# Square speed of light!

DrChinese
Gold Member
free_mind said:
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
Mentor
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.

russ_watters said:
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."

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.

DrChinese said:
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?

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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?

Entropy said:
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?

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
Staff Emeritus
free_mind said:
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

$$1\, foot * \frac{12\, inches}{1\, foot} = 12 inches$$

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

$$1\, foot^2 *\left( \frac{12\, inches}{1\, foot} \right)*\left( \frac{12\, inches}{1\, foot}\right) = 144 inches^2$$

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
Mentor
free_mind said:
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?

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pervect said:
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

$$1\, foot * \frac{12\, inches}{1\, foot} = 12 inches$$

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

$$1\, foot^2 *\left( \frac{12\, inches}{1\, foot} \right)*\left( \frac{12\, inches}{1\, foot}\right) = 144 inches^2$$

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.

Thank you for you exposition. But where do you get the idea that I considered that c is equal to c^2?

The problem is that 89875517873681764 m^2/s^2 have to be verified in a concrete experience. I don't know any experience where this have been verified. It is always said that the formula contributed to...; has been important to...; and so on... Entropy said in thread #31: "Millions of experiments have proven this. Just about every experiment ever to involve particle accelerators involves Einstien's equations." This means that 89875517873681764 m^2/s^2 has been achieved in terms that has been clearly verified?

In SI units you don't find m^2/s^2. How can we interpret this?
HallsofIvy said that (thread #12): "(...) mc2 have units of energy- that has physical significance (...)". Units of energy? What's this? The Energy results from m.c^2. not from c^2.

Pengwuino
Gold Member
free_mind said:
In SI units you don't find m^2/s^2. How can we interpret this?
HallsofIvy said that (thread #12): "(...) mc2 have units of energy- that has physical significance (...)". Units of energy? What's this? The Energy results from m.c^2. not from c^2.

You will never find a m^2/s^2 as a real value. You must add hte mass to get kg * m^2/s^2 which is energy. C^2 needs no interpretation because there is no such thing as a square meter per squared second. Its simply a value that needs a mass to go with it before it has any physical meaning

HallsofIvy
Homework Helper
free_mind said:
Thank you for you exposition. But where do you get the idea that I considered that c is equal to c^2?
You quoted quite a long section. Didn't you bother to read it first? No one said you thought "c is equal to c^2". The word used was "equivalent"- you were comparing c^2 to a speed, c. They are different in the same sense that an area ("square feet") is completely different from a length ("feet")

The problem is that 89875517873681764 m^2/s^2 have to be verified in a concrete experience. I don't know any experience where this have been verified. It is always said that the formula contributed to...; has been important to...; and so on... Entropy said in thread #31: "Millions of experiments have proven this. Just about every experiment ever to involve particle accelerators involves Einstien's equations." This means that 89875517873681764 m^2/s^2 has been achieved in terms that has been clearly verified?
E= mc2 has been verified repeatedly. That says nothing about "89875517873681764 m^2/s^2 has been achieved". Since there is no physical quantity that has units of m^2/s^2, I don't even know what you mean by "achieved".

In SI units you don't find m^2/s^2. How can we interpret this?
HallsofIvy said that (thread #12): "(...) mc2 have units of energy- that has physical significance (...)". Units of energy? What's this? The Energy results from m.c^2. not from c^2.
Yes, read your own quote here: I said mc^2, not c^2!! Once again, there is no physical quantity that has units of m^2/c^2.

russ_watters
Mentor
free_mind said:
The problem is that 89875517873681764 m^2/s^2 have to be verified in a concrete experience. I don't know any experience where this have been verified.....

In SI units you don't find m^2/s^2. How can we interpret this?
HallsofIvy said that (thread #12): "(...) mc2 have units of energy- that has physical significance (...)". Units of energy? What's this? The Energy results from m.c^2. not from c^2.
The formula for the area of a circle is pi*r^2. So what does the r^2 on its own give us? Nothing! You can't take part of an equation and expect that part to have a physical meaning all its own. That simply isn't how math/science works.

