Why don't photons possess mass?

[pmb_phy]

My textbook on relativity sais the usage of relativistic mass is pretty outdated and hardly anybody uses it anymore.

That is one person's opinion. There are others with the opposite opinion. E.g.

In defense of relativistic mass, T.R. Sandin, Am. J. Phys. 59 (11), November 1991

Concepts of Mass in Contemporary Physics and Philosophy, Max Jammer, Princeton Univ. Press, (2000)

Tell me something. How can something be called "outdated" when so many people use it in the modern relativity literature? E.g.

Apparatus to measure relativistic mass increase, John W. Luetzelschwab, Am. J. Phys. 71(9), 878, Sept. (2003).

Relativistic mass increase at slow speeds, Gerald Gabrielse, Am. J. Phys. 63(6), 568 (1995).

Especially when labs such as CERN have it on their websites? Any argument that I've seen to the contrary, i.e. that rel-mass is outdated or a bad idea etc., are all flawed. I've outlined it all another paper I wrote. If you want to read it then go to http://www.geocities.com/physics_world/ and click on On the concept of mass in relativity.

Here's a good example. From Black Holes and Time Warps: Einstein's Outrageous Legacy, Kip Thorne, page 82

..the inertia of every object must increase so rapidly that, as its speed approaches the speed of light, that no matter how hard one pushes on the object, one can never make it reach or surpass the speed of light.

Teach every student that mass does not increase with speed - He reads this book. What is he to make of this comment?

So when I wrote m, I thought everybody knew what I meant...

Not everyone apparently. I know what you mean but when someone asks the question about the mass of light then you shouldn't assume it. I.e. if someone says that since light has energy and E = mc² then why doesn't it have mass etc. then it should be apparent that they have relativistic mass in mind even though they may not know it.

Although the equivalence principle is very useful for making the analogy between gravity and accelleration, the equivalence is not complete: wouldn't it be better to only speak of a gravitational field in the presence of mass/energy. I believe a field (at least an EM field) is an invariant entity which is as real as a particle. Making it something you can transform away just by changing your coordinates seems pretty strange to me.

In physics one should not speak of that which can't be measured. The detection of photons is frame dependant as is the detection of radiation. The detection of the electric field depends on the frame of reference. I.e. for some obervers it exists and for other observers it doesn't. Same with magnetic field.

If you have Thorne's book then take a look at Box 12.5. It states

This startling discovery revealed that the concept of a real particle is relative, not absolute; that is it depends on the frame of reference. Observers in a freely-falling frames who plunge through the hole's horizon see no real particles outside the horizon, only virtual ones. Observers in accelerated frames, who by their acceleration, remain always above the horizon see a plethora of real particles.

I can't think of any distrbution but a spherical shell, but there (in spite of the gravitational potential) the field is (uniform) zero... Could you please describe such a distribution to me?

Sure. Please see - http://www.geocities.com/physics_world/gr/grav_cavity.htm

Pete

 

[da_willem]

Thank you for the example of a uniform gravitational field, it's a very elegant example!

[I think I heard you say in some other forum topic you were open to corrections to your site: judging on your formula for g I guess G=1 in this text? I think you forgot a minus sign after the second equality sign in (3), and where does R come from in your definition of g, I guess z=||r_1-r_2||?]

..the inertia of every object must increase so rapidly that, as its speed approaches the speed of light, that no matter how hard one pushes on the object, one can never make it reach or surpass the speed of light.

The increase of inertia for moving objects can also be explained by another definition of momentum p= \gamma mv, and not assigning the gamma to m, right?


[Do you know Marc Lange or his book: "The philosophy of physics"? Many of my thoughts about mass and fields are affected by this book, and I don't know how correct his work is...

E.g. Could you say: "because proper mass is Lorentz invariant, it has a continuous identity (this piece here is the same as that piece then)"?

And he tries to prove that an electric field has invariant mass, and is thus real and with that he tries to save spatiotemporal locality...]

 

[pmb_phy]

Thank you for the example of a uniform gravitational field, it's a very elegant example!

[I think I heard you say in some other forum topic you were open to corrections to your site: judging on your formula for g I guess G=1 in this text? I think you forgot a minus sign after the second equality sign in (3), and where does R come from in your definition of g, I guess z=||r_1-r_2||?]

Thanks. I'm unable to sit any longer this afternoon (bad back). I'll get back to you later today.

The increase of inertia for moving objects can also be explained by another definition of momentum p= \gamma mv, and not assigning the gamma to m, right?

That is not a description of inertia. Inertia is defined as that which mass describes. You can say that the particle will not be able to accelerate to c because the momentum will become infinite. I think that is what you're referring to.

[Do you know Marc Lange or his book: "The philosophy of physics"? Many of my thoughts about mass and fields are affected by this book, and I don't know how correct his work is...

No.

E.g. Could you say: "because proper mass is Lorentz invariant, it has a continuous identity (this piece here is the same as that piece then)"?

I don't know what is meant by "continuous identity."

And he tries to prove that an electric field has invariant mass, and is thus real and with that he tries to save spatiotemporal locality...]

What is the invariant mass of an electric field?

I'll get back later this afternoon as I mentioned. Until then keep this in mind; from The Evolution of Physics, Albert Einstein, Leopold Infeld, Simon & Schuster (1938) page 31

Physical concepts are free creations of the human mind, and are not, however it may seem, uniquely determined by the external world. In our endeavor to understand reality we are somewhat like a man trying to understand the mechanism of a closed watch. He sees the face and the moving hands, even hears its ticking, but has has no way of opening the case. If he is ingenious he may form some picture of a mechanism which could be responsible for all things he observes, but he may never be quite sure his picture is the only one which could explain his observations. He will never be able to compare his picture with the real mechanism and he cannot even imagine the possibility of the meaning of such a comparison. But he certainly believes that, as his knowledge increases, his picture of reality will explain a wider and wider range of his sensuous impressions. He may even believe in the existence of the ideal limit of knowledge and that it is approached by the human mind. He may call this ideal limit the objective truth.

Pete

 

[da_willem]

You can say that the particle will not be able to accelerate to c because the momentum will become infinite. I think that is what you're referring to.

That is what I was referring to.

What is the invariant mass of an electric field?

