Is increase in mass relative?

CGann

Let's say you're mass is 100kg and you're traveling at 0.9C. The relativistic effect on your mass is roughly 2.3 so you now have a mass of 230kg.

If you went whipping past me, would you appear to have a mass of 230kg or 100kg?

Seems like a no-brainer. But it keeps tripping me up!

TIA!

CG

ZapperZ

Please read this thread:

http://www.physicsforums.com/showthread.php?t=480839

Zz.

JesseM

In your frame I would now have a "relativistic mass" of 230kg, though my "rest mass" would still be 100kg. Most physicists prefer not to use the concept of relativistic mass these days, as it is potentially misleading; for example, although my relativistic mass does tell you the amount of force that would need to be applied in your frame to accelerate me a small amount in the same direction I am already traveling, if you wanted to accelerate me in some different direction I would no longer have the same resistance to acceleration as a 230kg object at rest in your frame.

pervect

There are several sorts of masses defined in special relativity - and more in General relativity. None of the sorts of masses defined in SR, though, turn out to have much, if anything, to do with "how much" a body gravitates. There also isn't any completely general definition of mass in General Relativity, which is to some extent compensated for by a larger number of less-general definitions that don't always apply or can't always be defined. Studying this issue (of how to define mass in GR) is one of the open problems of GR, though my personal opinion is that there are good reasons that will prevent any completely satisfactory general definition from ever being found.

The current state of the art in GR is that it is not "mass" that causes gravity, but the "stress-energy tensor", a more complex mathematical object with more than one component.

You might take a look at http://ajp.aapt.org/resource/1/ajpia...sAuthorized=no

This is probably the simplest attempt at an answer to your question. There are many hidden stumbling blocks here, and it is, alas mistake to assume that any of the masses defined in special relativity (including the so-called relativstic mass which isn't much used nowadays) has anything but the most general relationship to how a body gravitates. Substituting "relativistic mass" into some Newtonian equations (particularly, the Newtonian gravitational equation!) isn't in general any sort even a "good approximation".

A more advanced paper that talks about the "field" of an ultra-relativistic moving mass that you might find interesting (if it isn't too advanced) is http://arxiv.org/abs/gr-qc/0110032. However, it talks about what one can actually measure, which is not a "force" at all, but the Riemann tensor, which can very very roughly be thought of as the rate of change of the Newtonian field. In fact, one can equate parts of the Riemann tensor to Newtonian "tidal gravity". You can see from the text that the resulting "field" of a moving mass isn't spherically symmetric at all, and that any attempt to think of it as a spherically symmetric Newtonian field is doomed. It's interesting and educational to note that the electric field (which is much easier to define!) of an ultra-relativistic moving charge isn't spherically symmetric at all either. If you're really interested in understanding all the details, and have the necessary background in at least special relativity, I'd suggest studying the ultra-relativistic moving charge first, and coming back to the ultra-relativistic moving mass when you've understood the previous case.

Trenton

In a way a similar question to the mass in different frames of reference - but even more vexing!

Consider a black hole basking in the light of nearby stars which are moving slowly enough relative to the BH so that thier speed can be ignored.

First of all, how does one calculate the blue shift of the starlight at say the EH, and then further into the hole?

As the BH absorbs the light it aquires mass. But how much mass? If instead of a hole it were a lightwieght asteroid of the same size, the ambient starlight recieved over a year might have mass equivilent of say 1 gram. But the BH blueshifts the light to a much higher energy and therefore the mass of the infalling light is increased. Does the BH gain 1 gram or does it gain the increased amount?

Also I hear that at the EH time stops. I question this but assuming this is right, then surely incoming light is blueshifted to infinity?

JesseM

In relativity there is no coordinate-independent way to talk about the rate a clock is ticking. So while it's true that the rate a clock ticks relative to coordinate time in Schwarzschild coordinates will approach zero as the clock approaches the horizon, this is just a feature of how Schwarzschild coordinates work, in other coordinate systems around the same black hole like Kruskal-Szekeres coordinates, the rate a clock ticks relative to coordinate time remains finite as it crosses the horizon. Also note that even in special relativity, it is possible to construct non-inertial coordinate systems where the rate a clock ticks relative to coordinate time approaches zero as it gets close to some boundary, like the Rindler horizon of an accelerating observer who is using Rindler coordinates.

