I If Mass is relative, is "gravitational" mass relative?

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1. Mar 23, 2017

John Morrell

I was thinking through a thought experiment which led me to conclude that mass is relative. My thought process was basically this; there are two objects moving past each other. In the reference frame of either object, they are at rest and the other object is moving. Since the other object is moving, it has a higher kinetic energy than the reference object. This means that its mass, or the ratio of force to acceleration, would be higher for the 'moving' object than for the 'observer' object.

Now, if the inertial mass is relative, is gravitational mass also relative? And if now, does that mean that we can find the 'absolute mass' of a particle or body by measuring the gravitational force it exerts?

2. Mar 23, 2017

Smalde

Well, the definition of relative mass is only that: a definition. People used to set $m = \gamma m_0$ and then $E = mc²$, but more often then not one uses $E = m_0 \gamma c²$
Both relativistic force and relativistic acceleration have a factor gamma in them, and so the ratio of force to acceleration is independent of inertial frame.
But again it all depends on your choice where to place the gammas.

3. Mar 23, 2017

SiennaTheGr8

That doesn't follow, since kinetic energy depends on velocity, too. An increase in kinetic energy implies either an increase in mass or velocity (or both).

In fact, the mass that appears in the kinetic energy equation is not a function of velocity. It's invariant, which means all observers agree on its value no matter how fast they're traveling. That's true in Newtonian mechanics:

$K = \frac{1}{2}mv^2$,

and also in special relativity:

$K = (\gamma - 1)mc^2$,

where $\gamma = (1 - v^2 / c^2)^{-1/2}$. The $m$ doesn't depend on speed. It's possible (but unfashionable these days) to define a "relativistic mass" as $\gamma m$, and this quantity does increase with speed, but that's all due to $\gamma$.

The answer to your question about gravity is no. In Newtonian physics, it's $m$ (not $\gamma m$) that acts as the gravitational "charge." And in general relativity, gravity is the curvature of spacetime caused by the energy–momentum tensor, which is a more complicated object than a simple scalar like mass.

If the gravitational "charge" were dependent on speed, then a star could become a black hole only for certain observers traveling fast enough, while slower observers would say that it remains a star. That wouldn't make any sense!

4. Mar 23, 2017

Staff: Mentor

The problem with this is that all three of these things as you define them--mass, force, and acceleration--are frame-dependent. But there are other ways of defining all of these terms that make them not frame-dependent. In the case of mass, this definition is "rest mass". And you can write a relationship between the non-frame-dependent versions of all of these terms that is also non-frame-dependent.

So what you are discussing is not a matter of physics, it's a matter of words.

"Gravitational mass" is not a meaningful concept in GR. The source of gravity is the stress-energy tensor. The behavior of an object as a source of gravity is not frame-dependent.

In GR, gravity is not a force. It's spacetime curvature. Spacetime curvature is also not frame-dependent, so, as with mass, force, and acceleration above, you can write a non-frame-dependent relationship between a source of gravity (the stress-energy tensor) and spacetime curvature (the mathematical object that describes this in the relationship is called the Einstein tensor). This relationship is the Einstein Field Equation, which is the central equation of GR.

5. Mar 24, 2017

John Morrell

6. Mar 24, 2017

John Morrell

So it makes sense that gravity is a curvature of space, not a frame-variant quantity. But one thing I still don't understand is the argument against mass being dependent on a frame of reference. We define mass as the ratio between a force and an acceleration, right? I mean, mass has no other meaning as far as I know. And to an observer, it will take a much greater force to accelerate a body moving near relativistic speeds than it would to accelerate a body at rest with relation to the observer. A body moving at relativistic speeds will take less force to accelerate a given amount in its own reference frame than in another. Eventually, as the relative velocity approaches the speed of light, an observer would say that it takes nearly infinite force to accelerate the body any appreciable amount, while the body itself would not experience this effect.

