The Mystery of Photons and Gravity

[pixel01]

I've learned that photons have no mass, so why they are still influenced by black holes or other large celestial bodies?

 

[tiny-tim]

I've learned that photons have no mass, so why they are still influenced by black holes or other large celestial bodies?

Hi pixel01!

Two answers:

i] Photons have no rest-mass …

but they do have energy, and energy is mass.

ii] In general relativity, objects are not attracted by gravitation, they merely follow geodesics ("straight lines") in space-time.

So they don't need mass to be influenced by large bodies (though the amount of mass does affect which geodesic they follow).

 

[pixel01]

Hi pixel01!

Two answers:

i] Photons have no rest-mass …

but they do have energy, and energy is mass.

ii] In general relativity, objects are not attracted by gravitation, they merely follow geodesics ("straight lines") in space-time.

So they don't need mass to be influenced by large bodies (though the amount of mass does affect which geodesic they follow).

Thank you tiny-tim,
So what is the true mass of a photon? Say a photon beam at 400nm wavelength.

 

[Dale]

Thank you tiny-tim,
So what is the true mass of a photon? Say a photon beam at 400nm wavelength.

Which kind of mass do you mean by "true"?
Invariant (aka rest or proper)
Relativistic (aka inertial or transverse)
Longitudinal
Komar
ADM
Bondi

I would say that the "true" mass is the invariant mass which is 0 for a photon.

 

[pixel01]

Which kind of mass do you mean by "true"?
Invariant (aka rest or proper)
Relativistic (aka inertial or transverse)
Longitudinal
Komar
ADM
Bondi

I would say that the "true" mass is the invariant mass which is 0 for a photon.

I mean the moving-induced mass. The rest mass of photon is 0 already.

 

[Antenna Guy]

I mean the moving-induced mass. The rest mass of photon is 0 already.

That would be inertial mass.

$$\lambda=$$ wavelength

$$f=\frac{c}{\lambda}$$

$$h=$$ Planck's constant

$$E=hf$$

$$m=\frac{E}{c^2}$$

Regards,

Bill

 

[MeJennifer]

ii] In general relativity, objects are not attracted by gravitation, they merely follow geodesics ("straight lines") in space-time.

That would only be the case for imaginary test objects. General relativity has no background, objects that have mass or energy do not have worldines in spacetime they are the curvature of spacetime. In other words spacetime curvature is mass and energy and how they are distributed. By the way the curvature of spacetime can cause both attraction and repulsion.

 

[Fredrik]

That would only be the case for imaginary test objects. General relativity has no background, objects that have mass or energy do not have worldines in spacetime they are the curvature of spacetime. In other words spacetime curvature is mass and energy and how they are distributed. By the way the curvature of spacetime can cause both attraction and repulsion.

This is of course correct, but (as you already know) if the mass of the object is much less than that of the star that deflects its path, a test particle's path is an excellent approximation of the actual path.

(For those who don't know this already, a "test particle" is a particle that can move around in spacetime without contributing to space-time curvature. It's just a theoretical idea that we can use to approximately calculate the paths of real particles).

 

[lightarrow]

That would only be the case for imaginary test objects. General relativity has no background, objects that have mass or energy do not have worldines in spacetime they are the curvature of spacetime. In other words spacetime curvature is mass and energy and how they are distributed. By the way the curvature of spacetime can cause both attraction and repulsion.

Let's take a spherical, non rotating not charged, homogeneous star: there is curvature outside of it; does it mean the stress-energy tensor T is non zero there because of the gravitational energy only? Or the tensor T is zero there?

 

[robphy]

General relativity has no background, objects that have mass or energy do not have worldines in spacetime they are the curvature of spacetime. In other words spacetime curvature is mass and energy and how they are distributed.

I think a statement like
"spacetime curvature is mass and energy and how they are distributed."
needs some more precise justification... especially when using is.
Otherwise, I fear misconceptions that might arise from it.
In my opinion, "is associated with" seems more appropriate.

 

[Usaf Moji]

That would be inertial mass.

Wouldn't it also be gravitational mass?

 

[Antenna Guy]

Wouldn't it also be gravitational mass?

Only where the invariant/rest mass is zero. Otherwise, inertial mass is a component of gravitational/relativistic mass.

Regards,

Bill

 

[cryptic]

Only where the invariant/rest mass is zero. Otherwise, inertial mass is a component of gravitational/relativistic mass.

Regards,

Bill

Are inertial mass and gravitational mass not equal?

If You look at formula m=E/c² You can see that the kinetic energy of this mass is: E_kin=mc²/2=E/2.

 

[Antenna Guy]

Are inertial mass and gravitational mass not equal?

So long as one can distinguish between rest mass and inertial mass, I'm inclined to say no (clarif: they are not equal).

If You look at formula m=E/c² You can see that the kinetic energy of this mass is: E_kin=mc²/2=E/2.

I'm afraid I cannot. The "m" you are using might well be either rest mass or relativistic mass depending upon the usage - and relativistic kinetic energy can be described by:

$$KE=mc^2-m_0 c^2$$

where $m$ is relativistic mass, and $m_0$ is rest mass.

Relativistic mass is not generally inertial mass (i.e. v=0 with respect to your usage), and rest mass is not generally gravitational mass (i.e. a photon). However, I think it is true that relativistic mass is generally equivalent to gravitational mass.

Perhaps someone more adept than I could clarify/correct what I have said.

Regards,

Bill

 

[Usaf Moji]

I think it is true that relativistic mass is generally equivalent to gravitational mass.

Agreed.

 

[tiny-tim]

Welcome to PF!

Hi cryptic! Welcome to PF!

… If You look at formula m=E/c² You can see that the kinetic energy of this mass is: E_kin=mc²/2=E/2.

ah … you're using the Newtonian definition of KE, E = m₀v²/2.

The Einsteinian definition of KE differs, at low speeds, by an additive constant, m₀c²:

E = mc² = m₀c²/√(1 - v²/c²),
​

since that is approximately:

E = mc² = m₀c²(1 + v²/2c²) + …, = m₀c² + m₀v²/2 + ….​

So the Newtonian m₀v²/2 isn't half of the Einsteinian KE, it's the tiny (compared with m₀c²) second-order adjustment.

 

[Antenna Guy]

The Einsteinian definition of KE differs, at low speeds, by an additive constant, m₀c²:

E = mc² = m₀c²/√(1 - v²/c²),
​

I don't think there's anything "kinetic" about rest energy.

Regards,

Bill

 

[tiny-tim]

I don't think there's anything "kinetic" about rest energy.

Hi Bill!

I'm just using "KE" as the opposite of "PE".

 

[cryptic]

Hi cryptic! Welcome to PF!


ah … you're using the Newtonian definition of KE, E = m₀v²/2.

...

Hi,

but isn't it so that Photons have two parts of energy: oscillation energy and translation (kinetic ) energy? And oscillation ground state energy is quantum mechanically E=hf/2.

I think seriosus discussion is not possible because my posts disappeer after few minutes.

Regards

 

[Antenna Guy]

I'm just using "KE" as the opposite of "PE".

The bowliverse must be an interesting place...

Regards,

Bill