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I Tidal forces on a spring-mass system

  1. Feb 18, 2017 #1
    Say I have two masses connected by an unstretched massless spring at some height above a planet with a strong gravitational field.

    Once I let go, both masses would follow geodesics in spacetime towards the center of the planet. Because the masses travel radially, the spring would compress and store spring potential energy.

    Where does the spring energy come from? It doesn't come from a force since gravity isn't a force in GR.
    Classically, we can say that the cosine of the gravitational force vectors compresses the spring. Classically, it alines with F=-dU/dt. But in GR, it seems that the compression just happens out of nowhere, simply because space gets narrower.

    I thought about the potential energy due to height being converted into kinetic and spring energy as an explanation. But this sounds too classical for me.

    I also thought of the energy coming from spacetime itself. But I'm not so sure. If this were the case, then the surrounding spacetime would lose energy. This might make sense, as the mass falls, the spacetime it leaves behind becomes slightly less warped due to the two masses simply not being there. But then again, it enters a new region, and the masses simply warps that region as well. This reminds me somewhat of the rocket propulsion problem but instead of trading chemical PE for motion, it trades potential energy due to position in height for spring compression energy and kinetic energy. This reasoning is still seems a bit classical, so I'm not certain.
     
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  3. Feb 18, 2017 #2

    256bits

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    Would that be true - is towards the centre of the planet?
     
  4. Feb 18, 2017 #3
    Hmm well, the spring pushes back, so I guess the force on the masses from the spring would make the masses deviate from geodesic paths


    but initially, at the instant of letting go, it should be on geodesics. Also, if the spacetime warping is large enough, shouldn't the masses travel close to geodesic paths, so close to radial?
     
  5. Feb 18, 2017 #4

    256bits

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    Now, is that the answer. I don't know wrt to GR.
    Interesting question that you have proposed.
     
  6. Feb 18, 2017 #5

    Dale

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    This is a good question.

    Remember why gravity is considered to not be a force: it can be transformed away locally by a coordinate transform. One key word in that is "locally" which means over a region small enough that tidal effects are negligible. By construction, your scenario is large enough that tidal effects cannot be neglected, and the tidal forces cannot be transformed away.

    Additionally, as long as the spacetime is asymptotically flat you can define a conserved energy.
     
  7. Feb 18, 2017 #6

    PeterDonis

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    Actually, the spring would be stretched, not compressed; radial tidal gravity causes free-falling objects to separate, not converge.

    Also, the masses would not remain in free fall, because the spring would pull on them.

    From gravitational potential energy. At least, that's the best simple heuristic. It's no different than asking where the kinetic energy comes from when you drop a rock off a cliff.
     
  8. Feb 19, 2017 #7
    Ah, so gravity here is not fictitious. The equivalence principle doesn't hold.

    Since gravity cannot be transformed away, then what is gravity? It isn't a fundamental force in GR, its just movement in curved spacetime. It's hard to imagine gravity being made of other fundamental forces here. Is the tidal force from the constraints of the curved space?
     
  9. Feb 19, 2017 #8
    I meant seperated horizontally. So the left and right ones in the diagram, not the top and bottom ones.
    Screen Shot 2017-02-18 at 10.04.24 PM.png



    Yeah, I understand that it is from potential energy. But in the classical case, the energy is stored by a force from within the system.

    If I consider the two masses and the spring in between as an isolated system while excluding everything else, we see the spring compress, yet something external should be doing work on the system.
     
  10. Feb 19, 2017 #9

    PeterDonis

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    Ah, ok.

    No, you won't, because excluding everything else means excluding the Earth and its tidal gravity. In the presence of tidal gravity, the masses and the spring are not an "isolated system" any more, and reasoning as if they were will give wrong answers.
     
  11. Feb 19, 2017 #10
    I meant if we designate the system as mass and spring only, the earth would be external to the system and should influence it. Isolated system was a bad choice of words.

    If I drop a mass, I can approach it in two ways. Conservation where the mass and earth are part of the same system or the mass as its own system and the earth being external, doing work on the mass.
     
    Last edited: Feb 19, 2017
  12. Feb 19, 2017 #11

    Dale

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    That is correct.
    Spacetime curvature is essentially the same thing as tidal gravity.
     
  13. Feb 19, 2017 #12

    pervect

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    In order to answer "where the energy comes from", we'd have to localize the energy of the gravitational field. To answer "where", energy would need to have a location in the first place.

