# Photon dynamic mass

1. Feb 18, 2007

### DonnieD

i think E=mc^2
and E=hv so dynamic mass of photon is m=hv/c^2 ?? :surprised

2. Feb 18, 2007

### masudr

$E=m_0 c^2$ does not apply to photons.

The general equation is

$$E^2 = m_0^2c^4 + \vec{p}^2c^2.$$

As you can see, in the centre of momentum frame, where $\vec{p}=0$, the equation reduces to $E=m_0 c^2$, but there is no inertial frame where the photon is at rest.

Furthermore, $m_0 = 0$ for the photon, so the equation becomes

$$E=|\vec{p}|c$$.

3. Feb 18, 2007

### DonnieD

but the relativistic mass m=hv/c^2?? (not the rest mass m_0)

4. Feb 18, 2007

### cesiumfrog

Yes. It seems like this helps to understand momentum etc, but it is also equivalent to trivially expressing energy in different units.

5. Feb 18, 2007

### rbj

but the equation $E=m c^2$ works perfectly well for photons when

$$m = \frac{m_0}{\sqrt{1 - v^2/c^2}}$$

is the relativistic mass or the inertial mass ($m = p/v$) as observed by someone in a frame that is moving at velocity v with respect to the mass that is $m_0$ in its own frame.

and that is compatible with

$$E=m c^2$$
$$m = \frac{m_0}{\sqrt{1 - |\vec{v}|^2/c^2}}$$
and
$$\vec{p} = m \vec{v}$$

but i'm glad you were clear using the $m_0$ notation instead of just the "m" for rest mass.

Last edited: Feb 18, 2007
6. Feb 18, 2007

### pmb_phy

Yes. That is quite correct so long as you understand that the "m" you're using is inertial mass (aka "relativistic mass").

Pete

7. Feb 18, 2007

### pmb_phy

That is quite untrue. E = mc2 holds only in special circumstances such as isolated systems. IT wouldn't, say, work in a rod under stress. In such case the "relativistic mass" would have a different value than the energy would. For proof see the web page I constructed to prove this point and to give an example. See

http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm

Pete

8. Feb 18, 2007

### cesiumfrog

Your example depends on external forces (and worse, on comparing those in different reference frames, which for me sets off alarm bells: you'd be surprised how many long standing paradoxes are actually coordinate transformation errors), can you provide something more concrete?

9. Feb 18, 2007

### nakurusil

The photon is a massless particle.
Relativistic mass is a frowned upon concept.
Therefore it is a bad idea to speculate about the relativistic mass of the photon

10. Feb 18, 2007

### country boy

Right. For photons, just stick to energy and momentum.

Relativistic mass is, indeed frowned upon by purists. But purists may not be the ones who come up with the new ideas. It's still good to kick the concept around a bit.

11. Feb 19, 2007

### nakurusil

Correct.

Doesn't follow logically, why wouldn't the purists be the ones that come up with new ideas. Rigor and creativity go hand in hand, they aren't opposites.

12. Feb 19, 2007

### pmb_phy

I don't know where you got that comment that you quoted but I made no comment such as that in this thread.
And that was the point. I maqde that quite clear. Not all systems are closed you know. In fact it is the same problem (worked a bit differently) that Einstein published in either 1906 or 1907.
Now I'm curious; What exactly do you think a Lorentz tranasformation does??
I already gave you the most concrete example that can be given, its in the URL I gave you.

Its not like I'm saying something different than you can found in, say, Rindlers 1982 intro to SR text.

Pete

13. Feb 19, 2007

### pmb_phy

That's a good way to get into trouble when you calculate the mass-density of a gas of photons.
What exactly is a "purist"? Is it someone who frowns on rel-mass? Please defined the term "purist". Thanks.

Pete

14. Feb 19, 2007

### cesiumfrog

No, it's from the linked webpage you keep referring to and said you constructed. :rofl:

Sarcasm, eh? I really appreciate the more authorative reference since, as I'm not very clever, I don't trust myself to follow an online derivation as closely as I should. By more concrete, what I was actually thinking of is something like this:

If I gave you a spring (with known spring constant and unstressed rest mass) with an (ideal) mirror at each end, and a pair of lasers (to reflect a continuous stream of photons against the mirrors), could you demonstrate that relativistic mass is not equivalent to energy? If so then in what way would you measure the spring's energy versus relativistic mass?

