Can we deal with relativistic mass once and for all?

[Jan Nebec]

Hello!

I've been reading about relativistic mass for last few days and it leads me to even more confusion.
Supposing, we are assuming SR.

1. Why some people say that relativistic mass leads to confusion? As far as I learned, relativistic mass tells me the mass of an object, that is moving relative to me. So if I measure the mass of a moving electron, it would seem heavier, because I multiply it's rest mas with factor gama. I don't see any problems here?

2. Only problem that I see here is, that if someone uses relativistic mass instead of rest mass, he should carefully define velocity and frame of reference. Because for example, rest masses of 1kg and 10kg can both get relativistic mass of 100kg, so we have to define their velocity relative to observer to get full description.

3. However, relativistic mass can not be used to calculate gravitational attraction of an object! So it is useless in this respect. But I think it might be useful to describe it's inertial mass. But since we are told that inertial mass and gravitational mass are the same, I am confused, because increase of relativistic mass refers only to it's inertial mass. Am I right?

 

[DrStupid]

As far as I learned, relativistic mass tells me the mass of an object, that is moving relative to me.

No, it tells you the relativistic mass of an object, that is moving relative to you. The term "mass" is reserved for rest mass.

So if I measure the mass of a moving electron, it would seem heavier, because I multiply it's rest mas with factor gama.

Only if the speed remains constant.

I don't see any problems here?

The problem is that the property you are talking about is not called mass anymore. That's where the confusion comes from.

But since we are told that inertial mass and gravitational mass are the same

That's Newtonian mechanics. There is no such thing like gravitational mass in relativity.

 

[Jan Nebec]

No, it tells you the relativistic mass of an object, that is moving relative to you. The term "mass" is reserved for rest mass.
Only if the speed remains constant.
The problem is that the property you are talking about is not called mass anymore. That's where the confusion comes from.
That's Newtonian mechanics. There is no such thing like gravitational mass in relativity.

1. Well yes, I thought relativistic mass of that object moving relative to me.
2. Why if speed is constant? It would also work if an object accelerates. The object would seem harder and harder to accelerate, due to increase of relativistic mass.
3. How is it called then?

 

[DrStupid]

1. Well yes, I thought relativistic mass of that object moving relative to me.

Than you should write it. Mass and relativistic mass are different things.

2. Why if speed is constant? It would also work if an object accelerates. The object would seem harder and harder to accelerate, due to increase of relativistic mass.

And the force required to accelerate the object is only proportional to the relativistic mass if the speed remains constant (e.g. in circular motion). Otherwhise it doesn't even need to act parallel to the acceleration.

3. How is it called then?

It's usually called relativistic mass, but many physicists refuse to call it at all.

 

[Orodruin]

Yes, we can deal with relativistic mass once and for all. Physicists do so by not using it, since it is an antiquated concept.

I suggest that you read this Insight blog post.

 

[Nugatory]

I've been reading about relativistic mass for last few days and it leads me to even more confusion.

Which is consistent with people are saying about relativistic mass being confusing, no? The easy way to avoid that confusion is to not use the concept.

Why some people say that relativistic mass leads to confusion? As far as I learned, relativistic mass tells me the mass of an object, that is moving relative to me. So if I measure the mass of a moving electron, it would seem heavier, because I multiply it's rest mass with factor gamma. I don't see any problems here?

Mass is measured by applying a known force to the object, measuring the resulting acceleration, and then using $F=ma$ to calculate $m$ from the known $F$ and the observed $a$. But when you try this on a moving object, the "mass" that you calculate will only be equal to $\gamma{m_0}$ if the applied force is perpendicular to the direction of motion. If the force is parallel you will measure $\gamma^3m_0$, and for other directions you'll get something in between. So is the relativistic mass $\gamma{m_0}$ or $\gamma^3{m_0}$ or something in between? Confusion is in the eye of the beholder, but to me this is confusing.

and gravitational mass are the same, I am confused, because...

You are right. It is confusing. That's why we don't use it any more.