I know my language is not precise, but this seems like sound reasoning to me. The main reason I can think of for this not being true is if my understanding of special relativity doesn't apply to cases with force and acceleration, as some of you have maybe suggested.

7. Mar 24, 2017

John Morrell

Gamma is just a ratio between the effect of a force on the particle while at rest and the effect of a force on a particle with a given velocity. I don't see why this is in any way not a description of a very real and fundamental "relativistic mass". Could you explain?

8. Mar 24, 2017

vanhees71

Since 1908 it is known how to formulate special relativity in a clear and concise way such that the behavior of physical quantities under Lorentz (or Poincare) transformations is systematic and clear, and I don't understand, why the old complications seem still to be discussed on and on. In the short time between Einstein's paper of 1905, the physicists where very puzzled about this, but Minkowski came to the rescue with his famous talk on the four-dimensional formulation and the introduction of spacetime as the right mathematical framework to formulate SRT.

Without this, Einstein would never have been able to discover General Relativity, where these concepts are just made local, and spacetime became a "dynamical" pseudo-Riemannian (Lorentzian) manifold.

With that said, there is no doubt that (a) mass is a scalar, (b) total energy and momentum of a closed (!!!) system build a four-vector (the four-momentum vector), and (c) that the sources of the gravitational fields are all kinds of energy-momentum-stress distributions and not only the mass. This holds in GR and thus also in SR (which is just the approximation of GR, when gravity can be neglected).

9. Mar 24, 2017

SiennaTheGr8

Point taken about clocks running at different speeds for different observers (although you're mistaken about black holes).

But:

Clocks running at different rates for different observers is unintuitive, but it doesn't violate causality or imply that events themselves only happen for certain observers. If a star remained a star for some observers but became a black hole for others, we'd have a "doesn't make sense" problem of a different sort.

10. Mar 24, 2017

SiennaTheGr8

You can call $\gamma m$ the "relativistic mass" if you'd like. It's simply total energy in different units. But remember: total energy is the sum of rest energy $E_0$ and kinetic energy $E_k$, so if you call $\gamma m$ "relativistic mass," you must understand that you're talking about the sum of "rest mass" (the $m$ that everyone agrees on) and the "kinetic mass" (which is just kinetic energy in different units). Today, few physicists use the word "mass" to mean anything other than "rest mass."

You're mistaken about $\gamma$. There are a few ways to think about what $\gamma$ is, but fundamentally it's exactly what I said it is before: $(1 - v^2 / c^2)^{-1/2}$. You can think of it as the time-dilation factor ($t = \gamma t_0$, where $t_0$ is proper time). You can also think of it as the "kinetic energy factor" (in the sense that $E = \gamma E_0 = E_0 + E_k$).

The definition of mass you've cited ("ratio between a force and an acceleration") only holds in the classical limit.

In special relativity, momentum and velocity aren't directly proportional. Instead of being related by a constant multiplicative factor like they are in Newtonian physics (mass), they're related by a function of speed (total energy):

$\vec p c = E \vec \beta = \gamma E_0 \vec \beta$,

where $\vec \beta = \vec v / c$.

Force is the time-derivative of momentum. The time-derivative of that equation requires the chain rule, and you end up with three terms on the right side! When the rest energy (mass) is taken to be constant, that reduces to two terms (I'm using overdots to indicate $ct$-derivatives, not $t$-derivatives):

$\vec f = \gamma^2 (\beta \cdot \dot{\vec \beta}) E \beta + E \dot{\vec \beta}$,

or if you prefer using $m$, $\vec v$, and $\vec a$:

$\vec f = \gamma^3 \left( \dfrac{ \vec v \cdot \vec a }{c^2} \right) m \vec v + \gamma m \vec a$.

We're pretty damn far from $\vec f = m \vec a$.

In short, there simply is no such quantity as the "ratio between a force and an acceleration," because the velocity vector figures in there, too.

