Does mass become infinite near the speed of light?

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SUMMARY

The discussion clarifies that while relativistic mass increases with speed, it does not become infinite at the speed of light; instead, the kinetic energy of an object approaches infinity. The term "relativistic mass" is largely considered outdated and can lead to confusion. The invariant mass remains constant regardless of speed, and the increase in inertia is attributed to external energy input rather than speed itself. Observers in a spaceship traveling close to the speed of light perceive their mass as unchanged, while external observers note the increased inertia of the ship.

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  • Understanding of special relativity principles
  • Familiarity with the concept of invariant mass
  • Knowledge of kinetic energy and its relation to speed
  • Basic grasp of relativistic momentum
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  • #31
DrStupid said:
That applies to both rest mass and relativistic mass but only the latter is consistent with the classical definition of momentum and Lorentz transformation. Using rest mass instead requires a corresponding redefinition of momentum.
And I claim any attempt to define momentum as mv is ill conceived in SR.
 
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  • #32
PAllen said:
And I claim any attempt to define momentum as mv is ill conceived in SR.

The classical definition works as well as the relativistic definition. With the corresponding concepts of mass they are in fact identical.
 
  • #33
Shyan said:
I think textbook authors should abandon the concept of relativistic mass. Its of almost no use and produces too much trouble. I'm sure there can be other explanations for why nothing with mass can reach the speed of light. One thing that seems fine to me is to say that in SR, Newton's second law changes from \vec a=\frac{\vec F}{m} to \vec a=\frac{1}{\gamma m}(\vec F-\frac{\vec v \cdot \vec F}{c^2}\vec v).
I believe that Newton stated his second law as F = dp/dt
 
  • #34
Mass is just a word. In a sense, it's not wrong to use it to mean whatever mathematical term you want.

However.

Rest mass is invariant (like Newtonian mass) while relativistic mass is not. Rest mass is part of the source term for gravity (like Newtonian mass) while relativistic mass is not. Rest mass is the resistance of a body to a change of motion independent of that motion (like Newtonian mass) while relativistic mass is not.

There is no single relativistic mass that you can put into all relativistic equations to make them "look like" their Newtonian equivalents (see, for example, longitudinal and transverse mass).

Also, relativistic mass (as usually constituted) is ##P^0##, or the total energy of the object (give or take a factor of ##c##). Why not let energy be one thing and mass another, rather than having two terms for one thing?

@vanhees71 gave more modern reasons to prefer rest mass. I'd be interested to know if Newtonian mass emerges in a similar way from the Cartan's geometric formulation of Newtonian physics.
 
  • #35
From what I recall from one of Max Jammer's book on the history of the concept of Mass, "Concepts of Mass in Classical and Modern Physics", Newton's original concept of mass, based on even earlier roots, was "quantity of material".

Energy doesn't fit this definition, of course - it's not a material substance.

As has been discussed at length elsewhere (and seemingly ignored as much as it's discussed), relativistic mass is just another name for energy - which is different from the classical idea of "quantity of material".
 
  • #36
Ibix said:
Rest mass is invariant (like Newtonian mass) while relativistic mass is not.

That's the only reason to favor rest mas over relativistic mass. Everything else is far-fetched.
 
  • #37
DrStupid said:
That's the only reason to favor rest mas over relativistic mass. Everything else is far-fetched.
What's far fetched about defining:

Mass is the thing that scales momentum and kinetic energy. For a given speed, what determines momentum and KE is mass.

From that, if you measure at high speeds, you get that momentum is not mv, and KE is not mv2/2, as they appear to be at lower speeds. A calorimeter would suffice to measure the energy carried by projectile.

To me, this is utterly natural, and all else if far fetched. (What a useful, objective term: far fetched).

These definitions are natural in Newtonian physics. With them, the (ultimately approximate) relation mv is derived, not defined.
 
