What is relativistic mass and why it is not used much? - Comments

[Orodruin]

Orodruin submitted a new PF Insights post

What is relativistic mass and why it is not used much?

Continue reading the Original PF Insights Post.

 

[PAllen]

Just a comment that if you use 4-force instead of 3-force, you do recover a Newtonian like equation:

F = dp/dτ = d(m₀U)/dτ = m₀ dU/dτ = m₀A

where U is 4-velocity and A is 4-acceleration.

 

[Orodruin]

Just a comment that if you use 4-force instead of 3-force, you do recover a Newtonian like equation:

F = dp/dτ = d(m₀U)/dτ = m₀ dU/dτ = m₀A

where U is 4-velocity and A is 4-acceleration.

Indeed, this was previously an FAQ and the target audience is mainly people who are not familiar with 4-vectors. It was written due to the endless stream of people we get asking questions on the subject of changing mass in relativity.

It should be noted that if the system is not closed (e.g., an object absorbing external radiation), as in classical mechanics, you would obtain
$$
\frac{d(mU)}{d\tau} = m A + V \frac{dm}{d\tau},
$$
also analogous to the Newtonian case.

 

[m4r35n357]

Orodruin submitted a new PF Insights post

Could you clarify whether 'a' in your final equation represents proper or coordinate acceleration? I always have to rack my brains about this when reading about acceleration in SR ;)

 

[Orodruin]

Could you clarify whether 'a' in your final equation represents proper or coordinate acceleration? I always have to rack my brains about this when reading about acceleration in SR ;)

a is always the acceleration in a particular inertial frame for the purposes of this post. No knowledge about proper acceleration is assumed.

 

[Dale]

Just a comment that if you use 4-force instead of 3-force, you do recover a Newtonian like equation:

F = dp/dτ = d(m₀U)/dτ = m₀ dU/dτ = m₀A

where U is 4-velocity and A is 4-acceleration.

Yes. But usually by the time someone learns four-vectors they are not confused about relativistic mass vs. invariant mass.

In fact, I sometimes wonder if it isn't easier to just teach four-vectors than have such discussions at all.

 

[harrylin]

Well done Orodruin!

 

[Orodruin]

Well done Orodruin!

I think it should be pointed out that while I wrote the original FAQ, several PF science advisors contributed opinions and suggestions on how to refine it.

 

[Greg Bernhardt]

Great as always Orodruin!

 

[maline]

Another way of expressing SR kinematics in Newtonian terms (using 3-vectors in some one frame) is by phrasing the second law as F=dp/dt, and defining momentum by p=Ev/c^2, with E the total energy. The gamma factor can even be derived from this using the work/energy relation. Since these equations are simple (no square roots!), this has led me to speculate that such a definition of momentum should be seen as more "causative" than mv*gamma, although of course they are both frame dependent. This would give E/c^2, the relativistic mass, a status that would justify its onetime popularity. ;-)

 

[DrStupid]

and defining momentum by p=Ev/c^2, with E the total energy

That doesn't need to be defined. It can be derived from the classical definition of momentum.

 

[maline]

my point is that you can describe the dynamic laws of SR, in terms of three-vectors, without mentioning the gamma factor.

 

[PAllen]

my point is that you can describe the dynamic laws of SR, in terms of three-vectors, without mentioning the gamma factor.

A peculiarity of this approach (which, of course, does not make it wrong) is that the simple non-radiating case of forced motion is not simple at all. You have:

F = dp/dt = E' v + E v'

with all terms important for the simple non-radiating impulsive force (I use c=1). Meanwhile, with 4-vectors, the non-radiating case becomes the very simple:

F = m A

which has to do with the fact that the 4-vector approach sees gamma as an intrinsic feature velocity, by virtue of proper time. It is metric in nature rather than part of dynamics. In the language of 4-vectors, if we talk about velocity or acceleration divorced from any particular mass or energy, the gamma factors are built in.

 

[exponent137]

Can you write more about discreancy between special relativity and Newtonian gravity?
Maybe, if a very fast comet is flying close to earth, its gravitational force is proportional to Earth to m_0 and not to \gamma m_0
$m_0$ and not to $\gamma m_0$?
Is something else?

 

[DrStupid]

$m_0$ and not to $\gamma m_0$?
Is something else?

\left( {1 + \beta ^2 } \right) \cdot \gamma \cdot m_0

 

[Orodruin]

In general relativity gravitation is not coupled to mass only. Instead, the source of space-time curvature is energy, momentum, and stress. Only for weak gravitation and small velocities does it reduce to the Newtonian case where mass is the source of gravity. You simply cannot describe gravity in special relativity by changing the mass for the relativistic mass. There should be some threads here on that subject if you use the search function.