He says the (invariant) mass of an electric field is m_f = \sqrt{\frac{E_f}{c^4} - \frac{p_f^2}{c^2}} and in it's rest frame B=0 so the Poynting vecor is zero so it's momentum is zero, so the mass of the field is E_f / c^2. For the field energy he uses \int \frac{E^2}{8 \pi} dV

Although I read (parts of) your article in favor of relativistic mass, I also read a very good webpage confirming my own vision: http://www.wordiq.com/definition/Relativistic_mass

Among a lot of arguments I especially liked the following:

If one truly wishes to retain the notion that mass measures the "resistance" to acceleration, then mass can no longer be treated as a scalar quantity. This is because it is easier to accelerate something perpendicular to the direction of motion than parallel to the direction of motion. In effect, an object would have more mass in one direction than other

And the quote from Einstein himself:

“It is not good to introduce the concept of the mass M = m/(1 - v2/c2)1/2 of a body for which no clear definition can be given. It is better to introduce no other mass than ‘the rest mass’ m. Instead of introducing M, it is better to mention the expression for the momentum and energy of a body in motion.” – Einstein, in a 1948 letter to Lincoln Barnett

 

[pmb_phy]

He says the (invariant) mass of an electric field is m_f = \sqrt{\frac{E_f}{c^4} - \frac{p_f^2}{c^2}} and in it's rest frame B=0 so the Poynting vecor is zero so it's momentum is zero, so the mass of the field is E_f / c^2. For the field energy he uses \int \frac{E^2}{8 \pi} dV

That is not an invariant. Consider the energy, E₀, of the electric field between a parallel plate capacitor as measured in the rest frame of the capacitor. Let E be the energy as measured in a frame moving normal to the face of the plates. For your "mass" to be invariant the energy must transform as E = \gamma E_0[/tex]. However this is not the case. The energy transforms as E = E_0/\gamma[/tex]. For details please see<br /> <a href="http://www.geocities.com/physics_world/sr/rd_paradox.htm" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://www.geocities.com/physics_world/sr/rd_paradox.htm</a><br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Although I read (parts of) your article in favor of relativistic mass, I also read a very good webpage confirming my own vision: http://www.wordiq.com/definition/Relativistic_mass </div> </div> </blockquote>Sorry but that article is incorrect. It states<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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 people stopped referring to the relativistic mass altogether. <br /> <br /> Today, when physicists talk about the mass of an object they always mean the rest mass. </div> </div> </blockquote>That is simply incorrect as anyone can see from looking in many modern relativity texts and from online university lecture notes. Even in GR/cosmology texts/<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Among a lot of arguments I especially liked the following: <br /> <br /> And the quote from Einstein himself: </div> </div> </blockquote>Einstein was inconsistent with his definition of mass. On the one hand he said that its not good to use a velocity dependant mass. Such as mass is defined as the m in p = mv. Yet in his GR text he actuallyt uses this mass. Although when he uses it he uses it for small speeds it is not the rest mass that you've been using. He uses the definition for which the mass changes as a result of the particle being in a gravitational field. This mass, m, is defined as <br /> <br /> m = m_0 \frac{dt}{d\tau}<br /> <br /> For a time orthogonal spacetime (i.e. g<sub>0i</sub> = 0)<br /> <br /> m = \frac{m_0}{\sqrt{1 + 2\Phi/c^2 - v^2/c^2}}<br /> <br /> When v << c m becomes<br /> <br /> m \simeq \frac{m_0}{\sqrt{1 + 2\Phi/c^2}}<br /> <br /> This is not an invariant quantity. It depends on the gravitational potential. Yet this is the mass Einstein speaks of in the GR section of his text <b>The Meaning of Relativity</b>. The last edition of which was written much later than he wrote that letter to Lincoln Barnett. Einstein also stated that light has mass. That is something that you wouldn't gather from his letter to Lincoln Barnett either.<br /> <br /> By the way, there is a second part to that letter Einstein wrote to Barnett. It reads<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I do not agree with the idea that general relativity is geometrizing Physics or the gravitational field. The concepts of Physics have always been geometrical concepts and I cannot see why tghe g<sub>ik</sub> field should be called more geometrical than f.i. the electro-magnetic field or the distance between bodies in Newtonian mechanics. The notion comes probably from the fact that the mathematical origin of the g<sub>ik</sub> field is the Gauss-Riemann theory of the metrical continuum which we are won't to look at as part of geometry. I am convinced, however, that the distinction between geometrical and other kinds of fields is not logically founded. </div> </div> </blockquote>Do you agree with this part of that letter to? <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f600.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":biggrin:" title="Big Grin :biggrin:" data-smilie="8"data-shortname=":biggrin:" /> <br /> <br /> Pete

 

[pmb_phy]

That is not an invariant. Consider the energy, E₀, of the electric field between a parallel plate capacitor as measured in the rest frame of the capacitor. Let E be the energy as measured in a frame moving normal to the face of the plates. For your "mass" to be invariant the energy must transform as E = \gamma E_0[/tex]. However this is not the case. The energy transforms as E = E_0/\gamma[/tex]. For details please see<br /> <a href="http://www.geocities.com/physics_world/sr/rd_paradox.htm" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://www.geocities.com/physics_world/sr/rd_paradox.htm</a><br />

When I wrote this I had a particular example in mind, the one in this link. I have not considered, as of yet anyway, how the energy between the plates transforms when the capacitor is moving in a direction parallel to the capacitor plates, only then the capacitor is moving in a direction which is normal to the capacitor plates.<br /> <br /> Pete

 

[da_willem]

I don't have to book here, but I'll try to reproduce his reasoning. He said that when you compute the electric potential between two charged particles you bring together from infinity you get a negative number, but that is because in the initial situation the fields were not zero.

He computes the field energy between the two particles with a total electric field \vec{E}_1+\vec{E}_2:
E_f= \int \frac{ \vec{E}^2}{8 \pi} dV= \int \frac{ \vec{E}_1^2}{8 \pi} dV + \int \frac{ \vec{E}_2^2}{8 \pi} dV + 2 \int \frac{ \vec{E}_1 \cdot \vec{E}_2 }{8 \pi} dV

Where he assignes the first two terms to the respective particles, and the third term turns out to be exactly the binding energy. So now he has this positive energy assignable to the field, which he can assign a mass to and show the existence of the EM field and save spatiotemporal locality...

 

[da_willem]

That is simply incorrect as anyone can see from looking in many modern relativity texts and from online university lecture notes. Even in GR/cosmology texts/

Here a quote by John Baez:

relativistic mass is of no use at all in general relativity

If a particle accellerates it doesn't exert a larger gravitational pull does it?

Do you agree with this part of that letter to?

Does he refer to his search for a mathematical theory that explains every field as of geometrical nature, just like he did with the gravitational field? He never really succeeded, right? It would be a very elegant theory though...

 

[pmb_phy]

I don't have to book here, but I'll try to reproduce his reasoning. He said that when you compute the electric potential between two charged particles you bring together from infinity you get a negative number, but that is because in the initial situation the fields were not zero.

Whether the sign is positive or negative will depend on whether the signs are the same or whether they are opposite.

He computes the field energy between the two particles with a total electric field \vec{E}_1+\vec{E}_2:
E_f= \int \frac{ \vec{E}^2}{8 \pi} dV= \int \frac{ \vec{E}_1^2}{8 \pi} dV + \int \frac{ \vec{E}_2^2}{8 \pi} dV + 2 \int \frac{ \vec{E}_1 \cdot \vec{E}_2 }{8 \pi} dV

Where he assignes the first two terms to the respective particles, and the third term turns out to be exactly the binding energy.