  • #38
I'd say "the mass term in the stress energy tensor is not relativistic mass", "##P^0## already has a name" and "relativistic mass must mean more than one thing to make relativistic formulae look Newtonian" were simple statements of fact. All are reasons to favour "mass means invariant mass". See "if I move fast enough, do I turn into a black hole?", and "photons must have mass because they have energy and energy is mass" to name but two sources of confusion.
 
  • #39
PAllen said:
What's far fetched about defining:

Mass is the thing that scales momentum and kinetic energy. For a given speed, what determines momentum and KE is mass.

1. That's not sufficient to define a frame-independent mass. Relativistic mass also fits to this definition.
2. I was talking about the reason for the definition. Not about the definition itself.

PAllen said:
These definitions are natural in Newtonian physics. With them, the (ultimately approximate) relation mv is derived, not defined.

It's the other way around. p=m·v is the definition of momentum in classical mechanics and the properties of mass and the equation for kinetic energy result from this definition (among other things).

Ibix said:
"the mass term in the stress energy tensor is not relativistic mass"

In the stress-energy-tensor there is no mass term but an energy-term (more precisely energy density) which is proportional to relativistic mass and not to rest mass.

Ibix said:
"##P^0## already has a name"

And so does relativistic mass.

Ibix said:
"relativistic mass must mean more than one thing to make relativistic formulae look Newtonian"

Relativistic mass doesn't need to make relativistic formulae look Newtonian.

Ibix said:
"if I move fast enough, do I turn into a black hole?"

That's not a problem of relativistic mass but of Newton's law of gravitation. Newtonian gravity doesn't work in relativity - neither with relativistic mass nor with rest mass.

Ibix said:
"photons must have mass because they have energy and energy is mass"

With relativistic mass instead of rest mass this statement is correct. Such problems do not result from relativistic mass but from the existence of different concepts of mass - no matter which of them is preferred.

That's what I mean with far-fetched reasons. Rest mass is preferred because it is frame-independent in relativity - nothing more and nothing less.
 
  • #40
DrStupid said:
1. That's not sufficient to define a frame-independent mass. Relativistic mass also fits to this definition.
2. I was talking about the reason for the definition. Not about the definition itself.
No, it doesn't. By scaling energy, for example, I mean I fire a standard projectile at some speed, and test mass at the same speed. I find the energy ratio is the same independent of speed. I can measure the energy with a calorimeter. I define this speed independent ratio (for momentum or energy) as mass without recourse to weight. I can then find that weight is proportional to the same quantity.
DrStupid said:
It's the other way around. p=m·v is the definition of momentum in classical mechanics and the properties of mass and the equation for kinetic energy result from this definition (among other things).
I am proposing NOT to define momentum this way; it is NOT necessary to do it this way. You can certainly measure energy as a function of speed with just the mass definition given and find that the Newtonian formula is only approximate. I am not arguing that mine is the only sensible definition, but I strenuously arguing against any claim that it is unnatural, let along far fetched.
 
  • #41
Analogous to my energy example, one can measure momentum given some constant force that can be applied. Momentum is then defined by the time needed for the constant force to stop a body. Again one finds that the ratio of a test body's momentum to standard body's momentum (for objects at the same speed), is independent of speed. Mass is the scaling factor, for any given speed. Then, you can find, at high speeds, the mv is no longer an accurate formula for momentum.
 
  • #42
PAllen said:
I am proposing NOT to define momentum this way; it is NOT necessary to do it this way.

You can't change what already happened. Momentum has been defined that way in classical mechanics and the relativistic mass is a result of this definition.
 
  • #43
DrStupid said:
You can't change what already happened. Momentum has been defined that way in classical mechanics and the relativistic mass is a result of this definition.
Classical mechanics has been reformulated several different ways over its history. I am proposing a natural formulation in which energy and momentum expressions are derived rather than definitions. This redefinition is natural, even classically. Classically it would be expected to be equivalent to other formulations of classical mechanics. When two equivalent formulations become inequivalent with new measurements, you have a choice. All I claim is carrying over mv as a definition is NOT the only natural choice.
 