Anyway, on the subject of blueshift, an observer falling through a black hole will only see a finite blueshift as he crosses the horizon, he won't see the rate of clocks in the outside universe sped up to infinity. However a rotating black hole also has an inner "Cauchy horizon" beyond the outer event horizon (see the third diagram on this page), and there the blueshift is predicted to go to infinity in general relativity, which is the main reason physicists don't really trust general relativity's predictions about what goes on at the center of a rotating black hole, and tend to think we'll need a theory of quantum gravity to model what goes on beyond the Cauchy horizon.

pervect

Let's take the simplest case, where the black hole is in asymptotically flat space-time. This will happen automatically if one uses the usual Schwarzschild metric. Then the mass added to the black hole will be the "energy at infinity" of the infalling particle. This applies regardless of whether one considers the "mass" of the black hole to be the Komar mass, or the ADM mass, or the Bondi mass, or the mass parameter M that appears in the Schwarzschild metric, because all of these different sorts of masses agree and have the same value (well -almost, there is an issue if the infalling particle or light contributes "too much" to the mass-energy so that it becomes unreasonable to approximate it as a quasi-static system, making my remark somewhat of an oversimplification.)

It's rather interesting to note that the amount of mass added to a black hole by dropping one kilogram into it can be made arbitrarily low, and approaches zero as one approaches the event horizon. If you drop a kilogram into the black hole from infinity, the black hole gains 1kg. If you drop a kilogram into the black hole from near the event horizon, the black hole basically gains nothing - to be more precise, an arbitrarily small amount that approaches zero as one approaches the event horizon.

Let's consider that we have a 1kg mass at inifinity, and a very long, very strong, rope.

If you lower the 1kg further and further into the black hole on a rope, where does the energy go? Well, as the object is lowered into the black hole, it does work, "at infinity". In Newtonian terms, one would say that the mass loses energy due to the "gravitational binding energy" as its lowered into the field, and that this extracted energy is what does the work at infinity.

However, there's no solid definition in GR of "gravitational binding energy", except in the Newtonian limit. While we can usually (though not always) compute the mass of a system, we can't say exactly how it's distributed, or where it is. There are several possible ways of doing the computation, all of which give different answers as to "where" the mass is, though they all give the same end result.

ZapperZ

Whoa!

Do you guys think, from the way the OP presented his/her question, that he/she could understand all this? Really?

Zz.

JesseM

I don't think my original post was that complicated, and a lot of the subsequent posts were addressed to Trenton's question, not the OP. pervect's post #4 would probably be a bit over the head of CGann judging from the question though...

ZapperZ

I know you were. I'm just seeing this thread getting derailed VERY quickly in barely an hour after the original post, and it all started out with a rather simple question.

Zz.

harrylin

Indeed, as JesseM clarified, the increase in relativistic mass is relative.
(Now let's see if the OP still has issues!)

Note: although off-topic, it sounds as if JesseM made a mistake in the force calculation.
The equation for acceleration along x (the same direction as the motion):
F = gamma²*m*a = gamma³*m₀*a,
with m = relativistic mass and m₀ = rest mass.
However, for acceleration in the normal direction (perpendicular on the motion):
F = m*a = gamma*m₀*a

JesseM

Good point, I actually had it backwards--relativistic mass is resistance to acceleration in the direction perpendicular to the direction of motion, so if you want to accelerate it a small amount vertically when it's traveling horizontally, in that case (force needed)=(relativistic mass)*(acceleration). It's in the direction parallel to the motion that relativistic mass no longer gives the resistance to acceleration, so it doesn't really make sense as "inertial mass" in that direction (since inertial mass is defined in terms of resistance to acceleration). Sometimes physicists define a separate notion of "transverse mass" (equal to relativistic mass) and "longitudinal mass" (equal to rest mass times gamma^3, or relativistic mass times gamma^2), see here.

Trenton

I have a few thoughts about what might exist at or very near the center of black holes. In practice, there almost certainly is no such thing as a non-rotating black hole. It occurs to me that if the big bang were a singularity that became disrupted, its direction of rotation has something to do with the universe being comprised of matter rather than of antimatter. Anyway, more about that sort of thing later ...

I am having a more imeadiate problem with the EH. I would love to know how the formula for Rs was derived because there is a very serious problem with meerly evaluating the newtonian formula for escape velocity with v set to c. It only works for the proper speed (disance/proper time). Rs is the point at which an infalling object reaches a proper speed of c.

The velocity reached at EH is less than c when measured by any other means. Photons still overtake the infalling object and the photons are not blue shifted to infinity at this point.

