# Help with constant acceleration under special relativity

• B
Hi, I am trying to wrap my brain around special relativity equations but I'm struggling with the math. I am a computer programmer comfortable with the algebra and but never studied calculus or physics.

Say I am already travelling at velocity "v" where "v" is approaching the speed of light. I then, over a period of observer time "t", accelerate at a constant rate of "a". How do I then calculate the distance "d" I travelled over the period of time "t" (since I started accelerating)? Also how do I calculate my subjective time "T" that elapsed over time "t"?

I understand that if my velocity remains constant then:
T = t * √(1 - (v2 / c2))
but how does the acceleration affect my subjective time? I found another equation:
T = (c / a) * arsinh(at / c)
Can I simply add them together? I would assume not due to special relativity. How do I combine the formulas?

I also understand that if I were accelerating without any existing velocity:
d = (c2 / a) * (√(1 + (at / c)2) - 1)
but shouldn't my existing velocity be taken into consideration when calculating the distance that I cover? I don't see "v" anywhere in that equation. How to I add it?

Also can I just make "a" negative and run it through the formulas again afterwards to decelerate?

## Answers and Replies

Orodruin
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I believe you are missing some very fundamental concepts in relativity, namely the relativity of simultaneity and the relativity of distances. In SR, there is no absolute simultaneity and there is no unambiguous definition of "how far something travelled". The latter is true also in Galilean relativity (i.e., also in classical mechanics). You can only talk about how far something travelled in a given frame and this statement:
How do I then calculate the distance "d" I travelled over the period of time "t"
does not specify relative to what you want to know how far you travelled.

The same is true for velocities. It is non-sensical to just say "travelling at v" without specifying what that velocity is measured relative to.

I just watched MinutePhysics' Intro To Special Relativity series on YouTube and you're right... I was assuming I could define everything from a reference point (say Earth) but that loses all meaning very quickly. I've gotta rethink how I'm looking at the whole thing. Thanks for the quick answer though.

but how does the acceleration affect my subjective time?
$$v=\frac{at}{\sqrt{1+(\frac{at}{c}})^2}$$ where v=0 when t=0.

Orodruin
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$$v=\frac{at}{\sqrt{1+(\frac{at}{c}})^2}$$ where v=0 when t=0.
What do you think this is supposed to represent?

Hi.
McSeb knows how velocity affects IFR time and proper time of the moving body.
Here as its application velocity is not constant anymore but varies with proper acceleration and time.

Orodruin
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Hi.
McSeb knows how velocity affects IFR time and proper time of the moving body.
Here as its application velocity is not constant anymore but varies with proper acceleration and time.
Well, what you wrote down has no meaning unless you explain what you think it is supposed to represent. You cannot just go around throwing formulas around without understanding or explaining what they are supposed to say.

• jbriggs444
Assume a particle (A) is rest on the inertial coordinate system ##K## before time ##t_0##. At time ##t_0##the particle began to accelerate along x-axis (no rotation). For any ##{K_i}’##, at time ##{t_i}’## when the particle is rest relative to ##{K_i}’##, assume the acceleration of the particle is ##{a_A}’ ({t_i}’)=a_0## (a constant not change over ##{t_i}’##) measured on ##{K_i}’##. It's too tired to change the formula to latex these days, so use a picture.

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vanhees71
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What do you think this is supposed to represent?
It represents the solution for relativistic motion under constant proper acceleration. That means by definition that in the instantaneous rest frame the acceleration is constant
$$\dot{u}^{* \mu}=a^{* \mu}=\text{const}.$$
Since ##\dot{u}^{\mu} u_{\mu}=0## because of ##u^{*\mu}=(c,0,0,0)##, we have ##a^{* \mu}=(0,\alpha,0,0)##, where whitout loss of generality I have directed the constant acceleration in ##x^1## direction, in the lab frame we have, using the appopriate boost
$$\dot{u}^{\mu}=a^{\mu}=\frac{\alpha}{c} (u^1,u^0,0,0).$$
The solution is easily found by taking one more derivative wrt. proper time:
$$\ddot{u}^{\mu}=\frac{\alpha}{c} (\dot{u}^1,\dot{u}^0,0,0)=\frac{\alpha^2}{c^2} (u^0,u^1,0,0)=\frac{\alpha^2}{c^2} u^{\mu}.$$
From this we get
$$u^0=A \cosh(\alpha \tau/c)+B \sinh(\alpha \tau/c)$$
with integration constants ##A## and ##B##. Now assuming that at ##\tau=0## the particle is at rest, i.e., ##u^0=c##, we have ##A=c##, and since
$$u^1=\frac{c}{\alpha} \dot{u}^0=c \sinh(\alpha \tau/c) + B \cosh(\alpha \tau/c)$$
we get ##B=0##.
So we have
$$u^{\mu}(\tau)=c [\cosh(\alpha \tau/c),\sinh(\alpha \tau/c),0,0]$$
Integrating once more
$$x^{\mu}(\tau)=[c^2 \sinh(\alpha \tau/c)/\alpha,c^2 \cosh(\alpha \tau/c)/\alpha,0,0],$$
and thus for the lab velocity
$$v^1=c \frac{u^{1}}{u^0}=c \tanh(\alpha \tau)=\frac{\alpha t}{\sqrt{1+\alpha^2 t^2/c^2}}.$$

