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Problem involving fundamental principle

  1. Jun 20, 2012 #1
    Hello,

    I was unsure if I should post this or on the General Physics thread but since it's about Einstein's gravity, I'll post it here.

    All right, so the problem is as follows (from Surely You're Joking, Mr. Feynman!):

    I understand the question being asked. My question is, how you would go about to calculate it?
     
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  3. Jun 20, 2012 #2

    PeterDonis

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    Your question is very broad in scope; it would probably help if you would try to narrow it down a little. What do you already know that would be relevant to this problem, and how have you tried to approach it yourself?
     
  4. Jun 20, 2012 #3
    This has to do with proper time, I believe.
     
  5. Jun 21, 2012 #4

    PeterDonis

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    Yes, it does. That's what you want to maximize; it's what Feynman is calling "the time on your clock". The full equation for proper time in a gravitational field is rather complicated, but the following approximate formula, which is derived from the exact equation (which comes from General Relativity), should work fine for this problem:

    [tex]\tau = \int d \tau = \int_{0}^{T} \left[ g h - \frac{1}{2} \left( \frac{dh}{dt} \right)^{2} \right] dt [/tex]

    where h is your height above the ground (as a function of the time t elapsed on the ground clock), g is the acceleration due to gravity, and T is the time elapsed on the ground clock when you are supposed to return (1 hour)--you're assumed to start at time 0.

    The terms in the integrand should look somewhat familiar, btw, if you've had any Newtonian physics; but even if you haven't, you should be able to draw parallels with the two effects Feynman talks about: the first term, gh, gets bigger (more proper time elapsed) as you go higher, but the second gets bigger (less proper time elapsed, because it's being subtracted) as you go faster (dh/dt is your speed). (Btw, I made a key assumption here, that you are only moving vertically, not horizontally; this is because the clock on the ground is assumed to be stationary, so you just go straight up and come straight back down again to it.)

    The goal now is to figure out what function h(t) maximizes the value of the above integral, which gets into the calculus of variations:

    http://en.wikipedia.org/wiki/Calculus_of_variations

    I'll stop here to let you digest what I've said; please feel free to ask further questions once you have.
     
  6. Jun 21, 2012 #5
    About the units, is it safe to assume they are in MKS?

    Okay, so this is what I got from your post. Tau is the proper time; I am integrating Tau in respect to time from 0 to 3600 seconds, since that's the hour on Earth. You mentioned I wanted to maximize the proper time and it is so I can obtain how far the clock on the rocket can go. All right. So for h and dh/dt, namely, v(t), I will play around with values, meters and m/s, respectively? Or must I first figure out what function h(t) maximizes the height so that I can plug that in? I'm a bit confused with the methodology.
     
  7. Jun 21, 2012 #6

    PeterDonis

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    Any consistent system of units will work; if you're most comfortable with MKS, then that's fine.

    Yes.

    Remember it's not just values; h(t) and dh/dt are functions of time, so you need general expressions for them, such as [itex]h(t) = k t^{2}[/itex], where k is some constant.

    Yes, your goal is to find the function h(t) that maximizes the integral (not the height itself but the integral). You can try experimenting by just picking a function h(t), calculating its derivative, and plugging in to the integral to see what you get and how it varies as you change the function; but there's a quicker way.

    Did you read the web page I linked to on the calculus of variations? That gives you a general method for finding the function that maximizes an integral like this. In particular, the section on the Euler-Lagrange equation is applicable to this problem; it talks about how to maximize an integral that has this general form:

    [tex]\int L[x, f(x), f'(x)] dx[/tex]

    where the integrand L is called the "Lagrangian". The integral we have for your particular problem uses t instead of x, h(t) instead of f(x), and dh/dt instead of f'(x), but that's just notation. The web page shows how maximizing the above integral leads to the Euler-Lagrange equation:

    [tex]\frac{\partial L}{\partial f} - \frac{d}{dx} \frac{\partial L}{\partial f'} = 0[/tex]

    where the partial derivatives of L are taken by treating f(x) and f'(x) as independent variables, even though they are really connected. This will give you a differential equation for f(x), or in your particular problem, h(t); actually, it will give an equation for [itex]d^{2} h / dt^{2}[/itex], the second derivative of h, which you can integrate twice to find h itself.
     