To say things like 'all paths inside the EH point to the singularity' are in my view wrong, particularly if you make that all light-like paths.

Light always travels at c and therefore has infinite proper speed - and can escape from within the EH as long as it's path is along a very narrow cone. If it is a finite blueshift on the way in (except maybe at cauchy), it will be a finite redshift on the way out.

Only if there is a zero sized point at the center, will photons face an infinite red shift and not be able to escape.

JesseM

I'm not sure of the details but I assume the derivation would specific to general relativity, nothing to do with Newtonian escape velocity. For massive particles you can show that the proper acceleration needed to keep them at a constant radius goes to infinity as you approach the Schwarzschild radius, but for light I suppose you would need a different approach.
Why would you take a "view" that all physicists are wrong about this if you haven't studied the mathematics of general relativity in detail?? Beware the Dunning-Kruger effect...
What does proper speed have to do with whether it can escape? You need to calculate the coordinate speed of light in some coordinate system, which in general relativity will not necessarily be c (only in the "local inertial frame" of a free-falling observer, defined on an infinitesimally small patch of spacetime around that observer, are the laws of physics equivalent to those of inertial frames in special relativity, and even in special relativity light only has a coordinate speed of c in inertial frames, not non-inertial ones). In Schwarzschild coordinates the coordinate speed of a beam aimed outward would approach zero as the radius of emission approached the Schwarzschild radius. Meanwhile, if you use Kruskal-Szekeres coordinates, these coordinates do have the nice property that light always moves at the same constant coordinate speed in all directions (and light worldlines are at 45 degree angles from the vertical just like in Minkowski diagrams from special relativity), but in these coordinates the event horizon itself is moving outward at the same coordinate speed (so the event horizon is also a boundary line 45 degrees from the vertical). So it makes sense that light emitted on the horizon will stay on the horizon forever (unless it's aimed inwards rather than outwards), just the same way light emitted on the boundary of the future light cone of some event (and moving away from the location of the event) will stay on the boundary of the future light cone forever.
Nope, that sort of symmetry might seem intuitive but if you do the math it won't work out that way, light falling in from a finite radius only gets blueshifted a finite amount as it approaches the horizon, but if you consider the redshift seen by an observer at the same radius for light emitted close the horizon, then in the limit as the point of emission approaches zero distance from the horizon, the redshift seen by the distant observer approaches infinity.

Trenton

Pervect - A very interesting answer which I shall have to contemplate at length. There has to be some way in which what you are saying is broadly correct otherwise black holes would gain mass at a colossal pace (indeed at an infinite pace if there really is a zero sized point mass at the center - just one photon would do it in fact). Clearly they don't do this.

The other day I posted a question in another thread but did not get a response. I considered the eliptical orbit of a planet around a star. The orbit continually converts PE to KE and back again. Ignoring the star's radiation, the total mass-energy of the system should be a constant. Your answer appears to confirm this statement but only sort of.

The total gravity of the system should be that given by the actual mass of the star and planet, plus thier PE and KE. I am told that a hot object has more mass than an identical cold one, due to the vibrational energy, which is just KE - so I think the KE has to be added to the mass, and to be constant, so must the PE.

The trouble is, I can't for the life of me, figure out the mechanism by which PE contributes to mass-energy and thus to gravity. I am further baffled that 1Kg is added only if droped from infinity. I would have thought that while the closer to the hole the less was added, but would not have thought it possible that less than 1Kg could be added.

It seems to me that if you have say one solar mass in one place, the total mass-energy is one solar mass. If you have twp solar masses in one place the total mass-energy is two solar masses. But if you have two solar masses in two places you have more than two solar masses worth of mass-energy because there is PE. This must be connected to energy of free space (in GR terms rather than exotic stuff like zero-point energy).

Or have I completly lost the plot?

Trenton

Proper speed was mentioned because as far as I can make out, a proper speed of c would be reached by an object falling from infinity to the EH. This falls well short of a real speed of c.

In the case of matter, I proposed in an earlier post, that there may be a discrepency between velocity attained by an infalling object and that needed to escape. Normally these are considered the same at all points along the trajectory. However for dense stars where the speeds are such that relitivistic effects become significant, matter falling in does not have it's acceleration impeaded by mass aquisition due to speed while outgoing matter would be impeaded by this. This proposition may be invalid but I was trying to find a way in which it was valid to say matter could not escape not matter how fast it went.

