Problem involving fundamental principle

[Alcubierre]

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!):

You blast off in a rocket which has a clock on board, and there's a clock on the ground. The idea is that you have to be back when the clock on the ground says one hour has passed. Now you want it so that when you come back, your clock is as far ahead as possible. According to Einstein, if you go very high, your clock will go faster, because the higher something is in a gravitational field, the faster its clock goes. But if you try to go too high, since you've only got an hour, you have to go so fast to get there that the speed slows your clock down. So you can't go too high. The question is, exactly what program of speed and height should you make so that you get the maximum time on your clock?

I understand the question being asked. My question is, how you would go about to calculate it?

 

[PeterDonis]

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?

 

[Alcubierre]

This has to do with proper time, I believe.

 

[PeterDonis]

This has to do with proper time, I believe.

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:

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

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.

 

[Alcubierre]

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.

 

[PeterDonis]

About the units, is it safe to assume they are in MKS?

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

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.

Yes.

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?

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

Or must I first figure out what function h(t) maximizes the height so that I can plug that in?

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 into 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:

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

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:

$$\frac{\partial L}{\partial f} - \frac{d}{dx} \frac{\partial L}{\partial f'} = 0$$

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 $d^{2} h / dt^{2}$, the second derivative of h, which you can integrate twice to find h itself.

 

[Alcubierre]

Okay, I was thinking about doing it this way:

Using Lorentz transformation,

$t' = \gamma (t - \frac{vx}{c^2})$

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

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 $t$?

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

$L$ = $\frac{1}{2}$$m$$v^{2}$ + $mgh$,

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.

 

[PeterDonis]

Using Lorentz transformation

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.

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

$L$ = $\frac{1}{2}$$m$$v^{2}$ + $mgh$,

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

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

$$L = g h - \frac{1}{2} v^{2}$$

where v = dh/dt. And this L is what you want to plug into the Euler-Lagrange equation.

 

[Alcubierre]

Peter,

The Langrangian is

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

where h(t) = h, h'(t) = $\frac{dh}{dt}$. Do I have a $t$ 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:

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

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

$\frac{h'(t)}{gh - \frac{1}{2}(\frac{dh}{dt})^{2}} = C = constant$

$\Rightarrow$ h'(t) = $\frac{C}{gh - \frac{1}{2}(C)^{2}} := A$

$\Rightarrow$ $h(t) = Ax + B$

I have a feeling I am completely wrong and this is utter nonsense.

 

[PeterDonis]

The Langrangian is

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

where h(t) = h, h'(t) = $\frac{dh}{dt}$.

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.

Do I have a $t$ value?

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.

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).

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. 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:

$$\frac{1}{2} m v^{2} - m g h$$

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.

Moreover, I tried to plug in the Lagrangian into the Euler-Lagrange equation

...

I have a feeling I am completely wrong

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:

$$\frac{\partial L}{\partial h} - \frac{d}{dt} \frac{\partial L}{\partial h'} = 0$$

$$L = gh - \frac{1}{2} v^{2}$$

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.

$$\frac{\partial L}{\partial h} = g$$

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

So the Euler-Lagrange equation works out to

$$\frac{d^{2} h}{dt^{2}} = - g$$

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?

 

[Samshorn]

So the Euler-Lagrange equation works out to

$$\frac{d^{2} h}{dt^{2}} = - g$$

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?

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).

 

[Alcubierre]

Oh I do understand partial derivatives. To clarify, is this what you did?

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

$\frac{\partial L}{\partial h} - \frac{d}{dt} \frac{\partial L}{\partial h'} = 0$.

The first part yields,

$\frac{\partial L}{\partial h} = gh = g$.

The second yields

$\frac{d}{dt} \frac{\partial L}{\partial h'} = \frac{d}{dt} \frac{\partial L}{\partial v} = - \frac{1}{2}v^{2} = \frac{d}{dt}(-v) = - \frac{d^{2}h}{dt^{2}}$.