  8. Jun 22, 2012 #7
    Okay, I was thinking about doing it this way:

    Using Lorentz transformation,

    [itex]t' = \gamma (t - \frac{vx}{c^2})[/itex]

    where [itex]t'[/itex] is the time of the rocket, [itex]t[/itex] is the time on Earth, and [itex]\gamma[/itex] is [itex]\frac{1}{\sqrt{1 - (\frac{v}{c}})^{2}}[/itex]. Then I'd maximize this in relation to [itex]v[/itex] and [itex]x[/itex], where [itex]v[/itex] is the speed of the rocket and [itex]x[/itex] is the height, or distance from earth. I'd rewrite the Lorentz for velocity as a derivative of distance, [itex]\frac{dx}{dt}[/itex].

    The first derivative would give me the critical points when I set the x value equal to zero. Then, I'd take the second derivative and if it's positive, I'll have a minimum, and if it's a negative, a maximum. Wouldn't this method directly maximize the Lorentz transformation for a time [itex]t[/itex]?

    I guess doing the Lagrange would give me the same result, except I'd have to have the mass of the rocket [itex]m[/itex], and I wrote the Lagrange as,

    [itex]L[/itex] = [itex]\frac{1}{2}[/itex][itex]m[/itex][itex]v^{2}[/itex] + [itex]mgh[/itex],

    and I'd substitute the Lagrangian into the Euler-Lagrange.

    The only issue I'm having is finding a way to derive the Lorentz transformation. You see, I'm an incoming senior in high school and I only have AP Calculus AB under my belt. I'm teaching myself these things as I read articles and books, but I'm managing.

    Edit: I think I got how to derive this, but I will wait for you to check up on this to see if I am on the right path.
     
    Last edited: Jun 22, 2012
  9. Jun 22, 2012 #8

    PeterDonis

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    The Lorentz transformation equation you wrote down won't help you for this problem, because it only applies to a particular event on the rocket's worldline, i.e., a particular point (t', x'). It won't give you a *function* that tells you the time elapsed over the rocket's entire trajectory.

    You could try to use the Lorentz transformation as a starting point to derive an equation that tells you the effect of the rocket's motion, relative to the clock on the ground, on the time elapsed on the rocket's clock. However, that would only cover one of the two effects Feynman was talking about, the rocket's clock running slower as it goes faster. It won't tell you anything about the rocket's clock running *faster* when it is higher up in the gravitational field. To get the right answer you need to take both of these effects into account. That's what the integral I wrote down in post #4 does.

    What you have here is not the Lagrangian, it's the Hamiltonian--the total energy of the rocket, including its kinetic energy (first term) and potential energy (second term). To get the Lagrangian, you have to *subtract* the potential energy from the kinetic energy.

    Also, if you are working with the Lagrangian, you want to *minimize* it; this is the principle of "least action". Feynman's problem asked you to find the trajectory that *maximizes* the time elapsed on the rocket's clock. So you need to flip signs and subtract the kinetic energy from the potential energy, not vice versa.

    Finally, it turns out that you don't even need to know the mass of the rocket; you can divide it out and just work with the rocket's speed and height. This is because of the equivalence principle: one way of stating it is that all objects fall with the same acceleration in a gravitational field. So really all you need is the kinetic and potential energy "per unit mass".

    If you put together the three things I said just above, you will find that you come up with the integral I wrote down in post #4. And in post #6, I mentioned that the integrand in that integral is called the "Lagrangian"; strictly speaking, it's minus the "Lagrangian per unit mass". That integrand is what you want to plug into the Euler-Lagrange equation. In other words, you have

    [tex]L = g h - \frac{1}{2} v^{2}[/tex]

    where v = dh/dt. And this L is what you want to plug into the Euler-Lagrange equation.
     