  • #44
DrStupid said:
You can't change what already happened. Momentum has been defined that way in classical mechanics and the relativistic mass is a result of this definition.

Momentum is not universally defined as p=mv. In the Lagrangian formulation of classical mechanics, momentum is defined as the partial of the Lagrangian L(x, v) with respect to the velocity v [where x is a generalized coordinate and v=dx/dt is a generalized velocity]. This is actually a more general definition of momentum, one that works for angular momentum as well as linear momentum. This is a consequence of the Lagrangian forumaltion working with any choice of coordinates, it doesn't require any specialized coordinates. So if one uses angular coordinates with this formulation, one gets angular momentum, and when one uses linear coordinates, one gets linear momentum.
 
  • #45
The modern* approach to relativity is to work with coordinate-independent 4-dimensional vectors and tensors. Rest mass arises very naturally in this approach. The relevant equations are:

4-velocity:
$$ \textbf{V} = \frac{\mbox{d}\textbf{x}}{\mbox{d}\tau} $$
4-momentum:
$$ \textbf{P} = m \textbf{V} $$
4-force:
$$ \textbf{F} = \frac{\mbox{D}\textbf{P}}{\mbox{d}\tau} $$
where ##m## is rest mass and ##\tau## is proper time (both coordinate-independent invariants). What could be a more natural 4-d generalisation of 3-d Newtonian physics than that?*Actually it's been around for a long time now!
 
  • #46
pervect said:
Momentum is not universally defined as p=mv.

You are right. I should have been more precise: p=m·v is the definition in Newtonian mechanics.
 
Last edited:
  • #47
PAllen said:
All I claim is carrying over mv as a definition is NOT the only natural choice.

And all I claiming is that this definition has been carried over into special relativity before new formalisms were established.
 
  • #48
DrStupid said:
Relativistic mass doesn't need to make relativistic formulae look Newtonian.
Why this unhealthy obsession with making things look Newtonian. There is no good reason they should apart from in the Newtonian limit. Using relativistic mass is just a way of taking a part of a covariant object and splitting it into two quantities which do not transform covariantly. Now why would you want to do that - historical context aside.

DrStupid said:
That's not sufficient to define a frame-independent mass. Relativistic mass also fits to this definition.
No it does not. Velocity does not transform covariantly.

DrStupid said:
Relativistic mass doesn't need to make relativistic formulae look Newtonian.
But this is exactly what you have been arguing by wanting to hold on to p = mv!

DrStupid said:
And all I claiming is that this definition has been carried over into special relativity before new formalisms were established.
This I would just call a historical curiosity, not a good argument for using relativistic mass.
 
  • #49
Mass is a Casimir operator of the Poincare group. With the quantity that generates space-time translations, forming a Lorentz vector, called the four-momentum vector, it's defined as ##p_{\mu} p^{\mu}=m^2## (natural units, ##c=1##). That's the clearest theoretical definition I can think of within special relativity, and it clearly defines "invariant mass".
 
  • #50
Orodruin said:
Why this unhealthy obsession with making things look Newtonian.

I'm not aware of such an obsession.

Orodruin said:
No it does not. Velocity does not transform covariantly.

Nobody claimed something like that.

Orodruin said:
But this is exactly what you have been arguing by wanting to hold on to p = mv!

p=m·v is not a relativistic formula made look Newtonian but the original definition of momentum.

Orodruin said:
This I would just call a historical curiosity, not a good argument for using relativistic mass.

Nobody claimed that it is a good argument for using relativistic mass. It's just the origin of relativistic mass.
 
  • #51
DrStupid said:
p=m·v is not a relativistic formula made look Newtonian but the original definition of momentum

Yes, but this definition presumes you to have a notion of mass if you want to take it as a definition of momentum. There are two notions of mass in Newtonian mechanics, inertial mass and gravitational. Neither of those have a straight forward generalisation to special relativity. Gravitation is the domain of GR and inertial mass is direction dependent. If you want to keep the Newtonian version of the momentum formula you use it to define (relativistic) mass and not momentum and then you need to separately determine what momentum is.

