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

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Common Topics: mass, object, relativistic, special, concept

It happens that the term relativistic mass is used, in particular in the introductory text on special relativity. It should be noted that whether or not to use relativistic mass to a large extent is a matter of convention, convenience, and semantics as long as it is used properly and does not have any impact on the predictions of special relativity. This FAQ is intended to clarify the concept and explain why it has essentially fallen out of fashion in the scientific community.

## The concept of relativistic mass

In special relativity, the expressions for the total energy and momentum of an object can be written as
$$E = m_0 \gamma c^2, \quad \vec p = m_0\gamma \vec v,$$
where ##\gamma = \frac{1}{\sqrt{1-v^2/c^2}}## and ##m_0## is the invariant mass (or rest mass) of the object. If we introduce the relativistic mass
$$m = m_0\gamma$$
these expressions turn into
$$E = mc^2, \quad \vec p = m\vec v,$$
which many find more familiar since the first is an iconic relation in special relativity and the latter coincides with the classical expression for momentum.

## Shortcomings of relativistic mass

In classical mechanics, there are two different concepts of mass, gravitational mass, which determines the force on an object due to gravity (##\vec F = m\vec g##), and inertial mass, which determines the resistance to the acceleration of an object (##\vec a = \vec F/m##). Since special relativity cannot include Newtonian gravity, the concept of gravitational mass is moot. In addition, 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 generalization of either. In the rest frame of an object, we regain the relation between force and acceleration of classical mechanics, but this is well described by the invariant mass and there is no need to introduce relativistic mass for this purpose.
A common misconception that can be attributed to the concept of relativistic mass is that an object changes its internal structure by gaining mass when it travels at relativistic speeds. The object’s internal structure is independent of its velocity and it will always appear to be the same in its rest frame. The source of this confusion is that relativistic mass depends on the frame in which the object is observed and the concept of mass is typically regarded as a property of an object. See also our FAQ on the mass-energy equivalence.

Because of the above, relativistic mass is generally not used in scientific communication as quoting it would also require quoting the velocity of the object. Furthermore, as evidenced by ##E = mc^2##, the relativistic mass is simply proportional (and thus equivalent) to the total energy of an object, which is a concept far less liable to misunderstandings as energy is frame dependent also in classical mechanics.

## How kinematics are treated in special relativity

Instead of introducing the relativistic mass of an object, we simply consider the invariant mass ##m_0## to be the mass of an object. The total energy and momentum of an object is given by
$$E^2 = m_0^2 c^4 + p^2 c^2, \quad \vec p = m_0 \gamma \vec v.$$
The relativistic relation between the force acting on an object and its acceleration is given by
$$\vec F = \frac{d\vec p}{dt} = m_0 \frac{d(\gamma\vec v)}{dt},$$
which is equivalent to the definition in classical mechanics for small velocities. If the force is collinear with the velocity, this can be written as
$$F = m_0\gamma^3 a.$$
Note here that we would not recover ##F = ma## if we introduced relativistic mass. As ##\gamma \to \infty## when ##v \to c##, the faster an object is traveling, the more force is needed to accelerate it. This, or equivalently the fact that the total energy goes to infinity as ##v\to c##, is the more appropriate argument for why massive objects cannot be accelerated to the speed of light.

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58 replies
1. Dale says:
DrStupid

We get F=m·a in classical mechanics for closed systems. But that doesn't mean that we need to expect it outside classical mechanics as well.I disagree. In the classical domain f=ma is experimentally validated. So any theory which would generalize classical mechanics must reduce to f=ma in the appropriate limit. We therefore have a very strong expectation that we will see something like it in relativity, with some suitable modifications for the generalization.

Here the most appropriate generalization is the use of four vectors, in which case the familiar formulas hold and clearly reduce to the Newtonian expressions in the appropriate limit. The mass then is the invariant mass.

2. Orodruin says:
DrStupid

If you disagree with this definition than you are outside classical mechanics.Newton's formulation with his three laws is not the only formulation of classical mechanics. Nor is Principia a Bible that cannot be questioned or improved upon. Imagine a similar claim in quantum mechanics; that you have to adhere to the Schrödinger picture or you are not doing QM.

3. DrStupid says:
Dale

I disagree with this.Newton's definition II says "Quantitas motus est mensura ejusdem orta ex Velocitate et quantitate Materiæ conjunctim." That means p:=m·v. If you disagree with this definition than you are outside classical mechanics.

