Inertia

[Pete]

Request for opinion - Do you think there are any logical problems in defining the the term "inertia" to refer to the resistance of body to a change in mometum? The fact that inertial mass is defined as m = |p|/|v| seems to suggest this definition.

 

Pete

[Mr Skeptic]

Aren't inertial mass and inertia just synonyms?

[ajb]

Is that not how it is usually stated? What Mr Skeptic says is correct.

 

 

What I don't know is if the notion can be "extended" to include more general motions that do not obey Newton's laws.

[Pete]

Aren't inertial mass and inertia just synonyms?

The term inertia is a term which describes what mass is a measure of. But one doesn't say that "The inertial of a body is 12 grams". Some people like to say that inertia is a measure of a body's resistance to changes in velocity.

Is that not how it is usually stated? What Mr Skeptic says is correct.

Hard to say. At http://scienceworld.wolfram.com/physics/Inertia.html the definition is

 

inerrtia: The resistance to change in state of motion which all matter exhibits.

 

But at http://www.merriam-webster.com/dictionary/inertia the definition is

 

inertia - a property of matter by which it remains at rest or in uniform motion in the same straight line unless acted upon by some external force

 

At http://www.neutron.anl.gov/hyper-physics/inertia.html

 

inertia - inertia (în-ûr¹she)' date=' in physics, the resistance of a body to any alteration in its state of motion, i.e., the resistance of a body at rest to being set in motion or of a body in motion to any change of speed or of direction of motion.

What I don't know is if the notion can be "extended" to include more general motions that do not obey Newton's laws.

What kind of motion is that? Can you give me an example?

 

Newton used the term "quantity of motion" to refer to momentum. So in that sense I'm the above definition is what I'm thinking of (even though the author of that definition probably meant otherwise).

 

Pete

[ajb]

The "physics" definition is resistance to change of motion. The "maths" definition is resistance to change in momentum. I think the two are equivalent as momentum is a way of "quantifying" the motion. So,

 

[math]F(x(t)) = m a(x(t)) = m \frac{d^{2} x(t)}{dt^{2}}= \frac{d}{dt}p(x(t))[/math],

 

which is one of Newton's laws.

 

It is easy to think of motions that don't obey this. Consider a Lagrangian that is not quadratic in [math]\dot{x}[/math] or equivalently a Hamiltonian that is not quadratic in [math]p[/math]. The equations of motion you get are not those of Newtonian mechanics. Now, it may be true that "real" systems are always quadratic. But never mind that for now.

[Pete]

The "physics" definition is resistance to change of motion. The "maths" definition is resistance to change in momentum. I think the two are equivalent as momentum is a way of "quantifying" the motion. So,

 

[math]F(x(t)) = m a(x(t)) = m \frac{d^{2} x(t)}{dt^{2}}= \frac{d}{dt}p(x(t))[/math],

 

which is one of Newton's laws.

The reason I posted this in the relativity forum is because

 

[math]m a(x(t)) = m \frac{d^{2} x(t)}{dt^{2}}[/math]

 

doesn't hold. If it did then the two interpretations would be the same.

It is easy to think of motions that don't obey this. Consider a Lagrangian that is not quadratic in [math]\dot{x}[/math] or equivalently a Hamiltonian that is not quadratic in [math]p[/math]. The equations of motion you get are not those of Newtonian mechanics. Now, it may be true that "real" systems are always quadratic. But never mind that for now.

Seems to me that if this was an example of a "motion" then it would have to correspond to something physical. If not then such a Lagrangian cannot exist. In my opinion, a Lagrangian that has such a form but does not lead to an equation of motion which describes something physical is not a valid Lagrangian.

 

Pete

[ajb]

What does hold (in special relativity for sure) is a 4-vector version.

 

First we have to define a few things. I will assume you are happy with paths in Minkowski space-time as being parametrised by [math]\tau[/math]. Then we define the 4-velocity as

 

[math]U^{\mu}\frac{d x^{\mu}(\tau)}{d \tau}[/math],

 

it is just the vector tangent to the path [math]x^{\mu}(\tau)[/math].

 

Then, we can define the 4-momentum as

 

[math]P^{\mu} = m U^{\mu}[/math],

 

and then we define the 4-force to be

 

[math]f^{\mu} = \frac{d}{d \tau}P^{\mu}(\tau) = m \frac{d^{2}x^{\mu}(\tau)}{d \tau^{2}}[/math].

 

So again the mass is the "proportionality constant" between change of momentum and acceleration.

 

You can check that for small velocities this gives the usual Newton law. This then will give "m" to be the inertial mass from before.

[Pete]

What does hold (in special relativity for sure) is a 4-vector version.

Of course. That's basic SR and you need to explain the basics to me.

So again the mass is the "proportionality constant" between change of momentum and acceleration.