  10. Jun 23, 2012 #9
    Peter,

    The Langrangian is

    [itex] L[x, y, y'] = L[t, h(t), h'(t)] = gh - \frac{1}{2}(\frac{dh}{dt})^{2}[/itex]

    where h(t) = h, h'(t) = [itex]\frac{dh}{dt}[/itex]. Do I have a [itex]t[/itex] value?
    Also, you confused me with the minimize and maximize bit in post #8. You said I want to minimize the Lagrangian but maximize the function h(t) and h'(t). I do not understand.

    Moreover, I tried to plug in the Lagrangian into the Euler-Lagrange equation, but I do not know what I am doing, to be quite frank. What will that give me? This is what I did:

    [itex]\frac{\partial L[t, h(t), h'(t)]}{\partial h'(t)}[/itex] = [itex]\frac{h'(t)}{gh - \frac{1}{2}(\frac{dh}{dt})^{2}}[/itex] and [itex]\frac{\partial L[t, h(t), h'(t)]}{\partial h(t)} = 0[/itex]

    [itex]\frac{d}{dx}[/itex][itex]\frac{h'(t)}{gh - \frac{1}{2}(\frac{dh}{dt})^{2}} = 0[/itex]

    [itex]\frac{h'(t)}{gh - \frac{1}{2}(\frac{dh}{dt})^{2}} = C = constant[/itex]

    [itex]\Rightarrow[/itex] h'(t) = [itex]\frac{C}{gh - \frac{1}{2}(C)^{2}} := A[/itex]

    [itex]\Rightarrow[/itex] [itex]h(t) = Ax + B[/itex]

    I have a feeling I am completely wrong and this is utter nonsense.
     
  11. Jun 23, 2012 #10

    PeterDonis

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    Yes, that's right. Bear in mind, though, that actually this is minus the Lagrangian divided by the mass of the particle (since the mass appears in both the kinetic and the potential energy, we can divide it out and it won't matter to the answer). That's why we'll be maximizing instead of minimizing. See further comments below.

    You're not trying to do this for a single t value. You're trying to maximize an integral that runs over a range of t values, from 0 to T = 3600 seconds. So you don't want to look at things at just one t value; you want to look at things as a function of t, over the full range of the integral.

    You're not maximizing h'(t); you're just maximizing h(t) [edit: actually you're finding the function h(t) that maximizes the integral]. Once you have h(t), h'(t) is given, since it's just dh/dt; you can't choose it independently. The only reason you have to deal with h'(t) is that it appears in the integral you're trying to maximize. If it makes it easier, you can forget I said anything about minimizing and just focus on maximizing, since that's all that's needed for this problem. :redface: If so, just ignore the next few paragraphs.

    If you do want to have more about minimizing vs. maximizing, first recall the question Feynman asked: he wanted you to find the trajectory that maximizes the reading on the rocket's clock when it returns back to the clock on the ground. We came up with an integral that tells you what the time elapsed on the rocket's clock is, so finding the function h(t) that maximizes the value of that integral will give you the answer to Feynman's question.

    However, the integral we wrote down can also be interpreted in another way. Suppose we multiply the integral by -m; in other words, we flip the signs of both terms in the integrand and put the m back in that we factored out before. The integrand will look like this:

    [tex]\frac{1}{2} m v^{2} - m g h[/tex]

    which is just the object's kinetic energy minus its potential energy. This quantity is what is normally called the "Lagrangian" of the particle [edit: the term "Lagrangian" can have different meanings depending on context--it can mean just anything whose integral you want to maximize/minimize, or it can mean specifically the expression above; sorry about the confusing terminology]; and the integral of the Lagrangian [edit: meaning specifically the expression above this time] over a given trajectory for the object is called the "action" for that trajectory. It turns out that the question Feynman asked, what trajectory, what function h(t), maximizes the time on the rocket's clock, is equivalent to this question: what trajectory, what function h(t), minimizes the action? And one way of expressing the laws of physics as applied to the trajectory of the object is that the object will move on that trajectory which minimizes the action. This is called the "principle of least action", and it is an important principle in physics. We don't really need to go into it further here, but that's why you often see people talking about minimizing the action instead of maximizing the proper time (the time on the rocket's clock in this problem). As I said above, for this case, we can just focus on maximizing the proper time and forget about the minimizing altogether.

