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I Relativistic rocket - where is the relativistic mass?

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  1. Oct 28, 2016 #1
    A simple problem with a constant acceleration, ignoring mass of the fuel.

    The velocity of a rocket, which moves withe a constant acceleration g, is equal:
    v(t) = gt

    but I want to keep the const acceleration inside the rocket, not in the absolute sense.

    An acceleration has a dimension: L/T^2 [m/s^2];
    and in a moving rocket the meter is contracted gamma times,
    and a time is dilated gamma times too,
    so finally the absolute acceleration must be: ##g / \gamma^3##

    therefore:
    ##dv=\frac{g}{\gamma^3}dt##

    which can be easili integrated:
    ##\gamma^3dv = gdt\to\, v\gamma = gt\to\, \frac{v}{\sqrt{1-v^2}}=gt##

    finally:
    ##v(t) = \frac{gt}{\sqrt{1+(gt)^2}}##

    Is this a correct speed of the relativistic rocket?
     
  2. jcsd
  3. Oct 28, 2016 #2

    PeterDonis

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    Not in relativity.

    Read this:

    http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html

    I think the expression you end up with for ##v(t)## is consistent with what is in the article I linked to above; but your derivation is hand-waving, whereas that article does it using the correct relativistic starting point.
     
  4. Oct 28, 2016 #3
    What a hand-waving?

    It's rather obvious, that an absolute acceleration of the rocket should be equal to: g / gamma^3, to keep a constant 'gravity' inside the rocket.

    [m/s^2]
    the length contraction: +1, and dilation of time: +2 = 3.

    But it's strange somewhat for me, because there is nothing about the relativistic mass increase.
    Probably the relativistic mass is irrelevant.
     
  5. Oct 28, 2016 #4
    I can use a reversed version of the relativistic rocket, using the g = const in the absolute sense.

    In general: for an absolute acceleration a, the acceleration measured in the rocket is: a' = a x gamma^3, thus the standard work-energy equation is now:

    ##dE = Fdx = m a\gamma^3dx##
    but: ##a = dv/dt## thus:

    ##dE=m\gamma^3\frac{dv}{dt}dx=\gamma^3 vdv=m\gamma == mc^2 \gamma##

    The result is still correct without any relativistic mass usage!
     
  6. Oct 28, 2016 #5

    PeterDonis

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    You started with

    and then used the term "constant acceleration in an absolute sense" to describe this. That's not correct. The only "absolute" (in the sense of invariant) acceleration in relativity is proper acceleration, which is what you called "constant acceleration inside the rocket". There is no such thing as "acceleration in an absolute sense" other than that. So the concept you started from is not valid. It just happened to lead you to the correct answer (because ##v = g t## happens to be the non-relativistic limit of the correct formula).

    This is another example of the same conceptual issue: g / gamma^3 is not the relativistic version of "absolute acceleration inside the rocket". It's just the coordinate acceleration in a particular inertial frame when the proper acceleration is g.

    Yes. Relativistic mass is really an outdated concept; you don't need to use it at all.
     
  7. Oct 28, 2016 #6
    I know from the school and other, more advanced literature, the acceleration is the real quantity, probably because it's a second derivative.

    And if the relativistic mass is outdatet, then why it's still used in the relativity?

    For example: p = gamma mv, what is this?

    further:
    F-relativistic = dp/dt = ... what is this?
    a transversal vs longitudinal force - what is a purpose still to use these fantastic ideas?
     
  8. Oct 28, 2016 #7

    Ibix

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    ##p=\gamma mv## is the relativistic momentum. What we don't do is write ##p=mv## where ##m=\gamma m_0## is the relativistic mass. This is because all sorts of different "relativistic masses" are needed in different circumstances.
     
  9. Oct 28, 2016 #8

    PeterDonis

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    You may think you "know" that, but it isn't true. The only invariant acceleration in relativity is proper acceleration. Coordinate acceleration, the second derivative of position with respect to time, is not invariant; it can change depending on your choice of coordinates.

    If you have any references (textbooks or peer-reviewed papers) that you think support your claim, please post them.

    It almost never is in modern sources. If you are seeing it in sources you are using, they are probably older ones.
     
  10. Oct 28, 2016 #9
    The sense of an absolute quantity doesn't mean it's invariant automaticaly for any observer.
    The absolute means the thing is a real only, but the real things are not invariant, in the relativistic sense at least!

    What is invariant in the reality?
    Nothing is invariant, maybe a pure numbers only:
    for example: there are seven flies in my garden.
    The number 'seven' is preserved everywhere.
     
  11. Oct 28, 2016 #10
    We use term "mass" for ##m_0## only. That's all. It makes no difference if we write ##p=\gamma m_0 v## or ##p=mv## whith ##m=\gamma m_0##. Both variants are identical. And it doesnt matter how we name m or if we name it at all.
     
  12. Oct 28, 2016 #11

    Nugatory

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    "Invariant" has a precise meaning in relativity: A quantity is invariant if its value is the same no matter which frame you use to calculate it.
    That's the standard definition of force since Newton's day. ##F=ma## is the simplified version that we teach in high school physics when the students are not yet familiar with calculus, so not ready for the more complete and generally useful ##F=\frac{dp}{dt}##.
    A transversal vs longitudinal force - what is a purpose still to use these fantastic ideas?[/QUOTE]
    None. The advantage of the modern treatment of relativity based on invariants such as proper acceleration is that it eliminates the need to use these clumsy ideas.
     
  13. Oct 28, 2016 #12

    PeterDonis

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    None of this is physics. The physics question you asked has been answered. Thread closed.
     
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