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Gonzolo

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selfAdjoint

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Gravity curves spacetime near a ray of light becouse light has momentum and energy, and those gravitate, along with the mass the light doesn't have.Gonzolo said:

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What happens with black holes is different matter though..It isnt gravity, actually, the black hole wraps space time, and curves space in the shape of a tunnel, so light is actually travelling in straight line into that tunnel (into the black hole.)...

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Keep on chuggin !!

Vern

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As mentioned above, when it is said that light has no mass it is meant that light has zerophenylalanine said:If light has a mass, how can it travel at the speed of light?

There is more on this subject in the

Since inertial mass = passive gravitational mass (defined as that which is acted on by gravity) it follows that light has non-zero passive gravitational mass. Richard Feynman stated this in another way inAnd if it doesn't, why is it affected by gravity (e.g. black holes)?

One feature of this new law is quite easy to understand is this: In Einstein relativity theory, anything which has energy has mass -- mass in the sense that it is attracted gravitationaly. Even light, which has energy, has a "mass". When a light beam, which has energy in it, comes past the sun there is attraction on it by the sun.

Spacetime curvature is not the cause of light being deflected. It is a description of light being deflected in a non-uniform gravitational field. What you said here is equivalent in Newtonian language, to saying that objects fall because of tidal forces. That is an incorrect way of explaining why things fall using Newtonian physics. Similarly, spacetime curvature is another name for tidal forces. The gravitational deflection of light requires only the presences of a gravitational field. It does not require the presence of spacetime curvature. Light can be deflected in a uniform gravitational field and such a field has zero spacetime curvature.Gonzolo said:And it is affected by gravity because gravity curves space-time.

That pertains only to proper mass, not inertial mass.HIGHLYTOXIC said:Light doesnt have mass...Its mass is 0 otherwise, it wudnt have travelled at c...

Pete

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Gonzolo

Well... I understand it is not the most precise complete explanation, but it was short and quick, I did say it was the short answer... (and was the earliest :tongue2: )pmb_phy said:Spacetime curvature is not the cause of light being deflected. It is a description of light being deflected in a non-uniform gravitational field. What you said here is equivalent in Newtonian language, to saying that objects fall because of tidal forces. That is an incorrect way of explaining why things fall using Newtonian physics. Similarly, spacetime curvature is another name for tidal forces. The gravitational deflection of light requires only the presences of a gravitational field. It does not require the presence of spacetime curvature. Light can be deflected in a uniform gravitational field and such a field has zero spacetime curvature.

That pertains only to proper mass, not inertial mass.Pete

Strictly speaking, I understand that if there is a uniform gravitationnal field somewhere, it has to be inclined relative to "free space". Between both, there would be a curvature (or a discontinuity in the limit). Of course, I'm thinking with balls on a 2D sheet. I think I can give myself credit and say curvature is necessary "somewhere". (though I agree it doesn't have to be exactly where the light is.)

Tidal force? Are you saying I implied a third body (the Moon)? How is saying an object falls because of a force between the object and the Earth incorrect? What about the Lagrangian view?

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This is false, if something has zero rest mass then it must have zero relativistic mass because of SR equations.pmb phy said:As mentioned above, when it is said that light has no mass it is meant that light has zero proper mass (aka rest mass). It does not mean that it has zero inerrtial mass. Inertial mass (aka relativistic mass) is defined as the m in p = mv.

[tex]m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

If rest mass is zero, then relativistic mass must be zero as well because zero divided by anything is zero.

Relativistic mass isnt determined by p=mv, its determined by the equation above. p=mv is newtonian and isnt allowed to be used to calculate the momentum of light either because it would yield to zero momentum, therefore from the total energy equation for a massless particle moving at c...

[tex] E^2 = m_0^2c^4 + p^2v^2[/tex]

[tex] E^2 = (0)c^4 + p^2(c)^2[/tex]

[tex] E^2 = p^2c^2[/tex]

[tex] \frac{E^2}{c^2} = p^2[/tex]

[tex] p = E/c [/tex]

Where the energy is plank's constant times the frequency of the light.

