Would You Transgress To Assist An...
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No Gravity From Photons
47 Answers
A photon has energy. Why does it not generate a gravitational field? Energy = mass.
Answers
"Energy = mass" is sadly misleading. Photons have energy, but an individual photon has no mass. This isn't a contradictio
09:39 Wed 09th Nov 2022
OG: // Is momentum the product of mass and velocity? And a photon has zero mass? If something has no mass then no matter how fast it's shifting, that implies no momentum either. I'm sure it must be down to the subtle differences between Newtonian physics and Quantum physics. //
Very good question, and one I just completely glossed over in my first reply. It's entirely down to special relativity v. Newtonian mechanics, though -- quantum mechanics doesn't need to enter into the picture at this stage.
I'll go into more detail on this at some point in the coming days, but it's late and I just wanted to get something quick out before I go to sleep. It's a seriously important question, and the answer is deeply revealing, and I'm glad that you asked it.
Very good question, and one I just completely glossed over in my first reply. It's entirely down to special relativity v. Newtonian mechanics, though -- quantum mechanics doesn't need to enter into the picture at this stage.
I'll go into more detail on this at some point in the coming days, but it's late and I just wanted to get something quick out before I go to sleep. It's a seriously important question, and the answer is deeply revealing, and I'm glad that you asked it.
// If something has no mass then no matter how fast it's shifting, that implies no momentum either. //
I've tried to break it down into manageable bite size chunks.
Here we go Old_Geezer... Hold tight!
As you're well aware, Newtonian Mechanics describes the momentum of objects we see in our everyday lives eg. cars, cannonballs, motorbility scooters etc
The momentum is then calculated using the model p = mv kg m/s. Unfortunately the raw model breaks down at the quantum level.
Textbooks will refer to objects having 'invariant mass'. So if an object here on earth, say a piece of Iron, with a mass of 1kg, then that same piece of Iron will have exactly the same mass of 1kg on the moon, planet Pluto or anywhere else for that matter (pardon the pun).
Consider then the following explanation:
Most are familiar with E = mc² This is a special condition which derives from the general equation (GE);
E² = p²c² + m²c³c¹ (Apologies, since I do not know how to type c to the power of 4)
Where E is the total energy of the particle (Joules), p is the momentum of the particle (units given above), c is the speed of light (300,000,000 km/s in a vacuum), and m (kg) is the mass of the particle.
It is IMPORTANT TO NOTE the GE tells us, the total energy of a particle is a combination of its mass energy and its
momentum energy (WHICH IS NOT NECESSARILY RELATED TO ITS MASS!)
Should the particle be at rest, then it has no momentum, giving p = 0. Thus the equation reduces to E² = m²c³c¹ Taking the √ of both sides gives Einstein's famous SR equation E = m²c² where energy and mass are inextricably linked and completely interchangeable.
Now let us go back and address your query;
There is an interesting additional effect contained in the general equation. If a particle has no mass (m = 0) and is at rest (p = 0),
then the total energy is zero (E = 0). But an object with zero energy and zero mass is nothing at all. That is, IT DOES NOT AND CANNOT EXIST!
Therefore, if an object with no mass is to physically exist, it can never be at rest. Such is the case with light.
So when m = 0 the GE becomes E = pc. Since photons (particles of light) have no mass, they must obey E = pc and therefore get all of their energy from their momentum.
The momentum p can then be in direct relation to Planck's constant h. This is a direct association between energy and frequency. As a side note, other massless particles you might have come across are gluons.
Other texts may wish to explain in terms relativistic mass. Einstein in particular didn't like this. It suggests that mass is key to
momentum, when in fact it describes changes in a particle's energy. The term to my mind is somewhat of a misnomer.
I have done my best to avoid a mathematical treatise of the question posed. Inevitably at some point, an equation had to 'pop up'.
I await with much interest jim's explanation. He may well choose a two or even three pronged attack. Let's wait and see.
I've tried to break it down into manageable bite size chunks.
Here we go Old_Geezer... Hold tight!
As you're well aware, Newtonian Mechanics describes the momentum of objects we see in our everyday lives eg. cars, cannonballs, motorbility scooters etc
The momentum is then calculated using the model p = mv kg m/s. Unfortunately the raw model breaks down at the quantum level.
Textbooks will refer to objects having 'invariant mass'. So if an object here on earth, say a piece of Iron, with a mass of 1kg, then that same piece of Iron will have exactly the same mass of 1kg on the moon, planet Pluto or anywhere else for that matter (pardon the pun).
Consider then the following explanation:
Most are familiar with E = mc² This is a special condition which derives from the general equation (GE);
E² = p²c² + m²c³c¹ (Apologies, since I do not know how to type c to the power of 4)
Where E is the total energy of the particle (Joules), p is the momentum of the particle (units given above), c is the speed of light (300,000,000 km/s in a vacuum), and m (kg) is the mass of the particle.
It is IMPORTANT TO NOTE the GE tells us, the total energy of a particle is a combination of its mass energy and its
momentum energy (WHICH IS NOT NECESSARILY RELATED TO ITS MASS!)
Should the particle be at rest, then it has no momentum, giving p = 0. Thus the equation reduces to E² = m²c³c¹ Taking the √ of both sides gives Einstein's famous SR equation E = m²c² where energy and mass are inextricably linked and completely interchangeable.
Now let us go back and address your query;
There is an interesting additional effect contained in the general equation. If a particle has no mass (m = 0) and is at rest (p = 0),
then the total energy is zero (E = 0). But an object with zero energy and zero mass is nothing at all. That is, IT DOES NOT AND CANNOT EXIST!
Therefore, if an object with no mass is to physically exist, it can never be at rest. Such is the case with light.
So when m = 0 the GE becomes E = pc. Since photons (particles of light) have no mass, they must obey E = pc and therefore get all of their energy from their momentum.
The momentum p can then be in direct relation to Planck's constant h. This is a direct association between energy and frequency. As a side note, other massless particles you might have come across are gluons.
Other texts may wish to explain in terms relativistic mass. Einstein in particular didn't like this. It suggests that mass is key to
momentum, when in fact it describes changes in a particle's energy. The term to my mind is somewhat of a misnomer.
I have done my best to avoid a mathematical treatise of the question posed. Inevitably at some point, an equation had to 'pop up'.
I await with much interest jim's explanation. He may well choose a two or even three pronged attack. Let's wait and see.
⁴
;-)
Just reading through the 2nd time as it didn't click the first (I ought to read this on the PC monitor not a small mobile screen).
Having to take your word that something with no mass nor energy is nothing. It's not obvious that has to be so. And E = pc is all very well, but not yet clear where the p comes from. But maybe I'll get it this reading.
;-)
Just reading through the 2nd time as it didn't click the first (I ought to read this on the PC monitor not a small mobile screen).
Having to take your word that something with no mass nor energy is nothing. It's not obvious that has to be so. And E = pc is all very well, but not yet clear where the p comes from. But maybe I'll get it this reading.
Since Planck's constant is measured by (energy divided by frequency)
that is; Joules / hz or Joules seconds whichever you prefer.
Then rearranging p = hf/c it becomes;
cp = hf
The term on the right hand side is now in Joules (energy) and can be seen to be proportional to the particle's momentum .(p)
that is; Joules / hz or Joules seconds whichever you prefer.
Then rearranging p = hf/c it becomes;
cp = hf
The term on the right hand side is now in Joules (energy) and can be seen to be proportional to the particle's momentum .(p)
OK, O_G Let's consider another prong of attack.
Since cp = hf where h is a constant
Therefore the total energy of the
proton becomes dependent on f, which in turn is proportional to the product of cp
Consider an Alternator generating a pure sinusoidal output of 50 Hertz (50 cycles per second).
The frequency output from the Alternator relies on the revoultions its armature is turned. Think of the driving force of the Alternator shaft being the kinetic part of the equation i.e momentum p (the bit that is never allowed to rest in respect of a photon).
Hope this helps?
Since cp = hf where h is a constant
Therefore the total energy of the
proton becomes dependent on f, which in turn is proportional to the product of cp
Consider an Alternator generating a pure sinusoidal output of 50 Hertz (50 cycles per second).
The frequency output from the Alternator relies on the revoultions its armature is turned. Think of the driving force of the Alternator shaft being the kinetic part of the equation i.e momentum p (the bit that is never allowed to rest in respect of a photon).
Hope this helps?
// Apologies, since I do not know how to type c to the power of 4) //
Simply work in units where c = 1 and all your problems will be solved :)
One point that often gets overlooked -- and curse the days that E = mc² became the standard, because it tends to bake this view in -- is that the more physical formulation isn't E² = p² + m², but rather E² - p² = m². This innocent rearrangement is in fact crucial, for reasons that I hope this post will get to. Anyway, allow me to try a different, and probably still circular, tack. The short version is: it's just much more useful to redefine momentum away from mass x velocity than it is to not do that.
