Spacecraft With Solar Mass Energy Equivalent Kinetic Energy

[Devin-M]

One possible resolution of the disconnect might come from asking, where did all the energy come from to boost the shuttle to such a huge gamma factor?

So if you start with 1 star size clump of matter and one star sized clump of antimatter, and somehow convert almost all of the annihilation energy into 2 shuttles going in opposite directions, what happens to the gravity of the 2 star sized masses which “disappeared?”

 

[Devin-M]

We would have expected the original star to have a significant effect on the spacetime geometry around it, so it seems like we should also expect the two shuttles thrown off in the explosion to have a significant effect on the spacetime geometry surrounding them.

If it was approaching an everyday object could this lead to spaghettification from tidal forces? Imagine it passes a 100kg block of gelatin in perfect vacuum at a close approach of 100m. Is that gelatin staying in 1 piece?

 

[PeterDonis]

So if you start with 1 star size clump of matter and one star sized clump of antimatter, and somehow convert almost all of the annihilation energy into 2 shuttles going in opposite directions, what happens to the gravity of the 2 star sized masses which “disappeared?”

The stress-energy is now all in the shuttles, so that's where the "gravity" is. But, as I said, it will look very different from the gravity of a 2 solar mass star.

 

[PeterDonis]

If it was approaching an everyday object could this lead to spaghettification from tidal forces?

No. We are not talking about a black hole.

Imagine it passes a 100kg block of gelatin in perfect vacuum at a close approach of 100m. Is that gelatin staying in 1 piece?

Gelatin is a bad example because it has no structural strength to speak of, so even a very mild perturbation might cause it to fragment.

I would expect the general effect on objects that do have a reasonable structural strength to be a more or less impulsive (i.e., over a very short time) change in velocity as the shuttle passes. Since the shuttle is moving at only a tiny smidgen less than the speed of light, it will pass any ordinary object very quickly, so there will be no time for effects to build up.

 

[Vanadium 50]

very quickly

Of order an hour.

It will have comparable force to the sun at about 11 light-minutes coming in and 11 light minutes going out, making 22: 23 with rounding. You get half the solar effect at √2 farther out, and a quarter of the effect at a factor of 2, and so ojn. So the scale is around an hour.

We're talking 1/10,000 of Earth's orbital period, so we would expect 0.01% changes to the orbit. That's 10,000 miles. I have no idesa how to calculate what would happen if this happened, but I suspect it would wreck everybody's day.

 

[PeterDonis]

Of order an hour.

For passing through the inner solar system, yes. But for the scenario described in post #37, involving an ordinary object, it will pass in a fraction of a second.

 

[PeterDonis]

It will have comparable force to the sun

Will it? That's the question. It has a solar mass of kinetic energy, but it also has a solar mass of momentum, which should cancel at least a lot of the effect. At least, that was the intuitive guess I gave earlier in the thread. I realize the Olson & Guarino paper appears to be saying something different, but we haven't seen the details of the argument, and what they describe in their abstract is not a long distance effect like the Sun's field, it's an impulsive effect on test particles in or close to the path of the high speed object.

 

[Devin-M]

So comparing to an astronaut’s infall to a solar mass black hole, wouldn’t they expect to be spaghettified and moving close to speed of light when 100m away from the black hole’s event horizon?

Is there an intuitive answer why the tidal forces experienced by an astronaut should be less when the shuttle passes 100m away than the black hole case and not result in spaghettification?

 

[Vanadium 50]

but it also has a solar mass of momentum, which should cancel at least a lot of the effect.

It doesn't cancel it. It doubles it.

Olson & Guari do the full calculation, but you already know the answer. g₀₀ and g₁₁ are nearly equal and the only non-zero terms in the metric. What else has a metric like that? Light.

Just as light have a factor of 2 more bend than if you just consider g₀₀, so must the relativistic shuttle - the same equations have the same solution. And since momentum is conserved, you have twice the recoil on the earth.

You'd expect this approximation to be good to of order (m/E), and 2 is close to Olson & Guar's 1+β².

The question of what happens to the Earth only makes sense if this happens far away. If you drag two solar masses worth of "fuel" anywhere near the earth, of course the orbit will be highly perturbed. Further, if the "anti-shuttle" is anywhere near the shuttle, the total g₁₁ gets small and we are back to the Schwartzchild metric.