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My appologies. This will be a long post. Seems like this same topic comes up all the time! So I'll try to be clear and thorough.

_{0} is not an identity, it is an equality when the particle is a tardyon (i.e. a particle which always moves with v < c). It is this definition that I will refer to below. I will use the term "mass" to refer to inertial mass aka relativistic mass.

Mass is**defined** as the quantity *m* such that the quantity *m***v** is conserved. Define **momentum** as **p** = *m***v**. You could say that mass is defined so that momentum is conserved (It can then be shown that m = E/c^{2}). If the particle is a tardyon then one **derives** m = gamma*m_{0} on the basis of momentum conservation. It then turns out that the momentum of a tardyon in is

[tex]\bold p = \gamma m_0 \bold v[/tex]

Substitute the expression m = gamma*m_{0} into this expression for the momentum and you'll see that p = mv.

_{0}. If you want to use this equation for light then start with the definition m = p/v = p/c. Otherwise you're forgetting that when v->c the denominator goes to zero so that you have an improper(?) limit m = 0/0. This can be interpreted as saying that as v->c , m-> constant. This is how most relativists interpret this. E.g. see Rindler's latest text. Its in there.

m = p/v is the defining relation for relativistic mass. I urge you not to take my word for it but to look it up (see references below). I.e. Look in the relativity literature for yourself. You you can read the work of the people who are responsible for making this definition well known, namely Tolman and Lewis (Tolman's text covers this too). Max Jammer goes into detail on this in his new book on mass. Have you seen this book?

If you're really interested in this point then see the**Defining articles for relativistic mass**

* More recent publications on this subject*

**Special Relativity** A.P. French. French defines "inertial mass" (aka relativistic mass) as follows. From page 16

_{0}*v = mv. In any case relativistic mass is **defined** as the ratio m = p/v and therefore p = mv. Haven't you ever wondered why you can write momentum as the product of relativistic mass and velocity?

^{2} into that equation and you'll get p = mc and the definition still holds.

May I ask where you got the idea that relativistic mass is defined as above, i.e. m = gamma*m_{0}? That is a correct relationship, however it is not an identity. It is an equality, but only for tardyons.

Pete

You can define it that way if you wish. But I'm going by the definition which is almost always given in the relativity literature and how it was originally defined by Tolman and Lewis (kind of defined by Planck too in a limited sense). In that sense relativistic mass is not defined as you say, i.e. m = gamma*mArmoSkater87 said:This is false, if something has zero rest mass then it must have zero relativistic mass because of SR equations.

[tex]m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

Mass is

[tex]\bold p = \gamma m_0 \bold v[/tex]

Substitute the expression m = gamma*m

That equation was derived on the assumption that the particle travels less than the speed of light. As such the m in p = mv is m = gamma mIf rest mass is zero, then relativistic mass must be zero as well because zero divided by anything is zero.

Hardly anybody (probably nobody) in the relativity community has ever defined relativistic mass that way. Almost all (if not all) derivations of that relation that you'll find in the relativity literature will demonstrate this fact.Relativistic mass isnt determined by p=mv, its determined by the equation above.

That is incorrect. Why would you assume that Newton's laws are incorrect or don't apply to SR? Newton's second law, F = dp/dt, is still valid in SR. Newton's third law also applies for contact forces.p=mv is newtonian and isnt allowed to be used to calculate the momentum of light either because it would yield to zero momentum, therefore from the total energy equation for a massless particle moving at c...

m = p/v is the defining relation for relativistic mass. I urge you not to take my word for it but to look it up (see references below). I.e. Look in the relativity literature for yourself. You you can read the work of the people who are responsible for making this definition well known, namely Tolman and Lewis (Tolman's text covers this too). Max Jammer goes into detail on this in his new book on mass. Have you seen this book?