 

[Jan Nebec]

Which is consistent with people are saying about relativistic mass being confusing, no? The easy way to avoid that confusion is to not use the concept.
Mass is measured by applying a known force to the object, measuring the resulting acceleration, and then using $F=ma$ to calculate $m$ from the known $F$ and the observed $a$. But when you try this on a moving object, the "mass" that you calculate will only be equal to $\gamma{m_0}$ if the applied force is perpendicular to the direction of motion. If the force is parallel you will measure $\gamma^3m_0$, and for other directions you'll get something in between. So is the relativistic mass $\gamma{m_0}$ or $\gamma^3{m_0}$ or something in between? Confusion is in the eye of the beholder, but to me this is confusing.

You are right. It is confusing. That's why we don't use it any more.

Thank you very much! I think it's the best answer I could get so far!
How to prove, that calculated "mass" = $\gamma{m_0}$ if the applied force is perpendicular to the direction of motion? And how to prove the correlation when direction of the force is same as it's velocity vector?

 

[Orodruin]

Thank you very much! I think it's the best answer I could get so far!
How to prove, that calculated "mass" = $\gamma{m_0}$ if the applied force is perpendicular to the direction of motion? And how to prove the correlation when direction of the force is same as it's velocity vector?

Did you bother to read the Insight post I linked to in #5? It would have pointed you in the correct direction.

 

[Dadface]

Which is consistent with people are saying about relativistic mass being confusing, no? The easy way to avoid that confusion is to not use the concept.
Mass is measured by applying a known force to the object, measuring the resulting acceleration, and then using $F=ma$ to calculate $m$ from the known $F$ and the observed $a$. But when you try this on a moving object, the "mass" that you calculate will only be equal to $\gamma{m_0}$ if the applied force is perpendicular to the direction of motion. If the force is parallel you will measure $\gamma^3m_0$, and for other directions you'll get something in between. So is the relativistic mass $\gamma{m_0}$ or $\gamma^3{m_0}$ or something in between? Confusion is in the eye of the beholder, but to me this is confusing.

You are right. It is confusing. That's why we don't use it any more.

I'm trying to work out how we can confirm those conclusions experimentally. For example how do you apply a measurable perpendicular force to a moving object and measure the resulting acceleration? I'll have to think about it. I guess you can use some of the experimental techniques used when investigating circular motion.

 

[Ibix]

For example how do you apply a measurable perpendicular force to a moving object

Electron through a uniform magnetic field is the obvious one.

 

[Dadface]

Electron through a uniform magnetic field is the obvious one.

Of course - thank you. I don't know why but I was thinking of macroscopic objects.

 

[Ibix]

Of course - thank you. I don't know why but I was thinking of macroscopic objects.

Well, in principle a charged ping pong ball will work instead of the electron. It won't be as sensitive because the charge to mass ratio is lower, but the same physics is there. It's accelerating it to relativistic speeds that's really tricky...

 

[SiennaTheGr8]

Thank you very much! I think it's the best answer I could get so far!
How to prove, that calculated "mass" = $\gamma{m_0}$ if the applied force is perpendicular to the direction of motion? And how to prove the correlation when direction of the force is same as it's velocity vector?

To find the equation for the relativistic force, take the time-derivative of the equation for the relativistic momentum:

$\dfrac{d}{dt} \Bigl[ \vec p = \gamma m \vec v \Bigr]$.

Assuming constant $m$, you should end up with a vector equation that looks like this (in units where $c=1$):

$\vec f = \gamma^3 m (\vec v \cdot \vec a)\vec v + \gamma m \vec a$,

or equivalently (after dotting both sides with $\vec v$ and doing some algebra):

$\vec a = \dfrac{\vec f - (\vec f \cdot \vec v) \vec v}{\gamma m}$.

If $\vec f \perp \vec v$ (or if $\vec v = \vec 0$), then the numerator is just $\vec f$, and you get $\vec a = \vec f / (\gamma m)$.

If $\vec f \parallel \vec v$ (parallel or antiparallel, note), then the numerator is $\vec f / \gamma^2$ (hint: the unit vectors $\hat f$ and $\hat v$ can differ only by sign), so that $\vec a = \vec f / (\gamma^3 m)$.

