Is Relativistic Mass Still Relevant in Modern Physics Discussions?

[Aer]

So, if you want to talk about the "mass" of an object in relativity, you have to specify, or at least have it already be understood from context, which of these properties you really want to deal with. You can't have them all.

We were working with the assumption that the total energy (including kinetic) defined the mass of the system. And yes, it does appear problematic - but if you analyze things in the rest frame of the object, there is no problem.

 

[JesseM]

Now wait - a certain amount of energy is known to be used to accelerate the object. Since the force changes depending on the frame, does that mean the energy changes?

Presumably you can consider an infinitesimal acceleration from an infinitesimal input of energy, so you don't have to worry about this when defining resistance to acceleration in the object's own rest frame.

"total rest energy" is ambiguous.

Why? It's just the sum of the kinetic energy and rest masses of all the components with the potential energy between them as seen in the rest frame of the object as a whole.

 

[JesseM]

No evidence for what? In his last sentence before the Acknowledgments he writes:

"We can thus tell our students with confidence that kinetic energy has weight, not just as a theoretical expectation, but as an experimental fact."

Do you agree or disagree with this?

Does that not imply that the kinematic energy must be measured relative to the rest frame of the gravitational potential? Otherwise, what is the meaning of kinetic energy? The object must have motion relative to something to have kinetic energy.

Again, they're talking about the kinetic energy of parts of a compound object as seen in the compound object's rest frame. The introduction says:

The principle of equivalence—the exact equality of inertial and gravitational
mass—is a cornerstone of general relativity, and experimental tests of the universality
of free fall provide a large set of data that must be explained by any theory
of gravitation. But the implication that energy contributes to gravitational mass
can be rather counterintuitive. Students are often willing to accept the idea that
potential energy has weight—after all, potential energy is a rather mysterious
quantity to begin with—but many balk at the application to kinetic energy. Can
it really be true that a hot brick weighs more than a cold brick?

General relativity offers a definite answer to this question, but the matter is
ultimately one for experiment. Surprisingly, while observational evidence for the
equivalence principle has been discussed for a variety of potential energies, the
literature appears to contain no analysis of kinetic energy. The purpose of this
paper is to rectify this omission, by reanalyzing existing experimental data to look
for the “weight” of the kinetic energy of electrons in atoms. I will then try to
reconcile the results with the occasional (and not completely unreasonable) claim
that “objects traveling at the speed of light fall with twice the acceleration of
ordinary matter.”

Clearly the example of a hot brick vs. a cold brick involves a compound object whose parts can have greater or lesser kinetic energy in the object's rest frame, and likewise electrons in atoms are part of a compound object, so presumably he's talking about the gravitational mass (which by the equivalence principle is the same as inertial mass) of the atoms in their own rest frame, which can change as the kinetic energy of the electrons in this frame changes.

 

[EnumaElish]

Actually, isn't the m in E=mc^2 relativistic mass (if applied to other reference frames)? I thought the invariant mass version of the equation was E^2=p^2c^2+m^2c^4.

To my understanding, E=p^2c^2+m^2c^4=\gamma mc^2=m_r c^2; so yes, the m in E = mc² is relativistic mass. Aer would say that the quantity m_r should not be called "relativistic mass" because the second word of this definition starts with the 13^th letter of the contemporary English alphabet immediately followed by the 1^st letter of the said alphabet, and ends by repeating twice the 19^th letter of this symbolic collection. I guess an obvious alternative is to call it "shorthand for \gamma m."

 

[EnumaElish]

I wish I had the thread on hand, but the "relativistic mass" for a force from one frame to another was γ³m was it not? I think it was you who came up with this, though it may have been someone else - I forget.

You are referring to Doc Al's post under thread "Speed."

Anyway, in this thread, it was established that "relativisitic mass" was simply γm, where m is rest mass, was it not?

Yes, based on your own post and the following posts under the same thread.

 

[pervect]

I think pervect summed it up with his post pointing to: http://arxiv.org/PS_cache/gr-qc/pdf/9909/9909014.pdf which states there is no experimental evidence to back up the assertion.

The quoted reference says quite the opposite, actually.

 

[Aer]

The quoted reference says quite the opposite, actually.

I think the problem here is, we are referring to two very different things. Gravity effectively curves space which is why light bends in a gravitation field although there is no actual force on the photon. The same is applied to anything in a gravitation field, it follows the curvature of space, but since these objects have a so-called mass, we attribute a force to gravity.

Then we are talking about SR in which the curvature of space is defined as FLAT. And people like JesseM and others are saying that kinetic energy and the such contribute to an objects mass because of the force required on the body - and they refer back to gravity and GR. It is not the same thing, and any analysis of attributing a force to a moving body will show this as the definition of "relativistic mass" changes for different situations.

 

[JesseM]

I think the problem here is, we are referring to two very different things. Gravity effectively curves space which is why light bends in a gravitation field although there is no actual force on the photon. The same is applied to anything in a gravitation field, it follows the curvature of space, but since these objects have a so-called mass, we attribute a force to gravity.

Then we are talking about SR in which the curvature of space is defined as FLAT. And people like JesseM and others are saying that kinetic energy and the such contribute to an objects mass because of the force required on the body - and they refer back to gravity and GR. It is not the same thing, and any analysis of attributing a force to a moving body will show this as the definition of "relativistic mass" changes for different situations.

You don't need a detailed consideration of GR here, at least not when considering a small composite object that does not itself curve spacetime much. You can just consider the inertia of the object when you try to accelerate it in empty space using a non-gravitational force, and whatever its inertial mass in this situation, the equivalence principle tells you that this will be the same as its gravitational mass in the presence of a large object like a planet that does curve spacetime in a significant way. Or to put it another way, if the object is sitting on a scale in an elevator that is accelerating through empty space at 1G, the scale's reading should be the same as if the elevator was sitting on the surface of the earth, according to the equivalence principle.

Do you deny that in flat spacetime, SR predicts that the inertia of a compound object will be proportional to its total rest energy? Do you deny that the equivalence principle says there shouldn't be a difference between its inertial mass and its gravitational mass?

 

[Aer]

Your posts are getting more and more stupid every time you post.

You don't need a detailed consideration of GR here, at least not when considering a small composite object that does not itself curve spacetime much.

I am not considering the curvature of space by the object! Holy crap. The curvature of space is by the Earth and the object exists in this curvature.

You can just consider the inertia of the object when you try to accelerate it in empty space using a non-gravitational force, and whatever its inertial mass in this situation, the equivalence principle tells you that this will be the same as its gravitational mass in the presence of a large object like a planet that does curve spacetime in a significant way.

You must think that I don't know what the equivalence principle is and will let you get away with this retardation of the principle.

equivalence principle:states that there is no experiment a person could conduct in a small volume of space that would distinguish between a gravitational field and an equivalent uniform acceleration. This principle is the foundation of General Relavity.

your definition:the equivalence principle tells you that this [the inertia of the object when you try to accelerate it in empty space using a non-gravitational force] will be the same as its gravitational mass in the presence of a large object like a planet that does curve spacetime in a significant way.

What you may be referring to is the "weak equivalence principle" or "universality of free fall" because tests of the weak equivalence principle are those that verify the equivalence of gravitational mass and inertial mass (e.g. Dropping metal balls of different mass from the Tower of Pisa - a la Galileo).

