Does mass really increase with speed

[PeterDonis]

This is hard to imagine. The only layman interpretation I am aware of sounds like this: r is timelike inside the horizon, as it has only one direction, like time flows only in one direction. But the weirdness seems "only" to be a matter of the choosen coordinates and can be transformed away, you mentioned the Kruskal coordinates already.

The terminology of calling a coordinate "timelike" or "spacelike" is unfortunate since it doesn't really convey what's going one, especially if what looks like the *same* coordinate (r in this case) is said to be timelike in one coordinate chart (the interior Schwarzschild chart) and spacelike in others (ingoing Eddington-Finkelstein and Painleve). Here's what I think is a better way of looking at it:

A "coordinate" like r is really a shorthand way of referring to two different things. One is a set of surfaces in the spacetime: each surface is labeled with a unique value of the coordinate, and every event in the spacetime lies on one and only one of the surfaces. For example, in Schwarzschild spacetime, there is a set of surfaces of constant r that satisfies the above properties.

The second thing a coordinate refers to is a directional derivative: for example, r corresponds to $\partial / \partial r$, the rate of change of something in the "r direction". The thing to remember about this is to avoid the "second fundamental confusion of calculus" (I learned this term from George Jones, one of the mentors here, who pointed me at a reference to it in one of Roger Penrose's books): partial derivatives can change depending on what other variables are being held constant. So a coordinate defined as a directional derivative will depend on what other coordinates it is combined with in a specific chart.

You can probably see what's coming next: when you change coordinate charts, the two things above do not necessarily change together. For example, in all three of the coordinate charts for Schwarzschild spacetime that I mentioned above, the first aspect of the "r" coordinate is the same: i.e., the "r" coordinate in all three charts refers to the *same* set of surfaces of constant r. What changes from chart to chart is the directional derivative. This seems to be the usual convention for coordinate nomenclature: a given coordinate name, such as "r", is applied to a given set of curves; then the changes in the directional derivative between charts are captured by calling the coordinate "timelike" or "spacelike" in different charts, according to the direction the derivative points in.

As a concrete example, here's how things work out for all of the charts I have mentioned for Schwarzschild spacetime:

(1) The Schwarzschild chart. (Technically, there are actually two of these, exterior and interior, because the coordinates are singular on the horizon.) Outside the horizon, the directional derivatives look like this: $\partial / \partial t$ timelike; $\partial / \partial r$ spacelike; $\partial / \partial \theta$ spacelike; $\partial / \partial \phi$ spacelike. So a surface of constant t is a spacelike 3-surface; but a surface of constant r has one timelike and two spacelike dimensions. (I won't talk about surfaces of constant theta, phi here; angular coordinates work a little differently. The usual way of talking about them is just to say that, since the spacetime is spherically symmetric, we can think of it as a set of coordinate pairs (t, r), where each unique pair labels a 2-sphere, which is a spacelike 2-surface covering all possible values of theta, phi. So what I said above can be condensed to: outside the horizon, lines of constant t are spacelike, and lines of constant r are timelike, where each "line" is really a series of 2-spheres. The only exception is r = 0, which is a single point, and is not technically part of the spacetime because the curvature is infinite there--but that's a whole other post .)

Inside the horizon, the r and t derivatives switch directions: $\partial / \partial t$ is spacelike and $\partial / \partial r$ is timelike. This is what the common statements that "r is timelike inside the horizon" or "t is spacelike inside the horizon" refer to. You can also see that, inside the horizon, lines of constant *r* are now spacelike, and lines of constant *t* are now timelike. So the labeling of coordinates as "timelike" or "spacelike" will look backwards if you are looking at the lines of constant coordinate value instead of the directional derivatives.

(2) Ingoing Eddington-Finkelstein & Painleve charts. (I lump these together because they are the same in the aspects we're discussing; also I specify "ingoing" because there are also "outgoing" versions of these charts. I won't go into the difference here.) Outside the horizon, these are the same as the Schwarzschild exterior chart; $\partial / \partial T$ is timelike and the other three coordinate derivatives are spacelike. So (leaving out theta, phi again as above) lines of constant T are spacelike and lines of constant r are timelike. Note that we are using a different label, T, for the "time" coordinate because it refers to a different set of lines (or surfaces if we include the angular coordinates) than the Schwarzschild "t" coordinate does.