If you're really interested in this point then see the

The Principle of Relativity and Non-Newtonian Mechanics, R.C. Tolman, G.N. Lewis, Philosophical Magazine, 18, 510-523 (1909)

Non-Newtonian Mechanics: The Mass of a Moving Body, R. C. Tolman and G. N. Lewis, Philosophical Magazine, 18, (1912), pp 375-380

A popular SR text, one that is often used at MIT in fact, isThe Classical and Relativistic Concepts of Mass, Erik Eriksen and Kjell Vøyenli, Foundations of Physics, 6(1), 1976, pp 115-124.

Concepts of Mass in Contemporary Physics and Philosophy, Max Jammer, Princeton University Press (1999) Chapter Two, pages 41-61.

You can call p = mv "Newtonian momentum" or "Mrs. Buttersworth's momentum" but whatever you call it the fact remains the same - p = gamma*mBy inertial mass we mean the ratio of linear momentum to velocity.

Place E = mc[tex] p = E/c [/tex]

Where the energy is plank's constant times the frequency of the light.

May I ask where you got the idea that relativistic mass is defined as above, i.e. m = gamma*m

Pete

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Agreedpmb_phy said:Mass isdefinedas the quantitymsuch that the quantitymvis conserved. Definemomentumasp=mv. You could say that mass is defined so that momentum is conserved (It can then be shown that m = E/c^{2}). If the particle is a tardyon then onederivesm = gamma*m_{0}on the basis of momentum conservation. It then turns out that the momentum of a tardyon in is

[tex]\bold p = \gamma m_0 \bold v[/tex]

Substitute the expression m = gamma*m_{0}into this expression for the momentum and you'll see that p = mv.

Agreed, 0/0=undefined. I forgot that SR equations cant be used to tell what happens AT c.That equation was derived on the assumption that the particle travels less than the speed of light. As such the m in p = mv is m = gamma m_{0}. If you want to use this equation for light then start with the definition m = p/v = p/c. Otherwise you're forgetting that when v->c the denominator goes to zero so that you have an improper(?) limit m = 0/0. This can be interpreted as saying that as v->c , m-> constant. This is how most relativists interpret this. E.g. see Rindler's latest text. Its in there.

Thats strange because I though the word "relativistic" in general has to do with something close to the speed of light. For example "relativistic speeds" would be speeds close to c, and "relativistic mass" would be the mass as seen for something traveling close to c.Hardly anybody (probably nobody) in the relativity community has ever defined relativistic mass that way. Almost all (if not all) derivations of that relation that you'll find in the relativity literature will demonstrate this fact.

Perhaps we have two different definitions of relativistic mass? That may be cause the disagreement.That is incorrect. Why would you assume that Newton's laws are incorrect or don't apply to SR? Newton's second law, F = dp/dt, is still valid in SR. Newton's third law also applies for contact forces.

m = p/v is the defining relation for relativistic mass. I urge you not to take my word for it but to look it up (see references below). I.e. Look in the relativity literature for yourself. You you can read the work of the people who are responsible for making this definition well known, namely Tolman and Lewis (Tolman's text covers this too). Max Jammer goes into detail on this in his new book on mass. Have you seen this book?

Thanks, I'll check it out sometime.If you're really interested in this point then see theDefining articles for relativistic mass

More recent publications on this subject

A popular SR text, one that is often used at MIT in fact, isSpecial RelativityA.P. French. French defines "inertial mass" (aka relativistic mass) as follows. From page 16

Ok..now lets see, you say relativistic mass of light is p/v...in which case p=E/c. Since light moves at c, then its mass = E/c/c = E/c^2. That is not the "mass" of the light, it is in a sense its "potential" mass, or in other words the maximum mass into which that light can be converted into. In that sense and only in that sense can you say that light has "mass", but it really doesnt.You can call p = mv "Newtonian momentum" or "Mrs. Buttersworth's momentum" but whatever you call it the fact remains the same - p = gamma*m_{0}*v = mv. In any case relativistic mass isdefinedas the ratio m = p/v and therefore p = mv. Haven't you ever wondered why you can write momentum as the product of relativistic mass and velocity?