It is only the binding energy which contributes to the mass of the system. Not the energy of the fields of the point charges, that part is infinite. I.e. the first two integrals diverge. Only changes in energy yield a change in mass in situations like this. You can read more on this in

Electrostatic potential energy leading to a gravitational mass change for a system of two point charges, Timothy H. Boyer Am. J. Phys. 47, 129 (1979)

Electrostatic potential energy leading to an inertial mass change for a system of two point charges, Timothy H. Boyer Am. J. Phys. 46, 383 (1978)

So now he has this positive energy assignable to the field, which he can assign a mass to and show the existence of the EM field and save spatiotemporal locality...

That was a very poor way of looking at it and has led him to an erroneous result. For instance - What is the mass associated with the field energy of a single point particle?

Here a quote by John Baez:..

That's a load of bs. Whether this is useful or not is highly subjective. Perhaps he has never found it useful but I sure have. Different people study different aspects of GR.

If a particle accellerates it doesn't exert a larger gravitational pull does it?

What does acceleration have to do with this? We're talking about the dependence of mass on speed, not acceleration. If you're asking whether the gravitational pull is a function of the speed of the source then yes, absolutely

Measuring the active gravitational mass of a moving object, D. W. Olson and R. C. Guarino Am. J. Phys. 53, 661 (1985)

Abstract - If a heavy object with rest mass M moves past you with a velocity comparable to the speed of light, you will be attracted gravitationally towards its path as though it had an increased mass. If the relativistic increase in active gravitational mass is measured by the transverse (and longitudinal) velocities which such a moving mass induces in test particles initially at rest near its path, then we find, with this definition, that M_rel = \gamma(1+2\beta^2)M. Therefore, in the ultrarelativistic limit, the active gravitational mass of a moving body, measured in this way, is not M but is approximately 2\gamma M. <br />

I&#039;ve worked out a few examples of this myself. Unfortunately I got sick before I was able to update my website to include them. People such as Baez willk inevidably say &quot;Oh! That&#039;s not because of mass increase, that&#039;s because energy and momentum etc increase etc with speed&quot;. Rindler gets into this in his text. He places relativistic mass on the same footing as one would charge in EM. In fact there are equations in GR called the equations of gravitomagnetism which are almost identical to Maxwell&#039;s equations. Relativistic mass takes the place of charge and mass has been considered a &quot;gravitational charge&quot; even in Newtonian gravity. <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Does he refer to his search for a mathematical theory that explains every field as of geometrical nature, just like he did with the gravitational field? </div> </div> </blockquote>I dunno. Each person has a different opinion on this. What one person thinks is &quot;useful&quot; another person thinks of as &quot;useless&quot;. <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> He never really succeeded, right? </div> </div> </blockquote>I dunno. Try e-mailing him and asking. He&#039;s responded to me in e-mail often so he&#039;ll probably respond to you.<br /> <br /> Pete

 

[da_willem]

Whether the sign is positive or negative will depend on whether the signs are the same or whether they are opposite

He uses two opposite charges so their fields cancel when brought together

What does acceleration have to do with this? We're talking about the dependence of mass on speed, not acceleration. If you're asking whether the gravitational pull is a function of the speed of the source then yes, absolutely

I meant a particle accellerating thus requiring a greater speed; I shold have said that. And now I think about it offcourse a moving object produces a larger gravitational pull. As you think about it this way; invariant mass would be the one having no place in GR because it is not the source of a gravitational field, energy (proportional to relativistic mass) is?

Thanks for the Am. J. Physics articles (do you know if it's possible to get them delivered in Europe?). I'm still reading your article on the concept of mass (It's quite large you know...).

Try e-mailing him and asking. He's responded to me in e-mail often so he'll probably respond to you.

Sure. Do you have his mailadress; A.Einstein@hotmail.com or something??!

 

[pmb_phy]

..invariant mass would be the one having no place in GR because it is not the source of a gravitational field, energy (proportional to relativistic mass) is?

Yup. I agree to a certain extent.

Thanks for the Am. J. Physics articles (do you know if it's possible to get them delivered in Europe?).

If you have access to a university library then they should have it.

I'm still reading your article on the concept of mass (It's quite large you know...).

Unfortunately it was very large. I was hoping to keep it shorter when I wrote it but there is a lot to the subject and I chose thoroughness over shortness.

Sure. Do you have his mailadress; A.Einstein@hotmail.com or something??!

baez@math.ucr.edu or john.baez@ucr.edu

This is a photo of John - http://andrej.com/mathematicians/B/Baez_John.html

Reminds of of Mr. Kotter. :laughing:

Saying that relativistic mass has no role/place in relativity is pretty silly since it has always been there and must always be there. Saying it doesn't is like saying that momentum has no role/place in relativity. E.g. consider the force on a particle with proper mass m₀

\bold F = \frac{d}{dt}\left \frac{m_0 (d\bold r/dt)}{\sqrt{1-(d\bold r/dt)^2/c^2}}\right

Does v = dr/dt have a role? After all I never need to substitute v into any equation. But let's try it anyway.

\bold F = \frac{d}{dt}\left \frac{m_0 \bold v}{\sqrt{1-v^2/c^2}}\right

Does = \gamma = 1/\sqrt{1-v^2/c^2} have a role? After all I never need to substitute [m = \gamma m_0 into any equation. But let's try it anyway.

\bold F = \frac{d(\gamma m_0 \bold v)}{dt}

Does m = \gamma m_0 have a role? After all I never need to substitute m into any equation. But let's try it anyway.

\bold F = \frac{d(m\bold v)}{dt}

Does \bold p = m \bold v[/tex] have a role? After all I never need to substitute <b>v</b> into any equation. But let&#039;s try it anyway. <br /> <br /> \bold F = \frac{d\bold p}{dt}<br /> <br /> Seems that each time I defined a quantity the equations became simler. But I need not have defined serveral of those quantities, including velocity and momentum. Even force doesn&#039;t have to be defined. E.g. I can write the Lorentz &quot;force&quot; law as <br /> <br /> \frac{d}{dt} \frac{m_0 (d\bold r/dt)}{\sqrt{1-(d\bold r/dt)^2/c^2}} = q(\bold E + \bold v \times \bold B)<br /> <br /> I can express the law of conservation of momentum by saying that mv is conserved. This holds in all cases, not just for tardyons, when m = relativistic mass.<br /> <br /> I call the introduction of each term, i.e. v, m, p, gamma, f, etc. &quot;Economy of thought.&quot;<br /> <br /> Pete

 

[pmb_phy]

One can always formally define m=\sqrt{E^2 - \vec{p}\,^2} = \sqrt{p^2} but no physical interpretation in a rest frame can be obtained from it for a real photon (for which m=0). Then, for a virtual photon the mass defined in this way is perfectly legitimate.

Just a Note: When discussing quantum mechanics one has to keep in mind that quantities which are meaningful in classical relativity become meaningless in relativistic quantum mechanics and quantum field theory. The above relation defines proper mass as mentioned above. Relativistic mass becomes ill-defined in quantum mechanics, as does \bold p = \gamma m \bold v and \bold F = d\bold p/dt since each is a function of velocity and velocity is not a well-defined quantity.

