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Unit of acceleration

  1. Jul 30, 2015 #1
    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.
     
  2. jcsd
  3. Jul 30, 2015 #2

    bcrowell

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    The length can't be 2c, because that wouldn't have units of length.
     
  4. Jul 30, 2015 #3

    A.T.

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  5. Jul 30, 2015 #4

    Vanadium 50

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    This doesn't have units of velocity.
     
  6. Jul 30, 2015 #5

    Janus

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    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:

    [tex]v = c \tanh \left ( \frac{aT}{c} \right )[/tex]

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

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

    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
  7. Jul 30, 2015 #6

    Vanadium 50

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    Misspoke; it doesn't have units of acceleration. It does have units of velocity.
     
  8. Jul 30, 2015 #7

    PeterDonis

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    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.)
     
  9. Jul 30, 2015 #8

    Dale

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    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.
     
  10. Jul 30, 2015 #9
    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.
     
  11. Jul 30, 2015 #10
    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.
     
  12. Jul 31, 2015 #11
  13. Jul 31, 2015 #12
    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.
     
  14. Jul 31, 2015 #13
    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! :eek:
    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.

    Thanks for the answer.
     
  15. Jul 31, 2015 #14
    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
  16. Jul 31, 2015 #15
    Wait...
     
  17. Jul 31, 2015 #16
    Sinh-1 is not 1/sinh it's ArcSinH
    Okay. Correction.
     
  18. Jul 31, 2015 #17

    Nugatory

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    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.
     
  19. Jul 31, 2015 #18
    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/
     
  20. Jul 31, 2015 #19

    Dale

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    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.
     
  21. Jul 31, 2015 #20

    Janus

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    Then it accelerates at 0.2c/s
     
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