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I Effect of Special Relativity on spacescraft

  1. Jul 22, 2016 #1
    Let's say I have a Space craft traveling at 1% of speed of light and at rest speed it has an thrust of 600 kN with Specific impulse at 1.000.000s. We know we can calculate the relatavistic mass of the vessel with Eistein Special Relativity : γ = 1 / (1 - v2 / c2 )0.5 = 1.00005.

    Now the question is: what would be the effective thrust and Isp from an outside observer at rest speed?

    My initial though would be 1/ 1.00005 * 600 kN = 599.9699 but this seems too high as I heard Special Relativity only applies for bodies at constant speed, so I need something better

    According to this document, the relatavistic version of newton F = m * a would be
    F = (1 + γ^2 * v * vt) γ * m * a

    but how to calculate vt?
     
    Last edited: Jul 22, 2016
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  3. Jul 22, 2016 #2

    PeterDonis

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    I assume you mean an observer relative to whom the spacecraft is moving at 0.01c?

    I don't know where you "heard" this (you should always give references for statements like this), but it's wrong. SR works just fine for accelerating bodies, as long as there is no gravity present.

    The ##v## and ##v^t## in that formula are a column vector and a row vector, respectively; ##v^t## is the "transpose" of ##v##, what you get when you take a column vector and convert it into a row vector. The force ##F## in the formula is a matrix, not a number; it has to be because, as the article says, in SR the force is not always parallel to the acceleration.

    In your case, if we assume that the spacecraft's thrust is always in exactly the same direction as it is already traveling, then we can ignore those complications and just say that ##F = \left( 1 + \gamma^2 v^2 \right) \gamma m a##, where ##m## is the rest mass (so ##\gamma m## is the relativistic mass that you calculated). Since ##1 + \gamma^2 v^2 = \gamma^2##, this simplifies to ##F = \gamma^3 m a##.

    However, this doesn't necessarily answer your question. See follow-up post.
     
  4. Jul 22, 2016 #3
    I got is from Einstein specialrelativity for dumies
    I guess this is a bad source
     
  5. Jul 22, 2016 #4

    PeterDonis

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    What I assume you mean is, if we assume that the spacecraft itself feels a constant thrust of 600 kN for 1,000,000 seconds of time by the spacecraft's own clock, what thrust will the outside observer see the spacecraft producing, and for how long by the outside observer's clock?

    If that is what you mean, then the formula for force that you gave (with the clarifications I gave) is not enough by itself to answer this question. Why not? Because the spacecraft's speed as seen by the outside observer is not constant. So we first need to figure out how the spacecraft's speed changes.

    For simplicity I'll consider the case where the spacecraft starts out at rest relative to the outside observer, turns on its engine, and exerts 600 kN of thrust for 1,000,000 seconds by its clock. You haven't given the spacecraft's rest mass, so I'll assume that it's 60,000 kg, or 60 metric tons; that gives a proper acceleration (i.e., acceleration felt by the spacecraft's crew) of 10 N/kg = 10 m/s^2, or about 1 g.

    Now we use a very nice trick for describing the motion of the spacecraft in the outside observer's frame. The product of the proper acceleration ##a## (10 m/s^2) and the time ##\tau## by the spacecraft's clock gives what is called the "rapidity", ##w = a \tau##. The speed of the spacecraft, as seen by the outside observer, is the hyperbolic tangent of ##w##, ##v = \tanh w##. Using properties of hyperbolic functions combined with ##\gamma = 1 / \sqrt{1 - v^2}##, we find that ##\gamma = \cosh w## and ##\gamma v = \sinh w##. So now we can rewrite everything in terms of ##w##. The force becomes

    $$
    F = \gamma^3 m a = \cosh^3 w m a
    $$

    where ##ma## is just our 600 kN. The distance ##x## that the spacecraft covers in the outside observer's frame, and the time ##t## elapsed in that frame to cover that distance, are given by

    $$
    x = \frac{1}{a} \left( \cosh w - 1 \right)
    $$

    $$
    t = \frac{1}{a} \sinh w
    $$

    And the range of ##w## that the ship covers is simply zero to ##a \tau## for ##\tau## equal to 1,000,000 seconds, the time elapsed on the spacecraft's clock. These formulas should be sufficient to answer your question.
     
  6. Jul 22, 2016 #5

    PeterDonis

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    Yes, it is. A good rule of thumb is to look in actual textbooks or peer-reviewed papers. Pop science articles are highly variable and often can't be trusted, as you have found here.
     
  7. Jul 22, 2016 #6
    You explain it very well but I still don't fully understand how to apply it effectively. Could you please give an actual example putting these formula into practice?

    What would be the effective thrust of the vessel (with a thrust of 600 kN from it own point of reference) traveling at 0.01c for an outside observer at speed 0?
     
  8. Jul 22, 2016 #7

    PeterDonis

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    We have ##ma = 600 \text{kN}##, and ##\gamma = 1.00005## (the value you calculated earlier for a speed of 0.01c). Those values you can plug into the formula for ##F## given earlier (the one in terms of ##\gamma##, not ##\cosh w##).

