Problem involving fundamental principle

Samshorn

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.

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.

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.

Alcubierre

He does say he was wrong in some sense. The point of this thread was to maximize the programs of height and velocity and calculate what the proper time would be and I still haven't been able to do that because of the discord between you two. Peter provided me with a good idea on how to approach this (and even taught me, indirectly, how to solve Euler-Lagrange equations) and went through the steps enough so that I could do some thinking and work myself and still be able to solve it. Sam, you say your approach is better than Peter's "weal-slow limit," and it may be the case because you use a more explicit approach, but Peter did have in mind that I only have a single-variable calculus background and he was trying to make it easy for me to understand. I'm trying my hardest here to learn what I must to solve these problems, but giving a qualitative response as to how to solve isn't helping me much because I don't know what to do with your response below but look them up on Wikipedia or other sources and still be lost,


If you would show somewhat what you did so I could get started, that'd be great for me and for Peter to see how you did it yourself, if he hasn't already.

PeterDonis

Yes, that's true, and also the units are wrong; there should be a 1/c^2 in front. So the actual elapsed proper time would be:

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

where T = 1 hour is the time elapsed on the ground clock, as before. For solving the maximization problem that was posed in the OP, these differences don't matter; you still end up with the expression in brackets as what you need to apply the Euler-Lagrange equation to to obtain the function h(t) that maximizes the proper time. That's what I was focused on in my earlier posts.

I agree my wording wasn't clear in my original post. I did not mean that I just inserted g = acceleration of gravity in the equation I wrote down. I meant that in the course of deriving the equation I wrote down from the metric, the expression GM/R^2 appeared, which I then wrote "g" in place of, since it obviously equaled the acceleration of gravity. But that's not the same as just inserting g "from nowhere" in the equation. I didn't do that. I apologize if it wasn't clear from my original post that I didn't do that, but the fact remains that I didn't. I described how I derived my original equation from the metric, without making any assumptions about "g".

This is correct, but it misses my point. The "t" in the Schwarzschild metric is Schwarzschild *coordinate* time; but in the statement of the problem we are given time on the ground clock, not coordinate time. So I corrected the potential to be zero at h = 0, i.e., at the surface of the Earth. That lets the limits of integration be specified directly in terms of the given: 1 hour elapsed on the ground clock.

PeterDonis

Alcubierre, actually a fair bit of what Samshorn is describing is just to get to an equation like the one I originally wrote down (which only applies if the time is short enough that g can be considered constant), or a more complicated version when g is not constant, like what I wrote down in post #20. Basically there are three stages to the overall process, and I have only been talking about the third.

The three stages are:

(1) Determine the general physical law that governs gravity. This is what Samshorn is talking about here:

The "field equations" he is talking about are the Einstein Field Equations, which are the central equations of General Relativity. I was taking those as given for the purposes of this problem, but you can get a quick overview about them here:

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

(2) From the general physical law that governs gravity, find the particular solution that describes gravity around a spherically symmetric body (which we can approximate the Earth to be for this problem). This is what Samshorn is talking about here:

The "metric" Samshorn refers to is called the Schwarzschild metric, and I was also taking it as given for purposes of this problem, but you can read more about it here:

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

(3) Once you have the solution describing gravity around a spherically symmetric body, use it to determine what trajectory--what function h(t)--maximizes the time on the rocket's clock. This is what Samshorn is talking about here:

What I walked you through was an approximate version of the above, valid only if g can be considered constant. I'll let Samshorn give more specifics/references about the equations he would use for the case when g is not constant. He pointed out, correctly, that for this problem the Newtonian equations, which are simpler, are sufficient.

Alcubierre

Hello again, guys. I had some long personal businesses to take care, now I'm back and working on this again. So, Peter to be clear, I should maximize the integrand on post #37 or use a Newtonian equation? Also, I was having difficulties integrating that. I will post more about it tomorrow.

PeterDonis

Remember that the integral I wrote down in #37 is valid only for the case where g can be considered constant. As Samshorn pointed out a number of posts ago, for the specific case in question (elapsed time of one hour for the clock on the ground), g *can't* be considered constant, so the integral gets more complicated.

For the case where g can be considered constant, yes, you would maximize the integral in #37, which actually amounts to maximizing the expression in brackets--more precisely, you would find the function h(t) that maximizes the expression in brackets.

I'm not sure what you mean by "use a Newtonian equation".

Alcubierre

I still have not been able to figure this out. I stopped working on it for a while because I've been extremely busy. So to recap, when I apply calculus of variation (Euler-Lagrange) to the integrand on post #37, I get a(t) which I can integrate once to obtain v(t) and once more to obtain h(t). Then what do I do with those equations in respect to time, simply plug in a value for t or what? That's the part that is throwing me off a bit.

PeterDonis

If you want to answer the question posed in the OP (in the Feynman quote), the function h(t) *is* the answer.

If you want to find the actual numerical value of the height at a given time t, yes, plug that value of t into the function h(t). Note that, as I said in an earlier post, h(t) will have a term [itex]v_0 t[/itex], as well as the [itex]- 1/2 g t^2[/itex] term, corresponding to an initial upward velocity [itex]v_0[/itex]. To find the actual numerical value of [itex]v_0[/itex], you use the fact that [itex]h(t) = 0[/itex] when t = 1 hour; that lets you solve for [itex]v_0[/itex] in terms of [itex]g[/itex].

Alcubierre

So what is the difference between using 1/(c^2)[gh - .5gh'^2] (from post #37) and GM/(R^2) - .5gh'^2?

PeterDonis

I'm not sure what you actually meant to ask about with the second formula. If you meant this (note that I have inserted an [itex]h[/itex] in the first term and a factor of [itex]1 / c^2[/itex] in front to make the units the same as those in the first formula)...

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

...then this is just a rewrite of the first formula with [itex]GM/R^2[/itex] substituted for [itex]g[/itex]. But if you meant the alternate formula I gave in post #20, you've left out a term and got an extra factor of [itex]R[/itex] in the first term; it should be (correcting the units again):

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

If the latter is what you meant, the difference between the two formulas is that the first is only valid for short trajectories, while the second is valid generally. As Samshorn pointed out in a fairly early post in this thread, a one-hour time of flight, which is the actual number posed in the OP, is long enough that the first formula is not valid. However, as I pointed out in a somewhat later post, integrating the second formula is harder because of the h in the denominator of the second term.