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phenylalanine
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If light has a mass, how can it travel at the speed of light? And if it doesn't, why is it affected by gravity (e.g. black holes)?
Gonzolo said:The short answer is that light doesn't have a rest mass. Only relativistic mass. And it is affected by gravity because gravity curves space-time.
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.phenylalanine said:If light has a mass, how can it travel at the speed of light?
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 in the Feynman Lectures Vol -I page 7-11. The section is entitled Gravitation and RelativityAnd 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 doesn't have mass...Its mass is 0 otherwise, it wudnt have traveled at c...
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
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.
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*m0 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.ArmoSkater87 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]
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 m0. 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.If 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 isn't 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 isn't 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...
More recent publications on this subjectThe 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, is Special Relativity A.P. French. French defines "inertial mass" (aka relativistic mass) as follows. From page 16The 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*m0*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?By inertial mass we mean the ratio of linear momentum to velocity.
Place E = mc2 into that equation and you'll get p = mc and the definition still holds.[tex] p = E/c [/tex]
Where the energy is plank's constant times the frequency of the light.
pmb_phy said:Mass is defined as the quantity m such that the quantity mv is conserved. Define momentum as p = mv. You could say that mass is defined so that momentum is conserved (It can then be shown that m = E/c2). If the particle is a tardyon then one derives m = gamma*m0 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*m0 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 m0. 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 the Defining articles for relativistic mass
More recent publications on this subject
A popular SR text, one that is often used at MIT in fact, is Special Relativity A.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*m0*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?
Place E = mc2 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*m0? That is a correct relationship, however it is not an identity. It is an equality, but only for tardyons.
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 is Relativistic mass increase at slow speeds, Gerald Gabrielse, Am. J. Phys. 63, 568 (1995).ArmoSkater87 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.
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 let's 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 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....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.
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.
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.
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 don't 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 don't know all that much about SR, I've never learned SR anywhere except at home, in this forum the past few months.
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 amount 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.If light has mass how come, for example...a very heated block of iron (in a vacuum) doesn't 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/c2.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]?
Alkatran said:Gravity affects light, this has been proven. This means a force is being place on light to curve it. If light's mass was 0, any force would have an undefined effect (division by 0) but most likely infinite effect on it (the light would shoot towards the planet at c at even the smallest hint of gravity)
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.
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.
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*m0 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 mv is conserved. Define momentum as p = mv. You could say that mass is defined so that momentum is conserved (It can then be shown that m = E/c2). If the particle is a tardyon then one derives m = gamma*m0 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*m0 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 m0. 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 the Defining articles for relativistic mass
More recent publications on this subject
A popular SR text, one that is often used at MIT in fact, is Special Relativity A.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*m0*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?
Place E = mc2 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*m0? That is a correct relationship, however it is not an identity. It is an equality, but only for tardyons.
Pete
Chronos said:Tardyons are particles that possesses mass, hence, are forever limited to sub-light velocity. Tachyons are particles with imaginary mass [square root of -1] that are forever limited to travel at super-luminal velocities. Photons are massless particles that are forever limited to travel at c.
Mk 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.
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 amount 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.
Yes. That is correct. Or if the energy is E then the mass of the photon is m = E/c2
Pertaining to particles moving in an inertial frame of reference -Mk said:What are tardyons? Particles that have only rest mass? What kinds of tardyons are there? Surely more than only photons. What is the name for a particle not having rest mass? What particles are those?
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 of relativistic mass.pervect 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.
pmb_phy said:Pertaining to particles moving in an inertial frame of reference -
A tardyon is defined as a particle which always moves at v < c. You could also define them as particles for which E2 - (pc)2 > 0. Tardyons have positive proper mass (aka rest mass).
A tachyon is defined as a particle which always moves at v > c. You could also define them as a particle for which E2 - (pc)2 < 0. Tachyons have imaginary proper mass.
A luxon is define as a particle which always moves at v = c. You could also define them as a particle for which E2 - (pc)2 = 0. Luxons have zero proper mass.
A photon is not a tardyon.
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 existence of a tachyon.Mk said:So photons are luxons. Are there any particles that are tardyons or tachyons?
The mass of light does not affect its speed. According to Einstein's theory of relativity, the speed of light is a constant and is not influenced by its mass or the mass of the object it is traveling through.
Light has a constant speed because it is governed by the laws of physics, specifically the theory of relativity. According to this theory, the speed of light is the maximum speed at which anything in the universe can travel, and it remains constant regardless of the observer's frame of reference.
No, the mass of light does not affect gravity. According to Einstein's theory of general relativity, gravity is the result of the curvature of space-time caused by the presence of mass. Since light has no mass, it does not contribute to the force of gravity.
Yes, light can be affected by gravitational forces. According to the theory of general relativity, the path of light can be curved by massive objects, such as stars and black holes. This phenomenon is known as gravitational lensing.
Light does not have mass, therefore it does not have a mass-related energy. However, light does have energy in the form of photons, which can be converted into other forms of energy, such as heat or electricity.