Relativity (SR and GR) is a classical theory and as such photons really have no place in it in principle. However that doesn't mean that its not highly useful.

Then again this is yet another discussion about terminology and semantics. But such conversations are not a waste of time since Physical concepts are free creations of the human mind, and are not, however it may seem, uniquely determined by the external world. - Einstein

Refering to proper mass as "mass" and labeling it "m" instead of m[sub0[/sub] is just a way to simplify things in a certain context. E.g. in particle physics one studies the inherent properties of particles. It'd be exhausing to keep using the qualifier "proper" evertime you spoke of the proper mass of a particle. State what you mean once and be done with it - whatever simplifies things for you and your reader per your taste. However this would mean referring to the proper lifetime of a particle simply as the "lifetime". In relativity the lifetime of a particle is not the same thing as the particle's proper lifetime, yet its rare to see the proper lifetime actually called the "proper lifetime."

Pete

 

[humanino]

photons really have no place in [a classical theory] in principle

I would really appreciate if you elaborate on this. Since light has such a central role in GR, I don't understand this issue.
Is this merely the fact that individual quanta of light appear only in quantum phenomena ?
Or do you mean to say that light is entirely described by a wave in classical theory ?
Is it something else ?

Thank you in advance.

 

[pmb_phy]

I would really appreciate if you elaborate on this. Since light has such a central role in GR, I don't understand this issue.

Light, yes. Photons, no.

Is this merely the fact that individual quanta of light appear only in quantum phenomena ?

Yes.

Or do you mean to say that light is entirely described by a wave in classical theory ?

No. In quantum theory you cannot assign a velocity vector to a photon whereas in classical mechanics/relativity you can. In relativity you should think of a photon as a classical particle which moves at the speed of light, not as a quantum particle, wave-partilce duality and all, until there is a full-fledged theory of relativsitic quantum theory.

Note that I didn't say that it wasn't useful or that I wouldn't speak of photons in SR/GR. On the contrary, I often do. But to be honest I think of them as classical particles which move at v = c.

Are you familiar with the tex Exploring Black Holes? If so then seen the acknowledgment section (http://www.eftaylor.com/pub/front_matter.pdf)

Philip Morrison made several suggestions and convinced us not to invoke that weird quantum particle, the photon, in a treatment of the classical theory of relativity (except in some exercises).

That was what I had in mind.

Pete

 

[humanino]

Thanks for clear and fast answer Pete.

 

[pmb_phy]

Thanks for clear and fast answer Pete.

Glad to help. But that was just an opinion and I'm not attached to it 100%.

Photons are funny animals. Even some of the most well know physicists are not even convinced that they exist. For instance, Willis E. Lamb, Jr., Nobel Laureate, wrote a paper entitled Anti-photon, Applied Physics B, 60, 77-84(1995) in which he argues that photons don't exist. And as you probably know, Lamb shared the Nobel prize in 1955 for "his discoveries concerning the fine structure of the hydrogen spectrum."

As the abstract states

It should be apparent from the title of this article that that author does not like the word "photon," which dates from 1926. In his view, there is no such thing as a photon. Only a comedy of errors and historical accidents led to its popularity among physicists and optical scientists. etc.

Good reading and well worth your time.

Pete

 

[pmb_phy]

I'm still reading your article on the concept of mass (It's quite large you know...).

At this point I'd say toss in the circular filing cabinet. I've grown tired over the years of this topic and I've chosen to (1) delete the paper from my website and (2) say this only - Mass is what you define it to be and if you choose to defined it as I've explained then you can't go wrong.

Besides that I'd say don't bother with it. Its a waste of time worrying about a definition when you know what the definition is.

Pete

 

[pmb_phy]

Note: I won't be able to discuss this or anythiing else in the near, or even the foreseeable, future. Due to the problem of my herniated disk I have to spend more time off my feet and out of my chair and spend more time on the couch on my back. The internet is too much of a temptation and what very little time I spend on it is turned out to be far too much and is causing to much damage. to the disk. I will therefore be offline, at least until my back is all better and probably for a while, if not permanently.

Have a good one and it was a pleasure knowing you all.

Pete

 

[da_willem]

Note: I won't be able to discuss this or anythiing else in the near, or even the foreseeable, future. Due to the problem of my herniated disk I have to spend more time off my feet and out of my chair and spend more time on the couch on my back. The internet is too much of a temptation and what very little time I spend on it is turned out to be far too much and is causing to much damage. to the disk. I will therefore be offline, at least until my back is all better and probably for a while, if not permanently.

I'm sorry to hear that your situation is made worse by using the internet. Thanks though for your fast and accurate replies, and the discussions. I hope your back gets better and we will see you again on PF.

 

[nrqed]

Saying that relativistic mass has no role/place in relativity is pretty silly since it has always been there and must always be there. Saying it doesn't is like saying that momentum has no role/place in relativity. E.g. consider the force on a particle with proper mass m₀

\bold F = \frac{d}{dt}\left \frac{m_0 (d\bold r/dt)}{\sqrt{1-(d\bold r/dt)^2/c^2}}\right

Does v = dr/dt have a role? After all I never need to substitute v into any equation. But let's try it anyway.

\bold F = \frac{d}{dt}\left \frac{m_0 \bold v}{\sqrt{1-v^2/c^2}}\right

Does = \gamma = 1/\sqrt{1-v^2/c^2} have a role? After all I never need to substitute [m = \gamma m_0 into any equation. But let's try it anyway.

\bold F = \frac{d(\gamma m_0 \bold v)}{dt}

Does m = \gamma m_0 have a role? After all I never need to substitute m into any equation. But let's try it anyway.

\bold F = \frac{d(m\bold v)}{dt}

Does \bold p = m \bold v[/tex] have a role? After all I never need to substitute <b>v</b> into any equation. But let&#039;s try it anyway. <br /> <br /> \bold F = \frac{d\bold p}{dt}<br /> <br /> Seems that each time I defined a quantity the equations became simler. But I need not have defined serveral of those quantities, including velocity and momentum. Even force doesn&#039;t have to be defined. E.g. I can write the Lorentz &quot;force&quot; law as <br /> <br /> \frac{d}{dt} \frac{m_0 (d\bold r/dt)}{\sqrt{1-(d\bold r/dt)^2/c^2}} = q(\bold E + \bold v \times \bold B)<br /> <br /> I can express the law of conservation of momentum by saying that mv is conserved. This holds in all cases, not just for tardyons, when m = relativistic mass.<br /> <br /> I call the introduction of each term, i.e. v, m, p, gamma, f, etc. &quot;Economy of thought.&quot;<br /> <br /> Pete

