- #76

timmdeeg

Gold Member

- 1,109

- 82

- Thread starter lowemack
- Start date

- #76

timmdeeg

Gold Member

- 1,109

- 82

- #77

- 30,609

- 9,653

So you are postulating two rockets, each with a rest mass 59% the mass of Saturn, and each compressed so that they are less than half a meter long (so their centers can be 1 m apart without them touching). Wow. But ok, we'll go with that. It won't take long to see the problem.To simplify the scenario let's assume two hypothetical spherical symmetric mass distributions with identical radius r and identical rest mass m that can be superposed without any interaction except gravity. If both objects are at rest they shouldn't collapse to a black hole. That means

[itex]r > \frac{{4 \cdot G \cdot m}}{{c^2 }}[/itex]

If you need actual numbers let's take r = 1 m and m = 3.354·10^{26}kg (59% the mass of Saturn).

As I've said before, you left out two key part of the whole system. First, where does the energy come from to accelerate both these objects to 8.68% of the speed of light? You're talking about two objects each with 59% of the mass of Saturn. Have you calculated how much fuel they would need to have at the start? You have to *add* that fuel mass to the mass of the systems at the start.As the distance d (measured from center to center) can not be negative this is possible for

[itex]v > c \cdot \sqrt {1 - \left( {\frac{{4 \cdot G \cdot m}}{{r \cdot c^2 }}} \right)^2 }[/itex]

With the above mentioned values this minimum velocity would be 2,602·10^{7}m/s (8,68% the speed of light).

Second, some of that starting energy isn't contained in the rockets in the final state; it's contained in the rocket exhausts. You need to account for that as well. See below.

Ok, that at least clarifies that. Here are the correct equations for your scenario (I won't bother filling in actual numbers, the issue will be obvious without that).To avoid a resulting momentum or angular momentum it should be a head-on collision.

I have two rockets, each with a *payload* mass m, that are moving towards each other, so their combined momentum is zero. That means that the invariant mass of the two rockets combined is:

[tex]M_{total} = \frac{1}{c^{2}} E = 2 \gamma m[/tex]

where v is the velocity of each rocket (they're both the same, obviously). You have postulated that this combined invariant mass is sufficient for the rockets to be inside each other's Schwarzschild radius; that radius is

[tex]r_{S} = \frac{2 G M_{total}}{c^{2}} = \frac{4 G \gamma m}{c^{2}}[/tex]

But you have also postulated that the rockets were not within each other's Schwarzschild radius at the start, so their initial separation is greater than [itex]r_{S}[/itex], and their initial sizes are each less than 1/2 [itex]r_{S}[/itex].

To achieve that velocity, according to the rocket equation, each rocket with payload mass m must also start out with a fuel mass M given by:

[tex]\frac{M}{m} = \gamma \left( 1 + \frac{v}{c} \right) - 1[/tex]

That means each rocket starts out with a total mass M + m given by:

[tex]M + m = \gamma m \left(1 + \frac{v}{c} \right)[/tex]

*That* means each rocket starts out with a Schwarzschild radius, based on its own starting mass, of:

[tex]r_{s} = \frac{2 G \left( M + m \right)}{c^{2}} = \frac{2 G \gamma m}{c^{2}} \left( 1 + \frac{v}{c} \right)[/tex]

You will note that [itex]2 r_{s} > r_{S}[/itex], i.e,. the combined Schwarzschild radius of the two rockets at the start is *larger* than the Schwarzschild radius of the two rockets combined at the end. In other words, if the two rockets together are confined inside a black hole at the end, each rocket separately must have been confined inside a black hole at the start.

The above also implies, of course, that the combined invariant mass of the two rockets at the end is *not* the combined invariant mass of the whole system; there is still energy with a mass-equivalent of

[tex]\frac{4 G \gamma m}{c^{2}} \frac{v}{c}[/tex]

missing. This is the energy contained in the rocket exhaust.

- #78

- 1,895

- 332

As I said before that doesn't matter. To make any additional energy negligible you just have to keep its distant from the center great enough. Therefor we do not need to take it into account.As I've said before, you left out two key part of the whole system. First, where does the energy come from to accelerate both these objects to 8.68% of the speed of light?

