# Photon's mass

1. Jul 22, 2008

### pixel01

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

2. Jul 22, 2008

### tiny-tim

Hi pixel01!

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).

3. Jul 22, 2008

### pixel01

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

4. Jul 22, 2008

### Staff: Mentor

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

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

5. Jul 22, 2008

### pixel01

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

6. Jul 22, 2008

### Antenna Guy

That would be inertial mass.

$$\lambda=$$ wavelength

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

$$h=$$ Planck's constant

$$E=hf$$

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

Regards,

Bill

Last edited: Jul 22, 2008
7. Jul 22, 2008

### MeJennifer

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.

Last edited: Jul 22, 2008
8. Jul 22, 2008

### Fredrik

Staff Emeritus
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).

9. Jul 28, 2008

### lightarrow

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?

10. Jul 28, 2008

### robphy

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.

11. Jul 29, 2008

### Usaf Moji

Wouldn't it also be gravitational mass?

12. Jul 29, 2008

### Antenna Guy

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

Regards,

Bill

13. Jul 30, 2008

### cryptic

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.

14. Jul 30, 2008

### Antenna Guy

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

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

Last edited: Jul 30, 2008
15. Jul 30, 2008

### Usaf Moji

Agreed.

16. Jul 31, 2008

### tiny-tim

Welcome to PF!

Hi cryptic! Welcome to PF!
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.

17. Jul 31, 2008

### Antenna Guy

Re: Welcome to PF!

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

Regards,

Bill

18. Jul 31, 2008

### tiny-tim

Hi Bill!

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

19. Jul 31, 2008

### cryptic

Re: Welcome to PF!

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

20. Jul 31, 2008

### Antenna Guy

The bowliverse must be an interesting place...

Regards,

Bill