Light building the standard model

[Dale]

A few basic rules to follow, and you can model the behavior of just about any system of masses and forces you can imagine, with the only limitation being my knowledge of calculus.

Same with light. Just a few basic rules to follow and you can model the behavior of just about any system of masses and charges and EM fields you can imagine, with the only limitatin being your knowledge of calculus.

The rest of your post is basically nonsense. There is never any rule to science that nature should be completely described in 4 equations or less. These additional equations aren't added for the hell of it, but in order to match what is experimentally observed. If you don't like how complicated this universe is then you are certainly welcome to try to find a simpler one.

 

[Pengwuino]

Isn't that the experimental upper bound for the "rest mass" of a photon (not the relativistic mass)?

I thought I had written it down, but the upper limit of the photon mass is ~10^-51kg. That calculation should give a mass for that photon of ~ 10^-29kg or so.

 

[I like Serena]

I thought I had written it down, but the upper limit of the photon mass is ~10^-51kg. That calculation should give a mass for that photon of ~ 10^-29kg or so.

I checked back on the quoted articles from DaleSpam:
http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#photon_mass

and from you:
"Actually yah, I should point out my post was more of the theory side. Experimentally the upper limit on the photon mass is on the order of 10^-54 kg."
http://pdg.lbl.gov/2009/tables/rpp20...ggs-bosons.pdf

Interesting reads btw! Thanks for that! However, in DaleSpam's article I found:
"# Luo et al., “New Experimental Limit on the Photon Rest Mass with a Rotating Torsion Balance”, Phys. Rev. Lett, 90, no. 8, 081801 (2003).
A limit of 1.2×10−51 g (6×10−19 eV/c2)."Note the reference to "Photon Rest Mass", not just "mass" or "relativistic mass".
Relativistic mass (or perhaps I should say "energy equivalent mass") is not bounded as I understand it, it's just a matter of how much energy you put into the photon (how high its frequency is).

 

[DeG]

Could anyone give me the mathematics behind the blue/red shift of light as it travels through a gravitational potential difference?

 

[Dale]

Note the reference to "Photon Rest Mass", not just "mass" or "relativistic mass".
Relativistic mass (or perhaps I should say "energy equivalent mass") is not bounded as I understand it, it's just a matter of how much energy you put into the photon (how high its frequency is).

That is correct. When modern physicsts use the unqualified word "mass" they are generally referring to the invariant mass (aka rest mass). And specifically, that is the mass that is implied when we say that a photon is massless.

The term "relativistic mass" is not usually used any more, instead physicists usually simply speak of "total energy" which is the same as relativistic mass times c^2.

 

[Dale]

If I understand correctly (http://en.wikipedia.org/wiki/Four-momentum" ) we have $-||P||^2=m_0^2 c^2$ for a single object, which is zero for a photon.

Yes, where P is the four-momentum.

The box would have: $||P||^2=-(\sum E/c)^2$, since the total 3-momentum would be zero.

Yes. You can see this explicitly for a pair of photons in the CoM frame:
P1=(E/c,p,0,0)
P2=(E/c,-p,0,0)
||P1+P2||=||(2E/c,0,0,0)||=2E/c

So the invariant mass of the box with photons would be the sum of their "relativistic masses".

Plus the mass of the box, and only in the rest frame of the box, yes.

 

[DeG]

You say only in the rest frame of the box, does this imply that it wouldn't take a force to "accelerate," i.e change the motion, the photons in the box if you pushed the box? In other words, to someone moving the box it would seem as though the mass of the system is just the mass of the box, not the box plus the relativistic mass of the photons?

 

[I like Serena]

No.
To someone moving the box it would seem as if the mass of the system is the mass of the box plus the relativistic masses of the photons.

 

[Dale]

You say only in the rest frame of the box, does this imply that it wouldn't take a force to "accelerate," i.e change the motion, the photons in the box if you pushed the box? In other words, to someone moving the box it would seem as though the mass of the system is just the mass of the box, not the box plus the relativistic mass of the photons?

Sorry, occasionally the math-to-english translation gets fuzzy.