Then we put the results together,

$g - (- \frac{d^{2}h}{dt^{2}}) = 0$

$\frac{d^{2}h}{dt^{2}} = -g$.

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

If I integrate it twice, I obtain

$\int^{T}_{0}-g dt$

$= -gt = \frac{dh}{dt}$

and

$= \int^{T}_{0}-gtdt$

$= -g \frac{t^{2}}{2} = h$.

Then I plug this in the integrand,

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

I can integrate these separately and pull out the constants,

$=-2g \int^{T}_{0} t dt + \frac{1}{2}g \int^{T}_{0} \frac{t^{4}}{4} dt$

and I get,

$= -2g \frac{t^{2}}{2} + \frac{1}{2}g \frac{t^{5}}{20}|^{3600}_{0}$

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?

 

[Alcubierre]

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).)

You're right, the gravity is not a constant.

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).

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.

 

[Samshorn]

You're right, the gravity is not a constant.

Yes, and you understand that this invalidates all your calculations, right?

The point of this thread was to actually calculate the program of height and speed so that the proper time is maximized, just out of curiosity.

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.

 

[Alcubierre]

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?

 

[Samshorn]

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?

Yes.

If so, how did you do it?

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.

 

[Alcubierre]

That is quite a bit. I honestly have no idea where to begin.

 

[PeterDonis]

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

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

If I integrate it twice, I obtain

$\int^{T}_{0}-g dt$

$= -gt = \frac{dh}{dt}$

and

$= \int^{T}_{0}-gtdt$

$= -g \frac{t^{2}}{2} = h$.

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

$$h(t) = - \frac{1}{2} g t^{2} + v_{0} t + h_{0}$$

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

$$v_{0} = \frac{1}{2} g T$$

Then I plug this in the integrand

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 $v_{0}$ 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.

Is the last part correct, with the integration I mean?

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

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

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

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?

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.

 

[Alcubierre]

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.

 

[PeterDonis]

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.

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 $gh$ 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:

$$V(h) = \frac{GM}{R} - \frac{GM}{R + h}$$

where G is Newton's gravitational constant and R is the radius of the Earth. This expresses the same thing as the term $gh$ 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:

$$\frac{GM}{R} - \frac{GM}{R + h} - \frac{1}{2} \left( \frac{dh}{dt} \right)^{2}$$

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

 

[Alcubierre]

So the new integrand would be:

$$\frac{GM}{R} - \frac{GM}{R + h} - \frac{1}{2} \left( \frac{dh}{dt} \right)^{2}$$

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

The integrand being the Lagrange, and I'd do the same thing. All right, thank you for your patience. I will let you know if I have any trouble. Just a quick question before I attempt this, the integration of G and R would be something like Gt and Rt or would I factor or pull those out of the integral?

 

[PeterDonis]

To integrate that, I'd just normally integrate, well separately, and treat v-not and gravity as a number?

Yes, each term can be integrated separately with respect to t.

Which means I'd just tag a t term?

I'm not sure what you mean by this, but each term is just a constant times some power of t, so you just use the rule for integrating powers of the integration variable.

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.

See my post just prior to this one.

 

[PeterDonis]

The integrand being the Lagrange, and I'd do the same thing. All right, thank you for your patience. I will let you know if I have any trouble. Just a quick question before I attempt this, the integration of G and R would be something like Gt and Rt or would I factor or pull those out of the integral?

G, M, and R are all constants, but they won't occur separately; they'll only occur combined in one term, like the first. But what I wrote down isn't what you integrate, not yet; remember that you have to take the partial derivatives with respect to h and dh/dt of what I wrote down. I shouldn't have called it the "integrand"; it's really the "Lagrangian", just like $gh - 1/2 v^{2}$ was before.

 

[Alcubierre]

Oh yes, I know what you meant as the integrand, it is the Lagrangian. I was just asking how I'd integrate, or also, derive it, but I'll just treat them as a constant like you mentioned above.

