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Borgite

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- Thread starter Borgite
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- #1

Borgite

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- #2

Dickfore

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how do you define mass?

- #3

The_Duck

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- #4

Dickfore

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[tex]

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

[/tex]

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

[tex]

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

[/tex]

light may have mass:

[tex]

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

[/tex]

So, if [itex]\varepsilon > 1[/itex], we can have "massive" photons in a medium.

- #5

K^2

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- #6

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Please start by reading the FAQ thread in the General Physics forum.

Zz.

- #7

Dickfore

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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.

Can you prove your assertion?

- #8

K^2

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- #9

Dickfore

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_{i}=p/c. Gravitational mass is the tt component of the Stress-Energy tensor over c². So m_{g}=T_{tt}/c². Energy of a photon is pc, so we have m_{g}=pc/c²=p/c=m_{i}.

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

[tex]

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

[/tex]

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:

[tex]

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

[/tex]

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:

[tex]

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

[/tex]

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:

[tex]

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

[/tex]

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

- #10

QuantumClue

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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.

- #11

K^2

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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 TThese two quantities have different transformation properties with respect to Lorentz transformations.

- #12

Dickfore

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massless particle

So, it doesn't have mass. Good to know.

- #13

atyy

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

[tex]

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

[/tex]

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:

[tex]

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

[/tex]

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:

[tex]

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

[/tex]

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:

[tex]

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

[/tex]

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

The T

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

which will transform as the first component of a 4 vector (E/c,

I learnt this from harrylin.

Last edited:

- #14

Borgite

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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

- #15

Borgite

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Dickfore comments:

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:

- #16

- 35,995

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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

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

Zz.

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