15. Feb 19, 2007

### pmb_phy

What's with the attitude. I asked you where it came from. I wrote that web page years ago and I don't remember everything I wrote in all those web pages
Not at all. I appologize if my comments came off that way. I never post sarcastic comments intentionaly. I appologize if you thought of that as attitude.
I'm glad that reference was able to help you out. Here is the section of concern here. From page 150 in the section Relativistic Mechanics of Continua
That comment opened my eyes to the more difficult aspects of mass-energy. Through a long period of study I finally came to understand this. Its not that hard if you're skilled in the physics or if you have someone you trust walk you through it.
cesiumfrog! Please don't confuse cleverness with knowledge. There's no reason to assume that you're not very clever and simply lacking in the acquired skill required to understand the physics. This has been my experience throughout the last 25 years of my life. I doubt that you'd be posting in a physics forum if you weren't clever in the first place.
The relativistic mass of what? Of just the spring? This is a subtle question and details are required. Thanks.

Pete

Last edited: Feb 19, 2007
16. Feb 19, 2007

### cesiumfrog

What I'm trying to do is express your example from the website (which I find a bit too abstract-mathematical) in terms of a specific physical example (which assists my intuitive understanding). Do you agree your (unspecified) external forces can be replaced with streams of photons (of particular frequency and intensity, from a distant source, symmetric in the box's frame) reflecting off opposite sides of the box? And that the box can be replaced by a spring (of equal dimensions whilst compressed by the external forces), without compromising the example?

17. Feb 19, 2007

### pmb_phy

Pushing on the ends with your fingers is a good enough example, don't you agree? My fingers exert pressure as good as if not better than a stream of photons, but if a stream of photons is what you like as an example then who am I to second guess you on your iintuition?
Only under certain circumstances. If the force on a rod is such as to act towards pulling the rod apart then there is negative pressure present (called "tension") and a stream of photons can't do that. However in the case you stated the fact that the spring is compressed indicates that you're not asking about this situation but the situation in which the photons are acting to compress the spring.
Sure. In any case the sitations leads to the same conclusion and yes I can demonstraight it. However you told me that you are unable to follow this kind of derivation. How do you want me to respond beyond this?

Thanks

Pete

18. Feb 19, 2007

### country boy

Read again: "... purists may not be the ones who come up with the new ideas." They can, of course, come up with new ideas. The point is that non-purists mulling over seemingly superfluous notions like photon mass can lead (and has) to new insights like gravitational deflection of photons.

But if you want logic, see the philosophy pages.

19. Feb 19, 2007

### cesiumfrog

OK, sorry you're finding this tedious. Now in this situation, it naively seems to me that the relativistic mass of the compressed spring, in any moving frame, will be simply the rest mass of the compressed spring (ie. the rest mass of the relaxed spring, plus the mechanical potential energy) multiplied by the usual lorentz factor (for the relative velocity of the frame). Is this incorrect?

20. Feb 19, 2007

### rbj

not in every sense of the word. they do not have rest mass, but they do have inertial mass of $(h \nu)/c^2$. Peter Mohr and Barry Taylor (at least at one time they were at NIST) have (along with others) proposed that the definition of kilogram be changed from the one regarding the standard prototype in Paris to

The kilogram is the mass of a body at rest whose equivalent energy corresponds to a frequency of exactly (2997924582/66260693 ×1043) Hz.

another wording they have used is

The kilogram is the mass of a body at rest whose equivalent energy is equivalent to a collection of photons of frequencies that sum to exactly (2997924582/66260693 ×1043) Hz.

now, if you put that collection of photons into a hypothetical massless and perfectly mirrored box and put the box on a scale and put the standard kg prototype on the other platter of the scale, which way would it tip?

depends on whose face. not everybody is frowning.

i'm not speculating, i am making reference to an oft and recent out-of-favor convention that defines momentum as inertial mass times velocity. then there needs to be a differentiation in concept between inertial mass and invariant mass.

when i look up "Gravitational Red Shift" in my old 3rd semester physics textbook, it says:

"Although a photon has no rest mass, it nevertheless behaves as though it possesses the inertial mass
$$m = \frac{h \nu}{c^2}$$ ."

and then goes on to show how "inertial" energy ($m c^2$) is traded for potential energy (of a particle of said mass) and with that sum conserved, how inertial energy is less as the photon departs a star resulting in lower frequency and that this red shifting is not to be confused with doppler red shifting if the star is moving away.

it's not such a bad idea. i'm still at a loss to understand why so many insist that it is. there are at least a couple ways that photons behave as if they have mass of some sort.

Last edited: Feb 19, 2007
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