 

[Alcubierre]

Peter, this is what I got:

The Lagrangian is,

$$L = \frac{GM}{R} - \frac{GM}{R + h} - \frac{1}{2}(\frac{dh}{dt})^{2}$$

Plugging that into the Euler-Lagrange yields,

$$\frac{\partial L}{\partial h} = \frac{GM}{R} - \frac{GM}{R + h} = (-) \frac{GM}{(R + h)^{2}}$$

And,

$$\frac{d}{dx} \frac{\partial L}{\partial v} = - \frac{1}{2}v^{2} = \frac{d}{dx}(-v) = - \frac{d^{2}h}{dt^{2}}$$.

Thus,

$$(-) \frac{GM}{(R + h)^{2}} + \frac{d^{2}h}{dt^{2}} = 0$$

$$\frac{d^{2}h}{dt^{2}} = (-) \frac{GM}{(R + h)^{2}}$$

I put the negative sign in parenthesis because I am not sure if it is negative or positive, so I will change that once you take a look.

When I integrated that, I got

$$\frac{dh}{dt} = (-) \frac{GMt}{(R + h)^{2}} + v_{0}$$

and

$$h(t) = (-) \frac{GMt^{2}}{2(R + h)^{2}} + v_{0}t + h_{0}$$

Because h₀ = 0 when t = 0, it "goes" away. So when I put that in the integral I got,

$$\tau = \int^{T}_{0}\left[\left( (-) \frac{GMt^{2}}{2(R + h)^{2}} + v_{0}t \right) - \frac{1}{2} \left( (-) \frac{GM}{(R + h)^{2}}+ v_{0} \right)^{2}\right] dt$$

Again, the negative sign is in parenthesis because it is in question.
So what do you think?

 

[PeterDonis]

I put the negative sign in parenthesis because I am not sure if it is negative or positive

The negative sign should be there in the last equation for d^2h/dt^2. Taking the derivative of the GM/(R+h) term flips the sign, but there's a negative sign in front of that term in the Lagrangian, so the two negatives cancel. Asn they should, because we *want* the negative sign in the equation for d^2h/dt^2, since we want the rocket to be accelerated downward by gravity.

With the correct sign as above, everything to this point looks correct.

When I integrated that, I got

$$\frac{dh}{dt} = (-) \frac{GMt}{(R + h)^{2}} + v_{0}$$

and

$$h(t) = (-) \frac{GMt^{2}}{2(R + h)^{2}} + v_{0}t + h_{0}$$

Here's where it gets tricky. There is an h in the denominator of the integrand, so you can't just treat it as a constant like "g" was before. That makes the integral a lot harder if you want to try to do it directly. I don't have time to go into that now; the only comment I can make is that for this version of the problem, it's better to "cheat", in a way, by using the knowledge we got from the simpler version we've already done, which showed us that the trajectory that maximized the proper time was the free-fall trajectory. There are lots of different ways to figure out what the free-fall trajectory is, besides maximizing the Lagrangian integral. Samshorn talked about some of them. You can see some of the equations that result on these pages:

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

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

 

[Samshorn]

...the only comment I can make is that for this version of the problem, it's better to "cheat", in a way, by using the knowledge we got from the simpler version we've already done, which showed us that the trajectory that maximized the proper time was the free-fall trajectory. There are lots of different ways to figure out what the free-fall trajectory is, besides maximizing the Lagrangian integral. Samshorn talked about some of them.

That's exactly backwards. I explained how to derive the trajectory purely and explicitly from the condition that it maximizes the proper time, as requested. The approach you attempted did not relate to the proper time at all. You simply asserted - incorrectly - that the incremental proper time equals the integral of the classical Lagrangian of a particle in free-fall. But it doesn't, so your whole approach is false from the start. At best (after correcting your erroneous assumption of constant g), all you are doing is struggling to derive the classical trajectory of a particle in free fall. You're not giving a derivation from maximizing proper time, which is what the OP requested. So you have things exactly backwards.

