Q:

I read your statement about how light has momentum despite the fact that it has no mass. My question to you is regarding gravity in black holes. It is said that light can’t escape the enormous gravitational force in black holes; however, is it not true that gravity is directly proportional to the object’s MASS and inversely proportional to the distance between the two objects (Newtonian, I think). If so, light has no mass. So how would light be effected by this phenomenon??? Thanks for your enthusiasm in physics.
Dan Sweeney

- Dan Sweeney (age 16)

Thayer Academy, Braintree MA, USA

- Dan Sweeney (age 16)

Thayer Academy, Braintree MA, USA

A:

The use of words can make a lot of confusion. Unfortunately, the word "mass" has been used in two different ways in physics. One was the way Einstein used it in E=mc^{2}, where mass is really just the same thing as energy (E) but measured in different units. This is the same "m" that you multiply velocity by to find momentum (p), and thus is sometimes called the inertial mass. It's also the mass that provides the source of gravitational effects. Light has this "m" because it has energy. So it is indeed affected by gravity- not just in black holes but in all sorts of less extreme situations too. In fact, the first important confirmation of General Relativity came in 1919, when it was found that light from stars bends as it goes by the Sun.

The other way "mass" is often used, especially in recent years, is to mean "rest mass" or "invariant mass", which is sqrt(E^{2}-p^{2}*c^{2})/c^{2}. This is invariant because it doesn't change when you describe an object at rest or from the point of view of someone who says it's moving. Obviously that's a good type of "mass" to give when you want to make a list of masses of particles. For a light beam traveling in a single direction, E=pc, so this "m" is zero. There is no point of view from which the light is standing still!

However, once you consider light traveling in a variety of directions, the E's from the different parts just add up to give the total E but the vector p's don't. In fact the total p can be zero if there are beams traveling opposite ways. So for many purposes the older definition of m (the inertial mass) is more convenient than the invariant particle mass, since it's the inertial mass that's just the sum of the inertial masses of the parts. For light moving equally in all directions, like the light bouncing around inside a star, total p is zero, so both definitions just give m=E/c^{2}.

Mike W.

The other way "mass" is often used, especially in recent years, is to mean "rest mass" or "invariant mass", which is sqrt(E

However, once you consider light traveling in a variety of directions, the E's from the different parts just add up to give the total E but the vector p's don't. In fact the total p can be zero if there are beams traveling opposite ways. So for many purposes the older definition of m (the inertial mass) is more convenient than the invariant particle mass, since it's the inertial mass that's just the sum of the inertial masses of the parts. For light moving equally in all directions, like the light bouncing around inside a star, total p is zero, so both definitions just give m=E/c

Mike W.

*(published on 10/22/2007)*

Q:

i just have a question that, as in quantum physic theory, light has momentum...Do you think this is absolutely true or is not certainty?

- john (age 30)

london, uk

- john (age 30)

london, uk

A:

It's not just true in quantum physics. Even classical electromagnetism, as in Maxwell's equations, required that light have momentum. It's been measured in countless experiments. It's just plain true.

Mike W.

Mike W.

*(published on 11/08/2007)*

Q:

I think it was Einstine who proved that no mass can travel at the speed of light.As any object of any arbitrary mass when approximates the speed of light increases in mass.When it reaches the vecinity of speed of light its mass is so enormous that it requires infinite amount of energy to propel it.If this is indeed true, then light can have no mass as it could not travel in at the speed at 3*10^8km/sec.Am I correct?

- Nirendra Shrestha (age 27)

Kathmandu, Nepal

- Nirendra Shrestha (age 27)

Kathmandu, Nepal

A:

Since photons have zero rest mass they can move with the speed of light.

