Why is the Speed of Light Squared in E=mc^2?

[Entropy]

A nuclear explosion is not caused by the transformation of matter and energy.

Yes it is.

I already knew this connection, but the problem is that I don't know any pratical experience in which the "full" formula has been proved.

Millions of experiments have proven this. Just about every experiment ever to involve particle accelerators involves Einstien's equations.

 

[Swampeast Mike]

Could you explain what did you mean with "deception".

Seams to me that you have a clear explanation to my question; do you give the pleasure to know it?

In this context "deception" would mean pretending to be a different coalescence of mass in the universe.

Your question is universal as is my attempted explanation of a universe where time and space are relative. It is not however, and I repeat NOT, based upon faith.

 

[russ_watters]

The connection between atomic bombs and e=mc^2 is subtle. A nuclear explosion is not caused by the transformation of matter and energy. I already knew this connection, but the problem is that I don't know any pratical experience in which the "full" formula has been proved.

Clearly, you do not understand. See Doc Al's post #15(among others): Nuclear reactions are based on e=mc^2.

c^2 couldn't be interpreted as if it was a simple mathematic factor that result from a development of other equations. Einstein didn't choose the speed of light for "c" by chance.

He didn't choose it at all - he derived it.

There are chemical reactions where there are tiny mass differences as well. An example: When hydrogen and oxygen explosively combine to make water, the sum of the rest masses of the initial hydrogen and oxygen atoms is just a little bit less than the sum of the rest masses of the resulting water molecules.

Really? How much, exactly?

 

[Doc Al]

There are chemical reactions where there are tiny mass differences as well. An example: When hydrogen and oxygen explosively combine to make water, the sum of the rest masses of the initial hydrogen and oxygen atoms is just a little bit less than the sum of the rest masses of the resulting water molecules.

Yes! Even in chemical reactions E = m c^2 applies. Of course the amount of rest mass "converted" to energy is much less in chemical reactions than in nuclear reactions.

I know that E = m c^2does not require anything to reach the speed of light. Was an example. Einstein didn't say the particular situations where the formula will make sense. In your example, the complete formula application has not been proved.

What are you talking about? E = m c^2 is a result derived from Special Relativity, which has been exhaustively tested. Nuclear reactions, controlled or otherwise, have directly confirmed E = m c^2.

89875517873681764 m^2/s^2 is not only a number (a factor). This has a physics pratical sense. What it is?

Again, what are you talking about? We've given several examples. You've even supplied one. (What do you mean by a "practical" sense?)

 

[free_mind]

Clearly, you do not understand. See Doc Al's post #15(among others): Nuclear reactions are based on e=mc^2. He didn't choose it at all - he derived it. Really? How much, exactly?

Another example of this:

Chemical reactions involving spontaneous oxydation, in other words we are talking about burning.

 

[free_mind]

Yes it is.

NO, it isn't. Because to Einstein, realtivistic mass and energy are simply two different names for one and the same physical quantity. Read Einstein carefully and you will verify that this is as Iam saying.

Millions of experiments have proven this. Just about every experiment ever to involve particle accelerators involves Einstien's equations.

Humm. The formula has been partially proven. But I am talking from the beginning about a total proof.

 

[free_mind]

In this context "deception" would mean pretending to be a different coalescence of mass in the universe.

Your question is universal as is my attempted explanation of a universe where time and space are relative. It is not however, and I repeat NOT, based upon faith.

If you call "Faith" to the will of discover the truth freely without blarney anybody, YES is faith what determines my study, faith of getting the truth. The faith that you are talking about, you only could deduce it from my words with a great faith.

 

[russ_watters]

Humm. The formula has been partially proven. But I am talking from the beginning about a total proof.

There isn't total proof in science. But the equivalence of matter and energy is as close to completely proven as anything in science ever gets.

 

[free_mind]

There isn't total proof in science. But the equivalence of matter and energy is as close to completely proven as anything in science ever gets.

Don't interpret my words literally, they are not a mathematic formula :) . In science nor in nothing you have 100% . But show me an example where the E=mc^2 formula has been completely (not parcially and not because has been important to get conclusions in other observations, etc) proved.

 

[Pengwuino]

Don't interpret my words literally, they are not a mathematic formula :) . In science nor in nothing you have 100% . But show me an example where the E=mc^2 formula has been completely (not parcially and not because has been important to get conclusions in other observations, etc) proved.

Didn't someone already point that out to you a few times already? Big flash of light... mushroom cloud...

