Two ships again [lets beat that dead horse]

[nearc]

Two ships are traveling fairly close together side by side heading the same way on parallel paths near the speed of light [lets say v=.999C].

From the internal frame of reference [people in the ships] they can consider themselves as equivalent to being stationary, however from an external frame of reference [e.g. someone on a remote stationary point] we “see” length contraction of the ships, increased mass and time slowing down.

I think I understand all of the above, please chime in with any corrections I might need, however, what I fail to understand is how the distance between the two ships will change over time. Since from an external frame of reference the two ships have greater mass then what explains the ships not being more gravitationally attracted to each other? Is it because there is also a time dilation? Do these two aspects cancel out [i.e. increased mass and increased time]?

 

[phinds]

You are confusing remote observations with local reality. The people on the ships do NOT see any increase in mass or any time dilation --- those are artifacts of remote observation from a different frame of reference.

As far as the folks in the ships are concerned, they are sitting still and YOU are the one experiencing time dilation, etc.

 

[Dale]

what explains the ships not being more gravitationally attracted to each other?

The source of gravity in GR is not simply the energy (mass). It is the entire stress energy tensor which includes terms for momentum and pressure as well as energy. The ships gain energy, but also gain momentum. The combination of all those terms ensures that all invariants are agreed upon by all observers.

 

[Bill_K]

I don't see that this involves GR. The issue is simply, how does a Newtonian gravitational field transform under a Lorentz transformation.

 

[Dale]

I don't see that this involves GR. The issue is simply, how does a Newtonian gravitational field transform under a Lorentz transformation.

It doesn't. A Newtonian gravitational field is incompatible with the Lorentz transform. That is the whole reason that GR had to be developed.

The situation the OP mentioned is specifically one that demonstrates the incompatibility of Newtonian gravity with the Lorentz transform. It must be understood within the framework of GR.

 

[Bill_K]

It doesn't. A Newtonian gravitational field is incompatible with the Lorentz transform. That is the whole reason that GR had to be developed.

Wow, I disagree with that statement completely! GR was developed for totally different reasons, having nothing to do with the Lorentz transform.

There is certainly a regime in which it makes sense to talk about Newtonian gravity and SR together. But in the framework of SR you have to decide whether the gravitational field is going to be treated as a scalar, vector or tensor. And this depends in turn on whether the source is taken to be mass alone, energy-momentum 4-vector or stress energy tensor. The three choices differ primarily in how they transform under a Lorentz boost.

 

[Dale]

There is certainly a regime in which it makes sense to talk about Newtonian gravity and SR together.

Please provide a reference.

I don't think that the OP's scenario fits within that regime for precisely the reason that the OP mentioned.

 

[nearc]

The source of gravity in GR is not simply the energy (mass). It is the entire stress energy tensor which includes terms for momentum and pressure as well as energy. The ships gain energy, but also gain momentum. The combination of all those terms ensures that all invariants are agreed upon by all observers.

thanks for the reply, while i am working on learning the math of GR, I'm not at point yet where i can work everything out for myself, do you have a worked math example of what you propose or something similar?

if it is helpful we can add some more constraints we can allow the two ships to be sufficiently far apart such that they will take more than 1 billion years before they collide [in the internal reference frame from a stationary external point over 31 billions years]

 

[Dale]

thanks for the reply, while i am working on learning the math of GR, I'm not at point yet where i can work everything out for myself, do you have a worked math example of what you propose or something similar?

if it is helpful we can add some more constraints we can allow the two ships to be sufficiently far apart such that they will take more than 1 billion years before they collide [in the internal reference frame from a stationary external point over 31 billions years]

I don't have a worked out example. My comment comes directly from the Einstein Field Equations (http://en.wikipedia.org/wiki/Einstein_field_equations). Note that the "source term" in the EFE is the stress-energy tensor (http://en.wikipedia.org/wiki/Stress–energy_tensor), of which only one term represents energy/mass, and the rest represent momentum, pressure, and stress.

 

[pervect]

thanks for the reply, while i am working on learning the math of GR, I'm not at point yet where i can work everything out for myself, do you have a worked math example of what you propose or something similar?

if it is helpful we can add some more constraints we can allow the two ships to be sufficiently far apart such that they will take more than 1 billion years before they collide [in the internal reference frame from a stationary external point over 31 billions years]

I have half a worked example. But I'm not sure how much help it will be (asides from missing the most interesting part). More below.

But first, I'll suggest something simpler to get you ready. Have you looked at the electrostatic interaction between two moving charged particles as a clue?