Place E = mc^{2}into that equation and you'll get p = mc and the definition still holds.

Agreed, only for tardyons.May I ask where you got the idea that relativistic mass is defined as above, i.e. m = gamma*m_{0}? That is a correct relationship, however it is not an identity. It is an equality, but only for tardyons.

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Note: There are many threads in the SR/GR forum which cover everything we've spoken about in this thread and probably covers everything we could speak about in this thread. Its a much discussed topic in all physics forums and newsgroups.

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

I don't care for the term "relativistic mass". I prefer "inertial mass". In any case the definition of "relativistic mass" is the m in p = mv, same as inertial mass in Newtonian mechanics. So inertial mass applies to all speeds. Think of it like this - momentum also is sometimes called "relativistic momentum" and has different properties than momentum does at low speeds. I.e. In Newtonian mechanics the momentum will always be finite when the speed is finite. Not true for relativistic momentum.

**defined** as m = gamma*m_{0} but it is unclear to me whether you know where that relationship comes from, how it is derived and where it fits into the dynamics of relativistic particles.

In my view light has mass in all senses of the term mass. There are three senses of the term "mass" they are defined as

1) Inertial Mass - That which gives a body momentum.

2) Passive Gravitational Mass - That on which gravity acts.

3) Active Gravitational Mass - That which generates a gravitational field.

Light has all those properties.

Have you ever walked through a derivation of the formula m = gamma*m_{0}*v? If you do then you'll see that the "m" in the derivation is implicitly defined as the m in p = mv. Give it a try. Find a derivation and look at the starting assumptions and definition of "mass". You will find that these derivations start with the requirement that mv is conserved and then set out to find m as a function of v. The second part (finding m as a function of v) is a second requirement. If one starts with a photon and an electron and require mv be conserved then we can't assume that m is a function of v for the photon. We simply find m in terms of other quanties such as the freqeuncy, the momentum or the energy. I know of no SR text which uses of relativistic mass which doesn't say that light has relativistic mass. Three examples I know of off hand are Mould, Rindler, D'Inverno (all of which are both SR and GR) and French. In other texts which don't speak of relativistic mass, e.g. Foster and Nightingale, Ohanian, Taylor and Wheeler, they may still speak of the mass (or mass density) of radiation. This happens in the later two when there is electromagnetic energy whose total momentum is zero. E.g. two photons of equal momentum traveling in opposite directions. When evaluted in that frame is then called "rest energy" and labeled E_{0}. Using E^{2} - (pc)^{2} = m_{0}^{2} and letting p = 0 will yield E = E_{0} gives E_{0} = m_{0}c^{2}. No photon has this property. This is a property of a system of photons.

Also, don't be surprised if you see such well known texts, such as Jackson's EM text, which speaks of the center of mass of radiation.

As far as the mass of light - This has a long history, dating back before Einstein in fact. From my investigation into this subject it has become clear that it was Poincare who first spoke of the mass of radiation back in 1900. Einstein's first statement to this effect is in his article**The Principle of the Conservation of Motion of the Center of Gravity and the Inertia of Energy**, A. Einstein, Annalen der Physik 20 (1906): 627-233. Specifically he states this in the second section where he assigns a mass density to the electromagnetic field. He never seemed to waver on this point as evidenced in his 1916 GR paper and his and Infeld's book **The Evolution of Physics**.

Pete

ps - As you can tell, I love this topic and could go on forever on it.

Relativity also applies to low spead motion. In fact relativistic mass has even been measured for slowly moving particles. An article on this appeared in the American Journal of Physics. That article isArmoSkater87 said:Thats strange because I though the word "relativistic" in general has to do with something close to the speed of light. For example "relativistic speeds" would be speeds close to c, and "relativistic mass" would be the mass as seen for something traveling close to c.

I don't care for the term "relativistic mass". I prefer "inertial mass". In any case the definition of "relativistic mass" is the m in p = mv, same as inertial mass in Newtonian mechanics. So inertial mass applies to all speeds. Think of it like this - momentum also is sometimes called "relativistic momentum" and has different properties than momentum does at low speeds. I.e. In Newtonian mechanics the momentum will always be finite when the speed is finite. Not true for relativistic momentum.