So, effectively you are attributing a force to gravity just like I said. What is the force on a photon? You can only attribute a force to objects that already have "rest mass". You are replacing mass for acceleration in your definition of the equivalence principle and you can only do this when you do what I said above - attribute a force to gravity.

Don't even try to tell me that you know everything there possibly is to know about gravity! No scientist knows everyting there is possible to know about gravity - that is why there is debate on this issue in physics!

Back to what we do know about gravity: the acceleration is always the same, regardless of the mass of an object. In fact, an object can have no mass and will still experience the same acceleration (e.g. photons). This acceleration is due to the curvature of space, not an actual force.

Do you deny that in flat spacetime, SR predicts that the inertia of a compound object will be proportional to its total rest energy?

Of course not, if you say the total rest energy is the total rest mass form of energy.

Do you deny that the equivalence principle says there shouldn't be a difference between its inertial mass and its gravitational mass?

Like I elaborated on above, you are referring to the weak equivalence principle.

Do you deny that objects with relative motion (kinetic energy) do not behave as if they have more mass?

 

[JesseM]

Your posts are getting more and more stupid every time you post.

Again, Aer, please don't be a jerk.

I am not considering the curvature of space by the object! Holy crap.

I didn't say you were, I was just qualifying my own statement about the equivalence principle, since the argument wouldn't quite work for a very large object like a planet.

You can just consider the inertia of the object when you try to accelerate it in empty space using a non-gravitational force, and whatever its inertial mass in this situation, the equivalence principle tells you that this will be the same as its gravitational mass in the presence of a large object like a planet that does curve spacetime in a significant way.

You must think that I don't know what the equivalence principle is and will let you get away with this retardation of the principle.

equivalence principle:states that there is no experiment a person could conduct in a small volume of space that would distinguish between a gravitational field and an equivalent uniform acceleration. This principle is the foundation of General Relavity.

Seems to me that's exactly the same as my statement that "if the object is sitting on a scale in an elevator that is accelerating through empty space at 1G, the scale's reading should be the same as if the elevator was sitting on the surface of the earth, according to the equivalence principle" (assuming the elevator is considered to be a small volume of space).

your definition:the equivalence principle tells you that this [the inertia of the object when you try to accelerate it in empty space using a non-gravitational force] will be the same as its gravitational mass in the presence of a large object like a planet that does curve spacetime in a significant way.

I didn't say that this was the equivalence principle, just that it's a necessary consequence of it.

What you may be referring to is the "weak equivalence principle" or "universality of free fall" because tests of the weak equivalence principle are those that verify the equivalence of gravitational mass and inertial mass (e.g. Dropping metal balls of different mass from the Tower of Pisa - a la Galileo).

Huh? According to wikipedia the weak equivalence principle says that "The trajectory of a falling test body depends only on its initial position and velocity, and is independent of its composition." That is obviously not what I was talking about.

So, effectively you are attributing a force to gravity just like I said.

No I'm not. Read my statement about the elevator again, all I'm saying is that if the object is sitting on a scale in an elevator undergoing 1G acceleration in flat space, the reading must be the same as if the same elevator was sitting on the surface of the earth. Do you agree that an observer in a small elevator will not be able to experimentally distinguish whether he is undergoing 1G acceleration in flat space or whether he's at rest on the surface of the earth? If so, that's all you need to demonstrate that the reading on the scale will be the same, which means the inertial mass is the same as the gravitational mass.

What is the force on a photon? You can only attribute a force to objects that already have "rest mass".

Well, good thing we weren't talking about photons, we were talking about compound objects that have a rest frame. And I never said anything about treating gravity as a force, as you say, general relativity does not treat it as such.

You are replacing mass for acceleration in your definition of the equivalence principle

Again, huh? What specific quote are you referring to when you say I was "replacing mass for acceleration"?

Don't even try to tell me that you know everything there possibly is to know about gravity! No scientist knows everyting there is possibly to know about gravity - that is where there is debate on this issue in physics!

But I'm only talking about what general relativity predicts about what should happen. Anyway, most physicists expect that quantum gravity will replicate the predictions of general relativity in macroscopic domains where the spacetime curvature isn't too large.

Back to what we do know about gravity: the acceleration is always the same, regardless of the mass of an object. In fact, an object can have no mass and will still experience the same acceleration (e.g. photons). This acceleration is due to the curvature of space, not an actual force.

Sure, but different objects still have different gravitational masses, which can be measured by seeing the force they exert on a scale sitting in a gravitational field. And again, the equivalence principle shows the reading on the scale in a gravitational field must be the same as the reading on a scale in an elevator undergoing uniform acceleration (which in that case is measuring inertial mass).

Do you deny that in flat spacetime, SR predicts that the inertia of a compound object will be proportional to its total rest energy?

Of course not, if you say the total rest energy is the total rest mass form of energy.

What do you mean when you say "rest mass form of energy"? Do you agree that relativity predicts a compressed spring will have slightly more inertia than the same spring in its relaxed state, since it has a slightly larger rest energy? Do you agree that SR predicts a hot brick will have slightly more inertia than a cold one, again because it has a slightly higher rest energy?

Do you deny that the equivalence principle says there shouldn't be a difference between its inertial mass and its gravitational mass?

Like I elaborated on above, you are referring to the weak equivalence principle.

No I'm not, the quote from wikipedia shows that the weak equivalence principle is only about the trajectory of a falling object. I'm using the principle that all laws of physics should look the same in a small elevator undergoing uniform acceleration (in which a scale will measure inertial mass) as they do in the same elevator at rest in a gravitational field of equivalent strength (in which a scale will measure gravitational mass).

Do you deny that objects with relative motion (kinetic energy) do not behave as if they have more mass?

As jtbell pointed out, it's not so simple--an object in motion will be easier to accelerate in some directions then others. To avoid this issue, I'm only talking about the inertial mass of a compound, bound object in its own rest frame. In this frame, SR predicts that its inertial mass will be proportional to its total energy, and the equivalence principle predicts that its inertial mass must equal its gravitational mass. This is also noted by the authors of the paper pervect pointed to when they say "The principle of equivalence—the exact equality of inertial and gravitational mass—is a cornerstone of general relativity".

 

[Hurkyl]

What is the force on a photon?

Exactly what the definition of force says: the time derivative of momentum. Recall that an object does not need to have a nonzero rest mass in order to have nonzero momentum.

To my understanding, E=p^2c^2+m^2c^4=\gamma mc^2=m_r c^2; so yes, the m in E = mc² is relativistic mass.

You're forgetting that "E" could refer to a variety of things. For example:

E_rest = m_rest c²
E_total = γ m_rest c² = m_relativistic c²
E_kinetic = (γ - 1) m_rest c²

And, incidentally, you were looking for E² = (pc)² + (mc²)². (Where E is total energy, m is rest mass)

 

[JesseM]

She wasn't saying that's the definition: she was saying that the equivalence principle "tells you that". Since you can determine the inertia of an object by conducting an experiment in a small volume of space, she's right.

I'm a he. But otherwise, yeah, that's what I was saying--my statement wasn't supposed to be a definition of the equivalence principle, just a consequence of it.

 

[Hurkyl]

Doh, you caught it before I deleted it.

I thought I remembered you calling yourself Jessica once... apparently I was mistaken!

 

[learningphysics]

Aer, an example that has been posted by myself and others in this thread that you have not responded to:

A object heated up has greater rest mass than the same object cooled down. Do you agree with this or not?