You'll see that the transverse electric field changes - it intensifies. But there is a counterbalancing magnetic force. And the "force" isn't an invariant in this case, but the four-force is.

So, you might start thinking about the gravitational case, in a similar manner.

Onto the gravitational case:

The half-worked example I have is an outline of computing the Riemann curvature tensor of a moving observer.

But I've only computed part of the curvature tensor.

What you'll need to understand the half of what I've got is:

1) Enough familiarity with the Riemann tensor to understand the interpretation of it as a "force" per unit distance, a tidal force.

1a) It's also helpful to understand the limitations of this approach - specifically it's not the "real" definition of the Riemann, it just works very well under reasonable accelerations. We don't have any accelerations in this problem at all, so that's not an issue. Also, rotations (even slow ones) are not handled by this interpretation.

1b) The whole reason we use the Riemann is because the "Force" doesn't really transform like a tensor, and the Riemann does. So some care is advised! The goal is to get the Riemann "plugged into" your intuition. But if you're just starting, this might be hard/impossible. I'd recommend MTW ("Misner, Thorne, Wheeler - Gravitation") for more details on the "tidal force" interpretation of the Riemann, if you're ready for it - though they won't point out its limits.

2) The ins-and-outs of Orthonormal frame fields (which are a non-coordinate basis). A simple example - in polar coordinates you might have coordinates $r$ and $\theta$. You might also have unit vectors in the associated directions, typically called $\hat{r}$ and $\hat{\theta}$. Those vectors, because they are orthonormal, are handy to work with - they are called a non-coordinate basis. A coordinate basies would be $\frac{\partial}{\partial r}$ and $\frac{\partial}{\partial_{\theta}}$. Notation varies, the "hat" notation is used by MTW, though.

I'd guess I loose about 95% of the readers every time frame fields come up :-(.

For the posts

https://www.physicsforums.com/showpost.php?p=689706&postcount=2
https://www.physicsforums.com/showpost.php?p=690472&postcount=10 for a sketch

What's missing from the calculation is the part of the Riemann tensor that's analagous to the magnetic force / magnetic field in the electrostatic case. This is sometimes known as the magnetogravitic tensor. See http://en.wikipedia.org/wiki/Bel_decomposition

Unfortunatley, this is probably *the* key thing you really need.

The process of "decomposing" the Riemannn in this way is known as the Bel decomposition. The Bel decomposition is extremely helpful in understanding things, but it's an advanced topic and hard to find a good discussion about.

THe original papers are behind a paywall in French. There is a *very* short and not particularly detailed discussion in MTW's gravitation, which doesn't actually use the name "Bel Decomposition" so its hard to find.

One other thing - there is a new feature here in the strong field. THis is the "topogravitic" part of the tensor. You may or may not want to interpret this as a force. But you may need this to get all the numbers to finally balance out and fully understand the motion.

And another thing: It may or may not be possible to simplify this discussion enormously by considering the weak field. I've always thought that the terms involved were second order, making the linear approach a bit "iffy", but I should revisit that.

 

[Bill_K]

The half-worked example I have is an outline of computing the Riemann curvature tensor of a moving observer.

OMG pervect, you are totally off base too! I am beginning to lose faith. What is it they say: "To a man with a hammer, everything looks like a nail." Maybe that can explain the urge to make a GR problem out of something which is not.

All right if you must, but it is not about the Riemann tensor either. There are no tidal forces involved, only the attraction of one mass to another. Think of two geodesics, and realize you want to look at the Christoffel symbol, not the Riemann tensor.

 

[Dale]

Bill_K, Newtonian gravity is not consistent with special relativity. I don't know why you would suggest otherwise.

You are correct that there is no tidal gravity involved, but the question is why not? The naive attempt to merge Newtonian gravity and special relativity predicts that there would be. The question needs GR, IMO.

EDIT: actually, there is tidal gravity. Per the OP's scenario the ships are initially parallel and gravitationally attract until they collide. That is tidal gravity. So the question is how to reconcile the strength of the tidal gravity with the naive SR/Newtonian gravity combination, and the answer is you cannot, they are incompatible.

 

[Bill_K]

Bill_K, Newtonian gravity is not consistent with special relativity. I don't know why you would suggest otherwise.

