3 ms·

Can anyone explain electromagnetic momentum? I thought that em waves (photons) are massless. How can they have momentum (mass * velocity)?

nerd_stuff 11y ago
One derivation starts with E = mc^2, which is a special case of E^2 = m^2c^4 + p^2c^2 but with velocity set to zero. In the second equation if you instead set the mass to zero you get p = E/c (momentum equals energy divided by the speed of light) for massless particles. The deeper reasons are above my pay-grade.
gizmo686 11y ago
Because photons travel at the speed of light, we need to consider relativistic effects. Under special relativity, classical momentum (mass * velocity) is not actually conserved. To achieve conservation of mass, we consider the "inertial mass" of an object instead of the classical mass, where the intertial mass is given by: m'=ym y=1/sqrt(1-(v/c)^2) As you can see, y>1 and y approaches infinity as you speed approaches the speed of light. This is why we say that the mass of an object increases with its velocity, and that any object with mass traveling at the speed of light will have infinite mass. Now consider the momentum, p, of a photon. We have: p=m*y*v m=0 y=infinity v=c this gives us p=0*infinity, which is indeterminate, so we cannot use this equation to determine the momentum of a photon. Instead, we can use the engery-momentum relationship, which states: E^2 = (mc^2)^2 + (pc)^2 (This is a generalization of the famous E=mc^2 equation to also consider the momentum). In this equation m refers to the rest mass, not inertial mass. Because we are dealing with a photon, we have m=0, which gives us: E^2=(pc)^2 p=E/c Indicating that the momentum of a massless object is proportional to its energy.
- nitrogen 11y ago
  E^2 = (mc^2)^2 + (pc)^2 Is it just coincidence that this looks like the Pythagorean theorem?
  - ProcessBlue 11y ago
    Yes (but also no). The initial form of this equation is (mc^2)^2 = E^2 - (pc^2)^2 The Pythagorean theorem tells you the length of a 2D vector when you know x and y lengths. Here mc^2 is the length of the 4-momentum vector [E, pc^2] (where p is a standard 3D vector). However since our 4D spacetime is not Euclidian but Minkowskian the sign in the generalised Pythagoras theorem is a minus and not a plus.
  - jimmahoney 11y ago
    It's not entirely coincidence. Four dimensional space-time has a metric analogous to but not quite the same as euclidean space: the time part has the opposite sign from the space parts. ds^2 = (c dt)^2 - (dx^2 + dy^2 + dz^2) ds is an "invariant proper time" which has the same value in all frames of reference. Check any special relativity textbook for the details. Just as you can start with distance and then build up to momentum and energy in classical physics, in relativistic mechanics you can start with this metric and build up vectors in 4-space for velocity and energy/momentum. (Turns out that the time part is an energy while the space parts are momentum.) The upshot is that (m_0 c^2)^2 = E^2 - (p c)^2 which is the frame-invariant length of the energy-momentum 4-vector. (I think that m is better written as m_0, the rest mass, since the "m" notation sometimes means relativistic mass, which is different.)
  - evanb 11y ago
    No, it is not a coincidence [0]. It is a direct result of the fact that in relativity temporal and spatial information has to be unified into a single vector (a four-vector [1]). Since (in the Noether theorem sense [2]) energy is the conserved quantity related to symmetry of displacements in time and momentum is the conserved quantity related to symmetry under displacements in space, it is quite natural to unite them into one four vector, related to symmetry of displacements in spacetime. What's peculiar is that the temporal entry of the four vector gets the "opposite sign" for the Pythagorean theorem. That is, if you have a vector (t, x, y, z), then the "hypotenuse" (called an invariant) is t^2 - x^2 - y^2 - z^2 (up to a conventional choice of overall sign). What's surprising is that the hypotenuse is the thing that's the same for different observers. So, if you do a Lorentz transformation [3], t and the spatial entries will change, but the invariant combination won't. Everybody knows E = mc^2 is the relationship between a particle's mass and its energy. But what is its energy if it's moving? Certainly the particle gains energy the faster it moves, right? Yes. E = mc^2 is the 0-momentum version of a relativistic expression. Since E is temporal and momentum p is spatial, the four vector can be written (E, pc), which has an invariant E^2 - (pc)^2. We call this invariant the rest mass[4], up to some factors of c. Algebraically moving things around gives us the "Pythagorean" form. [0] https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relati... [1] https://en.wikipedia.org/wiki/Four-vector https://en.wikipedia.org/wiki/Four-vector [2] https://en.wikipedia.org/wiki/Noether%27s_theorem https://en.wikipedia.org/wiki/Noether%27s_theorem [3] https://en.wikipedia.org/wiki/Lorentz_transformation https://en.wikipedia.org/wiki/Lorentz_transformation [4] https://en.wikipedia.org/wiki/Invariant_mass https://en.wikipedia.org/wiki/Invariant_mass
- buster 11y ago
  What a nice and clear description! You should teach that stuff :)
- aroberge 11y ago
  "Relativistic Inertial mass" (m' = ym) is a bogus term invented so as to make the equations look the same as the classical ones in some limited circumstances - namely, when a force is acting in the same direction as the relative velocity. If a force is acting in a different direction, to obtain a relation of the form F=mdp/dt, you need a different* so-called relativistic inertial mass. I know of no one doing modern research in relativity using this obsolete concept.

