# Unit of acceleration

Tags:
1. Jul 30, 2015

### Stephanus

Dear PF Forum,
I have a problem concerning the unit of acceleration.
All this time in Relativity, we have been talking about c.
Either as a distance unit (light travel for 1 second) or as speed.
Sometimes we describe it as 300,000. Sometimes as 300,000,000 or 186,000.
The time unit is always second. No matter if you use English or metric
But this number is well understood.
It just hits me.
For an alien civilization. For a civilization on earth that uses other than English/metric just supposed they say that:
"The speed of light is 100,000,000 arm distance divided by swing time"
Supposed that their 1 arm distance is 0.5 meter
and their (pendulum) swing time is ... $\frac{1}{6} (our) seconds$ of course.
But acceleration is distance/(time * time). I'm having trouble with this term.

1. An object travels at 0.6c.
This I can understand perfectly well
2. The length is 2c.
This is 600,000 km.

-----------------------------------------------------------------
3 It accelerates at 0.5c
Now this is problem. I can't picture what is the acceleration?
-----------------------------------------------------------------

Two observers (A and B) are at rest.
Then B accelerates at 30,000km/sec2 for 100 seconds then stops accelerating. But it still travels steadily at 3,000,000km/sec after stop accelerating wrt A. Now that CAN'T be right is it?

Question:
What is B speed wrt A according to velocity addition formula?
$s = \frac{u+v}{1+uv}$ Remember u and v always in c unit.

2. Jul 30, 2015

### bcrowell

Staff Emeritus
The length can't be 2c, because that wouldn't have units of length.

3. Jul 30, 2015

### A.T.

4. Jul 30, 2015

Staff Emeritus
This doesn't have units of velocity.

5. Jul 30, 2015

### Janus

Staff Emeritus
something can't accelerate at 0.5c, it could accelerate at .5c per time unit ( 150,000,000 m/s^2 or 833333.333 arms/swing^2)
There are some special equations for use with acceleration:

$$v = c \tanh \left ( \frac{aT}{c} \right )$$

Where T is the time measured by the accelerated object and a the acceleration measured by the same.

You can use
$$T= \frac{c}{a} \sinh^{-1} \left ( \frac{at}{c} \right )$$

if you want to measure the time in the non-accelerated frame. Where t is the time measured in that frame.

So if the 100 sec is measured by B, then the final velocity is ~0.999999996c

100 seconds in the non-accelerated frame is equal to ~29.98 sec in the accelerated frame, so if that 100 sec is measured by the non-accelerated frame then the final velocity will be ~0.995c

Last edited by a moderator: May 7, 2017
6. Jul 30, 2015

Staff Emeritus
Misspoke; it doesn't have units of acceleration. It does have units of velocity.

7. Jul 30, 2015

### Staff: Mentor

You might find it easier to think of it as the rate of change of velocity, i.e., the units are velocity/time. So a 1 g acceleration corresponds to a change in velocity of 9.8 meters per second, every second.

Also, however, it's important to keep distinct two concepts that both are described with the word "acceleration". The one I just described is more precisely called "coordinate acceleration", and its behavior when we start talking about relativistic velocities is counterintuitive (see below).

The other concept is "proper acceleration", which it might be easier to think of as "felt force/rest mass". In other words, a 1 g proper acceleration is really a felt force equal to the object's weight when standing on the Earth's surface. This concept, unlike coordinate acceleration, carries over to relativity with no change in behavior in and of itself; what changes is its relationship to coordinate acceleration (see below).

Obviously, as you note, this can't happen. It's impossible to maintain a constant coordinate acceleration indefinitely in any inertial frame, since velocity can't reach the speed of light. What this is really telling you is that you need to think more carefully about exactly what you want B to do.

If what you want B to do is to constantly feel an acceleration equivalent to 30,000 km/sec^2 for 100 seconds, that's easy; just figure out what proper acceleration in g's that corresponds to, and multiply that by B's rest mass to get the force that must be applied. 1 g is about 0.01 km/sec^2, so what you've specified is a proper acceleration of about 3 million g's. That means B would need to feel a force equal to about 3 million times its "normal" weight ("normal" means "when at rest on the surface of the Earth) for 100 seconds. (This is the case that Janus gave formulas for.)

8. Jul 30, 2015

### Staff: Mentor

Not always. Often we use years. As in c is 1 light-year per year. Sometimes I use an approximation of c=1 ft/ns. In class once I had to calculate c in units of furlongs/fortnight.

You are free to choose whatever units you like. You are not required to use seconds for your unit of time.

9. Jul 30, 2015

### Stephanus

Of course no. But that's what we use.
X speed is 0.6. X travels for 2 seconds.
We usually say, X distance is 1.2. While what we mean is 1.2 ls.

10. Jul 30, 2015

### Stephanus

By the way, in space time diagram. Even x-axis is represented as unit of time. I'm not saying that you were incorrect (which very unlikely!). It just then when I want to know about acceleration in SR. I get stuck in this term.
$s = \frac{u+v}{1+uv}$ where we define u and v as factor of c.
Something travels at 60000 km/s, we say it travels at 0.2c
Something accelerates at 60000km/s^2, we say it accelerates at ...?
But there are many post' here. I'l read them one by one.

11. Jul 31, 2015

### Stephanus

12. Jul 31, 2015

### Stephanus

I see that you already corrected it.
But my point is.
Something accelerates at 150 000 km / sec 2. How you define 150 000 km / sec 2 in c unit.