*On* the horizon (these charts are nonsingular at the horizon, so this is meaningful here), $\partial / \partial T$ is *null* in both charts. ("Null" means it points in the same direction in spacetime as a light ray--an outgoing light ray, in this case.) However, the other three coordinate derivatives stay spacelike in this chart. So on the horizon, lines of constant T are still spacelike, but lines of constant r are null. In fact, that is one way of stating the *definition* of the horizon: it is a null line (of 2-spheres) of constant r.

Inside the horizon, $\partial / \partial T$ is spacelike; this means that lines of constant r are spacelike. This is why it's impossible to "hover" at a constant r inside the horizon: you would have to move on a spacelike line, i.e., faster than light. But $\partial / \partial r$ is *also* spacelike inside the horizon in this chart; in other words, *all four* coordinates are spacelike inside the horizon! This seems very weird, but that's how it is; what it is really telling you is that, to get a timelike vector at all inside the horizon, you have to combine $\partial / \partial T$ and $\partial / \partial r$ with opposite signs; for example, a future-directed timelike curve will have positive $\partial / \partial T$ and negative $\partial / \partial r$. This is just another way of saying that everything inside the horizon is forced to fall into the singularity. In Painleve coordinates, for example, an observer freely falling into the black hole from rest "at infinity" is described by the vector $\partial / \partial T - \sqrt{2M / r} \partial / \partial r$, where M is the mass of the hole in units where G = c = 1.

(3) The Kruskal chart. Here what we normally think of as "r" and "t" (or "T" in the Eddington or Painleve charts) are not coordinates at all: they are functions of the coordinates that are used to label curves. The actual coordinates T, X in the Kruskal chart don't have a straightforward physical interpretation, but they do have a key property that makes the chart nice for seeing the global structure of the spacetime: their directional derivatives work just like the ones for the standard Minkowski coordinates of special relativity. In other words, $\partial / \partial T$ is timelike everywhere, and $\partial / \partial X$ is spacelike everywhere, and their relationship is such that null curves (light rays) are always 45 degree lines in the chart.

In this chart, lines of constant r are hyperbolas outside and inside the horizon; and the horizon itself, r = 2M, is a null line, i.e., a 45-degree line. Actually, it is a *pair* of 45 degree lines in the "maximally extended" Kruskal chart, which is mathematically well defined but is not physically realistic (again, that's a whole other post); these lines are the asymptotes of the hyperbolas for r > 2M and r < 2M. For r > 2M, the hyperbolas are more vertical than horizontal, and for r < 2M, they are more horizontal than vertical, so it's easy to see how the nature of the r coordinate changes.

Lines of constant Schwarzschild t in the Kruskal chart are straight lines radiating from the origin (T = 0, X = 0, which corresponds to the point where the two horizon lines for r = 2M, the asymptotes of the r hyperbolas, cross). The exterior lines radiate to the left and right, and the interior lines radiate up and down. So again, it's easy to see how the nature of the Schwarzschild t coordinate changes from exterior to interior: the lines of constant t are obviously spacelike in the exterior and timelike in the interior.

Unfortunately, I don't know a simple way to describe how the lines of constant Painleve time or Eddington-Finkelstein time T (technically they aren't quite the same set of lines, but they're close) look on the Kruskal chart. But they are spacelike lines in both the exterior and interior regions.

 

[edpell]

The two rocket do have some gravitational attraction. Let us say that they start out separated by a distance L1 side by side. After 50 years Earth time we see through our telescope (at year 100 on Earth because the light had to return 50LYs) the distance is now L2 (a number smaller than L1). Likewise on the rockets the people (old people) see a distance L2. But they have aged only 1 year (pick gamma so this is so). The people on Earth conclude the gravitational force is low due to changing the separation by L1-L2 in 50 years where as the folks on the rocket conclude the force is larger due to a L1-L2 change in only 1 year. It seems like the rocket folks will think there is more force?

