# How can light have momentum if it has zero mass?

#### bcrowell

Staff Emeritus
Gold Member
How can light have momentum if it has zero mass?

The relativistic expression for the momentum of a massive particle is p=mγv. It's possible to get confused if one tries to apply this to a particle with zero mass, since it seems as though the result would have to be zero, and yet we know based on Maxwell's equations that light has momentum. (For example, let a light wave strike an ohmic surface perpendicularly. The electric field excites oscillating currents, and the magnetic field makes a force on these currents that is shown by the right-hand rule to be in the direction of propagation.)

The resolution of the apparent contradiction is that massless particles always travel at c, and γ approaches infinity as v approaches c. That means that mγv can't be evaluated simply by plugging in values for the variables. One could instead use calculus to take the appropriate limit. An easier approach is to use the relation m2=E2-p2 (in units with c=1), which relates a particle's mass, total mass-energy, and momentum. This equation is valid for both zero and nonzero values of m. The result for m=0 is p=E (or, in units with c≠1, p=E/c), which agrees with Maxwell's equations.

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bcrowell

#### afcsimoes

I don't understand this equation, maybe because it is written as text. Can you please edit your post and clarify it?

#### jtbell

Mentor
Is this more recognizable? $$p = m \gamma v = \frac{mv}{\sqrt{1-v^2/c^2}}$$

The letter in the middle is lower-case Greek "gamma".

#### afcsimoes

Is this more recognizable? $$p = m \gamma v = \frac{mv}{\sqrt{1-v^2/c^2}}$$

The letter in the middle is lower-case Greek "gamma".
Thank you. It's perfect. Previously gama was dificult to read correctly
Ir seams like Y

#### Kasim9

Is this more recognizable? $$p = m \gamma v = \frac{mv}{\sqrt{1-v^2/c^2}}$$

The letter in the middle is lower-case Greek "gamma".
According to this formula, the momentum of a light photon is infinity as v=c. This can't be right! De Broglie's formula is p = c/lambda which gives photons a definite momentum.

#### Orodruin

Staff Emeritus
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2018 Award
According to this formula, the momentum of a light photon is infinity as v=c. This can't be right! De Broglie's formula is p = c/lambda which gives photons a definite momentum.
The formula is valid for massive particles. Photons are massless. The universal relation is $E^2 = p^2c^2 + m^2 c^4$.

#### PWiz

According to this formula, the momentum of a light photon is infinity as v=c. This can't be right! De Broglie's formula is p = c/lambda which gives photons a definite momentum.
The formula actually gives a "correct" answer even for light, but it's not "useful". Here's why:
If you substitute $m_0 = 0$ and $v=c$ into the equation, you get an indeterminate, $p=\frac{0}{0}$ . This means that there is no unique answer (every number satisfies this equation) for the momentum if the only information we know is that the rest mass is zero and speed of propagation of the object in an inertial frame is $c$. This corresponds to reality - all photons have zero rest mass and travel at $c$ (as measured in inertial frames), but they all don't have identical momentum (they theoretically can have any value [classically at least, I don't know if QFT places any restrictions on its values]). Just the mass and velocity do not give us a definite answer in this case, so some other factor must come into picture (a variable not included in the equation $p=\gamma m_0 v$) which determines the momentum, and this is of course the frequency of the photon, which is included in the well known equation $E=hf$. Applying De Broglie's hypothesis gives us $p=\frac{hf}{c}$.

So in a strict sense, the equation $p=\gamma m_0 v$ doesn't give a "wrong" answer when applied to light, it just doesn't give a unique/useful answer.

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