# Is it not possible to travel in light speed?

1. Jan 25, 2008

### Seek45

Hi

I am just wondering if E = mc^2 then wouldn't it also be correct to say 1/2mv^2=mc^2 since E is kinetic energy. So in order for an object to travel near light speed, it has travel twice (1/2v^2=c^2) the speed of light according to the above formulae. But according E=mc^2 light speed is the limit as energy turns into mass. This lead to the idea that no matter how advanced things are, unless preventing energy turning to mass, it's not possible to travel close to speed of light. Or is the entire derived above formulae incorrect?

If it incorrect but is it really possible to travel close to light speed as a concept?

2. Jan 25, 2008

### JesseM

If m is the object's rest mass, the equation E=mc^2 only applies to an object at rest, for an object in motion with nonzero momentum p, the correct equation is $$E^2 = m^2 c^4 + p^2 c^2$$, and in relativity momentum is not mv but rather $$\gamma mv$$, where $$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$. With a little algebra, the above formula can be rearranged as $$E = \gamma mc^2$$. Since the "rest energy" is just $$mc^2$$, the kinetic energy in relativity is not (1/2)mv^2 but rather $$(\gamma - 1)mc^2$$, so that when you add the kinetic energy and the rest energy you get back the total energy $$\gamma mc^2$$. It's possible to show that in the limit as v gets very small relative to c, the relativistic kinetic energy $$(\gamma - 1)mc^2$$ approaches the Newtonian kinetic energy (1/2)mv^2.

3. Jan 25, 2008

### Seek45

Ah alright thanks, I had no idea that relativity equation also applies to E=mc^2. Just to digress a bit, I found textbooks explaining the equation in kinetic motion - such as turning energy into mass - without the relativistic equation. Wouldn't be conceptually incorrect for them to use the rest energy arrangement as it links special relativisty's concept of motion with non-special relativity of motion equation? I am not sure if I laid out my ideas correctly but I hope you can understand it.

4. Jan 25, 2008

### JesseM

What did they say, exactly? Do you mean that they were using E=mc^2 rather than E^2 = m^2c^4 + p^2c^2, or do you mean that they were using K.E. = (1/2)mv^2 rather than K.E. = (gamma - 1)mc^2? And what example of "turning energy into mass" were they using? It might help if you just quoted a paragraph or two from whatever section is confusing you...

5. Jan 26, 2008

### CJames

seek, to answer the original question about traveling at light speed, if you look at the equation

$$E = \gamma mc^2$$

by studying the definition

$$\gamma = 1/\sqrt{1-v^2/c^2}$$

you can see that as v approaches c, $$\gamma$$ approaches infinity, so the energy approaches infinity. It takes infinite energy for a particle to travel at c. That is, as you can also see from the equation, unless the mass is zero.

6. Jan 27, 2008

### Seek45

I believed I was confused when they linked Newtonian equation into quantum phenomena. Such as linking P = h/lamba with newtonian momentum to show the Law of Conservation of Momentum with photon applying to subatomic particles. Since it was photons I presumed they were talking about the speed of 3 x 10^8 m/s and with their De Broglie explaination they stated that lamba = h/mv. They tried to show the relationship between Newtonian equations with Quantum world as like the one shown above. They didn't really draw a clear line between Quantum principles with Newtonian's. As you have stated above that gamma must be incorporated in these equations such as special relativity's momentum, they (Quantum principles and Newtonian's) seemed more distincted from one another.

With E=mc^2 issue, yes, they didn't incorporate full equation and expected using E = mc^2 to give the full understanding of kinetic energy and mass being proportional when close light speed. As they explained also that mass energy changing to mass such as protons getting heavier when close to light speed according to them; they didn't use E =(gamma)mc^2 so bringing to the misconception of 1/2mv^2 = mc^2.