Dale, You asked for a reference. Flat space theories of gravity are a dime a dozen and trivial to write down. Here's a scalar theory: ◻φ = 4πT. Here's a tensor theory: ◻φ_μν = 4πT_μν. Not to say they are correct - obviously they are not. The scalar one predicts zero light deflection for example. But they are described by tensor equations, perfectly consistent with SR, and they reduce to Newtonian gravity in the static limit.

actually, there is tidal gravity. Per the OP's scenario the ships are initially parallel and gravitationally attract until they collide. That is tidal gravity.

Tidal gravity is the effect that an external gravitational field has on a swarm of test particles. We are talking about a single particle traveling in the field of the other, e.g. the Earth traveling in the field of the Sun. Which is simply the geodesic equation.

 

[Mentz114]

Dale, You asked for a reference. Flat space theories of gravity are a dime a dozen and trivial to write down. Here's a scalar theory: ◻φ = 4πT. Here's a tensor theory: ◻φ_μν = 4πT_μν. Not to say they are correct - obviously they are not. The scalar one predicts zero light deflection for example. But they are described by tensor equations, perfectly consistent with SR, and they reduce to Newtonian gravity in the static limit.
...

Yes, but those are not Newtonian gravity as I know it. Just because a theory reduces to Newtonian (φ = GM/r) in some limit does not mean it *is* Newtonian gravity.

I think you and Dalespam mean different things when you say 'compatible'. As you point out, these theories don't give good predictions. Does that not show that Newtonian gravity is 'incompatible' with SR ?

 

[Nugatory]

Can we assume that the mass of the "stationary observer" in the original post is negligible? I think so, because as OP phrased this question it was pretty clearly not about any effects that that observer might have had on the two ships. And if so... Start in the frame in which the center of mass of the two ships is at rest, and I have a lot of trouble seeing this as a GR problem.

Unless I'm really missing something, Bill_K had it right back in post #4: we have two modest-sized masses whose weak mutual attraction is described by Newtonian gravity and we've idealized them to points so there can be no tidal effects. Draw their (very slowly converging - see post #8) worldlines, assign coordinates using the center of mass frame, Lorentz transform these coordinates to those of a fast-moving observer, and be done.

 

[Bill_K]

Yes, but those are not Newtonian gravity as I know it. Just because a theory reduces to Newtonian (φ = GM/r) in some limit does not mean it *is* Newtonian gravity

Never said it was. They are possible relativistic extensions. And they are weak field, flat space, non-GR extensions. So the claim that GR is necessary is false.

 

[Mentz114]

Never said it was. They are possible relativistic extensions. And they are weak field, flat space, non-GR extensions. So the claim that GR is necessary is false.

Well, that is just a personal choice. If I wanted to analyse it fully I would say that GR is necessary.
You are assuming things in order to justify your choice of tools.

Unless I'm really missing something, Bill_K had it right back in post #4: we have two modest-sized masses whose weak mutual attraction is described by Newtonian gravity and we've idealized them to points so there can be no tidal effects. Draw their (very slowly converging - see post #8) worldlines, assign coordinates using the center of mass frame, Lorentz transform these coordinates to those of a fast-moving observer, and be done.

I'm sure the OP would be grateful if you could demonstrate.

 

[nearc]

just wanted to check in and say thanks for all the interest and help, I'm not absent, I'm just waiting for the dust to settle so i can pick a path to follow

 

[Jonathan Scott]

What's going on here? Why do we need to consider gravity at all?

If two objects are accelerating towards each other for any reason (which could be an elastic string between them, or magnets, or some sort of thrusters) then the way the acceleration transforms between frames of reference doesn't depend in any way on what is causing the acceleration. You can just look at the space and time dilation effects in the Lorentz transformation and see how the acceleration itself transforms.

You can then of course work backwards from that to say that any law of gravity consistent with relativity must have a force law consistent with that behavior (although that leaves plenty of options).

 

[Dale]

Not to say they are correct - obviously they are not.

Why are you trying to answer the OPs question using theories that you know are incorrect? You are doing the OP a grave disservice.

they are described by tensor equations, perfectly consistent with SR, and they reduce to Newtonian gravity in the static limit.

I think you mean the weak field limit, right? If not, please clarify. The OP has v=0.999 c which places this scenario outside the weak field limit.

Tidal gravity is the effect that an external gravitational field has on a swarm of test particles. We are talking about a single particle traveling in the field of the other, e.g. the Earth traveling in the field of the Sun. Which is simply the geodesic equation.

If you have two geodesics which are initially parallel and then intersect then you have tidal gravity. The shape and volume of a swarm of test particles around the ships will change, there is clearly tidal gravity in this scenario.

