Did I figure out what E=MC2 means?

[Lord Hakk]

here, last night roughly around 12:39 I was lying in bed thinking about E=MC2. Then I had what I thought was the answer to it. Based on what I knew about Einstein's time travel theory you know, going back in time requires going faster than the speed of light. so tell me if this is right but dont launch into a long description of mathamatics(im only 14)

E=MC2

v

10i X D

V

T

I= what is used to mesure mass.

D= double the speed of light

T=traveling back intime

[pgrmdave]

I'm not exactly sure what you're trying to say here, do you think you could elaborate a little bit?

[Lord Hakk]

I am trying to show Einstein's equation and what I think it means

[pgrmdave]

Well, I understand that you're trying to talk about Einstein's equation, but I'm unsure as to what you're trying to say. Are you saying that travelling at twice the speed of light means that you'd travel back in time? What is "10i X D"? Perhaps if you wrote out your thoughts I could understand you better.

[Lord Hakk]

i d and t are varyables and I say what they are

[pgrmdave]

What you have now does not make sense to me - 10 * any given unit of mass * 2 * the speed of light.

 

Are you trying to say that moving a mass at twice the speed of light moves it back in time? If so, they why twice, why 10 times the mass? How did you come to this understanding, what was your thought process?

[CraigD]

E=MC2 (usually written [math]E = MC^2[/math]) means

E = M times C times C,

not

E = M time C times 2.

…

E=MC2

v

10i X D

V

T

I= what is used to mesure mass.

D= double the speed of light

T=traveling back intime

Your “double the speed of light” term leads me to suspect you’re confusing multiplication by 2 ([math]C \times 2[/math]) with exponentiation by 2 ([math]C^2[/math]).

[InfiniteNow]

Also LH, don't get anyone wrong. These are hard concepts and it's good that you are thinking about them. Making mistakes is a great way to learn. :)

 

 

Verbally, my understanding of the equation is that energy and mass are the same, but have different "strengths." Itty bitty tiny teeny little pieces of mass have ENORMOUS energy, and this is reflected by the equation.

 

Even multiplying 0.0000023 (a pretty small number, right?) by c times c (which is just [math]C^2[/math]), when c equals roughly 300,000 kilometers per second, turns out to be:

 

0.0000023 x 300,000 x 300,000

 

Without the mass in the equation, it would be 300,000[math]^2[/math].

 

 

That equals 90,000,000,000 (holy guacamole that's a big number! :omg:)

 

Then, you mutiply the mass by that number and get this: 207,000

 

[EDIT]: My math above actually under-estimates the true energy. For more clarification, please check out Jay-Qu's reply to this in Post #11. [/EDIT]

 

 

So, even that itty bitty teeny weeny piece of mass (0.0000023 g) has huge power.

 

 

The equation also relates to acceleration, and as you move faster, you get "heavier."

 

Think about being in a sports car and someone accelerating with the pedal all the way to the floor. You get pushed back in your seat.

 

Well, as you approach the speed of light - c (roughly 300,000 km/s), you get heavier and heavier, and before you actually reach c, you weigh infinity!! :omg: That means it takes more energy to accelerate more than is available, and it just cannot (with current understanding and methods) be done.

 

 

Cheers. And keep up the excitement! :)

[UncleAl]

There is no mathematical prohibition of a massed body traveling at less than or greater than lightspeed. Given that division by zero is undefined (a massed body traveling at lightspeed) how do you pass from one regime to the other?

 

If angular momentum is conserved then time flows in one direction only. Consider a hollow right cylinder with inlet normal its base and outflow tangent to its side. Pump in water from the bottom and take it off at the side. No problem with even high speed flow. Now, reverse time. Doesn't work - hydraulic diode with no moving parts.

 

http://fourmilab.to/etexts/einstein/specrel/specrel.pdf

http://www.geocities.com/physics_world/sr/ae_1905_error.htm

http://www.physics.gatech.edu/people/faculty/finkelstein/relativity.pdf

Longitudinal and transverse mass

 

The general expression for energy in special relativity is (E=energy, m=rest mass, c=lightspeed, p=relativistic momentum, mv/sqrt[1-(v^2)/(c^2)] ).

 

E^2 = (m^2)(c^4) + (p^2)(c^2)

 

E=mc^2 by setting p=0, hence "m" is the rest mass. Lorentz transformation doesn't easily go into General Relativity, but the momentum four vector (consider the metric expression) does go well. Rearranging the above,

 

E^2 - (p^2)(c^2) = (m^2)(c^4)

 

E^2 - (c^2)(p^2) is the norm of the momentum 4-vector (E,p) (as x^2 - (c^2)(t^2) is the norm of the position 4-vector (t,x) ). (m^2)(c^4) is the norm in the rest frame where 4-momentum is (mc^2,0). Norms are conserved under Lorentz transformation; that is the 4-vector definition.