<br /> <br /> <br /> <br /> The way you present it, the definitions you choose at every step are totally arbitrary. So, to someone who does not know physics, it might sound like physicists just defined things as they wish with no physical motivation. That&#039;s of course very misleading.<br /> <br /> To make my point, consider the following step: <br /> <br /> <blockquote data-attributes="" data-quote="pmb_phy" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> pmb_phy said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> \bold F = \frac{d}{dt}\left \frac{m_0 \bold v}{\sqrt{1-v^2/c^2}}\right<br /> <br /> Does = \gamma = 1/\sqrt{1-v^2/c^2} have a role? After all I never need to substitute [m = \gamma m_0 into any equation. But let&#039;s try it anyway. </div> </div> </blockquote><br /> So why not define scoobydoo = \frac{\bold v}{\sqrt{1-v^2/c^2}} instead of defining gamma and then putting the gamma with the mass? After all, someone who would not already know some SR would probably find it more natural to put all the v dependence together. <br /> <br /> The reason of course is that the way things are grouped and the choice of new definitions are not at all arbitrary, as it might look from your post.<br /> Things are defined according to how they transform under certain symmetries. That&#039;s what we always do in physics, from simple nonrelativistic mechanics to particle physics and GR. Symmetry is the key point. Unfortunately it&#039;s not pointed out often. For example, the quantity m_0 is a useful concept because it&#039;s invariant under Lorentz transformations. The quantity \bold p = \gamma m_0 \bold v is a useful concept because the sum of this quantity is conserved in a collision (no matter what frame we are observing the collision from). So we give names to these quantities. the quantity &quot;scoobydoo&quot; I defined above has no invariance property so it&#039;s not useful to give it a special name. The same things is true for the quantity \gamma m_0. It has no interesting property so it should not really be given a special name. It&#039;s only if someone wants to keep using expressions familiar from nonrelativistic mechanics that there is some justification to treating this quantity in a special way (for example, if one really wants to keep using \bold p = m \bold v then one must define m = \gamma m_0). But there is no reason for wanting to keep using the nonrelativistic expressions. It&#039;s totally subjective. I think it makes more sense to accept the fact that expressions are different in SR than in classical mechanics and that the momentum is simply \gamma m_0 \bold v.<br /> <br /> Pat

 

[da_willem]

It's totally subjective. I think...

I think this is the whole issue, it's just a discussion over semantics. And to a large extent in physics you are free to make your own definitions to make the equations 'more elegant'.

But anyway, Peter is not going to reply, see a few posts ago:

Note: I won't be able to discuss this or anythiing else in the near, or even the foreseeable, future. Due to the problem of my herniated disk I have to spend more time off my feet and out of my chair and spend more time on the couch on my back. The internet is too much of a temptation and what very little time I spend on it is turned out to be far too much and is causing to much damage. to the disk. I will therefore be offline, at least until my back is all better and probably for a while, if not permanently.

 

[pmb_phy]

I'm sorry to hear that your situation is made worse by using the internet. Thanks though for your fast and accurate replies, and the discussions. I hope your back gets better and we will see you again on PF.

Thanks. It seemed to get a tad better - then it got much worse. I have to see a neurosurgeon next week. I may need surgery. I have a doctors appointment today and there is a computer next door. This is the exception to the rule though.

The point of my post that Pat responded too was that "has no role" is a meaningless thing to say pertaining to the concept or rest mass. Regarding the expression p = M(v)v. As we all know when this is the momentun of a particle then M(v) = \gamma m_0 \bold v and m₀ is an inherent property of the particle. This is often take to apply to everything and this may not be the case. Momentum may depend on the orientation of the body relative to its velocity.

There is an excellant example in

Mass renormalization in classical electrodynamics, David J. Griffith and Russell E. Owen, Am. J. Phys., 51(12), Dec. 1983.

The authors consider a dumbbell which consists of two identical charges held together by a rod. The momentum of the dumbbell has one value when the axis is parallel to the velocity and another value when perpendicular. In the arbitrary case where the axis makes an angle between those two values there is a range of values. And the momentum is not even parallel to the velocity in general!

[qote=Pat]The way you present it, the definitions you choose at every step are totally arbitrary.[/quote]Definitions are always "arbitrary" in that they are not set in stone by God. Perhaps Pat is unfamiliar with the saying

Physical concepts are free creations of the human mind, and are not, however it may seem, uniquely determined by the external world. - Albert Einstein

See http://www.geocities.com/physics_world/physical_concepts.htm for full quote and reference.

So we give names to these quantities. the quantity "scoobydoo" I defined above has no invariance property so it's not useful to give it a special name. The same things is true for the quantity . It has no interesting property so it should not really be given a special name.

That is quite untrue. You're considering this in a vaccum. Its not even clear what you mean by "no interesting property." Have you ever met two people who agreed on all points as to what is "interesting"? v is not invariant and yet I consider it useful. Consider something else besides a point particle. Consider a continuos mass distrubution. Then the mass of the distribution is the inetral of dM = (1/c²)T⁰⁰dV over the volume of the distribution. This integral is evaluated in the zero momentum frame. What is it that you think dM is? Consider the case of em radiation. Let all radiation in the region x>0 be moving in the +x direction and let all radiation in the region x<0 be moving in the -x direction. What do you think dM means in this case? I.e. when you're adding these quantities, what is it that you're adding?

Consider also that it is not m₀ that is the source of gravity, \gamma m_0 is the source. Thus a beam of photons all moving in the same direction will generate a gravitational field. Consider a sheet of matter in the z = 0 plane. If you consider the mass of the particles that make up the sheet as \gamma m_0 then when you consider the fact that it behaves as if the gravitational mass increases like \gamma m_0 then it becomes a very interesting quantity.

Pat - Do you think that v is an "interesting" quantity?

It's only if someone wants to keep using expressions familiar from nonrelativistic mechanics that there is some justification to treating this quantity in a special way..

- This comment makes no sense to me. People don't define things so as to look like the non-relativsitic case. It is simply a definition. Take as an example what happens in physics - Components of things like 3-force are actually what gets measured in the lab. These things are not invariant. The lifetime of a free neutron is not the same as the proper lifetime of 15 minutes. When the thing is moving in the lab then, on average, it lives longer when it is moving then when it is at rest. A moving rod is shorter than a stationary rod. That is physically measurable.

Its unwise to consider only the expression \bold p = \gamma m_0 \bold v as defining mass and to only consider it in a vaccum. There are extremely good reasons to define mass as m = \gamma m_0. Please don't be offended by this but its not like all these physicists for the last 100 years were so dense as not to think the way you do on this point. They've found by careful examination under various situations that the most reasonable thing to call "mass" as m = \gamma m_0. The way you define it doesn't always work. In fact it is invalid for a rod which is under stress and moving parallel to its length. That can be shown by looking at the stress-energy-momentum tensor.

Take care all and see you the next time I have a doctors appointment.

Pete

ps - Pat; to see more examples of where \bold p = \gamma m_0 \bold v fails, or where the definition of "invariant mass" of a system fails pleas see - http://www.geocities.com/physics_world/sr/invariant_mass.htm

There is also an imortant comment at the bottom of page 104 in Ohanian's text "gravitation and spacetime" regarding this. He explains the trouble with adding 4-momenta when the momenta are not constant and the particles have a spatial seperation

 

[pmb_phy]

Thanks. It seemed to get a tad better - then it got much worse. I have to see a neurosurgeon next week. I may need surgery. I have a doctors appointment today and there is a computer next door. This is the exception to the rule though.