 

[WannabeNewton]

If you have a family of test particles in an external non-uniform gravitational field and you see how their trajectories converge or diverge with respect to a reference point, then you are measuring the tidal gravity of that non-uniform field. If you are asking how two masses attract each other due to their own respective gravitational fields, you are just seeing how they respond to each others' fields as per the equations of motion coupled to the gravitational field. If you had a test particle falling in the gravitational field generated by some massive localized source, you wouldn't call that tidal gravity. Well here we just have two test particles falling in each others' fields.

 

[Dale]

we've idealized them to points so there can be no tidal effects.

Making them point masses makes it so there are no tidal strains in the ship from the opposite ship. That is not the same as there being no tidal effects. Each ships field exhibits tidal gravity, and, as the other ship moves through it, the local field changes.

Draw their (very slowly converging - see post #8) worldlines, assign coordinates using the center of mass frame, Lorentz transform these coordinates to those of a fast-moving observer, and be done.

Yes, you can calculate it that way. But the OP wants to know how to calculate it in the frame where they are moving fast. What theory of gravity can explain that? AFAIK, only GR. Newtonian gravity would predict that they collide in the moving frame much sooner than they do.

 

[Dale]

If you had a test particle falling in the gravitational field generated by some massive localized source, you wouldn't call that tidal gravity.

Yes, I would. Tidal gravity is spacetime curvature in GR, and the spacetime is curved by a massive localized source. Unless you are considering a small enough region of spacetime that there is no appreciable curvature then you are dealing with tidal gravity.

There is simply no way for two parallel geodesics to intersect without curvature and therefore tidal gravity.

 

[pervect]

I rather like the Christoffel symbol idea, it would be very interesting to work it out that way and compare the results. And it certainly would avoid a lot of words in trying to explain things. If the result turns out the same, I'd have to agree that I was making things overly complicated.

But let me say a few words about why I think I have a result to compare to, first :-)

There's a fairly simple geometric and experimental interpretation of what I'm proposing. We start out with two spaceships - I've semi-secretly made only one of the spaceships massive, and the other a "test spaceship". And our test spaceship is following a space-time geodesic. So in this particular case, it's proper acceleration is zero.

Now, we want the acceleration of our test spaceship, which is due to the massive one, "relative to infinity". So I've introduced an array of test spaceships, which are all also all following geodesics, each of which is instantaneously co-moving with it's neighbor. The array of spaceships extends outwards , towards infinity. The last test spaceship is "at infinity" and deemed not to be accelerating.

Because each neighboring pair is comoving, we can pretend (or define) that they are all comoving. If space-time were flat, there wouldn't be any ambiguity. Since space-time is curved, we actually can't compare velocities except locally :-(. But we've done the best we can.

Then we apply the geodesic deviation equation (not the geodesic equation) to get the relative acceleration between each pair of spaceships. From the spaceships point of view, it looks like we're just looking at the relative acceleration of gravity between the closer and the more distant spaceships, which we are our best approximation of comoving. Unsurprisingly, this is a) given by the geodesic deviation equation rather than the geodesic equation and b) equal to the integral of the tidal accelerations.

We're basically claiming that the total acceleration is the integral from where we're at to infinity of the tidal acceleration. Which is pretty obvious stated in those terms - but perhaps only actually true if space-time is flat.

If it really were a flat space problem, we should get the same results via the Christoffel symbol approach as via the above approach.

 

[WannabeNewton]

Yes, I would. Tidal gravity is spacetime curvature in GR, and the spacetime is curved by a massive localized source. Unless you are considering a small enough region of spacetime that there is no appreciable curvature then you are dealing with tidal gravity.

There is simply no way for two parallel geodesics to intersect without curvature and therefore tidal gravity.

There may be tidal gravity present in the space-time geometry determined by, say, a static massive star but there is no way you can determine this tidal gravity using nothing but a single test particle freely falling in this space-time; you need a congruence of time-like geodesics on some open subset of the space-time determined by the massive star whence you can consider a separation vector $\xi^{a}$ from a reference geodesic to a nearby one in the congruence, take the quantity $v^{a} = u^{b}\nabla_{b}\xi^{a}$ (where $u^{a}$ is the 4-velocity of the reference geodesic) which represents the rate of change of the separation vector along the reference geodesic, and then consider the acceleration $a^{a} = u^{b}\nabla_{b}v^{a}$. Regardless, tidal gravity is irrelevant for such a problem because all you care about is finding the trajectory of the single test particle, which comes straight out of the geodesic equation. Similarly, if you have two point particles interacting with each other gravitationally, you need only consider how their equations of motion couple to the space-time geometry jointly determined by them; tidal gravity is again irrelevant: http://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity

 

[Dale]

There may be tidal gravity present in the space-time geometry

That is what I was saying. It is there, and you know it is there by the fact that two parallel geodesics converge.