 

[Vanadium 50]

why the tidal forces experienced by an astronaut should be less than a solar mass black hole case at the same 100m distance?

Why do you think a fast moving object should behave the same as a slow heavy object. Does a bullet act like a boulder?

 

[PeterDonis]

comparing to an astronaut’s infall to a solar mass black hole

Which makes no sense because the two configurations are so different.

 

[PeterDonis]

It doesn't cancel it. It doubles it.

It doubles it in the narrow world tube surrounding the path of the shuttle, yes. That corresponds to doubling the bending of the path of a fast moving test particle by the Sun as compared to the Newtonian value, because the test particle passes very close to the narrow world tube of the Sun.

But does the field outside that world tube still fall off as the inverse square of the distance with an effective gravitational mass of $\gamma M \left( 1 + \beta^2 \right)$? That's the part I'm wondering about. It seems to me that it should fall off faster.

 

[Vanadium 50]

field outside that world tube

I don't know. I don't see an obvious limiting case that can be solved by symmetry and not the full solution. I suspect one or both bets are dotted into the line of sight direction at large distances.

Insofar as Ives-Stillwell can be used as an analogy, at large transverse distances you get a 1, not a 2.

 

[pervect]

The abstract of that paper, though, shows a factor of $\gamma$ in what they call the "active gravitational mass", i.e., taken at face value, they seem to be saying that the relativistic mass is the source of gravit, with an extra GR factor of $1 + \beta^2$ that comes in for similar reasons as in the analysis of light bending by the Sun.

However, they are defining "active gravitational mass" in a very narrow way: it is "measured by the transverse (and longitudinal) velocities which such a moving mass induces in test particles initially at rest near its path". That is not the same as the kind of "gravitational effects" that the OP is asking about.

I have so far been unable to find a non-paywalled version of the Olson & Guarino paper (for example, a preprint does not appear to be on arxiv.org), so I can't comment on the details of their reasoning. But I would be cautious about interpreting it with regard to this discussion.

I've seen a version of the full Olson & Guarino paper, but I'm not sure it's available anymore. So my comments are going mostly by memory.

However, I think the abstract is sufficient to answer the OP's question, and the answer is that the effect of the flyby (such as orbital pertubations) of such an ultra-relativistic object wil be simliar to a Newtonian flyby of an object with twice the mass. The factor is 2 because the flyby velocity beta is essentially unity.

The actual paper basically calculates the Newtonian and GR effects of a flyby on a cloud of dust, looking at the total induced velocity in the flat spacetime after the flyby.

While the actual flyby will cause additional GR effects such as the emission of gravitational waves that aren't present in the "dust" model, I don't think they'll be very significant. The end result of the flyby will be similar to a Newtonian flyby of not a 1x solar mass, but 2x a solar mass.

Interestingly enough, it's not a trivial calculation to find the effects of a Newtonian flyby. But I'll leave that to the reader.

Many people on PF don't like the fact that the paper's authors uses "relativistic mass" rather than "energy". But it's just a different name for the same thing, it doesn't affect the results. Since the paper compares the GR results to Newtonian results, and Newtonian physics uses mass, the author's choice to use "relativistic mass" rather than energy is understandable.

A non-rigorous popular level "explanation" of the factor of 2 is that light passing by a massive object deflects twice as much in GR as it does in Newtonian physics. The deflection of an ultra-relativistic particle moving at almost the speed of light is similar. The effect is essentially viewing this scenario from a set of coordinates where the ultra-relativistic particle is nearly stationary, and the large mass is moving instead.

 

[pervect]

While I don't recall how the paper did the calculation anymore, what comes to mind is considering the Schwarzschild solution in isotropic coordinate (t,x,y,z), perform a coordinatge transformation to cylindrical coordinates, (t, r, theta, z), then perform an additional coordinate transformation of z' = z - $\beta$ t to make the large mass "move" in the z direction in the transformed coordinates.

While one could also use isotorpic Schwarzschild (t,x,y,z) coordiantes directly, and do the z' = z-$\beta$ t coordiante transformation without first going to cylindrical, the cylindrical coordinates will be simpler as the problem has cylindrical symmetry. The basic insight is as my previous response suggested, it's just a messy coordinate change of the problem of the deflection of an ultra-relativistic particle. Then the particle follows a geodesic in the transformed coordinates.