 

[exponent137]

In general relativity gravitation is not coupled to mass only. Instead, the source of space-time curvature is energy, momentum, and stress. Only for weak gravitation and small velocities does it reduce to the Newtonian case where mass is the source of gravity. You simply cannot describe gravity in special relativity by changing the mass for the relativistic mass. There should be some threads here on that subject if you use the search function.

I searched these threads, and partly I understand better, but I do not find everything. As Dr. Stupid wrote above for the above example of Schwarchild geometry, factor $1+\beta^2$ means the main discrepancy between Newtonian gravity + SR vs. GR+"SR". Can you have still any concrete examples for this discrepancy. What is at big velocities and small gravity. (One example gave Dr. Stupid.)

 

[Orodruin]

Can you have still any concrete examples for this discrepancy.

The perhaps most famous differences between the predictions of Newtonian gravity and GR are:
- The perihelion shift of Mercury's orbit
- The amount of gravitational deflection of light passing by the Sun
In the first case, both the velocity and gravitational fields are relatively small so the effect is very small - it requires very good precision to make the measurement. The second case was one of the classic tests of GR and involves light, so the speeds involved are large (large means comparable to light speed).

 

[pervect]

I have a suspicion that a significant number of the readers here on PF are rather rusty on their calculus. (Some readers may not have had calculus at all - but trying to totally eliminate calculus from the exposition seems to me to require a totally different approach, if it's possible at all).

As a consequence of this (presumed) rustiness, simply mentioning that F = dp/dt may not entirely get through to such readers, who are still mentally attached to the idea of F = ma, both because of both familiarity and because it deosn't require them to try and remember their calculus.

A little more hand-holding, reviewing the classical perspective and explaining that the origin of F=ma was in fact F=dp/dt, might be helpful. As a consequence, it might be more clear when one explains that F=dp/dt no longer leads directly to F=ma in relativistic physics. If nothing else, it may inspire some readers to brush up on the necessary mathematical underpinings (this is probably rather optimistic). In other cases it may at least get across the idea of what's missing and preventing a fuller understanding of the issues.

I'm not sure of the details of how to actually go about writing all of this, but I thought I'd mention the idea in case someone was motivated.

 

[harrylin]

[..] As a consequence of this (presumed) rustiness, simply mentioning that F = dp/dt may not entirely get through to such readers, who are still mentally attached to the idea of F = ma, both because of both familiarity and because it deosn't require them to try and remember their calculus.

A little more hand-holding, reviewing the classical perspective and explaining that the origin of F=ma was in fact F=dp/dt, might be helpful. As a consequence, it might be more clear when one explains that F=dp/dt no longer leads directly to F=ma in relativistic physics. If nothing else, it may inspire some readers to brush up on the necessary mathematical underpinings (this is probably rather optimistic). In other cases it may at least get across the idea of what's missing and preventing a fuller understanding of the issues.

I'm not sure of the details of how to actually go about writing all of this, but I thought I'd mention the idea in case someone was motivated.

Some textbooks (such Alonso&Finn which I used as a student) present the SR derivations of acceleration from F=dp/dt; that's a useful stating point.

 

[vanhees71]

Sure, you have (in the usual (1+3)-dimensional formulation) for, e.g., a particle in an electromagnetic field (neglecting radiation corrections)
$$\frac{\mathrm{d} \vec{p}}{\mathrm{d} t}=q \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right ), \quad \vec{p}=m \frac{\vec{v}}{\sqrt{1-\vec{v}^2/c^2}}, \quad \vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t}.$$
Of course, I've used the invariant mass $m=\text{const}$ here. I never ever use something else as mass than this one and only invariant mass, but the (1+3)-formalism in a given inertial frame is often useful. For details on SR dynamics, also in manifestly covariant form, see my (unfinished) SRT FAQ:

http://fias.uni-frankfurt.de/~hees/pf-faq/srt.pdf p.25ff

 

[Erland]

"the resistance to acceleration depends on both the velocity of the object as well as the direction of the force and so relativistic mass cannot in general correspond to a generalisation of either."

I don't understand this remark. Can you develop it...
Why would we demand that relativistic mass should "correspond to" this or that?

 

[Orodruin]

Why would we demand that relativistic mass should "correspond to" this or that?

If relativistic mass is to be considered a proper generalisation of the mass in classical mechanics, we would expect it to exhibit certain properties - at least its defining properties. In the case of inertial mass, the defining property is resistance to acceleration and relativistic mass does not describe this property well.

