Discover the Relationship Between E=mc2 and Time with m=1kg and d=1m

[EM_Guy]

Although i have been warned by ADMIN, let's hope he takes a lenient view...

Let me start from the beginning...

KE=1/2mv2 was empirically found by firing projectiles into water.

It was found
1. KE is proportinal to mass of the projectile.
2. KE is proportinal to the square of speed of the projectile.

So they/he wrote KE=mv2 (1/2, the proportionality constant was added later on to give the equation a clean look, because previously the constant was somewhat cumbersome---you must be knowing all this)

But...my question is very silly rather utterly insane...

Why did he write mxv2 and not m+v2?
I know u may say this is common sense or that only multiplication depicts reality here.

But again...
Why does only multiplication express the reality here?

The 1/2 was not added later to give the equation a clean look.

Let's back up. Forget about relativity for a moment (or a few years).

As has been pointed out, the multiplication operation does not imply that either quantity is moving. The area of a rectangle is length times height. This does not mean that length is moving, nor does it mean that height is moving. You have to rethink what the multiplication operation is. If I have 9 groups of apples and 5 applies in each group, then I have 45 apples. Again, this does not mean that any apples are moving. It means that I have added together 9 groups of apples that have 5 applies in each group. 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 = 45. i.e. 9x5 = 45.

Now, work equals force times distance. (Well, work is the dot product of the force vector and the displacement vector). F = ma. So, W = mad.

From kinematics (study it!), if a = constant, then d = 0.5(vi + vf)/t. That is, a particle that is constantly accelerated from vi to vf will go d = 0.5*(vi+vf)/t. In this case, the acceleration a = (vf-vi)/t.

Plug the expression for a and d into the W = mad equation, and you see that the work done on an object turns out to be 1/2*m*vf^2 - 1/2*m*vi^2.

When work is performed on an object, its energy increases. In this case, its kinetic energy increases (not considering here potential energy, heat transfer, chemical energy, nuclear energy, etc.). The units of work and the units of energy are identical. (Study dimensional analysis). So, if the potential energy remains constant, the work done on an object equals the difference of the kinetic energies of that object.

Does any of this make sense? Note, we have not discussed E = mc^2 at all. But does any of this make sense?

You are seeing similarities between the KE = 1/2*m*v^2 and the E = mc^2 equations. But what folks are trying to tell you is that these are two very different equations. It turns out that this expression for kinetic energy (KE = 1/2*m*v^2) isn't exactly true, but only approximately true (and very close to true when objects are moving much slower than the speed of light). This, because the laws of physics are the same in all inertial frames of reference, and the speed of light is the same regardless of the inertial frame of reference of the observer. This gets into time dilation and length contraction and other very difficult concepts to understand.

If you really want to understand E = mc^2 (i.e. Einstein's theory of relativity), then you need to master kinematics, dynamics, the work-energy theorem, algebra, dimensional analysis, functional analysis, calculus, and electromagnetics.

One other comment. On the one hand, I applaud you for continuing to ask questions when you do not understand the answers. Keep asking why. Keep asking how. You should not just accept any scientific claim made without having understood the concept. But on the other hand, you can know and trust that the folks at these forums for the most part know what they are talking about. They are not omniscient, and they can make mistakes. But in general, what they are saying is true and right and reasonable. So pay careful attention and give serious thought to the things that are being said. In my reply alone, I think I'm giving you plenty to really think about. I have covered the nature of multiplication, kinematics, F = ma, W = Fd, and the derivation of the work - kinetic energy theorem. I think it is fair to say that any future questions you have on this topic should demonstrate that you have thought about these things.

 

[Deepak K Kapur]

Near as I can tell, neither of those is true. Where did you get them? The wiki does not include them and today, kinetic energy is derived mathematically.

Because that's what "proportional to" means.

For example, in a direct, linear proportion, doubling the independent variable doubles the dependent variable. In a square proportion, doubling the independent variable increases the dependent variable by a factor of four.

If you have two dependent variables, it should be easy to see that only multiplication enables both to act according to the definition of "proportion".

All that said, this has very little to do with your problem, unless this is the entry point into the next (first?) four years of your math education beyond arithmetic.

I too have studied proportionality and i too understand it.

Here, two factors (mass and speed/velocity) of a projectile act on water to raise its temperature. From experiments, we get the idea that speed factor plays a stronger (squared) part in raising the temperature. From the rise in temperature we get the idea about the role each factor plays in the KE of the projectile.

What is the need of multiplying these two factors?

When two factors (forces) act on a body, don't we add those forces? Why don't we multiply them?

 

[Dale]

I too have studied proportionality and i too understand it.

What is the need of multiplying these two factors?

I think you have a misunderstanding of both proportionality and units. Otherwise I don't see how you could ask this.

If f is proportional to g and f is also proportional to h then you cannot possibly have f=g+h. Proportional to g means that if g doubles then f doubles, but 2g+h is not equal to 2(g+h). Similarly for h. Only multiplication works since (2g)h=2(gh).

Also, the units don't work. You cannot add kg to m/s and you cannot set either of them equal to joules.

 

[Deepak K Kapur]

I think you have a misunderstanding of both proportionality and units. Otherwise I don't see how you could ask this.

If f is proportional to g and f is also proportional to h then you cannot possibly have f=g+h. Proportional to g means that if g doubles then f doubles, but 2g+h is not equal to 2(g+h). Similarly for h. Only multiplication works since (2g)h=2(gh).

Also, the units don't work. You cannot add kg to m/s and you cannot set either of them equal to joules.

In your case doesn't h get unnecessarily doubled when g is doubled?

 

[Deepak K Kapur]

The 1/2 was not added later to give the equation a clean look.

Let's back up. Forget about relativity for a moment (or a few years).

As has been pointed out, the multiplication operation does not imply that either quantity is moving. The area of a rectangle is length times height. This does not mean that length is moving, nor does it mean that height is moving. You have to rethink what the multiplication operation is. If I have 9 groups of apples and 5 applies in each group, then I have 45 apples. Again, this does not mean that any apples are moving. It means that I have added together 9 groups of apples that have 5 applies in each group. 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 = 45. i.e. 9x5 = 45.

Now, work equals force times distance. (Well, work is the dot product of the force vector and the displacement vector). F = ma. So, W = mad.

From kinematics (study it!), if a = constant, then d = 0.5(vi + vf)/t. That is, a particle that is constantly accelerated from vi to vf will go d = 0.5*(vi+vf)/t. In this case, the acceleration a = (vf-vi)/t.

Plug the expression for a and d into the W = mad equation, and you see that the work done on an object turns out to be 1/2*m*vf^2 - 1/2*m*vi^2.

When work is performed on an object, its energy increases. In this case, its kinetic energy increases (not considering here potential energy, heat transfer, chemical energy, nuclear energy, etc.). The units of work and the units of energy are identical. (Study dimensional analysis). So, if the potential energy remains constant, the work done on an object equals the difference of the kinetic energies of that object.

Does any of this make sense? Note, we have not discussed E = mc^2 at all. But does any of this make sense?

You are seeing similarities between the KE = 1/2*m*v^2 and the E = mc^2 equations. But what folks are trying to tell you is that these are two very different equations. It turns out that this expression for kinetic energy (KE = 1/2*m*v^2) isn't exactly true, but only approximately true (and very close to true when objects are moving much slower than the speed of light). This, because the laws of physics are the same in all inertial frames of reference, and the speed of light is the same regardless of the inertial frame of reference of the observer. This gets into time dilation and length contraction and other very difficult concepts to understand.

If you really want to understand E = mc^2 (i.e. Einstein's theory of relativity), then you need to master kinematics, dynamics, the work-energy theorem, algebra, dimensional analysis, functional analysis, calculus, and electromagnetics.

