# On the concept of mass

1. Jun 9, 2013

### levokun

The term "mass" is the source of many conflicting opinions among the authors writing on relativity theory.
Different authors denote by this term different concepts.
Quite often even the same author denotes by mass different concepts in his different writings.
For instance when introducing his famous diagrams Richard Feynman used the concept of invariant mass of a particle defined by equation $m^2=p^2$, where $p$ is four-momentum. But later in his Feynman Lectures on Physics he preferred to define mass by the equation $E=mc^2$. Thus defined mass $m$ obviously increases with increase of total energy $E$ and hence of speed of a particle.
The equation $E=mc^2$ usually referred to as the super-famous Einstein equation, though Einstein himself preferred another definition: $E_0=mc^2$, where $E_0$ is the rest energy, or energy of a particle at rest. To a certain extent the partial source of confusion was the term "rest mass" used by him.

Last edited by a moderator: Jun 9, 2013
2. Jun 9, 2013

### WannabeNewton

Is there a question here? The term 'mass' in SR most commonly refers to the rest mass of a particle, which is a Lorentz scalar.

3. Jun 9, 2013

### Staff: Mentor

Welcome to the forum.

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4. Jun 10, 2013

### yenchin

There is historical reason why people use relativistic mass, but
nowadays most people use $E=\gamma mc^2$, where $\gamma$ is the Lorentz factor. Here $m$ is the rest mass, as WannabeNewton said.

5. Jun 10, 2013

### Naty1

This discussion might give you some further perspective:

http://en.wikipedia.org/wiki/Rest_mass

In SIX EASY PIECES ( 1997), pages 87 to 91, Feynman seems to say repeatedly 'mass' DOES increase with speed. [He is using the older 'relativistic' mass..which includes kinetic energy.

For example:
and separately:
there is more than one definition of "mass" in relativity.

- invariant mass, or rest mass, or proper mass, which excludes the kinetic energy of the object's center of momentum
- relativistic mass, sometimes called inertial mass, which includes the kinetic energy of the object's center of momentum.

Feynman was referring to relativistic mass, but many users in this forum prefer the modern convention of referring to invariant mass, that is, rest mass. The prior post uses that convention.

Last edited: Jun 10, 2013
6. Jun 10, 2013

### dextercioby

My take : there's only one 'mass'. If one needs or feels the urge to put an adjective next to it, then <invariant> would be the best, but I don't advise it. In the community, when one speaks about mass, he either refers to the Newtonian definition (quantity of substance, as in chemistry), or to the 0-th component of the 4-momentum, in units where c=1 and the Minkowskian metric is mostry minus.

The subject in the original post has been pretty much exhausted by the Russian physicist Lev B. Okun. Specifically, he wrote a book, <Energy and Mass in Relativity Theory>, a summary of own articles and talks on the 2 concepts (amazon link: https://www.amazon.com/Energy-Mass-...UTF8&qid=1370900827&sr=1-2&keywords=Okun,+Lev) where he shared his teacher's (?) - Lev Landau - views and presented them from all sorts of angles. To a non-geometrist, Landau's textbook supplemented by Okun's writings serve as a solid foundation of Relativity Theory. Then you can, if you wish, add the somewhat abstract flavor of differential geometry.

Last edited: Jun 10, 2013
7. Jun 10, 2013

### e.chaniotakis

The relativistic increase in mass is due to the fact that at high speeds (u->c) the mechanics you use are relativistic.
The term invariant mass is the correct one -
a question: how would you define mass in a proper textbook format?
E.g I would say that mass is the energy of an object observed at its rest frame

8. Jun 11, 2013

### WannabeNewton

You can define it in a manifestly Lorentz invariant manner as $m^{2} = -p_{a}p^{a}$. In the instantaneous rest frame this reduces to $m = E$.

9. Jun 11, 2013

### Staff: Mentor

This is my preferred definition also. I also prefer the term "invariant mass" over the term "rest mass". Rest implies a frame where the entire system is at rest, which may not exist or may not be inertial. Invariant implies that any frame is fine for calculating it.

10. Jun 11, 2013

### WannabeNewton

Yeah invariant mass is definitely a better way to phrase it. I do agree with the OP though that the term "mass" in SR isn't always used consistently across different texts. For example in chapter 5 of Purcell's "Electricity and Magnetism", the term mass is constantly used to refer to the relativistic mass.

11. Jun 11, 2013

### Naty1

Entirely consistent with post#8, #9, I believe:

from other discussions in these forums...

Einstein:
Taylor & Wheeler:

And there is another good reason to view 'mass' as invariant mass: because that's the notion of gravitational curvature sourced from the stress energy momentum tensor.

