Why don't photons possess mass?

[pmb_phy]

pmb_phy : I don't see your point here. I think it has been posted several times that $$E^2 = \vec{p}\,^2 + m^2$$ and that the expressions you use are valid only in the rest frame where $$p=0$$.

I don't believe that you were paying close attention to what I was saying. I stated quite cleary above to da_willem that he and Sariaht were using "m" to mean different things. Sariaht is using "m" to refer to relativistic mass and da_willem was using it to mean proper mass. Also E refers to energy, not proper energy E₀. As da_willem is using "m" then it E does not equal mc², its E₀ = mc² where E₀ is E as measured in the rest frame of the particle.

The relation $E^2 - (pc)^2 = m^2 c^2$ is valid if an only if "m" is the proper mass of the particle. If, as Sariaht was using it (and how I use it) m is the relativistic mass of a particle. Defined as such the relation $E^2 - (pc)^2 = m^2 c^2$ is invalid. Also when m is the relativistic mass then the relation E = mc² is valid.

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 $$m=0$$).

I've never thought that defining "rest mass" in that way was a very good idea. Also one cannot legitimately define proper mass in that way since its a circular definition since the momentum of a tachyon is defined is defined in terms of proper mass. That is merely a relationship between energy, momentum and proper mass. It can't be taken as a definition of proper mass since that requires leaving momentum undefined - bad idea.

In a flat spacetime there is no gravitational field.

Depends on how one defines "gravitational field" and as I said when I posted that at least according to Einstein and according to Einstein the existence of a gravitational field is frame dependant and one can "produce" a gravitational field by chaning coordinates. To be precise - the presence of a gravitational field is dictated by the non-vanishing of the affine connection (when the spatial coordinates are Cartesian) and not the non-vanishing of the Riemann tensor.

Usually one defines the graviton field as a perturbation from the flat metric (even though it might not be the best convention : see Rovelli which defines the graviton field as the vierbein instead) ...if the metric is globally flat in A then their can't be any gravitational field in it.

That is a different notion than that given by Einstein. You're thinking of "gravitational field" as the non-vanishing of the Riemann tensor. I am not and Einstein did not. You can define "gravitational field" anyway you like but I have yet to see a good reason to define it in any other way than Einstein did, e.g. in his 1916 GR paper or any of his texts. In fact not everyone speaks of it that way in the GR literature.

Is this the meaning of your posts ?

No.

Pete

 

[humanino]

I still don't get it. I am not doing it on purpose, i am truly sorry, but I want to understand.

Why do you introduce the dinstiction between Energy and Proper energy whereas it is irrelevant to the the Photon ?

What kind of gravitational field could have a vanishing Riemann tensor and a non-vanishing affine connection ? The action for the gravitational is determined by the integral of the curvature : if the curvature is zero, there is non action, no Gravitational field ! Where am I wrong ?!

Thank you for help !

 

[humanino]

Sure. See Einstein's gravitational field at http://xxx.lanl.gov/abs/physics/0204044

I read the paper you are reffering here :
1 It does not seem to be submitted to any journal
2 The email address given here is hotmail address. Does not this guy has a position in an institute ?
3 In the aknowledgement, the author states that the people whose name is mentionned are not responsible for the possible mistakes in the paper. How should the reader interpet this ?
4 The view proposed in this paper are supposed to be Einstein's one, with opposition to other physisicists such as : Hawking and Thorne. I need better references to challenge such physicists names, not just pretending you have Einstein on your side. I see this paper's view as a very old fashioned one. The formalism is old fashionned. It looks like a student work !

I believe I am paying attention to your posts. I demand more argumentation. I was not doubting you have a real point until I read your reference.

 

[humanino]

The most convincing argument I saw so far for what the gravitational field is that is : how it should be mathematically represented, comes from Rovelli's last book, where he motivates the vierbein as the gravitational field. The vierbein in turn can be looked as a "square root" of the metric. Fermions require the introduction of the vierbein.

 

[pmb_phy]

Why do you introduce the dinstiction between Energy and Proper energy whereas it is irrelevant to the the Photon ?

It was you who commented on the applicability of the relation E = mc². I pointed out that relation is E₀ = mc² and, assuming m = proper mass, holds only for proper energy. I.e. you omitted the subscript denoting proper energy.

What kind of gravitational field could have a vanishing Riemann tensor and a non-vanishing affine connection ?