In the general case, where $\vec f_\perp$ and $\vec f_\parallel$ are the component vectors of $\vec f$ perpendicular and parallel to $\vec v$:

$\vec a = \dfrac{\vec f_{\perp}}{\gamma m} + \dfrac{\vec f_{\parallel}}{\gamma^3 m}$.

 

[vanhees71]

Well, now again you commit one of this lesser sins as introducing the relativistic mass by using an antiquated (in fact since about 110 years outdated) concept of what you call force. What you define as force is again not covariant. Rather use covariant four-vectors, and there's no more confusion (and it's even the logical way to generalize to concept of force from Newtonian to relativistic point-particle mechanics: time derivatives in Newtonian physics are naturally substituted by proper-time derivatives, not by coordinate-time derivatives in relativity):
$$p^{\mu} = m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}, \quad \frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=m \frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2}=K^{\mu},$$
where $K^{\mu}$ is the Minkoski-force four-vector. For the electromagnetic force on point charges (neglecting radiation reactions) is
$$K^{\mu}=\frac{q}{m c} F^{\mu \nu} p_{\nu}=q F^{\mu \nu} u_{\nu},$$
where
$$u_{\nu}=\frac{1}{c} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}$$
is the four-velocity unit vector of the particle (unit vector in the sense of Minkowski space, i.e., $\eta_{\mu \nu} u^{\mu} u^{\nu}=1$ with $(\eta_{\mu \nu})=\mathrm{diag}(1,-1,-1,-1)$.

 

[phinds]

Well, in principle a charged ping pong ball will work instead of the electron. It won't be as sensitive because the charge to mass ratio is lower, but the same physics is there. It's accelerating it to relativistic speeds that's really tricky...

I once watched a ping pong match between two expert Chinese players. I'm pretty sure they got it going that fast.

 

[SiennaTheGr8]

Well, now again you commit one of this lesser sins as introducing the relativistic mass by using an antiquated (in fact since about 110 years outdated) concept of what you call force. What you define as force is again not covariant. Rather use covariant four-vectors, and there's no more confusion (and it's even the logical way to generalize to concept of force from Newtonian to relativistic point-particle mechanics: time derivatives in Newtonian physics are naturally substituted by proper-time derivatives, not by coordinate-time derivatives in relativity):
$$p^{\mu} = m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}, \quad \frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=m \frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2}=K^{\mu},$$
where $K^{\mu}$ is the Minkoski-force four-vector. For the electromagnetic force on point charges (neglecting radiation reactions) is
$$K^{\mu}=\frac{q}{m c} F^{\mu \nu} p_{\nu}=q F^{\mu \nu} u_{\nu},$$
where
$$u_{\nu}=\frac{1}{c} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}$$
is the four-velocity unit vector of the particle (unit vector in the sense of Minkowski space, i.e., $\eta_{\mu \nu} u^{\mu} u^{\nu}=1$ with $(\eta_{\mu \nu})=\mathrm{diag}(1,-1,-1,-1)$.

(Just kidding.)

But the OP was clearly asking about 3-force, no?

 

[Jan Nebec]

To find the equation for the relativistic force, take the time-derivative of the equation for the relativistic momentum:

$\dfrac{d}{dt} \Bigl[ \vec p = \gamma m \vec v \Bigr]$.

Assuming constant $m$, you should end up with a vector equation that looks like this (in units where $c=1$):

$\vec f = \gamma^3 m (\vec v \cdot \vec a)\vec v + \gamma m \vec a$,

or equivalently (after dotting both sides with $\vec v$ and doing some algebra):

$\vec a = \dfrac{\vec f - (\vec f \cdot \vec v) \vec v}{\gamma m}$.

If $\vec f \perp \vec v$ (or if $\vec v = \vec 0$), then the numerator is just $\vec f$, and you get $\vec a = \vec f / (\gamma m)$.

If $\vec f \parallel \vec v$ (parallel or antiparallel, note), then the numerator is $\vec f / \gamma^2$ (hint: the unit vectors $\hat f$ and $\hat v$ can differ only by sign), so that $\vec a = \vec f / (\gamma^3 m)$.