Again, doesn't the error in your opening post cause you any concern about your line of reasoning?

r squared?!? what an atrocity!

pervect
Staff Emeritus
free_mind said:
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.

An atomic nucleus has the same energy (relativistic mass) before and after it fissions. However, they are not physically the same, they do not represent the same state of matter. They have the same energy, but they are not otherwise equivalent.

Having the same energy is different from being identical. So E=mc^2 is a conservation law, not a statement that matter and energy are identical.

pervect
Staff Emeritus
free_mind said:
Thank you for you exposition. But where do you get the idea that I considered that c is equal to c^2?

Go back to the origin of the thread, where you write

e=mc^2 is equivalent to e=m.89875517873681764 m/s
The speed of light is a constant. How this is possible in reality?

Is not possible to have a body traveling at this speed, because simply doesn't exists (or at least we don't discover it yet).

I already read several answers to this kind of question. But the answers are around the intrinsic mathematical need to square c in the formula e=mc^2. I am looking for a logical reason for this problem, because beyond mathematical reasoning there are the reality; many times the mathematic coincides with reality, but in this problem, is what it happens?

I look for help to eliminate my difficulty in understanding this. Thank you.

At this point, you are confusing c^2, which is velocity squared, with 'c', which is a velocity. You appeared at this point to be actually asking a question, rather than sitting on a soapbox, orating, which seems to be your current conversational "posture".

Dimensional analysis is the answer to your original question at the start of the thread. c^2 is not a speed, and should not be compared to a speed.

Hurkyl has also pointed out other instances of this sort of invalid comparison.

BTW, confusing c^2 with a velocity is a reasonably common mistake (not the most common, but it happens a fair amount). Probably the most famous instance was in a very old science fiction story called "Venus Equilateral" by George O Smith. It reputedly took a lot of convicing (in the letter columns of a magizine, this was in the days before the internet, and before my time ), but eventually George learned something from the exchange, and stopped calling c^2 a velocity.

I can only hope that similar learning will occur in this thread, too.

Anyway, I think I've said what I need to say to anyone who is listening, I think I'm going to take a short vacation from this thread, there are plenty of other interesting questions out there, it feels like I'm beating a dead horse more than engaging in a conversation at this point.

HallsofIvy said:
You quoted quite a long section. Didn't you bother to read it first? No one said you thought "c is equal to c^2". The word used was "equivalent"- you were comparing c^2 to a speed, c. They are different in the same sense that an area ("square feet") is completely different from a length ("feet")

That happened at my first thread. But why are you fixing your focus on that, taking in consideration that you can see in all my threads from that point that my positions didn't have nothing to do with the difference between m/s and m^2/s^2?

After all, until now, no one explain the meaning of m^2/s^2!

HallsofIvy said:
E= mc2 has been verified repeatedly. That says nothing about "89875517873681764 m^2/s^2 has been achieved". Since there is no physical quantity that has units of m^2/s^2, I don't even know what you mean by "achieved".")

C^2 is only a conversion factor with no particular meaning, that has resulted from mathematic deduction. This is what all accept!

HallsofIvy said:
Yes, read your own quote here: I said mc^2, not c^2!! Once again, there is no physical quantity that has units of m^2/c^2.

Sorry, you are right, was a mistake. But the units problem persists. I suppose that in you satement you want to said m^2/s^2 and not m^2/c^2, right? :)

pervect said:
Go back to the origin of the thread, where you write

At this point, you are confusing c^2, which is velocity squared, with 'c', which is a velocity. You appeared at this point to be actually asking a question, rather than sitting on a soapbox, orating, which seems to be your current conversational "posture".

About my posture you are wrong.

pervect said:
Dimensional analysis is the answer to your original question at the start of the thread. c^2 is not a speed, and should not be compared to a speed.

Hurkyl has also pointed out other instances of this sort of invalid comparison.

BTW, confusing c^2 with a velocity is a reasonably common mistake (not the most common, but it happens a fair amount). Probably the most famous instance was in a very old science fiction story called "Venus Equilateral" by George O Smith. It reputedly took a lot of convicing (in the letter columns of a magizine, this was in the days before the internet, and before my time ), but eventually George learned something from the exchange, and stopped calling c^2 a velocity.

