General relativistic comparison to Newtonian physics

[Myslius]

Let's say a small object orbits around massive object at some distance in perfectly circular orbit. In Newtonian physics the force acted upon orbiting body is:
Force = m1 * orbital_speed^2 / distance = G * (m1*m2) / r^2
for an object of 1kg orbiting a body of 2*30^kg at the distance of 1.521 * 10^11m the F is equal to:

F = G * (1kg *2*10^30) / (1.521 * 10^11m)^2 = 0.00576982944 Newtons

Orbital speed^2 = 0.00576982944 Newtons * 1.521 * 10^11m / (1kg)
Orbital speed = 29624.1634 m / s

So far so good, numbers matches perfectly.
Now, let's take an example of black holes where orbital speed is equal to c. First, we find an orbital distance:
distance = (29624.1634 m / s / c)^2 * 1.521 * 10^11m = 1485.18309 meters
In general relativity we have:
Radius of black hole = 2GM/c^2 = 2G(2*10^30)/c^2 = 2970.36619333 meters
We're interested in photon sphere here, where light orbits around black hole at speed of light, so the radius that we are interrested in is 2970.36619333 meters * 1.5 = 4455.54929 meters.

Now the difference between Newtonian physics and general relativity is:
4455.54929 meters / 1485.18309 meters = 3

Can you explain the factor of 3?

I know that this is (1 / sqrt(1 -2/3)^2 = 3, which is lorentz factor squared.

What's your understanding of the factor here? Shapiro delay, length contraction, time dilation, combination (multiplication of both)

How YOU understand it?

Reconciling general relativity with Newtonian physics:

Force = m1 * orbital_speed^2 / (distance * γ^2) = G * (m1*m2) / (r * γ )^2
where γ is equal to Lorentz factor.

This formula is now true for both, low velocities physics and physics near the black hole. But still, my question remains what exactly gets smaller. Is it the orbital speed, is it the distance, is the mass that's increasing? You could interpret it in many ways. Would like to hear your opinion.

 

[PeroK]

Can you explain the factor of 3?

How YOU understand it?

Reconciling general relativity with Newtonian physics:

There's no sense in trying to apply or reconcile Newtonian physics with light orbiting a black hole. It's valid to check that GR becomes Newtonian in the appropriate special cases, but Newtonian physics simply doesn't apply and cannot be generalised to all of GR.

 

[Myslius]

There's no sense in trying to apply or reconcile Newtonian physics with light orbiting a black hole. It's valid to check that GR becomes Newtonian in the appropriate special cases, but Newtonian physics simply doesn't apply and cannot be generalised to all of GR.

You're probably right.

 

[Ibix]

To get the Newtonian approximation you assume that all speeds are much less than $c$ and $r\gg R_S$. Neither of these assumptions are true for light orbiting a black hole at $1.5R_S$. So the explanation for the difference is simple: all the terms you neglected when you derived the Newtonian approximation were not small enough to neglect in this case.

 

[Vanadium 50]

There's no sense in trying to apply or reconcile Newtonian physics with light orbiting a black hole

Nor is there sense to writing "F = G * (1kg *2*10^30) / (1.521 * 10^11m)^2 = 0.00576982944 Newtons" with nine digits of accuracy. (especially since G isn't known that well) Extraneous precision is a sign of calculating without think whether the answer makes sense.

 

[Dale]

What's your understanding of the factor here? Shapiro delay, length contraction, time dilation, combination (multiplication of both)

How YOU understand it?

A better approach would be to use the parameterized post Newtonian expansion. In that test theory, which includes Newtonian gravity and all possible first-order deviations from it, each of the parameters has a standard meaning. You can derive your formula in the PPN expansion and see which term controls the difference between Newtonian gravity and GR.

 

[PeterDonis]

A better approach would be to use the parameterized post Newtonian expansion.

This expansion still assumes that $v / c$ is a small parameter, so it doesn't work for the photon sphere around a black hole.

 

[pervect]

One effect of GR that Newtonian physics does not capture at all is spatial curvature. This effect becomes more noticable at high velocities. It's unclear if this is the sole effect that matters in this case.

Trying to interpret spatial curvature as a force doesn't work.

Usually, ignoring spatial curvature causes a 2:1 error, for instance the notion that the deflection of starlight in GR is "twice the Newtonian value". However, the OP claims to have a 3:1 error if I'm understanding them correctly, so probably this isn't sufficient to explain things fully.

