolic tangent function. So “the rapidity increases linearly with the proper time,” describes a situation where a spaceship thinks it is under constant acceleration. This leads to a simple integral for the time dilation factor, $$\begin{align} {\mathrm dt\over\mathrm d\tau} &=\gamma = {1\over\sqrt{1-(v/c)^2}}\\ &={1\over\sqrt{1-\tanh^2(a \tau/c)}}=\cosh(a\tau/c),\\ t(\tau) &= {c\over a}\sinh\left({a\tau\over c}\right). \end{align}$$ Some hyperbolic trig formulas later and we can say $$\begin{align} \gamma(t) &= \sqrt{1+\left({a t/ c}\right)^2},\\ v(t) &= {at\over\sqrt{1+(a t/c)^2}},\\ x(t) &= {c^2 \over a} \left(\sqrt{1+(a t/c)^2} - 1\right).\end{align}$$ A quick Taylor expansion of the last gives $x(t)\approx\frac12 at^2$ for small times $t\ll c/a$, justifying an interpretation of $a$ as the acceleration the spaceship “feels.” This $\gamma(t)$ is the requested quantity, or you can rephrase as $\gamma =1+\frac{ax}{c^2}$ if preferred. Remarkably, $(\gamma - 1) mc^2$ is the relativistic kinetic energy, and this is just $m~a~x(t)$, the classical Work-Energy formula for a force which generates some constant acceleration $a$.

On the spaceship

Now consider that the inertial observer has left a clock at some place $x = L$ ticking away synchronized to their time $t$. We want to do a Lorentz boost into an inertial frame co-moving with the accelerated spaceship frame, when that frame’s clock shows a proper time $\tau$. This transform is ($w=ct$) $$\Delta w' = \gamma~(\Delta w -\beta~\Delta x),\\ \Delta x' = \gamma~(\Delta x -\beta~\Delta w).$$ We have to fill in these formulas and differences, so we know that $\gamma =\cosh\phi$, and that $\beta = v/c = \tanh\phi$, where $\phi=a\tau/c$, that $\Delta w' = c~(t'-\tau)$ while $\Delta w = c~(t-(c/a)\sinh\phi)$ and so on. And we want to first know what time is the clock showing right now, so what is $t(L, \tau)$ assuming $\Delta w'=0$. We find from the first equation that it shows$$ t = \left({L\over c} + {c\over a}\right)\tanh\left({\alpha \tau\over c}\right).$$ In other words on the spaceship, looking at the inertial frame, time in the inertial frame appears to tick at a positionally dependent rate, namely dependent on your distance from an imaginary plane at $x=-c^2/a.$ Clocks with that $L$ do not tick at all, $t=0$ no matter what $\tau$ hence $\phi$ is. This is an “event horizon” generated by the fact that the formula we had in the last section, $$x(t) = {c^2 \over a} \left(\sqrt{1+(a t/c)^2} - 1\right),$$ is a hyperbola with an asymptote $x = ct-(c^2/a).$ As the spaceship starts to go nearer and nearer the speed of light, some light pulses a finite distance behind it simply can never catch up to it, so it will never see any later ticks of those clocks behind it.

I like to tell people that this is basically the essence of special relativity. A lot of folks say that you shouldn't mix special relativity with acceleration, I tend to say that you should explain special relativity as the remarkable proposal that, it is a universal property of acceleration in our universe, that you can detect your acceleration $a$ and it's direction (say along the $x$-axis), and, when $a$ is nonzero, then clocks ahead of you by coordinate $x$ appear to tick at a rate $1+ax/c^2$ seconds per second, in addition to whatever Doppler effect you already expected. This is a different take on the “relativity of simultaneity” and in fact you can derive the whole Lorentz transformation from this little fact alone. This fact also immediately resolves several purported “paradoxes” like the Twin Paradox: the traveling twin saw the clocks on Earth tick slower just the same as Earth saw her clock tick slower, but she is aware of accelerating at some great distance from Earth, towards Earth, and during that time she must see Earth's clock tick much much faster than her own due to the tremendous distance between her and it, and this is why she agrees that her twin is older.

Answer 3

The problem here is that the phrase 'time dilation' has a particular meaning which refers to the difference in the time interval between two events in an inertial frame in which they occur in the same spot, and the the interval between the same events in another inertial frame in which they occur in different places. In that sense of the phrase 'time dilation', acceleration is totally irrelevant.

However, some people use the phrase time dilation in a more general and looser sense to refer to any case, such as a 'twin-paradox' set-up, in which the time experienced by one person differs from the time experienced by another when one person follows a straight path through spacetime and the other doesn't. If you google 'proper time' 'path' 'integral' you should soon find plenty of equations that allow you to calculate the effect of acceleration on proper time.