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Equivalence principle and Acceleration

  1. Oct 14, 2008 #1
    From equivalence principle, being on earth is equivalent to being on lift accelerating at 9.8 m/s2...so if we are in a lift and that accelerates at 9.8 m/s2 then at some time we will cross the velocity of light.. my questions are

    1) is earth accelerating ?
    2) what happens if the velocity crosses velocity of light?
    3) what is the equivalent velocity related to earth ,when we say "velocity of lift"...is it escape velocity of earth?
     
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  3. Oct 14, 2008 #2

    Fredrik

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    1. The coordinate acceleration of Earth is zero in some coordinate systems and non-zero in others. (I don't know what you really wanted to ask, but I have a feeling it was something else).

    2. It won't. A constant coordinate acceleration that goes on forever in an inertial frame isn't possible, and if it goes on for a finite time, doesn't feel like constant acceleration. If you want it to feel like constant acceleration forever, you have to make the proper acceleration 9.82 m/s2 at all times. That means that at any time, the coordinate acceleration in the co-moving inertial frame is 9.82 m/s2, but note that this is a different frame at each time. If your proper acceleration is constant, your world line will be a hyperbola (x2-t2=a2), and your speed will approach 1 (=c in the units I'm using), but never reach it.

    3. I don't understand the question.
     
  4. Oct 14, 2008 #3

    D H

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    Relativity experts: Please do not delve into a graduate-level exposition here. The OP doesn't understand the basics yet. Help him to walk before you help him to run.

    Spidey: You have misconstrued the meaning of equivalent here. It means that there is no experiment the people in that elevator car can conduct inside the elevator car that will tell the occupants whether they are at rest on the surface of a gravitational body or in an elevator car out in space that is undergoing a physical acceleration. In short, the effects of gravity and acceleration are equivalent in that the effects are completely indistinguishable.

    So, no the surface of the Earth is not undergoing a physical acceleration, at least not in the sense you are thinking. (Relativity experts: Please read the top notice.)

    A massive body will never reach the speed of light. A futuristic spaceship that can accelerate at 10 meters/second2 for centuries will not exceed the speed of light. The Newtonian concept with which you are familiar, acceleration = change in velocity divided by change in time, is only true for small velocities.

    I don't know what you mean in your third question. Please clarify.
     
  5. Oct 14, 2008 #4

    Fredrik

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    I think you're right DH. He doesn't know the basics, so my answers are probably too advanced, but I'm still going to post a link to a thread where I answered the question I think he really wanted to ask in #1. Here. Unfortunately that answer is more advanced, not less.
     
  6. Oct 14, 2008 #5

    D H

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    Spidey:

    To elaborate on what Fredrik and others said that other thread (linked in post #4), they are talking about something called the 4-acceleration. That is not the same as the second time derivative of the position of some object in three dimensional Euclidean space. For one thing, there is an extra dimension of time in the 4-acceleration. For another, space is not Euclidean in general relativity. The 4-acceleration in general relativity is much closer aligned with what a spring scale measures than what twice differentiating measurements from a ruler.
     
  7. Oct 14, 2008 #6
    Thanks for the info..i misunderstood the word "Equivalent".. . my third question becomes irrelevent now...what i meant is for accelerating lift, the lift has a changing velocity i.e. it has a velocity going upwards and so if we replace that lift with earth and if earth is accelerating then it will have velocity going upwards like lift..so i wanted to ask about this velocity...i thought this would be escape velocity..

    why i thought it as escape velocity is because for lift going upwards with velocity v,if we want to get out of lift we should go above the velocity of lift so that we can escape from lift..and i applied it to earth..i am not sure i clarified the third question...
     
  8. Oct 15, 2008 #7

    DrGreg

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    In relativity, velocities cannot simply be added to each other. Instead we have to use the formula

    [tex]\frac{u + v}{1 + uv/c^2}[/tex]​

    So, if you are experiencing a constant acceleration of 9.8 m/s2 (technically "proper acceleration"), after one second you are moving at 9.8 m/s, but after two seconds, your velocity is not 9.8 + 9.8 = 19.6 m/s, it is instead

    [tex]\frac{9.8 + 9.8}{1 + 9.8^2/c^2} = 19.59999999999997905565337 \mbox{m/s}[/tex]​

    Of course, that difference is tiny, but as time goes by, these tiny differences accumulate and become highly significant. The net effect is, you never go faster than the speed of light.

    If you do the calculus (and with an infinitesimal time change instead of the one second approximation of above), you get the accurate formula

    [tex]\left(x+\frac{c^2}{9.8}\right)^2 - c^2t^2 = \frac{c^4}{9.8^2}[/tex]​

    as Fredrik indicated. That answers question 2.

    The answer to 1 is "yes, relative to any endless stream of falling apples". These apples define a "local inertial frame" (subject to some technicalities). Any point on the surface of the earth accelerates relative to any of its local inertial frames ("undergoes proper acceleration").

    The answer to 3 depends on what height you released the apples from. But it never exceeds the speed of light. Remember there are lots of different inertial frames (all moving at constant velocities relative to each other) so there is no unique speed of anything -- it all depends which frame you choose.
     
  9. Oct 15, 2008 #8

    DrGreg

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    (Spidey, ignore this.)

    On a technicality, proper acceleration is the magnitude of 4-acceleration, and can be defined without reference to 4-acceleration at all - viz coordinate acceleration measured in a co-moving inertial frame.
     
  10. Oct 15, 2008 #9
    thank you all for clearing my doubt..

    i have yet another doubt with light..there is gravitational red and blue shift and because of this frequency is increased or decreased..light is also electromagnetic wave and so it has electric and magnetic components..is frequency the number of oscillations of electric and magnetic components?so if we say frequency is changed,does this mean the number of oscillations of electric and magnetic components is changed?
     
  11. Oct 15, 2008 #10

    Jonathan Scott

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    Gravitational red or blue shift means that the apparent frequency is different from what it was at the location where the light was emitted. In a static case (where no Doppler shift is also involved) it makes more sense to say that the frequency of a wave or signal of any sort (for example, the image of a light turning on and off, or a waving flag) is constant, but that it appears to vary with the observer's location because of local time dilation.
     
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