Relativistic Energy Visualisation



The well known formula

$$E^2=(pc)^2+(m_0c^2)^2$$ reflects that

$$pc$$

and

$$m_0c^2$$

are catheters in a triangle.

By recognizing that

$$pc=mcv$$

it is plain to see that as the velocity (v) approaches c, the alpha angle approaches vertical and thus

$$pc=mc^2$$

which indeed equals E.

Calculating alpha may be done by

$$tan(\alpha)=\frac{pc}{m_0c^2}=\frac{mcv}{m_0c^2}=\frac{\frac{m_0}{\sqrt{1-(\frac{v}{c})^2}}cv}{m_0c^2}$$

Now we can do some parameter eliminations like

$$\frac{pc}{m_0c^2}=\frac{\frac{1}{\sqrt{1-(\frac{v}{c})^2}}v}{c}$$

simplifying this expression we get

$$\frac{pc}{m_0c^2}=\frac{1}{\sqrt{(\frac{c}{v})^2-1}}$$

and finally we get

$$pc=\frac{m_0c^2}{\sqrt{(\frac{c}{v})^2-1}}$$

where it is quite easy to calculate pc with regard to number of rest energies, knowing only rest mass and velocity.

Visualisation of relativistic kinetic energy


Using vectors we may write total energy as

$$E=m_0c^2a_x+pca_y$$

which gives the magnitude of E as

$$|E|=\sqrt{(m_0c^2)^2+(pc)^2}$$

and while using

$$E_k=E-E_0$$

we may write

$$E_0=nm_0c^2a_x+npca_y$$

Ek is then

$$E_k=E-E_0=(1-n)m_0c^2a_x+(1-n)pca_y$$

where n<1 thus

$$E_k=m_{k0}c^2a_x+m_kvca_y$$

where m_k0<m_0 and m_k<m

The length of the E_0 vector is

$$E_0=\sqrt{(nm_0c^2)^2+(npc)^2}=n\sqrt{(m_0c^2)^2+(pc)^2}=m_0c^2$$

this means that

$$E_0=nE$$

where it is obvious that E_0 has less energy than the total energy, E.

The rather fascinating consequence of this is that Ek seams to have a "rest energy" of less than the actual rest energy, looking at pc the same happens here where it comes to the mass, this must happen because kinetic energy is calculateted by subtracting rest energy from E and the only way this can be done is by keeping the E-vector direction (but reversed) so that the pc-mass has the same proportion as the rest mass, otherwise substraction is impossible.