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# The force $F$ acting on a body with mass $m$ and velocity $v$ is the rate of changes of momentum: $F = (d/dt)(mv).$ If $m$ is constant, this becomes $F = ma,$ where $a = dv/dt$ is the acceleration. But in the theory of relativity the mass of a particle varies with $v$ as follows: $m = m_o/ \sqrt {1 - v^2/ c^2},$ where $m_o$ is the mass of the particle at rest and $c$ is the speed of light. Show that $F = \frac {m_o a}{(1 - v^2/c^2)^{2/3}}$

## $\frac{m_{0} a}{\left(1-v^{2} / c^{2}\right)^{3 / 2}}$

Derivatives

Differentiation

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##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

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### Video Transcript

you were given several interrelated equations having to do with the theory of relativity. And our goal is to show that this equation is true. So we're going to take the force equation. We were given forces the derivative of mass times velocity. We're going to substitute the mass equation in for em and we're going to find the derivative. Now, keep in mind that as we're finding the derivative are variable is not the r variable is Time t. So let's find the derivative of mass times velocity if the lot if masses actually am not over the square root of one minus B squared over C squared. We're multiplying that by the lost you were going to find the derivative of it. Okay, so because this is a quotient, we're going to use the quotient rule. So we have the bottom squared of one minus b squared over C squared times the derivative of the top and the derivative of the top is going to be m not times DVD t remember are variable is not VR variable his tee time and then we have minus the top am not V times a derivative of a derivative of the bottom and because the bottom has a square root, we're going to bring down the 1/2 and we're going to raise the inside to the negative 1/2 and then using the chain rule, we're going to take the derivative of the inside, which would be negative one. Oversee squared times to V. Times DVD t Remember Chain rule. These honor variable t is our variable, and this is all over the denominator squared and so were squaring the square root. So it's all over one minus b squared over C squared. Now, obviously, we have a lot of simplifying to Dio until we get to what we were hoping to show. And one of the things we can dio is replace the DVD T with a Remember A is DBT Acceleration is the rate of change of velocity. So we've to places where we can do that. And then we're going to be looking for other ways that we can kind of clean this up a little bit. For example, we have a 1/2 right here and down the line a ways it's being multiplied by two. So those will cancel. Okay, so let's see what we have now. So in the first term, we have, um not times a Remember, we replaced the DVD tea with a times the square root of one minus B squared over C squared. Then in the second term, we have plus Oh, yes, because we have two minus. We have minus a negative, so we could make it. Plus, I am not. Tons v times another V C. We have a V here and another V here. Sometimes to be squared times a times one minus B squared over C squared to the negative. 1/2. Let's see. Did we get everything? We still have this one over C squared. We haven't gotten that yet, so we'll write that as over C squared. And that whole thing is over. One minus B squared over C squared. All right. Next. It sure would be nice if we didn't have this negative exponents in here. And if we didn't have fractions Insider fraction. So the way we can deal with that is we could multiply the top and bottom by the special form of one one minus B squared over C squared to the 1/2 power. That's the same as the square root when you say it to the 1/2 power. Okay, so we're going to distribute that through both terms of the numerator. We're going to multiply it by this term, and we're going to multiply it by the last term as well. And so when we multiply the square root by the same quantity to the 1/2 it's just like having it to the first power. It's to the 1/2 multiplied by two. The 1/2 is to the first power. So I am not a times one minus b squared over C squared. And then when we multiply these two together, add the powers and you get the zero power. So essentially they cancel. So what we have left is Plus, I am not v squared a over C squared and that is all over. One minus B squared over C squared to the three halves power. Okay, the next thing we can do is distribute RM, not a and multiply it by one and multiply it by B squared over C squared. So we have am not times a minus am not times a times B squared over C squared. Plus, I am not a times B squared times a over C squared all over. One minus B squared over C squared to the three house power. And do you notice that these two terms here are opposites so we can cancel them? So what do we have left? We have remember, this was equal to the force we have. The force equals and not times a over one minus V squared over C squared to the three halfs power. And that is exactly what we were asked to show.

Oregon State University

#### Topics

Derivatives

Differentiation

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp