Exploring the Question: Does Light Have Mass?

[Champion]

I get confused because i am told it has mass but if it has mass wouldn't the laws be different and we would feel the weight of the sun light when we go outside?

 

[Dale]

Hi Champion, welcome to PF.

There have been experiments designed to answer exactly this question. All of them to date are consistent with the idea that http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html" .

 

[killjoy_scofiel]

light as particle is massless

 

[Champion]

what do you mean as particles? it take's other forms?

 

[pallidin]

This might help:

http://www.Newton.dep.anl.gov/askasci/phy00/phy00332.htm

 

[HallsofIvy]

what do you mean as particles? it take's other forms?

Yes, light acts both as a particle and as a wave. In fact, if you get small enough (the size of elementary particles) the very notion of "particle" or "wave" loses meaning.

By the way, your original argument, "we would feel the weight of the sun light when we go outside?" is invalid. If light had a very low weight, we wouldn't feel it. Do you feel the weight of the air?

 

[amppatel]

so if it doesn't have mass why can it not escape from a black hole? why does it experience gravity?

 

[Troponin]

Do you feel the weight of the air?

Wouldn't we lose the "feel" of air for a different reason? I can feel it when I drink from a straw. lol

Light has enough mass to feel the effects of a gravitational field and also exerts its own gravitational attraction, so it must have mass.
But, by definition, it can't have any rest mass.

At rest it would be massless, but it can't be at rest. At the speed of light, it shows the properties one would expect from mass.

 

[HallsofIvy]

No, light does not have mass. In the theory of relativity, "gravity" is a property of space around a massive object. Anything moving around a massive object has its trajectory different from a straight line whether it has mass or not.

I should warn you that you will stir up a nest of hornets if you refer to "mass" as other than "rest mass". Light has energy, not mass.

 

[Troponin]

No, light does not have mass. In the theory of relativity, "gravity" is a property of space around a massive object. Anything moving around a massive object has its trajectory different from a straight line whether it has mass or not.

I should warn you that you will stir up a nest of hornets if you refer to "mass" as other than "rest mass". Light has energy, not mass.

How do hornets have mass? I can't feel them.

So, mass always has energy so that total energy is concerved, but kinetic energy doesn't have mass? Isn't that just arguing semantics?




(Let it be known that I had no ill will when poking the hornet's next with a stick...just bored)

 

[jnorman]

light cannot escape a BH because the escape velocity is >C. alternatively, the photon cannot escape from BH because spacetime is warped to the extent that a straight line cannot extend past the EH, but is rather curved back upon itself.

light does not have measureable mass, per se, but as per E=MC2, the amount of energy of a given photon is equivalent to a fixed amount of mass.

 

[Dale]

How do hornets have mass? I can't feel them. ... (Let it be known that I had no ill will when poking the hornet's next with a stick...just bored)

Hmm, I am quite skeptical of this claim based on my own youthful experiences with bees and wasps.

So, mass always has energy so that total energy is concerved, but kinetic energy doesn't have mass? Isn't that just arguing semantics?

No, it is not semantics. I would highly recommend you look into the standard relativistic concept of the http://en.wikipedia.org/wiki/Four-momentum" . In geometric terms you can think of the mass of a particle as being the length (Minkowski norm) of its four-momentum vector, and the energy is just one component of the vector. So the distinction between the two concepts is not just semantic.

The other common usage of the word mass is "relativistic mass". This is the hornet's nest referred to earlier. The concept of "relativistic mass" is deprecated by most modern physicists precisely because there is no distinction (other than semantics) between "relativistic mass" and energy. So it is more clear to use the term energy when referring to "relativistic mass" and reserve the term mass to refer strictly to the invariant rest mass.

 

[Dale]

Light has enough mass to feel the effects of a gravitational field.

Think about this a little carefully. What does a satellite's orbital path depend on? So in the limit as m->0, what would you expect?

 

[Troponin]

Hmm, I am quite skeptical of this claim based on my own youthful experiences with bees and wasps.

Does this mean pain is analagous to mass? I never really felt them until they announced their presence in the unsavory way they're known to do.


The other common usage of the word mass is "relativistic mass". This is the hornet's nest referred to earlier. The concept of "relativistic mass" is deprecated by most modern physicists precisely because there is no distinction (other than semantics) between "relativistic mass" and energy. So it is more clear to use the term energy when referring to "relativistic mass" and reserve the term mass to refer strictly to the invariant rest mass.[/QUOTE]

Exactly, semantics. The terms hold more meaning and clarity when the term "mass" is limited to invariant rest mass...




All in fun, I just wanted to see what responses I'd get. Agitated assertions of one's point tend to bring out more definite answers. (An agitated "authority" also tends to talk down to the "moron" with use of simple analogies...which helps this particular moron gain a better understanding) lol

 

[Nickelodeon]

No, light does not have mass. In the theory of relativity, "gravity" is a property of space around a massive object. Anything moving around a massive object has its trajectory different from a straight line whether it has mass or not.

I should warn you that you will stir up a nest of hornets if you refer to "mass" as other than "rest mass". Light has energy, not mass.

Mass is structured energy and as light is also a form of structured energy it will show mass-like characteristics. However the wave structure of light dictates that the mass effect alternates between a photon and an anti-photon and so at the end of the day you end up with zero mass.

 

[Mentz114]

Matter, which makes up everything with mass, is very different from light. Arguing about whether light has mass is a dead end because it makes no difference to anything whatever you position you take.

Assigning matter-like properties to light in an attempt to understand it analogously with matter is also pointless. Maxwell's equations tell all.

 

[WCOLtd]

Can light have a frequency so that E = M?

I am guessing the answer would be yes; energy of any finite frequency can satisfy that equation as long as the time of exposure is sufficient.

Then let me be more clear;

If we assume that a maximum frequency of light exists (denoted by the Planck length); What if the observer was to accelerate towards the source of this light of maximum frequency?

Energy of the light must increase with the increase relative velocity, but the light's frequency can't go any higher. Would it be viable to assume that light packets condense into mass and decrease in velocity and in frequency since that is the only way energy can be conserved in this situation?

 

[Xyooj]

if it doesn't have mass then why does it bend in gravitational field?

hmmm...light cannot escape a black hole because the light is not bounce back so we can see it? what we see is what is not absorbed?

 

[jtbell]

if it doesn't have mass then why does it bend in gravitational field?

See post #8 in the Physics Forums FAQ (in the General Physics forum).

 

[atyy]

See post #8 in the Physics Forums FAQ (in the General Physics forum).

The excuse given in the post for not using the relativistic mass is that it is speed dependent. However, light always travels at the speed of light, so it happens to be the only sort of particle for which you might think the relativistic mass is also invariant, and therefore useful to put in a table. However, looking at the definition of relativistic mass in that post, if v=c, then the relativistic mass contains zero in the denominator. If a particle with rest mass greater than zero travels at the speed of light, its relativistic mass is infinite, which is nonsensical. One way to save the formula is to say that only particles with zero rest mass are allowed to travel at the speed of light. If a particle travels at the speed of light and its rest mass is zero, then its relativistic mass is indefinite (still nonsensical, but not as nonsensical as an infinite mass). I like this idea that light has some sort of indefinite mass, because then it makes sense that it should be attracted by gravity.

Strictly speaking, the special theory of relativity is not compatible with gravity. The indefinite mass of a photon is just a heuristic to see that a proper theory of gravity should predict the deflection of light. However, it is actually possible to construct coherent modifications of the special theory in which gravity doesn't bend light. So ultimately, that gravity bends light is based on experiment, not intellectual necessity.

 

[George Jones]

Can light have a frequency so that E = M?

I am guessing the answer would be yes; energy of any finite frequency can satisfy that equation as long as the time of exposure is sufficient.

Then let me be more clear;

If we assume that a maximum frequency of light exists (denoted by the Planck length); What if the observer was to accelerate towards the source of this light of maximum frequency?

Energy of the light must increase with the increase relative velocity, but the light's frequency can't go any higher. Would it be viable to assume that light packets condense into mass and decrease in velocity and in frequency since that is the only way energy can be conserved in this situation?

It seems that you're thinking of discrete spacetime. Some physicists have worked on this, but the Physics Forums Rules

https://www.physicsforums.com/showthread.php?t=5374

prohibit posters from posting their own speculations.

Overly Speculative Posts: One of the main goals of PF is to help students learn the current status of physics as practiced by the scientific community; accordingly, Physicsforums.com strives to maintain high standards of academic integrity. There are many open questions in physics, and we welcome discussion on those subjects provided the discussion remains intellectually sound. It is against our Posting Guidelines to discuss, in most of the PF forums, new or non-mainstream theories or ideas that have not been published in professional peer-reviewed journals or are not part of current professional mainstream scientific discussion.

If your ideas are based on published work, then you need to cite the publications. If your ideas are your own speculations, then Physics Forums is not the right place to promote your ideas.

 

[Dale]

if it doesn't have mass then why does it bend in gravitational field?

I like this idea that light has some sort of indefinite mass, because then it makes sense that it should be attracted by gravity.

so if it doesn't have mass why can it not escape from a black hole? why does it experience gravity?

Light has enough mass to feel the effects of a gravitational field

I don't know why any of you think that something needs mass to be deflected by gravity (aka passive gravitation). Under GR it is clear, but even under Newtonian gravity it should be clear: What is the Newtonian formula for the acceleration of a satellite of mass m located a distance r from a spherical planet of mass M? And what is the limit of the acceleration as m->0?

Spoiler

ma = f = GMm/r²
a = GM/r²

limit as m->0: a = GM/r²

 

[D Kennedy]

light does not have measureable mass, per se, but as per E=MC2, the amount of energy of a given photon is equivalent to a fixed amount of mass.

Would this mass not then be dependent upon the energy of the individual photon as opposed to fixed?

 

[Mentz114]

I don't know why any of you think that something needs mass to be deflected by gravity (aka passive gravitation). Under GR it is clear, but even under Newtonian gravity it should be clear: What is the Newtonian formula for the acceleration of a satellite of mass m located a distance r from a spherical planet of mass M? And what is the limit of the acceleration as m->0?

Well said, DaleSpam.

Is it not a bit iffy cancelling 0's ? It does show that matter doesn't couple to the gravitational field like charge to the electic field, where the force is proportional to the charge.

 

[jtbell]

ight does not have measureable mass, per se, but as per E=MC2, the amount of energy of a given photon is equivalent to a fixed amount of mass.

Would this mass not then be dependent upon the energy of the individual photon as opposed to fixed?

There are two different "kinds" of mass here, which have different properties:

https://www.physicsforums.com/showpost.php?p=1842796&postcount=5

 

[azzkika]

does not E = mc^2 imply a thing of zero mass has zero energy, therefore for energy to exist, it must have mass

 

[Hootenanny]

does not E = mc^2 imply a thing of zero mass has zero energy, therefore for energy to exist, it must have mass

No, E = mc² is not the complete equation. The complete relationship is:

$$E^2 = \left(pc\right)^2 + \left(m_0c^2\right)^2$$

Where m₀ is the rest mass (as jtbell said there are two types of mass, however when a Physicist says "mass" they nearly always mean "rest mass").

As can be seen from the full equation, it is possible for a particle to have zero mass but have non-zero energy.

 

[MeJennifer]

It seems that you're thinking of discrete spacetime. Some physicists have worked on this, but the Physics Forums Rules

https://www.physicsforums.com/showthread.php?t=5374

prohibit posters from posting their own speculations.

The person you refer to is asking an interesting question. Since when is asking a question equivalent to posting speculations?

This forum becomes more and more a cookie cutter 'don't think for yourself, just think what the professor tells you' platform.

 

[MeJennifer]

Can light have a frequency so that E = M?

I am guessing the answer would be yes; energy of any finite frequency can satisfy that equation as long as the time of exposure is sufficient.

Then let me be more clear;

If we assume that a maximum frequency of light exists (denoted by the Planck length); What if the observer was to accelerate towards the source of this light of maximum frequency?

Energy of the light must increase with the increase relative velocity, but the light's frequency can't go any higher. Would it be viable to assume that light packets condense into mass and decrease in velocity and in frequency since that is the only way energy can be conserved in this situation?

Excellent questions!

 

[DrGreg]

Can light have a frequency so that E = M?

I don't understand the question. Energy is measured in joules. Mass is measured in kg. They're different.

If we assume that a maximum frequency of light exists (denoted by the Planck length);

False assumption. The Planck length is not the "minimum possible distance"; it relates to uncertainty in measuring a distance.

When you measure the frequency of light you cannot measure its frequency exactly unless you take an infinite amount of time to do so (and the light itself must persist for an infinite amount of time without changing its frequency). The Planck length or Planck time relates to your uncertainty in measuring the frequency over a finite time interval.

 

[yuiop]

As Dalespan said, it is irrelevant whether a particle has mass or not to fall in a gravitational field according to GR.

To see this clearly consider the equivalence principle. Imagine a photon and a massive particle moving from left to right in flat space. They move in a straight line. Now imagine they are inside a rocket that is accelerating upwards. The photon and massive particle still move in a straight line relative to the flat space but to an accelerating observer inside the rocket the photon and massive particle follow a trajectory that from his point of view curves towards the floor. The photon does not require mass to move in a straight line and it does require mass to appear to curve towards the floor of the accelerating rocket. The same is true in a gravitational field by the equivalence principle. No passive gravitational mass is required. Put another way, particles (with or without mass) follow trajectories called geodesics that are determined by their velocities. The geodesics of particles in Schwarzschild geometry assume the particles have no mass. If the particles have significant active gravitational mass (i.e. they are themselves a source of gravity) then the geometry is no longer described by the exterior Schwarzschild metric because that assumes a vacuum and the presence of mass outside the central gravitational spherical mass (described by the interior Schwarzschild solution) changes the geometry.

It is also known that objects dropped from the same height simultaneously reach the floor simultaneously regardless of their individual masses and in GR this idea extends to objects with zero mass. So if we take an extreme example of dropping a stationary moon sized mass and a 1kg mass to a non rotating Earth sized planet with no atmosphere, they will land at the same time. Now if we drop the moon by itself and carefully time how long it takes to fall we will see that the Moon falls in less time than the 1kg object dropped by itself. This is because the planet is accelerating towards the mass of the Moon faster than the planet falls towards the 1 Kg mass. The active gravitational mass is important in this case and this is an example of a particle with significant mass changing the geometry. The large moon falling is not described by the Schwarzschild metric because it significantly changes the geometry.

In short, objects do not require mass to fall in GR and the Schwarzschild metric assumes falling test particles have no mass so it no mystery why a photon falls whether it has mass or not.

[EDIT] Also, as Dalespam mentioned, it can be seen from the Newtonian equation for gravitational acceleration GM/R^2 there is no variable for the mass of the falling object so a body with no mass can be accelerated downward even in Newtonian gravity.

The variable for the mass of the falling body (m) only appears in the Newtonian equation for the force of gravity GMm/R^2 but in GR no force is considered to be acting on a falling body. The Newtonian equation for gravitational acceleration GM/R^2 assumes a test particle with zero mass. There is a more complicated Newtonian formula for when the falling body has significant mass because you have to allow for the accleration of the attracting massive body towards the falling body and we get back to radially falling moon sized objects. So even in Newtonian physics, passive gravitational mass is not required for a body to fall. Passive gravitational mass (m) that appears in the gravitational force equation GMm/R^2 does however play a part in GR because when an object is not free falling it does experience a gravitational force that we measure as weight.

So Newtonian physics predicts that the gravitational force acting on a particle with zero rest mass is zero yet it also predicts that the particle will be accelerated downwards. That is in good agreement with GR. No passive gravitational mass or gravitational force is required for a particle to be accelerated downwards in GR or Newtonian physics!

 

[MeJennifer]

It is also known that objects dropped from the same height simultaneously reach the floor simultaneously regardless of their individual masses and in GR this idea extends to objects with zero mass.

That is simply untrue, both the mass of the planet and the mass of the object in question contribute to the time it takes for them to come together. Of course it is true that the small mass is negligible compared to the large mass but we are talking principles here.

 

[Dale]

It is also known that objects dropped from the same height simultaneously reach the floor simultaneously regardless of their individual masses and in GR this idea extends to objects with zero mass.

That is simply untrue, both the mass of the planet and the mass of the object in question contribute to the time it takes for them to come together. Of course it is true that the small mass is negligible compared to the large mass but we are talking principles here.

No MeJennifer, you are incorrect. Kev's statement is correct.

