Does mass become infinite near the speed of light?

[DrStupid]

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.

 

[PAllen]

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.

 

[Ibix]

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.

 

[DrStupid]

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.

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).

"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.

"$P^0$ already has a name"

And so does relativistic mass.

"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.

"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.

"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.

 

[PAllen]

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.

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.

 

[PAllen]

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.

 

[DrStupid]

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.

 

[PAllen]

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.

 

[pervect]

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.

 

[DrGreg]

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!

 

[DrStupid]

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.

 

[DrStupid]

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.

 

[Orodruin]

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.

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.

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!

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.

 

[vanhees71]

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".

 

[DrStupid]

Why this unhealthy obsession with making things look Newtonian.

I'm not aware of such an obsession.

No it does not. Velocity does not transform covariantly.

Nobody claimed something like that.

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.

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.

 

[Orodruin]

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.

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.

 

[DrStupid]

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.

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.

inertial mass is direction dependent

That would violate isotropy.

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.

 

[Orodruin]

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.

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.

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.

 

[DrStupid]

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.

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.

You cannot use one relation to define two new concepts.

Please read my posts more carefully. I do not use only one relation.

 

[Orodruin]

Please read my posts more carefully. I do not use only one relation.

This is what you implied when you said this:

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.

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.

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.

 

[DrStupid]

I said this:

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.

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.

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.

 

[Orodruin]

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?

 

[vanhees71]

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:
$$p_{\mu} p^{\mu}=m^2 c^2.$$

 

[DrStupid]

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:

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

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.

it results from the use of these equations

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!

 

[Orodruin]

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:

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.

 

[DrStupid]

This is explicitly claimed by you:

end of discussion