The point of my post that Pat responded too was that "has no role" is a meaningless thing to say pertaining to the concept or rest mass. Regarding the expression p = M(v)v. As we all know when this is the momentun of a particle then M(v) = \gamma m_0 \bold v and m₀ is an inherent property of the particle. This is often take to apply to everything and this may not be the case. Momentum may depend on the orientation of the body relative to its velocity.

There is an excellant example in

Mass renormalization in classical electrodynamics, David J. Griffith and Russell E. Owen, Am. J. Phys., 51(12), Dec. 1983.

The authors consider a dumbbell which consists of two identical charges held together by a rod. The momentum of the dumbbell has one value when the axis is parallel to the velocity and another value when perpendicular. In the arbitrary case where the axis makes an angle between those two values there is a range of values. And the momentum is not even parallel to the velocity in general!

Note; - That is what the authors conclude in that paper. I am not convinced that they are correct. However they do speak of a paradox regarding this problem so I suspect a piece of this puzzle is missing.

[qote=Pat]The way you present it, the definitions you choose at every step are totally arbitrary.

Definitions are always "arbitrary" in that they are not set in stone by God. Perhaps Pat is unfamiliar with the saying
See http://www.geocities.com/physics_world/physical_concepts.htm for full quote and reference.


That is quite untrue. You're considering this in a vaccum. Its not even clear what you mean by "no interesting property." Have you ever met two people who agreed on all points as to what is "interesting"? v is not invariant and yet I consider it useful. Consider something else besides a point particle. Consider a continuos mass distrubution. Then the mass of the distribution is the inetral of dM = (1/c²)T⁰⁰dV over the volume of the distribution. This integral is evaluated in the zero momentum frame. What is it that you think dM is? Consider the case of em radiation. Let all radiation in the region x>0 be moving in the +x direction and let all radiation in the region x<0 be moving in the -x direction. What do you think dM means in this case? I.e. when you're adding these quantities, what is it that you're adding?

Consider also that it is not m₀ that is the source of gravity, \gamma m_0 is the source. Thus a beam of photons all moving in the same direction will generate a gravitational field. Consider a sheet of matter in the z = 0 plane. If you consider the mass of the particles that make up the sheet as \gamma m_0 then when you consider the fact that it behaves as if the gravitational mass increases like \gamma m_0 then it becomes a very interesting quantity.

Pat - Do you think that v is an "interesting" quantity?

- This comment makes no sense to me. People don't define things so as to look like the non-relativsitic case. It is simply a definition. Take as an example what happens in physics - Components of things like 3-force are actually what gets measured in the lab. These things are not invariant. The lifetime of a free neutron is not the same as the proper lifetime of 15 minutes. When the thing is moving in the lab then, on average, it lives longer when it is moving then when it is at rest. A moving rod is shorter than a stationary rod. That is physically measurable.

Its unwise to consider only the expression \bold p = \gamma m_0 \bold v as defining mass and to only consider it in a vaccum. There are extremely good reasons to define mass as m = \gamma m_0. Please don't be offended by this but its not like all these physicists for the last 100 years were so dense as not to think the way you do on this point. They've found by careful examination under various situations that the most reasonable thing to call "mass" as m = \gamma m_0. The way you define it doesn't always work. In fact it is invalid for a rod which is under stress and moving parallel to its length. That can be shown by looking at the stress-energy-momentum tensor.

Take care all and see you the next time I have a doctors appointment.

Pete

ps - Pat; to see more examples of where \bold p = \gamma m_0 \bold v fails, or where the definition of "invariant mass" of a system fails pleas see - http://www.geocities.com/physics_world/sr/invariant_mass.htm

There is also an imortant comment at the bottom of page 104 in Ohanian's text "gravitation and spacetime" regarding this. He explains the trouble with adding 4-momenta when the momenta are not constant and the particles have a spatial seperation[/QUOTE]

 

[nrqed]

That is quite untrue. You're considering this in a vaccum. Its not even clear what you mean by "no interesting property." Have you ever met two people who agreed on all points as to what is "interesting"? v is not invariant and yet I consider it useful.

Then let me be more specific: I think that when constructing a theory, the starting point is a question of symmetries and invariance. What are the symmetries that we think our system (and the equations) must possess? I think that's the starting point. Then one builds quantities that have specific transformation properties under those symmetries. Equations must relate things that have the same transformation properties, so these are the "interesting" things to work with. Of course, one can measure anything one wants! But the theory will involve quantities which have specific transformation properties. That's the whole point of using tensors to build relativistically covariant equations. If we did not pay attention to transformation properties, we would be trying a lot of crazy equations that would prove useless in the end. Why didn't Dirac try writing down equations involving only P_0 separately, or just P_x and P_y, etc?? Because he knew his equation would prove up useless if it was not Lorentz invariant.

*Of course*, when you change the applications of a theory, the symmetries change and the useful quantities become different. What were useful quantities in classical, nonrealtivistic mechanics are no longer useful in SR, and what is useful in SR may have to be modify when going to GR.

All I was saying is that I the quantity \gamma m_0 is not a useful concept in SR. The mass m_0 is a useful quantity, and the quantity \gamma m_0 \vec{v} is useful as the zeroth component of the four-momentum, but not \gamma m_0.
If so, show me a relativistically invariant equation of SR which contains \gamma m_0 not as part of a four-vector!

Consider also that it is not m₀ that is the source of gravity, \gamma m_0 is the source.

So we are talking about GR or SR? I thought we were discussing SR. In any case, show me one of the fundamental eqs of GR which contain \gamma m_0 by itself (not as a component of a tensor) and then I will agree with you that it is a useful quantity to define in itself.

Pat - Do you think that v is an "interesting" quantity?

In classical mechanics, it is, indeed. Because the symmetries that are interesting are the ones under Galilean transformations and one is dealing with inertial frames. In that contex, the quantity d \vec v/ dt is an invariant and that's why it appears in Newton's second law. (of course, we could do it the other way around and talk about the equation in order to define inertial frames, etc but I am following here the symmetry -> laws of physics approach).

Tell me, if you are saying that it's totally up in the air what physical quantities we single out, then why do we put the three components v_x, v_y, v_z in a vector in the first place?? It's because the concept of a vector is invariant under rotation of the coordinate system whereas the individual components are not! Of course, one can measure the indivudual components in an experiment, but from a conceptual point of view, it makes more sense to group the 3 components in a vector. Is your point of view that grouping the 3 components ina vector a totally arbitrary definition??

What about things like d^3 \vec v/ dt^3? Why don't we use that quantity in classical mechanics? Of course, it *can* be measured. And anyone could come along and claim it's a useful quantity and so on.