Each mass causes the Earth to accelerate by some (different) small amount, but since they are dropped at the same time only the Earth's total acceleration matters.

 

[yuiop]

...

It is also known that objects dropped from the same height simultaneously reach the floor simultaneously regardless of their individual masses and in GR this idea extends to objects with zero mass.

That is simply untrue, both the mass of the planet and the mass of the object in question contribute to the time it takes for them to come together. Of course it is true that the small mass is negligible compared to the large mass but we are talking principles here.

Yes, the principle we are talking about is the equivalence principle.

Imagine a rocket of unladen mass of 1000 kgs with a payload of another 1000 kgs and a test mass of 1 kg near the nose. (Total mass =2001 kgs) Say the payload and test mass are released when the nose is moving at 0.6c. The released masses both continue at 0.6c while the rear of the accelerating rocket catches up with the freefalling masses. The rear (floor) of the rocket arrives at both the large mass and the small mass simultaneously. The discovery by Galileo that objects released together, fall at the same rate irrespective of their mass, over 500 years ago is still true today even in General Relativity.

Please note I was careful to use the word "simultaneously" (twice) in my statement "It is also known that objects dropped from the same height simultaneously, reach the floor simultaneously regardless of their individual masses" but the missing comma may have made the meaning unclear. Anyway, the fact the masses are released simultaneously is the key point, as Dalespam noted. I covered the case where objects are released one at a time in the earlier post to try and make the issue clear.

If the 1000 Kg payload of the accelerating rocket is released by itself the engine of the rocket has less mass to accelerate and the rocket accelerates faster towards the released payload than it would if the 1kg terst mass was releaed by itself. The equivalence principle shows that two masses released together, fall at the same rate, but masses dropped one at a time may fall at different rates with larger masses falling faster. The same is true in a gravitational field. The planet accelerates up towards the combined mass of the released falling objects when they are released together and the falling objects fall at an equal rate determined only by the mass and distance of the planet.


The full equation for the Newtonian gravitational acceleration is $$a = \frac{G(M + m)}{R^2}$$
ref http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation

When the mass (m) of the falling object is significantly smaller than the mass (M) of the gravitational body, the equation $$a = \frac{GM}{R^2}$$

is a reasonable aproximation, which is only exactly true in Newtonian physics when the mass of the falling mass (m) is exactly zero.

For the case of two objects (m2 and m3) falling together towards a large massive body (M), the acceleration of the planet towards the falling objects is:

$$a1 = \frac{G(m2+m3)}{R^2}$$

The acceleration of object m2 towards the original position of the planet is:

$$a2 = \frac{GM}{R^2}$$

The acceleration of object m3 towards the original position of the planet is:

$$a3 = \frac{GM}{R^2}$$

The total acceleration of object m2 towards the planet when the acceleration of the planet towards the object is taken into account is:

$$a2' = a2 +a1 = \frac{GM}{R^2} + \frac{G(m2+m3)}{R^2}$$

The total acceleration of object m3 towards the planet when the acceleration of the planet towards the object is taken into account is:

$$a3' = a3 +a1 = \frac{GM}{R^2} + \frac{G(m2+m3)}{R^2}$$

It can be seen that a2' = a3' and Galileo's claim that objects falling together, fall at the same rate regardless of their individual masses is true. It can further be seen that it is true that the equations for acceleration of a falling body shown above, are equally valid when the mass of the falling body is zero by setting the value of m2 or m3 to zero.

I am sure you will agree with the arguments stated above and that you simply misunderstood what I was getting at, due to a punctuation error on my part.

 

[MeJennifer]

No MeJennifer, you are incorrect. Kev's statement is correct.

Each mass causes the Earth to accelerate by some (different) small amount, but since they are dropped at the same time only the Earth's total acceleration matters.

The 3 mass centers form a triangle structure, what you are ignoring is the distance between the two small objects of different mass.

Take the extreme situation where two different masses are dropped at the same hight and at the same time, one on an arbitrary position over the planet and the other on the opposite side of that planet. Clearly the heavier mass will make contact with the planet before the lighter mass. By reducing the angle the effect is minimized but only if the centers of mass overlap is there no difference.

Again the difference is small but it is not zero.