Morten

*(published on 10/17/2009)*

Q:

MASS OF LIGHT IS NOT 0.
Light has mass. Generally,Mass is defined as the amount of substance or matter contained in the body.
Also , according to the Newton's gravitational law mass can be defined as a quantity which has gravitational property that is it can apply gravitational force to other body and also it can be influenced.
When light passes through the gravitational field of any heavenly body it deviates toward the heavenly body. So light has mass on the basis of Newton's gravitational law .
its mathematical interpretation is hinted below:
By the newton’s universal law of gravitation,
F=G(M1*M2)/R^2
where,
m1=mass of heavenly body
r=distance
f=gravitational force
m2=0=mass of light
If the mass of light is o then gravitational force would not be there. Since
F=G(M1*0)/R^2
experiments has shown that f is more than 1
thus we can conclude that light has mass
and momentum= mass * velocity
this simple logic also matches with broglie's law
thanks for the question

- GAURAB SEDHAIN (age 15)

NEPAL,NUWAKOT

- GAURAB SEDHAIN (age 15)

NEPAL,NUWAKOT

A:

You're on the right track in important ways, but it turns out life isn't quite so simple.

Yes, if we define the source of gravitational effects to be mass, we can see that light has mass in that sense. However, if you tried to use that force equation you gave, you'd calculate some bending of light in a gravitational field, but it would only be half the observed amount. General Relativity, which describes the distortion of space-time by mass and momentum, is needed to get the right answer.

(Your other definition, "Mass is defined as the amount of substance or matter contained in the body.", is not so useful. It just substitutes some words for others, and doesn't tell us what to expect to see in the world.)

Mike W.

Yes, if we define the source of gravitational effects to be mass, we can see that light has mass in that sense. However, if you tried to use that force equation you gave, you'd calculate some bending of light in a gravitational field, but it would only be half the observed amount. General Relativity, which describes the distortion of space-time by mass and momentum, is needed to get the right answer.

(Your other definition, "Mass is defined as the amount of substance or matter contained in the body.", is not so useful. It just substitutes some words for others, and doesn't tell us what to expect to see in the world.)

Mike W.

*(published on 05/28/2010)*

Q:

Light sails use the momentum of photons to provide propulsion for space flights. If a photon is reflected and hence loses momentum (to the sail) what effect is there on the reflected photon. Does its wavelength change? If so why dont we see a change in colour if we view things in a mirrow (single reflection) and why do LASERS work in spite of multiple reflections in the laser tube stimulating a single frequency of output light?

- Bill Unsworth (age 64)

Warrington, UK

- Bill Unsworth (age 64)

Warrington, UK

A:

Bill- I'm very sorry that this question somehow slipped between the cracks long ago.

When light bounces off an object it does impart momentum to it. In the simplest case, light bouncing off a very massive stationary object, the light imparts no energy to the object. The momentum imparted is twice that of the incoming light, since it just changes directions and thus changes the sign of its momentum.

A more interesting case looks at this from the point of view of somebody who says the big object is moving, for example away from the light source. From the new point of view, the light transferred energy to the object, since the momentum transfer times the velocity of the object isn't zero. Thus the reflected light has less energy and lower absolute value of momentum than the incoming light. You do in fact see a tiny change in color if you bounce light off moving objects. The effect is called laser Doppler velocimetry, or quasi-elastic light scattering. I used to do experiments of exactly that type.

The broadening of the frequency spectrum of the light emitted by atoms due to the Doppler shifts associated with their motions (in the mirror frame) is a real effect in laser technology. See

Mike W.

When light bounces off an object it does impart momentum to it. In the simplest case, light bouncing off a very massive stationary object, the light imparts no energy to the object. The momentum imparted is twice that of the incoming light, since it just changes directions and thus changes the sign of its momentum.

A more interesting case looks at this from the point of view of somebody who says the big object is moving, for example away from the light source. From the new point of view, the light transferred energy to the object, since the momentum transfer times the velocity of the object isn't zero. Thus the reflected light has less energy and lower absolute value of momentum than the incoming light. You do in fact see a tiny change in color if you bounce light off moving objects. The effect is called laser Doppler velocimetry, or quasi-elastic light scattering. I used to do experiments of exactly that type.