That is most likely the case.Perhaps we have two different definitions of relativistic mass? That may be cause the disagreement.

That depends on the definition of mass. According to how relativistic mass is defined, it is mass, just like any other mass, i.e. the ratio of momentum to speed - by definition!Ok..now lets see, you say relativistic mass of light is p/v...in which case p=E/c. Since light moves at c, then its mass = E/c/c = E/c^2. That is not the "mass" of the light, ...

I have to admit that I don't understand what you mean by this without first understanding how you think "relativistic mass" is defined. You seem to hold that m is...it is in a sense its "potential" mass, or in other words the maximum mass into which that light can be converted into. In that sense and only in that sense can you say that light has "mass", but it really doesnt.

In my view light has mass in all senses of the term mass. There are three senses of the term "mass" they are defined as

1) Inertial Mass - That which gives a body momentum.

2) Passive Gravitational Mass - That on which gravity acts.

3) Active Gravitational Mass - That which generates a gravitational field.

Light has all those properties.

Have you ever walked through a derivation of the formula m = gamma*m

Also, don't be surprised if you see such well known texts, such as Jackson's EM text, which speaks of the center of mass of radiation.

As far as the mass of light - This has a long history, dating back before Einstein in fact. From my investigation into this subject it has become clear that it was Poincare who first spoke of the mass of radiation back in 1900. Einstein's first statement to this effect is in his article

Pete

ps - As you can tell, I love this topic and could go on forever on it.

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Yes, that is exactly how I define as relativistic mass. The relationship come from Lorentz transformation. I dont know how that equation is derived, but do know how a similar one is, the equation for relativistic time. I love derivations so i'll make sure to find one online and take a look myself. Sorry if i've brought up a stupid arguement, but I really dont know all that much about SR, ive never learned SR anywhere except at home, in this forum the past few months.pmb phy said:I have to admit that I don't understand what you mean by this without first understanding how you think "relativistic mass" is defined. You seem to hold that m is defined as m = gamma*m0 but it is unclear to me whether you know where that relationship comes from, how it is derived and where it fits into the dynamics of relativistic particles.

So, from what you are saying light's mass is no different than the mass of the monitor im looking at right now. That bring up a question that I have...If light has mass how come, for example...a very heated block of iron (in a vacuum) doesnt lose mass from its emittion of photon?In my view light has mass in all senses of the term mass. There are three senses of the term "mass" they are defined as

1) Inertial Mass - That which gives a body momentum.

2) Passive Gravitational Mass - That on which gravity acts.

3) Active Gravitational Mass - That which generates a gravitational field.

Light has all those properties.

Also, I was wondering, does this "relativistic mass" of light have an actual numerical quantity? Or will this numerical quantity simply be [tex] \frac{h \nu}{c^2}[/tex]?

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When www.geocities.com starts working again I'll give you a web page I created which provides the derivation.ArmoSkater87 said:Yes, that is exactly how I define as relativistic mass. The relationship come from Lorentz transformation. I dont know how that equation is derived, but do know how a similar one is, the equation for relativistic time. I love derivations so i'll make sure to find one online and take a look myself.

Don't be sorry. Its how we learn. Sometimes by disagreeing with people we learn from the experience, either from us being wrong or from finding an enlightening way to describe the physics to those who are wrong. This is one topic that I've stuck with when I thought someone's response sounded incorrect but for which I unable to readily explain why. It took me awhile but I came up with the explanation that I was looking for.Sorry if i've brought up a stupid arguement, but I really dont know all that much about SR, ive never learned SR anywhere except at home, in this forum the past few months.

The block of ironIf light has mass how come, for example...a very heated block of iron (in a vacuum) doesnt lose mass from its emittion of photon?