    The Euler-Lagrange equation is something different from what you may be used to; it takes a bit of a shift in perspective to see what it's trying to tell you.

    Let me rewrite the Euler-Lagrange equation as it applies to this problem:

    [tex]\frac{\partial L}{\partial h} - \frac{d}{dt} \frac{\partial L}{\partial h'} = 0[/tex]

    [tex]L = gh - \frac{1}{2} v^{2}[/tex]

    The key thing now is that when taking the partial derivatives of L with respect to h and h', we pretend that h and h' are two separate variables, and when we take the partial derivative of L with respect to each one, we treat the other as being held constant. You said you have had Calculus AB, which I don't think covers functions of multiple variables; but it's actually pretty simple in this case. I'll write down the results; they should make sense when you've thought about them for a bit.

    [tex]\frac{\partial L}{\partial h} = g[/tex]

    [tex]\frac{d}{dt} \frac{\partial L}{\partial h'} = \frac{d}{dt} \frac{\partial L}{\partial v} = \frac{d}{dt} \left( - v \right) = - \frac{dv}{dt} = - \frac{d^{2} h}{dt^{2}}[/tex]

    So the Euler-Lagrange equation works out to

    [tex]\frac{d^{2} h}{dt^{2}} = - g[/tex]

    Do you see what this is telling you? And can you see how to integrate this twice to get the function h(t) that answers Feynman's question?
     
    Last edited: Jun 24, 2012
  12. Jun 24, 2012 #11
    It looks like you are assuming g (acceleration of gravity) is constant. For this problem, with the specified 1-hour trajectory, and with constant g, the path would reach a radial position more than three times the radius of the earth, where the acceleration of gravity is only about 1/9 the value at the earth's surface, so we can't assume constant g. Constant g would say the trajectory h(t) is a parabola, but of course the actual trajectory is a cycloid. (Strictly speaking, h(tau) is a cycloid; there's no simple characterization of h(t).)

    Also, if you're willing to take the acceleration of gravity for granted, then there's really no need for any calculation, because obviously the trajectory in question is just an object on a free ballistic trajectory subject to the acceleration of gravity at every point. In other words, your last equation is the definition of g for a free object. The point of Feynman's story was that the guy he was talking to didn't realize this, because he didn't immediately recognize that freefall trajectories in general relativity are precisely those that maximize proper time. So we know the answer without any calculation at all (assuming we know the shape of free fall trajectories).
     
    Last edited: Jun 24, 2012
  13. Jun 24, 2012 #12
    Oh I do understand partial derivatives. To clarify, is this what you did?

    The Lagrangian for this system is [itex]L = gh - \frac{1}{2} ( \frac{dh}{dt})^{2}[/itex] and we plugged that in the Euler-Lagrange which is,

    [itex]\frac{\partial L}{\partial h} - \frac{d}{dt} \frac{\partial L}{\partial h'} = 0[/itex].

    The first part yields,

    [itex]\frac{\partial L}{\partial h} = gh = g[/itex].

    The second yields

    [itex]\frac{d}{dt} \frac{\partial L}{\partial h'} = \frac{d}{dt} \frac{\partial L}{\partial v} = - \frac{1}{2}v^{2} = \frac{d}{dt}(-v) = - \frac{d^{2}h}{dt^{2}}[/itex].