- This comment makes no sense to me. People don't define things so as to look like the non-relativsitic case. It is simply a definition.

I just don't see any reason to introduce gamma m_0 other than to have the momentum equation look like the NR expression. It does not transform a scalar, a vector or any type of tensor for that matter under Lorentz transfo. The rest mass is an invariant, and the four-momentum transforms as a 4-vector. But splitting the four-momentum in a way (as a product of this "relativistic mass" times something else) such that the two pieces have no particular property under Lorentz transfos is not a ueful step, IMHO.

Take as an example what happens in physics - Components of things like 3-force are actually what gets measured in the lab. These things are not invariant. The lifetime of a free neutron is not the same as the proper lifetime of 15 minutes. When the thing is moving in the lab then, on average, it lives longer when it is moving then when it is at rest. A moving rod is shorter than a stationary rod. That is physically measurable.

Of course, one can measure anything in a lab. Including the x component of the momentum of an electron in any frame. By why didn't Dirac write an equation for the x component of the momentum of an electron? Because it is not a useful quantity to build the theory upon. And why? Because separating the x component hides the summetries underlying the theory! Of course one could work out separate equations for the x, y,z and t components but then one would realize that they are linked in a nontrivial way . And that means that they are not useful quantities from a theoretical point of view because the symmetries are not made explicit. Same thing for \gamma m_0.

Likewise for the length of the rod. Ofcourse, it is shorter in motion. But we don't have a separate equation for the length of the rod when it is moving at 0.1 c, at 0.2 c etc etc. The equations contain the proper length. If you want to fidn the length when it is moving at 0.1 c, you don;t look up an equation for that quantity, you use the Lorentz transfos to figure it out. So the useful quantity here is the proper length, not the length at 0.1 c. For the same reason, the useful information for a particle is its rest mass.

[
Its unwise to consider only the expression \bold p = \gamma m_0 \bold v as defining mass and to only consider it in a vaccum. There are extremely good reasons to define mass as m = \gamma m_0. Please don't be offended by this but its not like all these physicists for the last 100 years were so dense as not to think the way you do on this point.

I know it's an old debate. And I am not at my office so I can't give references but there are many people who do say that we should not teach the concept of relativistic mass anymore, that's it's an historical "faux-pas". It's a bit like using imaginary time in SR and GR. It used to be done but it is now realized that it's not conceptually a good approach (see for example MWT).

They've found by careful examination under various situations that the most reasonable thing to call "mass" as m = \gamma m_0. The way you define it doesn't always work. In fact it is invalid for a rod which is under stress and moving parallel to its length. That can be shown by looking at the stress-energy-momentum tensor.

The way "I define" it...you mean just m_0? Well, it should be clear if we are doing SR or GR. But in either case, all I am saying is that considering the quantity \gamma m_0 in isolation (and not as part of a four vector or a tensor) is incorrect. Of course, you can look at a certain mass distribution in a specific frame and say that it is something we can measure. All I am saying is that the meaningful quantity to discuss is the four-vector or the tensor, not a component in a specfic frame, for a particular mass distribution. That's what I mean by "useful" vs not useful quantities.

And I think that in teaching or presenting physics it's important to emphasize that anything can be measured and we can single out any combinations of of quantities we want, but the correct way to approach physics is through symmetries and that the useful quantities are the ones with specific transformation properties.

Take care all and see you the next time I have a doctors appointment.

Pete

Take care. I do very sincerely hope that you will be feeling better!


Pat

ps - Pat; to see more examples of where \bold p = \gamma m_0 \bold v fails, or where the definition of "invariant mass" of a system fails pleas see - http://www.geocities.com/physics_world/sr/invariant_mass.htm



There is also an imortant comment at the bottom of page 104 in Ohanian's text "gravitation and spacetime" regarding this. He explains the trouble with adding 4-momenta when the momenta are not constant and the particles have a spatial seperation

I'll look. However, my issue is not with the four-momentum, it's with singling out gamma m_0! But I'll look.

 

[pmb_phy]

Then let me be more specific: I think that when constructing a theory, the starting point is a question of symmetries and invariance. What are the symmetries that we think our system (and the equations) must possess?

To me the goal and purpose of a theory in the end is to describe nature and what we observe. What we observer are not invariant quantities.

But the theory will involve quantities which have specific transformation properties. That's the whole point of using tensors to build relativistically covariant equations.

The purpose of tensors in SR/GR is to meet the criteria of covariance in the laws of nature.

All I was saying is that I the quantity \gamma m_0 is not a useful concept in SR.

This is the point where opinion comes into it and we've left the area of calculation and prediction. In any case - Anything which is conserved is important. That is a pretty universal opinion. And \gamma m_0 is a quantity which is conser ved and that fact does not rely on the conservation of energy as many seem to think.

The mass m_0 is a useful quantity, ...

I agree. Proper time is important as is coordinate time. Coordinate time is as important as proper time and mass is just as important as mass. But you seem to keep arguing in a vacuum. There is more than one reason why (relativistic) mass is so often spoken of in the relativity literature. It has all the properties that one associates with mass, i.e. the inertial properties, the active and passive properties of gravitational mass. Proper mass has none of those.

..the quantity \gamma m_0 \vec{v} is useful as the zeroth component of the four-momentum, but not \gamma m_0.
If so, show me a relativistically invariant equation of SR which contains \gamma m_0 not as part of a four-vector!

That is incorrect. \gamma m_0 is the time component of 4-momentum. People replace it with inertial energy E but its more correct to use P⁰ = mc since it plays the role of time component perfectly since if dX is a spacetime displacement then dX⁰ = ct. So it t is the time component of dX then mc is the time component of P.

So we are talking about GR or SR? I thought we were discussing SR.

Why would you get that impression. Mass is mass. Relativity is relativity. There is no reason that I know of to define mass differently in SR than one does in GR. In fact it'd be a bad idea to do so in my opinion.

In any case, show me one of the fundamental eqs of GR which contain \gamma m_0 by itself (not as a component of a tensor) and then I will agree with you that it is a useful quantity to define in itself.

You're incorrectly demanding that mass must be a tensor in GR for it to be important. That's like saying that energy or pressure are not important in GR because they're components of tensors. That's your personal opinion. I don't expect you to agree with me and I don't expect you to think that its important. There's little use in wanting everyone to think in the exact same way. It'd be a pretty boring world and it would slow down the progress of physics.

However there are two equations which come to mind. One is in MTW. It reads

u*Tu = Mass

Where T is the energy-momentum tensor, u is the 4-velocity of the observer and "Mass" is what some people call "relativistic mass" or "mass-energy." MTW use the term "mass" in the section where they demonstrate the the energy-momentum tensor is symetric. This is in the beginning of the section in MTW where they speak of mass as the source of gravity

One can't really define these 4-tensors and 4-vectors without defining the components. I don't recall ever seeing someone define the energy-momentum tensor without ever speaking about the components. There are ways to speak of components as scalars. For example;

m = P*U/c²

Here m = mass (i.e. relativistic mass) of the particle whose 4-momentum is P as measured by an observer whose 4-velocity is U. Some people call this the energy as measured by the observer but it depends on what one calls the time component. As above u*Tu = Mass is the same thing. This is an observer-dependant invariant.

