# Physics Error in Larry Niven's "Neutron Star

• I
• PeterDonis
In summary, the light coming from distant stars is blueshifted when it is approaching a neutron star, and this is due to the gravitational pull of the neutron star.
PeterDonis
Mentor
TL;DR Summary
In the "Neutron Star" story by Larry Niven, he says the pilot of a spaceship on a hyperbolic orbit leading to a close approach to a neutron star sees light from distant stars blueshifted in all directions. However, this appears to be incorrect. A corrected calculation is given.
In the short story "Neutron Star" by Larry Niven, the narrator, Beowulf Shaeffer, is piloting a spaceship on a hyperbolic orbit that is supposed to make a close approach to a neutron star. He says at one point that he sees incoming light from distant stars blueshifted in all directions, and gives the neutron star's gravity as the reason. Here, we will do some calculations to see if this is actually correct, and will find that it isn't.

First, we need to get some parameters from the data given in the story. The first parameter is that the neutron star's mass is 1.3 solar masses. In geometric units, in which ##G = c = 1##, this is 1.92 kilometers. The second parameter is that the ship's speed relative to the star at closest approach is half the speed of light. (There are actually conflicting statements in the story about this, which we won't attempt to analyze in detail. The half lightspeed value is reasonable and sufficient for what we're going to calculate here.) The third parameter is the distance of closest approach; we will take this to be ##R = 8M##, which is 15.36 kilometers. (Again, there are conflicting statements in the story about this; this value is reasonable and sufficient for this discussion.)

We won't need to calculate the ship's orbit in detail; we can simply make the following observations on which to base simple calculations. First, when the ship first starts to fall towards the neutron star, we can assume that it is falling radially inward, to a good approximation, and at escape velocity. Second, at the point of closest approach to the neutron star, we can assume that the ship's trajectory is exactly transverse, i.e., perpendicular to a radial line through the star's center.

The general method of calculating the observed frequency shift of incoming light from distant stars makes use of the Schwarzschild metric, which I will take to be familiar enough that it need not be written down explicitly here, and consists of two elements: first, we calculate the gravitational blueshift that would be seen by a stationary observer (i.e., an observer "hovering" at rest at constant altitude above the star) at a given radius ##R##; second, we multiply that by a Doppler shift factor due to the ship's speed relative to the stationary observer. The Doppler shift factors we will consider are those for light coming from directly to the front of the ship (where the Doppler factor is a blueshift), directly to the rear of the ship (where the Doppler factor is a redshift of the same magnitude as the front blueshift), and to the side of the ship (where the Doppler factor is a redshift due to transverse Doppler, i.e., the usual SR time dilation due to the ship's speed).

We can capture the above in the following formulas for the factor by which the observed wavelength is changed from the source wavelength:

$$f_\text{front} = \sqrt{1 - \frac{2M}{R}} \sqrt{\frac{1 - v}{1 + v}}$$

$$f_\text{rear} = \sqrt{1 - \frac{2M}{R}} \sqrt{\frac{1 + v}{1 - v}}$$

$$f_\text{side} = \sqrt{1 - \frac{2M}{R}} \frac{1}{\sqrt{1 - v^2}}$$

Here ##f < 1## means a blueshift, and ##f > 1## means a redshift.

All we need now is to apply these formulas to the two cases described above.

Case 1: Radial infall at escape velocity. Here ##v = \sqrt{2M / R}##, which is a nice convenient value because its square appears in the first factor in all three formulas above. So the three formulas become:

$$f_\text{front} = \sqrt{1 - v^2 \frac{1 - v}{1 + v}} = 1 - v$$

$$f_\text{rear} = \sqrt{1 - v^2 \frac{1 + v}{1 - v}} = 1 + v$$

$$f_\text{side} = \frac{\sqrt{1 - v^2}}{\sqrt{1 - v^2}} = 1$$

So the light coming from the front is blueshifted, the light coming from the rear is redshifted, and the light coming from the side is not shifted at all (the two effects exactly cancel).