It's likely that one would only want to consider the linearized problem - so rather than using the full non-linear isotropic solution, one just considers the linearized version thereof.

 

[PeterDonis]

he effect is essentially viewing this scenario from a set of coordinates where the ultra-relativistic particle is nearly stationary, and the large mass is moving instead.

But in these coordinates, while we would certainly expect the particle to be deflected, we would not, it seems to me, expect the system of large masses (the solar system in this case) to be disrupted. We would expect its configuration to be basically the same after the flyby as before--because in this frame the particle has negligible effect on the spacetime geometry.

However, a flyby of either a one solar mass object or a two solar mass object (meaning, the total energy of the object taking into account that in the solar system frame it is moving at a tiny smidgen less than the speed of light) through the solar system would be expected to disrupt the solar system, not just deflect the object.

So we still have a disconnect here: in one frame (the object's rest frame), we expect no disruption of the solar system, but in another frame (the solar system rest frame), we do. So the intuitive reasoning described above must be wrong in at least one frame.

 

[Vanadium 50]

We would expect its configuration to be basically the same after the flyby as before--

Remember, the solar system is Lorentz contracted so that its width is smaller than a proton. An ultra-microscopic change in this frame ("basically the same") can correspond to a large change in the original frame.

In this case, planets move by of order 10,000 miles.

 

[PeterDonis]

the solar system is Lorentz contracted so that its width is smaller than a proton

It's contracted along the direction of motion, but not transversely. And it is flying by the shuttle in this frame at just a tiny smidgen less than the speed of light. So I'm not sure why there would even be an ultra-microscopic change to the solar system in this frame. It's certainly not obvious to me why planets would be expected to move transversely by 10,000 miles.

Of course, all these are intuitive arguments, not math. If I can't find an accessible copy of the Olson & Guarino paper online I will have to try to work through some math myself if I get a chance.

 

[Vanadium 50]

t's certainly not obvious to me why planets would be expected to move transversely by 10,000 miles.

So you have an argument that the initial motion is longitudinal. Not that it is small.

This is SR, not GR. You have the same issue in electron-proton scattering. In the electron's frame the proton is a static pancake. Nonetheless, quarks can still scatter.

 

[PeterDonis]

So you have an argument that the initial motion is longitudinal.

No, it's a reason for doubting that objects at significant transverse separations will have large motions induced. Obviously if the shuttle impacts the Sun or a planet head on we will have a significant change. But we're talking about a flyby, where the shuttle doesn't come close to the Sun or any of the planets. At least, that's how I'm interpreting the OP's question of whether one solar mass of kinetic energy is significant--he's asking whether it's significant at ordinary planetary distances, the way the Sun itself is.

This is SR, not GR.

No, it isn't. The whole point is that we have to consider the shuttle as being a non-negligible source of gravity, not just the solar system. Otherwise the answer is obvious: a test particle by definition cannot induce motion in other objects, no matter how high a gamma factor we boost it to.

You have the same issue in electron-proton scattering.

No, it's not at all the same. For electrons to scatter quarks inside protons, the electrons have to be shot inside the protons. AFAIK electrons passing by protons at significant transverse distances do not scatter quarks inside the protons, even if they have huge gamma factors.

 

[Vanadium 50]

I can tell you are getting angry, and I know where that leads. However, before I go, this has all been worked out in the 30's. The words to look up are "Breit frame". You can see how this works out in deeply inelastic scattering by hunting down a copy of George Sterman's lecture notes from various CTEQ summer schools. He's at Stony Brook, so that might help Google find them.

 

[PeterDonis]

The words to look up are "Breit frame".

I'll look it up when I get a chance.

 

[PeterDonis]

I can tell you are getting angry

No, just not convinced--or more precisely, not sure at this point which of the two intuitive arguments I've described, that give different answers, are giving the right answer. But I've already said that I need to take the time to work through the math when I get a chance. I think I have enough pointers at this point to do that when I'm able.

 

[PeterDonis]

I know where that leads

I'm not considering demodulating anybody, if that's what you mean.