One other comment. On the one hand, I applaud you for continuing to ask questions when you do not understand the answers. Keep asking why. Keep asking how. You should not just accept any scientific claim made without having understood the concept. But on the other hand, you can know and trust that the folks at these forums for the most part know what they are talking about. They are not omniscient, and they can make mistakes. But in general, what they are saying is true and right and reasonable. So pay careful attention and give serious thought to the things that are being said. In my reply alone, I think I'm giving you plenty to really think about. I have covered the nature of multiplication, kinematics, F = ma, W = Fd, and the derivation of the work - kinetic energy theorem. I think it is fair to say that any future questions you have on this topic should demonstrate that you have thought about these things.

Thanks for an elaborate reply as well as for the applaud.

But...
here i am talking about a particular case of multiplcation. I know multiplication does not always mean motion.
My question is about the interpretation of equations at hand ONLY.

Anyway, thanks.

 

[russ_watters]

In your case doesn't h get unnecessarily doubled when g is doubled?

g and h are not dependent on each other in dale's example. Let's make it really simple:
f = g + h

h = 1
g = 1

If I double h, by what factor does f increase?

 

[Deepak K Kapur]

g and h are not dependent on each other in dale's example. Let's make it really simple:
f = g + h

h = 1
g = 1

If I double h, by what factor does f increase?

Initially f=2, when h is doubled f=3. So, f increases by a factor of 1.5.

 

[russ_watters]

Initially f=2, when h is doubled f=3. So, f increases by a factor of 1.5.

Correct. So then, f is not proportional to h, right?

 

[Deepak K Kapur]

Correct. So then, f is not proportional to h, right?

I do know that what you are saying is mathematically right, but I am talking about the concepts.

Don't you find even an iota of sensibility in post 52?

 

[russ_watters]

I do know that what you are saying is mathematically right, but i am talking about the concepts.

Don't u find even an iota of sensibility in post 52?

No. In particular, you said in your first sentence that you understand proportionality, when clearly you do not*(the rest of the post was further demonstration of it). And now that your error in understanding how proportionality works is shown plain, instead of learning from that, you want to ignore the issue. I'm not sure what more we can do for you if you won't correct errors in your understanding that you know exist.

*Note: it is actually worse than even that: Dale's post contained some basic algebra that you didn't understand, which was why I needed to simplify it for you.

 

[Deepak K Kapur]

No. In particular, you said in your first sentence that you understand proportionality, when clearly you do not*. And now that your error in understanding how proportionality works is shown plain, instead of learning from that, you want to ignore the issue. I'm not sure what more we can do for you if you won't correct errors in your understanding that you know exist.

*Note: it is actually worse than even that: Dale's post contained some basic algebra that you didn't understand, which was why I needed to simplify it for you.

Actually, in Dale's example, I just wanted to say that 2 can be used with h also instead of g. That makes no conceptual sense.
Anyway, thanks a lot. You people have done a lot for me.

Actually, I just wanted to explore the relation of mathematics with reality.

 

[Dale]

Actually, in Dale's example, i just wanted to say that 2 can be used with h also instead of g. That makes no conceptual sense.

Of course it can be used with h also. As I said, f is proportional to g and f is proportional to h. Therefore if you double g you must double f (not 1.5 times) and if you double h you must double f. That is what it means to be proportional to both g and h.

 

[Dale]

In your case doesn't h get unnecessarily doubled when g is doubled?

No.

 

[russ_watters]

Anyway, thanks a lot. You people have done a lot for me.

Actuallly, i just wanted to explore the relation of mathematics with reality...

Fair enough. My recommendation to you at this point is that because you don't understand basic math, you are nowhere close to being able to understand how math relates to reality. You should therefore start by taking some basic math courses (algebra, geometry, trigonometry, calculus), then follow them up with some basic science groundwork courses that show how math is used in science.

Until you've completed such courses, you are wasting your time (and ours) trying to deconstruct how equations work.

Perhaps the more fundamental problem is you seem very unwilling to learn. You'll need to overcome that first.

This thread has run its course and is therefore locked.