# Acceleration of speed of light [closed]

If we consider the equation "$$E=mc^2$$" and if we differentiate the equation with respect to time, e.g. $$\frac{\mathrm dE}{\mathrm dt}=m\frac{\mathrm dc^2}{\mathrm dt},$$ we will get after differentiating, $$\frac{\mathrm dE}{\mathrm dt}=m2c\frac{\mathrm dc}{\mathrm dt},$$ This will be the equation after differentiation and if we see the term $$\mathrm dE/\mathrm dt$$, is the power term which is energy transfer per unit time. So from here we can find out the acceleration of speed of light which will be, $$\frac{\mathrm dc}{\mathrm dt}= \frac{\mathrm dE}{\mathrm dt}\cdot\frac{1}{2mc}.$$

• What is your question? – The Photon Oct 18 '19 at 15:09
• The speed of light doesn’t change. It is a constant. – G. Smith Oct 18 '19 at 15:43

When you write:

$$E = mc^2 \tag{1}$$

the $$m$$ in the equation is the rest mass, which is an invariant. The speed of light is also an invariant, so the time derivative of the right hand side is zero.

The problem is that you are differentiating an equation when both sides are constant. Mathematically you can do this, but physically it isn't very interesting since you just end up with zero equals zero.

Equation (1) is a special case of the more general equation:

$$E^2 = p^2c^2 + m^2c^4$$

where $$p$$ is the relativistic momentum, which is not a constant. So taking the time derivative is physically meaningful since $$dp/dt$$ is a force.