Yes. That is correct. Or if the energy is E then the mass of the photon is m = E/cAlso, I was wondering, does this "relativistic mass" of light have an actual numerical quantity? Or will this numerical quantity simply be [tex] \frac{h \nu}{c^2}[/tex]?

Pete

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Alkatran

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This means that light must have some mass, or else the equal/opposite force law breaks down (if an infinite force is placed on the light by the planet... )

Also, a light has energy, determined by it's frequency. E = mc^2, light has mass.

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Well, what is a force other than a change in momentum? And I don't believe anyone here is claiming that light has no relitivistic mass, it just lacks rest mass. This alludes to gravity having an "effect" on both rest and relitivistic mass. This makes sense even with newtonian definitions of force because an object needs not rest mass to have momentum, and "force" is just a change in momentum.Alkatran said:

This means that light must have some mass, or else the equal/opposite force law breaks down (if an infinite force is placed on the light by the planet... )

Also, a light has energy, determined by it's frequency. E = mc^2, light has mass.

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Sounds good to me.Alkatran said:Gravity affects light, this has been proven. This means a force is being place on light to curve it.

See

Its online at - http://xxx.lanl.gov/abs/gr-qc/9811052

Pete

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Wow! You're on fire! Several times too. The same question about light's mass comes up hundreds of times, physics forums should make a sticky on every forum about this, a nice thourough explination.pmb_phy said:My appologies. This will be a long post. Seems like this same topic comes up all the time! So I'll try to be clear and thorough.

You can define it that way if you wish. But I'm going by the definition which is almost always given in the relativity literature and how it was originally defined by Tolman and Lewis (kind of defined by Planck too in a limited sense). In that sense relativistic mass is not defined as you say, i.e. m = gamma*m_{0}is not an identity, it is an equality when the particle is a tardyon (i.e. a particle which always moves with v < c). It is this definition that I will refer to below. I will use the term "mass" to refer to inertial mass aka relativistic mass.

Mass isdefinedas the quantitymsuch that the quantitymvis conserved. Definemomentumasp=mv. You could say that mass is defined so that momentum is conserved (It can then be shown that m = E/c^{2}). If the particle is a tardyon then onederivesm = gamma*m_{0}on the basis of momentum conservation. It then turns out that the momentum of a tardyon in is

[tex]\bold p = \gamma m_0 \bold v[/tex]

Substitute the expression m = gamma*m_{0}into this expression for the momentum and you'll see that p = mv.

That equation was derived on the assumption that the particle travels less than the speed of light. As such the m in p = mv is m = gamma m_{0}. If you want to use this equation for light then start with the definition m = p/v = p/c. Otherwise you're forgetting that when v->c the denominator goes to zero so that you have an improper(?) limit m = 0/0. This can be interpreted as saying that as v->c , m-> constant. This is how most relativists interpret this. E.g. see Rindler's latest text. Its in there.

Hardly anybody (probably nobody) in the relativity community has ever defined relativistic mass that way. Almost all (if not all) derivations of that relation that you'll find in the relativity literature will demonstrate this fact.

That is incorrect. Why would you assume that Newton's laws are incorrect or don't apply to SR? Newton's second law, F = dp/dt, is still valid in SR. Newton's third law also applies for contact forces.

m = p/v is the defining relation for relativistic mass. I urge you not to take my word for it but to look it up (see references below). I.e. Look in the relativity literature for yourself. You you can read the work of the people who are responsible for making this definition well known, namely Tolman and Lewis (Tolman's text covers this too). Max Jammer goes into detail on this in his new book on mass. Have you seen this book?

If you're really interested in this point then see theDefining articles for relativistic mass

More recent publications on this subject

A popular SR text, one that is often used at MIT in fact, isSpecial RelativityA.P. French. French defines "inertial mass" (aka relativistic mass) as follows. From page 16

You can call p = mv "Newtonian momentum" or "Mrs. Buttersworth's momentum" but whatever you call it the fact remains the same - p = gamma*m_{0}*v = mv. In any case relativistic mass isdefinedas the ratio m = p/v and therefore p = mv. Haven't you ever wondered why you can write momentum as the product of relativistic mass and velocity?