    But it is not in general possible to localize gravitational energy. Therefore we can't answer the "where" question. For a textbook reference, try google search for the exact phrase "Why the energy of the gravitational field cannot be localized". You should get a hit on a textbook, Misner, Thorne & Wheeler's "Gravitation" from google books. This is the title of section $20.4 in said textbook.

    So the best answer we can give you is the one that's in the textbooks, the one that says that, contrary to your assumptions, it's not generally possible to localize gravitational energy.

    This is different from electromagnetism in flat space-time, where we can localize energy in the electromagnetic (EM) field. See all the threads about the energy, and rest mass, of a charge capacitor, for an example where we can localize energy in a charged capacitor, and that energy is stored in the field between the plates.

    (Some posters may not be convinced of this, but that's the textbook answer. I don't have a reference handy, alas, but it's widely known.)

    Unfortunately while this approach to energy being stored in fields works fine for E&M, it doesn't work for gravity, as we can't define a "field" of gravity that stores energy in the same manner as we do for electromagnetism.
     
  14. Feb 19, 2017 #13

    Dale

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    Yes, although as long as the spacetime is asymptotically flat we can say that there is a conserved amount of energy, even if we can't say where it is located.
     
  15. Feb 19, 2017 #14

    PAllen

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    Also, if metric is stationary, you can specify a well defined potential function; if the masses in this example are considered small enough to ignore their own fields (test masses), then the potential energy analysis is well posed for this simple case. This is also textbook.
     
  16. Feb 20, 2017 #15

    PeterDonis

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    But if you use the second approach, then the answer to your question "where does the energy come from?" is still "gravitational potential energy". That's where the Earth gets the energy that it applies as work to the mass-spring system.
     
  17. Feb 20, 2017 #16
    Well, it makes sense in that for a system consisting of mass, earth, we can define a quantity called potential energy and observe that this quantity is converted to other types of energy.

    But for the second approach, how can it come from gravitational potential energy if the system is single object? Potential energy is a property of multi body systems.


    Classically, we can consider a mass as a one object system and use the work energy theorem. W=change in KE. It turns out that W=F*distance gives raise to mgh which merely turns out to be the potential energy for the two body system of earth and mass. But it doesn't logically have to be that way. We can consider the earth and mass as a system, and define a potential energy called mgh and observe that this is numerically the same as the kinetic energy. I feel like the only reason both approaches give the same answer is that the existence of a force allows the work to be computed as mgh and that it is this coincidence that allows the same results.

    But without a defined force, can both approaches work in GR? Is work meaningful?
     
  18. Feb 20, 2017 #17
    For the capacitors, I can say that for a system consisting of both plates and the field between them, the potential energy of the system comes from the field, being that the formula has E and d. Even though the PE is a property of the system as a whole, it is in some sense localized since it has d in the formula. It's like how mgh doesn't literally give the potential energy of a object at a height, but rather the energy due to the configuration of the whole system, which turns out to depend on h.

    So in GR, if we define a system consisting of masses earth and spacetime itself, then the formula for potential energy wouldn't have terms that describe location?

    Are gravitational fields fictitious in GR? So the only thing is the curvature of spacetime, which is not a vector field with well defined direction. Then it would make sense that energy is not the same as for the capacitors.
     
  19. Feb 21, 2017 #18

    Dale

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    Potential energy is a property of systems with internal degrees of freedom and a time independent Lagrangian.
     
  20. Feb 21, 2017 #19
    I recall in classical mechanics, if a Lagrangian doesn't have a parameter, then certain aspects are conserved with respect to that parameter. So the Lagrangian in CM is L=K-V where K=.5mv^2 and V=mgh, none of which depends explictly on time. Then the lagrangian is time independent and energy is conserved.(i.e since L is constant, V+const=K for all time. L=const is just the change in energy. )

    What do you mean by degrees of freedom? Do you mean the number of generalized coordinates has to be finite?
     
    Last edited: Feb 21, 2017
  21. Feb 21, 2017 #20

    Dale

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    Hmm, I may have stated it wrong.

    What I meant is that the internal configuration of the system must be able to change. For example, in a hydrogen atom the orbital is an internal configuration.
     
  22. Feb 21, 2017 #21

    pervect

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    I would say that "gravitational fields" is somewhat ambiguous term, but one can draw an analogy between "gravitational fields" and the fictitious forces of Newtonian mechanics, which is what I think you mean when you say that gravitational fields are fictitious.

    I would say that the mathematical description of these fictitious not-quite-forces is via a mathematical entity called "Christoffel symbols". Textbooks don't speak to this point -seem to simply avoid using the "force" concept altogether (of course I haven't read every textbook in existence!). Which leaves students looking for an analogy to replace the familiar F=ma formula feeling rather lost.