Dale

From classical mechanics we expect ##vec{f}=mvec{a}## and ##vec{p}=mvec{v}## and ##vec{f}=dvec{p}/dt ## to all hold.We get F=m·a in classical mechanics for closed systems. But that doesn't mean that we need to expect it outside classical mechanics as well.

PAllen

To my mind, the Newtonian notion of inertia was tied to F=mA, with m constant.The Newtonean notation was F=dp/dt. That results in F=m·a if m is constant. But m was not defined to be constant in classical mechanics (in contrast to current meaning of the term "mass"). Such a definition wasn't required because the velocity dependence of m was already given by other basic requirement, including the transformation. Galilean transformation results in a constant mass for closed systems. Replacing Galilean transformation by Lorentz transformation turns m into the relativistic mass.

4. PAllen says:

To my mind, the Newtonian notion of inertia was tied to F=mA, with m constant. It is precisely the feature that this m and the m in universal gravitation were the same that led to notions of universal free fall that Newton was familiar with. The m in momentum has little to do with this. Further, Newton never defined m via momentum. He defined it via quantity of matter, which is not a very useful modern definition.

5. Dale says:
DrStupid

And relativistic mass is the proportionality factor between velocity and momentum, which is by definition the iniertial mass in classical mechancisI disagree with this. From classical mechanics we expect ##vec{f}=mvec{a}## and ##vec{p}=mvec{v}## and ##vec{f}=dvec{p}/dt ## to all hold. The question is how to generalize this. Using relativistic mass in place of ##m## does not generalize all of them (neither with three vectors nor with four vectors). Using invariant mass does (with four vectors).

6. Orodruin says:
DrStupid

The results are in agreement with all known independently reproducible experimental observations and nothing else decides between right or wrong in physics.So would the F = ma definition be if used it appropriately with the appropriate inclusion of terms in the variable-mass case on the force side (i.e., ##dot m vec v_{rm rel}##). I agree that the other option is cleaner, but for the purposes of the text, the distinction is not really relevant as that is not the definition that leads to severe misunderstandings of the basic theory – relativistic mass does just that.

I would agree with your comments if the text was longer and intended for university level students, but the thing is that it is not and you have to adapt your level to the target audience (and its typical attention span) even if that means not being as precise as you would like. The text is a very short argumentation for why physicists generally avoid talking about relativistic mass, intended for people who come to PF with some of the strong misconceptions that the concept induces.

7. DrStupid says:
Orodruin

The concept of lies-to-children is a crucial idea in education wherein we get people a bit closer to the right idea than they were before.F=dp/dt instead of F=m·a gets them as close to the right idea as currently possible. The results are in agreement with all known independently reproducible experimental observations and nothing else decides between right or wrong in physics. Of course the use of relativistic mas is not the optimal choice in relativity, but that doen't mean it is wrong.

8. Orodruin says:
SiennaTheGr8

Incidentally, I think we lose something linguistically by reserving "scalar" for Lorentz invariants. I find "number" a poor substitute for describing the type of quantity that, say, total energy is—it would be easy to infer that a quantity that is a "number" doesn't have a unit. (And anyway, total energy is invariant under spatial rotation, so it is a scalar in that context, isn't it?)This is not an issue only for scalars but for many other mathematical concepts as well. Just take such a thing as a "vector". In relativity we generally assume that it means a 4-vector or, when restricting ourselves to rotations, 3-vectors. However, a "vector" generally is just an element of a vector space, which would also include tensors or any other type of elements of vector spaces (such as function spaces etc). If we would wish to be precise, we would say that energy is a scalar under rotations, but not a Lorentz scalar.

9. Orodruin says:
DrStupid

F=m·a as defining relation for inertial mass actually is in that category.I don't understand this answer. This is exactly what I was saying in my post.

DrStupid

That shouldn't be mentioned, unless in order to correct this misconception.I disagree again. The concept of lies-to-children is a crucial idea in education wherein we get people a bit closer to the right idea than they were before. If you wish to clear that up, fine, but that is not the goal of the text and in this context counter-productive. The goal of the text is to clarify a different lie that in my experience is responsible for many more severe misunderstandings. For some time when this text was written, every other thread in the relativity forum was filled with misconceptions based on relativistic mass and trying to apply it in statements such as F = ma. You will not find the same type of misconceptions regarding the definition of mass in classical mechanics. Partially because you could just as well define it through F = ma and the variable mass case would look slightly different, but still contain the same physics. (I am not saying it is a good idea, just that you could do it.)