    However, that only gives you ##F## for one particular instant, the instant at which the vessel is traveling at 0.01c relative to the outside observer. Since your original question included specific impulse, i.e., a time, you need to know more than just the thrust at one particular instant. You need to know how things vary with time. That's why I gave more general formulas for how things change with time.
     
  9. Jul 22, 2016 #8
    Well for your information, I will calculate the effective thrust every frame (which range between 1/30 sec to 100000000 sec).
    I underatand it becomes inaccurate at higher time frames but that's ok.

    So from my understanding ma = F / y3

    F = 600.000 * 1.000.000 * 9.80665 * 0.5 = 294,1995 GJ

    ma = 294,1995 / 1.00005 3 = 294.15553

    the observable thrust would therefore be 0.99985 * 600 kN= 599.91 kN

    to ensure the same amount of fuel is used, the Isp would have to be adjusted by the same factor

    Correct?
     
  10. Jul 22, 2016 #9

    PeterDonis

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    No, ##ma## is constant; it is the thrust that the spacecraft actually feels, which is 600 kN. ##F## is the thrust that the outside observer sees the spacecraft exerting.

    I have no idea where you are getting this from. ##F = \gamma^3 m a##. If ##\gamma = 1.00005## and ##ma = 600##, then we have ##F = 1.00015 * 600 = 600.09##.

    No, the Isp seen by the outside observer is the total time that the thrust is exerted by the spacecraft, as seen in the outside observer's frame. To calculate that, you need to calculate ##t## using the formula I gave earlier, for the value ##w = a \tau## corresponding to ##\tau = 1,000,000##. To calculate that, you need to know ##a##, not just ##ma##, which means you need to know ##m##, the rest mass of the spacecraft. I assumed that was 60 metric tons earlier. Note also that that formula for ##t##, like all the formulas I gave earlier, assumes that the spacecraft starts from rest in the outside observer's frame. I have not verified that the spacecraft will actually reach a speed of 0.01c within 1,000,000 seconds of its own time at the thrust given (doing that also requires knowing ##a## and hence ##m##).
     
  11. Jul 22, 2016 #10
    huh, I'm confused. I though the vessel would accerate less the closer it neared the speed of light, making it impossible to go faster than it. How can the Force be actualy be higher the faster you travel?

    Edit: That would mean you pull out 0.09 kN from nowehere. It needs to slow down by 0.09 / 600 = 0.015 % to maintain the law of conservation
     
    Last edited: Jul 22, 2016
  12. Jul 22, 2016 #11

    PeterDonis

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    "Accelerate" is ambiguous. The proper acceleration--the acceleration actually felt by the spacecraft and its crew--is constant; it is 600 kN divided by the rest mass of the spacecraft. The coordinate acceleration in the outside observer's frame--the rate at which the spacecraft's speed in that frame changes with time in that frame--does decrease as the vessel speeds up. See below.

    "Force" is also ambiguous. The force felt by the spacecraft is constant--600 kN, the thrust exerted by the engines. But you were asking about the force as measured by the outside observer. It turns out that I was mistaken in saying that increases; it is also constant! I had misinterpreted the formula in the Usenet Physics FAQ article you linked to; it was using ##a## to mean ##dv/dt##, the rate of change of the vessel's speed with respect to time in the outside observer's frame. So let's re-calculate.

    We have the speed ##v## of the vessel in the outside observer's frame given by ##v = \tanh w##. We also have, from the formula I gave earlier for ##t##, ##w = \sinh^{-1} a t##. (These formulas aren't affected by my misinterpretation of the article that I just referred to.) Putting these together gives (note that I am here using units where ##c = 1##, as I did in previous posts; at the end I'll show what happens when we put the factors of ##c## back in):

    $$
    v = \frac{at}{\sqrt{1 + a^2 t^2}}
    $$

    We then simply take the derivative of this with respect to ##t## to get, after some algebra:

    $$
    \frac{dv}{dt} = \frac{a}{\left( 1 + a^2 t^2 \right)^{3/2}}
    $$

    where ##dv / dt## is the vessel's coordinate acceleration in the outside observer's frame. Notice that this decreases as ##t## increases.

    We can go further and rewrite the above in terms of ##\gamma##, by observing that ##\gamma = \cosh w = \sqrt{1 + \sinh^2 w} = \sqrt{1 + a^2 t^2}##. This means that

    $$
    v = \frac{at}{\gamma}
    $$

    and

    $$
    \frac{dv}{dt} = \frac{a}{\gamma^3}
    $$

    If we then substitute this into the (properly interpreted) formula from the article, which is ##F = \gamma^3 m dv/dt##, we get ##F = m a##! That is, the force as measured by the outside observer is constant, and the same as the force felt by the spacecraft crew. (Here I am using ##a## to mean the proper acceleration, and ##dv/dt## to mean the coordinate acceleration, to avoid confusion.) What is going on here is that the decrease in ##dv/dt## as the ship speeds up is exactly compensated by the increase in mass of the ship as seen by the outside observer, plus the effects of time dilation. (See the discussion of Isp below.) Sorry for the mixup on my part in misstating this before; it's been a while since I worked through this from the outside observer's viewpoint.

