Recovering Newton's energy conservation law for an Earth's lab

[epovo]

TL;DR Summary

Schutz derives the expression for p0 , but it's unclear to me why it should be conserved.

I'm looking at Schutz 7.4 where first he obtains the following expression for a geodesic:

$$ m \frac {dp_\beta} {d\tau} = \frac 1 2 g_{\nu\alpha,\beta } p^\nu p^\alpha $$

This means that if all the components of $g_{\nu\alpha }$ are constant for a given $\beta$, then $p_\beta$ is also constant along the geodesic.

Then, he gives an example for a lab on earth, using the weak gravity metric $ds^2 = -(1+2\phi)dt^2 + (1-2\phi)(dx^2 + dy^2 +dz^2)$. After some work and approximation he arrives at

$$ - p_0 \approx m + m\phi + \frac {(p^x)^2 + (p^y)^2 + (p^z)^2} {2m} $$

Nice. We get the mass, potential energy and kinetic energy.
But how is this an example of the geodesic expression above for $\beta = 0$ as the text suggests? A lab on Earth is surely not following a geodesic. I suppose that $g_{\nu\alpha }$ are constant on the lab, but the expression that ensures that $p_0$ would then be constant was derived only for geodesics. So that would not work for a lab on Earth, right?
I am confused

 

[Dale]

Conserved quantities are not conserved only on geodesics. They are conserved on all valid paths.

 

[anuttarasammyak]

But how is this an example of the geodesic expression above for β=0 as the text suggests? A lab on Earth is surely not following a geodesic.

In high school physics say we pop up a ball, the quantity
$$mgh+\frac{1}{2}mv^2$$ or $$mc^2+mgh+\frac{1}{2}mv^2$$
is constant along the path or geodesic. I think the description says it.

 

[epovo]

My problem is the way the thing is presented. The lab on Earth is given as an example of
$$ m \frac {dp_\beta} {d\tau} = \frac 1 2 g_{\nu\alpha,\beta } p^\nu p^\alpha $$
Which in turn was obtained from $p^\alpha p^\beta_{;\alpha}=0$, i.e. a geodesic.
Maybe the result above is valid for any trajectory, but Schutz did not certainly derive it in general. Hence my confusion.

 

[anuttarasammyak]

If forces other than gravity apply on a body, its trajectory is not geodesic. I think these forces should be considered in energy conservation in general.

 

[Ibix]

Summary:: Schutz derives the expression for p0 , but it's unclear to me why it should be conserved.

So that would not work for a lab on Earth, right?

I don't have Schutz to hand, but isn't he talking about a ball in free fall in a lab? So the equation is not for the motion of the lab but for the motion of a ball in a coordinate system where the lab is at rest? That latter is certainly what the chosen metric implies.

 

[Ibix]

Just got home and checked 7.4 and I think I'm right. From just before equation 7.30 in Schutz (emphasis added):

The frame in which the metric components are stationary is special, and is the usual "laboratory frame" on Earth. Therefore $p_0$ in this frame is related to the usual energy defined in the lab, and includes the particle's gravitational potential energy

 

[vanhees71]

Conserved quantities are not conserved only on geodesics. They are conserved on all valid paths.

What do you mean by that? Usually a conserved quantity is meant to be conserved for the solutions of the equations of motion, i.e., in this case along the solutions for the motion along a geodesic, because that's the equation for a particle under the influence of gravitation only (i.e., force free in the sense of general relativity).

That's the usual meaning of a conservation law in the sense of Noether's theorem in point-particle mechanics: The conserved quantities are conserved "on shell" and in general not for arbitrary "paths".

 

[Dale]

in this case along the solutions for the motion along a geodesic,

In this case, yes. But not in general. The equation of motion for a mass bouncing on a spring is not on a geodesic but has a conserved energy. The two are distinct concepts.

 

[PeterDonis]

Conserved quantities are not conserved only on geodesics. They are conserved on all valid paths.

No, this statement is too strong.

In the case under consideration, the conserved quantity is energy at infinity. That is conserved on geodesic paths, but if, for example, we slowly lower an object in the gravity field, its energy at infinity will not be conserved; it will decrease (and we can extract the energy at infinity that the object gives up as work).