    Then we put the results together,

    [itex] g - (- \frac{d^{2}h}{dt^{2}}) = 0[/itex]

    [itex] \frac{d^{2}h}{dt^{2}} = -g [/itex].

    I do see what it's telling me. The second derivative of h is the gravitational pull to the Earth, or [itex]-g[/itex].

    If I integrate it twice, I obtain

    [itex]\int^{T}_{0}-g dt[/itex]

    [itex]= -gt = \frac{dh}{dt}[/itex]

    and

    [itex]= \int^{T}_{0}-gtdt[/itex]

    [itex]= -g \frac{t^{2}}{2} = h[/itex].

    Then I plug this in the integrand,

    [itex] \tau = \int d \tau = \int^{T}_{0} [g(-gt) - \frac{1}{2}(-g \frac{t^{2}}{2})^{2}]dt[/itex]

    I can integrate these separately and pull out the constants,

    [itex]=-2g \int^{T}_{0} t dt + \frac{1}{2}g \int^{T}_{0} \frac{t^{4}}{4} dt[/itex]

    and I get,

    [itex]= -2g \frac{t^{2}}{2} + \frac{1}{2}g \frac{t^{5}}{20}|^{3600}_{0}[/itex]

    Is the last part correct, with the integration I mean? I plugged in the 3600 and I got a very large number. Moreover, another conceptual question, and perhaps you mentioned it but I missed it. The second derivative of h, that is the acceleration but more explicitly, the velocity and the height that it yields, those are simply the functions, right? So to maximize those functions, I plug the functions in the proper time integral (above, which gave me a large number) or do I have to find a constraint equation so I can maximize the h(t) and v(t) with a Lagrange multiplier?
     
  14. Jun 24, 2012 #13
    You're right, the gravity is not a constant.

    I do know this, he explains why he asked afterwards. The point he was trying to make is that people don't apply what they learn to real things. He said, I quote, "I don't know what's the matter with people: they don't learn by understanding, they learn by some other way — by rote or something. Their knowledge is so fragile!" The point of this thread was to take the question further and actually calculate the program of height and speed so that the proper time is maximized, just out of curiosity.
     
  15. Jun 24, 2012 #14
    Yes, and you understand that this invalidates all your calculations, right?

    Maximizing the proper time for the Schwarzschild metric, for a one-hour round trip, the object needs to be moving at about 4.7 miles/sec (i.e., about 67% of escape velocity) on its upward starting point at the earth's surface, and it must follow a cycloidal path of radius versus (proper) time, reaching a maximum radial distance (from the earth's center) of about 7365 miles (i.e., about 3400 miles above the earth's surface, remembering the earth's radius is about 3964 miles). The parametric angles of the starting and ending points are +-1.494 radians.
     
  16. Jun 24, 2012 #15
    The assumption with my calculations was that gravity was a constant thus making the rocket follow a parabolic trajectory. However, I forgot that the rocket will go into space and gravity is weaker there, so gravity was only "constant" until a certain height for a certain amount of time. My question is, how would I treat gravity in this problem?

    Are the programs of height and speed that you gave me in the second paragraph of your post the answer I have been looking for? If so, how did you do it?
     
  17. Jun 24, 2012 #16
    Yes.

    From the equivalence principle find the field equations as the simplest covariant equations involving derivatives of no greater than second order. From the field equations derive the metric for a spherically symmetrical field (to represent the earth's gravity). By definition, the timelike geodesics of this manifold maximize (or extremize) proper time. Apply the calculus of variations to the metric to determine the equations of geodesic paths in this manifold. Then consider a purely radial geodesic path, and integrate the geodesic equations to show that the radial position versus proper time is a cycloid (coincidentally, just as it is in Newtonian theory). Then plug in the conditions of the stated problem (one hour round trip from the earth's surface) to determine the parameters of the trajectory.

    Before you tackle this, I'd suggest you work out the ballistic radial trajectory for Newtonian mechanics. Since the question concerns very slow motion in very weak gravity, the trajectory is virtually the same as in general relativity.
     
