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Energy and mass are interrelated. As everything has energy could any object be massless? For example a photon is a packet of energy but still it is considered to be a massless particle. Why is it so?

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marked as duplicate by ACuriousMind, Emilio Pisanty, Prahar, John Rennie, JamalS Apr 3 '15 at 8:25

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    $\begingroup$ no, energy and mass are not always interrelated..photon has no mass but it has energy. If it had mass it could not move with the speed of light i guess. $\endgroup$ – Žarko Tomičić Apr 2 '15 at 15:25
  • $\begingroup$ i know it's an anachronism, but photons do not have rest mass (now called invariant mass). they have momentum and speed, and if you divide the former by the latter, you get a quantity of dimension: mass. sometimes we call that "relativistic mass". for velocities less than $c$, it's $$ m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} $$ where $m_0$ would be the "rest mass". $\endgroup$ – robert bristow-johnson Apr 2 '15 at 23:38
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There is only one mass. Lets make this clear. The concept of "relativistic mass" is not really a useful concept in my opinion. The invariant mass, or simply the mass, is defined as (in natural units, so $c = 1$):

$$E^2 - p^2 = m^2$$

The reason this is a much more useful definition for a mass, is because this quantity is Lorentz invariant, meaning it has the same value in every reference frame. If you define mass in any other way you are going to run into unnecessary trouble.

For the photon, this invariant mass is assumed to be 0, so its energy, $E$, gets a contribution only from the momentum of the photon, hence $E = p$. There are justifications for why we assume the photon has zero mass. The photon only has 2 degrees of freedom; the longitudinal polarisation does not exist precisely because the photon is massless. We also have other reasons to believe the photon is massless. Some laws of electromagnetism would have to be modified as well if the photon isn't massless, an example of which would be Coulomb's law. Hence Coulomb's law provides a good test of the photon mass (refer to this paper) which has been assigned the upper limit of $m ≲ 10^{−14}$ eV/$c^2$.

For other particles this is not the case; since they also possess this intrinsic mass they get contributions to the energy from that quantity as well and therefore $E^2 = p^2 + m^2$.

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  • $\begingroup$ Can this invariant mass be negative? $\endgroup$ – becko Apr 2 '15 at 18:30
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    $\begingroup$ Since the invariant mass appears in the formula squared, negative and positive mass would have the same meaning. By convention we choose to make it positive. (Although I think when you introduce general relativity you have to start distinguishing between positive and negative masses.) $\endgroup$ – Harry Johnston Apr 2 '15 at 20:46
  • $\begingroup$ @becko if you're asking if $E$ can be greater than $p$ in this equation, I believe the answer is no because $E$ varies with $p$, but I could be wrong. $\endgroup$ – Señor O Apr 2 '15 at 21:33
  • $\begingroup$ @becko:negative mass doesnt exist(not found yet) because that would mean the object with negative mass would move in the opposite direction of applied force and in the opposite direction of normal gravity. $\endgroup$ – Paul Apr 3 '15 at 1:44
  • $\begingroup$ @Paul, well, actually a very small negative test mass will behave the same in reaction to gravity as a very small positive test mass because the weird effects will cancel out. However, for larger masses that actually effect the gravitational field, you end up with unbounded acceleration and some other weird, counter-intuitive effects... $\endgroup$ – k_g Apr 3 '15 at 1:50
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Defination of mass is : $$m^2=E^2-p^2$$ Where $m$,$E$ and $p$ are mass,total energy and momentum of the object respectively. Its not necessary that if a photon has energy it should have mass because an object or a particle can have energy due to its momentum alone or its mass alone or due to its both mass and momentum and thus a photon has energy due to its momentum alone.(it doesnt have mass)

Putting $m=0$ in the energy momentum relation (the above equation) we find that $E=p$,that is a photon has energy equal to its momentum without any mass.

Here I have taken unit in which $c=1$.

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Photons never have zero mass. As you point out, they have energy, therefore they have mass - usually referred to as "relativistic mass".

Photons are said to have zero "rest mass", ie no mass when they are at rest (stationary). But photons are never stationary, so this really has no physical meaning. That they would have no mass at rest is required by relativity or else their energy when travelling at c would be infinite. As this isn't the case, they are notionally assigned zero mass at rest.

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  • $\begingroup$ But isn't relativistic mass defined as the product of rest mass and the Lorentz factor? $\endgroup$ – mattecapu Apr 2 '15 at 17:33
  • $\begingroup$ @mattecapu: When it's not singular, sure. $\endgroup$ – Charles Apr 2 '15 at 20:02
  • $\begingroup$ My feeling is that the assignment of null rest mass to photons can have physical meaning, at least in a sense. With mass-energy equivalence, this is essentially the statement that rest-photons do not exist. $\endgroup$ – David H Apr 2 '15 at 21:12
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    $\begingroup$ @mattecapu, i would define relativistic mass as the momentum of the body or particle divided by its speed. for a photon: $$ E = \hbar \omega = p c $$ so relativistic mass is $$ m = \frac{p}{c} = \frac{\hbar \omega}{c^2} $$ the reason the rest mass (or "invariant mass") of photons are zero is because $$ m_0 = m \sqrt{1-\frac{v^2}{c^2}} $$ when $v=c$, then $m_0 = 0$ no matter what finite value $m$ has. $\endgroup$ – robert bristow-johnson Apr 2 '15 at 23:47
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    $\begingroup$ Relativistic mass is such an ugly & complicating concept. Modern treatments of relativity use "rest mass" and "total energy". $\endgroup$ – rob Apr 3 '15 at 2:18

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