In some reference frame (and in units where c=1) if you have a box with a four-momentum
$$P_B=(m,0,0,0)$$

Enclosing two photons with four-momenta
$$P_1=(E,-E,0,0)$$
$$P_2=(E,E,0,0)$$

Then the mass of the box is:
$$|P_B+P_1+P_2|=|(m+2E,0,0,0)|=m+2E$$

So the invariant mass is the sum of the mass of the box and the total energy (aka relativistic mass) of the photons in the rest frame of the box. However, in another reference frame:

$$P'_B=\left(\frac{m}{\sqrt{1-v^2}}, \frac{m v}{\sqrt{1-v^2}}, 0,0\right)$$
$$P'_1=\left(E\sqrt{\frac{1+v}{1-v}},-E\sqrt{\frac{1+v}{1-v}},0,0\right)$$
$$P'_2=\left(E\sqrt{\frac{1-v}{1+v}},E\sqrt{\frac{1-v}{1+v}},0,0\right)$$

So

$$m+2E \ne m + E\sqrt{\frac{1+v}{1-v}} + E\sqrt{\frac{1-v}{1+v}}$$

Therefore the invariant mass is not the sum of the mass of the box and the total energy of the photons in another frame.

The photons do contribute to the inertia of the box in all reference frames, but not necessarily by an amount equal to the "relativistic mass" of the photons.

 

[DeG]

I see. Thank you.

 

[I like Serena]

Yes, thanks for the clarification!


Apparently it matters whether the box you have is filled with photons, or filled with sand.
In one frame the 2 boxes will have the same invariant mass and the same energy.
But in another frame, the 2 boxes still have the same invariant mass, but different energies.


Since I fiddled around with the formulas I do have a couple of questions/comments.

Shouldn't the multipliers you have in P'₁ and P'₂ be interchanged?
And I think the final formula should have a couple of √(1-v²) in it, since I think we're comparing relativistic masses (energies) both in the other frame here.

 

[Dale]

Apparently it matters whether the box you have is filled with photons, or filled with sand.
In one frame the 2 boxes will have the same invariant mass and the same energy.
But in another frame, the 2 boxes still have the same invariant mass, but different energies.

The above conclusion still holds for sand.

Since I fiddled around with the formulas I do have a couple of questions/comments.

Shouldn't the multipliers you have in P'₁ and P'₂ be interchanged?

Oops, you are correct , my apologies. The correct four-momenta should be:
$$P'_1=\left(E\sqrt{\frac{1-v}{1+v}},-E\sqrt{\frac{1-v}{1+v}},0,0\right)$$
$$P'_2=\left(E\sqrt{\frac{1+v}{1-v}},E\sqrt{\frac{1+v}{1-v}},0,0\right)$$

And I think the final formula should have a couple of √(1-v²) in it, since I think we're comparing relativistic masses (energies) both in the other frame here.

That is just a matter of algebra.
$$\frac{1+v}{\sqrt{1-v^2}}=\frac{1+v}{\sqrt{(1-v) (1+v)}}=\sqrt{\frac{1+v}{1-v}}$$

 

[I like Serena]

And I think the final formula should have a couple of √(1-v²) in it, since I think we're comparing relativistic masses (energies) both in the other frame here.

That is just a matter of algebra.
$$\frac{1+v}{\sqrt{1-v^2}}=\frac{1+v}{\sqrt{(1-v) (1+v)}}=\sqrt{\frac{1+v}{1-v}}$$

That's not what I meant.
I meant that instead of:

$$m+2E \ne m + E\sqrt{\frac{1+v}{1-v}} + E\sqrt{\frac{1-v}{1+v}}$$

it should be
$$\frac {m+2E} {\sqrt{1-v^2}} \ne \frac {m} {\sqrt{1-v^2}} + E\sqrt{\frac{1+v}{1-v}} + E\sqrt{\frac{1-v}{1+v}}$$
That is, in another frame, the mass-energy of a box with sand is not the same as the mass-energy of the box with photons.