Last edited:
• SiennaTheGr8
Orodruin
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It represents the solution for relativistic motion under constant proper acceleration. That means by definition that in the instantaneous rest frame the acceleration is constant
$$\dot{u}^{* \mu}=a^{* \mu}=\text{const}.$$
...
Well, yes, but my point is that it is far from clear from #4 how this answers the OP's question or, for that matter, how it relates to the quoted part of the OP. In particular, the quoted section was
but how does the acceleration affect my subjective time?
which I can only take to mean "proper time" and which is not even included in the equation posted in #4. My view is that a post should be helpful to the OP rather than just throwing out a formula without any explanation. In particular when, as is the case here, the OP is struggling with more basic concepts and needs to get those straight first.

• vanhees71
vanhees71
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Well, yes, but my point is that it is far from clear from #4 how this answers the OP's question or, for that matter, how it relates to the quoted part of the OP. In particular, the quoted section was

which I can only take to mean "proper time" and which is not even included in the equation posted in #4. My view is that a post should be helpful to the OP rather than just throwing out a formula without any explanation. In particular when, as is the case here, the OP is struggling with more basic concepts and needs to get those straight first.
I must admit, I didn't get the OP's question first (maybe because it's so hard to read, not using LaTeX), but I think my previous explanation of the formula given in #4 also answers the OP's question.

Of course, it's not "subjective time", because subjective time has no place in physics. To the contrary nowdays physical time is the most accurately definable physical quantity known, and that's why the definitition of the SI base units (taking legal effect very likely already next year) starts from the definition of the second. Anyway, this is off-topic again.

So the OP asked about the definition of proper time in terms of coordinate time of an inertial reference frame. This relation, by definition, is
$$\tau=\int \mathrm{d} t \sqrt{1-\vec{v}^2/c^2},$$
where I assume the trajectory of the particle given in the inertial reference frame (I also invoked the notion of ##c## as the speed of light, because maybe natural units are confusing more at this stage of discussion than helping).

Of course, for ##\vec{v}=\text{const}## you simply get
$$\tau=t \sqrt{1-\vec{v}^2/c^2}.$$
However, for the trajectories, not constant in time, you have to do the integral for the given motion, and my derivation in #9 shows the formula claimed in #1. I edit my posting to get ##c## in place again. The upshot is that
$$t=\frac{c}{\alpha} \sinh\left (\frac{\alpha \tau}{c} \right) \; \Rightarrow \; \tau=\frac{c}{\alpha} \text{arsinh} \left (\frac{\alpha t}{c} \right).$$
You can of course as well get this by using the integral with the velocity given in #4. Reinvoking ##c## it reads
$$v(t)=\frac{\alpha t}{\sqrt{1+(\alpha t/c)^2}}.$$
This gives
$$1-v^2/c^2=\frac{1}{1+\alpha^2 t^2/c^2}.$$
Integrating leads again to the above given formula.

Why do you have an arrow over v in the definition of proper time?
You surely are not taking the square root of a vector, are you?

vanhees71
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##\vec{v}^2## is a scalar.

##\vec{v}^2## is a scalar.

You are using a very anti-conventional notation for a scalar.
The latex-like notation that creates the quote above shows "\vec{v}^2## is a scalar".

Orodruin
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You are using a very anti-conventional notation for a scalar.
The latex-like notation that creates the quote above shows "\vec{v}^2## is a scalar".
There is absolutely nothing unconventional about this notation. It is standard notation for ##\vec v \cdot \vec v##, which is a scalar.

• vanhees71, robphy and PeterDonis