The broadening of the frequency spectrum of the light emitted by atoms due to the Doppler shifts associated with their motions (in the mirror frame) is a real effect in laser technology. See

Mike W.

*(published on 07/19/2010)*

Q:

How can a particle have momentum when it does not have mass? Is this because some particles like photons can travel close to or at the speed of light and then that gives them mass through the lorenz transformations?

- Bob (age 19)

Seattle, WA

- Bob (age 19)

Seattle, WA

A:

You're pretty much right. In order for a particle without rest mass to have momentum, it must actually travel at exactly the speed of light. This lets the rest mass

(sqrt(E^{2}-p^{2}c^{2})/c^{2}) be zero even for non-zero p, so long as E=pc.

Mike W.

(sqrt(E

Mike W.

*(published on 06/19/2011)*

Q:

I have a follow up question about "amount" of bending of light in a gravitational field:
You mentioned before that "you'd calculate some bending of light in a gravitational field, but it would only be half the observed amount. General Relativity, which describes the distortion of space time by mass and momentum, is needed to get the right answer"
I heard this (=half the observed amount) several times but could you show how to calculate the amount of light bent by "the sun" according to
1)Newtonian physics
2)General Relativity
and finally show
3)the actual observed values
and make a comparison?

- Anonymous

- Anonymous

A:

For the Newtonian or Special Relativistic calculation, it's easiest to treat the light as a particle, although a wave description would also work. The light is traveling at speed c. Say that the gravitational field g is at right angles to the velocity. (That's the part of g that gives curvature.) The radius of curvature (this is just simple geometry) is then c^{2}/g.In other words, for each little distance dx traveled, the direction changes by an angle dθ = gdx/c^{2} (in radians).

To get the total curvature as the light goes by the sun (say with closest approach distance L from the center) we have to integrate dθ over the whole path, taking into account the changing distance from the sun and the changing angle between the acceleration and the velocity. So long as you make the (excellent) approximation that the total curvature is very small you can just add the curvatures on the nearly uncurved path. If I've done the integral correctly, that gives θ=2GM/Lc^{2 }for the total curvature.

Now what happens in GR? I'm a little more comfortable with a wave picture here. That part of the curvature we just calculated basically corresponds to the change in kinetic energy when something falls in a field. Energy is just the same thing as frequency, quantum mechanically. So what that part consists of is just that the part of the wave closer to the sun is higher frequency, with the wavefronts spaced closer together. Try drawing some wavefronts spaced a little closer together on one side and farther apart on the other, remembering that the wave is propagating at right angles to the fronts. You'll see the wave curve.

Why did I go through that wave exercise? In GR, the gravitational distortion affects not only the time part of spacetime (the part we just treated) but also the space part. In a standard coordinate choice, there's extra space, more path length, for the parts of the wave nearer the sun. Projection of that curved space onto a flat plane makes waves which have equal spacing nearer and farther from the sun look like they are more tightly spaced near the sun, since the flat space doesn't have that extra stretch in there. So that's the source of the extra bending. This second bending is just as big for massive particles as for light, but for particles moving slowly compared to c, the first component of the bending is so much bigger that it's hard to see this second part.

Why are the two components of equal importance for light? In dividing the effect up into the two parts, we've implicitly assumed a conventional coordinate system in which the speed of light is isotropic and everywhere constant. [The previous answer I'd given here was messed up./mw] I can now do the calculation, and it works out. It looks like the time-like part of the effect comes mainly from the region where the light gets closest to the sun. The space-like part seems to come mostly from two regions, where the light is approaching and leaving that region of closest approach. The reason for that is that the peculiar lengths (in this choice of coordinates) are only along the radial direction, not along the tangential direction. Unfortunately I don't yet have a deep enough feel for the calculation to summarize simply why the two parts have to come out equal.

As for the observational results, there are many references in .

Mike W.