The first point is that physics is generally the study of things that we can actually measure. Moreover, because the world is complicated, the drive is always to find things that we can easily predict as well as measure. And the easiest thing to predict, and then compare to measurements, is anything that stays constant through the whole mess of complexity. Early in history, two such quantities were found, namely energy and momentum. It's well-known that "energy cannot be created or destroyed, only transferred between different forms". Likewise, the total momentum, here meaning mass times velocity, was seen to stay constant. So let's take it as a given that:
1. momentum= mass times velocity is a useful quantity, because
2. it is conserved in physical processes eg in collisions.
Now we need to confront ourselves with an interesting observation. Photons do indeed have no mass (here we don't care that this is because they're travelling at the speed of light, we've just measured it). Therefore, we can suppose, they have no momentum. But what happens, for example, when an electron absorbs/emits a photon? This happens all the time (eg Compton scattering, which we observe all the time in radiotherapy as a medical application). The answer is that the electron's momentum changes. But we argued that momentum is conserved. How can this be?
There are two choices here:
1. We can abandon the idea that momentum is conserved -- or at least we can just assume that it isn't conserved in this case.
2. We can relax the definition of momentum so that photons can have it after all, and the conservation law still applies with this new view.
Which is more useful? Answer: 2! If, as Zebu's formula says, you associate the momentum with the frequency, then you can perfectly account for Compton scattering by applying conservation of momentum to predict the change in the frequency of the light.
In that sense, the definition momentum = mass times velocity was only introduced first because we weren't able to appreciate truly what momentum was when we first came across it. It works for objects with mass, travelling at low speeds, but doesn't extend far enough to be useful any more at high speeds or for objects without mass.
* * * *
As to E² - p² = m² : again, you can apply the same idea. Physicists hunt for quantities that are fixed. In this case, the question is: "what is one value that all observers, no matter how fast they are moving, will agree on for a moving particle?" It's precisely the combination
E² - p² that everybody measures the same way, and it's precisely that conversation that is identified with the mass of a particle. For photons, the mass equals zero, so that this reduces to E = p. For anything else, at least one observer will see the particle at rest, so that for that observer they can say that E = m (or mc², if they want to be boring).
I have no intention of proving that this quantity indeed stays constant, but it's useful to bear in mind that the natural way the equation should appear, E² - p² = m², tells us far more about where this equation comes from than any other arrangement.
More later. But I hoped this would act as a useful extra to Zebu's approach.
Simply work in units where c = 1 and all your problems will be solved :)
One point that often gets overlooked -- and curse the days that E = mc² became the standard, because it tends to bake this view in -- is that the more physical formulation isn't E² = p² + m², but rather E² - p² = m². This innocent rearrangement is in fact crucial, for reasons that I hope this post will get to. Anyway, allow me to try a different, and probably still circular, tack. The short version is: it's just much more useful to redefine momentum away from mass x velocity than it is to not do that.
The first point is that physics is generally the study of things that we can actually measure. Moreover, because the world is complicated, the drive is always to find things that we can easily predict as well as measure. And the easiest thing to predict, and then compare to measurements, is anything that stays constant through the whole mess of complexity. Early in history, two such quantities were found, namely energy and momentum. It's well-known that "energy cannot be created or destroyed, only transferred between different forms". Likewise, the total momentum, here meaning mass times velocity, was seen to stay constant. So let's take it as a given that:
1. momentum= mass times velocity is a useful quantity, because
2. it is conserved in physical processes eg in collisions.
Now we need to confront ourselves with an interesting observation. Photons do indeed have no mass (here we don't care that this is because they're travelling at the speed of light, we've just measured it). Therefore, we can suppose, they have no momentum. But what happens, for example, when an electron absorbs/emits a photon? This happens all the time (eg Compton scattering, which we observe all the time in radiotherapy as a medical application). The answer is that the electron's momentum changes. But we argued that momentum is conserved. How can this be?
There are two choices here:
1. We can abandon the idea that momentum is conserved -- or at least we can just assume that it isn't conserved in this case.
2. We can relax the definition of momentum so that photons can have it after all, and the conservation law still applies with this new view.
Which is more useful? Answer: 2! If, as Zebu's formula says, you associate the momentum with the frequency, then you can perfectly account for Compton scattering by applying conservation of momentum to predict the change in the frequency of the light.
In that sense, the definition momentum = mass times velocity was only introduced first because we weren't able to appreciate truly what momentum was when we first came across it. It works for objects with mass, travelling at low speeds, but doesn't extend far enough to be useful any more at high speeds or for objects without mass.