Also, please note that the "simpler derivation" you attempted did not "show us that the trajectory that maximized the proper time was the free-fall trajectory". First, your derivation didn't have anything to do with proper time (aside from an erroneous assertion). Second, and more importantly, free-fall trajectories (i.e., timelike geodesics) are DEFINED as the paths that maximize proper time, so there was nothing to "show". That was the whole point of Feynman's story - that many people (who have learned things by some kind of rote) will go off and struggle with the problem without even realizing that what they are doing is deriving the geodesic equations. Maybe the OP was just trolling to illustrate Feynman's point?

 

[TrickyDicky]

That was the whole point of Feynman's story - that many people (who have learned things by some kind of rote) will go off and struggle with the problem without even realizing that what they are doing is deriving the geodesic equations.

And it's a really good point Feynman makes, that can be made extensive to science in general wrt people just preaching what they learned by rote, not really understanding it.

 

[PeterDonis]

That's exactly backwards. I explained how to derive the trajectory purely and explicitly from the condition that it maximizes the proper time, as requested.

Hm, yes, you're right, you were talking about doing it using the geodesic equation. Sorry for the mixup on my part.

The approach you attempted did not relate to the proper time at all. You simply asserted - incorrectly - that the incremental proper time equals the integral of the classical Lagrangian of a particle in free-fall.

Please go back and re-read all my posts. You are correct that the expression $gh - 1/2 v^{2}$ is not the classical Lagrangian, strictly speaking. It's minus the classical Lagrangian divided by the particle's mass. I said that explicitly in at least one previous post, and it seems like the OP understands the distinction.

But minus the classical Lagrangian divided by the particle's mass also happens to be the equation for the elapsed proper time of the particle following the trajectory h(t), provided that g can be considered constant over the range of height covered. That can be shown directly from the metric.

You pointed out, correctly, that g can't be considered constant for the problem as specified, which I hadn't spotted because I hadn't run the actual numbers. If g can't be considered constant, then the integral of minus the Lagrangian divided by the mass gets more complicated, as I noted in my last post.

At best (after correcting your erroneous assumption of constant g), all you are doing is struggling to derive the classical trajectory of a particle in free fall. You're not giving a derivation from maximizing proper time, which is what the OP requested.

The Euler-Lagrange equation is a general method for finding the maximum of any integral of the given general form. So given an integral that yields elapsed proper time, which the one I gave does for the case where g can be considered constant, applying the Euler-Lagrange equation gives you the function h(t) which maximizes the proper time.

Also, please note that the "simpler derivation" you attempted did not "show us that the trajectory that maximized the proper time was the free-fall trajectory".

Yes, it did; it showed that the function h(t) which maximizes the proper time integral has the property $d^{2}h / dt^{2} = - g$, for the case where g can be considered constant. That's equivalent to saying that the particle falls freely. If you use the more complicated expression that's valid when g can't be considered constant, you still get the same expression for $d^{2}h / dt^{2}$, just with the "g" dependent on h on the RHS.

free-fall trajectories (i.e., timelike geodesics) are DEFINED as the paths that maximize proper time, so there was nothing to "show". That was the whole point of Feynman's story - that many people (who have learned things by some kind of rote) will go off and struggle with the problem without even realizing that what they are doing is deriving the geodesic equations. Maybe the OP was just trolling to illustrate Feynman's point?

Even if that's the definition, it's still a worthwhile exercise, IMO, to see how the answer can arise by different routes. But I'll let the OP comment on whether the exercise was worthwhile and what he was trying to illustrate.

 

[PeterDonis]

But minus the classical Lagrangian divided by the particle's mass also happens to be the equation for the elapsed proper time of the particle following the trajectory h(t), provided that g can be considered constant over the range of height covered. That can be shown directly from the metric.

