Does Nuclear Reaction Energy Contribute to the Existence of Infinity?

[Bible Thumper]

Is it true that 0.1% of the mass in an uncontrolled nuclear chain reaction gets converted into energy via E=Mc^2? And if this is so, then does the mass technically go to infinity in accordance with the Lorentz transformation before turning into energy?

 

[JesseM]

Is it true that 0.1% of the mass in an uncontrolled nuclear chain reaction gets converted into energy via E=Mc^2? And if this is so, then does the mass technically go to infinity in accordance with the Lorentz transformation before turning into energy?

This isn't a case of a sublight particle being accelerated to lightspeed, but of one type of particle being annihilated and another being created, akin to what happens when antimatter and matter annihilates and produces high-energy photons. Particle creation/annihilation is something that is allowed by the rules of quantum field theory.

 

[Bible Thumper]

This isn't a case of a sublight particle being accelerated to lightspeed, but of one type of particle being annihilated and another being created, akin to what happens when antimatter and matter annihilates and produces high-energy photons. Particle creation/annihilation is something that is allowed by the rules of quantum field theory.

Particle creation/annihilation takes place at the event horizon, where the laws of physics may be broken. Radiation is the inevitable result of such activities.
Can it be argued that infinity is still playing a role somehow, since there are the same conditions set up that also exist at that place where physical laws may be broken (event horizon)?
Can we apply the part of the Lorentz transformation that says mass tends to infinity when it turns into energy and get away with it, without introducing the particle/antiparticle pair?

 

[JesseM]

Particle creation/annihilation takes place at the event horizon

No, it happens everywhere in quantum field theory. Particle creation/annihilation at the event horizon is just used as a way of explaining Hawking radiation.

where the laws of physics may be broken.

Physicists don't think GR breaks down at the event horizon, but only in the vicinity of the singularity. And it's not that all laws of physics fail there, it's that a theory of quantum gravity (which GR would just be an approximation to, the way Newtonian gravity is an approximation to GR) would be needed to deal with what's happening there.

 

[Bible Thumper]

This isn't a case of a sublight particle being accelerated to lightspeed, but of one type of particle being annihilated and another being created, akin to what happens when antimatter and matter annihilates and produces high-energy photons. Particle creation/annihilation is something that is allowed by the rules of quantum field theory.

Now I went here:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html
And scrolled down to the third frame, under Relativistic Mass. I inputted the value of "1" for the speed of light, and M returned infinity. To this day I am having a hard time trying to get around this fact of Einstein's equation!
HELP!

 

[Hurkyl]

I inputted the value of "1" for the speed of light,

Actually, you inputted the value "1 c" for velocity.

and M returned infinity.

Actually, it returned "Infinity * m₀". (In particular, it's an indeterminate form when m₀=0)

To this day I am having a hard time trying to get around this fact of Einstein's equation!

Why? What's the problem?

 

[atyy]

Now I went here:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html
And scrolled down to the third frame, under Relativistic Mass. I inputted the value of "1" for the speed of light, and M returned infinity. To this day I am having a hard time trying to get around this fact of Einstein's equation!
HELP!

I tried it and it said m=(infinity)m_o.

Would it make you happy if m_o=0?

 

[JesseM]

Now I went here:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html
And scrolled down to the third frame, under Relativistic Mass. I inputted the value of "1" for the speed of light, and M returned infinity. To this day I am having a hard time trying to get around this fact of Einstein's equation!
HELP!