 

[PeterDonis]

Let us say that they start out separated by a distance L1 side by side. After 50 years Earth time we see through our telescope (at year 100 on Earth because the light had to return 50LYs) the distance is now L2 (a number smaller than L1).

Is the distance "side by side" perpendicular to the direction of the rockets' motion? If so, it won't appear to change as viewed from Earth (or from the rockets, of course) due to the rockets' motion.

 

[edpell]

Yes side by side. There is a small gravitational pull. Both rocket have some mass, rest mass or rest mass time gamma we can argue about but either way they have mass and gravitational attraction and over 50 years even a small attraction adds up.

 

[DrStupid]

I think that this scenario you have postulated needs to be nailed down more precisely. How about giving some actual numbers? You don't need to give many; just the following:

(1) The rest mass of the rockets. (Just one number, we'll assume it applies to both rockets.)

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

$r > \frac{{4 \cdot G \cdot m}}{{c^2 }}$

If you need actual numbers let's take r = 1 m and m = 3.354·10²⁶ kg (59% the mass of Saturn).

Deformations due to the tidal forces shall be neglected.

(2) The distance between the rockets whey they are supposedly inside each other's Schwarzschild radius.

Due to the assumption above they will never be inside each other's Schwarzschild radius and if they are at rest they even can not be inside the Schwarzschild radius of the entire system. But if the bodies are moving with an identical absolute value v of the velocities this will happen for

$d < 2 \cdot \left( {r_s - r} \right)$

with the common Schwarzschild radius

$r_S = \frac{{4 \cdot G \cdot m}}{{c^2 \cdot \sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$

As the distance d (measured from center to center) can not be negative this is possible for

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

With the above mentioned values this minimum velocity would be 2,602·10⁷ m/s (8,68% the speed of light).

(3) The relative directions the rockets are traveling in; this will fix the combined momentum of the rockets.

To avoid a resulting momentum or angular momentum it should be a head-on collision.

 

[timmdeeg]

PeterDonis, thank you for your valuable post 71#, which improved my notion regarding the switching coordinates. It also motivated me to look into "General Relativity from A to B" by Robert Geroch. I bought this book long time ago, perhaps you know it. And I will need to digest your post furthermore.

 

[PeterDonis]

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

$r > \frac{{4 \cdot G \cdot m}}{{c^2 }}$

If you need actual numbers let's take r = 1 m and m = 3.354·10²⁶ kg (59% the mass of Saturn).

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.

As the distance d (measured from center to center) can not be negative this is possible for

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

With the above mentioned values this minimum velocity would be 2,602·10⁷ m/s (8,68% the speed of light).

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.

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.

To avoid a resulting momentum or angular momentum it should be a head-on collision.

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).

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:

$$M_{total} = \frac{1}{c^{2}} E = 2 \gamma m$$

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

$$r_{S} = \frac{2 G M_{total}}{c^{2}} = \frac{4 G \gamma m}{c^{2}}$$

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 $r_{S}$, and their initial sizes are each less than 1/2 $r_{S}$.

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:

$$\frac{M}{m} = \gamma \left( 1 + \frac{v}{c} \right) - 1$$

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

$$M + m = \gamma m \left(1 + \frac{v}{c} \right)$$

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

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

You will note that $2 r_{s} > r_{S}$, 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

$$\frac{4 G \gamma m}{c^{2}} \frac{v}{c}$$

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

 

[DrStupid]

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?

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.

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.

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.

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.

 

[PeterDonis]

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.

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:

You could also use solar sails or something similar to accelerate the object (just to give you another example).

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.

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.

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.

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.

 

[PeterDonis]

Continuing from my last post, we are considering how to analyze DrStupid's scenario using GR, since the scenario implicitly assumes that gravity is not negligible and so SR is not valid:

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.

 

[DrStupid]

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.

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.

 

[DrStupid]

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.

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".

 

[timmdeeg]

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".

It seems, your expectation is in agreement with http://ajp.aapt.org/resource/1/ajpias/v53/i7/p661_s1?isAuthorized=no .

 

[DrStupid]

It seems, your expectation is in agreement with http://ajp.aapt.org/resource/1/ajpias/v53/i7/p661_s1?isAuthorized=no .