I should clarify one point here. Strictly speaking, integrating minus the classical Lagrangian divided by the mass gives the *difference* between the elapsed proper time along the trajectory h(t) and the elapsed time at some constant height. The formula $gh - 1 / 2 v^{2}$, or the more complicated formula in h for the case where g is not constant, gives the difference relative to h = 0, i.e., relative to the clock on the ground. For the purpose of answering Feynman's question, finding the function h(t) that maximizes the difference is sufficient; but to actually find the elapsed time, the time on the clock on the ground needs to be added back in. Apologies for not making that clear earlier.

Edit: also, to calculate the actual elapsed time, the formula in h needs to be divided by $c^{2}$, so the units are correct.

 

[Samshorn]

But minus the classical Lagrangian divided by the particle's mass also happens to be the equation for the elapsed proper time of the particle following the trajectory h(t)...

No it isn't. You'd see this if you actually tried to go from the metric line element to that classical Lagrangian. Please note that changing the sign and dividing by the particle's mass are not the issue. Those are trivial. The issue is that the proper time is the square root of the quadratic line element, whereas the Lagrangian is essentially proportional to the quadratic line element itself - not to the square root. So they are not the same things, nor are they even proportional to each other. So it's incorrect to claim that the classical Lagrangian represents d tau.

That can be shown directly from the metric.

No it can't, because it's false. What CAN be shown is that extremizing the classical Lagrangian yields (approximately) the same paths as maximizing the elapsed proper time, but this does not imply that the classical Lagrangian is proportional to the elapsed proper time (which it isn't). And of course, showing that these two things yield the same trajectories is precisely what is needed here, and it's precisely what we do when we derive the geodesic equations from the metric. By leaving out this derivation, and simply taking for granted that the paths yielded by the classical Lagrangian are the ones that maximize proper time, you are simply assuming the very thing you've been asked to demonstrate.

Please go back and re-read all my posts. You are correct that the expression $gh - 1/2 v^{2}$ is not the classical Lagrangian, strictly speaking. It's minus the classical Lagrangian divided by the particle's mass. I said that explicitly in at least one previous post, and it seems like the OP understands the distinction.

Again, that's not the issue. The difference between the metric expression for d tau and the classical Lagrangian for a free-falling particle is greater than you realize. You would see this if you actually try to go from one to the other. Furthermore, by assuming the classical Lagrangian of a free particle in a gravitational field with acceleration g, you have already assumed the solution. I mentioned this in my first post. The definition of "g", the acceleration of gravity, is d^2r/dt^2, so this already gives the equation of the trajectory.

You pointed out, correctly, that g can't be considered constant for the problem as specified, which I hadn't spotted because I hadn't run the actual numbers. If g can't be considered constant, then the integral of minus the Lagrangian divided by the mass gets more complicated, as I noted in my last post.

Sure, but we still have (by definition) d^2r/dt^2 = g, recognizing that g = -m/r. This is nothing but the geodesic equation (in the classical approximation). Solving this gives the cycloid trajectory. My point is that you aren't showing how this emerges from maximizing the proper time in a spherically symmetrical metric field, you are simply assuming it, and then disguising the assumption by submerging it in some irrelevant manipulations of the classical Lagrangian.

 

[PeterDonis]

The issue is that the proper time is the square root of the quadratic line element

Agreed.

whereas the Lagrangian is essentially proportional to the quadratic line element itself - not to the square root.

Can you clarify the relationship? The OP may not understand what that means.

By leaving out this derivation, and simply taking for granted that the paths yielded by the classical Lagrangian are the ones that maximize proper time, you are simply assuming the very thing you've been asked to demonstrate.

Actually, what I did has no necessary connection to the classical Lagrangian. I derived the expression I wrote down for dtau by taking the square root of the line element, as you say. I probably should have made it clearer in my original post that that was what I was doing.

Furthermore, by assuming the classical Lagrangian of a free particle in a gravitational field with acceleration g, you have already assumed the solution.