But no particle with nonzero rest mass can move at the speed of light. A photon (or any other particle moving at the speed of light) has zero rest mass, so if you try to use the relativistic mass formula you get 0*infinity which is undefined. But the formulas of relativity like the relativistic mass formula or the time dilation formula are only supposed to give the values for things in different inertial frames (which always move at less than c relative to one another), they're not meant to give useful answers when v=c. I don't know if physicists even talk about the "relativistic mass" of photons (most physicists prefer to avoid the concept of relativistic mass altogether and just talk about relativistic momentum and energy), but here's one way you could do it: if the relativistic mass of an object is M we have the formula for its energy E=Mc^2 (which is equivalent to the more common formula E^2 = m^2*c^4 + p^2*c^2, where m is the rest mass and p is the relativistic momentum p = m*v/sqrt[1 - v^2/c^2]), and from quantum mechanics we also have the formula for the energy of photons E=hf where h is Planck's constant and f is the frequency, so you could set these equal and get Mc^2 = hf, or M = hf/c^2.

 

[Bible Thumper]

But no particle with nonzero rest mass can move at the speed of light. A photon (or any other particle moving at the speed of light) has zero rest mass, so if you try to use the relativistic mass formula you get 0*infinity which is undefined. But the formulas of relativity like the relativistic mass formula or the time dilation formula are only supposed to give the values for things in different inertial frames (which always move at less than c relative to one another), they're not meant to give useful answers when v=c. I don't know if physicists even talk about the "relativistic mass" of photons (most physicists prefer to avoid the concept of relativistic mass altogether and just talk about relativistic momentum and energy), but here's one way you could do it: if the relativistic mass of an object is M we have the formula for its energy E=Mc^2 (which is equivalent to the more common formula E^2 = m^2*c^4 + p^2*c^2, where m is the rest mass and p is the relativistic momentum p = m*v/sqrt[1 - v^2/c^2]), and from quantum mechanics we also have the formula for the energy of photons E=hf where h is Planck's constant and f is the frequency, so you could set these equal and get Mc^2 = hf, or M = hf/c^2.

Thanks for the elaborate answer.
I think I'm just having problems with this quote of yours:

they're not meant to give useful answers when v=c.

Now why not? Where is the limit set in the equations Einstein wrote that says v musn't equal c? There's not even an implied limit, as far as I can see, anyway.
About what you stated earlier, which was this:

This isn't a case of a sublight particle being accelerated to lightspeed, but of one type of particle being annihilated and another being created, akin to what happens when antimatter and matter annihilates and produces high-energy photons. Particle creation/annihilation is something that is allowed by the rules of quantum field theory.

I took this to mean that at near v=c, things get uncertain and statistical. Is this correct? Is it correct to say that as the mass becomes infinite as it equals c is impossible to observe, just as it's impossible to observe virtual particle pairs?

 

[Dale]

Where is the limit set in the equations Einstein wrote that says v musn't equal c?

In the part where you get division by zero when v equals c.

 

[JesseM]

Now why not? Where is the limit set in the equations Einstein wrote that says v musn't equal c? There's not even an implied limit, as far as I can see, anyway.

Because the equations are meant to translate between different inertial reference frames, and there are no inertial frames with a relative velocity of c. Also, as DaleSpam said, if you try to plug in v=c you get division by zero.

I took this to mean that at near v=c, things get uncertain and statistical. Is this correct?

No, where would you get that idea? In any case "near v=c" isn't really meaningful in an absolute sense, anything that's moving at close to c in one inertial frame is at rest in some other inertial frame, and the laws of physics are identical in every inertial frame.

Is it correct to say that as the mass becomes infinite as it equals c is impossible to observe, just as it's impossible to observe virtual particle pairs?

No particle with nonzero rest mass ever reaches c in any inertial frame, and a particle traveling close to c shouldn't be any harder to observe than any other particle.

 

[Bible Thumper]

No, where would you get that idea?

I got the idea from you. This is what you said:

One type of particle is being annihilated and another being created. Particle creation/annihilation is something that is allowed by the rules of quantum field theory.

Then to that statement of yours, which introduced uncertainty and statistics (AKA quantum field theory), I replied:

I took this to mean that at near v=c, things get uncertain and statistical. Is this correct?

Do things get uncertain and statistical?

 

[Bible Thumper]

In the part where you get division by zero when v equals c.