Case 2: Transverse motion at closest approach. Here ##v = 1/2## and ##R = 8M##, and our three formulas become:

$$f_\text{front} = \sqrt{1 - \frac{1}{4}} \sqrt{\frac{2 - 1}{2 + 1}} = \frac{1}{2}$$

$$f_\text{rear} = \sqrt{1 - \frac{1}{4}} \sqrt{\frac{2 + 1}{2 - 1}} = \frac{3}{2}$$

$$f_\text{side} = \frac{\sqrt{1 - \frac{1}{4}}}{\sqrt{1 - \frac{1}{4}}} = 1$$

Here the results are the same qualitatively as above: light coming from the front is blueshifted, light coming from the rear is redshifted, and light coming from the side (which here means light coming radially inward) is not shifted at all. In fact, the numerical values we obtained are the same as the ones we would get from case 1 if we plugged in ##v = 1/2##. There is no deep significance to this; it is just a fortuitous consequence of choosing ##R = 8M## as the closest approach distance. However, it does imply that the qualitative behavior we have described should remain the same during the entire infall process.

I invite any comments from readers, particularly if anyone sees any issues with the above calculations.

Lluis Olle
You don’t need to do any hard calculations to arrive at the result. Simply note that the speed of the trajectory in the local IRF is not bounded from above except by the speed of light. This immediately leads to the conclusion that some light from infinity will be blueshifted and some redshifted.

If I had to do the computation I would do so using invariants of both the observer and null worldlines.

Lluis Olle and vanhees71
PeterDonis said:
the numerical values we obtained are the same as the ones we would get from case 1 if we plugged in ##v = 1/2##. There is no deep significance to this; it is just a fortuitous consequence of choosing ##R = 8 M## as the closest approach distance.
Actually, on thinking this over, I realized that ##R = 8M## as the closest approach distance is not coincidental. Since I am assuming that the infall is at escape velocity, the closest approach distance will be the value of ##R## at which the escape velocity is half the speed of light. Since ##v_e = \sqrt{2M / R}##, that gives ##R = 8M##. So I could have collapsed both cases into one.

hutchphd and vanhees71
PeterDonis said:
He says at one point that he sees incoming light from distant stars blueshifted in all directions, and gives the neutron star's gravity as the reason.
I didn't read the story. In fact, the only books I read from Larry Niven are "Ringworld" and some other also about Ringworld.

When "he" says incoming light blueshifted in all directions... is that literally, or is some kind of onboard computer doing the measures... and then I guess is comparing with how he "saw" the stars before approaching the neutron star? Because at first glance, seems that the doppler shifting due gravity should be pretty small, compared with the effect of the speed at 0.5c. I don't know if the character has some kind of super-powers in his eyes, because if not the bluesifying would have to be huge to be detectable with the eyes.

Nevertheless, very well explained. Congrats. Even I can understand (not going into the details of metric and so on), what you're doing, and even you get some values.

Lluis Olle said:
When "he" says incoming light blueshifted in all directions... is that literally
The narrator is describing what he observes. He is using the ship's instruments, so he probably has some kind of frequency or wavelength readout to help. But he also says the stars literally look bluer to his eyes.

Lluis Olle said:
at first glance, seems that the doppler shifting due gravity should be pretty small, compared with the effect of the speed at 0.5c.
The math is right there in the OP to this thread. If you assume motion at escape velocity, the gravitational shift and the Doppler shift due to speed are equal in magnitude for all values of ##R##. The only variable factor is direction, which affects how the two effects combine.