    I would say that by setting up a local orthonormal basis (coordiantes t,x,y,z) one can more-or-less regard ##\Gamma^x{}_{tt}, \Gamma^y{}_{tt}, \Gamma^z{}_{tt}## as the mathematical entities most similar to forces. They aren't vectors, though - they're not even quite tensors - so they are not really forces.

    To try and expand on this a bit more - and to show some of it's limitations - assume for the time being that the other Christoffel symbols not mentioned are zero. Then in the absence of any non-gravitatioanl forces we can write one of the geodesic equations (the one for the x-component) as:

    $$\frac{d^2 x}{d \tau^2} + \Gamma^x{}_{tt} \left( \frac{dx}{d\tau} \right) ^2 = 0$$

    Furthermore, if we have non-gravitational forces (say electric fields) this is where they'd be introduced. The right hand side wouldn't be zero anymore, but would have a mangitude equal to the sum of the non-gravitational forces. (I'm not sure offhand if it'd have a plus or minus sign, to be honest).

    Taking the Newtonian limit, at low speeds ##t = \tau##. So we have a relationship between coordinate acceleration and something that's rather like a "force" that sums to zero, rather simiiar to F = ma. It's actually more like mF = ma, but of course "gravitatioanl forces" are proportional to the mass of the object being acted on, so that's OK.

    The limitations on this are worth mentioning. The first is that it isn't relativistic, because we've assumed that t=##\tau##. But Newtonian forces aren't relativistic either. One might digress into a discussion of four-forces here, but I think that would derail the thread too much. So I'll stick with pointing out that we're only considering the low speed limit.

    I will add though that it might be a good idea to revisit how special relativity handles forces with four-vectors, if one has the necessary background.

    The second limitation is more serious. What if the other Chrirstoffel symbols are not zero? Consider for instance ##\Gamma^x{}_{xt}##. This adds a term to the geodesic equation equal to ## \Gamma^x{}_{xt} \left( \frac{dx}{d\tau} \right) \left( \frac{dt}{d\tau} \right)##. One can perpahs interpret this as a velocity dependent force, somehwat similar to a coriolis force in Newtonian mechanics, another "fictitious force". But what do we make of ##\Gamma^x{}_{xx}##? The analogy becomes more and more strained as we try to take it more seriously - so perhaps it's best not to pursue it beyond a motivational level.

    Christoffel symbols are probably not familiar to many readers of this thread. I suppose a few digressions would be useful as background, some of which I haven't done myself. One is to further explore how "fictitious forces" work in a tensor formulation of Newtonian mechanics, to solidify understanding the difference between "fictitous forces" due to a non-inertial frame of reference, and "real forces".

    The other I've already mentioned, that's to explore how "forces" are treated in special relativity via four-forces.
     
  23. Feb 24, 2017 #22
    Very interesting. I'm not so
    I'm not familiar with the math of tensor calculus so the Christoffel symbols are not that clear to me.

    So let the system be a single mass system, i.e a mass directly above a planet, with the planet external to the system. The mass is moving along a geodesic because the planet has no actual pull( no forces). But somehow curved spacetime imparts energy(i.e does "work") on the mass.

    according to this site:
    http://physics.stackexchange.com/questions/212167/what-is-a-christoffel-symbol
    " The Christoffel symbols measure the degree to which an observer following a straight line in coordinate space is not in free fall."


    This means that locally, we can say that the symbols vanish along a geodesic. So in this case, how does a single mass gain energy if the christoffel symbols are zero? Or maybe on a geodesic, energy is actually somehow conserved. As the mass moves along the geodesic downwards, it gains kinetic energy but also loses its speed through time. So maybe the tradeoff is balanced in such a way, that the velocity of a object through spacetime, not space, is constant along local geodesics. Is that correct?



    I think I understand that for tidal forces, the christoffel symbols are not zero because its no longer local, so it makes sense that the symbols coming from the planet can do some sort of work to change to energy of the mass.
     
  24. Feb 24, 2017 #23

    Dale

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    You have to be a little careful here. The Christoffel symbols vanish if the coordinates* form geodesics. The Christoffel symbols depend on the choice of coordinates, not on any given worldline.

    So for example, suppose that you have an object which is resting on the surface of the earth. Such an object is not on a geodesic. However at any event along the worldline (say noon today at that object's location) you can construct coordinates where the Christoffel symbols are 0. In those coordinates the object is undergoing coordinate acceleration which is equal to its proper acceleration. This is also called a momentarily comoving inertial frame.

