Unit of Acceleration: Calculating Speed with Velocity Addition Formula

[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?
-----------------------------------------------------------------And instead of asking question number 3. I'd like to ask about this question.
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.

 

[bcrowell]

2. The length is 2c.
This is 600,000 km.

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

 

[A.T.]

No matter if you use English or metric

Are asking about the geometric unit system?
https://en.wikipedia.org/wiki/Geometrized_unit_system

 

[Vanadium 50]

It accelerates at 0.5c

This doesn't have units of velocity.

 

[Janus]

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.

Which would make 1 m/s equal to 1/3 arms/swing.

But acceleration is distance/(time * time). I'm having trouble with this term.

http://Why? it stands for distance per sec per sec or change in velocity per sec. In arm distance and swing time 1 m/sec^2 works out to be equal to 0.0555 arms per swing per swing.[/PLAIN]

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?
-----------------------------------------------------------------

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)

And instead of asking question number 3. I'd like to ask about this question.
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.

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

 

[Vanadium 50]

This doesn't have units of velocity.

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

 

[PeterDonis]

acceleration is distance/(time * time). I'm having trouble with this term.

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).

B accelerates at 30,000km/sec2 for 100 seconds then stops accelerating.

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.)

 

[Dale]

The time unit is always second.

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.

 

[Stephanus]

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

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.

 

[Stephanus]

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

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.

 

[Stephanus]

Are asking about the geometric unit system?
https://en.wikipedia.org/wiki/Geometrized_unit_system

No.
I asked about unit of acceleration concerning the speed of light.

 

[Stephanus]

3 It accelerates at 0.5c

This doesn't have units of velocity.

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.

 

[Stephanus]

But acceleration is distance/(time * time). I'm having trouble with this term.

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.

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 something 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.

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,

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

Thanks for the answer.

 

[Stephanus]

something can't accelerate at 0.5c, it could accelerate at .5c per time unit

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

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.

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 $v = \tanh(0.1T)$?

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.

Can we simplify the equation into this?
$T = \frac{sinh^{-1}(at)}{a}$

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

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.

 

[Stephanus]

Wait...

 

[Stephanus]

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

 

[Nugatory]

Something accelerates at 150 000 km / sec 2. How you define 150 000 km / sec 2 in c unit.

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.

 

[PWiz]

Btw, glancing at Janus answer, I see something 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.

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/

 

[Dale]

Something accelerates at 60000km/s^2, we say it accelerates at ...?

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.

 

[Janus]

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.

Then it accelerates at 0.2c/s

 

[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.

 

[Nugatory]

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 $V = tanh(at)$?

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.

One more thing.
In ST diagram, shouldn't the picture rotated 90 degree?

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.

 

[Stephanus]

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.

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.

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.

Oh.

 

[PWiz]

I think I should have changed the title, "What is the equation for acceleration in SR"

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

 

[Stephanus]

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

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

 

[Stephanus]

For a nice derivation of this formula and others, go to http://www.lecture-notes.co.uk/susskind/special-relativity/lecture-6/relativistic-kinematics/

I have clicked your link. I'm new in SR, so there are many things that I don't understand. I'll try to study them.
I want to ask about this equation first.

Position $X^\mu \rightarrow (\tau,0,0,0)$

The position of a given time $X^\mu$ is $(\tau,0,0,0)$. But if I look at this coordinate $(\tau,0,0,0)$ isn't it a vertical line? Or an event at the time axis? I imagine something like $X^\mu = (0,X^\mu,0,0)$
Or $(\tau,0,0,0)$ has other meaning?

 

[PWiz]

But if I look at this coordinate (τ,0,0,0)(\tau,0,0,0) isn't it a vertical line? Or an event at the time axis?

It is. What is the world line for an object in its own rest frame? The object would describe itself as being at rest in its frame (with the object itself the origin of the frame), so the spatial components have to be 0.

 

[Stephanus]

It is. What is the world line for an object in its own rest frame? The object would describe itself as being at rest in its frame (with the object itself the origin of the frame), so the spatial components have to be 0.

So?
What does $X^\mu \rightarrow (\tau,0,0,0)$ mean?
The position coordinate is the time (at rest) $\tau$ multiplied by V?

 

[Nugatory]

The position of a given time $X^\mu$ is $(\tau,0,0,0)$. But if I look at this coordinate $(\tau,0,0,0)$ isn't it a vertical line? Or an event at the time axis? I imagine something like $X^\mu = (0,X^\mu,0,0)$
Or $(\tau,0,0,0)$ has other meaning?

Strictly speaking, $(\tau,0,0,0)$ is a single point. However, you can construct a line (possibly curved, possibly straight) by plotting all the points $(\tau,0,0,0)$ defined by all the possible values of $\tau$.

Whether that line is straight and vertical is just a matter of choosing where on your piece of paper you're going to put the dot corresponding to $(\tau,0,0,0)$ for a particular value of $\tau$ - and that is just a matter of where you've decided to draw your axes. Draw a slanting t-axis and the line of all points $(\tau,0,0,0)$ for all possible value of $\tau$ will be slanting, draw a curved t-axis and the line will be curved. None of this has any physical significance.

(For an analogy, consider drawing a picture of a landscape that includes a pair of railroad tracks heading off into distance... Every point on the railroad tracks will correspond to a point on the sheet of paper on which you're drawing the picture... on the paper the lines corresponding to the railroad tracks will not be parallel even though the real tracks are. If I were drawing a map or a satellite-eye view, I'd get a deifferent picture in which those lines didn't intersect, but that's just a choice I made about how I'd draw my picture).

 

[Stephanus]

Whether that line is straight and vertical is just a matter of choosing where [..]

By "Vertical" perhaps I didn't express myself clearly. $(\tau,0,0,0)$ is a point (as you said) perpendicular to the coordinate origin (0,0,0,0) dimension.
I talked in 2 D (1 time, 1 spatial) dimension such as ST diagram. And of course, a point $(\tau,0)$ is not a (vertical) line. A vertical line would be a line that you track from $(\tau,0)$ to $(0,0)$.
And it's NOT even vertical !. It's a line in 4D dimension. Hmhhhh. What language did I use?
Still reading the rest of your post,...