Can Lagrangian mechanics be applied to motion in an expanding universe

[Buzz Bloom]

TL;DR Summary

Since L = T - V, and T equals the kinetic energy (KE) of a particle whose trajectory is to be calculated, how is KE defined since some of its motion will be due to the expanding universe?

Summary: Since L = T - V, and T equals the kinetic energy (KE) of a particle whose trajectory is to be calculated, how is KE defined since some of its motion will be due to the expanding universe?

My understanding may well be wrong, but it is the following.

if a particle is stationary at point Y with respect to (WRT) comoving coordinates (CC) , then its motion WRT a reference point X which is also stationary WRT CC is entirely due to Hubble's law, and this motion does not contribute to the particle's KE relative to X. If the distance between X and Y is D, then the the velocity of the particle as seen by an observer at X is H₀D. (H₀ is Hubble's constant.) However, this quantity does not contribute to the KE of the particle as seen by the observer at X.

Is the above correct? If it is not, please post an explanation of how this KE is related to the calculation of a trajectory using Lagrangian mechanics. If it is correct, then is the following also correct?

If a particle of mass M is moving at velocity v relative to the point it occupies which is stationary WRT CC then the particle has
KE = (1/2)Mv²
as seen by an observer stationary WRT CC at any point.

Is this correct? If not please post any help you might have WRT this topic which you have conveniently available.

If it is correct, I would appreciate some help in understanding how to use L = T-V to set up an equation for the trajectory of a particle in an expanding universe satisfying Hubble's law. To keep the problem simple, assume no gravitating bodies and that any changes in H₀ are insignificant.

 

[Buzz Bloom]

I am surprised to see this thread moved to this forum (from Other Physics Topics) since I am unaware of the influence of relativity on the problem I am discussing. I am looking forward to seeing an explanation of how relativity interacts with the problem.

 

[fresh_42]

I moved it because the main subject are reference frames. I doubt that "how is KE defined since some of its motion will be due to the expanding universe" can be seen this way.

 

[PAllen]

Expanding universe is explicitly a model in GR (it is a specific family of pseudo-Riemannian manifolds). Therefore, a valid analysis of anything in the context of expanding universe requires GR.

 

[Buzz Bloom]

I confess that my experience with Lagrangian mechanics is weak, which is why I am seeking help. My thought about this topic is that the metric (with a single spatial dimension) for a flat universe with Hubble's law with a constant value for H is:
is

[1] ds² = - c²dt² + a(t)²dx²​

where

[2] a(t) = e^Ht.​

For t = 0, a(t) = 1.
The assumed initial conditions for a particle at t=0 are its position

[3] x(0) = x₀,​

and it's velocity WRT CC is

[4] v(0) = v₀.​

If my assumptions in post #1 are correct, then the KE of the particle is a constant

[5] KE = (1/2) Mv².​

I am not sure whether the velocity is

[6a] v = dx/dt or​

[6b] v = a(t)dx/dt,​

although I think [6a] is the proper choice.

If the universe is not expanding, and a has the constant value 1, the trajectory is straight forward.

[7] x(t) = x₀ + v₀t.​

I need help in setting up the equation of motion assuming [1] and [2], and I do not know how to use the Lagrangian to do this.

Any help would be much appreciated.

 

[haushofer]

Velocity is coordinate dependent. Do you want to define it for comoving observers or not?

 

[Buzz Bloom]

Velocity is coordinate dependent. Do you want to define it for comoving observers or not?

Hi haushofer:

Thanks for responding.

I want to assume that observers are stationary WRT CC. My primary concern is the assumption I made in post #1 regarding whether the KE includes, as an observed velocity component, the Hubble velocity HD.

Regards,
Buzz

 

[PeroK]

if a particle is stationary at point Y with respect to (WRT) comoving coordinates (CC) , then its motion WRT a reference point X which is also stationary WRT CC is entirely due to Hubble's law, and this motion does not contribute to the particle's KE relative to X.

One problem is that this is not how kinematics is handled in GR. Vectors, for example, are a local quantity. There are no position vectors and the velocity of a particle is only defined locally. The locally measured energy of a particle is given by the inner product of the observer's four velocity and the particle's four momentum.

In general, you cannot directly take the inner product of two vectors defined at different points in spacetime.

 

[Buzz Bloom]

One problem is that this is not how kinematics is handled in GR. Vectors, for example, are a local quantity.