PeterDonis said:
on a hyperbolic orbit
PeterDonis said:
If you assume motion at escape velocity
At escape velocity the classical orbit is parabolic.

vanhees71
Orodruin said:
At escape velocity the classical orbit is parabolic.
Yes, strictly speaking that is true.

vanhees71
Lluis Olle said:
Because at first glance, seems that the doppler shifting due gravity should be pretty small
The reason Shaeffer is doing 0.5c is gravitational potential energy converted to kinetic - exactly the same source as the blueshift. So you'd expect a fairly simple relationship between the gravitational and kinematic frequency shifts, and probably similar values, even without the explicit calculation.

vanhees71
Orodruin said:
At escape velocity the classical orbit is parabolic.
at 0.5c? ummm

Lluis Olle said:
at 0.5c? ummm
At ##R = 8M##, the escape velocity is 0.5 c, yes.

Even for radial free-fall from infinity into a Schwarzschild black hole, the optical effects are actually extremely similar to what you would expect simply from aberration plus kinematic Doppler shifts in flat spacetime. If you look at some of the animations on youtube that were done by people who understand GR and did calculations, this is what you see.

PeterDonis said:
At ##R = 8M##, the escape velocity is 0.5 c, yes.
I mean that at speed of 0.5c, the trajectories would not be exactly hyperbolic, parabolic or elliptical, but would have some corrections due to the SR. I don't know if it makes any difference.

Lluis Olle said:
I mean that at speed of 0.5c, the trajectories would not be exactly hyperbolic, parabolic or elliptical, but would have some corrections due to the SR. I don't know if it makes any difference.
The idea that geodesics in a four dimensional space time should be conic sections is not even well formed.

One would want to reduce things to a two dimensional space before talking about hyperbolas, parabolas and ellipses. But which space-like two dimensional slice would you choose? It would then be nice if that two dimensional space were flat. But how will one find a flat two dimensional slice of a curved four dimensional spacetime?

@PeterDonis has used phrasing which makes these concerns irrelevant.

Yes, that's what I said... that the trajectories would not be a conic section. Who introduced the "concept" that the classic orbit would be "parabolic" because the scenario is at escape velocity is not me, but some other. I just quoted that art 0.5c the trajectories wouldn't be conic sections.

Lluis Olle said:
Yes, that's what I said... that the trajectories would not be a conic section. Who introduced the "concept" that the classic orbits would be "parabolic" because the scenario is at escape velocity it's not me, but some other. I just quoted that art 0.5c the trajectories wouldn't be conic sections.
The point is that they would not be trajectories in a flat two dimensional space at all. So the question of whether they are or are not conic sections does not arise.

jbriggs444 said:
which space-like two dimensional slice would you choose?
The spacelike slices that are orthogonal to the timelike Killing vector field of the spacetime. That is what one is doing when one uses Schwarzschild coordinates, as I am doing in the analysis of this problem. Since the spacetime is static, all such slices are the same, so it doesn't matter which one you pick.

Lluis Olle said:
the trajectories would not be a conic section
In the sense that they would not be conic sections in a flat Euclidean 3-space, that is true, because the spacelike slices described above are not flat Euclidean 3-spaces.

However, it is standard terminology in relativity to use the terms "ellipse", "parabola", and "hyperbola" to describe the same families of orbits as those terms describe in Newtonian physics, i.e., bound orbits, unbound orbits just at escape velocity, and unbound orbits at greater than escape velocity. Those families of orbits still exist even though the spacelike slices described above are not Euclidean 3-spaces.

Lluis Olle said:
corrections due to the SR.
No. There are no "SR corrections" involved, and what I described above has nothing to do with speed or any effects of SR due to speed. It has to do with the spacelike slices not being flat, as above.

vanhees71 and jbriggs444
Lluis Olle said:
Yes, that's what I said... that the trajectories would not be a conic section. Who introduced the "concept" that the classic orbit would be "parabolic" because the scenario is at escape velocity is not me, but some other. I just quoted that art 0.5c the trajectories wouldn't be conic sections.
No, you just read my post the wrong way. My point was that the classical trajectory would be parabolic. The orbit is not classical at relativistic speeds. The idea of a conical section describing the motion was first given in the OP.

vanhees71
PeterDonis said:
No. There are no "SR corrections" involved, and what I described above has nothing to do with speed or any effects of SR due to speed. It has to do with the spacelike slices not being flat, as above.
Ok.