 

[pervect]

I think the problem here is, we are referring to two very different things. Gravity effectively curves space which is why light bends in a gravitation field although there is no actual force on the photon. The same is applied to anything in a gravitation field, it follows the curvature of space, but since these objects have a so-called mass, we attribute a force to gravity.

Gravity curves/distorts space-time, not just space. It's true that one can view small objects (test masses or test light rays) as following geodesics in space-time. However, the various Christoffel symbols with time components that describe "curved"/distorted space-time can be reasonably interpreted as "forces" - for instance, the Christoffel symbol \Gamma^x{}_{tt} can be regarded as a static force in the 'x' direction, equivalent to the Newtonian gravitational force. Similarly the sum of the Christoffel symbols \Gamma^x{}_{yt}+\Gamma^x{}_{ty} can reasonably regarded as a coriolis force in the 'x' direction due to motion in the 'y' direction. (Because of symmetry concern, both of the symbols in the above sum are equal).

Light beams near a massive body curve as a result of multiple Christoffel symbols, the closest English translation to the math in my opinion is to say that part of their curvature is due to "forces" (Christoffel symbols which include time components) - the other part of the curvature of light is due to Christoffel symbols _without_ time components, which can be regraded as the curvature of _space_ (not space-time, because none of these Christoffel symbols have any time components).

The exact mathematical expression is the geodesic equation

<br /> \frac{d^2 x^i}{d \tau^2} + \Gamma^i{}_{jk} \left( \frac{dx^j}{d \tau} \right) \left( \frac{dx^k}{d \tau} \right) = 0<br />

Then we are talking about SR in which the curvature of space is defined as FLAT. And people like JesseM and others are saying that kinetic energy and the such contribute to an objects mass because of the force required on the body - and they refer back to gravity and GR. It is not the same thing, and any analysis of attributing a force to a moving body will show this as the definition of "relativistic mass" changes for different situations.

Kinetic energy does contribute to the invariant mass of a system of particles, even in SR.

Consider a closed system of particles that do not interact with the outside universe, but only with each other, and which interact with each other only when the occupy the same point in space at the same time (no fields).

Note that this simple model can be generalized to include particles that interact via fields, but making this generalization requires including the momentum and energy stored in the fields. One can alternatively model the fields as an exchange of fictitious particles.

Calculate (in geometric units where c=1) the quantity E^2 - p^2, where E is the total energy of the system of particles in some frame, and p is the total momentum of the system of particles in the same frame. You will find that this quantity is frame independent (for a _closed_ system), and is, by defintion, the invariant mass of the system.

You will find that the invariant mass of the system of particles is NOT the sum of the invariant masses of its components. The invariant mass of the system includes contributions due to the kinetic energy of the particles.

You will have to pull some clever tricks to model the equivalent of a solid sphere containing at hot gas which is _not_ expanding without the use of fields, but it's possible. You will have to devise an exchange system of fictitious particle which mimics tension to pull this off.

 

[EnumaElish]

You're forgetting that "E" could refer to a variety of things. For example:

E_rest = m_rest c²
E_total = γ m_rest c² = m_relativistic c²
E_kinetic = (γ - 1) m_rest c²

And, incidentally, you were looking for E² = (pc)² + (mc²)². (Where E is total energy, m is rest mass)

That should have been E² as you indicate. Also, the equation I was thinking about was E = m c^{2} + K = \gamma m c^2 as stated in Aer's class notes. Here m = m_rest, K is kinetic energy and \gamma={1}/{\sqrt{1-({v}/{c})^{2}}}. If one defines m_\text{relativistic} = \gamma m_\text{rest} then E_total = m_relativisticc². It can be argued that m_relativistic is not "mass," it is simply m + K/c². And if one were to define m_relativistic as "mass" then one also has to remember that unlike rest mass, m_relativistic is directional.

 

[pervect]

There are some simple SR thought experiments that can illustrate the connection between energy and "passive" gravitational mass, via the equivalence principle. The idea is to consider what sort of forces are required to counteract the "gravity" of an accelerating space-ship, and to apply the equivalence principle.

Experiment #1

Conisider a rocket accelerating with a proper acceleration equal to a. (A proper acceleration is the acceleration measured by an instantaneously co-moving observer, i.e. an obsrever in an inertial frame moving at the same speed as the rocket, with the same velocity, that is not accelerating).

Suspend a stationary particle with invariant mass m and charge q with an electric field so that it's acceleration relative to the rocket is zero. Show that m*a = q*E, where E is the electric field, i.e. E = m*a/q

----rocket----->
       m-->E

The rocket accelerates to the right. The mass m is suspended by an electric field that also points to the right, so that it does not accelerate relative to the rocket.


Experiment #2

Consider the same experiment, except that mass m is not stationary. Everything else remains the same, the rocket accelerates at the same rate, and the electric field points in the same direction as the first figure.

a) Suppose the particle is moving in a direction that's at right angles to the rocket's trajectory. The particle is moving with velocity v. Show that the electric field required to keep the particle from accelerating relative to the rocket becomes E = gamma*m*a/q, where gamma = 1/sqrt(1-(v/c)^2)

I'll omit the detailed calculations for now. People who get stuck might research "transverse mass". Perhaps someone else would like to post the detailed calculations.

b) Suppose the particle is moving in the same direction as the spaceship is accelerating with velocity v Show that the electric field required to keep the particle from accelerating relatie to the rocket is the same as case a), i.e. E=gamma*m*a/q

hint:

Relative to the comoving inertial obsrver, we require the relativistic difference v(t+dt) - a*dt = v(t) in order that the particles velocity stay constant relative to the rocket. This means that the v(t+dt) is the relativistic sum of v(t) and a*dt, i.e.

v(t+dt) = (v(t)+a*dt)/(1+v(t)*a*dt/c^2)

Show that in the limit of small dt, this implies that dv/dt = (1-(v/c)^2)*a

 

[Paulanddiw]

This is an interesting thread. When I measure my weight, I am a rest (v=0), so I'd expect my weight to be the sum of the rest masses of the particles that make up my body.

 

[pervect]

This is an interesting thread. When I measure my weight, I am a rest (v=0), so I'd expect my weight to be the sum of the rest masses of the particles that make up my body.

If you fuse two deuterium atoms to make a helium atom, the sum of masses of the two deuterium atoms is not the same as that of the resulting helium atom.

The same principle is at work with chemical binding energies and with the heat energy generated by chemical processes that is at work with nuclear binding energies and the energies (of various forms) that are generated by nuclear processes.

However, the magnitudes are a lot different - the chemical binding energies are so small that they do not make any practical difference to the mass of a body, and so are routinely ignored.

 

[pervect]

Experiment #2

Consider the same experiment, except that mass m is not stationary. Everything else remains the same, the rocket accelerates at the same rate, and the electric field points in the same direction as the first figure.

a) Suppose the particle is moving in a direction that's at right angles to the rocket's trajectory. The particle is moving with velocity v. Show that the electric field required to keep the particle from accelerating relative to the rocket becomes E = gamma*m*a/q, where gamma = 1/sqrt(1-(v/c)^2)

I'll omit the detailed calculations for now. People who get stuck might research "transverse mass". Perhaps someone else would like to post the detailed calculations.

I'll fill in the detailed calculations.

Let the direction that the particle moves in be the x direction, and let the direction that the spaceship accelerates be the y-direction.