I do not need that because I never defined a specific method for the acceleration of the objects. It was your idea to think about rockets. So if there would rise problems from this method it is not my fold. But even with rockets there are no problems that couldn't be solved. Just make them bigger than their own initial Schwarzschild radius and let the objects release at sufficient distance to push the rockets far enough out of the way.Second, some of that starting energy isn't contained in the rockets in the final state; it's contained in the rocket exhausts. You need to account for that as well. See below.

But you do not need any rockets. You could also use solar sails or something similar to accelerate the object (just to give you another example). The specific method and the energy needed for the acceleration is irrelevant for this Gedankenexperiment.

- #79

- 30,609

- 9,653

What I was calling the "rocket equation" is a simple consequence of the conservation of energy and momentum; those laws must be obeyed by any method of accelerating objects. All that changes is the specifics of how the energies in that equation are assigned to parts of the system. For example:I do not need that because I never defined a specific method for the acceleration of the objects. It was your idea to think about rockets.

In this case the energy that pushes the solar sails still needs to come from somewhere; the "fuel mass" that I was calling M would reside at the energy source for the solar sail, and would gradually be expended as the sails were accelerated, so that what I was calling the energy of the "rocket exhaust" would now be the energy contained in the radiation that pushed the sails. But the total energy of the system would still be the same. Also, in order to keep the total momentum zero, the source of the radiation that pushes the sail has to move in the other direction, to cancel out the radiation's momentum; so some of the energy that I was calling "rocket exhaust" would actually become kinetic energy of the radiation source. I'm pretty sure that in this case the total energy of the system would have to be even higher than in the rocket case.You could also use solar sails or something similar to accelerate the object (just to give you another example).

That is already accounted for in the equation I gave. In the final state, all that is present in each final object is the "payload" with rest mass m, which is what you are calling the "object" itself; all of what I was calling the "fuel" with original mass M is no longer there, it's been converted into energy, some of which is now contained in the "object" and some of which is contained in what I called the "rocket exhaust". The point in the case of the rocket is that, when the rocket is sitting on the launch pad, *all* of that mass is in the same place.But even with rockets there are no problems that couldn't be solved. Just make them bigger than their own initial Schwarzschild radius and let the objects release at sufficient distance to push the rockets far enough out of the way.

It's true that the latter point does *not* apply, strictly speaking, in the case of a solar sail or some similar method; the source of the radiation that pushes the sail could be anywhere, in principle. However, I have thought of yet another factor that we have not yet taken into account. Since we are talking about Schwarzschild radius and objects being inside it, we are implicitly assuming that the gravity of the amount of total mass contained in those objects in their final state is not negligible. That means that SR does not really apply to this scenario, since SR assumes that gravity is negligible.

So a correct analysis of this scenario requires GR; i.e., it requires taking into account the curvature of spacetime produced by the system as a whole. Doing that changes things a lot. I'll put that in a separate post.

- #80

- 30,609

- 9,653

Consider first a simpler case where the proper acceleration of both bodies is zero; we just have two objects, each with total mass 1/2 M_0 (half of the total final invariant mass in your scenario), separated by some distance r which is greater than the Schwarzschild radius associated with M_0, and initially at rest. Their mutual gravity will cause them to fall into each other; at some point, they will be separated by *less* than the Schwarzschild radius associated with M_0, and they will form a black hole. This is just a stripped-down version of the spherically symmetric collapse of a star, as in the classic Oppenheimer-Snyder paper of 1939.

There is a technical point here: since the bodies will acquire kinetic energy as they fall, their starting rest masses will be *less* than 1/2 M_0; how much less depends on how far apart they are at the start. Their *total* energy at the start is still 1/2 M_0, but not all of that energy will be rest mass. The difference can be thought of as the "gravitational potential energy" of each body in the field of the total invariant mass M_0. From a distance much greater than the initial separation of the bodies, the system as a whole will look like a single mass M_0.

The case where the two bodies are accelerated towards each other, by rockets or solar sails, or whatever, clearly can't change the final conclusion; the bodies will still form a black hole. The only difference is that, since the source of energy that accelerates the bodies may not be part of the final system, the initial invariant mass of the system (i.e., the two bodies) may be *less* than M_0, assuming the final invariant mass of the black hole that is formed is M_0. (In other words, things like the energy in the radiation that pushes the solar sails, or in the rocket exhaust, or in the momentum of the radiation source, are not part of the "system", so the system can exchange energy with other systems, whereas in the first case of purely freely falling bodies, the system was isolated and its total energy could not change.)