To be precise that is not true. What is true is that proper mass is the "proportionality constant" between change of 4-momentum and 4-acceleration. After all 4-velocity is not the time rate of change of 4-position. Its the rate of change of 4-position per unit proper time' date=' not [i']time[/i]. The 4-vector approach cannot work in all possible cases. It only works in special cases. The quantity that fully describes mass is the stress-energy-momentum tensor.

You can check that for small velocities this gives the usual Newton law. This then will give "m" to be the inertial mass from before.

Thanks' date=' but I was overly aware of that 23 years ago.

 

Note: Needless to say, I'm trying to avoid discussions which have to deal with [i']mass[/i] for obvious reasons and am thus I'm only interested in inertia. Using 4-velocity one can't even speak of an object at rest so I'm avoiding 4-vectors, escpecially since they have only a limited usage. Thanks.

 

Pete

Edited July 18, 2008 by Pete

[ajb]

From what I know, the 4-vector description here is fine for particles. It is when you want extended objects that the energy-momentum tensor is required. So, for fluids or fields etc, you do need to do more.

 

 

As I am sure you know the notion of a mass in general relativity is much more complicated. This is something I know even less about!

[Pete]

A friend of mine pointed out an error in the interpretation that I started with. He explains

I think this makes no sense. If you apply a force to a particle or a system,

you get a change of momentum, and the amount of this change depends on the magnitude of the force and on the length of time it acts, but it does not depend on the mass of the system. So merely looking at momentum changes produced by a force does not tell you the inertia.

I guess its best to think of inertia as that quantity that defines momentum for a given velocity.

From what I know, the 4-vector description here is fine for particles.

That depends on how the term particle is defined. If it is possible for an object to have internal stesses, regardless of its size, then a 4-descrition can't work in general.

 

Pete

[ajb]

Particle is a point mass, nothing internal going on.

[Pete]

Particle is a point mass, nothing internal going on.

I assume that you mean "point object" since a photon is considered a particle and I imagine that you'd say that it has no mass at all, correct?

 

What do you mean by "point" mass/object? Do you mean an object whose size is zero? The term particle has also been defined as an indivisible unit of matter. The two are not neccesarily the same.

 

In classical mechanics the term particle applies to objects which can be considered small with respect to the system being considered (e.g. see Newtonian Mechanics, A.P. French, pages 1-4). E.g. an atom will behave like a particle under various circumstanecs. But if one looks closely it can't be considered a particle. A planet can be considered a particle if one is only concerned with celestial motion. I don't think that it has ever been proven that an elementary particle such as an electron has zero size. I doubt that one can locate an electron with infinite precision and thus speaking of it as being a "point" is not quite clear. String theory actually supposes that objects like electons are not pointlike, that they are like strings and thus have dimensions.

 

In classical mecanics it is troublesome to think of point charged particles due to its infinite self energy. So one may argue that true point particles are those which have zero size this becomes a serious problem in classical mechanics. And the subject I wish to addess here is classical mechanics, not quantum mechanics.

 

Question: Would you consider a photon or an EM wave to have inertia?

 

Pete

Edited July 18, 2008 by Pete

[Mr Skeptic]

Question: Would you consider a photon or an EM wave to have inertia?

 

Definitely. It couldn't have momentum without having inertia, by definition.

[ajb]

As we are talking about classical special relativistic systems, it is fine to think of particles as a point. I tend not to use the notion of rest mass and relativistic mass etc. Mass means (in the old school language) rest mass.

 

Generally inertia is something to to with acceleration, the photon is not really accelerated as it always travels at c. However, there is a recoil force on an atom when it loses energy via a photon. So, I am not really sure if one thinks of a photon as having inertia.

 

In general relativity, a better notion could be geodesic deviation. That is how near by geodesics separate or converge. I don't know if you could define the "inertia" of a photon to be the geodesic deviation of null geodesics. I believe this has been looked at, but I have no idea what is known/accepted/common knowledge to experts etc. Of course, this really boils down to calculating the curvature.

[Pete]

As we are talking about classical special relativistic systems, it is fine to think of particles as a point.

In such a case all charged particle's will have an infinite electromagnetic mass. If a particle has no charge then one has to accelerate it by other means, e.g. by scattering other particles off of it or, if it has a magnetic moment, by placing it an an inhomogenous magnetic field and delfect it using the neutron's magnetic moment.

I tend not to use the notion of rest mass and relativistic mass etc. Mass means (in the old school language) rest mass.

I take it that you never consider things other than particles, would that be correct? In relativity it is know that stress has inertia (its the purpose of this thread to investigate what that means).

Generally inertia is something to to with acceleration, the photon is not really accelerated as it always travels at c.

I question that interpretation, hence the purpose of this thread. E.g. French defines inertial mass as m = p/v and as such it must have inertia, yet it is unable to accelerate. I believe that this is the case in any place in which one defines the mass of a tardyon.