Any gravitational field with no tidal forces, E.g. a uniform gravitational field. The metric of a uniform gravitational field is

$$ds^2 = c^2(1+gz/c^2)^2dt^2 - dx^2 - dy^2 - dz^2$$

The Christoffel symbols are calculated at - http://www.geocities.com/physics_world/uniform_chris.htm. The only non-vanishing Christoffel symbols are

$$\Gamma ^0_{03} = \frac {g}{c^2}\frac{1}{1 + gz/c^2}$$
$$\Gamma ^3_{00} = -\frac {g}{c^2}\frac{1}{1 + gz/c^2}$$

where g = gravitational acceleration at z = 0. According to the equivalence principle a uniform gravitational field is equivalent to a uniformly accelerating frame of reference in flat spacetime. One can obtain the metric for the uniform g-field by transforming from an inertial frame in flat spacetime to a uniformly accelerating frame of reference. The Christoffel symbols will change from zero to the above values but the Riemann tensor will remain zero as it must (i.e. if a tensor vanishes in one coordinate system then it vanishes in all coordinate systems). The uniform gravitational field was the very first gravitational field that Einstein considered in his theory of general relativity. In fact one form of the equivalence principle is stated as follows

Einstein's Equivalence Principle: A uniform gravitational field is equivalent to a uniformly accelerating frame of reference.

A few examples of a gravitational field with zero spacetime curvature from the general relativity/cosmology literature are the gravitational field of a vacuum domain wall and a straight cosmic string. E.g. see

Gravitational Field of Vacuum Domain Walls, Alexander Vilenkin, Phys. Lett. 133B, page 177-179

Gravitational field of vacuum domain walls and strings, Alexander Vilenkin, Phys. Rev. D, Vol 23(4), (1981), page 852-857

Cosmic strings: Gravitation without local curvature, T. M. Helliwell, D. A. Konkowski, Am. J. Phys. 55(5), May 1987, page 401-407

(Note: The author of the Ma. J. Phys. article uses the term "local curvature" in the title of this paper but he is referring to Riemann = 0)

The action for the gravitational is determined by the integral of the curvature : if the curvature is zero, there is non action, no Gravitational field ! Where am I wrong ?!

What does the action have to do with gravitational acceleration?

The view proposed in this paper are supposed to be Einstein's one, with opposition to other physisicists such as : Hawking and Thorne.

It is a fact that Hawking and Thorne are in opposition to Einstein on this point. But then again they're in opposition to other GRist such as Tolman too.

In any case this is not about what Hawking and Thorne think. Its about what Einstein thought. You haven't supported your claim that Einstein associtated the non-vanishing of the gravitational field with the non-vanishing of the Riemann tensor. Would you care to support your claim?

Its funny that you should mention Thorne since it was Thorne who pointed out to me that the Riemman tensor for a uniform gravitational field is zero. In fact he was the one who sent me the references to the paper by Vilinken for the domain wall which also has zero spacetime curvature (in the region of the gravitational field outside the wall itself). When I asked him about this and his position on gravity = curavture he simply told me that it all depends on how you define gravity.

And is also not in opposition to the GR historian/GR expert Dr. John Stachel, Boston University, former head of the Einstein Papers project. His paper is referenced in that article. Did you even read the paper you're now complaining about? It was intented to explain all the details of why I said this since its too long to get into in detail in a forum.

Try reading The Foundation of the General Theory of Relativity, A. Einstein, Annalen der Physik, 1916 and/or The Meaning of Relativity, A. Einstein, Princeton University Press.

It's pretty clear what Einstein meant when he wrote in a letter to Max von Laue (Einstein to Max von Laue, September 1950)

... what characterizes the existence of a gravitational field from the empirical standpoint is the non-vanishing of the components of the [affine connection], not the vanishing of the [components of the Riemann tensor]. If one does not think in such intuitive (anschaulich) ways, one cannot grasp why something like curvature should have anything at all to do with gravitation. In any case, no rational person would have hit upon anything otherwise. The key to the understanding of the equality of gravitational mass and inertial mass would have been missing.

That comment cannot be taken in any other way.