In the case of light, it makes perfect sense for a photon to stay on the boundary of a future light cone but what I don't get is the idea that the EH can be compared to a light cone. In what sense is the EH moving out (at c or any other speed)? It moves out only when the mass is added to as is clear from the equation.

In any case if time no longer progresses at EH is it not better simply to state the photons stay where they are because they are frozen in time? It is irrelevent how fast something is going if it is in a locale where time dilation has become infinite and time has stopped?

In any case why would outgoing light stop at the EH? The curvature gets more severe with r < Rs so an outgoing photon would never get to the EH.

I shall now answer your question concerning my 'view' that all physicists are wrong about this. I don't take that view because I am not qualified to do so but I do pose the question. I pose the question because I know that where humans are concerned it is due dillegence to do so - and I am qualified to say that because I have studied the psychology!

I am currently studying the math of GR but it is very difficult to square the math with the numerous written descriptions of objects falling into a black hole etc. I am not going to pretend I am any good at math but I never had this problem with any other aspect of physics. The math and the descriptions were for the most part, complimentry.

I do need to see a full derivation of the Rs equation. Hopfully in that will be meanings I have so far failed to garner.

JesseM

Proper speed can only be defined relative to a choice of coordinate system, it refers to the rate of change in coordinate position relative to proper time. In what coordinate system do you think that statement would be true? Wouldn't be true in Schwarzschild coordinates, for example.
In the local inertial frame of any freefalling observer (see discussion of local inertial frames here if not familiar with the concept), the event horizon would be moving outward at c at the moment the observer crosses the horizon. And in Kruskal-Szekeres coordinates which I was discussing before, the event horizon is moving outwards as coordinate time increases, at exactly the same coordinate speed as a photon (and as I mentioned, Kruskal-Szekeres coordinates have the nice property that light moves at the same coordinate speed everywhere, unlike in Schwarzschild coordinates or most other GR coordinate systems). See qualitative features of the Kruskal-Szekeres diagram from the wiki article to get a basic idea of how these coordinate systems work, and you can also take a look at the diagrams from the textbook gravitation which I scanned and posted here.

Keep in mind that the "size" of any surface, whether the event horizon or the surface of the Earth, is totally dependent on the type of coordinate system you use, and in general relativity there are no global "preferred" coordinate systems like inertial frames in special relativity (although as noted above you can still have locally inertial frames in SR), you can use any coordinate system you want and the equations of general relativity will work equally way in each one as long as you define the metric correctly in that coordinate system. See this article on "diffeomorphism invariance" for more on the complete freedom you have to define coordinate systems however you like in general relativity. It might seem counterintuitive that just by switching coordinate systems a spherical surface can go from having a fixed radius to an expanding one, but in general relativity neither version is more "correct" than the other.
The Schwarzschild radius equation is specifically for the radius in Schwarzschild coordinates.
As I said in my first reply to you in post #6, there is no objective sense in which time "no longer progresses at the EH", that's just a feature of Schwarzschild coordinates. At any arbitrary boundary, say the plane dividing two halves of your room, one could define a coordinate system where the ratio of proper time to coordinate time approaches zero as an object approaches the boundary.
Inside the horizon there is no outgoing light, a light beam sent in any spatial direction will always have a decreasing Schwarzschild radius. You can see this in terms of the Kruskal Szekeres diagram if you include surfaces of constant Schwarzschild radius inside the horizon (they look like hyperbolas), but it may be easier to visualize in another coordinate system called Eddington-Finkelstein coordinates, which uses the same radial coordinate as Schwarzschild coordinates but defines the time coordinate differently. From the textbook Gravitation, a diagram showing light cones for events both outside and inside the horizon (the vertical cylinder):



And another diagram which explicitly shows the worldlines of light rays bounding the light cones, both light rays which were aimed "inward" (the straight lines at 45 degrees) and light rays which were aimed "outward" (the curved lines, which you can see continue to approach the singularity inside the horizon):



Fair enough, it's good to ask questions about claims as long as you don't start out with the preconception that all the scientists who make the claim are wrong...
As I said I'm not sure where to find a derivation of the fact that the Schwarzschild radius has properties of an "event horizon" like not allowing light to escape (what little I know about general relativity is mostly conceptual), but in general I've heard that Exploring Black Holes is a good book that introduces the mathematics of black holes without requiring the level of knowledge of a normal textbook on general relativity.