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Definition time

  1. Sep 24, 2010 #1
    This is a continuation of the thread Derivation of proper time of acceleration in SR

    We have definitions for proper time, proper speed, proper acceleration, coordinate speed, coordinate time, and coordinate acceleration.

    1) What is the definition ofproper distance?

    2) If proper speed (or velocity) is [itex] v_p = v* \gamma[/itex] then it is conceivable for [itex] v_p [/itex] to exceed the speed of light. Is that true? If so, what physical meaning does that have??? Is there a limit on [itex] v_p < c?[/itex]. Seems like there should be in the sense that there should be NO FR for a given observation in which [itex] v \geq c [/itex]

    Likewise, proper acceleration or [itex]a_p = a\gamma^3[/itex] also can be [itex] > c [/itex] but I have been told that there is no limit on acceleration other than that the coordinate velocity be [itex] < c [/itex]. Is that valid?
     
    Last edited: Sep 25, 2010
  2. jcsd
  3. Sep 24, 2010 #2

    JesseM

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    The proper distance between two events with a spacelike separation is the distance between them in the frame where they occur simultaneously.
    Proper speed is the distance you travel in some specific (arbitrary) inertial frame per unit of the traveler's own proper time. For example, suppose I want to define proper speed relative to the Earth's rest frame for a rocket traveling at 0.8c relative to the Earth. Then to reach a destination 20 light-years away in the Earth frame, the rocket would take 20/0.8 = 25 years to get there in the Earth frame, but the rocket's own clock would be slowed by a factor of 0.6 so the rocket would only experience a time of 25*0.6=15 years to reach the destination. In this case the "proper speed" is 20 light-years / 15 years = (4/3)c. As you can see from the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken] it is quite possible to reach destinations which are a vast distance away in the Earth's frame in fairly short amounts of onboard time, like getting to the center of the galaxy 30,000 light years away in only 20 years of onboard time, accelerating at 1G for half the trip and then decelerating at 1G for the second half.
    That doesn't really make sense, speed and acceleration have different units so you can't say an acceleration is "greater than c" or "less than c" any more than you can say a mass or time interval is "greater than c" or "less than c". But if you're wondering about the physical meaning of proper acceleration, proper acceleration at any point on an object's worldline is just the instantaneous coordinate acceleration in the inertial frame where they are instantaneously at rest at that point, and the proper acceleration also corresponds to the G-force that would be felt by an observer on board the accelerating rocket (which could be measured with an accelerometer).
    Yes, at any given moment the acceleration can be arbitrarily high, but your coordinate acceleration in some inertial frame must decrease before you reach a speed of c in that frame.
     
    Last edited by a moderator: May 4, 2017
  4. Sep 24, 2010 #3

    DrGreg

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    If you think about it long enough, celerity (=proper speed) is what a car's speedometer measures -- the revolutions of a wheel measure distance along the ground, but the time is the car's own time.

    You can't compare apples with oranges so it makes little sense to compare a celerity with a speed. The celerity of light is infinite*, so the celerity of any mass is always smaller than the celerity of light.

    *To be pedantic and strictly rigorous about this, it would be more accurate to say the celerity of light is undefined and the celerity of any massive object is unbounded, diverging to infinity as the speed tends to c.
     
  5. Sep 24, 2010 #4
    I am not sure Dr Greg?

    The ground is length contracted during movement, so the speedometer would measure velocity wrt. the length contracted road. No?

    Celerity would be proper time with respect to 'map distance'.

    Am I wrong?
     
  6. Sep 24, 2010 #5
    That is why in the twins paradox the "traveling twin" ages slower than the sit-on-his-butt twin - which you calculated for me me many months ago. Of course, that poor slob who goes to a distant galaxy and returns in a fairly short time finds all his friends and family are a hell of a lot older than he is - and he can't go back.

    You're right - acceleration and velocity are two different animals. The restriction in each FR is that the velocity must not equal or exceed c.
     
    Last edited by a moderator: May 4, 2017
  7. Sep 24, 2010 #6
    Yes, you are absolutely right, [tex]v_p=v \gamma(v)[/tex] implies [tex]c_p= c \gamma(c)=\infty[/tex]

    It makes no sense to compare celerity with speed and to declare that celerity can exceed speed of light. It is like saying that tomatoes are redder than potatoes are elliptical. :-)
     
  8. Sep 25, 2010 #7
    Then, is proper speed

    [tex] v_p = v \gamma ?[/tex]

    Then, what does celerity mean? )-:
     
  9. Sep 25, 2010 #8
    It basically means how far you travel using a chart divided by the travelers proper time.

    Example:

    Suppose a space traveler wishes to go to location X far away from Earth.

    Now, he checks the latest edition of Google space charts and verifies it is 1 light year away. He wants to get there in 0.75 years, how fast does he have to travel?

