Exploring the Question: Does Light Have Mass?

[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 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.

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:rofl: 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

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