* * * *
As to E² - p² = m² : again, you can apply the same idea. Physicists hunt for quantities that are fixed. In this case, the question is: "what is one value that all observers, no matter how fast they are moving, will agree on for a moving particle?" It's precisely the combination
E² - p² that everybody measures the same way, and it's precisely that conversation that is identified with the mass of a particle. For photons, the mass equals zero, so that this reduces to E = p. For anything else, at least one observer will see the particle at rest, so that for that observer they can say that E = m (or mc², if they want to be boring).
I have no intention of proving that this quantity indeed stays constant, but it's useful to bear in mind that the natural way the equation should appear, E² - p² = m², tells us far more about where this equation comes from than any other arrangement.
More later. But I hoped this would act as a useful extra to Zebu's approach.
// subtle differences between Newtonian physics and Quantum physics. //
Anyone embarking on a course of QM will find the differences a bit more than subtle!
Take these two giants of the Quantum world. Both were theoretical physicists and spent their whole lives trying to fathom the mysteries of QM. This is what they had to say;
If quantum mechanics hasn't profoundly shocked you, you haven't understood it yet. -- Niels Bohr.
“I think I can safely say that nobody understands quantum mechanics." -- Richard Feynman
jim
As is typical of you, another lucid account in providing this board with an interesting alternative. Thank you!
Anyone embarking on a course of QM will find the differences a bit more than subtle!
Take these two giants of the Quantum world. Both were theoretical physicists and spent their whole lives trying to fathom the mysteries of QM. This is what they had to say;
If quantum mechanics hasn't profoundly shocked you, you haven't understood it yet. -- Niels Bohr.
“I think I can safely say that nobody understands quantum mechanics." -- Richard Feynman
jim
As is typical of you, another lucid account in providing this board with an interesting alternative. Thank you!
// I will say, from first reading, it suddenly becomes clear I, and I suspect many more, no longer have a good idea what momentum is. //
No, I think your understanding of it is likely still fine, if you think of momentum as the "oomph" that a thing has, and when it hits you makes you hurt more if there's a lot of it. It's just when it comes to quantifying it that you have to update your "momentum = mass times speed" definition, because even things without mass can still have an oomph that works in much the same way.
Actually, a separate point, that neither Zebu nor I touched on yet, is that even if a thing has mass, it turns out that you have to update the formula from mass times speed to include a "super-mega-high-speed enhancement premium boost" that only kicks on noticeably when you're travelling at something approaching the speed of light. The more technical name for it is the Lorentz factor, but, yeah, it's a super-mega-high-speed enhancement premium boost.
Most people's conception of physics is reduced to what happens on Earth, at slow speeds, when everything is visible. It shouldn't be a surprise to learn that this misses out on a *lot* of details relevant at high speeds, or at low masses, or at ultra-high masses for that matter, when General Relativity screws up everything all over again.
No, I think your understanding of it is likely still fine, if you think of momentum as the "oomph" that a thing has, and when it hits you makes you hurt more if there's a lot of it. It's just when it comes to quantifying it that you have to update your "momentum = mass times speed" definition, because even things without mass can still have an oomph that works in much the same way.
Actually, a separate point, that neither Zebu nor I touched on yet, is that even if a thing has mass, it turns out that you have to update the formula from mass times speed to include a "super-mega-high-speed enhancement premium boost" that only kicks on noticeably when you're travelling at something approaching the speed of light. The more technical name for it is the Lorentz factor, but, yeah, it's a super-mega-high-speed enhancement premium boost.
Most people's conception of physics is reduced to what happens on Earth, at slow speeds, when everything is visible. It shouldn't be a surprise to learn that this misses out on a *lot* of details relevant at high speeds, or at low masses, or at ultra-high masses for that matter, when General Relativity screws up everything all over again.
Yes, I think of momentum as mass times "velocity", but I "correct" the velocity by dividing it by the square root of (1-v squared), using Jim's notation.
I prefer Jim's notation, but perhaps the old notation shows better what is happening. Divide the velocity by the square root of (1 - (v/c) squared). This gives a sensible, and correct, result for all speeds less than the speed of light. At the speed of light the momentum of a massless photon becomes zero times infinity, which it is almost possible to believe can be a finite number!
I prefer Jim's notation, but perhaps the old notation shows better what is happening. Divide the velocity by the square root of (1 - (v/c) squared). This gives a sensible, and correct, result for all speeds less than the speed of light. At the speed of light the momentum of a massless photon becomes zero times infinity, which it is almost possible to believe can be a finite number!