However, there is a recoil force on an atom when it loses energy via a photon. So, I am not really sure if one thinks of a photon as having inertia.

Always nice to test the boundaries of one's basic concepts, yes?

In general relativity, a better notion could be geodesic deviation. That is how near by geodesics separate or converge. I don't know if you could define the "inertia" of a photon to be the geodesic deviation of null geodesics.

Geodesics deviation applies to locally inertial frames too and one doesn't assign inertia to such frames.

 

Pete

Edited July 19, 2008 by Pete

[ajb]

Honestly inertia is not a concept I have thought about much or looked into at all. Other than the mass of particles. My background is quantum field theory and geometry rather than classical general relativity. At least that is my excuse!

 

Thinking about the idea of the inertia of a photon, the closes thing I can think of is it's energy, as [math]p =\frac{E}{c}[/math], thus any change in momentum is directly proportional to the change in energy.

 

Does the graviton have inertia? Again I am not even sure what that means.

[Pete]

Honestly inertia is not a concept I have thought about much or looked into at all.

I'd wager that most physicists don't look into it past physics 101

Other than the mass of particles. My background is quantum field theory and geometry rather than classical general relativity. At least that is my excuse!

In this thread I'm sticking within special relativity. The mechanics of contniuos media is addressed in special relativity. That's where it can get tricky.

Thinking about the idea of the inertia of a photon, the closes thing I can think of is it's energy, as [math]p =\frac{E}{c}[/math], thus any change in momentum is directly proportional to the change in energy.

That works for a photon. So I guess you're saying that a photon has inertia due to its energy?

 

How do you think you'd define mass density if you wanted/needed to?

Does the graviton have inertia? Again I am not even sure what that means.

Oooo! Good question. Could a graviton interact with other kinds of particles, e.g. can they scatter off other particles?

 

Pete

[ajb]

For photons I am thinking (please note the inverted commas) "[math]E = m[/math] " and [math]p = E[/math] (in geometric units).

 

Gravitons interact will all other matter and each other due to GR being non-linear. They are massless and so travel at c. I am thinking there energy would also be some kind of measure of the "inertia".

 

I don't ever remember inertia being discussed at all in my undergraduate studies in physics. Most of my graduate studies have been on QFT and then geometry. Any thing I have read about GR does not discuss inertia in any detail. Inertia does not seem to be a concept used widely past meaning "mass".

 

Define mass density. I think that is tricky in general. In classical mechanics, it is easy. I am not sure about in special relativity as mass does not have an intrinsic meaning, other than in the rest frame.

Edited July 19, 2008 by ajb

[Pete]

Define mass density. I think that is tricky in general. In classical mechanics, it is easy. I am not sure about in special relativity as mass does not have an intrinsic meaning, other than in the rest frame.

I don't follow. Why should mass need to have an intrinsic meaning in order for mass density to be defined?

 

Pete

[ajb]

Thinking that maybe it is not very well defined in all inertial frames, i.e. invariantly. Maybe I am wrong, but that is my first impression.

 

The "correct" replacement is the energy-momentum tensor.

 

The closed thing I can think of is something like

 

[math]\rho = \sqrt{-g} \: T_{\mu \nu}X^{\nu}X^{\mu}[/math],

 

with [math]X[/math] being a time-like vector.

 

I think [math]\rho[/math] would have the interpretation as the "mass-energy density". Well, for sure, it defines (mathematically) an object that can be integrated over the manifold.

Edited July 19, 2008 by ajb

[Pete]

Thinking that maybe it is not very well defined in all inertial frames, i.e. invariantly. Maybe I am wrong, but that is my first impression.

Why do you think it should be invariant? Is this related to your preferance to speak only in terms of proper mass?

The "correct" replacement is the energy-momentum tensor.

I was speaking in terms of the stress-energy-momentum tensor in the first place because mass densities cannot be be defined with anything other than a second rank tensor.

 

Pete

[ajb]

Sorry, I edited my last post a little.

 

If something is not formulated invariantly, then it can have no "intrinsic" meaning.

 

I think the stress energy-momentum tensor is the same as the stress-energy-momentum tensor.

 

[math]T_{\mu \nu} = \frac{-1}{\sqrt{-g}}\frac{\delta S}{\delta g^{\mu \nu}}[/math]

[Pete]

Sorry, I edited my last post a little.

 

If something is not formulated invariantly, then it can have no "intrinsic" meaning.

Meaningful in what sense? Even if there is no "intrinsic" meaning there is a physical meaning. I.e. we know that time intervals and spatial intervals are not inot intrinsic yet they are physically very meaningful. One needs to be careful about this too. For instance I can define the term mass of particle as measured by observer by

 

m_obs = P*U_obs/c²

 

where U_obs is the observer's 4-velocity and P is the 4-momentum of the particle and m_obs is the relativistic mass of the particle as measured by the observer. This is a geometric quantity and is invariant in the sense that it remains unchanged by an arbitrary Lorentz transformation. Different observers measure different masses.