 

[free_mind]

Yes! Even in chemical reactions E = m c^2 applies. Of course the amount of rest mass "converted" to energy is much less in chemical reactions than in nuclear reactions.


What are you talking about? E = m c^2 is a result derived from Special Relativity, which has been exhaustively tested. Nuclear reactions, controlled or otherwise, have directly confirmed E = m c^2.


Again, what are you talking about? We've given several examples. You've even supplied one. (What do you mean by a "practical" sense?)

Until now I saw examples of parcial pratical application of the formula E=mc^2 .

 

[free_mind]

Didn't someone already point that out to you a few times already? Big flash of light... mushroom cloud...

Study the Einstein theory and you will find that Einstein's politics played a more decisive role in the story of the atomic bomb than his physics.

 

[Pengwuino]

Study the Einstein theory and you will find that Einstein's politics played a more decisive role in the story of the atomic bomb than his physics.

... Someones political feelings does not change the laws of physics. E=mc^2 was fully demonstrated by the atomic bomb. How exactly can you dispute that? What is the "part" that is explained that isn't the whole of the theory?

 

[Doc Al]

Study the Einstein theory and you will find that Einstein's politics played a more decisive role in the story of the atomic bomb than his physics.

Why not study Einstein's theory to find out how E = m c^2 is derived and what it means?

 

[free_mind]

... Someones political feelings does not change the laws of physics. E=mc^2 was fully demonstrated by the atomic bomb. How exactly can you dispute that? What is the "part" that is explained that isn't the whole of the theory?

Ok. I will try to be synthetic taking in account the place where we are.

The strength of the nuclear bond depends on the number of neutrons and protons involved. It varies in such a way that binding energy is released both in splitting up a heavy nucleus into smaller parts and in fusing light nuclei into heavier ones. This, as well as the chain reaction phenomenon, explains the immense power of nuclear bombs.

Einstein's formula it's all about different kinds of energy. Sure, there are some radioactive decay processes following nuclear fission, and, if so inclined, one can view the decay of a neutron decaying into a slightly lighter proton as a transformation of rest energy into other energy forms. But these additional processes contribute a mere 10 per cent of the total energy set free in nuclear fission. The main contribution is due to binding energy being converted to other forms of energy - a consequence not of Einstein's formula, but of the fact that nuclear forces are comparatively strong, and that certain lighter nuclei are much more strongly bound than certain more massive nuclei.

 

[free_mind]

Why not study Einstein's theory to find out how E = m c^2 is derived and what it means?

The mathematic meaning could be far from the pratical world. Don't you know many situations in which this happens?

Mathematics couldn't be seen as an absolute concept to explain the world, notwithstanding the enormous help that gave and will give in our understanding of the world .

 

[pervect]

NO, it isn't. Because to Einstein, realtivistic mass and energy are simply two different names for one and the same physical quantity. Read Einstein carefully and you will verify that this is as Iam saying.

It is true that for single point particles, relativistic mass and energy are the same. It's also true that nuclear reactions transform matter into energy. I don't see why you are advancing the first point as a counter-argument to the second - both are true.

 

[free_mind]

It is true that for single point particles, relativistic mass and energy are the same. It's also true that nuclear reactions transform matter into energy. I don't see why you are advancing the first point as a counter-argument to the second - both are true.

Einstein would say that in a system where there is energy (E), it automatically has the relativistic mass m=E/c2; whenever a system has the mass m, you need to assign it an energy E=mc2. Once the mass is known, so is the energy, and vice versa. In that context, it makes no sense to talk about the "transformation of mass into energy" - where there's one, there's the other.

 

[selfAdjoint]

Einstein would say that in system where there is energy (E), it automatically has the relativistic mass m=E/c2; whenever a system has the mass m, you need to assign it an energy E=mc2. Once the mass is known, so is the energy, and vice versa. In that context, it makes no sense to talk about the "transformation of mass into energy" - where there's one, there's the other.

Some energy you can do work with. But a mass just sitting at rest (you do know the formula e=mc^2 is only true in the rest frame don't you?) can't do any work. In order to transform the frozen form of energy we call matter into the fluid kind that can do work we need some specific physical transformation to take place.

 

[russ_watters]

The mathematic meaning could be far from the pratical world. Don't you know many situations in which this happens?

Name one relevant to this thread.