Then you are denying that energy is being created from mass in nuclear chain reactions?

 

[Dale]

No. How did you come to that mistaken conclusion?

 

[Bible Thumper]

No. How did you come to that mistaken conclusion?

Because. The hyperphysics link says that when you input c for v, you get infinity for M as a result. This means that, according to the relativistic mass equation, the only way you're going to get E is by accelerating that v to c. And, as the link demonstrated, when you accelerate V to c, you get infinity for M.
Therefore,

In the part where you get division by zero when v equals c.

This quote of yours proves you deny that energy is being created from mass in accordance with E=MC2, because you said you had a problem with getting division by zero when v equals c.

1. E=MC2
2. That means, M must equal C to get the E
3. In order for M to get to C, it must become infinite (infinite mass), according to the Relativistic Mass equation on www.hyperphysics.com that I posted earlier.

 

[Dale]

No, it just proves that you don't know what you are talking about.

The equation e=mc² means that mass is a form of energy (e.g. you can measure the mass of a particle in kilograms or electron volts with a proportionality constant of c² for converting between the two). It does not in any way imply that matter needs to be accelerated to c in order to "get the E". It already has/is the E.

 

[Bible Thumper]

No, it just proves that you don't know what you are talking about.

The equation e=mc² means that mass is a form of energy (e.g. you can measure the mass of a particle in kilograms or electron volts with a proportionality constant of c² for converting between the two). It does not in any way imply that matter needs to be accelerated to c in order to "get the E". It already has/is the E.

My G-d, I'm dealing with another Fredrik (read here for my encounter with Fredrik: Relativistic Rod and Hole ( 1 2 3 4 5 ... Last Page)!
Look. We know this:

1. Before the chain reaction, there exists nothing but a certain amount of U235 mass
2. During the chain reaction, something happens, and that something has to be related to E=MC2.
3. After the chain reaction, there exists less mass than what we started with.
4. It's a well-known fact of life that that mass went bye-bye in the form of energy.

I'm sorry, DaleSpam, but I cannot get it any more simple than this. I believe I made it illiterate-proof, in fact (but that depends on subsequent posting).
What is known is that an E=MC2 process took place. Do you agree that an E=MC2 process took place (Y/N)?
Knowing that an E=MC2 process took place, we have to analyze the mechanism by which that M turned into E (M stands for mass and E stands for energy).
Since the only equation we may go by is the relavistic mass equation I posted earlier, we have to presume the energy got to be energy by a direct mass-to-energy conversion somehow.

But how?

The equation tells us the energy got that way by the mass (labeled as "v" in the equation--"v" stands for the velocity of the mass). Since the only way the mass can get to E is through a v=c phenomena, we easily and dullardly conclude that there was an acceleration of that mass to the speed of light that took place in order for the E (energy) to be generated.
The only way that can happen (for the v of the mass to turn into the e of the energy) is if the mass, labeled as v, went to infinity.

 

[JesseM]

Knowing that an E=MC2 process took place, we have to analyze the mechanism by which that M turned into E (M stands for mass and E stands for energy).
Since the only equation we may go by is the relavistic mass equation I posted earlier, we have to presume the energy got to be energy by a direct mass-to-energy conversion somehow.

But how?

Because it's permitted by the rules of quantum field theory for some set of particles to annihilate and create new particles in their place, with the rule that the total energy beforehand must be equal to the total energy beforehand (there are additional rules governing these creation/annihilation events too of course, but that's the only one that's relevant here). This is not the same as a particle being accelerated to light speed, so the relativistic mass equation isn't relevant.

The equation tells us the energy got that way by the mass (labeled as "v" in the equation--"v" stands for the velocity of the mass). Since the only way the mass can get to E is through a v=c phenomena,

It's just not true that the "only way the mass can get to E is through v=c", that's a wrong idea of yours that doesn't come from anything in relativity. Quantum field theory allows discontinuous creation/annihilation events, and although this is a quantum phenomenon and not specifically predicted by SR alone, it also doesn't violate SR in any way.

