Question

E=MC^2 ?

Answer 1

That is correct. What specifically are you wishing to know about it?

Answer 2

jus say me in general about it
pls.

Answer 3

It relates the energy of a particle to its mass, and is a consequence of the theory of relativity. What it tells us is that mass \(m\) and energy \(E\)are equivalent, and that a little bit of mass is equal to a large amount of energy due to teh multiplication factor of the speed of light \(c\) squared. It is this principle that permits power generation in nuclear power plants, the destructive energy of nuclear weapons, as well as the life giving energy of our sun.

\(E=mc^2\) is only one part of the full equation which is far more interesting than that equation. This specific part only deals with particles at rest (the rest energy or rest mass), but there is also a part that deals with the object moving. The full equation is \[E=\sqrt{(pc)^2+(mc^2)^2}\]where \(p\) is the momentum of an object. There are three interesting things that comes from this equation. The first we have dealt with, is that of a particle at rest \(p=0\), meaning resulting in the realisation that energy and mass are equivalent. The second is that particles without mass, will still have a momentum, meaning that they will be able to move other particles about by colliding with them. Photons (particles of light) have zero rest mass, but are able to exert forces upon other objects through collision.

Last but by far not the least, is that the equation predicts the anti-matter. the reason being comes from teh mathematics. If you take teh square root of any quantity, it will have two solutions. for example the square root of 4 i.e. \(\sqrt{4}\) will equal +2 and -2, because \((-2)^2=4\) as does \((+2)^2=4\). Therefore, the mathematics predicts that for a given momentum and rest mass, that there should exist particles with negative energy. Indeed, it was the Nobel prize winning physicist Paul Dirac who first took the second solutions seriously and predicted anti-matter a full year or two before it was experimentally observed.

Answer 4

@jonny--> why we have to square C?! why cant it be E=MC ? what's the purpose of squaring C?

Answer 5

The c is squared because that will put into units of energy. Mass multiplied by velocity gives units of momentum, but mass multiplied by velocity squared gives units of energy.

Answer 6

well..why cant we go faster than light..?!

Answer 7

It all boils down the consequences of the constancy of the speed of light. No matter what speed you do, you will always measure the speed of light to be the same. The consequences of this are the phenomena of length contraction, time dilation, and teh increase of inertial mass. it is this latter that is important in explaining why we cannot go faster than light.

As you increase your speed (exerting energy to do so, or in the parlance of physicists "doing work"), your mass increases, meaning more work is needed to accelerate it faster, which increases your mass more, meaning even more work is needed to accelerate you and so on. It is probably best illustrated using the equation for relativistic kinetic energy.

\[E_k=\frac{mc^2}{\sqrt{1-v^2/c^2}}-mc^2\]

Recall that the work done in accelerating a particle of mass \(m\) from a velocity \(v_1\) to \(v_2\) is the difference in kinetic energy i.e. \(W=E_{k2}-E_{k1}\). But look at the bottom line of the equation above. When \(v=c\) the bottom line collapses to zero, and thus you are dividing the top line by zero. Any number divided by zero results in infinity (mathematicians prefer to say it is undefined). Physically what happens is that as \(v\) increases toward \(c\), the kinetic energy increases rapidly until at \(v=c\) the kinetic energy becomes infinite.

Remember we said that the work done was \(W=E_{k2}-E_{k1}\). Well you can see that to push an object from \(v_1\) to \(v_2\) requires an infinite amount of work (and hence an infinite amount of energy applied to the particle) since \(W=\infty-E_{k1}=\infty\), since \(E_{k1}) will always be finite, and infinity minus a finite number will always leave you with infinity.

Thus you cannot push a particle with mass to the speed of light, meaning you cannot accelerate beyond the speed of light. Photons can travel at the speed of light, because they do not have mass, and indeed they will only ever travel at teh speed of light, as any energy you give them to change their speed will just change their frequency instead. (light slowing in an optical medium is a different phenomena).

Answer 8

Say speed of light be 30Mph..Space shuttle is travelling at the speed of 29.9Mph..why cant we give more energy to the space shuttle to overcome the mass obtained while travelling near the speed of light?! is it Possible?

Answer 9

See above. It isnt possible, as it would take an infinite amount of energy to push it to the speed of light. Even if you were to transform every piece of matter in the universe to energy, you still wouldn't have enough. this would be true regardless of the rest mass of the particle. As long as the rest mass is finite (i.e. has a non-zero mass) then it will always become infinite at the speed of light. Indeed, in the above equation for relativistic kinetic energy, the term \[\frac{m_0}{\sqrt{1-v^2/c^2}}\] is the relativistic mass, where m is the rest mass of the particle. So you can see that as you approach c \[m'=\frac{m_0}{\sqrt{1-v^2/c^2}}\rightarrow\infty\]which is what pushes the kinetic energy to infinity.

Answer 10

WoW! =) ThankS bRo! =)