But if you want more on this then all you need to do is go to the library and read How Einstein Discovered General Relativity: A Historical Tale With Some Contemporary Morals, J.J. Stachel, General Relativity and Gravitation, Proceedings of the 11th International Conference on General Relativity and Gravitation, (Stockholm,Cambridge University Press, Jul 6-12, 1986). As Stachel writes on page 202

Within a few year years, Lavi-Civita, Weyl and Cartan generalized the Christoffel symbols to the concept of affine connection. This concept served to make the relationship between the mathematical representations of various concepts much clearer. Just because it is not a tensor field, the connection field provides adequate representation of the gravitational-cum-inertial field required by Einstein's equivalence principle. There is no (unique) decomposition of the connection field into an inertial plus gravitational tesor.

Since Stachel is probably the world's leading authority on the history of general relativity, as well as being a noted GRist as well, I'm very comforatable with everything I've explained.

In any case this is all written in Einstein's work. All one has to do is pick it up and read it. Why would you think otherwise? On what basis do you hold that the Riemann tensor must vanish when there is no gravitational field?

A uniform gravitational field is defined as a gravitational field with no spacetime curvature. For this definition and a derivation of the metric based on this definition please also see Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173.

Pete

 

[pmb_phy]

Although this is getting way off topic I'm going to address one more point - that of da_willem

As I understand it the metric (as determinded by the energy content of space) is the 'replacement' of gravitational potential in general relatvity.

It can be shown (See - http://www.geocities.com/physics_world/gr/grav_force.htm) that the components, G_k of the gravitational force, G, on a particle moving in a gravitational field can be expressed in terms of the gravitational potentials $g_{\alpha\beta}$ as

$$G_k = \frac{1}{2}mg_{\sigma\beta,k}v^{\sigma}v^{\beta}$$

where $v^{\sigma} = (1, \bold v)$ and $m = \gamma m_0$. Its possible, as is the case for a uniform gravitational field, for G to be finite and non-zero and yet Riemann = 0. In the special case where the metric has constant spatial components then we can express the above in terms of $\Phi = (g_{00} - 1)c^2/2$ as

$$\bold G = -m \nabla \Phi$$

The metric for a uniform gravitational field is (Ref. - Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173)

$$ds^2 = c^2(1+gz/c^2)^2dt^2 - dx^2 - dy^2 - dz^2$$

for which Reimann = 0 but for which G is not zero.

Pete

 

[humanino]

Dear Pete : the examples you provide to illustrate/motivate the assertion that the gravitational field is not identical to curvature of spacetime

a uniform gravitational field

Gravitational Field of Vacuum Domain Walls, Alexander Vilenkin, Phys. Lett. 133B, page 177-179

Gravitational field of vacuum domain walls and strings, Alexander Vilenkin, Phys. Rev. D, Vol 23(4), (1981), page 852-857

Cosmic strings: Gravitation without local curvature, T. M. Helliwell, D. A. Konkowski, Am. J. Phys. 55(5), May 1987, page 401-407

(Note: The author of the Ma. J. Phys. article uses the term "local curvature" in the title of this paper but he is referring to Riemann = 0)

seem to me to be specific or pathological examples. I certainly do not consider relevant the example of a constant gravitational field. Of course, by assuming a constant gravitational field, one can obtain many sorts of inconsistent results, but this is only due to the fact that a gravitational field is never constant. This is non-physical. There is a local approximation in which the gravitational field looks constant : this is only a local approximation.

The other examples you provide (as well as those provided in the reference physics/0204044) are either : non-physical, or at least the relevance of which is far from being proven : examples such as the Vacuum Domain Walls are nice mathematical constructions, but certainly not physical situations proven to occur in Nature. Or, the other examples such as Cosmic strings, are too specific situations : it is clear that GR is not perfectly suited to describe cosmic strings, but we are able to investigate them with GR, even though it requires re-constructing the geometrical shape of spacetime around the string "by hand", that is we have no general method to systematically handle those structures. At least, I consider the examples provided so far unable to motivate the assertion "the gravitational field is not equivalent to the curvature".

What does the action have to do with gravitational acceleration?

You were referring to the possibility that in an empty region of spacetime, a spurious gravitational-like field could artificially appear due to a pathological choice of coordinates. Otherly said : if gravitation is equivalent to acceleration, one could construct fake gravitational fields by choosing fuzzy coordinates. I was answering here that it seems to me, this pseudo-paradox has been understood now for a while. This assertion is wrong, and it is most easily seen with the lagrangian formulation :
(1) the integral of the curvature vanishes on any subdomain of A
(2) the gravitational field is zero on A
Assertion (1) is equivalent to assertion (2). This the way I understood GR so far, and I never faced any inconsistency.