Then the y component of the momentum (in relativistic units where c=1) in an inertial co-moving frame is

<br /> p_y = \frac{m v_y }{\sqrt{1-v_x^2 - v_y^2}}<br />

where v_x = v is the velocity of the particle in the x direction, and v_y = 0 at t=0. Because the spaceship is accelerating, v_y will be a function of time, even though its inital value is zero.

The y-component of the force on the particle is just F = qE = dp_y / dt

Now
<br /> \frac{dp_y}{dt} = \left(\frac{dp_y}{dv_y}\right)\left( \frac{dv_y}{dt}\right)<br />

But we know that d v_y/d_t = a

We can differentiate the expression of p_y as a function of v_y, and make the above substitution for d v_y/dt to get

<br /> \frac{d p_y}{dt} = \frac{m a (1-v_x^2)}{(1-v_x^2 - v_y^2)^{\frac{3}{2}}}<br />

We wish to evaluate this expression at t=0.
Substituting v_x = v and v_y = 0 and simplifying yields the final result.

F = q*E = gamma*m*a

from which E = gamma*m*a/q follows directly.

 

[Aer]

Your posts are getting more and more stupid every time you post.

Again, Aer, please don't be a jerk.

I was merely providing my expert analysis in trends.

I am not considering the curvature of space by the object! Holy crap.

I didn't say you were, I was just qualifying my own statement about the equivalence principle, since the argument wouldn't quite work for a very large object like a planet.

You are referring to gravitational mass! That is what the weak equivalence principle is all about. See http://www.npl.washington.edu/eotwash/equiv.html , gravitational mass is only mentioned in the Newton analysis and the weak equivalence princple. Nowhere is "gravitational mass" mentioned in the GR (i.e. equivalence principle) interpretation. Notice all you need to know is highlighted in red on the page, perhaps specially for you: spacetime itself is curved.[/color]

This is what you and pervect do not understand. Gravity is not a force in General Relativity. The idea of gravity as a force is a Newtonian concept that Einstein abondanded with his General Relativity Theory.

equivalence principle:states that there is no experiment a person could conduct in a small volume of space that would distinguish between a gravitational field and an equivalent uniform acceleration. This principle is the foundation of General Relavity.

Seems to me that's exactly the same as my statement that "if the object is sitting on a scale in an elevator that is accelerating through empty space at 1G, the scale's reading should be the same as if the elevator was sitting on the surface of the earth, according to the equivalence principle" (assuming the elevator is considered to be a small volume of space).

There is no experiment that can be done that would detect a difference between the two - however, gravity is not a force. That is the entire point of the equivalence principle. To explain how gravity is like a force, yet it is not. It is merely the curvature of spacetime. All objects (with and without mass) follow the same curvature. Now if you had a light particle bouncing back and forth on a scale in your accelerating frame, it is not going to measure a mass on the scale. However, photons still follow the curvature of spacetime created by gravity.

I didn't say that this was the equivalence principle, just that it's a necessary consequence of it.

But it is not a necessary consequence because all objects regardless of mass with follow the same path from the curvature of spacetime as defined in General Relativity. The gravitational mass and inertial mass equivalence is explained by the "weak equivalence principle" which Galileo proved. Again, check http://www.npl.washington.edu/eotwash/equiv.html .

Huh? According to wikipedia the weak equivalence principle says that "The trajectory of a falling test body depends only on its initial position and velocity, and is independent of its composition." That is obviously not what I was talking about.

This only proves what I've thought all along. You have no idea what the hell you are talking about.

So, effectively you are attributing a force to gravity just like I said.

No I'm not.

Yes you are.

And I never said anything about treating gravity as a force, as you say, general relativity does not treat it as such.

Good, you are learning.

Sure, but different objects still have different gravitational masses, which can be measured by seeing the force they exert on a scale sitting in a gravitational field. And again, the equivalence principle shows the reading on the scale in a gravitational field must be the same as the reading on a scale in an elevator undergoing uniform acceleration (which in that case is measuring inertial mass).

You still don't get that this has no relevance to the relativistic mass subject we are discussing.

Do you agree that SR predicts a hot brick will have slightly more inertia than a cold one, again because it has a slightly higher rest energy?

Thermal energy is kinetic energy on the atomic level (not subatomic level which is quantum physics). Since kinetic energy has no effect on an objects rest mass, neither will thermal energy. And yes I realize there is a long history of assuming thermal energy is considered apart of the rest energy in E₀=mc².

All this talk about gravitational mass is useless. It has nothing to do with relativistic mass which is the issue here. You say that relativistic mass is useful. I say it is not useful. agrees with me:

In the earlier years of relativity, it was the relativistic mass that was taken to be the "correct" notion of mass, and the invariant mass was referred to as the rest mass. Gradually, as special relativity gave way to general relativity and found application in quantum field theory, it was realized that the invariant mass was the more useful quantity and scientists stopped referring to the relativistic mass altogether.

The accepted usage in the scientific community today (at least in the context of special relativity) considers the invariant mass to be the only "mass", while the concept of energy has replaced the relativistic mass. In popular science and basic relativity courses, however, the relativistic mass is usually presented, most likely due to its conceptual simplicity.[/url]

Just in case you need a summary: Kinetic energy does not add to the mass of an object, relativistic or not because the concept of "relativistic mass" is wrong.

 

[JesseM]

You are referring to gravitational mass! That is what the weak equivalence principle is all about. See http://www.npl.washington.edu/eotwash/equiv.html , for example:

Other predictions

* The equivalence of inertial mass and gravitational mass: This follows naturally from freefall being inertial motion.

Or see the section on the equivalence principle from the hyperphysics website:

Experiments performed in a uniformly accelerating reference frame with acceleration a are indistinguishable from the same experiments performed in a non-accelerating reference frame which is situated in a gravitational field where the acceleration of gravity = g = -a = intensity of gravity field. One way of stating this fundamental principle of general relativity is to say that gravitational mass is identical to inertial mass.

The physicists who wrote the http://arxiv.org/PS_cache/gr-qc/pdf/9909/9909014.pdf that we were discussing earlier also describe the GR equivalence principle in this way in the introduction:

The principle of equivalence—the exact equality of inertial and gravitational mass—is a cornerstone of general relativity, and experimental tests of the universality of free fall provide a large set of data that must be explained by any theory of gravitation. But the implication that energy contributes to gravitational mass can be rather counterintuitive.

Anyway, all that I'm really talking about here is the reading on a scale for an object placed in a gravitational field. If you don't want to call this reading "gravitational mass", you don't have to, although I think most physicists would, even in the context of general relativity. All I'm saying is that this reading will always be the same as the reading for the same object sitting on a scale which is undergoing uniform acceleration in deep space, a reading which is usually understood to be a measurement of the "inertial mass". Regardless of whether you disagree about the usage of terminology, do you disagree with this physical claim about the readings of scales in different settings? If you don't disagree with me on any physical question, then all you're doing is quibbling over the standard meaning of certain terminology, and the quotes above suggest you're wrong about that anyway.

This is what you and pervect do not understand. Gravity is not a force in General Relativity. The idea of gravity as a force is a Newtonian concept that Einstein abondanded with his General Relativity Theory.

Of course I understand this, and I'm sure pervect does too. I never said anything about gravity being a force in GR. If you think that talking about an object's gravitational mass implies you're treating gravity as a force, I disagree (and I think the quotes above suggest physicists would disagree too), and I never intended that implication.