Hi Perok:

Thanks for responding.

I am confused here about what "local quantity" means. Does it mean that using GR kinematics an observer at position X which is stationary WRT CC, cannot caculate the trajectory of a particle at a distant initial position Y with a specified initial velocity relative to Y (which is also assumed to be stationary WRT CC)?

Regards,
Buzz

 

[PeroK]

Hi Perok:

Thanks for responding.

I am confused here about what "local quantity" means. Does it mean that using GR kinematics an observer at position X which is stationary WRT CC, cannot caculate the trajectory of a particle at a distant initial position Y with a specified initial velocity relative to Y (which is also assumed to be stationary WRT CC)?

Regards,
Buzz

Define relative velocity.

 

[Buzz Bloom]

Define relative velocity.

Hi Perok:

https://www.thefreedictionary.com/Relative+velocityRelative velocity:

the velocity with which a body approaches or recedes from another body, whether both are moving or only one.​

https://www.thefreedictionary.com/velocityVelocity:

2. Physics​

A vector quantity whose magnitude is a body's speed and whose direction is the body's direction of motion.​

https://www.thefreedictionary.com/speedSpeed:

1. Physics The rate or a measure of the rate of motion, especially:​

a. Distance traveled divided by the time of travel.​

b. The limit of this quotient as the time of travel becomes vanishingly small; the first derivative of distance with respect to time.​

I certainly hope that this definition (as well as the definitions of terms in which terms are defined) make sense in the context of my previous post.

BTW: Since I am assuming that the space relevant to the problem in this thread is one dimensional, then perhaps in the case the velocity might possibly be considered to a scalar rather than a vector.

Regards,
Buzz

 

[PeroK]

Hi Perok:

https://www.thefreedictionary.com/Relative+velocityRelative velocity:

the velocity with which a body approaches or recedes from another body, whether both are moving or only one.​

https://www.thefreedictionary.com/velocityVelocity:

2. Physics​

A vector quantity whose magnitude is a body's speed and whose direction is the body's direction of motion.​

https://www.thefreedictionary.com/speedSpeed:

1. Physics The rate or a measure of the rate of motion, especially:​

a. Distance traveled divided by the time of travel.​

b. The limit of this quotient as the time of travel becomes vanishingly small; the first derivative of distance with respect to time.​

I certainly hope that this definition (as well as the definitions of terms in which terms are defined) make sense in the context of my previous post.

BTW: Since I am assuming that the space relevant to the problem in this thread is one dimensional, then perhaps in the case the velocity might possibly be considered to a scalar rather than a vector.

Regards,
Buzz

Okay, but unless you are actually learning GR, what is the point of the question?

In GR vectors are local quantities, so you cannot directly define a relative velocity as $v_a - v_b$, if a and b are not at the same local region of spacetime.

 

[Buzz Bloom]

Okay, but unless you are actually learning GR, what is the point of the question?

Hi Perok:

I indicated in post #2 that I did not think GR was relevant to the problem. Apparently the moderator felt that the expanding universe assumption required GR, even for the problem I defined.

I tried to explain the point of the question in post #1, but apparently I failed to explain it adequately. I want to be able to calculate the trajectory of a particle assuming one spatial dimension, and assuming an expanding universe in terms of Hubble's law. I assume the observer of the particle is stationary WRT CC. I would expect to represent the trajectory in terms of a differential equation based on Lagrangian mechanics. Complexities arise because it is not clear whether the Hubble velocity HD relates to the particle's KE, or not. To keep the issue simple, I also assumed that there are no gravitational bodies involved. This implies no potential energy, so the Lagrangian is only the particle's KE.

I also do not know if the approach of using the metric [1] in post #5 (to represent Hubble's law) works OK with using Lagrangian mechanics, or not.

Regards,
Buzz

 

[PeroK]

Hi Perok:

I indicated in post #2 that I did not think GR was relevant to the problem. Apparently the moderator felt that the expanding universe assumption required GR, even for the problem I defined.

I tried to explain the point of the question in post #1, but apparently I failed to explain it adequately. I want to be able to calculate the trajectory of a particle assuming one spatial dimension, and assuming an expanding universe in terms of Hubble's law. I assume the observer of the particle is stationary WRT CC. I would expect to represent the trajectory in terms of a differential equation based on Lagrangian mechanics. Complexities arise because it is not clear whether the Hubble velocity HD relates to the particle's KE, or not. To keep the issue simple, I also assumed that there are no gravitational bodies involved. This implies no potential energy, so the Lagrangian is only the particle's KE.