 

[pmb_phy]

You're really taking this thread off topic. This thread is about the mass of photons. Not the definition of gravitational field. If you must keep this up then I suggest that you do this in PM so as not to take this thread more off topic than it already is.

Dear Pete : the examples you provide to illustrate/motivate the assertion that the gravitational field is not identical to curvature of spacetime ... seem to me to be specific or pathological examples.

That's quite meaningless. The term "pathological" does not apply in this context. The only meaning that can be given to your use of the term would be "abnormal" and all corodinate systems are equally normal in general relativity. It appears to me that you've simply chosen a term with negative connotations to describe something you don't like or have never considered before. But feel free to call it what you like. It does not detract from the fact that, as Einstein defined it, it is a gravitational field and this is about how Einstein defined it.

Of course, by assuming a constant gravitational field, one can obtain many sorts of inconsistent results,..

Care to provide proof? If so then please do so in PM. I don't care to waste more space here on something as off-topic as this.

Even in Newtonian gravity an example of a uniform gravitational field which can be generated with a finite distribution of mass can readily be found. There is no basis for claiming that a uniform g-field is non-physical.

In any case, as Einstein said, you can "produce" a gravitational field merely by changing coordinates. To be exact, in his 1916 GR paper Einstein wrote

It will be seen from these reflexions that in pursuing the general theory of relativity we shall be led to a theory of gravitation, since we are able to "produce" a gravitational field merely by changing the system of co-ordinates.

Sorry but the rest of your comments are nothing more than comments based on your idea of what a gravitational field should be. This is simply off this side topic, i.e. as how Einstein defined it. It was already apparent that you don't like the way Einstein defined it.

In any case you've missed the point. The point was that the criteria for the presence of a gravitational fields is the non-vanishing of the the Christoffel symbols and not the non-vanishing of the Riemman tensor.

As far a "physical" you have not provided proof that domain walls have never existed and you have not provided proof that straight cosmic strings don't exist nor have your provided proof that a uniform gravitational field does not exit and you have not proved that a uniformly accelerating frame of reference is not equivalent to a uniform gravitational field.

The existence of a gravitational field depends on the choice of coordinates. That is the essence of the equivalence principle - i.e. the gravitational field can be transformed away at any point in spacetime, curvature cannot - one can only choose a region of spacetime small enough so that you can ignore the effects of tidal forces - and the interpretation of this point is pretty debated in the relativity community.

But as I've said, you can feel free to define it as you wish. But as I've also said that I was telling you how Einstein defined it. And the presence of a gravitational field does not require the presence of spacetime curvature.

Let me put this in the most simplest terms I can - As defined by Eintein (not Hawking etc) If there is a gravitational field at a point P in spacetime then it does not imply that the Riemann tensor vanishes at P. Also, the existence of the gravitational field is dependant on the choice of spacetime coordinates.

Do you care to prove that Einstein defined it as being different from what I've explained to you? If so then please do so in PM.

 

[humanino]

please follow the discussion

Dear Pete,

I wish to continue this discussion, at least I would like to go deeper in understanding your considerations. Since this is off-topic in this forum, I will post a new thread in the General physics forum.

 

[pmb_phy]

Dear Pete,

I wish to continue this discussion, at least I would like to go deeper in understanding your considerations. Since this is off-topic in this forum, I will post a new thread in the General physics forum.

Please don't. The subject is general relativity so if you wish to start a new thread then I recommend that you place it in the relativity forum.

 

[humanino]

Sorry ! That's too late. A moderator could move the thread form General physics to S&G Relativity.

 

[pmb_phy]

Dear Pete : the examples you provide to illustrate/motivate the assertion that the gravitational field is not identical to curvature of spacetime ...seem to me to be specific or pathological examples.

One last comment - It takes logic to prove an assertion is true. But it takes just one example to prove it's false.

I've provided two examples (vacuum domain walls and something strange involving strings) and that's twice as much as was needed to prove the gravity=curvature assertion false.

Pete

 

[pmb_phy]

Sorry ! That's too late. A moderator could move the thread form General physics to S&G Relativity.

You understand, don't you, that you can delete a thread you started and place it in another forum by yourself, right?