So you are correct that it is possible for two bodies, neither of which is a black hole, to come together (maintaining zero net momentum) to form a black hole, and I was wrong to think that was not possible. However, my answer to the original question, does the acceleration of bodies (in the sense of acceleration felt, or "proper acceleration") depend on their velocity, is still no. In both the cases I just described, the acceleration the bodies feel is specified by the scenario, and there are no constraints on what we can specify. In the first case, the proper acceleration of the bodies is always zero; in the second, it is whatever the acceleration source (rocket, solar sail, whatever) produces, and we can specify it to produce any acceleration we want, in principle, including a constant one.

- #81

- 1,895

- 332

As I said before the size rockets of the rockets can be much grater than their oown Schwarzshild radius. You mentioned that this already accounted for in the equation you gave but I do not see it there.The point in the case of the rocket is that, when the rocket is sitting on the launch pad, *all* of that mass is in the same place.

Last edited:

- #82

- 1,895

- 332

Of course this is possible but that's not the key point. Is is even possible for fast bodies that doesn't form a black hole if they come together with negligible relative velocity? I would say it is and that would mean that in this special scenario the velocity of the bodies affect their gravity even with your definition of the "amount of gravity produced".So you are correct that it is possible for two bodies, neither of which is a black hole, to come together (maintaining zero net momentum) to form a black hole, and I was wrong to think that was not possible.

- #83

timmdeeg

Gold Member

- 1,109

- 82

It seems, your expectation is in agreement with http://ajp.aapt.org/resource/1/ajpias/v53/i7/p661_s1?isAuthorized=no [Broken].I would say it is and that would mean that in this special scenario the velocity of the bodies affect their gravity even with your definition of the "amount of gravity produced".

Last edited by a moderator:

- #84

- 1,895

- 332

Thanks for this link. I was already aware that the gravitational mass of a photon must be twice its inertial mass (due to deflection of light in gravitational fields resulting from GR or observed at the edge of the sun) but I wasn't sure whether this can be shown for relativistic bodies with rest mass too. We live and learn.It seems, your expectation is in agreement with http://ajp.aapt.org/resource/1/ajpias/v53/i7/p661_s1?isAuthorized=no [Broken].

Last edited by a moderator:

- #85

- 30,609

- 9,653

The article is behind a paywall so I can't read the text, but I strongly suspect that the authors are using confusing terminology. Calling the effect they are describing an "increase in active gravitational mass" would only be justified if the Newtonian formula for the "force" of gravity were correct. It isn't. See below.It seems, your expectation is in agreement with http://ajp.aapt.org/resource/1/ajpias/v53/i7/p661_s1?isAuthorized=no [Broken].

The experimental result you are referring to for the deflection of light by the Sun is well known, of course, but it doesn't mean what you think it means. It is true that, if I do a naive Newtonian calculation of how much the light should be deflected, by dividing the light's energy by c^2 and plugging into the Newtonian formula for "acceleration due to gravity", I get an answer that is half the deflection that is actually observed. That is because gravity is not described by the Newtonian formula; it's described by the GR formula, which is the Einstein Field Equation. So trying to draw deductions from what the Newtonian formula says is not correct.Thanks for this link. I was already aware that the gravitational mass of a photon must be twice its inertial mass (due to deflection of light in gravitational fields resulting from GR or observed at the edge of the sun) but I wasn't sure whether this can be shown for relativistic bodies with rest mass too. We live and learn.

In particular, the deflection result for light does not mean that the light's "gravitational mass" is twice its "inertial mass"; to justify any such interpretation, you would have to first specify how we are to measure the light's "inertial mass" in such a way that that relationship always holds. Saying that the light's inertial mass is its energy divided by c^2 won't work, because there are other scenarios where the energy divided by c^2 is the *same* as what you are calling the "gravitational mass". (For example, put some light in a box with reflecting walls whose mass is negligible; the externally measured gravitational mass of the box will be the total energy of the light divided by c^2. This will also be its inertial mass if you try to push it and measure the ratio of applied force to acceleration.)