 

[Dale]

That's not what I meant.
I meant that instead of:

$$m+2E \ne m + E\sqrt{\frac{1+v}{1-v}} + E\sqrt{\frac{1-v}{1+v}}$$

it should be
$$\frac {m+2E} {\sqrt{1-v^2}} \ne \frac {m} {\sqrt{1-v^2}} + E\sqrt{\frac{1+v}{1-v}} + E\sqrt{\frac{1-v}{1+v}}$$
That is, in another frame, the mass-energy of a box with sand is not the same as the mass-energy of the box with photons.

No, what I wrote is correct. Remember, this is a response to DeG's question regarding my statement that the mass of the box with photons is the sum of the relativistic masses of the photons plus the mass of the box, but only in the rest frame of the box.

The mass of the box with photons was determined to be $m+2E$ in the rest frame. Since mass is invariant, that is the mass in all frames, so the left hand side is correct. Similarly, the mass of the box is m in all frames, and only the relativistic mass (total energy) of the photons is different in the moving frame. So the right hand side is also correct, and not equal to the left hand side.

 

[I like Serena]

Hmm, the question of DeG was: "to someone moving the box it would seem as though the mass of the system is just the mass of the box, not the box plus the relativistic mass of the photons?"So what happens if we push against the box with some force until the box is accelerated to some speed v?

Then to accelerate only the box, you would need an impulse $\frac {mv} {\sqrt{1-v^2}} \approx mv$.
But to accelerate the box with photons, wouldn't you need a bigger impulse?
Wouldn't that impulse be:
$$\frac {mv} {\sqrt{1-v^2}} + E\sqrt{\frac{1+v}{1-v}} - E\sqrt{\frac{1-v}{1+v}}\approx (m + 2E)v$$
In other words, the "mass" that needs to be accelerated is not just the box, but it is approximately the box plus the relativistic masses of the photons in the rest frame?
EDIT:
And now that I think some more about it, don't we have:
$$\frac {mv} {\sqrt{1-v^2}} + E\sqrt{\frac{1+v}{1-v}} - E\sqrt{\frac{1-v}{1+v}} = \frac {(m+2E)v} {\sqrt{1-v^2}}$$
So the force to accelerate the box with photons in the rest frame, is the same as the force to accelerate a box with sand.

Finally, this looks more like the mass-energy equivalence that I was looking for and expecting!

 

[Dale]

Hmm, the question of DeG was: "to someone moving the box it would seem as though the mass of the system is just the mass of the box, not the box plus the relativistic mass of the photons?"

Which is the question (about mass) that I answered.

So what happens if we push against the box with some force until the box is accelerated to some speed v?

Then to accelerate only the box, you would need an impulse $\frac {mv} {\sqrt{1-v^2}} \approx mv$.
But to accelerate the box with photons, wouldn't you need a bigger impulse?
Wouldn't that impulse be:
$$\frac {mv} {\sqrt{1-v^2}} + E\sqrt{\frac{1+v}{1-v}} - E\sqrt{\frac{1-v}{1+v}}\approx (m + 2E)v$$
In other words, the "mass" that needs to be accelerated is not just the box, but it is approximately the box plus the relativistic masses of the photons in the rest frame?

This is a related, but slightly different question (about force and acceleration). However, the answer is somewhat complicated in relativity.

It turns out that using ordinary non-relativistic vectors leads to a fairly complicated relationship between force and acceleration that cannot be expressed using a simple single number like "relativistic mass". Specifically, as you say "the mass that needs to be accelerated" is different in the direction parallel to and perpendicular to the velocity. See the section on "Transverse and Longitudinal Mass" here:
http://en.wikipedia.org/wiki/Mass_in_special_relativity#The_relativistic_mass_concept

In general, I would not recommend using ordinary vectors to work with forces and accelerations in special relativity. You are much better off using four-vectors where the equivalent of f=ma holds at all speeds.
http://en.wikipedia.org/wiki/Four-force
http://en.wikipedia.org/wiki/Four-acceleration
http://en.wikipedia.org/wiki/Invariant_mass

 

[I like Serena]

This is a related, but slightly different question (about force and acceleration). However, the answer is somewhat complicated in relativity.

