What is the kinetic energy of an object traveling at the speed of light?

[joeyjo100]

When an object is traveling at the speed of light, c, what is its kinetic energy?
Is it the objects mass multiplied by the speed of light squared, according to Einsteins special relativity?
Or is it the objects mass multiplied by its velocity (speed of light) squared, divided by 2, according to classical mechanics?


I am aware no object with mass can go the speed of light, but let's have a little fun and assume that they can.


I'm not very good at physics, so please don't rip me apart :S

 

[Doc Al]

When an object is traveling at the speed of light, c, what is its kinetic energy?

Massive objects cannot travel at the speed of light.

Is it the objects mass multiplied by the speed of light squared, according to Einsteins special relativity?
Or is it the objects mass multiplied by its velocity (speed of light) squared, divided by 2, according to classical mechanics?

For an object moving very fast (but still less than the speed of light), neither formula is correct. The second one (½mv²) is an approximation that is good for everyday speeds that are small compared to light speed. The first one doesn't make much sense. (mc² is the rest energy of some mass.)

I am aware no object with mass can go the speed of light, but let's have a little fun and assume that they can.

It doesn't work that way. If you want a real physics answer, you have to stick within the known boundaries of what's possible. (Otherwise, how can you expect physics to give you an answer?)

 

[michael879]

E=mc^2 is the rest energy of a mass. The full kinetic energy is given by [tex]\gamma m_0c^2[/tex]. [tex]\gamma = (1-v^2/c^2)^{-1/2}[/tex] and blows up as v -> c. So the kinetic energy would diverge as a particles velocity approached the speed of light.

On a side note, the E=mc^2 equation that everyone quotes so often has 2 meanings. If u call m the "relativistic mass" (which is rly not good practice because of all the confuson it causes), then this equation gives the correct total relativistic energy. When the m is taken as the rest mass (as it should be), this is simply the rest energy of the particle. If you take the limit v << c in the full equation, you end up with something like: [tex]E = 1/2 m_0v^2 + m_0c^2[/tex], which gives the classical Newtonian answer, plus some constant term that ends up being the energy contained in the mass.

 

[Calimero]

When an object is traveling at the speed of light, c, what is its kinetic energy?
Is it the objects mass multiplied by the speed of light squared, according to Einsteins special relativity?
Or is it the objects mass multiplied by its velocity (speed of light) squared, divided by 2, according to classical mechanics?

Neither. Massive object traveling at the speed if light would have infinite kinetic energy, therefore no massive object can't travel at the speed of light.