There is no experiment that can be done that would detect a difference between the two - however, gravity is not a force. That is the entire point of the equivalence principle. To explain how gravity is like a force, yet it is not. It is merely the curvature of spacetime. All objects (with and without mass) follow the same curvature. Now if you had a light particle bouncing back and forth on a scale in your accelerating frame, it is not going to measure a mass on the scale.

A box containing a photon bouncing back and forth between mirrored walls would weigh a little more than an empty box on a scale undergoing uniform acceleration in deep space, and the increase in inertial mass of the box should be equal to the energy of the photon (as measured in the box's rest frame).

But it is not a necessary consequence because all objects regardless of mass with follow the same path from the curvature of spacetime as defined in General Relativity. The gravitational mass and inertial mass equivalence is explained by the "weak equivalence principle" which Galileo proved. Again, check http://www.npl.washington.edu/eotwash/equiv.html .

OK, I admit I was a bit fuzzy on the weak vs. strong equivalence principle definition, that page suggests that in general relativity, the weak principle (that inertial mass and gravitational mass are equal, ie that the reading on a scale would be the same whether the scale was undergoing uniform acceleration in deep space or at rest with respect to a gravitational field) would just be a special case of the strong equivalence principle (that the results of all experiments would be the same whether an observer was undergoing uniform acceleration in deep space or at rest with respect to a gravitational field). But even though the weak equivalence principle predates relativity, it is still a part of relativity as suggested by the quotes I gave above, and it is in fact an obvious consequence of the strong equivalence principle. So again, my original physical argument, that if inertial mass is proportional to total energy than gravitational mass (ie the reading on a scale for a bound system in a gravitational field) must be too, still stands.

Do you agree that SR predicts a hot brick will have slightly more inertia than a cold one, again because it has a slightly higher rest energy?

Thermal energy is kinetic energy on the atomic level (not subatomic level which is quantum physics). Since kinetic energy has no effect on an objects rest mass, neither will thermal energy. And yes I realize there is a long history of assuming thermal energy is considered apart of the rest energy in E₀=mc².

Kinetic energy certainly has an effect on the rest mass of a compound object, provided you use the standard definition of "rest mass" for compound objects. But to avoid quibbling over definitions, do you agree that relativity predicts the inertia of a compound object whose parts have greater total kinetic energy in the compound object's rest frame (like a hot brick) will be larger than than the inertia of the same compound object when its parts have lesser total kinetic energy in the compound object's rest frame (like a cold brick)?

All this talk about gravitational mass is useless. It has nothing to do with relativistic mass which is the issue here. You say that relativistic mass is useful. I say it is not useful.

My argument is not about the utility of relativistic mass at all, it's just about the fact that the measured weight/inertia of a compound object is proportional to its total energy in its own rest frame (which is not the same as the sum of the relativistic masses of all its components in this frame, since there may be potential energy involved as well). You have been denying this for a long time--I originally jumped into this thread just to question the following statement by you:

Then we have people like pmb_phy claiming a contained gas's weight is a measure of the rest mass of the particles PLUS the kinetic energy they possess. UMMM - NO! That's wrong, the weight is only a measure of the rest mass of the particles and nothing more.

Do you now admit that you've been wrong all along, and that the weight (as measured by a physical scale) of a bound system is not just a measure of the rest mass of the particles that make up the system, but is in fact a measure of the system's total energy (including kinetic and potential) in its own rest frame, at least according to the theory of relativity?

agrees with me:

In the earlier years of relativity, it was the relativistic mass that was taken to be the "correct" notion of mass, and the invariant mass was referred to as the rest mass. Gradually, as special relativity gave way to general relativity and found application in quantum field theory, it was realized that the invariant mass was the more useful quantity and scientists stopped referring to the relativistic mass altogether.

The accepted usage in the scientific community today (at least in the context of special relativity) considers the invariant mass to be the only "mass", while the concept of energy has replaced the relativistic mass. In popular science and basic relativity courses, however, the relativistic mass is usually presented, most likely due to its conceptual simplicity.[/url]

Just in case you need a summary: Kinetic energy does not add to the mass of an object, relativistic or not because the concept of "relativistic mass" is wrong.

I don't know what you mean by the "mass of an object"--as always, I would say that the "rest mass" of a compound object is defined to be equal to its total energy in its own rest frame, and this is not in general the same thing as the sum of the "relativistic masses" of its components, so the quote above is irrelevant, since I haven't been defending the use of "relativistic mass" in the first place. I provided many quotes earlier in this thread to show that this was indeed the standard way of defining "rest mass" for a compound object whose parts may be in motion relative to one another, and you provided zero to support your claim that I was wrong. In any case, regardless of issues of terminology, it is certainly true that according to relativity, the weight (as measured by a physical scale) of a compound object is proportional to its total energy in its own rest frame, regardless of whether the scale is accelerating in free space or is at rest in a gravitational field.

 

[Aer]

Well, it is mentioned on the wikipedia entry on general relativity, for example:

Or see the section on the equivalence principle from the hyperphysics website:

In both instances, the context of the "weak equivalence principle" is meant.

This follows naturally from freefall being inertial motion.


And of course, what I originally said: "Tests of the weak equivalence principle are those that verify the equivalence of gravitational mass and inertial mass. These experiments demonstrate that all objects fall at the same rate when the effect of air resistance is either eliminated or negligible."

But this is beside the point - gravitational mass has nothing to do with our discussion of relativistic mass!

This is what you and pervect do not understand. Gravity is not a force in General Relativity. The idea of gravity as a force is a Newtonian concept that Einstein abondanded with his General Relativity Theory.

Of course I understand this, and I'm sure pervect does too.

Well, of course you are wrong about pervect, but I'll accept that you understand if you say so. Anyway, from pervect "However, the various Christoffel symbols with time components that describe 'curved'/distorted space-time can be reasonably interpreted as 'forces'". See, he thinks there is some 'interpreted forces' involved with gravity.

A box containing a photon bouncing back and forth between mirrored walls would weigh a little more than an empty box on a scale undergoing uniform acceleration in deep space, and the increase in inertial mass of the box should be equal to the energy of the photon (as measured in the box's rest frame).

There is no reason to believe this! A photon is massless. It doesn't make sense to say a photon has 'no mass in its rest frame' because the the frame at v=c is not defined in relativity. However, a photon does have energy which is a result from quantum mechanics NOT relativity. For relativity, E=γmc² and γ=1/0 which is undefined. Relativists use E²=(pc)²+(mc²)² which is derived from E=γmc² and p=γmv. They then take the QM result that pc=hf and plug it into E²=(pc)²+(mc²)² with m=0 to get E=hf. Nowhere do they say that a photon has relativitic mass in order to explain the energy. In fact, relativistic mass m_r is undefined for a photon since m_r=γm = 1/0*0=0/0=undefined!

So again, my original physical argument, that if inertial mass is proportional to total energy than gravitational mass (ie the reading on a scale for a bound system in a gravitational field) must be too, still stands.

And what value does this argument even have with relativistic mass? You seem to have gone off topic just a bit.

Kinetic energy certainly has an effect on the rest mass of a compound object, provided you use the standard definition of "rest mass" for compound objects.

I don't accept the defintion of rest mass you've provided - it's really that simple. You think all energy is included in the rest mass when in fact we've shown with the photon example that to think of energy and mass interchangably is very wrong.