It turns out that using ordinary non-relativistic vectors leads to a fairly complicated relationship between force and acceleration that cannot be expressed using a simple single number like "relativistic mass". Specifically, as you say "the mass that needs to be accelerated" is different in the direction parallel to and perpendicular to the velocity. See the section on "Transverse and Longitudinal Mass" here:
http://en.wikipedia.org/wiki/Mass_in_special_relativity#The_relativistic_mass_concept

In general, I would not recommend using ordinary vectors to work with forces and accelerations in special relativity. You are much better off using four-vectors where the equivalent of f=ma holds at all speeds.
http://en.wikipedia.org/wiki/Four-force
http://en.wikipedia.org/wiki/Four-acceleration
http://en.wikipedia.org/wiki/Invariant_mass

Thanks for the links which provide interesting reading material.
Btw, as you can see I avoided calculation of force or acceleration, restricting myself to impulse, which is the change in momentum (since I was not entirely sure how force or acceleration would turn out ).
Aren't my results correct?

 

[Dale]

Oh, I didn't read carefully enough. Let me look again.

For impulse a more useful link would have been:
http://en.wikipedia.org/wiki/Four-momentum

The conservation of four-momentum contains the concepts of conservation of energy, momentum, and mass, all in one nice neat mathematical package.

 

[I like Serena]

The conservation of four-momentum contains the concepts of conservation of energy, momentum, and mass, all in one nice neat mathematical package.

Yep!
That's why I picked this one to work things out.
It keeps things nicely neat and simple, and shows the relation with classical mechanics, whereas force and acceleration tend to become messy.
And btw, doesn't the relation f = dp/dt hold?
In my example the average force would be the impulse divided by the time to accelerate the box.

 

[Dale]

OK, so looking back at your post I have made a more relevant reply. For a massive system the four-momentum is a four-vector with a length given by the rest mass and a direction given by the four-velocity.

So, assuming that you want the four-velocity to be the same for both the empty and the photon-carrying boxes after the impulse then you can just use a basic "similar triangles" argument to show that the norm of the four-impulse is larger for the photon-carrying box by a ratio equal to the ratio of the (invariant) masses:
$$\frac{m+2E}{m}$$

If you boost that into any frame then the similar triangles argument still holds, as you would expect from the fact that the norm is frame invariant. So in all frames the ratio of the norms of the impulses depends on the ratio of the masses of the systems.

 

[I like Serena]

I'm still puzzling on relativistic masses.
I can see from wiki entries that the concept is avoided, and that there are separate parallel and transversal forms.
As it is I can distinguish 6 types of masses.

Let's say we have 4-momentum P = (E, p), where p is the 3-momentum, v is the velocity to a (distant) inertial observer.
I'm setting c=1 for ease of notation.
Let furthermore f = dp/dt be the 3-force.The 6 types of masses I see, are:

1. The rest mass m₀, which is also the invariant mass for a single object.

2. The invariant mass, which is the norm of the 4-momentum P. This one is the same for all observers.

3. The energy-mass E, which is the first component of P.

4. The momentum-speed-ratio $\frac {|\boldsymbol p|} {|\boldsymbol v|} = \gamma m_0$.

5. The parallel force-acceleration-ratio $\frac {f_\parallel} {\dot v_\parallel}$.

6. The transversal force-acceleration-ratio $\frac {f_\perp} {\dot v_\perp}$.Numbers 5 and 6 are messy and probably best avoided.
They can be derived from the time-derivative of the momentum.

Number 3 is ambiguous, since we saw with the box of photons, that it matters to an observer whether we are talking about a box of photons or a box of sand.

The interesting thing is that number 4, the momentum-speed-ratio, appears to match perfectly with the old relativistic mass concept, which appears to be consistent (but not the same) in all frames of reference, and which obeys the rules of classical mechanics as long as it is only being used in combination with momentum and impulse.Please correct me if I'm wrong, because I would really like to know.