Pete

 

[humanino]

Yes Pete ! I hear your argumentation, and I find it really intersting. I had elementary logic too in my math lectures
I am not trying to prove you wrong. I am trying to convince myself that indeed, "the gravity=curvature assertion false" as you said. Although, I really have not been convinced (yet?) by the examples you provided.

I hope you will participate the following of the discussion.

 

[da_willem]

That is incorrect. If one is using m to mean proper mass then E₀ = mc², not E = mc².

Pete

But when p=0 there is no distiction between energy and proper energy, so you could equally well write E=mc^2 with m proper mass, right?!

If, as Sariaht was using it (and how I use it) m is the relativistic mass of a particle. Defined as such the relation is invalid. Also when m is the relativistic mass then the relation E = mc2 is valid.

My textbook on relativity sais the usage of relativistic mass is pretty outdated and hardly anybody uses it anymore. So when I wrote m, I thought everybody knew what I meant...

The existence of a gravitational field depends on the choice of coordinates. That is the essence of the equivalence principle - i.e. the gravitational field can be transformed away at any point in spacetime, curvature cannot

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.

Even in Newtonian gravity an example of a uniform gravitational field which can be generated with a finite distribution of mass can readily be found

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?

 

[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 [itex]E = \gamma E_0[/tex]. However this is not the case. The energy transforms as [itex]E = E_0/\gamma[/tex]. For details please see
http://www.geocities.com/physics_world/sr/rd_paradox.htm

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

Sorry but that article is incorrect. It states

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.

Today, when physicists talk about the mass of an object they always mean the rest mass.

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/

Among a lot of arguments I especially liked the following:

And the quote from Einstein himself:

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

$$m = m_0 \frac{dt}{d\tau}$$

For a time orthogonal spacetime (i.e. g_0i = 0)

$$m = \frac{m_0}{\sqrt{1 + 2\Phi/c^2 - v^2/c^2}}$$

When v << c m becomes

$$m \simeq \frac{m_0}{\sqrt{1 + 2\Phi/c^2}}$$

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 The Meaning of Relativity. 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.

By the way, there is a second part to that letter Einstein wrote to Barnett. It reads

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_ik 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_ik 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.

Do you agree with this part of that letter to?

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 [itex]E = \gamma E_0[/tex]. However this is not the case. The energy transforms as [itex]E = E_0/\gamma[/tex]. For details please see
http://www.geocities.com/physics_world/sr/rd_paradox.htm

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.

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 [itex]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 $2\gamma M$.

I'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 "Oh! That's not because of mass increase, that's because energy and momentum etc increase etc with speed". 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's equations. Relativistic mass takes the place of charge and mass has been considered a "gravitational charge" even in Newtonian gravity.

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?

I dunno. Each person has a different opinion on this. What one person thinks is "useful" another person thinks of as "useless".

He never really succeeded, right?

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

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 [itex]\bold p = m \bold v[/tex] have a role? After all I never need to substitute v into any equation. But let's try it anyway.

$$\bold F = \frac{d\bold p}{dt}$$

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't have to be defined. E.g. I can write the Lorentz "force" law as

$$\frac{d}{dt} \frac{m_0 (d\bold r/dt)}{\sqrt{1-(d\bold r/dt)^2/c^2}} = q(\bold E + \bold v \times \bold B)$$

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.

I call the introduction of each term, i.e. v, m, p, gamma, f, etc. "Economy of thought."

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 [itex]\bold p = m \bold v[/tex] have a role? After all I never need to substitute v into any equation. But let's try it anyway.

$$\bold F = \frac{d\bold p}{dt}$$

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't have to be defined. E.g. I can write the Lorentz "force" law as

$$\frac{d}{dt} \frac{m_0 (d\bold r/dt)}{\sqrt{1-(d\bold r/dt)^2/c^2}} = q(\bold E + \bold v \times \bold B)$$

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.

I call the introduction of each term, i.e. v, m, p, gamma, f, etc. "Economy of thought."

Pete

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's of course very misleading.

To make my point, consider the following step:

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

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.

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.

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.
Things are defined according to how they transform under certain symmetries. That's what we always do in physics, from simple nonrelativistic mechanics to particle physics and GR. Symmetry is the key point. Unfortunately it's not pointed out often. For example, the quantity $m_0$ is a useful concept because it'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 "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 $\gamma m_0$. It has no interesting property so it should not really be given a special name. 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 (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'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 $\gamma m_0 \bold v$.

Pat