Place E = mc^{2}into that equation and you'll get p = mc and the definition still holds.

May I ask where you got the idea that relativistic mass is defined as above, i.e. m = gamma*m_{0}? That is a correct relationship, however it is not an identity. It is an equality, but only for tardyons.

Pete

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Chronos

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Photons are not tachyons? What are some examples of tachyons?Chronos said:

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There's already a sci.physics.faq on the topicMk said:Wow! You're on fire! Several times too. The same question about light's mass comes up hundreds of times, physics forums should make a sticky on every forum about this, a nice thourough explination.

http://math.ucr.edu/home/baez/physics/Relativity/SR/light_mass.html [Broken]

One may notice some slight differences between the FAQ version above and Pete's version - not enough to affect any experimental results, just about semantics and usage.

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Interesting, I think i'll measure the weight of a lightbulb and then measure it again a few years later. That should be enough time for a significant change.pmb phy said:The block of iron will loose mass. In fact this was how Einstein first derived E = mc2. He started with a body at rest in the inertial frame S'. The body emits radiation in equal and opposite directions and thus, for momentum to be conserved, the body remains at rest in S. He then analyzes this from a frame S' which is moving relative to S. The total amout of momentum of radiation in this frame is non-zero. For momentum to be conservd in this frame the momentum of the emitting body must have decreased. Since the velocity didn't change the mass of the body must have changed. He shows that the change in mass is in accorded with E = mc2. I.e. if the energy of the radiation is dE then the mass decreases by the amount dm where dm = dE/c2.

So then what mass-energy conversion is going on when an electron and positron annihilate to create two gamma rays, if both the pair, and the 2 photons have mass.Yes. That is correct. Or if the energy is E then the mass of the photon is m = E/c2

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Pertaining to particles moving in an inertial frame of reference -Mk said:

A

A

A

A photon is not a tardyon.

If the light bulb is absorbing energy at the same rate that its radiating energy then its mass would remain constant. But consider the magnitudes we're speaking of. The amount of mass radiated by a 100 watt light bulb that has been turned on for 10,000 years is 0.35 grams.ArmoSkater87 said:Interesting, I think i'll measure the weight of a lightbulb and then measure it again a few years later. That should be enough time for a significant change.

That post of mine was not intended to be an FAQ. It was strictly intended to accurately define the terminology used in this thread, namely an accurate and precise definition ofpervect said:One may notice some slight differences between the FAQ version above and Pete's version - not enough to affect any experimental results, just about semantics and usage.

The conversion is in the composition of the particles and masses which make up the system and not in the actual value of the total mass of the system. Mass is conserved in such reactions. In fact if we're speaking of pure SR (inertial frame, no gravity etc) then the total mass of any closed system is conserved.ArmoSkater87 said:So then what mass-energy conversion is going on when an electron and positron annihilate to create two gamma rays, if both the pair, and the 2 photons have mass.

Pete

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So photons are luxons. Are there any particles that are tardyons or tachyons?pmb_phy said:Pertaining to particles moving in an inertial frame of reference -

Atardyonis defined as a particle which always moves at v < c. You could also define them as particles for which E^{2}- (pc)^{2}> 0. Tardyons have positive proper mass (aka rest mass).

Atachyonis defined as a particle which always moves at v > c. You could also define them as a particle for which E^{2}- (pc)^{2}< 0. Tachyons have imaginary proper mass.

Aluxonis define as a particle which always moves at v = c. You could also define them as a particle for which E^{2}- (pc)^{2}= 0. Luxons have zero proper mass.

A photon is not a tardyon.

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Any particle which moves at a speed less than light is a tardyon. For example, protons, electrons, muons, etc. are all tardyons because they all move at speeds less than the speed of light. Nobody has ever detected a particle which was moving at a speed greater than light so therefore nobody has ever detected the existance of a tachyon.Mk said:So photons are luxons. Are there any particles that are tardyons or tachyons?

:tongue2:

Pete