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.

 

[PeterDonis]

It seems, your expectation is in agreement with http://ajp.aapt.org/resource/1/ajpias/v53/i7/p661_s1?isAuthorized=no .

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.

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.

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.

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.

 

[PeterDonis]

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.

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:

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.

 

[timmdeeg]

My problem is to understand why "active gravitational mass of a moving object" isn't a priori in contradiction with 'mass is invariant'.

 

[PeterDonis]

My problem is to understand why "active gravitational mass of a moving object" isn't a priori in contradiction with 'mass is invariant'.

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").

 

[DrStupid]

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.

What else? Gravitational mass is defined by Newtons law of gravitation.

Saying that the light's inertial mass is its energy divided by c^2 won't work

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.

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

 

[PeterDonis]

What else? Gravitational mass is defined by Newtons law of gravitation.

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.

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

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".

 

[DrStupid]

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.

I am aware of this problem but there is no other definition of 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.

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.

 

[PeterDonis]

I am aware of this problem but there is no other definition of gravitational mass.

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.

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.

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).

 

[DrStupid]

Then why bring up the term?

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).

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's not what we are talking about because:

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).

 

[lvirany]

This is where Minkowski differs from Einstein & Lorentz. Minkowski explains the observed length contraction without having to use the concept of mass. Space-time as measured by the moving observer is uniformly dilated in a sheet-like way through the concept of 'proper space' as well as 'proper time'.

 

[PeterDonis]

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).

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.

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.

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 principle this is true, to have a fully self-consistent solution one must take into account the "self-interaction" of anybody 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.

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.

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.

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.

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).

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).

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).

 

[DrStupid]

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).

What would be the reading on an free falling accelerometer?

 

[PeterDonis]

What would be the reading on an free falling accelerometer?

Zero.

 

[DrStupid]

Zero.

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

 

[PeterDonis]

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

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.

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.

 

[DrStupid]

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.

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.

 

[PeterDonis]

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.

Fine. What does that have to do with whether looking at such a scenario using coordinate acceleration is *required*, as opposed to one possible method but not the only one?

 

[DrStupid]

What does that have to do with whether looking at such a scenario using coordinate acceleration is *required*, as opposed to one possible method but not the only one?

Nothing.

 

[timmdeeg]

My problem is to understand why "active gravitational mass of a moving object" isn't a priori in contradiction with 'mass is invariant'.

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.

Ok, there is no contradiction. And it is probably simply wrong to compare 'invariant mass' (a term within SR) with 'active gravitational mass' (GR).

But how about this reasoning:

A heated body has increased mass and thus an increased gravitational field due to the increased kinetic energy of the particles from which it is composed. A heated body weighs more. Now let's imagine a cold spherical mass M, which's radius exceeds 2GM very slightly. Would it form a black hole on heating (assuming the coefficient of thermal expansion low enough)?

The szenario of DrStupid is much different, but kinetic energy is involved as well.

 

[PeterDonis]

A heated body has increased mass and thus an increased gravitational field due to the increased kinetic energy of the particles from which it is composed.

Yes. But bear in mind that this assumes that the net momentum of the particles composing the mass is unchanged by the heating process (for example, it could be heated by radiation falling on it from all directions in a spherically symmetric manner). This is similar to the stipulation in DrStupid's scenario where the rockets' momenta are equal and opposite so they sum to zero.

Now let's imagine a cold spherical mass M, which's radius exceeds 2GM very slightly. Would it form a black hole on heating (assuming the coefficient of thermal expansion low enough)?

In principle, yes, you can cause an object to collapse into a black hole by heating it. Technically, the exact scenario you describe cannot be realized because it is impossible to have a body in stable equilibrium with a radius less than 9/8 of the Schwarzschild radius (i.e., 9/4 GM / c^2). So a body whose radius exceeds 2 GM / c^2 only very slightly would not be stable, it would already be collapsing into a black hole. But you could take a body that was just at the stable limit and add heat to it, and that would push it "over the edge" into collapsing (because its radius would now be slightly *less* than the minimum for stability for its new, slightly higher mass).