Mass and energy are only interchangable on the quantum level. So this excludes thermal energy as well as kinetic energy within compound objects.

But to avoid quibbling over definitions, do you agree that relativity predicts the inertia of a compound object whose parts have greater total kinetic energy in the compound object's rest frame (like a hot brick) will be larger than than the inertia of the same compound object when its parts have lesser total kinetic energy in the compound object's rest frame (like a cold brick)?

Any inertia of the parts within the compound object have no bearing on the whole. Your logic is flawed.

My argument is not about the utility of relativistic mass at all, it's just about the fact that the measured weight/inertia of a compound object is proportional to its total energy in its own rest frame

Which would have to include the relativistic masses of all the objects contained within! Energy and mass are only interchangable on the quantum level. It was thought long ago that they were interchangable on the macro level because there was no concept of the quantum level. Modern relativity has revised this thinking. Kinetic energy, thermal energy, potential energy all have no bearing on the mass of a particle. The only mass is the rest mass - period and this mass is only interchangable at the quantum level. So you and anyone you can find to support your position, is wrong and applying outdated thinking.

Do you now admit that you've been wrong all along

NO! You obviously need to do a little reading of modern relativity.

I don't know what you mean by the "mass of an object"

Of course you don't!

The mass of any object is a function of the gravitational field the object creates. Or rather, the strength of the gravitational field an object produces is a function of the object's mass. What clearer definition do you need?

I would say that the "rest mass" of a compound object is defined to be equal to its total energy in its own rest frame

And you and everyone you can find to support this, would be wrong.

Think of it this way: you can put 10 free particles in a volume of space that are all whirling around with great velocity. They do not create any greater curvature of the spacetime around them than if they were just at rest in spacetime because the curvature each particle creates is a function of the rest mass of each. Now put a box around them and call it a compound object. You'll have to assume that the box is massless compared to the particles (these are very heavy particles!). Now the curvature of spacetime around them has not increased any yet the rest mass, by your definition, of the compound object (box with particles inside) is much greater than the rest mass of the sum of the rest mass of each particle combined. And of course, all this makes absolutely no sense whatsoever unless "mass" is the ill-defined "relativistic mass" in which case we are not talking about "rest mass" at all.

If you cannot understand this, you are hopeless.

It shouldn't have even taken such a drawn out discussion.

But nevertheless, I predict you are going to post another ignorant reply.

 

[learningphysics]

What astounds me about this thread is that Aer, you're making claims and disagreeing with concepts that all relativists accept.

This thread really has nothing to do with the use of relativistic mass. It is simply JesseM desperately trying to help you understand a simple point: The inertia of a compound object in its rest frame is proportional to its total energy in its rest frame... This is equivalent to the definition of rest mass as total energy in the rest frame divided by c^2.

You have a hot brick at rest... the force you have to exert to accelerate it is greater than the force required to accelerate the same brick with the same acceleration when it's cold at rest. Do you agree or disagree?

 

[Aer]

What astounds me about this thread is that Aer, you're making claims and disagreeing with concepts that all relativists accept.

All relativists? Maybe retarded relativists...

This thread really has nothing to do with the use of relativistic mass. It is simply JesseM desperately trying to help you understand a simple point: The inertia of a compound object in its rest frame is proportional to its total energy in its rest frame... This is equivalent to the definition of rest mass as total energy in the rest frame divided by c^2.

No, this thread is all about relativistic mass. Read the OP! Also, this discussion stems from JesseM's very first post in this thread:

Then we have people like pmb_phy claiming a contained gas's weight is a measure of the rest mass of the particles PLUS the kinetic energy they possess. UMMM - NO! That's wrong, the weight is only a measure of the rest mass of the particles and nothing more.

Are you sure about that? the last question from this FAQ says that the apparent inertia of a black box filled with a gas will increase as the temperature increases, which I would think would mean the weight would increase as well

rest mass of the particles PLUS the kinetic energy = relativistic mass!

You have a hot brick at rest... the force you have to exert to accelerate it is greater than the force required to accelerate the same brick with the same acceleration when it's cold at rest. Do you agree or disagree?

Of course I do not agree with this outdated concept. You must remember that mass and energy were thought to be interchangable on the macro-level long ago. That is not the case today as it was realized it is only valid at the quantum level. However, there are still misguided fools out there such as yourself.

 

[learningphysics]

All relativists? Maybe retarded relativists...

No, this thread is all about relativistic mass. Read the OP! Also, this discussion stems from JesseM's very first post in this thread:


rest mass of the particles PLUS the kinetic energy = relativistic mass!

Of course I do not agree with this outdated concept. You must remember that mass and energy were thought to be interchangable on the macro-level long ago. That is not the case today as it was realized it is only valid at the quantum level. However, there are still misguided fools out there such as yourself.

You really don't have a clue do you? You haven't even worked through the math or gone through any of the derivations...

Find me one physicist with a phd that doesn't say that the rest mass of a hot brick is greater than the rest mass of the same brick when it's cold.

That wikipedia quote you used just shows you really don't understand what JesseM has been saying, and neither do you understand the quote.

 

[JesseM]

In both instances, the context of the "weak equivalence principle" is meant.

Call it whatever you want. My original point was just that the equivalence principle of general relativity implies that the weight of an object in a gravitational field must be proportional to the inertial mass, and since special relativity predicts that the inertia of a compound object is proportional to its total energy in the object's own rest frame, the equivalence principle alone is enough to tell us that the weight of a compound object in a gravitational field is also proportional to its total energy, without any further consideration of the details of GR.

But this is beside the point - gravitational mass has nothing to do with our discussion of relativistic mass!

You seem to have this confused idea that I have been trying to make some point about relativistic mass. I haven't--go back and look at my previous posts on this thread, my point was always focused on inertia and weight being proportional to total energy for a compound object, and also the terminological issue that "rest mass" is traditionally defined as total energy for a compound object. The discussion has dragged on for as long as it has because you keep making statements which seem to contradict this, such as this, from post #9:

Then we have people like pmb_phy claiming a contained gas's weight is a measure of the rest mass of the particles PLUS the kinetic energy they possess. UMMM - NO! That's wrong, the weight is only a measure of the rest mass of the particles and nothing more.

Or this, from post #21:

the mass of an object is its total energy in its rest frame

should be "the mass of an object is the sum of all its constituents' rest masses".

Or post #24:

Just to be clear, are you claiming for sure that the inertia of a black box filled with gas won't appear to increase when the temperature increases, or are you just not certain either way?

OK - let me state that I cannot be certain, but according to mass as it is defined, the answer would be that the mass of the gas would not appear to increase.

Or post #28:

No, not if you use the definition given by Tom Roberts, where the "rest mass" of a composite object is defined as its total energy (divided by c^2, presumably) in the center-of-mass frame.

Very well, then his definition of "rest mass" is not the proper definition of "rest mass"

Or post #29:

that says that the inertia of a composite object (its resistance to being accelerated) will be a function of its total energy, not just the energy of the rest mass of all the constituent particles:

The acceleration of an object is only properly measured in it's rest frame which implies the total energy is the rest energy.

Or post #41:

I'm confident that the same force will not accelerate your car as quickly as if the objects in your car were moving slower (in the center-of-mass frame of the car), ie the inertia of the car will be different, since two expert sources have said this is true.

