Just a question about Kinetic energy at the speed of light.

[KiwiK]

Hi I'm new here and hopefully this is in the right place.

 

Alright this may seem like a stupid question, However it is one that's been bugging me quite a bit.

 

So E=[math]\Delta[/math]mc[math]^{2}[/math]

 

And E[math]_{K}[/math]=[math]\frac{1}{2}[/math]mv[math]^{2}[/math]

 

Does this mean that if an object is to go from rest to the speed of light it will "loose" half of it's mass?

 

also substituting between the equations for an omject at the speed of light would give us the equation

 

E[math]_{K}[/math]=[math]\frac{1}{2}[/math]E

 

So kinetic energy is half of the total energy for that object and does this mean that mass constitutes the other half?

 

I hope you understand what I mean and thanks.

[insane_alien]

you can't use Ek=0.5mv^2 above about 0.1c as newtonian mechanics becomes far too inaccurate.

 

better to use einsteinian mechanics. and objects appear to gain mass, not lose it.

[ajb]

The equation [math] E_{K}= \frac{1}{2}mv[/math] is not the correct expression in special relativity. This is the root of your misunderstanding.

 

One can define a "relativistic kinetic energy" as

 

[math] T = mc^{2} \left( \frac{1}{1-u^{2}/c^{2}}-1 \right)[/math]

 

as measured in a comoving frame of speed [math]u[/math].

 

If [math]u << c[/math] then you can use the binomial theorem to expand the square root and get

 

[math]T \approx \frac{1}{2}mv^{2}[/math].

 

So, what about as [math]v \rightarrow c[/math]? You the kinetic energy tends to infinity!

[KiwiK]

The equation [math] E_{K}= \frac{1}{2}mv[/math] is not the correct expression in special relativity. This is the root of your misunderstanding.

 

One can define a "relativistic kinetic energy" as

[math] T = mc^{2} \left( \frac{1}{1-u^{2}/c^{2}}-1 \right)[/math]

as measured in a comoving frame of speed [math]u[/math].

 

If [math]u << c[/math] then you can use the binomial theorem to expand the square root and get

 

[math]T \approx \frac{1}{2}mv^{2}[/math].

 

So, what about as [math]v \rightarrow c[/math]? You the kinetic energy tends to infinity!

 

As that makes much more sense, thank you both