For a lab at rest on the Earth's surface, however, its worldline is an orbit of the timelike Killing vector field of the spacetime (if we idealize the Earth as stationary), and energy at infinity is conserved along such orbits. I don't have the Schutz book, but I would guess that property is what he is implicitly relying on.

 

[PeterDonis]

The equation of motion for a mass bouncing on a spring is not on a geodesic but has a conserved energy.

That's because the motion of the mass bouncing on the spring is stationary--simple harmonic motion. (We are assuming zero damping.) So there is an analogue of a timelike Killing vector field involved, and the conserved energy is the analogue of "energy at infinity" for this timelike KVF.

 

[epovo]

ok, I am quoting from Schutz here:
"For instance, suppose we have a stationary gravitational field. Then a coordinate field can be found in which the metric components are time independent, and in that system $p^0$ is conserved [...]. The system in which the metric components are stationary is special, and is the usual 'laboratory frame' on Earth."
This follows from the Eq on the first post,

$$ m \frac {dp_\beta} {d\tau} = \frac 1 2 g_{\nu\alpha,\beta } p^\nu p^\alpha $$

If that eqn is valid for any trajectory, then the trajectory of the lab is included and I have no problem. But as I said, the derivation was done for geodesics only.

 

[Dale]

That's because the motion of the mass bouncing on the spring is stationary--simple harmonic motion. (We are assuming zero damping.) So there is an analogue of a timelike Killing vector field involved, and the conserved energy is the analogue of "energy at infinity" for this timelike KVF.

Yes, that is the key. The symmetry is the important thing, not the geodesic path.

 

[Dale]

Then a coordinate field can be found in which the metric components are time independent, and in that system p0 is conserved

So here is where he is bringing in the symmetry. It is this time independence that leads to the conservation law.

 

[PeterDonis]

suppose we have a stationary gravitational field

Ah, this is what I suspected. "Stationary" means there is a timelike Killing vector field; that is what allows there to be a conserved energy in the first place.

The symmetry is the important thing, not the geodesic path.

More precisely:

To have a conserved energy that applies to test objects traveling on worldlines, there must be a timelike Killing vector field.

If there is a timelike Killing vector field, then there is a conserved energy for test objects traveling on two types of worldlines: (1) geodesics; and (2) orbits of the timelike Killing vector field.

 

[PeterDonis]

the derivation was done for geodesics only

Looking at the equation, it does look like it comes directly from the geodesic equation, specialized to an appropriate choice of coordinates. But, as you note in the OP, orbits of the timelike Killing vector field in a stationary spacetime, such as the worldline of a lab at rest on Earth, are not geodesics.

As I said before, I don't have Schutz's book, but I suspect @Ibix was correct in post #7 and the trajectory Schutz is talking about, along which there is a conserved energy, is the trajectory of an object in free-fall in the lab, not the lab itself. The fact that $g_{\nu \alpha , \beta} = 0$ for $\beta = 0$ is not a property of the lab's worldline; it's a property of the choice of coordinates that is being made--those coordinates are specialized to the timelike Killing vector field. The lab's worldline is then chosen so it is at rest in those coordinates.

What is confusing is that, as I have noted in previous posts, it also happens to be the case that there is a conserved energy along the lab's worldline, because that worldline is an orbit of the timelike Killing vector field (that's why the lab is at rest in the chosen coordinates). But you can't derive that fact from the geodesic equation, because the lab's worldline is not a geodesic. You have to derive it another way. It doesn't seem like Schutz is talking about that at all.

 

[Ibix]

As I said before, I don't have Schutz's book, but I suspect @Ibix was correct in post #7 and the trajectory Schutz is talking about, along which there is a conserved energy, is the trajectory of an object in free-fall in the lab, not the lab itself.

Here's the relevant portion of Schutz (2nd ed), section 7.4, p179:

The geodesic equation can thus, in complete generality, be written:$$m\frac{\mathrm{d}p_\beta}{d\tau}=\frac 12g_{\nu\alpha,\beta}p^\nu p^\alpha\quad (7.29)$$We therefore have the following important result: if all the components $g_{\alpha\nu}$ are independent of $x^\beta$ for some fixed index $\beta$, then $p_\beta$ is a constant along any particle's trajectory.