I also do not know if the approach of using the metric [1] in post #5 (to represent Hubble's law) works OK with using Lagrangian mechanics, or not.

Regards,
Buzz

I would say that the classical Lagrangian mechanics requires modification even for SR. The classical concept of kinetic energy is superseded by the energy-momentum four-vector.

What does expanding universe mean in classical mechanics or SR? It only makes sense in GR.

 

[Buzz Bloom]

I would say that the classical Lagrangian mechanics requires modification even for SR. The classical concept of kinetic energy is superseded by the energy-momentum four-vector.

What does expanding universe mean in classical mechanics or SR? It only makes sense in GR.

Hi Perok:

I would expect that the metric [1] (I will repeat it below) captures the essence of Hubble's law since it specifies that the dx constituent of the metric with its multiplier a(t) implies a flat universe is expanding according to Hubble's law (assuming H is a constant).
[1] ds² = - c²dt² + a(t)²dx²
where
[2] a(t) = e^Ht.
Note a(0) = 1.

Regards,
Buzz

 

[PAllen]

Hi Perok:

I would expect that the metric [1] (I will repeat it below) captures the essence of Hubble's law since it specifies that the dx constituent of the metric with its multiplier a(t) implies a flat universe is expanding according to Hubble's law (assuming H is a constant).
[1] ds² = - c²dt² + a(t)²dx²
where
[2] a(t) = e^Ht.
Note a(0) = 1.

Regards,
Buzz

Space is flat in your 2d metric for the trivial reason that intrinsic curvature is impossible in 1d. However your 2d metric has curvature, and this means that relative velocity of separated bodies is undefined - it would vary based on path along which they were parallel transported.

The upshot is that even for your simple case you need GR techniques. Meanwhile, you are proposing to use Newtonian formulas which are already inadequate in SR.

A correct Lagrangian to use to find inertial motion from some initial conditions, for a test body, is simply the proper time integral. Extremizing it gets you the relevant geodesic. Potential and kinetic energy need not be involved.

 

[Buzz Bloom]

A correct Lagrangian to use to find inertial motion from some initial conditions, for a test body, is simply the proper time integral. Extremizing it gets you the relevant geodesic.

Hi Paul:

Thank you for your post.

I looked up

https://en.wikipedia.org/wiki/Proper_time ,​

and found (as I understand it) that I need to write the integral of ds/c between the strarting point a and the end point b of a trajectory of a particle. Then the trajectory is the path that extremizes this integral. If this is correct, then my remaining question is: can I use my metric [1], making the integrand:

(dt² - (a(t)²/c²)dx²)^1/2 ?​

Is this correct?

Regards,
Buzz

 

[PAllen]

Hi Paul:

Thank you for your post.

I looked up

https://en.wikipedia.org/wiki/Proper_time ,​

and found (as I understand it) that I need to write the integral of ds/c between the strarting point a and the end point b of a trajectory of a particle. Then the trajectory is the path that extremizes this integral. If this is correct, then my remaining question is: can I use my metric [1], making the integrand:

(dt² - (a(t)²/c²)dx²)^1/2 ?​

Is this correct?

Regards,
Buzz

Yes, your integral is correct. There are tricks you can use to solve it more easily, but that is separate from the concept.

You can also solve the variation for a starting event and initial 4-velocity rather than using two events. In fact, generally, you just derive the geodesic equation using Euler-Lagrange, and then this diff eq can be used for all types of initial conditions.

 

[vanhees71]

I moved it because the main subject are reference frames. I doubt that "how is KE defined since some of its motion will be due to the expanding universe" can be seen this way.

Also the "expanding universe" is specifically general relativistic. Of course the least-action principle is the most safe ground you can start with. Of course it's no longer the $T-V$ recipy of non-relativistic mechanics anymore.

The most simple case is the free fall of a test particle in a given "background spacetime", like in this case the Friedmann-Lemaitre-Robertson-Walker metric describing the "expanding universe" on the large scale average. There one convinient form of the action, valid for both massive and massless test particles, reads
$$L=-\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu},$$
where the dot refers to the derivative with respect to an arbitrary world parameter $\lambda$. Since it's homogeneous to 2nd degrees in $\dot{x}^{\mu}$ and not explicitly $\lambda$ dependent you automatically have $g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=\text{const}$ and thus $\lambda$ is automatically affine. For the massive case you can make $\lambda=\tau$ by setting the constant to 1. For the massless case it's 0 and $\lambda$ is of course only defined up to a scale, but it's irrelevant, because all you care about are the world lines you get by solving the Euler-Lagrange equations, describing the motion of the test particles.