 

A similar idea holds for the electric and magnetic field. I.e. there are quantities known as electric field as measured by observer/I] and magnetic field as measured by observer which are both (observer dependant) 4-vectors.

I think the stress energy-momentum tensor is the same as the stress-energy-momentum tensor.

They mean the exact same thing. I use the more appropriate term stress-energy-momentum tensor because it more fully decribes this tensor, i.e. the components are energy, momentum and stress, not just energy and momentum.

Thinking that maybe it is not very well defined in all inertial frames, i.e. invariantly. Maybe I am wrong, but that is my first impression.

 

The "correct" replacement is the energy-momentum tensor.

 

The closed thing I can think of is something like

 

[math]\rho = \sqrt{-g} \: T_{\mu \nu}X^{\nu}X^{\mu}[/math],

 

with [math]X[/math] being a time-like vector.

 

I think [math]\rho[/math] would have the interpretation as the "mass-energy density". Well, for sure, it defines (mathematically) an object that can be integrated over the manifold.

Different [math]X[/math] gives different [math]\rho[/math]'s. Normally the time-like vector is the 4-velocity of an observer and in that sense is a scalar. If the [math]X[/math] is intrepreted to be a unit vector which defines a coordinate system then the quantity becomes coordinate dependant and non-invariant in that sense.

 

The [math]\rho[/math] as you have defined it will equal the energy density and not the mass density. Mass density is defined as the ratio of momentum density to velocity. Even if you let v -> 0 it will be different than energy density. Remember that E = mc² does not always hold.

 

What is the purpose of [math] \sqrt{-g}[/math] in that definition?

 

I seem to have gotten off track of the purpose of this thread, i.e. inertia. Sorry.

 

A word on invariance here - The components of a tensor quanity are defined as the scalars which result by inputing unit 4-vectors/1-forms which define the coordinate system. Although this is still a map of 4-vectors/1-forms to numbers they are still scalars in that sense. However they are not invariants in the sense that a change in coordinate system requires a change in unit vectors.

 

Pete

Edited July 19, 2008 by Pete
multiple post merged

[ajb]

[math]\rho[/math] as I have defined it is independent of any coordinates used, but of course depends on the observer via the [math]X[/math] as your rightly say.

 

Thing's don't have to be observer independent, as such, but they should not depend on the coordinates use. In special relativity, where we are only considering linear changes of coordinates this is equivalent to Lorentz invariance. (as a passive transformation)

 

 

The [math]\sqrt{-g}[/math] is needed so that we have a tensor density which (at least formally) can be integrated over the space-time manifold. You can either include it here or in the definition of your integration measure. To me it makes more sense to construct tensor densities. (I am sure you know this.)

 

How is the mass-density related to the energy-momentum tensor?

 

[math]T_{\mu \nu} = T(\partial_{\mu}, \partial_{\nu})[/math] depends on the coordinates employed. Again I am sure we both know this.

[Pete]

[math]\rho[/math] as I have defined it is independent of any coordinates used, but of course depends on the observer via the [math]X[/math] as your rightly say.

 

Thing's don't have to be observer independent, as such, but they should not depend on the coordinates use.

The term observer in SR is normally taken to be synonymous with coordinate system in the sense that different observers correspond to different inertial frames of reference.

The [math]\sqrt{-g}[/math] is needed so that we have a tensor density which (at least formally) can be integrated over the space-time manifold.

Please elaborate. Why can't components be intergrated over spacetime. MTW defines mass-energy density as

 

[math]\rho = T(U, U)[/math]

 

where [math]U[/math] is the observer's 4-velocity. MTW didn't include [math]\sqrt{-g}[/math] in their deifnition.

You can either include it here or in the definition of your integration measure. To me it makes more sense to construct tensor densities. (I am sure you know this.)

Not really. I never payed much attention to tensor densities. Its one of those things that are on my back burner. As I recall they are used to keep integrals invariant under coordinate transformations, right? I've never seen them used in the texts that I learned relativity from, not that they aren't important.

How is the mass-density related to the energy-momentum tensor?

[math]\rho^i = T^{0i}/v^i[/math]

 

Different orientations lead to different directions of momentum for this reason. I.e. momentum is not parallel to velocity in general. Only for closed systems, isolated systems or point (structureless) particles.

[math]T_{\mu \nu} = T(\partial_{\mu}, \partial_{\nu})[/math] depends on the coordinates employed. Again I am sure we both know this.

 

Sure. That's what I meant when I wrote components of a tensor quanity are defined as the scalars which result by inputing unit 4-vectors/1-forms which define the coordinate system.

 

Pete

Edited July 20, 2008 by Pete