 

[renormalize]

While I don't recall how the paper did the calculation anymore, what comes to mind is considering the Schwarzschild solution in isotropic coordinate (t,x,y,z), perform a coordinatge transformation to cylindrical coordinates, (t, r, theta, z), then perform an additional coordinate transformation of z' = z - $\beta$ t to make the large mass "move" in the z direction in the transformed coordinates.

It's likely that one would only want to consider the linearized problem - so rather than using the full non-linear isotropic solution, one just considers the linearized version thereof.

I've been thinking along similar lines. But I question your suggested coordinate transformation. Wouldn't it be more proper to perform a Lorentz-boost to get the mass moving in the $z$-direction, à la what Jackson does in Classical Electrodynamics (chap. 11) to get the EM field of a uniformly moving point charge? And just as the pure Coulomb field picks up a magnetic component when boosted (due to mixing of the $(t,z)$ coordinates), I believe a boost of the diagonal Schwarzschild metric field must similarly lead to non-removable off-diagonal terms. So the calculation and interpretation of the boosted metric and its geodesics is apt to be non-trivial. But I do agree that using the linearized Schwarzschild solution may be easier and should suffice if we concentrate on the behavior of geodesics in the "far-field", i.e., those with a large-enough impact parameter $b$.

 

[PeterDonis]

Wouldn't it be more proper to perform a Lorentz-boost

There is no such thing as a global Lorentz boost in a curved spacetime.

 

[PeterDonis]

the pure Coulomb field picks up a magnetic component when boosted

Yes, but the reason there are off-diagonal components of the EM field tensor when it is boosted is that there are off-diagonal components of the EM field tensor before it is boosted. The EM field tensor is antisymmetric, so it always has off-diagonal components, even for a "pure Coulomb" field.

The metric tensor in Schwarzschild coordinates, by contrast, is diagonal, and a simple "boost" in one direction will not necessarily add any off-diagonal components. You would have to do the actual math to see.

 

[renormalize]

There is no such thing as a global Lorentz boost in a curved spacetime.

Maybe in a generic curved spacetime, but they do exist in specific spacetimes. As far as I can tell by reading https://arxiv.org/pdf/gr-qc/9805023.pdf, Lorentz transformations can be implemented by first expressing the Schwarzschild (and more generally the Kerr) metric in the Kerr-Schild (KS) form $g_{\mu\nu}\left(x\right)=\eta_{\mu\nu}+2H\left(x\right)l_{\mu}l_{\nu}$ , where $\eta$ is a flat background metric:

and then defining the boost velocity $\mathbf{v}$ relative to that background metric:

The result for the boosted KS Schwarzschild metric turns out to be:

Note that the four non-vanishing components of the null-vector $l$ manifestly give the boosted metric off-diagonal terms, and I'm doubtful that there exists a coordinate transform that returns that metric to a Schwarzschild-like diagonal form. Regardless, the explicit result (23) makes it straightforward, if tedious, to compute the Christoffel symbols and examine the geodesics followed by a test mass in boosted Schwarzschild spacetime. But for now that's above my pay grade. Maybe it's time to break out the computer algebra?

 

[PeterDonis]

Maybe in a generic curved spacetime, but they do exist in specific spacetimes.

No. There is no such thing as a global Lorentz boost in any curved spacetime. There can't be, because the definition of a Lorentz transformation is that it preserves the Minkowski metric on the spacetime--but a curved spacetime's metric is not the Minkowksi metric.

The transformations in the paper you referenced are expanding the usage of the term "Lorentz boost" to apply to coordinate transformations in asymptotically flat spacetimes that, heuristically, "look like" Lorentz boosts at infinity (where spacetime is flat--but infinity is not actually part of the physical spacetime, it's a mathematical artifact used to aid in computations). But that doesn't make them actual Lorentz boosts. It just means the authors of the paper (and other similar papers in the literature) are using terminology in a way that experts have no problem properly interpreting, but that can be misleading if you don't understand what they're actually doing.

 

[PeterDonis]

I'm doubtful that there exists a coordinate transform that returns that metric to a Schwarzschild-like diagonal form

I don't know why you would be. Any valid coordinate transformation is invertible.

 

[renormalize]

No. There is no such thing as a global Lorentz boost in any curved spacetime. There can't be, because the definition of a Lorentz transformation is that it preserves the Minkowski metric on the spacetime--but a curved spacetime's metric is not the Minkowksi metric.