The standard, full, solution in GR is to get rid of the idea that gravity is a force, and to instead note that orbits are geodesics, and to forget about "forces" and start studying the geodesic equations. Unfortunately, this is likely to involve more math than the OP is ready for. But it will give the correct answers, in a uniform paradigm that is not a hodge-podge frankenstein monster built up of incompatible and different theories that are "stitched together".

Another motivational solution might be to use an approach based on Fermat's principle. From the wiki article of the same name:

Fermat's principle states that the path taken by a ray between two given points is the path that can be traversed in the least time. In order to be true in all cases, this statement must be weakened by replacing the "least" time with a time that is "stationary" with respect to variations of the path

This involves another effect that does not involve interpreting gravity as a force, typically (though somewhat misleadingly) called gravitational time dilation.

 

[Dale]

This expansion still assumes that $v / c$ is a small parameter, so it doesn't work for the photon sphere around a black hole.

Yes, but it should still differ in weaker fields, it isn’t like it is an effect that just suddenly appears at the photon sphere. I am assuming of course that there is a first order term that would control this, if it is only a second order effect then it won’t show up in the PPN.

 

[Myslius]

However, the OP claims to have a 3:1 error if I'm understanding them correctly, so probably this isn't sufficient to explain things fully.

No you misunderstood completely, I'm claiming that the formula gives the same result as GR, when orbital speed is between 0 and c. The factor 3 comes from spacetime curvature, when distance is 1.5r_s. The number is way lower for speeds below c, and it reduces to Newtonian gravity formula when v >> 0. Also this has nothing to do with deflection angle.

 

[PeterDonis]

i'm claiming that the formula gives the same result as GR, when orbital speed is between 0 and c.

What formula are you referring to?

The factor 3 comes from spacetime curvature, when distance is 1.5r_s.

What are you basing this on?

 

[Myslius]

escape velocity at 1.5 r_s is c/sqrt(3/2), lorentz factor is 1 / sqrt(1 -2/3)

 

[PeterDonis]

escape velocity at 1.5 r_s is c/sqrt(3/2)

What does this have to do with anything? We're not talking about objects on an escape trajectory; we're talking about objects in bound circular orbits.

lorentz factor is 1 / sqrt(1 -2/3)

What does this Lorentz factor mean, physically?

 

[Myslius]

everything, it tells how much space is curved, it determines angle of curvature, how much object experiences gravitational time dilation. In special relativity case it determines how much length contracts for external observer, how much mass increases. But you already know that. What's the point of asking that?

 

[PeterDonis]

everything, it tells how much space is curved, it determines angle of curvature, how much object experiences gravitational time dilation.

No. All of this is wrong.

In special relativity case it determines how much length contracts for external observer, how much mass increases.

This is true for the Lorentz factor in SR, but it has nothing whatever to do with the comparison you are trying to make between Newtonian and GR values.

But you already know that.

No, I know no such thing. See above.

What's the point of asking that?

To see what your answer was. Now that you have answered, it is apparent that you have made some errors which need to be corrected.

First, let's look at the actual math of Newtonian gravity vs. GR for the case in question, a spherically symmetric gravitating massive body with an object in a free-fall circular orbit around it. For the Newtonian case, the orbital velocity of this object is given by:

$$
v = \sqrt{\frac{GM}{r}}
$$

where $G$ is Newton's gravitational constant, $M$ is the mass of the body, and $r$ is the circumference of the orbit divided by $2 \pi$. (Note that I did not say that $r$ is the "radius" of the orbit--the reason why will be apparent in a moment.)

For the GR case, the orbital velocity of the object is given by:

$$
v = \sqrt{\frac{GM}{r - r_s}}
$$

where $r_s = 2 G M / c^2$ is the Schwarzschild radius corresponding to the mass $M$. You are considering the case where the massive body is a black hole, so $r_s$ will also be the "areal radius" of the hole's horizon, i.e., $r_s = \sqrt{A / 4 \pi}$, where $A$ is the horizon area. However, this $r_s$, and more generally the "radius" $r$ that appears in the above formulas, in the GR case is not the physical radius to the "center". For an ordinary massive body like a planet or star, this is because space is not Euclidean (more precisely, "space" as seen by observers at rest relative to the body is not Euclidean), so the physical distance to the center is not the circumference of the circle divided by $2 \pi$, it is larger. For a black hole, there is no physical "center" at all--the locus $r = 0$ is not a place in space, it is a moment of time (which is only accessible by falling through the hole's horizon). That is why I defined $r$ as I did above--because that definition of "radius" is the only one that allows the GR case to be compared with the Newtonian case.

