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An equation for momentum, $$p = \gamma mv, \tag{1}$$ then for photon we get some indeterminate momentum which is 0/0.

But the formula $E=(m^2c^4+p^2c^2)^{1/2}$ gives for photon $$p=E/c .\tag{2}$$ And from equation $E=hc/\lambda$, we get an energy which is finite and then using (2) we get the momentum.

If equation (1) gives photon's momentum in an indeterminate form which is not true, then should (1) be termed valid for all particles?

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  • $\begingroup$ $(1)$ is only valid for $v\neq c$. $\endgroup$
    – Wood
    Commented Oct 30, 2016 at 9:12
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    $\begingroup$ (1) is not the definition of momentum. $\endgroup$
    – Robin Ekman
    Commented Oct 30, 2016 at 10:06
  • $\begingroup$ I think it is the relativistic momentum. But if you do not take it as momentum, then according to the formula you have to introduce relative mass which also gets an indeterminate form $\endgroup$
    – Mockingbird
    Commented Oct 30, 2016 at 10:59
  • $\begingroup$ It is the correct expression for the relativistic momentum of a massive particle. But that is a derived result, it is not a definition. The definition of momentum is $\partial L/\partial x$, or the quantity conserved according to Noether's theorem because of translation invariance. $\endgroup$
    – Robin Ekman
    Commented Oct 30, 2016 at 11:18
  • $\begingroup$ @ a curious mind Is this edit now legitimate for this question to be not termed duplicate? $\endgroup$
    – Mockingbird
    Commented Oct 31, 2016 at 19:08

2 Answers 2

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I am not an expert of this subject but my understanding is that the equations like

$p=\gamma m_0v$

or

$E=\gamma m_0c^2$

are not considered to give the correct idea but they give quick estimations for the massive particles.

I think that equation 2 is the correct form and equation 1 is the compact form that is derived from equation 2 for massive particles.

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  • $\begingroup$ Which is another way of saying that it doesn't work for photon. $\endgroup$
    – Mockingbird
    Commented Oct 30, 2016 at 11:16
  • $\begingroup$ @Mockingbird yes you are right. $\endgroup$
    – hsinghal
    Commented Oct 30, 2016 at 13:29
  • $\begingroup$ They are not estimates. They are exactly correct for massive particles. They just are not general enough to include massless particles. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Oct 30, 2016 at 17:56
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Or is it for photon (1) doesn't work?

The four-momentum for a massive particle is

$$p^\mu = mU^\mu$$

where

$$U^\mu=\frac{dx^\mu}{d\tau}$$

is the (time-like) four-velocity of the particle (the unit tangent vector of the particle's time-like world line). The components of the four-momentum are

$$m(\gamma c, \gamma \vec v) = (E/c, \gamma m \vec v)$$

Note carefully that the 'length' of the (time-like) four-momentum is (proportional to) the rest energy - the time component of the four-momentum in a coordinate system in which the particle is at rest

$$p_\mu p^\mu = m^2\left((\gamma c)^2 - (\gamma v)^2 \right) = (mc)^2 = (E_0/c)^2$$

from which we get the relativistic energy-momentum relationship

$$(E/c)^2 - \gamma^2 m^2 v^2 = (E_0/c)^2 \rightarrow E^2 = (pc)^2 + (E_0)^2 = (pc)^2 + (mc^2)^2 $$

However, the world line of a photon is light-like (null) and thus there is no four-velocity for a photon; $U^\mu$ does not exist for a null world-line.

Nonetheless, the (light-like) four-momentum for a photon exists and is given by

$$p^\mu = (E/c,\vec p)$$

and since it is light-like, its 'length' is zero (null)

$$p_\mu p^\mu = (E/c)^2 - p^2 = 0 \rightarrow E^2 = (pc)^2$$

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