By the way, it's not even clear what you mean by "the only way mass can get to E"--"energy" doesn't refer to a type of stuff distinct from ordinary matter, it's a property of all particles much like momentum, particles moving slower than the speed of light have energy just like photons have energy, if you think it makes sense to say photons "are" energy while sublight particles aren't, you're confused about what the term means. As I said, for an object moving slower than light its energy is given by the equation E^2 = m^2*c^4 + p^2*c^2, where m is the rest mass and p is the relativistic momentum p = mv/sqrt(1 - v^2/c^2). For an object moving at light speed, the rest mass m is always zero so p is undefined in this equation, so you must instead use the quantum equation E = hf, where f is the frequency and h is Planck's constant.

 

[Bible Thumper]

If you think it makes sense to say photons "are" energy while sublight particles aren't, you're confused about what the terms energy and mass means.

Does everyone else agree? Is energy and mass indistinguishable in this regard?

 

[JesseM]

Does everyone else agree? Is energy and mass indistinguishable in this regard?

Indistinguishable in what regard? I just said that energy was a property (as is mass), not a type of substance.

 

[Dale]

the only way the mass can get to E is through a v=c phenomena

Here is your error. The mass does not need to go "through a v=c phenomena" to "get to E". It already has that energy even when it is at rest (v=0).

Let's consider the http://hyperphysics.phy-astr.gsu.edu/Hbase/nucene/u235cs.html#c1". The fission products have a mass of 191 MeV less than the fission reactants. Let's assume that all of that energy goes into the kinetic energy of one of the produced neutrons. The resulting neutron would have a total energy of 1.13 GeV, in other words a velocity of 0.56 c. The entire amount of mass converted to energy is barely sufficient to take the lightest product to half of c, let alone all the way to c. The energy did not come from accelerating any mass to c, it was there in the rest mass of the reactants from the beginning.

 

[Bible Thumper]

Here is your error. The mass does not need to go "through a v=c phenomena" to "get to E". It already has that energy even when it is at rest (v=0).

Let's consider the http://hyperphysics.phy-astr.gsu.edu/Hbase/nucene/u235cs.html#c1". The fission products have a mass of 191 MeV less than the fission reactants. Let's assume that all of that energy goes into the kinetic energy of one of the produced neutrons. The resulting neutron would have a total energy of 1.13 GeV, in other words a velocity of 0.56 c. The entire amount of mass converted to energy is barely sufficient to take the lightest product to half of c, let alone all the way to c. The energy did not come from accelerating any mass to c, it was there in the rest mass of the reactants from the beginning.

Don't you think this is like reducing fission to another energy-generating event like combustion? Is this correct? Should we be regarding the process of energy-generation in fission as we do in combustion?

 

[atyy]

Don't you think this is like reducing fission to another energy-generating event like combustion? Is this correct? Should we be regarding the process of energy-generation in fission as we do in combustion?

You hit the nail on the head. Yes.

Edit: Find the energy of combustion reaction, calculate the change in mass. Then find the energy of a nuclear reaction, and calculate the change in mass. This may tell you why it is an extremely good approximation to pretend that mass does not change in combustion.

 

[Bible Thumper]

You hit the nail on the head. Yes.

Edit: Find the energy of combustion reaction, calculate the change in mass. Then find the energy of a nuclear reaction, and calculate the change in mass. This may tell you why it is an extremely good approximation to pretend that mass does not change in combustion.

DaleSpam? JesseM? Do you guys form a consensus on this?
Is fission just a form of energy-generation like combustion that just happens to work on the atomic and sub-atomic level?

And if this is so, how can a combustion process prove E=MC2 like fission can?

 

[matheinste]

Hello Bible Thumper

In the energy to mass equation think of c as just a number, a conversion factor. Nothing has to move or accelerate to make the equation true. You start with a mass and multiply it by c and the result is a certain amount of energy.