13. Jul 31, 2015

### Stephanus

It's acceleration regarding to Relativiy that concerns me.
Supposed A travels at 3000 km/s. B at 6000 km/s.
You just can't add the velocity: $s = \frac{3000+6000}{1+3000*6000} = 0.0050000$
This is how you do should do it. I'm positive sure that you already know: $s = \frac{0.01+0.02}{1+0002} = 0.029994$ Multiply 0.02994 by c, you'l have 8998.2 km/sec
What I want to know is, how can we do that with acceleration?
Supposed A is accelerating at 3000km/s2 for 3 seconds.
You just can't at-ing the formula to find the velocity. $V = at = 9000km/s$
And I imagine something like this.
$s = \frac{0.01+0.01}{1+0.0001} = 0.019998c = 5999.4 km/sec$
Do that again for three times.
$s = \frac{0.01+0.019998}{1+0.002} = 0.029992c = 8997.601 km/sec$
But why should we 'slice' it in seconds?
Ahhh, the integral. Perhaps I should do integral. Have to give it more time. But we integral-ing acceleration to get distance not velocity!
Btw, glancing at Janus answer, I see someting like sinh, hyperbolic isn't it? I remember playing around with Minkoswki, keep boosting the worldline, it forms something like hyperbolic curve. Should take a look at it more deeply/closely.

That is the concept. Yes. Should take a look at Janus' answer.

14. Jul 31, 2015

### Stephanus

Yes I knew, what I mean is 0.5c/second. But it's just doesn't make any sense if wee integral it to get distance or if we AT it to get velocity

We often describe c is 1.
1. Can we use $v = \tanh(aT)$ for determining velocity?
2. Something is accelerating at 30000 km / sec2. Is this how we calculate the velocity $v = \tanh(0.1T)$?

Can we simplify the equation into this?
$T = \frac{sinh^{-1}(at)}{a}$
I'm sorry. I calculate the final velocity using wolfram alpha. It matches.
But when I calculate the time using wolfram alpha it doesn't match. Where did I go wrong?
$sinh(at) = sinh(0.1 * 100) = sinh(0.1) = 11013.23287$
Putting this number into the equation.
$T = \frac{1}{11013.23282 * 0.1} = 0.000907999$ can't see it near 29.98 seconds.

Correction: Sinh-1 is not 1/Sinh but it's ArcSinh.
Clear now. The mistake is mine.

Last edited: Jul 31, 2015
15. Jul 31, 2015

### Stephanus

Wait...

16. Jul 31, 2015

### Stephanus

Sinh-1 is not 1/sinh it's ArcSinH
Okay. Correction.

17. Jul 31, 2015

### Staff: Mentor

When we say that $c=1$, we're just deciding to use light-seconds instead of meters to express distances (whenever we're using seconds to express times - If we were using years for time we'd use light-years for distance). One light-second is 300,000 km, so an acceleration of 150000 km/sec^2 can also be written as .5 ls/sec^2.

18. Jul 31, 2015

### PWiz

Actually the hyperbolic expression applies for constant proper acceleration. Proper acceleration is the acceleration 'felt' by an object. When $v<<c$, the coordinate acceleration of this accelerating frame as measured by an inertial frame will be approximately equal to the proper acceleration felt by the object.

However, as $v$ increases, we know that the coordinate acceleration measured must reduce (even if the proper acceleration stays constant), since the object would otherwise reach light speed, which is not possible. In fact the actual equation relating these accelerations together is $a= (1-\frac {v^2}{c^2})^{\frac{3}{2}} \alpha$ , where $a$ is the coordinate acceleration and $\alpha$ is the proper acceleration.
For a nice derivation of this formula and others, go to http://www.lecture-notes.co.uk/susskind/special-relativity/lecture-6/relativistic-kinematics/

19. Jul 31, 2015

### Staff: Mentor

You would either say 60000 km/s^2 or about 6000000 g. There isn't a universal natural scale for acceleration as far as we know.

20. Jul 31, 2015

### Janus

Staff Emeritus
Then it accelerates at 0.2c/s

21. Jul 31, 2015

### Stephanus

I think I should have changed the title, "What is the equation for acceleration in SR"
It's not that I just asking withouth thinking. It's just that I'm really lost with the integral of velocity addition in SR.
I see that many misunderstood my question. Which is all my fault, mea culpa.
Janus' answer is the one that I'm looking for. $V = c tanh(\frac{at}{c})$ C is still the factor.
Can we rewrite the equation to $V = tanh(at)$?
One more thing.
https://en.wikipedia.org/wiki/Hyperbolic_function#/media/File:Hyperbolic_functions-2.svg

In ST diagram, shouldn't the picture rotated 90 degree?

Thanks for any valuable responds so far.

22. Jul 31, 2015

### Staff: Mentor

Yes, if you choose units in which $c=1$. Measure time in seconds and distances in light-seconds, and then the equation will read $v=\tanh(at)$. Measure time in seconds and distance in kilometers, and you'll need the 300000 factors there.
Whether the time axis is up or sideways is just a convention; it's right either way. The time-vertical convention everyone here uses may be more common, but you'll often see the other one used, and it's not hard to make the mental adjustment.

23. Jul 31, 2015

### Stephanus

Thanks a lot. It encourages me to study SR next time. If $v = \tanh(at)$ is incorrect, then there's a gap (big one) that I have to study again.
Oh.

24. Jul 31, 2015

### PWiz

If you go through the link I gave in my previous post, you will understand where the hyperbolic function comes from.

25. Jul 31, 2015

### Stephanus

Oh sorry, I haven't clicked it. Wild guess: circle is c2 = x2+y2, hyperbole: c2 = y2-x2Ok, I'll click it.