In the special case that the force is parallel or anti-parallel to the velocity, we obtain:

$\vec f = \gamma^2 E \dot{\vec \beta} = \gamma ^3 m \vec a$,

and we can call $\gamma^3 m$ the ratio between force and acceleration. But note that if the force is perpendicular to the velocity, we instead get:

$\vec f = E \dot{\vec \beta} = \gamma m \vec a$.

I hope this helps correct some of your misconceptions.

11. Mar 24, 2017

SiennaTheGr8

An object's inertia (its resistance to acceleration) does increase with speed, BUT:
• it also depends on the direction of the force relative to the object's velocity,
• and mass (the invariant $m$ from Newtonian physics) is only one contributor to it.
Even if you define a "relativistic mass" as $\gamma m$, which varies with speed, that quantity is not the object's resistance to acceleration (except in the special case that the force is perpendicular to the object's velocity).

This is one of the arguments against using the concept of relativistic mass. Yes, it makes the relativistic momentum equation look like the classical one:

$\vec p = m_{rel} \vec v = \gamma m \vec v$,

but it often gives people the mistaken idea that maybe $\vec f = m_{rel} \vec a$, or that $E_k = m_{rel}v^2 / 2$.

12. Mar 24, 2017

DrStupid

There is no "gravitational mass" in relativity. Your question can be answered for very limited conditions (e.g. using Newton's law of gravitation for relativistic trajectories of a point mass in the far field of another point mass) but in general it is simply pointless.

Yes, that's how spring scales work.

That just means that this quantity depends on velocity but not that it doesn't exists. Of course it is possible to define M(v) with F=M·a. (The usefulness of such a quantity is another question.)

13. Mar 24, 2017

DrStupid

The invariant mass is a modern concept. It is correct that the mass as used by Newton is invariant in classical mechanics. However, it turns into the frame-dependent relativistic mass in relativity.

14. Mar 24, 2017

SiennaTheGr8

And what exactly is this quantity $M(v)$ that always serves as the ratio between force and acceleration? I contend that there is no such quantity, as I've discussed above.

I disagree. Perhaps it's a matter of interpretation, but certainly it's the case that Einstein's original paper on the mass–energy equivalence demonstrated that $E_0 = mc^2$, where $E_0$ is rest energy and $m$ is precisely the $m$ that appears in the classical $E_k = mv^2/2$.

Personally, I just don't use "mass" at all. Everything is much simpler for me when I use only "energy" terms (total energy $E$, kinetic energy $E_k$, rest/proper energy $E_0$).

15. Mar 24, 2017

pervect

Staff Emeritus
There are at least two general approaches one can take to special relativity. One approach winds up with "mass" being "relative", which I'm interpreting as being what I'd call "observer dependent". But then force turns out to be observer-dependent too. Along with distance, and time, and acceleration. So pretty much everything becomes observer dependent, and it becomes rather a mess to understand anything.

The other approach attempts to make everything observer-independent. The observer dependent concept of distance gets replaced with the observer-indepenent concept of "proper distance". Similarly, observer-dependent time gets replaced with "proper time", and acceleration gets replaced with "proper acceleration". Mass gets replaced with "rest mass" or "invariant mass". Force gets replaced too, but not quite as simply - the observer dependent three-force gets replaced with an observer independent four-force, but this requires adding another dimension to the concept of "force".

This still doesn't explain gravity, which goes beyond the realm of SR into GR. But it's pretty hopeless to understand GR without understanding SR first. It seems to me that the idea of changing the definitions slightly to make all the quantities observer-independent is much more fruitful for understanding than to try to untangle all the observer-dependent quantities using the "old" definitions. Popluarizations tend to take the other approach, but I don't think this leads to a real understanding of special relativity in the long run.

The path to this approach starts with understanding "proper time", which can be regarded as the time one measures on a wristwatch. This is contrasted with coordinate time, which, while observer-dependent, still has a useful role in physics, which, however, I won't get into at this point.