  18. Jun 24, 2012 #17
    That is quite a bit. I honestly have no idea where to begin.
     
  19. Jun 25, 2012 #18

    PeterDonis

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    Yes, you've got it. Everything up to here is correct.

    Yes, although I would recommend including the integration constants when you write these down. So after two integrations, you would have

    [tex]h(t) = - \frac{1}{2} g t^{2} + v_{0} t + h_{0}[/tex]

    where [itex]v_{0}[/itex] and [itex]h_{0}[/itex] are the two integration constants. Since h = 0 when t = 0 (the rocket starts out on the ground), you can immediately see that [itex]h_{0} = 0[/itex]. Since h = 0 when t = T (1 hour) also, you can plug t = T into the above and solve for [itex]v_{0}[/itex] to get

    [tex]v_{0} = \frac{1}{2} g T[/tex]

    You can do that if you want to see what the actual value of the maximized proper time is. But you don't have to to know that the function h(t) that is written down above is the one that maximizes the proper time. The Euler-Lagrange equation already guarantees that. So the function h(t) that is written down above *is* the answer to Feynman's question.

    The point Feynman was trying to make, of course, as Samshorn pointed out, is that the rocket doesn't have to fire its engines at all, after the initial thrust that gives it its upward velocity of [itex]v_{0}[/itex] at time t = 0. After that it can just free-fall until it returns to the ground. In other words, the path that maximizes proper time is the free-fall path.

    I think you mixed up h and dh/dt when plugging in to the integrand. When I do it, I get this:

    [tex]\tau = \int_{0}^{T} \left( gh - \frac{1}{2} v^{2} \right) dt = \int_{0}^{T} \left[ g \left( - \frac{1}{2} g t^{2} + v_{0} t \right) - \frac{1}{2} \left( - gt + v_{0} \right)^{2} \right] dt[/tex]

    which is straightforward, if a bit tedious, to integrate.

    I think I answered this above, but to repeat: the function h(t) that is obtained above *is* the function that maximizes the integral. You're not trying to maximize the function(s), you're trying to maximize the integral, and once you have the function h(t) as derived above, you're basically done. You can plug the function into the integral and plug in numbers to get an actual numerical value for the maximized integral, as above, but the maximizing is already done once you've got the function h(t) above.
     
  20. Jun 25, 2012 #19
    Although I mixed up the terms, was the integration correct?

    But what about the point the guy above made, stating that the gravity is not a constant because the trajectory is not parabolic, rather it is cycloidal due to the rocket reaching a radial position more than three times the radius of the Earth in the hour.
     
    Last edited: Jun 25, 2012
  21. Jun 25, 2012 #20

    PeterDonis

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    Actually, if the problem covers enough range of height that "g" changes (which this one does, I apologize for not pointing that out earlier), then g itself has to be treated as a function. What Samshorn is describing is the most general way of doing that; but he is also using the fact that we already know what trajectory maximizes the proper time--the free-fall trajectory. So what he is really giving you is a way to determine the free-fall trajectory in the general case.

    Another way you could try to work the general case would be to recognize that the term [itex]gh[/itex] in the integrand represents the potential energy per unit mass of the rocket, as I mentioned before. To account for g varying, it's sufficient to just substitute a more general expression for potential energy per unit mass:

    [tex]V(h) = \frac{GM}{R} - \frac{GM}{R + h}[/tex]

    where G is Newton's gravitational constant and R is the radius of the Earth. This expresses the same thing as the term [itex]gh[/itex] did, the difference in potential energy per unit mass between the ground and the rocket's height, but it does it in a way that is valid for any height. So the new integrand would be:

    [tex]\frac{GM}{R} - \frac{GM}{R + h} - \frac{1}{2} \left( \frac{dh}{dt} \right)^{2}[/tex]

    Then you would go through the same process with the Euler-Lagrange equation that we did before.
     
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