    *I think that the correct terminology would be "the integral curves of the timelike coordinate basis" rather than "the coordinates", but that seems cumbersome.
     
  25. Feb 25, 2017 #24

    pervect

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    I think that would depend on the particular choice of viewpoint and coordinates you made. So if you happen to use a viewpoint / coordinate system centered on the planet - which is what pretty much the default choice - then you can more or less describe things this way. But when you try to treat all possible viewpoints equally, you run into problems.

    One can usually say that the energy of the system of the planet + the other mass is constant. (I say usually because one needs auxillary conditions, like asymptotic flatness, to say even this much. For the purposes of the discussion, lets just assume from here on that these conditions are met).

    But unlike Newtonian theory, it's not possible in general to assign this much energy to the planet, that much energy to the mass, and the remainder to some sort of field.

    And unless one can do this, one can't say, in complete generality, that the planet is doing work on the mass. Because one needs a concept of energy that is localized to do this.

    If you adopt the Newtonian viewpoint, you can say this, and GR reduces to the Newtonian viewpoint in the appropriate limit. But if you take advantage of the full flexibility of GR, you can't use the chain of logic you've been asking about.

    The only textbook explanations I have for why you can't do this involve tensor math. Tensor math tells you how things transform when you change coordinates. And it tells you how to set things up so you can make statements that don't depend on your coordinate choice. So, I guess you're sort of stuck without the math, at least I don't see at the current time any way to answer your questions in a totally satisfactory manner without using math that you don't have.

    The good news is that if you make the usual and conventional coordinate choices, centered on the heavy planet, you can continue to think that the planet gives energy to the falling mass. The bad news is that you are stuck with a not-very-satisfying explanation that requires you to adopt a specific viewpoint.

    Energy in GR is an advanced topic. So to learn about how GR views things, you wouldn't want to start with energy.

    Yes.

    This is almost exactly the argument I quoted earlier (I think it was in this thread) from MTW which describes why it's impossible to localize energy in GR. So you want to do the familiar thing, and assign energy to the planet, the falling mass, and the field. And it just isn't working out. It has been noticed before that this doesn't work out, and the resolution is more radical than one might really like.

    See also my remarks about the viewpoint - one way of interpreting your question is asking about the viewpoint of the falling mass - which is the viewpoint where the Christoffel symbols are zero. The point is that there isn't a view of energy that's "independent of viewpoint" in the manner one expects.

    Energy in GR is a very technical topic. But it might be rewarding to read a bit about Noether's theorem, and some of the history of energy in GR. A somewhat historical paper that might give you some insight is "E. Noether's Discovery of the Deep Connection Between Symmetries and Conservation Laws", <<link>>[/PLAIN] [Broken]

    The following quote in particular shows why I think this paper might be of some interest.

    So to recap what I think are the key points. Energy in GR is a tricky and advanced concept - there's a lot of work on the topic, we do have some partial answers, so it's not like there is nothing out there.

    Part of the issue here is that it's hard to discuss without tensors, which you aren't familiar with. Tensors answer questions about how things act when you change coordinates (which I've also called viewpoints in this thread). It's rather abstract, but it's powerful. The basic idea is that when a theory is independent of viewpoint, one can adopt the most convenient viewpoint to work a problem. This is sometimes called "covariance".

    Tensors describe the conditions that are necessary to present a problem in a manner that's independent of the particular choice of viewpoint. Errors that arise from not understanding tensors tend to be of two sorts. One sort is not necessarily an error, it involves not taking advantage of the existence of different viewpoints to choose the simplest one to analyze a problem. One needs the confidence that it's possible to analyze a problem from any viewpoint.

    The other issue involves violating the rules that make this possible, the rules that make physics independent of viewpoints. The technical langauge here is usually called "general covariance", we want our formulation of physical laws to be generally covariant.

    One way of describing tensors is to say that they encode physics in such a way that it's possible to mathematically demonstrate that the results one get don't depend on viewpoint/coordinate choices.

    Then in this langauge, we can say that energy in GR is not a tensor, it is not something that is defined independently of viewpoint/coordinates.
     
    Last edited by a moderator: May 8, 2017
  26. Feb 28, 2017 #25
    That makes sense. So because we can construct coordinates where the symbols are 0 at any particular instant, we need to do an integral of sorts so that "on average", something balances to zero? And there that would be the timelike coordinate basis.
     
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