For instance, suppose we have a stationary (i.e. time-independent) gravitational field. Then a coordinate system can be found in which the the metric components are time independent, and in that system $p_0$ is conserved. Therefore $p_0$ (or, really, $-p_0$) is usually called the 'energy' of the particle, without the qualification 'in this frame'. Notice that coordinates can also be found in which the same metric has time-dependent components: any time-dependent coordinate transformation from the 'nice' system will do this. In fact, most freely falling locally inertial systems are like this, since a freely falling particle sees a gravitational field that varies with its position, and therefore with time in its coordinate system. The frame in which the metric components are stationary is special, and is the usual 'laboratory frame' on Earth. Therefore $p_0$ in this frame is related to the usual energy defined in the lab, and includes the particle's gravitational potential energy, as we shall now show. Consider the equation$$
\begin{array}{rclr}
\vec p\cdot\vec p&=&-m^2=g_{\alpha\beta}p^\alpha p^\beta&\\
&=&-(1+2\phi)(p^0)^2+(1-2\phi)[(p^x)^2+(p^y)^2+(p^z)^2]&\quad (7.30)
\end{array}

It continues overleaf, deriving the equation in @epovo's OP. I can type that up if necessary, but I don't think I need to. I think this is enough to see that Schutz talks about the "laboratory frame" and "the particle" as distinct things. He's using the geodesic equation because he's talking about the energy of a freely falling particle, and he's working in a coordinate system ("the lab frame") which uses a weak-field approximation to Schwarzschild coordinates. Objects at rest in that frame, indeed, do not follow geodesics. But freely falling objects do, since the inertial/non-inertial character of a motion is invariant.

 

[epovo]

Thank you all. I think my problem is solved now.

 

[vanhees71]

In this case, yes. But not in general. The equation of motion for a mass bouncing on a spring is not on a geodesic but has a conserved energy. The two are distinct concepts.

Yes, but always the conservation law refers to "on-shell" solutions of the equations of motion. That was my point. In Noether's theorem you have a symmetry of the variation of the action, leading to a constraint on the Lagrangian (or Hamiltonian) valid for all trajectories/fields, defining what it means that a one-parameter Lie group is a symmetry. Then a conservation law follows for the corresponding generator of this Lie symmetry transformation for the solutions of the equations of motion.

By definition energy is the corresponding conserved quantity if there is translation invariance in some temporal direction. In GR for free motion (i.e., for motion of a text particle under the sole influence of gravity) this can be of course reformulated as a symmetry of spacetime, i.e., the existence of a time-like Killing vector. The corresponding "energy conservation law" then also holds for the solutions of the equation of motion, which in this case is the geodesic of the test particle.

If you have additional forces you need an additional corresponding symmetry of the corresponding action, i.e., the translation invariance in direction of the Killing vector. E.g., for a charged particle in Reissner-Norström spacetime energy is conserved, because in standard coordinates $g_{\mu \nu}$ don't depend on (coordinate) time and thus the Lagrangian for the EoM,
$$L=\frac{m}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\mu}-q \dot{x}^{\mu} A_{\mu},$$
where
$$A^0=Q/(4 \pi r), \quad A^k=0$$
is the electromagnetic four-potential of the field of the charged spherically symmetric body (or black hole), doesn't explicitly depend on time either, from which you derive that the Hamiltonian is conserved along the trajectories (solutions of the equations of motion).

 

[Dale]

@vanhees71 I am not sure why you are trying to argue with me. I never said otherwise. I was never talking about off shell, you are the only one bringing that up. It is weird to have you quote me and then argue about things unrelated to my actual statements.

If you want to make a point about on or off shell to the OP then do so but leave me out of it.

 

[vanhees71]

Then I misunderstood your statement in #2:

Conserved quantities are not conserved only on geodesics. They are conserved on all valid paths.