 

[DrChinese]

Einstein's formula it's all about different kinds of energy. Sure, there are some radioactive decay processes following nuclear fission, and, if so inclined, one can view the decay of a neutron decaying into a slightly lighter proton as a transformation of rest energy into other energy forms. But these additional processes contribute a mere 10 per cent of the total energy set free in nuclear fission. The main contribution is due to binding energy being converted to other forms of energy - a consequence not of Einstein's formula, but of the fact that nuclear forces are comparatively strong, and that certain lighter nuclei are much more strongly bound than certain more massive nuclei.

There is no difference in the basic principle regardless of what part of the atom it comes from, and I believe you know that already. Einstein's formula still applies and is calculated exactly the same way.

So my question is, what are you really asking here in this forum? Are you trying to get the answer to a question (somehow I am beginning to doubt this) or are you trying to make a specific statement? If you are trying to make a statement, perhaps you could move forward to that.

 

[russ_watters]

Frankly, the critical units error in the opening post seems to be the whole point of the thread - but even after it was pointed out, free_mind doesn't seem to want to drop the line of reasoning that error started. free_mind appears to not want to let go of the idea that c^2 is a speed.

 

[free_mind]

Name one relevant to this thread.

Generically speaking the difference between pure mathematics and applied mathematics. I know that there are several cases where pure mathematics became applied mathematics. But I am not so optimistic as Nikolai Lobachevsky: "There is no branch of mathematics, however abstract, which may not someday be applied to the phenomena of the real world."

 

[free_mind]

Some energy you can do work with. But a mass just sitting at rest (you do know the formula e=mc^2 is only true in the rest frame don't you?) can't do any work. In order to transform the frozen form of energy we call matter into the fluid kind that can do work we need some specific physical transformation to take place.

The specific physical transformation results from the other forms of energy. The main energy set free in a nuclear fission results from binding energy converted to other forms of energy.

 

[free_mind]

There is no difference in the basic principle regardless of what part of the atom it comes from, and I believe you know that already. Einstein's formula still applies and is calculated exactly the same way.

So my question is, what are you really asking here in this forum? Are you trying to get the answer to a question (somehow I am beginning to doubt this) or are you trying to make a specific statement? If you are trying to make a statement, perhaps you could move forward to that.

The scope that underlies my participation in this forum, is mainly related with the difficulty in define the exact characterization of c^2 and it's qualitative meaning, on the other hand, the inexistence of an empirical demonstration where the Einstein formula has been completely proven.

HallsofIvy about c^2 said: "No one has said that "speed squared" has any particular physical significance. You can if you like think of it as J/kg (...)". Where in this affirmation is the scientific precision?

 

[Entropy]

The main energy set free in a nuclear fission results from binding energy converted to other forms of energy.

The binding energy adds to the atom's mass. Why do you think a He-4 nucleus doesn't have the same mass as two protons and two neutrons?

 

[free_mind]

The binding energy adds to the atom's mass. Why do you think a He-4 nucleus doesn't have the same mass as two protons and two neutrons?

Please, read the thread #45.

 

[Entropy]

Please, read the thread #45.

I did. You don't understand. Binding energy contributes to an atom's mass. Two protons and two neutrons have a smaller mass than a He-4 nucleus. If you don't believe me, look up the masses of [free] two protons and [free] two neutrons and compare it to the mass of a He-4 nucleus. That change in mass is because when these 4 nucleons join to form a single nucleus, known as nuclear fusion, mass from these nucleons is converted directed into energy. The change in mass when plugged into E = mc^2 will yield the exact energy released in the TOTAL reaction. That proves the equation is right. (Note: a more common form of neclear fusion involves H-2 and two protons, which happens in the sun, but it is more complex because it results in the production of positions and this other is reaction is simplier and works for these purposes).

 

[pervect]

The scope that underlies my participation in this forum, is mainly related with the difficulty in define the exact characterization of c^2 and it's qualitative meaning, on the other hand, the inexistence of an empirical demonstration where the Einstein formula has been completely proven.

HallsofIvy about c^2 said: "No one has said that "speed squared" has any particular physical significance. You can if you like think of it as J/kg (...)". Where in this affirmation is the scientific precision?

There isn't any particular difficulty in the characterization of c^2 and it's meaning. The only difficulty is in your understanding.

We've given you a very simple analogy - a foot is different than a foot^2.

For more advanced understanding, we've given you links to the theory of dimensional analysis, which addresses the topic of whether c is different than c^2 more precesisly than the simple, easy-to-understand analogy does or can.