One can take something like the Faraday tensor and contract it with the 4-velocity of an observer and get the electric field 4-vector. Take the scalar product of the electric field 4-vector with a basis vector and get a component of the electric field. Notice that all of this is tensor manipulation, i.e. contractions of tensors etc.

For more on the electric field 4-vector see Wald or Thorne and Blanchard's text

http://www.pma.caltech.edu/Courses/ph136/yr2002/chap01/0201.2.pdf

You've been speaking of GR/SR completely from the geometric view and as if it were the only view that there is. There is also the 3+1 view that is used quite often as is the analytic treatment of tensors rather than the geometric treatment of tensors.

I've snipped the rest since its more of the same, i.e. more of "any number which is not a scalar is useless/not important." I see no use in going down that road since its all semantics.

I know it's an old debate. And I am not at my office so I can't give references but there are many people who do say that we should not teach the concept of relativistic mass anymore, that's it's an historical "faux-pas".

And there are many people who say that one should teach relativistic mass and that it is incorrect to claim that it is an historical "faux-paux". It is certainly no such thing. All arguements I've seen to back that claim up are flawed.

Some examples of authors whose texts use relativistic mass Wolfgang Rindler, Richard A. Mould, Ray D'Inverno, Bernard F. Schutz (see his new book), etc. (These are authors whose texts where published after 1991).

It's a bit like using imaginary time in SR and GR. It used to be done but it is now realized that it's not conceptually a good approach (see for example MWT).
[/qupte]
Why do you refer me to MTW? They themselves use relativistic mass although they call it "mass" or "mass-energy" rather than "relativistic mass."

Take care. I do very sincerely hope that you will be feeling better!

Thanks. I go into surgery Monday. Wish me luck folks.

I'll look. However, my issue is not with the four-momentum, it's with singling out gamma m_0! But I'll look.

Its all semantics anyway. To date nobody has stated anything that can be considered fact as to why one view is correct and the other incorrect. However I have seen people incorrectly think that E/c² = p/v in all cases. That is an incorrect assumption. This is invalid in some instances when a body is under stress.

Up to this point its been philosophy more than relativity.

Pete

 

[da_willem]

Thanks. I go into surgery Monday. Wish me luck folks.

Good luck! I hope this goes well, and you'll feel better afterwards...

 

[humanino]

Yes, good luck Pete ! We appreciate your discussions. I hope you feel better soon.

 

[marlon]

Good luck Pete, and get well soon.

We will be waiting for your return man...

regards
marlon

 

[Tom McCurdy]

GL Pete
it should go fine

 

[Sariaht]

If mass is energy, but energy is not mass cause it's a wave allthough particles are waves and have mass then I may very well be a wave moving through space in the speed of light with a hundred procent risk of turning into dust in the big nothing.

"Must be some cind of forcefield"

 

[Mwyn]

I love subatomic particles I find it a fascinating topic one of "the most' interesting in my opinion but there has been a couple questions that have made me ponder about the true nature of them...
1.are particles pieces of matter? or actual waves?
2.if they are made from actual particles or pieces of matter then are quarks composted of something smaller?
3.if they are not then how can it be possible for it to be fundamental?
4.& Finally if particles are actual waves then what are they waves of? and what prevents us from being see through if we are not made out of actual matter?

If anybody can answer any of these questions you would of helped me out big time and I thank you very much.

 

[misogynisticfeminist]

to Mwyn:

1. Its like asking a hamephrodite (sp.) whether he/she is male or female.

2. Strings perhaps, but i don't think anyone in the physics community ever said that quarks are indeed the most fundamental, and absolutely ruled out anything more fundamental than that.

3. Waves basically carry energy from one place to another. Can't really answer the fourth question though...

: )

 

[marlon]

I love subatomic particles I find it a fascinating topic one of "the most' interesting in my opinion but there has been a couple questions that have made me ponder about the true nature of them...
1.are particles pieces of matter? or actual waves?
2.if they are made from actual particles or pieces of matter then are quarks composted of something smaller?
3.if they are not then how can it be possible for it to be fundamental?
4.& Finally if particles are actual waves then what are they waves of? and what prevents us from being see through if we are not made out of actual matter?

If anybody can answer any of these questions you would of helped me out big time and I thank you very much.

Particles and waves are two different descriptions of the same thing. There are dual. This means that when trying to describe a particle you can look at them as "little bullets" that bounce on some detector giving a "click"-sound for example. A particle is caracterised by stuff like momentum and energy. Now de Broglie found out that momentum p is equal to h/l ; where l is a wavelength and h is the socalled Planck constant. Energy E is equal to hv ; where v is the frequence. Now l and v are quantities describing a wave. So particles are not waves of something. What you can do when you want to describe a particle with energy E and momentum p is "construct" the equantion of a wave with v and wavelength l. You are describing the same thing but with another "language", if you will. So what this wave is, is of no relevance. The only thing that matters is that when you measure some wave (ie : you determin the v and the l) you are allowed to say that you measured a particle with certain mass, E and p...

This way of working is the fundament of quantummechanics.

Quarks are up till now the most fundamental particles that feel the strong force. This force keeps baryons and mesons together and it also binds an atomic nucleus together. Ofcourse it is possible that some more fundamental "thing" is found. Strings are a result of the attempt to unify all known interactions : gravitation, electromagnetic force, weak force (alpha beta and gamma decay) and the strong force. yet this is still a very young and unstable modell.

marlon

 

[marlon]

Well yes, the photon does not acquire a Higgs mass, but that's somewhat secondary.. The most fundamental reason is that it is a gauge boson, and the U(1) gauge symmetry of electromagnetism is both a global symmetry and crucially a local one. Mathematically, that implies it has zero mass.

INDEED HAELFIX, YOU WROTE THE ONLY GOOD ANSWER TO THE QUESTION OF THIS THREAD!


THIS IS THE REASON WHY PHOTONS DO NOT HAVE MAS!
The U(1)-symmetry is never broken. For this reason, there ain't no interaction with the Higgs-field...


marlon

 

[Mwyn]

So basically the question I want to know is that If I could shrink myself into a quantum universe were atoms are the size of basket balls. (highly unlikely yet metaphorically), what would I see? would I see orbs of energy? orbs that look and feel like marbles? or would they be orb like waves?

 

[selfAdjoint]

More like the waves. All the elementary particles in the Standard Model are treated as geometrical points, and no experients exist that give them firm non-zero dimensions, so you wouldn't see orbs. What you might see (assuming of course that you can "see" at this scale) is blobs with a clear gradient of blobness from the center out to trail off at the edges, and a max somewhere in the middle.

 

[Mwyn]

I have another question though? would energy itself or the stuff that bonds particles together be made from smaller particles too? like how electricity is composited of electrons, are the atoms blobs of matter or energy?