I didn't assume that at all. As I said, I derived an expression for the proper time by taking the square root of the line element. I did note that the formula I came up with was equal to minus the classical Lagrangian divided by the mass, but I'll agree that the connection between the two is not as simple as I made it seem. I did not mean to imply that I was taking the classical Lagrangian as a starting point; I wasn't. I made no use of it at all in my actual derivation.

I mentioned this in my first post. The definition of "g", the acceleration of gravity, is d^2r/dt^2, so this already gives the equation of the trajectory.

That's not the definition of g, at least not if you do the derivation by taking the square root of the line element; "g" is then $GM / R^{2}$, where R is the radius of the Earth.

My point is that you aren't showing how this emerges from maximizing the proper time in a spherically symmetrical metric field, you are simply assuming it

No, I'm not. I've explained what I did above, but just to recap quickly:

(1) I took the square root of the line element for a purely radial trajectory;

(2) I expanded the square root and retained only leading order terms;

(3) I subtracted out the constant potential at the surface of the Earth;

(4) I wrote r = R + h, where R is the radius of the Earth, and made the approximation h << R (which, as you noted, is not actually valid for the problem as stated, but would be valid for a short enough trajectory);

(5) I wrote "g" for $GM / R^{2}$;

(6) I left out the constant factor $1 / c^{2}$ in the formulas. This changes the units to "energy per unit mass", or "action per unit mass" when integrated with respect to time, but doesn't change anything else;

(7) I wrote down the integral of the resulting expression with respect to time, and applied the Euler-Lagrange equation to it, which finds the function h(t) that maximizes the integral. This will be the function h(t) that maximizes the proper time; and to find the actual proper time, the units will need to be corrected and the (constant) time elapsed on the ground clock (which I effectively subtracted off by subtracting the constant potential at the Earth's surface) will need to be added back.

Nowhere in the above did I use the classical Lagrangian; I merely commented on the similarities of the formula I obtained to the classical Lagrangian. I just did a derivation from the line element. If you want to say that's a roundabout way of deriving the geodesic equation, that's fine. But it's certainly not the same as *assuming* the geodesic equation or the classical Lagrangian; the only thing I *assumed* was the metric.

 

[Samshorn]

For ease of reference, here's what you wrote in post #4, where you introduced your derivation:

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:

τ = ∫dτ = ∫ [gh − 1/2(dh/dt)^2]dt

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.

So you clearly defined, or at least stated, that g is the acceleration of gravity. Also, there's no indication of how you go from the expression for proper time to this expression (which I think is what the OP was asking for), you just asserted it. Obviously once you get to that expression, which is the classical Lagrangian for a free-falling particle, and have identified g as the acceleration of gravity, i.e., d^2h/dt^2 = -g, you're done. That's the geodesic equation for the trajectory.

I derived the expression I wrote down for dtau by taking the square root of the line element, as you say. I probably should have made it clearer in my original post that that was what I was doing... If you do the derivation by taking the square root of the line element; "g" is then $GM / R^{2}$, where R is the radius of the Earth.

Well, you stated in your original post that your "g" represented the acceleration of gravity (see above), so it's understandable that people would think that's what you meant.

By the way, isn't there something odd about your use of "g" (=m/r^2) here? You supposedly derived the expression from the square root of the metric, but that gives the factor (1 - 2m/r)^(1/2), and hence to the first approximation 1 - m/r, and the variable part is just -m/r. At no point does the quantity m/r^2 appear in this derivation. In other words, the metric gives the gravitational potential directly as -m/r, and yet you wrote the gravitational potential not as -m/r, but rather as g*r, where you defined g = -m/r^2. It's hard to see why anyone would write it like that if they were deriving it from the metric. Do you see what I mean?

And when you go on to call "g" the acceleration of gravity, well... Also, the use of h instead of r is strange, as is the fact that you omitted any discussion of deriving it from the square root of the metric to maximize proper time until just now, even though that's precisely what the OP requested...

I've explained what I did above, but just to recap quickly:

I don't disagree with what you outlined there, other than the use of the past tense. What you're describing now is not what you presented previously. (See the quote above.) Essentially what you're describing now is what I outlined for deriving the geodesic equation from the metric line element, although you stop short of actually deriving the cycloid solution.