GR explains the "increase" in deflection of ultra-relativistic particles as a consequence of the spacetime curvature produced by the mass of the "source" object. The Newtonian formula only captures a part of the effects of that curvature, the "static" part, i.e., the part analogous to the Coulomb force in electromagnetism. But there is an additional effect analogous to the magnetic force in electromagnetism, which only appears when an object is moving relative to the source (or, equivalently, when the source is moving relative to the object); in the limit when the speed of the relative motion approaches the speed of light, this "magnetic" effect becomes equal in magnitude to the static effect. That's why light and ultrarelativistic particles deflect more.

I said "analogous to" the Coulomb and magnetic forces above, but it's important to keep in mind one crucial difference: the objects being deflected (the light or the ultrarelativistic particles) feel *zero* acceleration; they are in free fall. So the answer to DrStupid's question about acceleration depending on velocity is still no.

Last edited by a moderator:

- #86

- 30,609

- 9,653

On re-reading, I should note that the increased deflection of ultra-relativistic particles by a large mass is often attributed to the Shapiro time delay effect (i.e., gravitational time dilation close to a mass, meaning that an object just grazing the mass spends a longer time there as seen from far away), or the space curvature caused by the mass (meaning that the objects have to travel through a larger distance), or some combination of the two. For example, see Garth's post on PF here:GR explains the "increase" in deflection of ultra-relativistic particles as a consequence of the spacetime curvature produced by the mass of the "source" object. The Newtonian formula only captures a part of the effects of that curvature, the "static" part, i.e., the part analogous to the Coulomb force in electromagnetism. But there is an additional effect analogous to the magnetic force in electromagnetism, which only appears when an object is moving relative to the source (or, equivalently, when the source is moving relative to the object); in the limit when the speed of the relative motion approaches the speed of light, this "magnetic" effect becomes equal in magnitude to the static effect. That's why light and ultrarelativistic particles deflect more.

https://www.physicsforums.com/showpost.php?p=842496&postcount=12

Also see Ned Wright's page here:

http://www.astro.ucla.edu/~wright/deflection-delay.html

The explanation I gave, adding a "magnetic" component to the effective "force" seen by an object moving relative to the mass (I put "force" in quotes because, as I noted before, objects moving under this "force" feel zero acceleration), is a different way of saying the same thing; all of these explanations refer to the same underlying mathematics.

- #87

timmdeeg

Gold Member

- 1,109

- 82

- #88

- 30,609

- 9,653

Because you're using the wrong definition of "active gravitational mass"; you're plugging numbers into the Newtonian formula for gravitational "force" and trying to read off what the "active gravitational mass" is by applying F = ma, but the Newtonian formula for F is not correct; it doesn't fully describe the actual "force" exerted by a massive object.

When you plug numbers into the correct formula for the "force" (i.e., adding in the "magnetic" force that I referred to, which is predicted by GR but is *not* predicted by Newtonian theory), you find that the "active gravitational mass" you deduce for the object via "F = ma" is equal to its inertial mass, as it should be. (This is all bearing in mind, as I noted before, that this "force" is not felt--the object in question is in free fall.)

However, as I also noted, viewing gravity as a "force" is not the recommended way to view it in GR, because even after adjusting the formula for the "force" as above, you still have to be careful about other formulas like "F = ma"; the straightforward interpretation of that formula in Newtonian terms does not work in the relativistic case. It turns out to be easier to discard the idea of gravity as a "force" altogether and view things in terms of spacetime curvature; in those terms you would predict the trajectory a particle moving at high speed relative to a gravitating mass by looking at the curvature of space and time caused by the mass, and viewing the particle's trajectory as a geodesic (the analogue to a straight line) in that curved spacetime. This gives the same answer as the "force" viewpoint (when we use the correct covariant formulas for "F" and "a").

- #89

- 1,895

- 332

What else? Gravitational mass is defined by Newtons law of gravitation.Calling the effect they are describing an "increase in active gravitational mass" would only be justified if the Newtonian formula for the "force" of gravity were correct.