DrStupid said:
Nobody claimed that it is a good argument for using relativistic mass. It's just the origin of relativistic mass.
I know its origin. It does not make it better as an archaic concept which has largely been abandoned.
 
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  • #52
Orodruin said:
Yes, but this definition presumes you to have a notion of mass if you want to take it as a definition of momentum.

No, the notion of mass is not required for but given by such basic definitions and assumptions.

Orodruin said:
There are two notions of mass in Newtonian mechanics, inertial mass and gravitational. Neither of those have a straight forward generalisation to special relativity.

The classical inertial mass (quantity of matter) is given by Def. 2 (p=m·v), Lex 2 (F=dp/dt), Lex 3 (F1+F2=0), isotropy, principle of relativity and the transformation. This implicit definition works with both Galilean and Lorentz transformation but the results are different.

The gravitational mass is given by Newton's law of gravitation together with the other conditions mentioned above. That doesn't work with Lorentz transformation.

Orodruin said:
inertial mass is direction dependent

That would violate isotropy.

Orodruin said:
If you want to keep the Newtonian version of the momentum formula you use it to define (relativistic) mass and not momentum and then you need to separately determine what momentum is.

I can use it to define mass and momentum.
 
  • #53
DrStupid said:
No, the notion of mass is not required for but given by such basic definitions and assumptions.
Sorry, but this is just silly. If you do not have a notion of mass, defining momentum in terms of it is a vacuous statement. Make up your mind. You either define momentum through the relation or you define mass through the relation - you cannot have it both ways.

DrStupid said:
That would violate isotropy.

But you do not have isotropy! The mass is moving! This is a well-known result in special relativity and was known very early on. This is the very result that led to the concepts of transversal and longitudinal (relativistic) mass - both now defunct.

DrStupid said:
I can use it to define mass and momentum.
No, you can't. You cannot use one relation to define two new concepts. What you can do is to use the second law to define momentum (it is what force is the change in and the momentum of something with zero velocity is zero). Then you can define mass through ##p = mv##. You do lose out on ##\vec F = m\vec a## (this is the mass which would be directionally dependent) and ##E_k = mv^2/2## if you do so. There is no a priori reason of preferring one over the other.
 
  • #54
Orodruin said:
If you do not have a notion of mass, defining momentum in terms of it is a vacuous statement.

No, it isn't because the definition of momentum doesn't exist in vacuum. I already mentioned the additional conditions which also needs to be fulfilled.

Orodruin said:
This is the very result that led to the concepts of transversal and longitudinal (relativistic) mass - both now defunct.

Don't confuse relativistic mass with longitudinal mass.

Orodruin said:
You cannot use one relation to define two new concepts.

Please read my posts more carefully. I do not use only one relation.
 
  • #55
DrStupid said:
Please read my posts more carefully. I do not use only one relation.
This is what you implied when you said this:
DrStupid said:
p=m·v is the definition in Newtonian mechanics
This statement by itself does not define mass and momentum. With regards to the rest it needs to be taken as a whole and you need the definition of momentum in terms of the second law before doing p = mv if you want p = mv to define mass.

DrStupid said:
Don't confuse relativistic mass with longitudinal mass.
By longitudinal, I mean transversal and vice versa. The relativistic mass is equal to the transversal mass, but this has to do with how force relates to acceleration - which is the alternative definition of inertial mass.

DrStupid said:
No, it isn't because the definition of momentum doesn't exist in vacuum. I already mentioned the additional conditions which also needs to be fulfilled.
Then you are not defining momentum through p = mv, you are defining mass through that relation. Either way works but you simply cannot define two quantities with one relation.
 