 

[DrStupid]

In the case of inertial mass, the defining property is resistance to acceleration and relativistic mass does not describe this property well.

That's not a problem of relativistic mass. Invariant mass isn't better either. In both cases you need to know the velocity to get the force for a specific acceleration or vice versa. I would go the other way around: resistance to acceleration is not suitable as defining property in relativity.

 

[Dale]

Invariant mass isn't better either.

It is if you generalize to four vectors

 

[Orodruin]

That's not a problem of relativistic mass.

Of course it is. It is an obvious problem if you are looking to use the same type of argumentation as in the non-relativistic case. The point of the argument as stated is to see why relativistic mass is not a good generalisation of any non-relativistic mass. Once you have established that, you can figure out what kind of quantity you should be looking at in relativity and look at the 4-vector relations as alluded to by Dale. It turns out that there is a proportionality constant between the 4-velocity and the 4-momentum, which is the mass (or equivalently, rest energy) of the system. Taking the non-relativistic limit, this can then be identified with the inertia and thereby establishing the equivalence between an object's rest energy and its inertia in its rest frame.

 

[DrStupid]

It turns out that there is a proportionality constant between the 4-velocity and the 4-momentum, which is the mass (or equivalently, rest energy) of the system.

And relativistic mass is the proportionality factor between velocity and momentum, which is by definition the iniertial mass in classical mechanics (and it apperas to be equivalent to the total energy in relativity).

Relativistic mass is just not frame-independent. That's the only reason to favor invariant mass over relativistic mass in relativity. Everything else is far-fetched.

 

[Orodruin]

And relativistic mass is the proportionality factor between velocity and momentum, which is by definition the iniertial mass in classical mechanics (and it apperas to be equivalent to the total energy in relativity).

This is quite misleading. On one side of your equation you are taking the projection of a 4-vector onto the spatial directions and on the other you are normalising the corresponding quantity by the time-component. The proportionality factor if you do the relativistically appropriate thing and compare apples with apples instead of oranges is the inertial mass.

Part of the point is that momentum does not vary linearly with velocity in relativity.

 

[SiennaTheGr8]

The problem with "relativistic mass" is entirely pedagogical. If you already know what you're doing and like to use it, there's really no issue (except that other people who know what they're doing might look at you funny if you use it in conversation).

... resistance to acceleration is not suitable as defining property in relativity.

I would agree with this.

In Newtonian mechanics, the turns of phrase "measure of inertia" and "resistance to acceleration" refer to mass. There's not much room for ambiguity here.

In special relativity, these concepts become ambiguous or even meaningless.

First, "resistance to [3-]acceleration": in general, there simply is no multiplicative factor (let alone a constant of proportionality) that tells you how an object will accelerate under the influence of a given 3-force. Rather, an object's total energy tells you how it will accelerate under the influence of a particular pairing of 3-force and 3-velocity.

As for "measure of inertia," well, that might mean all sorts of things. Maybe the multiplicative factor relating 3-velocity and 3-momentum? That's total energy (which isn't a constant of proportionality!). Maybe just the frame-independent part of that quantity? That's rest energy (mass). Maybe the multiplicative factor relating 3-acceleration and 3-force? That doesn't exist (see above). Maybe the multiplicative factor relating proper acceleration and proper force (i.e., relating 3-acceleration and 3-force in the object's rest frame)? That's rest energy (mass)—and this one's actually a constant of proportionality.

If we're talking about 4-vectors, though, we can restore meaning to these concepts. An object's rest energy (mass) is its measure of "spacetime inertia," if you will. It's the object's resistance to 4-acceleration under the influence of a 4-force. The more rest energy (mass) an object has, the more it resists change to its (direction of) 4-velocity. It's the constant of proportionality relating 4-acceleration to 4-force, and also 4-velocity to 4-momentum.

 

[Herman Trivilino]

Why would we demand that relativistic mass should "correspond to" this or that?

We wouldn't. The point being made is that some learners have that expectation. If you can generalize the Newtonian expression for momentum $\vec{p}=m \vec{v}$ by replacing $m$ with relativistic mass, it creates the expectation that you ought to be able to do the same with the Newtonian expressions for force $\vec{F}=m \vec{a}$ and kinetic energy $K=\frac{1}{2}mv^2$. Which of course you can't!

Moreover it creates the expectation that the increase in relativistic mass of an object is associated with some change in a property of the object, when in fact it's due to the geometry of spacetime.