Matheinste.

 

[JesseM]

DaleSpam? JesseM? Do you guys form a consensus on this?
Is fission just a form of energy-generation like combustion that just happens to work on the atomic and sub-atomic level?

It depends what you mean by "like combustion". Nuclear reactions always involve the creation of high-energy photons, whereas I'm not sure if all the chemical reactions called "combustion" involve photon emissions, though I think most would.

And if this is so, how can a combustion process prove E=MC2 like fission can?

The total energy of a multiparticle system bound together by forces like an atomic nucleus or a molecule depends not just on the rest energy and kinetic energy of the particles, but also on their potential energy (the difference between potential energy in the bound state and potential energy when all the particles are far apart is known as 'binding energy'). In a reaction that splits apart or fuses such multiparticle systems, there is a change in potential energy which has to be compensated for, either by adding kinetic energy to the systems, or by the creation of new particles whose energy makes up the difference. If you look at the total energy of the all the reactant molecules and the total energy of the product molecules, any difference in energy should be exactly compensated for either by a change in kinetic energy (the system heats up or cools down) or by the energy of any photons that were emitted or absorbed. The difference in energy will be much, much smaller than in a nuclear reaction, though.

 

[Dale]

DaleSpam? JesseM? Do you guys form a consensus on this?
Is fission just a form of energy-generation like combustion that just happens to work on the atomic and sub-atomic level?

And if this is so, how can a combustion process prove E=MC2 like fission can?

Yes, atyy is correct. This concept is known as http://en.wikipedia.org/wiki/Binding_energy" . The resulting mass defect applies for any bound masses whether the binding is gravitational, chemical, or nuclear, or any other form of binding energy you could concieve of.

 

[Bible Thumper]

Yes, atyy is correct. This concept is known as http://en.wikipedia.org/wiki/Binding_energy" . The resulting mass defect applies for any bound masses whether the binding is gravitational, chemical, or nuclear, or any other form of binding energy you could concieve of.

Thanks, DaleSpam. And sorry about equating you with Fredrik earlier, but I digress...

 

[Dale]

That's ok, I have nothing against Fredrik, even when we disagree.

 

[Naty1]

I have always wondered why such equations don't have limits set such as "valid only when v is less than c.

Is fission just a form of energy-generation like combustion that just happens to work on the atomic and sub-atomic level?

No fission or fusion is fundamentally different from simple "combustion"...It would be better to compare fission/fusion with chemical reactions: the former involved nuclear energies and nuclear mass changes, the latter does not...

from http://en.wikipedia.org/wiki/Chemical_reaction

Classically, chemical reactions encompass changes that strictly involve the motion of electrons in the forming and breaking of chemical bonds,...

 

[Dale]

No fission or fusion is fundamentally different from simple "combustion"...It would be better to compare fission/fusion with chemical reactions: the former involved nuclear energies and nuclear mass changes, the latter does not

The mass of combustion products is slightly less than the mass of combustion reactants.

 

[Bible Thumper]

The mass of combustion products is slightly less than the mass of combustion reactants.

But the best, ideal experiment to verify things is one carried out in space, free from atmospheric gasses and other catalysts.

 

[Tac-Tics]

Why are people so fascinated by infinity? Does infinity exist in the universe? Of course it does. It's the number of steps required to divide a continuous line into all its points. Its the (non-inclusive) upper limit to the amount of force, momentum, and mass of an object. It's the number of posts required to convince a troll on the Internet he's wrong. You don't need physics to reify infinity and if you're not using good physics to begin with, you're never get closer to understanding the truth behind reality. (And "never" is just another way of saying that it will take infinitely long.)

 

[atyy]

Why are people so fascinated by infinity? Does infinity exist in the universe? Of course it does. It's the number of steps required to divide a continuous line into all its points.

And continuous lines exist?

In fact, the answer is that correct application of the formula being discussed does not yield infinity.