In the general case, where $\vec f_\perp$ and $\vec f_\parallel$ are the component vectors of $\vec f$ perpendicular and parallel to $\vec v$:

$\vec a = \dfrac{\vec f_{\perp}}{\gamma m} + \dfrac{\vec f_{\parallel}}{\gamma^3 m}$.

Great thanks for this explanation! It's so nice to see how people here are helping others!

 

[pervect]

Hello!

I've been reading about relativistic mass for last few days and it leads me to even more confusion.
Supposing, we are assuming SR.

1. Why some people say that relativistic mass leads to confusion? As far as I learned, relativistic mass tells me the mass of an object, that is moving relative to me. So if I measure the mass of a moving electron, it would seem heavier, because I multiply it's rest mas with factor gama. I don't see any problems here?

If you're only assuming SR, and you are assuming a point particle (which an electron is to a very good approximation), there aren't any severe problems. You may start to have problems in your understanding if you attempt to apply your ideas to situations that involve extended bodies that are not point particles, or in situations where special relativity is not good enough and you need General relativity.

Given the happy situaton of SR + point particles, the main problem you have is that some authors are not going to share your liking for relativistic mass, and that they will talk about the mass of something and mean the invariant mass. So you have to be aware that some authors may use the term "mass" differently than your preference.

If you are not in the happy situation of dealing with SR and point particles, there are things you need to understand that you probalby don't currently understand. . And I've written about them, but what I've written may have gotten lost in the vast amount of other stuff that's been written. However, there's no real need to repeat what I've already written unless you are interested, can't dig it out of what I've already written, or have questions about what I've written. Having questions about what I've written is quite possible, since what I've written are posts, not a textboos, and you probably need a textbook to fully understand all the implications of what I've written about. But we are starting to pick up a lot of "if's" here.

2. Only problem that I see here is, that if someone uses relativistic mass instead of rest mass, he should carefully define velocity and frame of reference. Because for example, rest masses of 1kg and 10kg can both get relativistic mass of 100kg, so we have to define their velocity relative to observer to get full description.

Yes, I'd agree.

3. However, relativistic mass can not be used to calculate gravitational attraction of an object! So it is useless in this respect. But I think it might be useful to describe it's inertial mass. But since we are told that inertial mass and gravitational mass are the same, I am confused, because increase of relativistic mass refers only to it's inertial mass. Am I right?

If you are talking about calculating gravitational attraction, you are getting into realms where you probably need general relativity. The reason that general relativity was created was that special relativity wasn't really compatible with Newtonian gravity when one looked at the issue closely. But this is going to be a big project, to understand GR.

But let's stick to inertial mass. Realizing that you don't want to talk about GR yet, you may still want to talk about the inertial mass of objects in special relativity. You're most likely talking about situations in which you no longer have point particles, but extended bodies in SR. There are some things that you need to understand here that you probably don't. One of the thigns I wrote about these issues earlier is in https://www.physicsforums.com/threa...lso-change-with-velocity.943333/#post-5969483. I do have concerns about how complete and understandable what I wrote is, but it's good jumping-off place for my concerns about what you're probably missing about dealing with extended bodies in special relativity. The executive summary is that you really need the stress-energy tensor to deal properly with extended bodies in special relativity, and you probably don't have knowledge of it.

It may not be complete enough for you to understand my point :(, but it's a starting point for the issues that I'm trying to point out to you about objects in special relativity that aren't point particles.

 

[Jan Nebec]

But let's stick to inertial mass. Realizing that you don't want to talk about GR yet, you may still want to talk about the inertial mass of objects in special relativity.

Okay, so for now, just for a concept, can I have an explanation that only inertial mass (that is equal to this confusing relativistic mass) increases with velocity, because an object is harder and harder to accelerate when it approaches the speed of light?

 

[pervect]

Okay, so for now, just for a concept, can I have an explanation that only inertial mass (that is equal to this confusing relativistic mass) increases with velocity, because an object is harder and harder to accelerate when it approaches the speed of light?

If you're still asking about the gravitational effects of a moving object, I can say that the problem you'll have is that substituting relativistic mass into the Newtonain formula to caculate the effects of a relativistic flyby will give different answers as compared to using the theory of general relativity. Using this quasi-Newtonian approach (which doesn't have any sound basis) one will UNDER estimate the effects of the relativistic flyby by a factor of roughly 2:1. Thus rapidly moving objects are even better at causing gravitational effects than their relativistic mass increase would suggest.