10. DrStupid says:
Orodruin

You had been using the word "scalar" with the meaning of a Lorentz scalar in a previous post and in the context of relativistic mass not being a Lorentz scalar. This made your post appear to simply state that ##M## was not a Lorentz scalar, but still a number.I still do not see why my post suggests that M is a number – even if you read "scalar" as lorentz scalar. Nevermind, it was just a misunderstanding that has been cleared.

Orodruin

It is clearly within the lies-to-children category.F=m·a as defining relation for inertial mass actually is in that category. That shouldn't be mentioned, unless in order to correct this misconception.

11. SiennaTheGr8 says:

Incidentally, I think we lose something linguistically by reserving "scalar" for Lorentz invariants. I find "number" a poor substitute for describing the type of quantity that, say, total energy is—it would be easy to infer that a quantity that is a "number" doesn't have a unit. (And anyway, total energy is invariant under spatial rotation, so it is a scalar in that context, isn't it?)

12. Orodruin says:
DrStupid

Can you explain what exactly let you think that (in order to avoid it the next time).You had been using the word "scalar" with the meaning of a Lorentz scalar in a previous post and in the context of relativistic mass not being a Lorentz scalar. This made your post appear to simply state that ##M## was not a Lorentz scalar, but still a number.

DrStupid

And such misconceptions shouldn't be suported by using by using them without immediate correction.I strongly disagree. With this mentality you would get absolutely nowhere in high-school physics. It is clearly within the lies-to-children category.

13. DrStupid says:
Orodruin

Your previous post seemingly suggests that it is a number.Can you explain what exactly let you think that (in order to avoid it the next time).

Orodruin

"Scalar" in relativity is typically reserved for frame-independent scalar quantities.Would my statement be wrong for lorentz scalars?

Orodruin

The point is that laymen try to use relativistic mass in expressions such as F = ma and K = mv^2/2 or think that a moving object's inner structure changes.And such misconceptions shouldn't be suported by using by using them without immediate correction.

14. Orodruin says:
DrStupid

$M = left( {I_3 + frac{{v cdot v^T }}{{c^2 – v^2 }}} right) cdot frac{m}{{sqrt {1 – frac{{v^2 }}{{c^2 }}} }}$

edit: typo correctedWhere M is obviously a matrix and not a number. Your previous post seemingly suggests that it is a number. "Scalar" in relativity is typically reserved for frame-independent scalar quantities.

DrStupid

That's what I'm talking about. This sentence reads like a confirmation for F=m·a as the defining relation of inertial mass in classical mechanics. This is simply not the case.I believe you are missing the mark. At that point it would be detrimental to the Insight to start talking about how Newton-2 looks for variable mass systems in classical mechanics (which would be another Insight altogether). @Mister T has it figured out in #30. The point is that laymen try to use relativistic mass in expressions such as F = ma and K = mv^2/2 or think that a moving object's inner structure changes.

15. DrStupid says:
Orodruin

Actually, this is false. 3-force and 3-acceleration are not necessarily parallel in relativity.$M = left( {I_3 + frac{{v cdot v^T }}{{c^2 – v^2 }}} right) cdot frac{m}{{sqrt {1 – frac{{v^2 }}{{c^2 }}} }}$

edit: typo corrected

16. DrStupid says:
Orodruin

I believe the […] makes it pretty clear that what is discussed in the sentence refers to classical mechanics.That's what I'm talking about. This sentence reads like a confirmation for F=m·a as the defining relation of inertial mass in classical mechanics. This is simply not the case. It should be pointed out that this equation is valid for constant inertial mass only – no matter if you take it as a definition or derive it from F=dp/dt. Whith this knowledge it would be obvoius that is must not be used with a variable mass.

17. Orodruin says:
DrStupid

Of course there is an M(v) with F=M·a, but it is almost ever different from relativistic mass.Actually, this is false. 3-force and 3-acceleration are not necessarily parallel in relativity.