The quantities are NOT conserved on all valid paths but only on the solutions of the equations of motion, be they geodesic (motion under influence of gravity only) or under the influence of forces. That was all I wanted to stress, because it's important. You don't need to be offended so easily!

 

[Sagittarius A-Star]

Recovering Newton's energy conservation law for an Earth's lab

Classical mechanics contains energy conservation. But it is interesting, that Newton did not know, what energy is. In his book Principia, he defined the conserved quantities mass and 3-momentum, but not energy. He did not make use of the oberservation of Leibniz, that $m v^2$ is conserved in many mechanical systems (without discovering the factor $1/2$ for kinetic energy).

Source:
https://en.wikipedia.org/wiki/Vis_viva

 

[Dale]

NOT conserved on all valid paths but only on the solutions of the equations of motion

@vanhees71 This is not quantum mechanics. Only paths that are solutions to the equations of motion are “valid paths”

You don't need to be offended so easily!

I am tired of being unnecessarily lectured by you simply because you repeatedly fail to recognize that I might use a word differently that you do. It is exceptionally tiresome.

Next time you think you want to lecture me, please ask for clarification instead, and be prepared to hear and accept that I use some word differently from your approved manner.

 

[PeterDonis]

The quantities are NOT conserved on all valid paths but only on the solutions of the equations of motion

And not even all of those. See my post #10.

 

[vanhees71]

@vanhees71 This is not quantum mechanics. Only paths that are solutions to the equations of motion are “valid paths”

I am tired of being unnecessarily lectured by you simply because you repeatedly fail to recognize that I might use a word differently that you do. It is exceptionally tiresome.

Next time you think you want to lecture me, please ask for clarification instead, and be prepared to hear and accept that I use some word differently from your approved manner.

I thought it's about the action principle and there the valid paths are not onlx the solutions of the eqs. of motion but all where the action functional is defined. I've not mentioned QT at all.

 

[anuttarasammyak]

I thought it's about the action principle and there the valid paths are not onlx the solutions of the eqs. of motion but all where the action functional is defined.

Say forces other than gravity which apply on a body satisfy the constraint
$$\mathbf{F}\cdot\mathbf{dl}=0$$
, they would not harm energy conservation. We can design slide for children in various shapes to get same final speed.

 

[Dale]

not onlx the solutions of the eqs. of motion but all where the action functional is defined

Why would you consider those to be valid paths? No physical object can take them. What sort of degradation of the word “valid” makes a physically impossible path “valid” in classical mechanics? Do you have some authoritative reference that asserts that this is the official meaning of a “valid” path in classical mechanics?

To be clear. Once again we are having a pointless semantic discussion. We have no disagreement on the physics but just what constitutes a “valid” path.

 

[PeterDonis]

Why would you consider those to be valid paths?

Because he's talking about using the principle of least action, where you derive the equation of motion by extremizing the action integral. In the action integral, all paths are valid, not just those that satisfy the equation of motion.

Doing this is just as valid in classical mechanics as in quantum mechanics; in fact much of the nineteenth century development of classical Newtonian mechanics was the development of the Lagrangian approach. The use of the term "on shell" in post #19 is IMO misleading, though, because I've never seen that term used in the context of classical mechanics, only in the context of quantum mechanics.

 

[PeterDonis]

I thought it's about the action principle

Discussion of the principle of least action in this context is probably off topic for this particular thread. The OP was not asking a general question about conservation laws, but a specific question about a specific scenario which can be analyzed just fine using the geodesic equation and the properties of Killing vector fields.

 

[vanhees71]

Say forces other than gravity which apply on a body satisfy the constraint
$$\mathbf{F}\cdot\mathbf{dl}=0$$
, they would not harm energy conservation. We can design slide for children in various shapes to get same final speed.

Sure, and example is the magnetic part of the electromagnetic force on a point particle.

 

[vanhees71]

Why would you consider those to be valid paths? No physical object can take them. What sort of degradation of the word “valid” makes a physically impossible path “valid” in classical mechanics? Do you have some authoritative reference that asserts that this is the official meaning of a “valid” path in classical mechanics?

To be clear. Once again we are having a pointless semantic discussion. We have no disagreement on the physics but just what constitutes a “valid” path.