  • #56
I said this:

DrStupid said:
The classical inertial mass (quantity of matter) is given by Def. 2 (p=m·v), Lex 2 (F=dp/dt), Lex 3 (F1+F2=0), isotropy, principle of relativity and the transformation.

Discussions make no sense if such an important information is ignored.

Orodruin said:
By longitudinal, I mean transversal and vice versa. The relativistic mass is equal to the transversal mass, but this has to do with how force relates to acceleration - which is the alternative definition of inertial mass.

Longitudinal and relativistic mass are completely different things. We should discuss them separately.

Orodruin said:
Then you are not defining momentum through p = mv, you are defining mass through that relation. Either way works but you simply cannot define two quantities with one relation.

See above.
 
  • #57
DrStupid said:
Discussions make no sense if such an important information is ignored.
But this is something you added later! If you considered it so important, why did you not state it from the beginning?
 
  • #58
Well there is progress in physics from Galilei/Newton to Einstein/Minkowski, and for a good reason we nowadays define momentum using the fundamental discovery of the relation between conservation laws and symmetries of physical laws by Noether to unanimously define the observables. This concept works for both classical and quantum physics.

From this perspecitve the most simple way to (heuristically) derive energy and momentum of a free particle in special relativity is to define an action, built with the velocity ##\vec{v}=\mathrm{d} \vec{x}/\mathrm{d} t## only. This action should be Lorentz invariant, and the unique way to write it thus is
$$A[x]=-m c^2 \int \mathrm{d} t \sqrt{1-\dot{\vec{x}}^2/c^2}$$
with ##m=\text{const}## the (invariant) mass of the particle and ##c## the limiting speed of Minkowski space (aka speed of light in vacuo). From this the momentum follows from Noether's theorem applied to spatial translation invariance:
$$\vec{p}=\frac{\partial L}{\partial \dot{\vec{x}}}=\frac{m \dot{\vec{x}}}{\sqrt{1-\dot{\vec{x}}^2/c^2}}$$
and energy from Noether's theorem applied to temporal translation invariance:
$$E=H=\dot{\vec{x}} \cdot \vec{p}-L=\frac{m \dot{\vec{x}}^2}{\sqrt{1-\dot{\vec{x}}^2/c^2}}+m c^2 \sqrt{1-\dot{\vec{x}}^2/c^2}=\frac{m c^2}{\sqrt{1-\dot{\vec{x}}^2/c^2}}.$$
It's no surprise that due to the Lorentz covariance of the formalism
$$(p^{\mu})=(E/c,\vec{p})$$
builds a Lorentz four-vector, and mass is a scalar, independent of the velocity of the particle:
$$p_{\mu} p^{\mu}=m^2 c^2.$$
 
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  • #59
Orodruin said:
But this is something you added later! If you considered it so important, why did you not state it from the beginning?

No, not this way! These are some of my previous statement:

DrStupid said:
Alternatively mass could be defined implicit by other laws and definitions.

DrStupid said:
You can't simply choose a definition because it must be consistent with related laws and definitions which are universally accepted at the same time.

DrStupid said:
it results from the use of these equations

DrStupid said:
p=m·v is the definition of momentum in classical mechanics and the properties of mass and the equation for kinetic energy result from this definition (among other things).

I highlighted the plural for you. I made clear from the beginning that there is not only one relation that leads to relativistic mass. So don't try to blame me for your faults!
 
  • #60
DrStupid said:
I highlighted the plural for you. I made clear from the beginning that there is not only one relation that leads to relativistic mass. So don't try to blame me for your faults!
Seriously, chill. Just because you claimed p = mv was the definition of momentum when you are using it as the definition of mass is no reason to get upset. The fact remains that you cannot define two concepts with one relation. This is explicitly claimed by you:
DrStupid said:
I can use it to define mass and momentum.
but later in the post where you specify which definition of Newtonian mechanics you use it is clear that you have already defined momentum! Hence, p = mv is not your definition of momentum.

Apart from that, I suggest you read vanhees recent post.
 
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