I can only hope that similar learning will occur in this thread, too.

I am always searching for something to learn. My principle is: we can learn so much with the ones that don't know nothing.

pervect said:
Anyway, I think I've said what I need to say to anyone who is listening, I think I'm going to take a short vacation from this thread, there are plenty of other interesting questions out there, it feels like I'm beating a dead horse more than engaging in a conversation at this point.
I knew that m.m/s.s is equivalent to m^2/s^2, this is basic! But even if I had written correctly my doubt about velocity will be underlying my question. But you can see that no one in this forum characterized m^2/s^2. So, this units are something that no one knows what it is! :)

About your another comment, I am not going to say nothing, because you will beat me in experience. :)

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russ_watters said:
The formula for the area of a circle is pi*r^2. So what does the r^2 on its own give us? Nothing! You can't take part of an equation and expect that part to have a physical meaning all its own. That simply isn't how math/science works.

Again, doesn't the error in your opening post cause you any concern about your line of reasoning?
r^2 is geometrically explainable, but c^2 with it's extraordinary units it's not so easy to explain.

Entropy said:
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).

Did you understand that I was talking about the nuclear bombs?

Pengwuino said:
You will never find a m^2/s^2 as a real value. You must add hte mass to get kg * m^2/s^2 which is energy. C^2 needs no interpretation because there is no such thing as a square meter per squared second. Its simply a value that needs a mass to go with it before it has any physical meaning

It seems that anything multiplied by mass in Einstein formula will have a physical meaning, probably even potatoes :)

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russ_watters said:
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.

That means that you don't agree with my affirmation? This is a common knowledge between mathematicians; it is similar to fundamental research (or basic research) and applied research.

russ_watters said:
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?

If you read all my posts you will understand that they are different from the first. I inserted lots of threads, but you fixed you atention at the first one; if you read you will see that the other threads point to different problems.

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robphy
Homework Helper
Gold Member
free_mind said:
I knew that m.m/s.s is equivalent to m^2/s^2, this is basic! But even if I had written correctly my doubt about velocity will be underlying my question. But you can see that no one in this forum characterized m^2/s^2. So, this units are something that no one knows what it is! :)

free_mind said:
r^2 is geometrically explainable, but c^2 with it's extraordinary units it's not so easy to explain.

I'm curious.
Do you also have a problem with the units in Newton's Gravitational constant?
$$\mbox{gravitational constant}\ G = 6.67 \times 10^{-11}\rm m^3 kg^{-1} s^{-2}$$

russ_watters
Mentor
free_mind said:
r^2 is geometrically explainable
How? What physical significance does it have?
...but c^2 with it's extraordinary units it's not so easy to explain.
There is nothing at all extrordinary about the units of C^2. They appear in many, many different equations -from other kinetic energy equations, to the many forms of Bernoulli's equation. Do you have a similar problem accepting Bernoulli's equation?
If you read all my posts you will understand that they are different from the first. I inserted lots of threads, but you fixed you atention at the first one;
Yes, I know - the reason I'm focusing on that first post is that you appear to be unwilling to acknowledge the error! Others, too. You're ducking and weaving, trying to press your point witout acknowledging the errors in it as they are addressed - quickly dropping each point and moving to the next as its flaws are exposed.
That means that you don't agree with my affirmation?
You said you had examples - are you willing to provide any or not?

Let me give you one: black holes. Black holes were derived mathematically and were peculiar enough that people doubted their existence. But not anymore. Today, their existence is an observational fact. However, it would not be a surprise to me at all if you didn't accept that....
This is a common knowledge between mathematicians...
As a matter of fact, there is considerable debate over the physical reality of mathematical derivations, with virtually all of the difficulty coming from people who don't want to accept the physical reality of what is being described in the equations - ie, you in this thread. Much of it had to do with QM - QM made a great many people uncomfortable about its implications. The debate largely died down because every time a new piece of evidence was found, it confirmed the physical reality of the equations.

Today, largely because of the success of QM, when new implications of a theory derived mathematically, they are not as easily dismissed as mathematical peculiarities because of discomfort.

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