To get the total curvature as the light goes by the sun (say with closest approach distance L from the center) we have to integrate dθ over the whole path, taking into account the changing distance from the sun and the changing angle between the acceleration and the velocity. So long as you make the (excellent) approximation that the total curvature is very small you can just add the curvatures on the nearly uncurved path. If I've done the integral correctly, that gives θ=2GM/Lc

Now what happens in GR? I'm a little more comfortable with a wave picture here. That part of the curvature we just calculated basically corresponds to the change in kinetic energy when something falls in a field. Energy is just the same thing as frequency, quantum mechanically. So what that part consists of is just that the part of the wave closer to the sun is higher frequency, with the wavefronts spaced closer together. Try drawing some wavefronts spaced a little closer together on one side and farther apart on the other, remembering that the wave is propagating at right angles to the fronts. You'll see the wave curve.

Why did I go through that wave exercise? In GR, the gravitational distortion affects not only the time part of spacetime (the part we just treated) but also the space part. In a standard coordinate choice, there's extra space, more path length, for the parts of the wave nearer the sun. Projection of that curved space onto a flat plane makes waves which have equal spacing nearer and farther from the sun look like they are more tightly spaced near the sun, since the flat space doesn't have that extra stretch in there. So that's the source of the extra bending. This second bending is just as big for massive particles as for light, but for particles moving slowly compared to c, the first component of the bending is so much bigger that it's hard to see this second part.

Why are the two components of equal importance for light? In dividing the effect up into the two parts, we've implicitly assumed a conventional coordinate system in which the speed of light is isotropic and everywhere constant. [The previous answer I'd given here was messed up./mw] I can now do the calculation, and it works out. It looks like the time-like part of the effect comes mainly from the region where the light gets closest to the sun. The space-like part seems to come mostly from two regions, where the light is approaching and leaving that region of closest approach. The reason for that is that the peculiar lengths (in this choice of coordinates) are only along the radial direction, not along the tangential direction. Unfortunately I don't yet have a deep enough feel for the calculation to summarize simply why the two parts have to come out equal.

As for the observational results, there are many references in .

Mike W.

*(published on 08/23/2011)*

Q:

How can a massless object such as light have a momentum? I understood from previous answers that mass can be seen in two different ways one of them is "internal mass", or the other way of saying it is energy the one we could see in a famous equation E=mc^2. Well, knowing from Einstien's studies that this equating was build basing on m=0 (where m is a mass of the photon, particle of light) I can clearly see a paradox in it.

- Maria K. (age 15)

Los Angeles, California, USA

- Maria K. (age 15)

Los Angeles, California, USA

A:

I don't see any paradox.

**p**=M**v**, where M is the *inertial* mass. The total energy is also given by E=Mc^{2,} with the same M.

The energy is also given by E=sqrt(mc^{4}+**p**^{2}c^{2}) where m is the rest mass. When m=0, E=pc.

Mike W.

*(published on 05/06/2015)*

Q:

Actually gravity works on everything that has energy, not mass

- Sandeep Kumar Dash (age 12)

kalpakkam, Tamil Nadu, India

- Sandeep Kumar Dash (age 12)

kalpakkam, Tamil Nadu, India

A:

True. But since *inertial *mass and energy are *the same thing*, just measured in different units, there's no distinction. As we make clear above, rest mass is *not* required for gravitational interactions. That's why light is affected by gravity. I'm surprised that you thought we were saying something different.

Mike W.

*(published on 07/18/2015)*

Q:

Thought experiment: two identical boxes hanging on each side of a very accurate balance. Both have a tiny hole on a side wall. shining a photon into one of the box would change the weight of the box as the photon has added some energy to the box??

- Tony (age 50)

Hong Kong

- Tony (age 50)

Hong Kong

A:

Yes, but it's really a small effect. Say it's a visble photon, you get about 4*10^{-33 }gm extra weight.

Mike W.

*(published on 02/01/2016)*