    We can also see that

    $$
    \frac{d\gamma}{dt} = \frac{a^2 t}{\sqrt{1 + a^2 t^2}} = a v
    $$

    Putting this together with the equation for ##dv/dt## tells us that

    $$
    \frac{d \left( \gamma v \right)}{dt} = v \frac{d\gamma}{dt} + \gamma \frac{dv}{dt} = a v^2 + \frac{a}{\gamma^2} = a
    $$

    So we can rewrite our equation for ##F## as ##F = m d\left( \gamma v \right) / dt = m dp/dt##, where ##p## is the momentum of the spacecraft in the outside observer's frame. This formula is useful because it generalizes, without any modification, to the case where ##F## and ##p## are vectors, i.e., to the case where the force is not parallel to the velocity.

    Finally, what about Isp? Unlike the force above, that is different for the outside observer vs. the spacecraft and its crew. The formula I gave for ##t## above applies: for ##\tau = 1,000,000##, and assuming a spacecraft rest mass of 60 tons, so ##a = 10##, we have (adding factors of ##c## where needed to make the units come out right--I was using units where ##c = 1## in the previous formulas):

    $$
    t = \frac{c}{a} \sinh \left( \frac{a \tau}{c} \right) = \frac{1}{10} 299,792,458 \sinh \left( \frac{10,000,000}{299,792,458} \right) = 1,000,185
    $$

    So the ship's fuel burn will last an extra 185 seconds according to the outside observer. Also, the ship's speed when the burn finishes, according to the outside observer, will be (again adding back the factors of ##c##):

    $$
    \frac{v}{c} = \frac{at}{\sqrt{c^2 + a^2 t^2}} = 0.033344
    $$

    So the ship, starting from rest, would indeed reach (and exceed) 1% of the speed of light during the burn.
     
  13. Jul 23, 2016 #12
    Alright, I understand now that calculating Force is not very useful for my point of reference as it stays the same

    But what it is essential for you to understand is that I'm dealing with a space simulation (KSP) that has no concept if time dilation. It processes everything in time frames of 0.02 seconds and higher (for time warp)

    What I need is the amount of acceleration within a specific amount of time in the coordinate system at rest, which you call dv / dt (=delta velocity divided by delta time)

    Now let's say we want to calculate the proper acceleration for a vessel currently traveling at 0.01c with rest mass 600 mT and 600 kN, with delta time = 0.02 seconds (normal time progression)

    So my acceleration a = 600,000 kg / 600,000 newton = 1 m/s2

    Now rewriting your formula dv / dt = a / y3 to dv = (a / y3) dt

    Enter our data dt = 0.02 and get:

    dv = (1 / 1.000053) 0.02 = 0.019997 m/s speed increase in the coordinate system at rest.

    Although from the vessel point of view, it would still accelerate at 1 m/s and for t = 0.02 that would be 0.02 m

    But in our simulation, everything runs in the same time speed

    This would mean we have to slow down the vessel in the opposite direction it is traveling.

    Now we could achieve this either by applying a force opposite in the direction it is traveling or by reducing the effective thrust by 0.015%

    In both vases would reduce the Effective Thrust (for the simulation)

    However, reducing the effective thrust will reduce the amount of effective fuel used proportionally by the same amount.

    But now I have a contradiction as the reduced fuel spend is more than the reduced amount of fuel spend due to standard time dilation at 0.01c (0.005%)

    Perhaps we need to slow down everything on the vessel by 0.015% ?

    Now I'm confused.
     
  14. Jul 23, 2016 #13

    PeterDonis

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    So you have to do everything in a single inertial frame, the frame of the "observer at rest". No problem.

    Ok, so the rest mass of the spacecraft is 10 times larger than I had assumed. No problem.

    At an instant of time at which the spacecraft is moving at 0.01c in the frame of the observer at rest, yes. But note that, if the spacecraft starts from rest in this frame (speed 0), and its proper acceleration is only 1 m/s, instead of the 10 m/s I assumed, the spacecraft will run out of fuel before it reaches 0.01c.

    But the vessel is time dilated relative to the frame of the observer at rest, so a time step of 0.02 s in the latter frame is only 0.02 / gamma = 0.019999 s in the vessel's frame (assuming the vessel starts the time step moving at 0.01c in the frame of the observer at rest). So from the vessel's point of view, it accelerates at 1 m/s for 0.019999 s.

    Which means you have to do the simulation in the frame of the observer at rest. The simulation can't tell you anything from the vessel's point of view; you'll have to calculate that yourself, separately.

    No, it means the simulation can't tell you anything about what happens from the vessel's point of view. See above.
     
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