What I understood reading you insight is that the blueshift comes from: the effect of the mass of the neutron star, that makes for a non-flat geometry (let me say that the "effect" comes from GR), and the relative speed, that sounds to me pretty SR related at first glance.

PeterDonis said:
first, we calculate the gravitational blueshift that would be seen by a stationary observer (i.e., an observer "hovering" at rest at constant altitude above the star) at a given radius R; second, we multiply that by a Doppler shift factor due to the ship's speed relative to the stationary observer.

Orodruin said:
No, you just read my post the wrong way. My point was that the classical trajectory would be parabolic. The orbit is not classical at relativistic speeds. The idea of a conical section describing the motion was first given in the OP.
:)

I think that is what the Larry Niven story tells. Don't know, I didn't read.

Nevertheless, is a good story worth to read? Ringworld was indeed very good, and I enjoyed a lot.

Lluis Olle said:
What I understood reading you insight is that the blueshift comes from: the effect of the mass of the neutron star, that makes for a non-flat geometry (let me say that the "effect" comes from GR), and the relative speed, that sounds to me pretty SR related at first glance.
First, the whole point of the OP is that there is not a blueshift for light coming in from all directions (contrary to what is said in the story). For motion at escape velocity, there is a blueshift of light coming from the front, a redshift of light coming from the rear, and no shift at all for light coming from the side.

Second, the two effects that I decomposed the shift into are, first, the gravitational blueshift seen by an observer hovering at rest relative to the central mass at a fixed altitude (which is the same for light coming from all directions), and second, the Doppler shift due to the speed of the actual free-falling observer relative to a "hovering" observer that is momentarily co-located (which is direction dependent). The gravitational blueshift part can be thought of as an effect of non-flat spacetime geometry; but that's not the same as the non-flat spatial geometry of a spacelike slice of constant Schwarzschild coordinate time. The formula I used for the Doppler shift part is indeed the standard SR formula; but that is not a correction to the shape of the orbit; in fact it has nothing to do with the shape of the orbit.

Lluis Olle
Lluis Olle said:
I think that is what the Larry Niven story tells.
The story does indeed use the term "hyperbolic orbit". For this discussion, I am interpreting that term in the way I described in post #17. However, to make the math simpler, I assumed an orbit at exactly escape velocity, rather than at greater than escape velocity; that's why @Orodruin correctly pointed out that "parabolic" would be a more strictly accurate description than "hyperbolic"--where, again, I am interpreting the terms as I described in post #17. (@Orodruin specifically said the "classical" orbit, meaning Newtonian; that's why I explained the usage of those terms that I am adopting, which continues to be valid in relativity.)

Lluis Olle and vanhees71
Lluis Olle said:
is a good story worth to read?
I would say yes if you enjoyed whatever Niven you have read so far.

Lluis Olle
PeterDonis said:
I would say yes if you enjoyed whatever Niven you have read so far.
I'm a total Niven fan, and also a fan of the Niven/Pournelle books like the Mote and Heorot books.

Niven does try to keep his physics somewhat consistent, but one has to turn a blind eye sometimes when the plot requires otherwise. The stasis field has all the problems of 'unobtanium' for instance, and makes me cringe.
Other times he's corrected things. The ringworld was shown to be orbitally unstable, and in the first sequel, he introduced a mechanism where it could be corrected (and the non-functionality of which became the plot of the book).
In Neutron star, Beowulf blackmails the puppeteers because he knows their secret homeworld has no moons due to lack of their knowledge of tides. It is totally unbelievable that a race more advanced than humans would have no concept of second-order gravitational effects. Also, tides are unnoticable on most planets where the moons are but distant asteroids.

