What approximation simplifies relativistic force to Newtonian acceleration?

[Jontafin410]

Newton's Second law of motion F=ma, where m is the mass of the object that undergoes an acceleration a due to to an applied force F. This law is accurate at low speeds. At high speeds, we use the corresponding formula from Einstein's theory of relativity
F=m$$\frac{d}{dt}$$($$\frac{v(t)}{\sqrt{1-\frac{v(t)}{c}}$$)

Where v(t) is the velocity function and c is the speed of light. Compute

^{v^2}

What has to be "ignored" to simplify this expression to the acceleration a=v'(t) in Newton's second law?

 

[lanedance]

do you mean
$$F = m \frac{d}{dt}\frac{v(t)}{\sqrt{1-\frac{v(t)}{c}}}$$

what do you think?

 

[mike1967]

ignore the speed of light?

 

[Char. Limit]

At low speeds, is the ratio of v to c significant?

do you mean
$$F = m \frac{d}{dt}\frac{v(t)}{\sqrt{1-\left(\frac{v(t)}{c}\right)^2}}$$

what do you think?

Fixed it for you.

 

[mike1967]

no =) it such a small number it can be neglected. so ignore the ration. thank you

 

[Char. Limit]

Yep!

 

[Jontafin410]

Yea the equation thing on the sight is confusing. What would the derivative of that equation be?

 

[Char. Limit]

Well, firstly, I believe that it is $F=ma\gamma$, where $$\gamma=\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$$. But I'm not certain.

 

[Matterwave]

$$F=\frac{dp}{dt}=\frac{d(\gamma mv)}{dt}=m\frac{d}{dt}(\frac{v}{\sqrt{1-\frac{v^2}{c^2}}})}\neq\gamma ma$$

You can't simply pull the gamma out of the derivative because it's v-dependent. You need to take the derivative. You can use the chain rule and the division rule.