As I must use Newtons law of gravitation to determine the gravitational mass it makes sense to use his definition of inertial mass too. That leads to m=E/c² in relativity and this works in every case including for light.Saying that the light's inertial mass is its energy divided by c^2 won't work

Of course you can also use rest mass and relativistic momentum but that leads to the same result for the gravitational mass.

- #90

- 30,609

- 9,653

You do know that Newton's theory of gravity is wrong, right? That it is experimentally falsified? Including his law of gravitation? So if you are using his laws to define "gravitational mass", you are defining something that is going to give you false predictions in regimes where his laws are known to be wrong. Particles moving at or near the speed of light is one such regime.What else? Gravitational mass is defined by Newtons law of gravitation.

Again, only if you insist on using Newton's (wrong) definition. If you use the correct relativistic formulas, you get that "inertial mass" always equals "gravitational mass", in so far as those terms even have useful definitions. Or you can recognize that this whole issue is irrelevant in GR, and calculate everything using spacetime curvature without ever having to worry about "inertial mass" or "gravitational mass".Of course you can also use rest mass and relativistic momentum but that leads to the same result for the gravitational mass.

- #91

- 1,895

- 332

I am aware of this problem but there is no other definition of gravitational mass.You do know that Newton's theory of gravity is wrong, right? That it is experimentally falsified? Including his law of gravitation? So if you are using his laws to define "gravitational mass", you are defining something that is going to give you false predictions in regimes where his laws are known to be wrong.

That makes no sense because in GR there is no such thing like "gravitational mass". In GR the source of gravitation is not mass but the stress-energy tensor. If you talk about gravitational mass you are talking about Newton's law of gravitation (even if you are not aware of it). If you do not want to refer to Newton's law of gravitation you must not talk about gravitational mass.If you use the correct relativistic formulas, you get that "inertial mass" always equals "gravitational mass", in so far as those terms even have useful definitions.

- #92

- 30,609

- 9,653

Then why bring up the term? As I've already noted, the effect you were trying to describe, bending of light by the Sun, can be described without even talking about "gravitational mass" or "inertial mass" at all.I am aware of this problem but there is no other definition of gravitational mass.

Well, you were the one who brought up the term "gravitational mass"; I put the term in scare-quotes precisely because of the issue you describe. A better way of expressing the point I was making would be to say that the trajectory of a body that is in free fall is independent of the rest mass of the body. The trajectory does depend on the body's initial velocity relative to the source of gravity, but given two objects of different rest mass with the same initial velocity, they will both follow the same trajectory (as long as no other forces are acting).That makes no sense because in GR there is no such thing like "gravitational mass". In GR the source of gravitation is not mass but the stress-energy tensor. If you talk about gravitational mass you are talking about Newton's law of gravitation (even if you are not aware of it). If you do not want to refer to Newton's law of gravitation you must not talk about gravitational mass.

- #93

- 1,895

- 332

Try to quantify the effect of velocity on gravity without it and you will see why. Gravitational mass is a well defined value that allows to describe this effect with a few words and a very simple equation (see the abstract quoted by timmdeeg).Then why bring up the term?

That's not what we are talking about because:The trajectory does depend on the body's initial velocity relative to the source of gravity, but given two objects of different rest mass with the same initial velocity, they will both follow the same trajectory (as long as no other forces are acting).

1. It applies to trajectories in static gravitational field only but a real gravitational field will be influenced by the bodies and as this interaction depends on the mass two objects with different mass will not follow the same trajectory.

2. We are not talking about two different bodies in the same static gravitational field but about one body in the dynamic gravitational field of another body moving with different velocities.

3. If we want to talk about two bodies with different velocities in an almost static gravitational field we should not compare their trajectories but their accelerations (not 4-accelerations).

- #94

- 7

- 0

- #95

- 30,609

- 9,653

As an abstraction used to simplify the understanding of one particular phenomenon, I have no real objection; I personally would not use the term "increased gravitational mass" to describe what's going on; I would say that the force exerted by the moving massive body is not a pure Newtonian static force but has a "magnetic" component, as I said before. But that interpretation leads to the same equation as is given in the abstract, so it's an issue of terminology, not physics.Try to quantify the effect of velocity on gravity without it and you will see why. Gravitational mass is a well defined value that allows to describe this effect with a few words and a very simple equation (see the abstract quoted by timmdeeg).