I'd be more than happy to see these sources. This would do nothing but undermine the foundations of SR and probably neccessitate modifications to GR.

I could keep going, but you get the point. You have been consistently denying both my physical claim that the inertia and weight of a compound object is proportional to its total energy, and also my terminological claim that the standard definition of "rest mass" for a compound object is its total energy in the object's rest frame. Neither of these points has jack squat to do with relativistic mass! I am happy to accept the judgement of most physicists that "relativistic mass" is more confusing then helpful and should therefore be avoided in physics discussions, and in fact I've pointed this out on other threads prior to this one--see this post, for example.

Well, of course you are wrong about pervect, but I'll accept that you understand if you say so. Anyway, from pervect "However, the various Christoffel symbols with time components that describe 'curved'/distorted space-time can be reasonably interpreted as 'forces'". See, he thinks there is some 'interpreted forces' involved with gravity.

Well, I'm not a GR expert, perhaps there is an alternate way to interpret the mathematics, or perhaps pervect is just saying that the effects of curved spacetime come to resemble forces in some limit (as they must, since GR is supposed to reduce to Newtonian gravity in certain limits). I'm remembering something that physicist Kip Thorne says on p. 397 of his book Black Holes and Time Warps:

Is spacetime really curved? Isn't it conceivable that spacetime is actually flat, but the clocks and rulers with which we measure it, and which we regard as perfect in the sense of Box 11.1, are actually rubbery? Might not even the most perfect clocks slow down or speed up, and the most perfect of rulers shrink or expand, as we move them from point to point and change their orientations? Wouldn't such distortions of our clocks and rulers make a truly flat spacetime appear curved?

Yes.

[He then goes on to describe how a field that changes the length of rulers and speed of clocks could give exactly the same predictions as the usual curved spacetime picture in the case of a black hole]

What is the real, genuine truth? Is spacetime really flat, as the above paragraphs suggest, or is it really curved? To a physicist like me this is an uninteresting question because it has no physical consequences. Both viewpoints, curved spacetime and flat, give precisely the same predictions for any measurements performed with perfect rulers and clocks, and also (it turns out) the same predictions for any measurements performed with any kind of physical apparatus whatsoever. For example, both viewpoints agree that the radial distance between the horizon and the circle in Figure 11.1, as measured by a perfect ruler, is 37 kilometers. They disagree as to whether that measured distance is the "real" distance, but such a disagreement is a matter of philosophy, not physics. Since the two viewpoints agree on the results of all experiments, they are physically equivalent. Which viewpoint tells the "real truth" is irrelevant for experiments; it is a matter for philosophers to debate, not physicists. Moreover, physicsts can and do use the two viewpoints interchangeably when trying to deduce the predictions of general relativity.

...

The curved spacetime paradigm is based on three sets of mathematically formulated laws: Einstein's field equation, which describes how matter generates the curvature of spacetime; the laws which tell us that perfect rulers and perfect clocks measure the lengths and the times of Einstein's curved spacetime; and the laws which tell us how matter and fields move through curved spacetime, for example, that freely moving bodies travel along straight lines (geodesics). The flat spacetime paradigm is also based on three sets of laws: a law describing how matter, in flat spacetime, generates the gravitational field; laws describing how that field controls the shrinkage of perfect rulers and the dilation of the ticking rates of perfect clocks; and laws describing how the gravitational field also controls the motions of particles and fields through flat spacetime.

...

The exemplars of the curved spacetime paradign include the calculation, found in most relativity textbooks, by which one derives Schwarzschild's solution to the Einstein field equations, and the calculations by which Israel, Carter, Hawking, and others deduced that a black hole has no "hair." The flat spacetime exemplars include textbook calculations of how the mass of a black hole or other body changes when gravitational waves are captured by it, and calculations by Clifford Will, Thibault Damour, and others of how neutron stars orbiting each other generate gravitational waves (waves of shrinkage-producing field).

...

The flat spacetime paradigm's laws of physics can be derived, mathematically, from the curved spacetime paradigm's laws, and conversely. This means that the two sets of laws are different mathematical representations of the same physical phenomena, in somewhat the same sense as 0.001 and 1/1000 are different mathematical representations of the same number. However, the mathematical formulas for the laws look very different in the two representations, and the pictures and exemplars that accompany the two sets of laws look very different.

As an example, in the curved spacetime paradigm, the verbal picture of Einstein's field equation is the statement that "mass generates the curvature of spacetime." When translated into the language of the flat spacetime paradigm, this field equation is described by the verbal picture "mass generates the gravitational field that governs the shrinkage of rulers and the dilation of the ticking of clocks." Although the two versions of the Einstein field equation are mathematically equivalent, their verbal pictures differ profoundly.

It is extremely useful, in relativity research, to have both paradigms at one's fingertips. Some problems are solved most easily and quickly using the curved spacetime paradigm; others, using flat spacetime. Black-hole problems (for example, the discovery that a black hole has no hair) are more amenable to curved spacetime techniques; gravitational-wave problems (for example, computing the waves produced when two neutron stars orbit each other) are most amenable to flat spacetime techniques. Theoretical physicists, as they mature, gradually build up insight into which paradigm will be best for which situation, and they learn to flip their minds back and forth from one paradigm to the other, as needed. They may regard spacetime as curved on Sunday, when thinking about black holes, and as flat on Monday, when thinking about gravitational waves. ... Since the laws that underlie the two paradigms are mathematically equivalent, we can be sure that when the same physical situation is analyzed using both paradigms, the predictions for the results of experiments will be identically the same. We thus are free to use the paradigm that best suits us in any given situation.

A box containing a photon bouncing back and forth between mirrored walls would weigh a little more than an empty box on a scale undergoing uniform acceleration in deep space, and the increase in inertial mass of the box should be equal to the energy of the photon (as measured in the box's rest frame).

There is no reason to believe this! A photon is massless. It doesn't make sense to say a photon has 'no mass in its rest frame' because the the frame at v=c is not defined in relativity. However, a photon does have energy which is a result from quantum mechanics NOT relativity. For relativity, E=?mc² and ?=1/0 which is undefined. Relativists use E²=(pc)²+(mc²)² which is derived from E=?mc² and p=?mv. They then take the QM result that pc=hf and plug it into E²=(pc)²+(mc²)² with m=0 to get E=hf. Nowhere do they say that a photon has relativitic mass in order to explain the energy. In fact, relativistic mass m_r is undefined for a photon since m_r=?m = 1/0*0=0/0=undefined!

Again, I said nothing about the photon's relativistic mass, only that its total energy (which, as you say, is given by E=hf) contributes to the inertia/weight of the box, according to relativity. Any physicist would agree that this is what relativity predicts.

So again, my original physical argument, that if inertial mass is proportional to total energy than gravitational mass (ie the reading on a scale for a bound system in a gravitational field) must be too, still stands.

And what value does this argument even have with relativistic mass? You seem to have gone off topic just a bit.

Again, look back over my old posts, you'll see I was never focused on anything related to "relativistic mass", my focus has always been on pointing out that you are making claims about weight/inertia that disagree with the predictions of relativity, and also that you are not using the standard definition of "rest mass" for compound objects. You didn't give me a definite answer, do you agree or disagree "that if inertial mass is proportional to total energy than gravitational mass (ie the reading on a scale for a bound system in a gravitational field) must be too"?

Mass and energy are only interchangable on the quantum level. So this excludes thermal energy as well as kinetic energy within compound objects.