This is an important point in the derivation of Noether's theorem, and I thought you were discussing energy conservation in the context of Noether's theorem and thus the least-action principle. It's not by chance that Noether's great paper about the different theorems about the connection between symmetries and conservation laws was written in addressing precisely the puzzling question of "energy conservation" in GR, because it's the (in my opinion up to today) final answer: Global energy conservation holds if there is a corresponding translation invariance of the (variation of) the action. For the motion of a test particle in GR (no matter whether there are additional interactions than gravity or not) it's a sufficient condition that you can define coordinates, where the time-like coordinate doesn't occur in the Lagrangian.

 

[Dale]

Because he's talking about using the principle of least action, where you derive the equation of motion by extremizing the action integral. In the action integral, all paths are valid, not just those that satisfy the equation of motion.

I am not the only one that uses this terminology: “it is obvious why a solution that gives a minimum would be a valid path“. http://hep.physics.wayne.edu/~harr/courses/5210/w08/lecture08.htm

So again, this is a semantic question and my use of “valid” is valid. And furthermore my post that started this inane discussion was not even specifically about the action principle. You would have to go out of your way to read it like that and out of your way to deliberately misinterpret it as something that I am in need of correction about.

To me it doesn’t make sense to call the other paths valid. Only the one that extremizes the action is valid. The integral goes over other paths but why should we consider a path “valid” merely because the integral examines it and discards it?

In the context of QM there are people that call non-stationary paths “valid”, but I haven’t seen it in classical mechanics. And even if someone has, it still doesn’t make it more than a semantic choice.

I don’t need the lectures from either you or @vanhees71 when I make different semantic choices than you. If I mess up the actual physics then please correct me, but we have now hijacked this thread with an unnecessary slew of semantics.

 

[vanhees71]

You mess up the physics when using standard terms in a different meaning! This was an important mathematical point to clarify and not hijacking the thread, because it's the essence to answer the question asked about energy conservation in general relativity!

 

[Dale]

You mess up the physics when using standard terms in a different meaning!

This is not a “standard term” with only the @vanhees71-approved meaning. You immediately go into lecture mode to correct anyone who uses some word the slightest different from you. If someone doesn’t say it exactly the “@vanhees71-approved” way then they “mess up the physics”.

I neither need nor welcome another pointless semantic debate. Just because you use a word a specific way does not mean I need to.

It is insulting for you to continually place yourself in the role of my physics teacher, I don’t need your physics instruction. You should just assume that if there is something that you feel like lecturing me about then it is probably merely a semantic difference.

This was an important mathematical point to clarify and not hijacking the thread

Then make the important mathematical point directly to the OP and leave me out of it. I am not wrong simply because I use a word differently and I am not in need of correction or instruction from you.

Or at the least spend a few minutes to identify the semantic issue and point out that I am using a word differently than you.

 

[PeterDonis]

You mess up the physics when using standard terms in a different meaning!

Just so you know it isn't just @Dale who has issues with this viewpoint: this claim of yours is way, way, way too strong.

First, ordinary language is vague and ordinary language terms can have multiple meanings in physics discussions. If you really want to not mess up the physics, the proper way to ensure that is to use math. It is not a good strategy to use vague ordinary language terms with your preferred meaning and then pretend that's the only possible meaning of those vague ordinary language terms.

Note that the issue is not that your interpretation of "valid" in this context is unreasonable; it is reasonable. But so is @Dale's interpretation of "valid" in this context, as he explains it in post #32 (and note that he gives a reference to support it as well). So we have (at least) two reasonable interpretations of an ordinary language term in a physics context which are not consistent with each other. So talking as though someone who uses a different reasonable interpretation from your reasonable interpretation is "messing up the physics" is not, well, reasonable.

Second, when you find other SAs or Mentors disagreeing with you, it is (a) very impolite, and (b) a very unreasonable Bayesian prior, to assume that they reason they are disagreeing with you is that they do not understand the physics and need to have you explain it again. It is much, much more likely that the issue is just terminology. It would be very helpful if you would try to approach future disagreements with other SAs or Mentors with that mindset.