So you cannot treat the problem purely as a SR problem, you need a theory of gravity also. And Newtonian gravity cannot be used because it doesn't work outside of the weak field limit, which the OP's scenario is. It isn't a question of measuring the tidal gravity, but its presence requires the use of something more than SR.

 

[WannabeNewton]

Ah I see what you were trying to say; you were simply talking about the presence of tidal gravity + its global effects and not whether it was relevant to the calculation of the trajectories correct? I certainly agree then.

 

[Bill_K]

Why are you trying to answer the OPs question using theories that you know are incorrect? You are doing the OP a grave disservice.

Not to mention betrayal of the entire human race.

The point is that Newtonian theory has more than one possible extension to relativistic speeds, and each of them is self-consistent and contained within SR. Gravity was confirmed to be a tensor only after experimental confirmation, e.g. the deflection of light.

they are described by tensor equations, perfectly consistent with SR, and they reduce to Newtonian gravity in the static limit.

I think you mean the weak field limit, right? If not, please clarify. The OP has v=0.999 c which places this scenario outside the weak field limit.

No, I mean the static, aka slow motion limit, which must be Newtonian gravity in any case. The OP's scenario extends this to relativistic velocities, which is still weak field. (i.e. linear, GR is nonlinear.)

 

[Bill_K]

Suppose you have two ships each of rest mass m, a proper distance d apart, perpendicular to the line of sight. Suppose they are receding together at velocity V in the observer's frame.

1) Scalar theory, ◻φ = 4π T. In this theory ("Nordstrom's first theory"), the Newtonian potential φ is a relativistic scalar, and gravity couples only to rest mass. The force equation is

DP_μ/Dτ = m φ_,μ

Splitting this up into 3-d components P_μ = (γm, γmv) where c = 1,

d(γmv_i)/dτ = d(mv_i)/dt = m φ_,i

Since both the separation and the relative velocity are perpendicular to the recessional motion, both v_i and φ_,i are unchanged under the Lorentz boost with velocity V to the observer's frame.

2) Vector theory, ◻φ_μ = 4π J_μ, This theory is the gravitational analog of electromagnetism, in which the Newtonian potential is the 0-component of a relativistic vector quantity. The force equation is

DP_μ/Dτ = (φ_ν,μ - φ_μ,ν)J_ν

where J_μ = m V_μ.

3-d components:

d(γmv_i)/dτ = d(mv_i)/dt = γm (φ_0,i + (φ_j,i - φ_i,j) vj)

Defining E_i = φ_0,i and B_i = (φ_j,k - φ_k,j)ε_ijk we get the ("Lorentz") force equation

dv_i/dt = E_i + ε_ijk B_k v_j

The Lorentz invariance of this result is obvious since it was derived from a Lorentz invariant equation, but can also be demonstrated using the transformation of the 3-d quantities E and B under a Lorentz boost with velocity V to the observer's frame:

E_i' = γ(E_i + ε_ijk B_k V_j)
B_i' = γ(B_i - ε_ijk E_k V_j)

and so on.

3) Tensor theory, ◻φ_μν = 4π T_μν. Here the Newtonian potential is the 00-component of a relativistic tensor. This theory closely resembles linearized GR. The force equation is

DP_μ/Dτ = (φ_μσ,ν + φ_μν,σ - φ_νσ,μ) T_νσ

where T_μν = m V_μV_ν.

Taking 3-d components:

d(γmv_i)/dτ = d(mv_i)/dt = γ²m (φ_00,i + (φ_0i,j - φ_0j,i)v_j + (φ_ij,k + φ_ij,k - φ_ij,k)v_jv_k)

Defining E_i = φ_00,i, B_ij = φ_0j,i - φ_0j,i and H_ijk = φ_ij,k + φ_ij,k - φ_ij,k , we get the force equation

dv_i/dt = E_i + B_ij v_j + H_ijk v_j v_k.

 

[Mentz114]

Suppose you have two ships each of rest mass m, a proper distance d apart, perpendicular to the line of sight. Suppose they are receding together at velocity V in the observer's frame ...

Thanks, Bill_K. Well worth closer study.