After going through all this, one still has the problem of gravity, which involves some more adjustments to one viewpoints as one changes from special relativity to general relativity. This leads to relaization that gravity is something that's more complex than a force. The cannonical example of this is "gravitational time dilation". Ordinary "forces" don't cause time dilation, but gravity must. Exploring this leads to some rather advanced mathematics, the mathematical requirements to understand special relativity are modest once one is willing to think of things in observer-independent terms. The mathematical requirements to understand GR are quite advanced.

16. Mar 24, 2017

John Morrell

First of all, thanks everyone! The more we discuss this the more I realize that a lot of things which I assumed were fundamental are actually just approximations. That's been really helpful. My initial question about "gravitational mass" has been totally cleared up, so thanks there.

I like the way that several people have explained mass in saying that it's easier to just look at an objects total energy instead of its rest mass + its kinetic energy. i think what I can take from this is that
a) there is no such thing as "relative gravitational mass"; gravity is a stretching of space-time which depends in part on the invariant m which does not depend on the observer.
b) F=ma is a Newtonian approximation, and in a more rigorous model force and acceleration are related to the total energy, the direction of velocity, and the direction of the acceleration.
c) This means that energy and force are relative to the frame of reference.

Does this all sound correct?

17. Mar 24, 2017

SiennaTheGr8

Almost entirely correct.

a) Correct, there is no such thing as "relative gravitational mass," and the invariant $m$ (rest energy $E_0$ in different units) contributes to the energy–momentum tensor.

b) Yep, except replace "direction of acceleration" with "magnitude of velocity" (speed). Also, "total energy" is correct and indeed the way I prefer to view things, but it's certainly not incorrect to focus on mass instead (because $E = \gamma mc^2$). To nitpick, I don't think "rigorous" is the best word there—Newtonian physics is perfectly rigorous within its domain of applicability.

c) Yes, total energy is frame-dependent because $E = E_0 + E_k$, and $E_k$ is frame-dependent (it is even in classical physics, when $E_k = mv^2 / 2$). Force and acceleration are also frame-dependent in special relativity, but they aren't in classical physics!

Caution: you may occasionally hear people say that acceleration is "absolute" in special relativity, in contrast with velocity, which is relative. What they mean by this is that if something is accelerating, then there is no inertial frame in which it isn't accelerating. That doesn't mean that all inertial observers will agree on the magnitude of the acceleration; just that it isn't zero.

18. Mar 25, 2017

vanhees71

NO!!!!! Again $m$ in relativity is invariant by definition! Everything else is a 110 year old misunderstanding, which has been corrected sinc then.

It is this invariant $m$ which is the mass also in the non-relativistic limit. Take the case of one particle moving in an external (e.g., electromagnetic) field. The Newtonian laws are valid in the instaneous rest frame of the particle, and there your $m_{\text{rel}}=m$. There is no need for $m_{\text{rel}}$ anywhere!

The use of a three-tensor-valued mass is even worse, and it's not needed either. For a covariant formulation of classical mechanics, see

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

19. Mar 25, 2017

Staff: Mentor

Getting back to your original question as stated above, the concept of "gravitational mass" is not well-defined in general relativity. You have to specify how, exactly, you want to measure the gravitational effect of a moving object.

This sort of question has come up here a number of times over the years. The forum's search feature doesn't work well for this kind of search, so I resorted to Google:

Here are a couple of threads, that I remember seeing before, that provide "an" answer (with equations) for "gravitational mass". Pay particular attention to posts by 'pervect'. There are probably other suitable threads, but I'm about to go out for the day, so feel free to sift through the search results further.

https://www.physicsforums.com/threa...n-relativistic-mass-or-just-rest-mass.651612/ (2012)

https://www.physicsforums.com/threa...oving-object-versus-stationary-object.687833/ (2013)

Last edited by a moderator: May 8, 2017
20. Mar 25, 2017