 

[Buzz Bloom]

Yes, your integral is correct. There are tricks you can use to solve it more easily, but that is separate from the concept.

You can also solve the variation for a starting event and initial 4-velocity rather than using two events. In fact, generally, you just derive the geodesic equation using Euler-Lagrange, and then this diff eq can be used for all types of initial conditions.

Hi Paul:

I found the following at

https://en.wikipedia.org/wiki/Euler–Lagrange_equation .​

What I think I want to do is first consider L as the integral you said was OK in my post #17. Then I would like to use your suggestion to use a "starting event and initial 4-velocity", but I need a bit of help with that. If I assume the velocity is WRT CC, I want to calculate how the velocity changes as the particle moves to a different position.

One assumption I conjectured (assuming the metric [1]) is that if a particle's velocity WRT CC has a specific value V relative to a specific reference point X, the the particle's velocity WRT CC is also V relative any other reference point. Can you tell me whether or not this conjecture is correct? If it is not, can you explain why or post a reference that explains why?

If the conjecture is correct, can you help me set the boundaries for the integral so that the lower boundary represents the particle's position X₀ and velocity V WRT CC, and the upper boundary is another point X₁?

Regarding the Euler-Lagrange equation above, in the context of the Integral for L, I assume q = x(t), and q-dot = dx(t)/dt. Is this correct? Or is q = a(t)x(t)?

I would much appreciate any help you can give me.

Regards,
BuzzBTW, preceding my effort to make the correct calculation I am discussing in this thread, I made a conjecture that led me to a solution, but I have no confidence that it is correct. This solution is:
V(x₁) = V(x₀) + H(D₁ - D₀).
Here H is the Hubble constant, and D represents the distance of the particle from an observer stationary WRT CC. I am hoping I will either confirm that calculation based on my conjecture is correct, or I will learn why it isn't.

 

[Buzz Bloom]

Also the "expanding universe" is specifically general relativistic. Of course the least-action principle is the most safe ground you can start with. Of course it's no longer the $T-V$ recipy of non-relativistic mechanics anymore.

The most simple case is the free fall of a test particle in a given "background spacetime", like in this case the Friedmann-Lemaitre-Robertson-Walker metric describing the "expanding universe" on the large scale average. There one convinient form of the action, valid for both massive and massless test particles, reads
$$L=-\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu},$$
where the dot refers to the derivative with respect to an arbitrary world parameter $\lambda$. Since it's homogeneous to 2nd degrees in $\dot{x}^{\mu}$ and not explicitly $\lambda$ dependent you automatically have $g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=\text{const}$ and thus $\lambda$ is automatically affine. For the massive case you can make $\lambda=\tau$ by setting the constant to 1. For the massless case it's 0 and $\lambda$ is of course only defined up to a scale, but it's irrelevant, because all you care about are the world lines you get by solving the Euler-Lagrange equations, describing the motion of the test particles.

Hi vanhees:

I do appreciate your post as an attempt to help me understand what I need to understand in order to solve the thread's problem in which I am interested. Unfortunately, most the concepts you introduce in your post are far over my head.

Regards,
Buzz

 

[PAllen]

Hi vanhees:

I do appreciate your post as an attempt to help me understand what I need to understand in order to solve the thread's problem in which I am interested. Unfortunately, most the concepts you introduce in your post are far over my head.

Regards,
Buzz

Well, one you should be able to get is that @vanhees71's suggested Lagrangian is one of the tricks I mentioned to make the problem easier. While not true for all variational problems, it is true here that you can vary the square of proper time instead of proper time. (I do not currently recall the justification for this, but I am sure others here can explain it off the top of their heads; I would have to research it a bit). If you are willing to accept the easier Lagrangian, your problem will be simpler.

 

[vanhees71]

For SRT I've shown it here:

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf
For geodesics in GRT you can simply show that the "square form" is equivalent to the "root form" of the Lagrangian for the choice of an affine worldline parameter (for massive particles of course the most intuitive choice is the proper time) since it leads to the same equation of motion for the geodesic.