My huge gripe was the avoidable gaffe in Integral Trees where the trees are floating in winds that move at different relative speeds. They're held vertical by tidal forces and are balanced by high winds in opposite direction at their ends where the foliage is.
They wander sometimes too close to the edge where the atmosphere is thinner. When this happens, the (huge) tree breaks in the middle (a process that takes months) and, as the book describes, the one further out is lost into the low pressure region and the inner one moves to the dense region where it grows a new outer end.
OK, that's totally wrong on several levels. Primarily, it is the outer one that is saved since it drags on the air that causes its orbital speed to climb or drop, bringing it back into the mainstream. The half on the thicker air is the one lost to the low pressure area. Second problem is that without the balance of the other end, the half that is saved will careen straight through the torus and out the other side to die away from its mate. No, if the trees are to same themselves, they need to do it by temporarily increasing foliage on the thin end (or having at autumn at the thick end), something that can be replaced quickly once the tree is re-centered in its orbital radius.

PeterDonis said:
First, the whole point of the OP is that there is not a blueshift for light coming in from all directions (contrary to what is said in the story). For motion at escape velocity, there is a blueshift of light coming from the front, a redshift of light coming from the rear, and no shift at all for light coming from the side.

Second, the two effects that I decomposed the shift into are, first, the gravitational blueshift seen by an observer hovering at rest relative to the central mass at a fixed altitude (which is the same for light coming from all directions), and second, the Doppler shift due to the speed of the actual free-falling observer relative to a "hovering" observer that is momentarily co-located (which is direction dependent). The gravitational blueshift part can be thought of as an effect of non-flat spacetime geometry; but that's not the same as the non-flat spatial geometry of a spacelike slice of constant Schwarzschild coordinate time. The formula I used for the Doppler shift part is indeed the standard SR formula; but that is not a correction to the shape of the orbit; in fact it has nothing to do with the shape of the orbit.
In this article, the author uses SR and the Lagrangian (not GR) to calculate an approximate orbit, and its result is the classical orbit plus a correction due to SR.

Even if it's approximate, I think you can most of the times say "the classical orbit plus a correction due to SR", but perhaps in some extreme cases the correction would be bigger that the "classical" term itself.

PeterDonis said:
First, the whole point of the OP is...
I think there's a typo, should read "PD" ¿?

Lluis Olle said:
the author uses SR and the Lagrangian (not GR) to calculate an approximate orbit, and its result is the classical orbit plus a correction due to SR.
Yes, and as the abstract to that article notes:

All three characteristics are qualitatively correct, though suppressed when compared to
more accurate general-relativistic calculations
In other words, while you can sort of wave your hands and get the same qualitative features of orbits this way, you can't get the accurate numerical answers that GR gives you.

The reason is simple: as the abstract notes, the "SR" correction is for the relativistic kinetic energy. But there is no correction to the potential energy. That doesn't make sense from an order of magnitude perspective, because for free-fall motion in a stationary spacetime, the kinetic and potential energy are both of the same order of magnitude. So one would not expect to get numerically accurate answers if only the kinetic energy correction is included.

The abstract then notes:

This model is improved upon by including relativistic gravitational potential energy
Which amounts to now using GR instead of SR.

Lluis Olle and vanhees71
Lluis Olle said:
I think there's a typo, should read "PD" ¿?
No. "OP" means the original post in this thread.

PeterDonis said:
TL;DR Summary: In the "Neutron Star" story by Larry Niven, he says the pilot of a spaceship on a hyperbolic orbit leading to a close approach to a neutron star sees light from distant stars blueshifted in all directions. However, this appears to be incorrect. A corrected calculation is given.
...
I invite any comments from readers, particularly if anyone sees any issues with the above calculations.
I missed this when originally posted - my only comment is thank you very much for working through this! Neutron Star is one of my favourites and I have wondered how 'hard' (in the sense of consistent rather than difficult) the science was. I hope I can find it on the bookshelf, I don't think I have re-read it this century.

Thanks again - an early candidate for my favourite post of 2023!

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