To be perfectly frank, it doesn't seem like you are trying very hard here. I agree with V50: you can do better. Especially if you are a teacher.Deepak K Kapur said:Well...i was expecting such a response...
I am a teacher myself but never snub anyone who tries to understand something honestly, even if the quality of questions posed is dismal...
I also don't threat such a person...
russ_watters said:I'm sorry, but since the units don't match, that's just mathematical gibberish. I don't mean to sound insulting here, but how much schooling have you had? What is the highest level of math and science you've completed? In the US, we typically start with a course at age 14 that teaches the basic tools of science, including dimensional analysis. I've you've passed that level, you'll need to go back and brush-up on it. Here's a link discussing it:
http://www.chem.tamu.edu/class/fyp/mathrev/mr-da.html
Because it was derived to mean something else. Did you read the wiki link I posted? I guess it's unfortunate that the proportionality constant happens to be C (squared), but that's all it is in this equation: a proportinality constant. C happens to be the speed of light,
russ_watters said:To be perfectly frank, it doesn't seem like you are trying very hard here. I agree with V50: you can do better. Especially if you are a teacher.
Not just a chuckle, u can have a hearty laugh...DarioC said:Ah I think I get the concept problem that you are having. The C in the formula is just a short way of saying that the calculation of the conversion of matter to energy in a non-moving mass (like in a nuclear reactor ) requires the use of a constant that is equal to the speed of light (because of it's derivation). The formula has nothing to do with velocity or a moving object-- as used.
Perhaps if you could think of the letter C in this formula as just being a number (C stands for constant, chuckle) that happens to be the same as the speed of light (because of it's derivation) you would stop thinking of it meaning that the mass object in the formula must be moving.
Sorry about he liberties I'm taking here guys, I use to write tech manuals and some of them had to get very basic. Old habit.
DC
Now that it got to this point, I can only have one advice for you. Put aside this question and start learning physics the right way. Then after learning a bit of special relativity, you'll find out that why you are wrong and what is the right way of thinking about this. Keep in mind that you can't expect to have a good understanding of something if you're not willing to put enough effort in learning it.Deepak K Kapur said:Not just a chuckle, u can have a hearty laugh...
But for me the math involved is very strange ( or u may take it as a proof of my foolishness)...
Two same quantities [mass and speed ( whether constant or not)] are involved in the same mathematical relation of multiplication...and... in one case the operation of multiplication says that the object moves and in the second it says it doesn't...u may feel okay with it but i can't.
Mathematics represents nature...what kind of weird representation is this...i wonder.
There is a popular ghazal ( a type of song) in India. It has a couplet...i translate it below...DarioC said:Uug, too early in the AM for this stuff.
I just pulled out my book on relativity by Albert himself, published in 1931 and from a little quick review I think that you may well have a good point with your fixation (chuckle) with the velocity of light part of the formula. I refreshed my brain that the mc squared is actually a part of the calculations involving the energy and equivalent mass of light when dealing with different inertial frames. As in: I'm moving relative to you.
Strangely enough I myself have never seen in all my books anything that I identify as a connection between those formulas and the energy given off by a nuclear reaction. The only thing that occurs to me is that most of the energy given off by an atomic bomb is initially light. So your moving/non-moving point is better taken by me now.
As the gentleman said above, you are going to have to dig in on the basics. You will really like what Poincare (sp?) and Lorentz have to say.
Good luck
DC
So, I don't know of another one with speed, but for other equations with similar issues, I gave you an example of torque vs work, that you didn't appear to notice. For that matter, then, any equation with distance in it (circumference or area of a circle, area of a square, etc.) is different than torque and work even though all of them include distance. There is also the equation for weight (and hydrostatic pressure), which uses gravitational acceleration even though the object isn't accelerating (there is a good reason for that, though).Deepak K Kapur said:Can you give another example of an equation (other than E=mc2), that has mass multiplied by velocity/speed but does not involve motion of mass?
You shouldn't feel OK with it. Your argument is so wrong that you cannot help but feel confused.Deepak K Kapur said:Two same quantities [mass and speed ( whether constant or not)] are involved in the same mathematical relation of multiplication...and... in one case the operation of multiplication says that the object moves and in the second it says it doesn't...u may feel okay with it but i can't.
Although i have been warned by ADMIN, let's hope he takes a lenient view...russ_watters said:I think the argument was 10% less bad than that, dale: he wasn't saying that when multiplied by anything it represented motion, only when multiplied by speed (or distance?). Not quite as wrong, but still pretty wrong. That's why I like the example of work vs torque: exactly the same units but one involves motion and the other doesn't.
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.Deepak K Kapur said:KE=1/2mv2 was empirically found by firing projectiles into water...
...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)
Because that's what "proportional to" means.Why did he write mxv2 and not m+v2?
You are completely wrong. Please stop saying this and read the wiki article on mass-energy equivalence I provided. The words "rest mass" or energy literally appear in it a hundred times.Scott Cox said:Hello DC
Whoever told you that Einsteins equation applies only to stationary objects doesn't really understand the meaning of it. After all c is the speed of light so the equation is only useful when things are moving very fast.
Fredrik said:In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is ##\frac 1 2 mv^2##. In special relativity, it's ##\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}##. Note that this is equal to ##mc^2## if and only if ##v=0##.
Deepak K Kapur said: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?
I too have studied proportionality and i too understand it.russ_watters said: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.
Deepak K Kapur said:I too have studied proportionality and i too understand it.
I think you have a misunderstanding of both proportionality and units. Otherwise I don't see how you could ask this.Deepak K Kapur said:What is the need of multiplying these two factors?
In your case doesn't h get unnecessarily doubled when g is doubled?DaleSpam said: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.
EM_Guy said: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.
g and h are not dependent on each other in dale's example. Let's make it really simple:Deepak K Kapur said:In your case doesn't h get unnecessarily doubled when g is doubled?
Initially f=2, when h is doubled f=3. So, f increases by a factor of 1.5.russ_watters said: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?
Correct. So then, f is not proportional to h, right?Deepak K Kapur said:Initially f=2, when h is doubled f=3. So, f increases by a factor of 1.5.
I do know that what you are saying is mathematically right, but I am talking about the concepts.russ_watters said:Correct. So then, f is not proportional to h, right?
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.Deepak K Kapur said: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?
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.russ_watters said: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.
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.Deepak K Kapur said: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.
No.Deepak K Kapur said:In your case doesn't h get unnecessarily doubled when g is doubled?
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.Deepak K Kapur said:Anyway, thanks a lot. You people have done a lot for me.
Actuallly, i just wanted to explore the relation of mathematics with reality...