I believe the last sentence is an oblique [as I see it] reference to the fact that an increase of velocity of a center of mass is NOT associated with an increase in gravitational attraction; in other words, the SET [the source of gravity] acts in such a way as to reflect invariant mass.
You can tell relative velocity should not change gravity because gravitational curvature produced by an object" is frame-invariant; it doesn't matter what your state of motion is relative to the object....hence one does NOT want to get other frame dependent energy sources, such as kinetic energy, mixed up with 'mass'...nor 'gravity'....

12. Jun 11, 2013

### levokun

It is more consistent to use the term mass for both massive and massless particles than to use the term rest mass for massless particles, which are never at rest.

13. Jun 11, 2013

### ZapperZ

Staff Emeritus
Welome to the Forum, Lev. It's a pleasure to have you here.

In case you are interested, I've made several references to your paper on this topic in a number of discussions in this forum. This is one such example:

On the other hand, Frank Wilczek has written a lengthy discussion (and not easy to follow) on the origin of mass.

http://arxiv.org/abs/1206.7114

It may not follow in your theme for this thread here, but I'm curious as to what you make of it.

Cheers!

Zz.

14. Jun 11, 2013

### Naty1

from Wilczek's Conclusions:

15. Jun 13, 2013

### levokun

Cheers, Zz!

It is nice to see on this thread people who can explain more clearly than I can
how the term mass should be used.
As for the lengthy article by Wilczek, it is obviously lacking
references to the papers of Steve Weinberg

Best regards
Lev

16. Jun 22, 2013

### codelieb

Lev,

In The Feynman Lectures on Physics Feynman does not precisely "define mass by the equation $E=mc^2$." In fact he defines velocity-dependent mass as a modification (made by Einstein) of Newton's law, starting on the very first page of the book, page 1-1 of Volume I, where it says,

For example, the mass of an object never seems to change: a spinning top has the same weight as a still one. So a “law” was invented: mass is constant, independent of speed. That “law” is now found to be incorrect. Mass is found to increase with velocity, but appreciable increases require velocities near that of light. A true law is: if an object moves with a speed of less than one hundred miles a second the mass is constant to within one part in a million. In some such approximate form this is a correct law.​

And again when special relativity is introduced, in chapter 15,

Newton’s Second Law, which we have expressed by the equation

$F = d(mv)/dt$,

was stated with the tacit assumption that m is a constant, but we now know that this is not true, and that the mass of a body increases with velocity. In Einstein’s corrected formula $m$ has the value

$m = m_0 / \sqrt(1 − v^2/c^2)$,

where the “rest mass” $m_0$ represents the mass of a body that is not moving and $c$ is the speed of light, which is about $3×10^5$ km· sec$^{−1}$ or about $186,000$ mi · sec$^{−1}$.

For those who want to learn just enough about it so they can solve problems, that is all there is to the theory of relativity—it just changes Newton’s laws by introducing a correction factor to the mass.​

(From this he eventually derives $\Delta E = \Delta (mc^2)$, at the end of the chapter.)

I have only found one clue that might explain why Feynman used the (pedagogically poor) velocity-dependent mass as the basis for his lectures on special relativity in FLP, in Jagdish Mehra's biography, The Beat of a Different Drum, which includes interviews of Feynman Mehra made only a few weeks before Feynman's death. In one of these interviews Feynman says,

As for the lectures on physics, I have put a lot of thought into these things over the years. I've always been trying to improve the method of understanding everything. I had already tried to explain the results of relativity theory in my own way to my girlfriend, Arline, and then I used the same explanations in my lectures. These things are very personal, my own way of looking at things and I recognize them. I did everything—all of it—in my own way.​

Arline was Feynman's first wife. They married when they were young - Feynman was 23 - but they had already been girlfriend and boyfriend for 10 years. It is probable that Feynman studied relativity theory at age 13 or 14 (when he was studying calculus) and he would probably have studied from a book borrowed from a public library, one of the old books (this would have been 1933 or 1934) that use velocity-dependent mass. Furthermore, even if Feynman at that tender age realized mass was an invariant 4-scalar, it is very unlikely he would try to explain relativity theory to his non-scientist girlfriend in terms so abstract as Minkowski spacetime. It's more likely he would choose the less correct but more appealing to "common sense" way of teaching relativity theory, involving $m(v)$ . Thirty years later, when Feynman composed his freshmen lectures on relativity, it seems he simply ignored what he had learned about mass since he was 13, and for emotional reasons reconstructed his old lectures in Arline's memory (who died of tuberculosis not long after they were married, while Feynman was working at Los Alamos on the first atom bomb). At least that would explain the inconsistency between his more mature work, in which mass is a constant, and his freshmen relativity lectures, in which $m(v)$ plays a pivotal role.