18. Orodruin says:
DrStupid

Wouldn't it be a good idea to clear this misconception from the beginning? The sentence "[…] inertial mass, which determines the resistance to acceleration of an object (##vec a = vec F/m##)" reather sounds like a confirmation.You have taken the quote out of context. I believe the […] makes it pretty clear that what is discussed in the sentence refers to classical mechanics. So yes, it should be a reaffirmation of the reader’s assumed knowledge of Newton-2.

19. DrStupid says:
SiennaTheGr8

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.At least not a scalar. Of course there is an M(v) with F=M·a, but it is almost ever different from relativistic mass.

Orodruin

The typical target audience of the text will consider ##F=ma## as the defining relation of inertial mass and that simply does not generalise to relativity.Wouldn't it be a good idea to clear this misconception from the beginning? The sentence "[…] inertial mass, which determines the resistance to acceleration of an object (##vec a = vec F/m##)" reather sounds like a confirmation.

20. SiennaTheGr8 says:
Orodruin

I think this is the crucial point. The text is not written as an argument for this kind of person. The typical target audience of the text will consider ##F=ma## as the defining relation of inertial mass and that simply does not generalise to relativity. This point is made in the text where I specifically discuss this discrepancy with the relation to the time derivative of momentum. You cannot have F=ma and p=mv at the same time if m depends on v.Yes, and to be clear, my comment was meant as a general one on relativistic mass, not as a response to the Insights article.

21. Orodruin says:
SiennaTheGr8

If you already know what you're doingI think this is the crucial point. The text is not written as an argument for this kind of person. The typical target audience of the text will consider ##F=ma## as the defining relation of inertial mass and that simply does not generalise to relativity. This point is made in the text where I specifically discuss this discrepancy with the relation to the time derivative of momentum. You cannot have F=ma and p=mv at the same time if m depends on v.

I just find this argument

DrStupid

That's not a problem of relativistic mass. Invariant mass isn't better either.a bit off the mark. It is rather clear that it is a problem if you expect to recover classical results, as pointed out in #30. That invariant mass suffers from the same issue just provides a straw man argument. Once we have established that the concept has issues, we can start the consideration of what the more appropriate quantity to consider should be.

22. Mister T says:
Erland

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.

23. SiennaTheGr8 says:

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).

DrStupid

… 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.

24. Orodruin says:
DrStupid

And relativistic mass is the proportionality factor between velocity and momentum, which is by definition the iniertial mass in classical mechancis (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.

25. DrStupid says:
Orodruin

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 mechancis (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.

26. Orodruin says:
DrStupid

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.

27. Dale says:
DrStupid

Invariant mass isn't better either.It is if you generalize to four vectors

28. DrStupid says:
Orodruin

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.

29. Orodruin says:
Erland

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.

30. Erland says:

“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?

31. vanhees71 says:

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:

32. harrylin says:
pervect

[..] 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.

33. pervect says:

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.

34. Orodruin says:
exponent137

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).

35. exponent137 says:

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.)

36. Orodruin says:

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.

37. DrStupid says:
exponent137

##m_0## and not to ##gamma m_0##?
Is something else?$left( {1 + beta ^2 } right) cdot gamma cdot m_0$

38. exponent137 says:

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?

39. PAllen says:
maline

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.

40. maline says:

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

41. DrStupid says:
maline

and defining momentum by p=Ev/c^2, with E the total energyThat doesn't need to be defined. It can be derived from the classical definition of momentum.

42. maline says:

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. ;-)

43. Orodruin says:
harrylin

Well done Orodruin! :smile: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.

44. Dale says:
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(m[SUB]0[/SUB]U)/dτ = m[SUB]0[/SUB] dU/dτ = m[SUB]0[/SUB]A

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.

45. Orodruin says:
m4r35n357

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.

46. m4r35n357 says:
Orodruin

Orodruin submitted a new PF Insights postCould 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 ;)

47. Orodruin says:
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(m[SUB]0[/SUB]U)/dτ = m[SUB]0[/SUB] dU/dτ = m[SUB]0[/SUB]A

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
$$frac{d(mU)}{dtau} = m A + V frac{dm}{dtau},$$
also analogous to the Newtonian case.

48. PAllen says:

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

F = dp/dτ = d(m[SUB]0[/SUB]U)/dτ = m[SUB]0[/SUB] dU/dτ = m[SUB]0[/SUB]A

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