This doesn't explain the "why" of it, just what the theories predicts.

Let's be specific about exactly how we are measuring the effects of the relativistic flyby. We imagine a test environment with a cloud of test particles (too light to have much gravity of their own), that are all at rest in empty space. We perform a flyby with a massive relativistically moving object. Then, using techniques like doppler radar, we find the velocities induced by the cloud of test particles by the flyby.

There are theoretical issues (like the Shapiro time delay) that would affect the interpretation of our doppler radar results if we tried to get more detailed answers as to what happens during the flyby. Rather than attempt to solve these interpretation isssues, we sidestep them by performing the velocity measurements in the empty , flat space-time of special relativity before the flyby, and after the flyby, and don't worry about what is happening during the flyby, neatly avoiding some long and technical discussions about curved space-time. It is of course easiest to have the velocities all zero before the flyby, which is what I assumed, doppler measurement before the flyby would just confirms that the velocities are indeed zero.

For a paper discussing this approach, see http://adsabs.harvard.edu/abs/1985AmJPh..53..661O

See also my previous OF posts mentioning the same paper. You should be able to find them with PF search. I find:

https://www.physicsforums.com/threa...-pulse-of-light-to-e-c-2.906468/#post-5709053
https://www.physicsforums.com/threads/general-relativity-vs-Newtonian-gravity.896367/#post-5638860
https://www.physicsforums.com/threads/centripetal-acceleration-of-photons.901519/#post-5675883

there are more. Note the tie-ins with the "extra" deflection of light in GR.

This doesn't have much to do with previous point, which in a more specific example would involve calculating the ratio of force/acceleration of a very long rigid rod, and noticing that one gets different values for the ratio of acceleration / force depending on whether one pushes the rod from behind, or pulls it from front. The distribution of stresses in the rod affects its acceleration/force ratio (which I would assume you would call the 'mass' of the rod). There are some discussions of this in one of Rindler's textbooks https://www.amazon.com/dp/0198567324/?tag=pfamazon01-20, though the details of what Rindler writes are not the same as my presentation.

 

[Jan Nebec]

If you're still asking about the gravitational effects of a moving object...

No, I understand that SR doesn't include gravity and that mass multiplied with gamma factor must not be used to calculate gravitational attraction of an object. If I stay only in SR, can I say that relativistic mass is equal to inertial mass but gravitational mass stays same as rest mass?

 

[Orodruin]

No, I understand that SR doesn't include gravity and that mass multiplied with gamma factor must not be used to calculate gravitational attraction of an object. If I stay only in SR, can I say that relativistic mass is equal to inertial mass but gravitational mass stays same as rest mass?

There is no gravitational mass in SR.

 

[DrStupid]

I understand that SR doesn't include gravity and that mass multiplied with gamma factor must not be used to calculate gravitational attraction of an object. If I stay only in SR, can I say that relativistic mass is equal to inertial mass but gravitational mass stays same as rest mass?

Newton's law of gravitation must not be used under relativistic conditions. That's not a question of relativistic mass or rest mass. It's the law itself that fails.

 

[SiennaTheGr8]

Okay, so for now, just for a concept, can I have an explanation that only inertial mass increases with velocity, because an object is harder and harder to accelerate when it approaches the speed of light?

It's often tricky to convert the math into words. I think you more or less understand the basic principle here, but there's some imprecision and ambiguity in the way you've phrased this.

The main problem is the term "inertial mass." You obviously don't mean the invariant mass $m$. Do you mean the "relativistic mass" $\gamma m$ (aka "total energy" in different units)? If so, then yes, that quantity increases with speed, though I wouldn't say it's because an object gets harder to accelerate as it approaches the speed of light. Rather, it increases with speed because kinetic energy contributes to total energy. Put differently, $\gamma$ increases with speed, so $\gamma m$ increases with speed, too.