Mike Gottlieb
Editor, The Feynman Lectures on Physics, New Millennium Edition

Last edited: Jun 22, 2013
17. Jun 22, 2013

### atyy

It seems just a question of naming, so perhaps he also had in mind another lesson his father taught him?

http://www.haveabit.com/feynman/2

Last edited: Jun 22, 2013
18. Jun 27, 2013

### levokun

Dear Michael,

Your reply added two important statements about $m(v)$ which were missing in my original thread, namely:

1) the inconsistent and illegal attempt to keep for the relativistic particles the non-relativistic definition of momentum $p=mv$;

2) the emotional personal reason for Feynman in memory of his first wife Arline to use $m(v)$
via Newtonian definition of momentum as was stressed on page 488 of the book of Mehra.

I agree with what you have written about the drawbacks of this approach.What I cannot agree with is that you (and Feynman whom you quote) ascribe the concept of velocity-dependet mass $m(v)$ to Einstein. In fact Einstein criticized this concept, though not irrerprochably consistently. (See the article by Carl Adler "Does mass depend on velocity Dad?" which was published in 1987, but was not noticed by Feynman in the last year of his life.) In 1921 Einstein had cast his famous equation in the form $E_0=mc^2$, where $E_0$ is the energy of a particle at rest. But he never wrote it in the invariant form $p^2=m^2$. Note that in this form it was first written in 1941 in the first edition of textbook "Field Theory" by Lev Landau and Eugene Lifgarbagez. However for unknown reasons they preferred not to introduce notation $E_0$ for the rest energy. This tradition of avoiding $E_0$ is kept in the modern (tenth) edition of their course.

Best regard.
Lev

19. Jun 27, 2013

### codelieb

Dear Lev,

Feynman (and his coauthors) ascribe $m(v)$ to Einstein in The Feynman Lectures on Physics. I do not. (I am aware of Einstein's ambivalence towards $m(v)$, mostly through proofreading your papers on the concept of mass, and the research I conducted on your behalf in the Einstein Papers at Caltech.)

But are you telling me that before 1941 there was not a single publication (book or article) in which it was stated that mass is the magnitude of the 4-momentum (or something equivalent, like E^2 - p^2 = m^2)? I find that remarkable. The basis for this was laid out by Minkowski in 1908. Surely, it must have been recognized long before 1941? I suppose one could only come to your conclusion by examining every publication between 1908 and 1941 that discusses relativity theory... but that would probably be impossible, so I think you must either be assuming that you have looked at a sufficiently large sample to infer your conclusion (of course such an inference can never be certain) or you are repeating what you have read elsewhere, in which case I would like to know your source.

Best regards,
Mike

Last edited: Jun 27, 2013
20. Jun 27, 2013

### harrylin

If you search this forum then you'll find a number of interesting discussions on that topic, with quite some references. And the physics FAQ gives IMHO a fair summary:
http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html
What do you think?

21. Jun 27, 2013

### atyy

Since Einstein did not criticize it consistently, why can the concept not be ascribed to Einstein in certain of his writings? Presumably you are thinking of http://www.fourmilab.ch/etexts/einstein/E_mc2/www/ ?

22. Jun 27, 2013

### Phy_Man

You may not have noticed it but Lev B. Okun is the person who created this thread.

23. Jun 27, 2013

### Phy_Man

Just a quick note on terminology - The proper mass of a particle is more properly referred to as simply a scalar. The term Lorentz scalar refers to any quantity which remains invariant under a Lorentz transformation. Any quantity which remains invariant under a general spacetime transformation is referred to simply as a scalar. Likewise any quantity which remains invariant under an orthogonal transformation (i.e. a "rotation" of the coordinates about the origin) is called a Cartesian scalar. Thus distance and volume are Cartesian scalars.

24. Jun 27, 2013

### WannabeNewton

The rest mass of a time-like particle is a Lorentz scalar because it transforms under the trivial representation of the proper Lorentz group. The Lorentz transformations preserve the inner product between vectors hence $p_a p^a$ transforms under the identity element.

25. Jun 27, 2013

### WannabeNewton

This depends entirely on what you mean by invariant. Isometries preserve the metric tensor under the induced pullback hence the metric tensor remains invariant under isometries in that sense. It is obviously not a scalar field. If by "general space-time transformation" you mean arbitrary diffeomorphisms that are endomorphisms of space-time then this is also not true because diffeomorphisms can move points around: $f:\mathbb{R}\rightarrow \mathbb{R},x \mapsto x + a$ does not leave invariant the scalar field $g:\mathbb{R}\rightarrow \mathbb{R},x \mapsto x^2$ because $g(f(x)) = g(x + a) = (x + a)^{2}\neq x^{2} = g(x)$. If you mean coordinate transformations then yes scalar fields are the ones which unequivocally remain the same under coordinate transformations (this is because coordinate transformations are a special class of diffeomorphisms called passive diffeomorphisms-they fix points, they don't move them around).

Last edited: Jun 27, 2013