@PeterDonis thank you for the correction; I acknowledge my confusion on this point.

Could you advise me on proper/better terminology? For some specific curved spacetime, coordinatized by a specific collection of four quantities $x$, consider the transformation (as gleaned from the paper cited in post #63): $x\rightarrow x'=\Lambda\,x,\:g\left(x\right)\rightarrow g'\left(x'\right)=\Lambda^{-1}g\left(x\right)\Lambda,\:\Lambda\in\text{SO}\left(1,3\right)$. In lieu of using terms like Lorentz "boost", "rotation", etc., what nomenclature would you suggest to label this type of transformation, i.e., how best to distinguish this operation from the usual sense of Lorentz transforms that preserve the flat metric but don't exist globally in curved spacetime?

 

[PeterDonis]

Could you advise me on proper/better terminology?

Unfortunately I'm not aware of any terminology that's consistent in the literature for particular classes of coordinate transformation in curved spacetimes. One just has to be aware that if terms like "Lorentz boost" are used they can't possibly mean the same transformation that they would mean in flat spacetime, so one has to look at the actual properties of the transformation (such as, in the case we discussed, the fact that it is restricted to asymptotically flat spacetimes and "looks like" an ordinary Lorentz boost at infinity).

 

[Devin-M]

If the shuttle passed a 100kg block of ballistic gelatin at a close approach of 100m, would the gelatin essentially not “feel” the gravitational wave until after close approach, since the gravitational wave and shuttle would be moving at essentially the same speed?

 

[PeterDonis]

If the shuttle passed a 100kg block of ballistic gelatin at a close approach of 100m, would the gelatin essentially not “feel” the gravitational wave until after close approach, since the gravitational wave and shuttle would be moving at essentially the same speed?

Yes.

 

[pervect]

But in these coordinates, while we would certainly expect the particle to be deflected, we would not, it seems to me, expect the system of large masses (the solar system in this case) to be disrupted. We would expect its configuration to be basically the same after the flyby as before--because in this frame the particle has negligible effect on the spacetime geometry.

However, a flyby of either a one solar mass object or a two solar mass object (meaning, the total energy of the object taking into account that in the solar system frame it is moving at a tiny smidgen less than the speed of light) through the solar system would be expected to disrupt the solar system, not just deflect the object.

So we still have a disconnect here: in one frame (the object's rest frame), we expect no disruption of the solar system, but in another frame (the solar system rest frame), we do. So the intuitive reasoning described above must be wrong in at least one frame.

What happens in the spaceship frame did has me puzzled a bit. It appeared at first glance to contradict the paper, but it turns out there is no conflict. I see that Vanadium made similar comments.

I will note that the paper I cited, and my analysis below is for dust, not planets, so I will not address that point that you raised.

However, what happens for the dust case in the frame of the spaceship is interesting, and I think I have a handle on it. The ultra-relativistic dust flies by the spaceship, which has some very small mass, so the dust is deflected only slightly in the frame of the spaceship. Essentially, it should be deflected by twice the Newtonian deflection due to the small, but non-zero, gravity of the spaceship. It's well known that light deflects twice as much in GR as in Newtonian physics, and the difference between the geodesics of light and the geodesics of ultra-relativistic particles is negligible.

Thus we expect in the space-ship frame that the dust will be deflected by some small angle $\theta$, winding up with a velocity in the longitudinal direction of $\beta \cos \theta$ and a transverse velocity of $\beta \sin \theta$, where we can use the usual GR formula to compute $\theta$.

However, to transform from the space-ship frame to a frame where the dust was initially stationary, we essentially need to do a relativistic velocity subtraction.

The velocity subtraction formula with a transverse and longitudinal components is a bit messy, wiki gives it in https://en.wikipedia.org/wiki/Velocity-addition_formula

The relevant formula is that the longitudinal direction is "x" in the wiki analysis, and the transverse direction is "y".

The formula wiki gives is then

$$ u^\prime_y = u_y \frac{\sqrt{1- \beta^2}}{1-\beta \beta \cos \theta}$$

which with $\cos \theta ~ 1$ should basically mean the transverse velocity in the dust frame should be larger by a factor of approximately $\gamma$ than its small value in the spaceship frame.