Does Light Really Have Mass?

[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 acceleration 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 height 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.

 

[rbj]

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. Again the difference is small but it is not zero. In effect the Earth will accelerate more in the direction of the heavier mass.

yeah, this one got me, too, Dale. i guess the Earth tips a little toward the heavier mass as it accelerates downward. almost the same as angels dancing on the head of a pin.

 

[yuiop]

The 3 mass centers form a triangle structure, what you are ignoring is the distance between the two small objects of different mass. Again the difference is small but it is not zero. In effect the Earth will accelerate more in the direction of the heavier mass.

What you are describing is a tidal effect and the equivalence principle can always be broken by considering tiadl effects. In a wide accelerating rocket everything falls parallel to the acceleration axis of the rocket. In the real gravity of a spherical mass everything falls towards to the centre of the gravitational body. An extreme example would be to drop two objects with different masses simultaneously from the same height but on opposite sides of the planet and then there will be difference due to the planet accelerating towards the more massive falling object.

Generally speaking when talking about the equivalence principle we consider a region that is localised enough that tidal effects are insignificant and objects are considered to be (near enough) falling parallel to each other.

So re-stated, the statement should be "two objects released from the same height at the same time, that are close enough to each other that their falling paths are considered to be parallel, will fall at the same rate, irrespective of their individual masses".

Better?

Although you are technically correct Jennifer, your comments are not really adding anything to issue of whether a particle such a light, has to have mass in own right, in order to be affected by gravity




picky :p

 

[MeJennifer]

An extreme example would be to drop two objects with different masses simultaneously from the same height but on opposite sides of the planet and then there will be difference due to the planet accelerating towards the more massive falling object.

While you wrote this posting I was actually updating my posting with exactly such an example.

 

[Dale]

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 height 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.

OK MeJennifer

 

[atyy]

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.

This statement requires:
1) An object has inertial mass
2) An object has gravitational mass
3) Inertial mass equals gravitational mass

In the statement, "the mass of a photon is zero", is the inertial or gravitational mass being referred to?

 

[rbj]

This statement requires:
1) An object has inertial mass
2) An object has gravitational mass
3) Inertial mass equals gravitational mass

In the statement, "the mass of a photon is zero", is the inertial or gravitational mass being referred to?

in my opinion neither. but i am not one of those who would say "the mass of a photon is zero" without qualification. i would say instead "the rest mass of a photon is no larger than something like 10^-55 kg and is most likely zero". the inertial mass is, from what i can tell, the scaler quantity that one multiplies the velocity vector of some body by to get the momentum vector.

 

[rbj]

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

E^2 = \left(p c\right)^2 + \left(m_0 c^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.

actually, Hoot, E = mc² is complete if the m in E = mc² is the "relativistic mass" (the term you real physicists want to deprecate), not the rest mass. I'm glad to see that you used m₀ in your notation to differentiate it from the relativistic mass and so our notation agrees on symbols.

m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}

the reason why the rest mass of the photons (or any conceptual particle that moves at speed c) is zero is because

m_0 = m \sqrt{1 - \frac{v^2}{c^2}} \rightarrow 0

if v \rightarrow c.

at least, that's the simple way i look at it. the way i look at it is that i derive

E^2 = \left(p c\right)^2 + \left(m_0 c^2\right)^2

from these other facts.

photons have energy, then they have relativistic or inertial mass of m=E/c². then if you multiply by their supposed velocity of c, you get momentum of p=E/c. plug that p into the equation above and that m₀ into the equation above and E/c² into the remaining m, and i think you'll get equality.

 

[Hootenanny]

actually, Hoot, E = mc² is complete if the m in E = mc² is the "relativistic mass" (the term you real physicists want to deprecate), not the rest mass.

Of course that is true. However, I was emphasising the point that although there are two types of "mass", when a physicist says "mass" without any qualification they are almost always referring to the invariant mass. Perhaps I should have said "an alternative representation" instead of "full equation".

 

[MeJennifer]

Although you are technically correct Jennifer, your comments are not really adding anything to issue of whether a particle such a light, has to have mass in own right, in order to be affected by gravity

In a curved spacetime, which is any spacetime that contains mass or energy, light will follow the straightest possible path. It has absolutely nothing to do with it having mass or not.

 

[yuiop]

In a curved spacetime, which is any spacetime that contains mass or energy, light will follow the straightest possible path. It has absolutely nothing to do with it having mass or not.

Ok, we are in agreement here then

 

[yuiop]

This statement requires:
1) An object has inertial mass
2) An object has gravitational mass
3) Inertial mass equals gravitational mass

In the statement, "the mass of a photon is zero", is the inertial or gravitational mass being referred to?

Hi atyy,

To avoid hijacking this thread, I have replied to your post in a different thread here https://www.physicsforums.com/showpost.php?p=1870718&postcount=15 as my reply is not specifically about photons.

...

E^2 = \left(p c\right)^2 + \left(m_0 c^2\right)^2

from these other facts.

photons have energy, then they have relativistic or inertial mass of m=E/c². then if you multiply by their supposed velocity of c, you get momentum of p=E/c. plug that p into the equation above and that m₀ into the equation above and E/c² into the remaining m, and i think you'll get equality.

I put a similar argument in this post here: https://www.physicsforums.com/showpost.php?p=1870574&postcount=13

 

[samalkhaiat]

Let me surprise you all by saying that Maxwell's equations which describe light, do admit massive solutions. However, such massive "light" solution can always be gauged away completely. Therefore these solutions are not physical ones.


sam

 

[Count Iblis]

The mass of a single photon is zero. The mass of two or more photons can be nonzero, because mass is the total energy in the zero momentum frame. So, if you have two photons with equal energy moving in the opposite direction, then the mass is twice the energy of a single photon.

The mass of zero photons can be nonzero as well. If you take 6 square mirrors of mass m and glue them together to form a cube, then the cube will have a mass of slightly more than 6 m, even if there is only a vacuum inside. This is due to the vacuum energy of the elecromagnetic field inside the cube. So, zero photons can have more mass that a single photon.

 

[atyy]

The mass of a single photon is zero. The mass of two or more photons can be nonzero, because mass is the total energy in the zero momentum frame. So, if you have two photons with equal energy moving in the opposite direction, then the mass is twice the energy of a single photon.

The mass of zero photons can be nonzero as well. If you take 6 square mirrors of mass m and glue them together to form a cube, then the cube will have a mass of slightly more than 6 m, even if there is only a vacuum inside. This is due to the vacuum energy of the elecromagnetic field inside the cube. So, zero photons can have more mass that a single photon.

It took me a while to figure out what you were saying, but true in a very precise sense indeed.

 

[atyy]

The mass of a single photon is zero.

Oh wait, I didn't think this one statement through. Does a single photon have a zero momentum frame?

Edit: OK, I understand - I should have said - true in several very precise senses indeed.

 

[yuiop]

...
The mass of zero photons can be nonzero as well. If you take 6 square mirrors of mass m and glue them together to form a cube, then the cube will have a mass of slightly more than 6 m, even if there is only a vacuum inside. This is due to the vacuum energy of the elecromagnetic field inside the cube. So, zero photons can have more mass that a single photon.

I am curious. Does the introduction of one photon to our initially dark vacuum box destroy the vacuum energy? Wouldn't the box have a total mass of box+vacuum+photon?

 

[yuiop]

Oh wait, I didn't think this one statement through. Does a single photon have a zero momentum frame?

I guess another way to define rest mass is total energy minus kinetic energy and for a photon total energy = kinetic energy so its rest mass is unambiguously zero by that definition.

 

[Phrak]

The mass of a single photon is zero. The mass of two or more photons can be nonzero, because mass is the total energy in the zero momentum frame. So, if you have two photons with equal energy moving in the opposite direction, then the mass is twice the energy of a single photon.

The mass of zero photons can be nonzero as well. If you take 6 square mirrors of mass m and glue them together to form a cube, then the cube will have a mass of slightly more than 6 m, even if there is only a vacuum inside. This is due to the vacuum energy of the elecromagnetic field inside the cube. So, zero photons can have more mass that a single photon.

No need to go that far. A single photon standing wave has zero momentum such that m_0 =E / c^2

In fact, it's hard to pin-down the long list of ideal conditions required for massless photons...

 

[Phrak]

On second thought, doesn't

m_0^2 = E^2 - p^2 or more precisely

m_0 = p_\mu

serve in particle physics as the Dirac delta function serves in quantum mechanics; each physical impermissible, but serving as the basis of theory?

The idealization in the first case is an unrealizable photon with energy precisely equal to momenutum, and in quantum mechanics an unrealizable wavefunction with either precise momentum or precise position.

 

[Phrak]

m_0 = p_\mu

I'm compelled to correct this before things get too far along.

m_0 U^\mu = p^\mu

where \ U^\mu is the four velocity, ( \ U^\mu U_\mu = -1 ) so that

m_0^2 = -p_\mu p^\mu