    The answer is simple: he needs a celerity (or proper velocity) of 1 1/3 since 1/0.75=1 1/3.

    A celerity of 0.75 translates into a velocity as follows:

    [tex]v = \frac{w}{\sqrt{1 +w^2}} = \frac{0.75}{\sqrt{1 +0.75^2}} = 0.6[/tex]

    So he needs a velocity of 06c and relativity takes care of the rest!
     
    Last edited: Sep 25, 2010
  10. Sep 25, 2010 #9
    well put
     
  11. Sep 25, 2010 #10
    yes


    Proper speed and celerity are two different names for the same thing.
     
  12. Sep 25, 2010 #11
    Passionflower -

    I am having a brain dump )-:

    where does this equation below (from your post) come from?

    [tex]v = \frac{w}{\sqrt{1 +w^2}} = \frac{0.75}{\sqrt{1 +0.75^2}} = 0.6[/tex]

    I don't remember it.

    In addition (this is for anyone) I have a hypothetical simple problem.

    An astronaut departs Earth (point E.) He/she accelerates to 0.6c over a two year period.

    He/she then travels at 0.6 c for 6 years and then decelerates to 0 over the next two years to a planetary system at Star S.

    The return is the opposite:

    acceleration from S from 0 to 0.6 over two years,

    steady state coasting at 0.6 c for six years

    deceleration to 0 at E over two years.

    This is a perfectly symmetrical "loop" hence the astronaut will be younger than the cohort of people who were the same age as he/was when he/she left and the speed was relativistic (0.6 c over a total of 12 years + the accleration/deceleration phases.)

    I need to figure out:

    a) We know by Earth time or lab time as starthaus often chooses to refer to it, this astronaut was gone for 20 years. How much older is this astronaut in his/her own or rocket ship time ?

    b) How far did he/she actually travel (one way or both ways) both in distance as measured by Earth and distance as measured by te rocket ship.

    c) Assume, of course, no effect of gravity from other planetary or celestial bodies.

    This acceleration is about [itex] 3 m/sec^2 [/itex] which is 1/3 gravity. This would be "do-able."
     
    Last edited: Sep 25, 2010
  13. Sep 25, 2010 #12
    w is the same [tex]v_p[/tex]

    So, if you invert:

    [tex]w=v \gamma(v)[/tex] you get Passion's formula.
     
    Last edited: Sep 25, 2010
  14. Sep 25, 2010 #13
    starthaus - "I" - what?

    I know that for 12 "earth-years" the astronaut coasted for 2x0.6x6 = 7.2 light-years of earth distance. The proper time of the astronaut would thus be (6 + 6) yr x 0.8 = 9.6 years for this period of "coasting" at 0.6 c. (That is because [itex] \gamma = 0.8 [/itex] for a v of 0.6.) This confirms the notion that the proper velocity of 7.2/9.6 = 0.75 c is greater than the Earth coordinate velocity of 7.2 ltyr/12 yr = 0.6 (given.)

    Now I have to fish for the correct formulas in the prior thread on Derivation of Proper Time for Acceleration in SR to pick up the elapsed distances and proper times for the acceleration/deceleration phases. I will assume that we will start off at some acceleration [itex] a [/itex] in S (the Earth frame) which will decay over the two earth-years to 0.6 c.

    There will be then 6 years of coasting. Then I will assume that at 8 earth years out there will be a deceleration from 0.6 c to zero over two years (terminus at 10 years,) and then an acceleration from 0 to 0.6 c over two years starting at 10 years out with again a 6 year period of coasting and finally a deceleration from 0.6 c to 0 starting at 18 years out and ending up at 20 years. We have two periods of acceleration (0 [itex]\Rightarrow[/itex] 0.6 c) and two periods of deceleration (0.6 c [itex]\Rightarrow[/itex] 0.)
    We don't know the distances during these four periods nor do we know the proper times during these periods.

    Fishing for those formulas in that 100+ post thread will be a real bear. I will not object if I can get some input on that as I have no textbook. I don't even know what textbook would carry it. I think I have split the problem down to simple terms enough.

    If nothing else, confirmation that

    a) [itex] v_p = 0.75 c [/itex] during the coasting phase and that my proper times and distances for those coasting phases are correct.
    b) My division of the trip into the above segments is correct

    would be most helpful and prevent me from going down a blind alley.

    This problem would be a piece of cake in Newtonian Physics even with a decaying accleration/deceleration..
     
  15. Sep 25, 2010 #14
    ...but you know the elapsed coordinate time . I must have shown you this calculation several times already.
     
  16. Sep 25, 2010 #15
    starthaus -

    You did - in earlier Sept. and maybe before that. I downloaded it then. Now that I've been through hyperbolic functions, I will seek to understand it and calculate the answers to my questions. I will get back when I am done with a "short" version of what I have done. I had to print it out. I can't read it on-line.

    stevmg
     
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