The comparison above is of orbital velocity at a given $r$, but you appear to want to compare the Newtonian and GR values of $r$ for a given orbital velocity. We can obtain those by simply inverting the above formulas; for the Newtonian case, we get:

$$
r_N = \frac{GM}{v^2}
$$

and for the GR case we get

$$
r_R = \frac{GM}{v^2} + r_s = GM \left( \frac{1}{v^2} + \frac{2}{c^2} \right)
$$

The ratio $r_R / r_N$, which is what you appear to be interested in, is now easily found:

$$
\frac{r_R}{r_N} = 1 + \frac{r_s}{r_N} = 1 + 2 \frac{v^2}{c^2}
$$

Note that the above ratio is not $1$ for any nonzero value of $v$; it increases quadratically in $v$, and gives a ratio of $3$ at $v = c$.

Note also that there is no mention of either escape velocity or Lorentz factor in any of the above. For a given value of $v$, we can of course compute the gamma factor $1 / \sqrt{1 - v^2 / c^2}$, which gives the gamma factor of the object in the circular orbit relative to an observer who is "hovering" at rest relative to the gravitating massive body at the same $r$. But this has nothing whatever to do with the comparison between the Newtonian and the GR case (the Lorentz factor of course is meaningless in Newtonian physics anyway).

Finally, note that the ratio above is not the same as "spatial curvature", which has nothing whatever to do with any comparison between Newtonian and GR values of anything; nor is it the same as gravitational time dilation.

 

[Ibix]

it tells how much space is curved

No.

it determines angle of curvature

I don't think that's a thing.

how much object experiences gravitational time dilation

No.

how much length contracts for external observer

Yes.

how much mass increases

No. Unless you mean relativistic mass, in which case you should say "relativistic mass", since "mass" unqualified usually means the invariant mass. And be prepared to be told that the concept of relativistic mass was deprecated decades ago.

 

[PeterDonis]

how much length contracts for external observer

Yes.

I'm not so sure. The length contraction in question is relative to a "hovering" observer at the same $r$, as I pointed out in my post just now. I don't think that's what the OP meant by "external observer".

 

[Ibix]

I'm not so sure. The length contraction in question is relative to a "hovering" observer at the same $r$, as I pointed out in my post just now. I don't think that's what the OP meant by "external observer".

Well, OP does begin that sentence with "[in] special relativity case", which I took to mean that the comments in that sentence were referring to flat spacetime. If so, that comment is correct. I accept that other interpretations are possible.

 

[Myslius]

https://wikimedia.org/api/rest_v1/media/math/render/svg/6913f70aac912e2a144c98036c7d8b96abf90ae7

• 
  is the proper time between two events for an observer close to the massive sphere, i.e. deep within the gravitational field
• 
  is the coordinate time between the events for an observer at an arbitrarily large distance from the massive object (this assumes the far-away observer is using Schwarzschild coordinates, a coordinate system where a clock at infinite distance from the massive sphere would tick at one second per second of coordinate time, while closer clocks would tick at less than that rate)

I can't fill the cup that's already full. PeterDonis, Ibix, some things in your comments are true, some are wrong. However the attutude of NO NO NO, WRONG, ALL THIS IS WRONG throws me off of discussing further.

 

[PeterDonis]

• 
  is the proper time between two events for an observer close to the massive sphere, i.e. deep within the gravitational field
• 
  is the coordinate time between the events for an observer at an arbitrarily large distance from the massive object (this assumes the far-away observer is using Schwarzschild coordinates, a coordinate system where a clock at infinite distance from the massive sphere would tick at one second per second of coordinate time, while closer clocks would tick at less than that rate)

These times are for observers at rest relative to the massive gravitating body, not in free-fall circular orbits about it. So they are irrelevant to the comparison you were trying to make.

Note also that coordinate times are not "for" any particular observer; they are just coordinate labels. There is no "time between events for an observer" if that observer's worldline does not pass through those events.

some things in your comments are true, some are wrong.

No, everything I said is correct, and I showed my work to confirm that.

Everything @Ibix said is correct as well.

However the attutude of NO NO NO, WRONG, ALL THIS IS WRONG throws me off of discussing further.

If you say things that are wrong, you should expect to be told you are wrong. If you are unable to discuss further after being told that, then you should re-evaluate your own attitude.

 

[Myslius]

No, everything I said is correct, and I showed my work to confirm that.

"No, everything I said is correct", That's very healthy attitude while learning. There's no point to pinpoint what's wrong.

 

[PeterDonis]

"No, everything I said is correct", That's very healthy attitude while learning. There's no point to pinpoint what's wrong.

Since you are apparently unable to have any further productive discussion, this thread is closed.