# The Mass of Light

My question deals with how the mass of light effects its speed. First question, does light actually have mass? If so how can you determine its mass. Does the mass of light impact its ability to move at a fixed velocity.

how do you define mass?

The way special relativity defines mass, light has no mass. It is for this reason that it travels at the same special speed in all reference frames: any massless object must always travel at the speed of light. You can define the mass of something in special relativity by looking at how its energy varies as a function of its momentum: you have energy $$E = \sqrt{p^2c^2 + m^2c^4}$$ where p is the momentum of the object and m the mass. Light is observed to obey E = pc; you can see that this agrees with setting m = 0 in the first equation.

Note that, however, if light propagates in a medium (eg. water) it is actually a collective polarization effect of the whole medium. The properties of the medium (at long wavelengths) are described by the dielectric constant $\varepsilon(\omega)$, which, in general, is a complex function of (circular) frequency. The quasiparticles associated with this collective polarization and propagation of electromagnetic waves obey a dispersion relation:

$$\varepsilon(\omega) \left(\frac{\omega}{c}\right)^{2} - k^2 = 0$$

According to the formula given above and the wave-particle dualism equations:

$$E = \hbar \omega, \; \vec{p} = \hbar \vec{k}$$

light may have mass:

$$m = \frac{1}{c} \sqrt{\left(\frac{E}{c}\right)^{2} - p^{2}} = \frac{\hbar \omega}{c^{2}} \sqrt{\varepsilon^{2}(\omega) - 1}$$

So, if $\varepsilon > 1$, we can have "massive" photons in a medium.

K^2
Why is it that hard to throw in the word "rest mass" in there? Light has no rest mass. It has inertial mass, and it has gravitational mass. (These last two have to be the same for all objects by invariance principle.)

ZapperZ
Staff Emeritus
My question deals with how the mass of light effects its speed. First question, does light actually have mass? If so how can you determine its mass. Does the mass of light impact its ability to move at a fixed velocity.

Zz.

Why is it that hard to throw in the word "rest mass" in there? Light These last two (inertial mass and gravitational mass) have to be the same for all objects by invariance principle.

K^2
Inertial mass is ratio of momentum to velocity. For light, that's mi=p/c. Gravitational mass is the tt component of the Stress-Energy tensor over c². So mg=Ttt/c². Energy of a photon is pc, so we have mg=pc/c²=p/c=mi.

Inertial mass is ratio of momentum to velocity. For light, that's mi=p/c. Gravitational mass is the tt component of the Stress-Energy tensor over c². So mg=Ttt/c². Energy of a photon is pc, so we have mg=pc/c²=p/c=mi.

But, the $T_{t t}$ component transforms as the product $A^{0} B^{0}$ of two 4-vectors with respect to Lorentz transformations, i.e.:

$$T'_{t t} = \gamma^{2} \left[T_{t t} - \beta (T_{t x} + T_{x t}) + \beta^{2} T_{x x} \right]$$

Momentum is a 3-vector in the direction of the velocity of the particle. When you mentioned ratio between two vectors, I guess you mean the coefficient of proportionality between the two vectors. This coefficient for a particle with a nonzero (rest) mass is:

$$\frac{m}{\sqrt{1 - v^{2}/c^{2}}}$$

This expression is not applicable for massless particles, so we have to somehow modify it. And, indeed, this is the (total relativistic) energy of the particle divided by $c^{2}[/tex]. Thus, your definition of inertial mass reduces to: $$m_{\mathrm{i}} = \frac{E}{c^{2}} = \frac{P^{0}}{c}$$ The last equation identifies this "inertial mass" as the time component of the energy-momentum 4-vector. It transforms with respect to Lorentz transformations as: $$P'^{0} = \gamma (P^{0} - \beta P^{1})$$ These two quantities have different transformation properties with respect to Lorentz transformations. Therefore, equating them is nonsense. My question deals with how the mass of light effects its speed. First question, does light actually have mass? If so how can you determine its mass. Does the mass of light impact its ability to move at a fixed velocity. I was going to talk about the massless photon in another thread, but apparently the thread is old, so I will talk about it here when I have wrote it up. It involves the concept of a Higgs field. A little reading on it might suffice until then. K^2 Science Advisor These two quantities have different transformation properties with respect to Lorentz transformations. They transform differently for general stress energy tensor, but we are talking about stress energy tensor of a massless particle. If you perform substitution for Ttx, you'll see that the two quantities transform the same way. massless particle So, it doesn't have mass. Good to know. atyy Science Advisor But, the [itex]T_{t t}$ component transforms as the product $A^{0} B^{0}$ of two 4-vectors with respect to Lorentz transformations, i.e.:

$$T'_{t t} = \gamma^{2} \left[T_{t t} - \beta (T_{t x} + T_{x t}) + \beta^{2} T_{x x} \right]$$

Momentum is a 3-vector in the direction of the velocity of the particle. When you mentioned ratio between two vectors, I guess you mean the coefficient of proportionality between the two vectors. This coefficient for a particle with a nonzero (rest) mass is:

$$\frac{m}{\sqrt{1 - v^{2}/c^{2}}}$$

This expression is not applicable for massless particles, so we have to somehow modify it. And, indeed, this is the (total relativistic) energy of the particle divided by [itex]c^{2}[/tex]. Thus, your definition of inertial mass reduces to:

$$m_{\mathrm{i}} = \frac{E}{c^{2}} = \frac{P^{0}}{c}$$

The last equation identifies this "inertial mass" as the time component of the energy-momentum 4-vector. It transforms with respect to Lorentz transformations as:

$$P'^{0} = \gamma (P^{0} - \beta P^{1})$$

These two quantities have different transformation properties with respect to Lorentz transformations. Therefore, equating them is nonsense.

The Ttt component of the stress energy tensor is energy density, whereas the E in E=pc is energy. Would the transformation work out if we considered a wavepacket and integrated the energy density over a volume to get energy? Similar to section 8 of http://www.fourmilab.ch/etexts/einstein/specrel/www/ which states

$$\frac{\rm E'}{\rm E} = \sqrt{\frac{1-v/c}{1+v/c}}.$$

which will transform as the first component of a 4 vector (E/c,p), if we taken into account E=pc.

I learnt this from harrylin.

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Thank you all, I believe I have some understanding to my original question.

My second question is how Extreme Gravity may effect light. If you shine a bright light near the event horizon on a black hole, could the gravity of the black hole grab and bend the light? What I do not grasp if this is possible, is how can an item, with no mass, be moved?

In retrospect, if you shine a bright light on a fan (powered of course). The light remains on the same area of the fan. If the gravity of a black hole is strong enough to take form, how can its effect on light be any different than the fan?

Should I be referring to a pool of oil in a whirlpool motion instead of a fan?

If my analogy is incorrect, please clarify. Thank you again

It may be possible that the response from Dickfore answers my quesion but I would appreciate more clarity

Note that, however, if light propagates in a medium (eg. water) it is actually a collective polarization effect of the whole medium. The properties of the medium (at long wavelengths) are described by the dielectric constant LaTeX Code: \\varepsilon(\\omega) , which, in general, is a complex function of (circular) frequency. The quasiparticles associated with this collective polarization and propagation of electromagnetic waves obey a dispersion relation:

ZapperZ
Staff Emeritus
Thank you all, I believe I have some understanding to my original question.

My second question is how Extreme Gravity may effect light. If you shine a bright light near the event horizon on a black hole, could the gravity of the black hole grab and bend the light? What I do not grasp if this is possible, is how can an item, with no mass, be moved?

In retrospect, if you shine a bright light on a fan (powered of course). The light remains on the same area of the fan. If the gravity of a black hole is strong enough to take form, how can its effect on light be any different than the fan?

Should I be referring to a pool of oil in a whirlpool motion instead of a fan?

If my analogy is incorrect, please clarify. Thank you again

Again, this is covered in one of the entry in the FAQ thread on why light is affected by gravity. Please read it!

Zz.