 

Or, bludgeon it as per textbook treatments: Body with rest mass "m" traveling without acceleration. Position vector "x" at time "t." It is then acted upon by a constant external force F.

 

Its relativistic mass at time "t" is m/sqrt[1 - (v^2)/c^2)] as per Special Relativity. v = dx/dt as per Newton:

 

F = d/dt(m/sqrt[1 - (v^2)/(c^2)] times v)

 

Multiply both sides by dx/dt. Fdx = dE where E = total energy

 

dE/dt = v d/dt(mv/sqrt[1 - (v^2)/(c2)])

 

Set u = v/sqrt[1 - (v^2)/(c^2)] to condense a lot of writing,

 

dE/dt = mv du/dt

 

We also know that:

 

u^2 = (v^2)/[1 - (v^2)/(c^2)]

 

Therefore,

 

u^2 - (u^2)(v2)/c^2 = v^2

 

v^2 = (u^2)/[1 + (u^2)/(c^2)]

 

Given v>0,

 

v = u/sqrt[1 + (u^2)/(c^2)]

 

Substituting backwards,

 

dE/dt = mv du/dt = mu/sqrt[1 + (u^2)/(c^2)] du/dt

 

Given (dE/dt)/(du/dt) = dE/du we obtain,

 

dE/du = mu/sqrt[1 + (u^2)/(c^2)]

 

Take the indefinite integral to get E,

 

E = (mc^2)sqrt[1 + (u^2)/(c^2)] + K

 

where "K" is the constant of integration. Now comes the fun part!

 

E = (mc^2)sqrt[1 + (v^2)/{(c^2)-(v^2)}] + K = (c^2)m/sqrt[1 - (v^2)/(c^2)] + K

 

"M" is the observed mass (Lorentz transformation of the rest mass "m") of the body as per Special Relativity,

 

M = m/sqrt[1 - (v^2)/(c^2)]

 

Substitute, hence

 

E = Mc^2 + K

 

where "K" is that constant of integration. The energy equivalent of a body is in direct proportion to its observed mass. The constant of proportionality is (lightspeed)^2. Given conservation of energy and conservation of mass, E = mc^2 is the correct expression for all observed masses. "K" is zero, the correction for potential energy, etc.

[InfiniteNow]

As always Al, you gently break down the issue into terms even a 3 year old could understand. :cup:

 

 

That's going to take ME some time to figure out! :cup:

[Jay-qu]

when c equals roughly 300,000 kilometers per second,

 

yes, it does but in SI units (needed in the equation) its 3x10^8 m/s. Squaring this you get 9x10^16, so that big number is infact actually a whole lot bigger.

[InfiniteNow]

yes, it does but in SI units (needed in the equation) its 3x10^8 m/s. Squaring this you get 9x10^16, so that big number is infact actually a whole lot bigger.

:cup:

 

 

Many thanks. I was trying to make it SO simple, that I actually did it wrong.

 

 

More beer please. :cup:

 

:cup:

[Qfwfq]

And you could at least bother to use the shining new LaTeX that good ol' Alex gave us, if you want someone to bother going through all those equations.

[arkain101]

One thing,

 

Time travel is not what you expect. There is not traveling through time. There is only an ability to manipulate the change that occurs in the molecular level of material.

 

Everything is constantly time traveling, or in other words everything is always changing at different rates.

 

Something that travels near the speed of light will not dissapear, it will remain in your time observation, however, the input you recieve from this object will change and the change of the change of the object will change, but nothing will dissapear, nothing will travel other dimensions in the sense of a worm hole to some other world.

 

Yet we have no idea what will happen to consciousnes under these circumstances.

[anglepose]

ok then then ive made a little diagram of time cones just to get you thinking it might not be explained very welll so ask if you want a part of it explained

[ATTACH]664[/ATTACH]

[anglepose]

also would time travel just let you see the past or could you feel it...

thatll get you thinking

[Janus]

Based on what I knew about Einstein's time travel theory you know, going back in time requires going faster than the speed of light.

 

Let's start right here. Einstein's theory does not say that if you go faster than light you go back in time. Though you hear this idea quite often, it is due to a misunderstanding. A person hear's that as an object approaches the speed of light, time slows down for it. He then assumes that if you extend that idea to speeds greater than light this trend would cause time to go backward for the object, IOW, the object would go back in time. The problem is that this extension is not correct.

 

Unfortunately, you hear this idea repeated quite often, especially on TV and in movies. Some character in a movie, purportedly a respected scientist says, "As we all know, Einstein's theory says..." . The problem is that the person who wrote those lines never studied the Theory, he's just going on what he heard somewhere else, and he just assumed it was true. The misconception just keeps being repeated.

 

As to what E=mc² actually means, it has been touched on above, but it bears repeating.

It means that in an piece of matter with a mass equal to "m" there is a potential energy equal to that mass times the speed of light squared (c x c)