But the abstraction does not generalize well, and it certainly does not qualify IMO as a "fundamental property" of objects that needs a fully general explanation. It's just a particular abstraction that happens to work well in a particular restricted set of cases.

In principle this is true, to have a fully self-consistent solution one must take into account the "self-interaction" of any body with its own field. This raises the same issues that it does in electromagnetism: for a "point particle" the self-interaction is infinite. Most of the time we can avoid this issue altogether by idealizing all bodies but one as "test bodies" whose mass is negligible and whose effect on the overall field is therefore also negligible. That is the idealization I understood us to be using in this discussion. Even if we consider the body that is the source of the field to be moving, the other bodies in the scenario we are considering, as I understand it, are still "test bodies" in this sense.1. It applies to trajectories in static gravitational field only but a real gravitational field will be influenced by the bodies and as this interaction depends on the mass two objects with different mass will not follow the same trajectory.

In practice, we find that bodies as large as planets appear to follow geodesics in the overall background field of the solar system. By that I mean that we can compute their trajectories without having to know their individual masses, just the overall mass that produces the field. So any "self-interaction" effects are not enough to disturb the trajectories even of objects of significant size in this particular case.

There are cases (e.g., binary pulsars) where we do see significant effects due to interaction between two massive bodies, but the key piece of evidence for such interaction is the emission of gravitational waves by the system as a whole, and consequently the gradual inspiral of the two objects towards each other, which in at least one case has been measured for (IIRC) more than 30 years and matches the predictions of GR. This effect is not even predicted at all by Newtonian theory, which predicts that such binary systems should maintain the same orbital parameters indefinitely.

In other words, in the rest frame of the body producing the gravitational field, you are talking about two different "test bodies" of negligible mass with different initial velocities, rather than two different "test bodies" with different masses but the same initial velocity. Fair enough.2. We are not talking about two different bodies in the same static gravitational field but about one body in the dynamic gravitational field of another body moving with different velocities.

Why? What makes these "accelerations" (which are just coordinate accelerations in a particular frame and have no direct physical meaning) the right things to compare, as opposed to 4-accelerations which correspond to a direct physical observable (e.g., the reading on an accelerometer).If we want to talk about two bodies with different velocities in an almost static gravitational field we should not compare their trajectories but their accelerations (not 4-accelerations).

Please note, I'm not asking why we *can* talk about these coordinate accelerations. I don't disagree that we can. But you are saying we *should* talk about them, which to me means that there is something physically fundamental about them, something that has to appear in *any* physical model of what's going on. I disagree; I can give a physical model that explains everything without ever using these coordinate accelerations, but only using 4-accelerations (and other covariant or invariant geometric objects).

- #96

- 1,895

- 332

What would be the reading on an free falling accelerometer?What makes these "accelerations" (which are just coordinate accelerations in a particular frame and have no direct physical meaning) the right things to compare, as opposed to 4-accelerations which correspond to a direct physical observable (e.g., the reading on an accelerometer).

- #97

- 30,609

- 9,653

Zero.What would be the reading on an free falling accelerometer?

- #98

- 1,895

- 332

Correct. And an observable that is alway zero does not provide any useful information.Zero.

- #99

- 30,609

- 9,653

An "observable" that is always zero because it is identically zero conveys no useful information, yes. But an observable that is zero precisely because some physical condition is met, and could just as well be nonzero if that condition is not met, certainly does convey useful information. Free fall is a definite physical state of motion; you can enter and leave it at will, simply by shutting off your rocket or turning it back on again, and seeing that your accelerometer reading changes accordingly.Correct. And an observable that is alway zero does not provide any useful information.

Put another way, 4-acceleration is zero for a freely falling object, but not all objects are freely falling, so the fact that 4-acceleration is zero for a particular object does, in fact, convey useful information.

- #100

- 1,895

- 332

But leaving free fall requires a force and I am afraid as soon as we do that your next question would be "Where does this force comes from and what about the involved energies?" Therefore I prefer a setup with gravitational interactions only.Free fall is a definite physical state of motion; you can enter and leave it at will, simply by shutting off your rocket or turning it back on again, and seeing that your accelerometer reading changes accordingly.