 

[Tac-Tics]

And continuous lines exist?

They sure do. I may have to denote them using mathematic notation, and surely I can't point at them with my finger and show they are, in fact continuous, but they are as real as mass or a gravitational field. (Real doesn't necessarily mean you can point to it!)

 

[Dale]

Does infinity exist in the universe? Of course it does. ... It's the number of posts required to convince a troll on the Internet he's wrong.

 

[DrGreg]

To be pedantic, the official mainstream*[/color] mathematical view is that infinity doesn't exist as a number (although, of course, it does exist as an abstract concept). Any statement that includes the word (or symbol for) "infinity" is, strictly speaking, a shorthand convention for a rather more precise statement.

For example, when we say "there are an infinite number of points in a line", that is a shorthand for the more precise statement "there is no upper bound for the number of distinct points we can select from a line".

Given \gamma = 1 / \sqrt{1 - v^2/c^2}, we never ought never to say \gamma = \infty when v = c, but we can say \gamma \rightarrow \infty as v \rightarrow c. This is itself a mathematical shorthand for the more precise statement that, for any (large) number \gamma_0 &gt; 0 it is possible to find a velocity v₀ such that, whenever v₀ < v < c, then \gamma &gt; \gamma_0.

____
*[/color]I say "mainstream" because there may be some esoteric branches of "modern" maths that take a different view.

 

[Tac-Tics]

*[/color]I say "mainstream" because there may be some esoteric branches of "modern" maths that take a different view.

You could simply leave it as this:

Infinity is not a real number. The real numbers are a logical extension of the rational numbers, of course, but they do not include any value greater than any other or less than any other. The real number system are the ones that best correspond to our intuition. They have an extremely mature theory behind them. Science has implicitly made the assumption that all measurements are approximations of real numbers since the advent of physics. They are what people mean by "number" in an informal sense an overwhelming part of the time. Your test scores are real numbers. Your IQ is plotted against a real bell curve. Currency is handled electronically as floating point numbers (computer approximations of reals).

But reals do not have infinitely large quantities! Infinity is something that must be added in later, but it is not so straightforward until you understand the idea of a limit!

 

[JesseM]

*[/color]I say "mainstream" because there may be some esoteric branches of "modern" maths that take a different view.

Infinity can be treated as a number (as opposed to just a limit) in complex analysis, which is not really so esoteric as mathematics goes (it's basically just calculus applied to functions on the complex numbers)

 

[Max™]

Does everyone else agree? Is energy and mass indistinguishable in this regard?

More or less.

Relativity actually suggests that everything, even space-time, may only be distinguishable as energy gradients.

The proven aspect though, is that yes, Mass is just a lot of Energy tangled up in one location. Thus REST mass.

I have always wondered why such equations don't have limits set such as "valid only when v is less than c.

They do, they're just hidden inside of Quantum Mechanics.

 

[Naty1]

Originally Posted by Naty1

No fission or fusion is fundamentally different from simple "combustion"...It would be better to compare fission/fusion with chemical reactions: the former involved nuclear energies and nuclear mass changes, the latter does not

The mass of combustion products is slightly less than the mass of combustion reactants.

I agree, but I thought the basic difference is that chemical reaction energies come from changes in the electron orbit energies rather than changes in the nucleus, which is true for fission/fusion...is that accurate?

But do typical combustion changes involve nuclear changes which chemical reactions usually do not?? never thought about that...wood to carbon via a campfire..yea, I guess carbon could have a different nuclear makeup than "wood" ...?

 

[Dale]

I agree, but I thought the basic difference is that chemical reaction energies come from changes in the electron orbit energies rather than changes in the nucleus, which is true for fission/fusion...is that accurate?

Correct, the binding force involved in chemical reactions is EM, the binding forces involved in nuclear reactions are the strong and weak nuclear forces. Any bound system shows a mass deficit which is proportional to the binding energy. The principle is the same, but the magnitude is obviously much different.