But I sense that what you're really getting at is the relationship between force and acceleration in general (or, perhaps, the relationship between force and acceleration given a specified angle between the force and velocity vectors). If that's the case, then you must accept that force and acceleration in SR aren't related by a simple multiplicative factor. They are related by the vector equations I gave in my post above:

$\vec f = \gamma^3 m (\vec v \cdot \vec a)\vec v + \gamma m \vec a$

and

$\vec a = \dfrac{\vec f - (\vec f \cdot \vec v) \vec v}{\gamma m} = \dfrac{\vec f_{\perp}}{\gamma m} + \dfrac{\vec f_{\parallel}}{\gamma^3 m}$

(with $c=1$).

If you want to be able to point to a single quantity and say "this is the thing that relates the force and acceleration 3-vectors," then you'll have to settle for a matrix (or rather, a pair of matrices that are inverses of each other), but it makes no sense to speak of a matrix "increasing," and it would be poor nomenclature to call that matrix the "inertial mass." Best not to use the term "inertial mass" in SR at all, in my opinion.

I might rewrite your statement like this:

Under a given force applied at a given angle relative to an object's velocity, the object's resultant acceleration depends not only on the object's (invariant) mass, but also on the object's speed at the moment of application—the greater the speed, the smaller the magnitude of acceleration. This is indeed related to the fact that nothing can be accelerated to the speed of light, but the principle actually holds regardless of the given angle between force and velocity (i.e., all else being equal, a given object is always harder to accelerate when it's moving faster, even if you're trying to accelerate it in the direction opposite to its current direction of motion).

Someone please correct me if I've gotten anything wrong.

 

[pervect]

No, I understand that SR doesn't include gravity and that mass multiplied with gamma factor must not be used to calculate gravitational attraction of an object. If I stay only in SR, can I say that relativistic mass is equal to inertial mass but gravitational mass stays same as rest mass?

Since SR as a theory doesn't include gravitation, why do you think that "gravitational mass" is even defined in the theory?

I get the impression that my previous post wasn't helpful at all to you, since you didn't respond to any of the points I tried to make in it :(. I'm at a loss as to how to attempt to fix this, unfortunately, it seems we are talking past each other.

 

[PeterDonis]

Supposing, we are assuming SR.

In which case there is no gravity. And therefore much of what you are trying to discuss can't be discussed.

 

[PAllen]

I'll just add that given that that SR is a theory of 4 dimensional spacetime, the natural quantities it deals with are scalars and 4-vectors, not the 3-vectors of Newtonian physics. In this case, there is a single scalar quantity relating 4-force to 4-acceleration, and that is simply invariant mass not relativistic mass. This is the only possible notion of a scalar inertial mass in SR. As noted by others, if you consider 3-vectors (which are not vectors under Lorentz transform, they don't represent 4-d geometrical objects at all), then scalar inertial mass is impossible, and certainly it is not relativistic mass. You have, in effect, a different inertial mass for every direction of force application in relation to velocity, if you consider 3-force and 3-acceleration.

 

[vanhees71]

(Just kidding.)

But the OP was clearly asking about 3-force, no?

Well, then you should tell them the better notion right away, which is Minkowski force, i.e., a (spacelike) four-vector, and that I did.

 

[atyy]

I'll just add that given that that SR is a theory of 4 dimensional spacetime, the natural quantities it deals with are scalars and 4-vectors, not the 3-vectors of Newtonian physics. In this case, there is a single scalar quantity relating 4-force to 4-acceleration, and that is simply invariant mass not relativistic mass. This is the only possible notion of a scalar inertial mass in SR. As noted by others, if you consider 3-vectors (which are not vectors under Lorentz transform, they don't represent 4-d geometrical objects at all), then scalar inertial mass is impossible, and certainly it is not relativistic mass. You have, in effect, a different inertial mass for every direction of force application in relation to velocity, if you consider 3-force and 3-acceleration.

Scalar inertial mass is possible for 3-force, as discussed, for example in the Feynman lectures: http://www.feynmanlectures.caltech.edu/I_15.html.

 

[atyy]

No, I understand that SR doesn't include gravity and that mass multiplied with gamma factor must not be used to calculate gravitational attraction of an object. If I stay only in SR, can I say that relativistic mass is equal to inertial mass but gravitational mass stays same as rest mass?