The analogy alone should be enough to at least make you think about why you assume that c is equivalent to c^2. Given that a foot is not equivalent to a foot^2, why should you asusme that a velocity (c) is equivalent to a velocity^2 (c^2)?

The one thing we haven't done (yet) is to spoon-feed you some of the elements of dimensional analysis. I'll try that in a bit, but I do get the feeling that you aren't really hear to learn stuff, you don't seem to be listening very much. Rather than listening, you seem to be making a bunch of more or less unfounded statements, and then attempting to defend them. Anything that doesn't agree with your unfounded claims seems to get mostly ignored.

Before I start, I'm going to ramble on a bit about the role of mathematics in physics. Mathematics is not a hinderance, as you seem to think. It is an esesential tool. Mathematics does not cause errors in understanding. Mathematics greatly helps to eliminate errors. It *is* possible for errors to creep in in spite of mathematics. This happens when one makes incorrect assumptions. Mathematics is a codified form of logic, so it helps to insure that the conclusiosn follow from the premises. It can't necessarily find errors in the fundamental assumptions. It can greatly aid in ensuring that the conclusions follow from the premises.

Enough about mathematics, let's go back to spoon-feeding you some dimensional analysis.

The idea of dimensional analysis is that every phhysical quantity contains two parts: a number, which gives the magnitude of the quantity, and a unit, which describes how the quantity transforms under scale changes.

Scale changes are when one uses different units - like feet, instead of inches, or seconds instead of minutes.

So let's go back to feet and feet^2. There are 12 inches in a foot, so the rules of dimensional analysis tell us that if we have one foot, and we transform it so that it's units are in inches, we get 12 inches. These two expressions represent the same physical quantity, i.e. 1 foot is 12 inches.

The same rules tell us that if we have one square foot, when we transform to inches we get 144 square inches.

You can check this out for yourself if you really want to - take a square foot, and see how many square inches are in it.

Note that the rules of transformation are totally different for square feet than they are for feet.

This is why we say that feet^2 are different than feet. It also means that we can't directly compare quantites in feet and quantities in square feet in any meaningful way. Because the quanties transform differently, the result of comparing the number part of the quantites will not give the same result in different units (remember, every physical quantity has two parts - a number, and a unit).

These rules can be written down very concisely by treating units as quantites which 'cancel out' in fractions.

See for instance

http://www.chemistrycoach.com/use.htm

<br /> 1\, foot * \frac{12\, inches}{1\, foot} = 12 inches<br />

Feet appear in the numerator and denominator once, and "cancel out", leaving inches.

<br /> 1\, foot^2 *\left( \frac{12\, inches}{1\, foot} \right)*\left( \frac{12\, inches}{1\, foot}\right) = 144 inches^2<br />

Feet appear twice in the numerator and denominator. Both feet "cancel out", leaving inches^2.

Now, let's apply this to velocity.

Suppose we have a velocity of 1 foot/second. The rules of dimensional analysis say that this transforms to 12 inches/second.

Now let's say we have a velocity^2 of 1 foot^2 / second^2. The rules of dimensional anaysis say that this transforms to 144 inches^2/second^2.

Now you can see why a velocity^2 is different than a velocity. The numerical value transforms in a completely different manner when we change units (i.e feet to inches, in this example).

We can also use dimensional analysis to transform the seconds into minutes

60 feet/minute == 1 foot/second
3600 feet^2/minute^2 = 1 foot^2/second^2.

Knowing how to transform both the "feet" (distance), and the "seconds" (time) in the velocity gives us all the information we need to transform a velocity from feet/seconds to any other units we desire. (Furlongs per fortnight, for an extreme example).

This is dimensional analysis in a nutshell. To recap, a physical quantity consists of two parts: a number, AND a unit. Two quantites can be compared directly only if they have the same units. c and c^2 do not have the same units, so they cannot be compared dirrectly.

 

[russ_watters]

Generically speaking the difference between pure mathematics and applied mathematics. I know that there are several cases where pure mathematics became applied mathematics.

Not good enough. You say its a general principle, then say you have a specific example: give us your specific example, or admit you have none.

You're claiming that SR is wrong because it hasn't been "completely" proven. Don't you see the contradiction there?

And again, free_mind - doesn't your error in your first post concern you?

edit: So in your first post, you demonstrated that you don't understand the math of units, and in further posts, you demostrated that you don't understand the scientific method (the invalidity of the concept of "completely proven"). With such severe misunderstandings out in the open (you seem to have acknowledged the first, at the very least), don't you think you should take a step back and consider the validity of your opinion?