I have always said that I wish to avoid talking about the "mass" of a compound object since we can't even agree on how that's defined, and just talk about physical questions like what a scale will read when you put that compound object on top of it. Relativity predicts that the reading on the scale will be proportional to the sum of the rest masses of all the components plus the sum of their kinetic and potential energies, as seen in the compound object's rest frame. I don't know, and don't really care, whether you'd call this an "interchange of mass and energy". But if you disagree with this claim about the reading on a scale, then you're just ignorant of the predictions of relativity. I'm sure you could do a purely classical relativistic calculation to show this, like considering a box full of ball bearings which are bouncing around in a box and imparting little impulses (which could be analyzed using relativistic kinematics) to the scale whenever they hit the box's floor, and then calculating the average reading on the scale as a function of the average velocity of the ball bearings in the box's rest frame.

But to avoid quibbling over definitions, do you agree that relativity predicts the inertia of a compound object whose parts have greater total kinetic energy in the compound object's rest frame (like a hot brick) will be larger than than the inertia of the same compound object when its parts have lesser total kinetic energy in the compound object's rest frame (like a cold brick)?

Any inertia of the parts within the compound object have no bearing on the whole. Your logic is flawed.

It's not just my logic, it's also Einstein's--remember, the example of a hot brick weighing more than a cold one was from one of his papers.

My argument is not about the utility of relativistic mass at all, it's just about the fact that the measured weight/inertia of a compound object is proportional to its total energy in its own rest frame

Which would have to include the relativistic masses of all the objects contained within!

No, there is no need to make use of "relativistic mass" when calculating the total energy of the compound object. One could just calculate E = \sqrt{m^2 c^4 + p^2 c^2} for each component and then add them up, along with whatever potential energies are involved.

I don't know what you mean by the "mass of an object"

The mass of any object is a function of the gravitational field the object creates. Or rather, the strength of the gravitational field an object produces is a function of the object's mass. What clearer definition do you need?

As I understand it, all forms of energy contribute to the curvature of spacetime--so a hot planet would produce a slightly stronger gravitational field than a colder but otherwise equivalent planet, and likewise a compressed spring would produce a slightly stronger gravitational field than a relaxed version of the same spring.

I would say that the "rest mass" of a compound object is defined to be equal to its total energy in its own rest frame

And you and everyone you can find to support this, would be wrong.

How can "everyone" be wrong about an issue of how a term is defined? The symbol-string R-E-S-T-M-A-S-S has no inherent meaning, it means whatever physicists choose it to mean. Even if you have defined "rest mass" for a single particle, that does not lead you to a single unique definition of "rest mass" for a collection of particles which do not share a common rest frame, you have to make a choice of how you want to define rest mass for such a compound object. What I am saying is that the standard definition used by physicists is that it means the total energy of the compound object in its rest frame, and I provided a number of credible sources to show that this is the standard definition.

Think of it this way: you can put 10 free particles in a volume of space that are all whirling around with great velocity. They do not create any greater curvature of the spacetime around them than if they were just at rest in spacetime because the curvature each particle creates is a function of the rest mass of each.

As I understand it, general relativity says that all forms of energy contribute to something called the "stress-energy tensor" which determines the curvature of spacetime. For example, see this post by physicist John Baez where he's discussing how kinetic energy and potential energy contribute to the stress-energy tensor--at the end he says, in response to a comment by someone else on the group:

>It would seem that only "kinetic" energy contributes
>to gravitation in GR. Is that correct?

No, both kinetic and potential energy contribute.

And of course, all this makes absolutely no sense whatsoever unless "mass" is the ill-defined "relativistic mass" in which case we are not talking about "rest mass" at all.

I have said over and over that it is total energy in the compound object's rest frame, not "relativistic mass", which determines the inertia and which is defined as the compound object's "rest mass". Do you think the notion of total energy is ill-defined?

 

[JesseM]

rest mass of the particles PLUS the kinetic energy = relativistic mass!

It's true that you can call this the "relativistic mass" if you wish, but there's no need to do so. You could also just use the equation E^2 = m^2 c^4 + p^2 c^2, where m is the rest mass and p is the relativistic momentum, and you will get the same answer for the total energy as if you had used the equation E = Mc^2 where M is the "relativistic mass". Remember, the dispute over "relativistic mass" is just an aesthetic one about whether the term is misleading or not, it's not as if any calculation involving relativistic mass will actually give different results from an analogous calculation that doesn't, and any statement involving relativistic mass can be replaced with an equivalent one involving only concepts like rest mass, relativistic momentum and energy.

Of course I do not agree with this outdated concept.

Again, it's only "outdated" for aesthetic reasons, it's not as if physicists using relativistic mass made any different predictions about the results of any actual physical experiments than physicists who don't. Both would agree that the inertia of a compound object is equal to total energy, regardless of whether they calculate this by summing E = \sqrt{m^2 c^4 + p^2 c^2} + P or by summing E = M c^2 + P for each component (P is the potential energy of each component).

 

[Garth]

The term 'relativistic mass' requires the addition of 'kinetic energy' to 'rest mass'. As kinetic energy is frame dependent this is a frame dependent (3+1)D space + time perspective. In this perspective the energy required to accelerate a body to high velocity reappears as an apparent increase in inertial mass according to E = mc²- "the faster it goes the harder it is to push".

The term 'mass' to mean simply 'rest mass' as an invariant quantity is consistent with the 4D energy-momentum of a particle being conserved and frame invariant. This is a frame independent 4D space-time perspective. Therefore this may seem the more 'pure' form to a relativist working with frame independent space-time. The energy used to accelerate an object to high velocity is redefined and absorbed by the time dilation suffered by that object as observed by the 'stationary' observer. (The 'gamma' factor)

However physics is actually done by a physicist locked into one particular frame, or another, and not 'frame independent'. Therefore from the perspective of a real experimenter the use of 'relativistic mass' may be the more obvious to use. It is a matter of choice; so long as you know the full implications of the perspective you select both systems should be equivalent.

Garth

 

[learningphysics]

The term 'relativistic mass' requires the addition of 'kinetic energy' to 'rest mass'. As kinetic energy is frame dependent this is a frame dependent (3+1)D space + time perspective. In this perspective the energy required to accelerate a body to high velocity reappears as an apparent increase in inertial mass according to E = mc²- "the faster it goes the harder it is to push".

The term 'mass' to mean simply 'rest mass' as an invariant quantity is consistent with the 4D energy-momentum of a particle being conserved and frame invariant. This is a frame independent 4D space-time perspective. Therefore this may seem the more 'pure' form to a relativist working with frame independent space-time. The energy used to accelerate an object to high velocity is redefined and absorbed by the time dilation suffered by that object as observed by the 'stationary' observer. (The 'gamma' factor)

However physics is actually done by a physicist locked into one particular frame, or another, and not 'frame independent'. Therefore from the perspective of a real experimenter the use of 'relativistic mass' may be the more obvious to use. It is a matter of choice; so long as you know the full implications of the perspective you select both systems should be equivalent.

Garth

The thread has long since stopped being one about the use of 'relativistic mass' but one that contests what relativity actually physically predicts.

A hot brick at rest according to special relativity has greater rest mass than a cold brick at rest (the same brick after being heated up while leaving its center of mass motionless). This is being contested by Aer as being an outdated concept of special relativity.