 

[TheTommy1]

Correct me if I'm wrong, but my understanding is that light bends around the sun, not because light has mass but because the mass of the sun is so great it "warps" space/time and that light is following this "warped" space/time.

 

[I like Serena]

Correct me if I'm wrong, but my understanding is that light bends around the sun, not because light has mass but because the mass of the sun is so great it "warps" space/time and that light is following this "warped" space/time.

Yes, this is true and it is what is stated in general relativity theory.

If we try to explain it with classical mechanics, the mass of the sun exerts a force of gravity on the photons (whose mass is still irrelevant).
Apparently, as pengwuino stated, the resulting deviations of the photons are off by a factor of about 2.

I'm afraid the discussion in this thread has spun a bit away from the OP.

 

[cragar]

Correct me if I'm wrong, but my understanding is that light bends around the sun, not because light has mass but because the mass of the sun is so great it "warps" space/time and that light is following this "warped" space/time.

Light also has its owns Gravitational field.

 

[Dale]

Light also has its owns Gravitational field.

True, but it is not significant in this case.

 

[elegysix]

How do we know light has its own gravitational field? This certainly was never mentioned in my physics classes so far... and I've not seen any mention of it with electromagnetic waves... Where can I find information on this?

 

[cragar]

Well relativity says that mass energy or pressure bends space-time and light has energy.
If a positron and an electron collide they produce a photon or 2, I am not sure if its 1 or 2, but the electron and positron have mass and I am sure you would agree they have a Gravitational field. So if energy is conserved why would the gravitational field go away. It would seem weird. Even tho the Electric and magnetic field go away. But its something think about. Google gravitational field of light.

 

[Dale]

How do we know light has its own gravitational field? This certainly was never mentioned in my physics classes so far... and I've not seen any mention of it with electromagnetic waves... Where can I find information on this?

It is called a pp-wave spacetime:
http://en.wikipedia.org/wiki/Pp-wave_spacetime

 

[HallsofIvy]

This is a major part of what I am questioning. From classical mechanics alone, I see no reason to believe this. We know classical mechanics works. So starting with classical mechanics, How do you figure that these things happen?

Because, in some extreme cases, classical mechanics doesn't work. Specifically, the Michaelson-Morely experiment gives results different from classical mechanics and so do experiments in the photo-electric effect.

 

[elegysix]

So light has its own gravitational field... even more reason for me to think it has mass. Lol.
Just out of curiosity, does anyone know what the calculated mass from its gravitational field would be? Is this the same as the mass calculated from radiation pressure?

I know that your immediate response is that this is pointless to do, but humor me please lol

 

[cragar]

Are you asking what rest mass a particle would have that had an equal G field to a photon. I think they call it Gravitational effective mass. I am sure you could do it with E=mc^2. If light had mass it seems like it would pulverize other things because it is traveling so fast could you imagine the kinetic energy. When we collide protons together in accelerators after the collision we can have the sum of the rest masses of the particles greater than the 2 protons, we are turning kinetic energy into mass. Thats why we build bigger accelerators to find more massive particles. If you think light having its own G field is weird, you could ask does the G-field itself have its own field or does an E or B field have a Gravitational field. Are gravitational waves affected by gravity.

 

[elegysix]

Anyone know the most important experiments done with light? I want to review them.

 

[I like Serena]

Anyone know the most important experiments done with light? I want to review them.

Well, a few that spring to mind:

Relativity theory:
Michelson-Morley experiment
Deflection of light by the Sun
Gravitational redshift

Quantum theory:
Young's double-slit experiment
Photoelectric effect
Compton effect

 

[Dale]

So light has its own gravitational field... even more reason for me to think it has mass. Lol.

Did you miss the entire thread?

Just out of curiosity, does anyone know what the calculated mass from its gravitational field would be? Is this the same as the mass calculated from radiation pressure?

I know that your immediate response is that this is pointless to do, but humor me please lol

It is not pointless, but it is not easy to do either, and it depends very strongly on the exact setup. See the link I posted earlier and google "pp-wave spacetime". You will get several examples, but the math is somewhat intense.