nerd_stuff 11y ago

One derivation starts with E = mc^2, which is a special case of E^2 = m^2c^4 + p^2c^2 but with velocity set to zero. In the second equation if you instead set the mass to zero you get p = E/c (momentum equals energy divided by the speed of light) for massless particles. The deeper reasons are above my pay-grade.

gizmo686 11y ago

Because photons travel at the speed of light, we need to consider relativistic effects. Under special relativity, classical momentum (mass * velocity) is not actually conserved. To achieve conservation of mass, we consider the "inertial mass" of an object instead of the classical mass, where the intertial mass is given by: m'=ym y=1/sqrt(1-(v/c)^2) As you can see, y>1 and y approaches infinity as you speed approaches the speed of light. This is why we say that the mass of an object increases with its velocity, and that any object with mass traveling at the speed of light will have infinite mass. Now consider the momentum, p, of a photon. We have: p=m*y*v m=0 y=infinity v=c this gives us p=0*infinity, which is indeterminate, so we cannot use this equation to determine the momentum of a photon. Instead, we can use the engery-momentum relationship, which states: E^2 = (mc^2)^2 + (pc)^2 (This is a generalization of the famous E=mc^2 equation to also consider the momentum). In this equation m refers to the rest mass, not inertial mass. Because we are dealing with a photon, we have m=0, which gives us: E^2=(pc)^2 p=E/c Indicating that the momentum of a massless object is proportional to its energy.

nitrogen 11y ago
E^2 = (mc^2)^2 + (pc)^2 Is it just coincidence that this looks like the Pythagorean theorem?
- ProcessBlue 11y ago
  Yes (but also no). The initial form of this equation is (mc^2)^2 = E^2 - (pc^2)^2 The Pythagorean theorem tells you the length of a 2D vector when you know x and y lengths. Here mc^2 is the length of the 4-momentum vector [E, pc^2] (where p is a standard 3D vector). However since our 4D spacetime is not Euclidian but Minkowskian the sign in the generalised Pythagoras theorem is a minus and not a plus.
- jimmahoney 11y ago
  It's not entirely coincidence. Four dimensional space-time has a metric analogous to but not quite the same as euclidean space: the time part has the opposite sign from the space parts. ds^2 = (c dt)^2 - (dx^2 + dy^2 + dz^2) ds is an "invariant proper time" which has the same value in all frames of reference. Check any special relativity textbook for the details. Just as you can start with distance and then build up to momentum and energy in classical physics, in relativistic mechanics you can start with this metric and build up vectors in 4-space for velocity and energy/momentum. (Turns out that the time part is an energy while the space parts are momentum.) The upshot is that (m_0 c^2)^2 = E^2 - (p c)^2 which is the frame-invariant length of the energy-momentum 4-vector. (I think that m is better written as m_0, the rest mass, since the "m" notation sometimes means relativistic mass, which is different.)
- evanb 11y ago
  No, it is not a coincidence [0]. It is a direct result of the fact that in relativity temporal and spatial information has to be unified into a single vector (a four-vector [1]). Since (in the Noether theorem sense [2]) energy is the conserved quantity related to symmetry of displacements in time and momentum is the conserved quantity related to symmetry under displacements in space, it is quite natural to unite them into one four vector, related to symmetry of displacements in spacetime. What's peculiar is that the temporal entry of the four vector gets the "opposite sign" for the Pythagorean theorem. That is, if you have a vector (t, x, y, z), then the "hypotenuse" (called an invariant) is t^2 - x^2 - y^2 - z^2 (up to a conventional choice of overall sign). What's surprising is that the hypotenuse is the thing that's the same for different observers. So, if you do a Lorentz transformation [3], t and the spatial entries will change, but the invariant combination won't. Everybody knows E = mc^2 is the relationship between a particle's mass and its energy. But what is its energy if it's moving? Certainly the particle gains energy the faster it moves, right? Yes. E = mc^2 is the 0-momentum version of a relativistic expression. Since E is temporal and momentum p is spatial, the four vector can be written (E, pc), which has an invariant E^2 - (pc)^2. We call this invariant the rest mass[4], up to some factors of c. Algebraically moving things around gives us the "Pythagorean" form. [0] https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relati... [1] https://en.wikipedia.org/wiki/Four-vector https://en.wikipedia.org/wiki/Four-vector [2] https://en.wikipedia.org/wiki/Noether%27s_theorem https://en.wikipedia.org/wiki/Noether%27s_theorem [3] https://en.wikipedia.org/wiki/Lorentz_transformation https://en.wikipedia.org/wiki/Lorentz_transformation [4] https://en.wikipedia.org/wiki/Invariant_mass https://en.wikipedia.org/wiki/Invariant_mass
buster 11y ago
What a nice and clear description! You should teach that stuff :)
aroberge 11y ago
"Relativistic Inertial mass" (m' = ym) is a bogus term invented so as to make the equations look the same as the classical ones in some limited circumstances - namely, when a force is acting in the same direction as the relative velocity. If a force is acting in a different direction, to obtain a relation of the form F=mdp/dt, you need a different* so-called relativistic inertial mass. I know of no one doing modern research in relativity using this obsolete concept.