If you want to say that's a roundabout way of deriving the geodesic equation, that's fine.

Yes, what you're outlining now is a roundabout way of deriving the geodesic equation from the condition of maximizing the proper time. What you presented previously wasn't.

 

[PeterDonis]

So you clearly defined, or at least stated, that g is the acceleration of gravity.

Which is equal to GM/R^2 at the surface of the Earth. I already agreed that I should have been clearer about the initial presentation. See further comments below.

Obviously once you get to that expression, which is the classical Lagrangian for a free-falling particle

Only it isn't; it's minus the classical Lagrangian divided by the mass of the particle. And you yourself have already pointed out that the relationship of this expression with the classical Lagrangian is not as simple as I made it seem.

By the way, isn't there something odd about your use of "g" (=m/r^2) here? You supposedly derived the expression from the square root of the metric, but that gives the factor (1 - 2m/r)^(1/2), and hence to the first approximation 1 - m/r, and the variable part is just -m/r.

Yes, but r is not height above the Earth's surface; it's distance from the Earth's center. The equation I wrote down has h, height above the Earth's surface. As you noticed:

Also, the use of h instead of r is strange

Not if you're trying to answer the question Feynman actually asked: how to maximize the proper time on the rocket clock, given that an hour elapses *on the ground clock*. The integral I wrote down has "t" as the time of the ground clock, *not* Schwarzschild coordinate time. That means you need to adjust the "zero" of the potential to the radius of the Earth, which is the radius of the ground clock; and it also means you want to change variables from r to h, so that your distance variable is distance from the ground. That's what steps 3 and 4 in what I posted last time are for. When you do all that, you end up with a term GM/R^2 times h; so the "g" pops out, and you might as well recognize that. That's what step 5 in what I posted last time is for.

as is the fact that you omitted any discussion of deriving it from the square root of the metric to maximize proper time until just now, even though that's precisely what the OP requested...

I thought the OP was asking about the maximization part, not the part about how you derive the initial expression that then gives you the integral to maximize. I would be glad to post more details about how the equation I wrote down gets derived from the metric if the OP requests it.

Yes, what you're outlining now is a roundabout way of deriving the geodesic equation from the condition of maximizing the proper time. What you presented previously wasn't.

What I outlined in my last post *is* what I presented previously, just with a clearer explanation of the steps. See above.

 

[Samshorn]

What I outlined in my last post *is* what I presented previously, just with a clearer explanation of the steps. See above.

No it isn't. Remember you wrote

τ = ∫dτ = ∫ [gh − 1/2(dh/dt)^2]dt

but now you agree (I think) that that isn't right. That expression isn't the elapsed proper time along the trajectory, it's an approximation for the DIFFERENCE between the proper time along the trajectory and the proper time on the ground. So the expression you were insisting represented dtau really doesn't at all. Also, as noted before, you defined g as the acceleration of gravity, thereby begging the question.

Which is equal to GM/R^2 at the surface of the Earth.

Remember the thread of the conversation. ME: "You said g was the acceleration of gravity". YOU: 'I said no such thing. g is -m/r^2." ME: "Well, here's a quote where you said g was the acceleration of gravity". YOU: "Yes, which equals -m/r^2". And so it goes.

The point is that you said g was the acceleration of gravity, which means d^2h/dt^2 = g for the object in question, which is already the geodesic equation.

Not if you're trying to answer the question Feynman actually asked: how to maximize the proper time on the rocket clock, given that an hour elapses *on the ground clock*.

To the order of approximation that you are working, the weak-slow limit, I think that distinction is insignificant. In other words, the height of the apogee that you compute is going to be essentially the same, whether you say the trip was 1 hour on the Earth clock or 1 hour on the ballistic clock. If you really were trying to work it out exactly, you would have to do it the way I described, rather than making all those weak-slow approximations.