One can consider relativistic theories of gravity in SR (flat spacetime), for example, Nordstrom's theory (a coherent relativistic theory of gravity, but falsified by observations) and the quantum theory of spin-2 fields (whose classical regime is general relativity) - both of those flat spacetime theories can be reformulated as theories of curved spacetime, just as General Relativity (GR) is usually presented.

That the inertial mass of SR is energy can motivate the idea that energy is source of gravity in GR. However, the relationship is heuristic, and the equivalence principle has to go through several reformulations in the transition from Newtonian gravity to general relativity: from (i) inertial mass = gravitational mass in Newtonian gravity; to (ii) acceleration is locally approximately like gravity in GR to (iii) the principle of minimal coupling in GR.

 

[vanhees71]

Now we even try to discuss GR, where several posters in this thread still struggle with SR, and this most due to the fact that some still insist on using outdated concepts. Unfortunately even Feynman did so in the famous Feynman lectures, which are among my most highly favored general theory books, but in this point they are bad. There are many excellent books of physicists as eminent as Feynman that contain unfortunate approaches that do not help students but confuse them. That happens to any textbook writer from time to time. That's why one should read not only one book but many.

The best way to avoid trouble with what's called "equivalence principle" you can, if you have the minimum of necessary math in order to do GR, formulate it in a very simple form:

Spacetime is a 4D-Lorentzian manifold (i.e., a pseudo-Riemannian space with the fundamental form of signature (1,3)). This implies that around any point $x$ there exists a map defining coordinates $x^{\mu}$ such that
$$g_{\mu \nu}(x) \mathrm{d} x^{\mu} \mathrm{d} x^{\nu} = \eta_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}.$$
Such frames are called Galileian and are the best approximation of inertial frames possible in the presence of gravity, which cannot be described adequately by special-relativistic models but only in GR. Physically they can be realized by free-falling bodies that can be considered pointlike. An example is the International Space Station, freely falling in the gravitational field of the Earth (solar system). To a very good approximation the astronauts do not observe gravity and thus are to a good approximation in such a local inertial (or Galilean) reference frame.

 

[PAllen]

Scalar inertial mass is possible for 3-force, as discussed, for example in the Feynman lectures: http://www.feynmanlectures.caltech.edu/I_15.html.

Sorry, I see nothing there that gets around the fact that the ratio of 3 force to 3 acceleration is dependent on angle between force and velocity, thus not a scalar. He uses the word inertia only in one paragraph, and does not analyze different directions of force application. Seems not at all to support your claim.

 

[PAllen]

Despite what I have written on this thread earlier, I now propose that you can define a plausible scalar inertial mass for 3-force using 4-force as a model. Just as in 4-force, for the case of no mass/energy flows, invariant mass gives the ratio between 4-force and 4-accleration, you can reduce this to 3 dimensions and time derivatives (rather than proper time derivatives) by using the notion of celerity (the spatial part of 4-velocity, i.e. γv). Then, define time rate of change of celerity (celeration anyone??). Then 3-force over invariant mass gives celeration. Since celerity goes to infinity as as v approaches c, no matter how large the rate of change of celerity, you can't get bigger than infinite celerity, so speed remains below c. This attaches the limiting factor to the behavior of proper time (or spacetime geometry), not mass.

Then, to get coordinate acceleration from celeration, you divide the celeration component parallel to velocity by γ³, and the component orthogonal to velocity by γ, with none of the derivation for this relying on mass or momentum. These give you the velocity parallel and orthogonal components of coordinate acceleration.

 

[atyy]

Sorry, I see nothing there that gets around the fact that the ratio of 3 force to 3 acceleration is dependent on angle between force and velocity, thus not a scalar. He uses the word inertia only in one paragraph, and does not analyze different directions of force application. Seems not at all to support your claim.

That fact is derived as a consequence of Eq 15.10 (and the preceding equation).

 

[atyy]

Here is another reference illustrating that the relativistic mass can be considered an inertial mass:
http://www.scholarpedia.org/article/Nordtvedt_effect by Kenneth Nordvedt
"The subsequent theory of Special Relativity generalized this connection by asserting that every form of energy within a body contributes to the body's inertial mass; M=E/c²."