Why Does 1/2mγv^2 Fail to Accurately Calculate Relativistic Kinetic Energy?

[Ascendant78]

Homework Statement

Show that 1/2mγv^2 does not give the correct kinetic energy.

Homework Equations

1/2mγv^2

γ = 1/(sqrt(1-v^2/(c^2)))

The Attempt at a Solution

Well, since the classical mechanics version of kinetic energy was the integral of momentum with respect to v, I felt I could simply explain that the v^2 in the denominator of γ must be integrated as well. However, the integration that I get from applying this to the equation is different than the actual equation for relativistic kinetic energy.

I'm assuming the issue is when it comes to relativistic kinetic energy, you cannot integrate the relativistic momentum with respect to v and get the correct answer, but can someone tell me why and a better approach to showing why their equation of 1/2mγv^2 is invalid?

 

[collinsmark]

Homework Statement

Show that 1/2mγv^2 does not give the correct kinetic energy.

Homework Equations

1/2mγv^2

γ = 1/(sqrt(1-v^2/(c^2)))

The Attempt at a Solution

Well, since the classical mechanics version of kinetic energy was the integral of momentum with respect to v, I felt I could simply explain that the v^2 in the denominator of γ must be integrated as well. However, the integration that I get from applying this to the equation is different than the actual equation for relativistic kinetic energy.

I'm assuming the issue is when it comes to relativistic kinetic energy, you cannot integrate the relativistic momentum with respect to v and get the correct answer, but can someone tell me why and a better approach to showing why their equation of 1/2mγv^2 is invalid?

Do you have to start at some specific starting point in your calculations?

I mean, you know that the correct kinetic energy is $\left( \gamma - 1 \right) mc^2$. Is it acceptable to simply show that $\frac{1}{2}\gamma mv^2$ is not equal to that?

 

[Ascendant78]

Do you have to start at some specific starting point in your calculations?

I mean, you know that the correct kinetic energy is $\left( \gamma - 1 \right) mc^2$. Is it acceptable to simply show that $\frac{1}{2}\gamma mv^2$ is not equal to that?

I think they are looking for some type of derivation, but the chapter that came before this problem didn't give us anything like that to work with. The only thing I could come up with was integrating the momentum with respect to v. Other than that, I really haven't been able to come up with much of anything else. How else can we derive relativistic kinetic energy?

 

[collinsmark]

I think they are looking for some type of derivation, but the chapter that came before this problem didn't give us anything like that to work with. The only thing I could come up with was integrating the momentum with respect to v. Other than that, I really haven't been able to come up with much of anything else. How else can we derive relativistic kinetic energy?

It's quite a bit tougher to derive the $K.E. = \left( \gamma - 1 \right) mc^2$, rather that just prove that it's not equal to $\frac{1}{2}\gamma mv^2$.

Are you sure you are suppose to derive the correct formula? If the problem is only asking you to show that it's not equal to $\frac{1}{2}\gamma mv^2$, that would be a lot easier than deriving it.

But if you are going to derive it, I might suggest first deriving the relativistic 3-force, by taking the time derivative of the 3-momentum:

$$\vec f = \frac{d}{dt} \left\{ \gamma m \vec v \right\}$$

and note that both $\gamma$ and $\vec v$ are functions of $t$, so you'll have to use the chain rule to take the derivative.

And then noting (similar to the Work-Energy theorem)

$$\vec f \cdot \vec v = \frac{d (K.E.)}{dt}$$

and integrate that.

It's not a simple calculation and there may be tricks that I left out of my outline. It's possible, but not particularly easy.

 

[Ascendant78]

It's quite a bit tougher to derive the $K.E. = \left( \gamma - 1 \right) mc^2$, rather that just prove that it's not equal to $\frac{1}{2}\gamma mv^2$.

Are you sure you are suppose to derive the correct formula? If the problem is only asking you to show that it's not equal to $\frac{1}{2}\gamma mv^2$, that would be a lot easier than deriving it.

But if you are going to derive it, I might suggest first deriving the relativistic 3-force, by taking the time derivative of the 3-momentum:

$$\vec f = \frac{d}{dt} \left\{ \gamma m \vec v \right\}$$

and note that both $\gamma$ and $\vec v$ are functions of $t$, so you'll have to use the chain rule to take the derivative.

And then noting (similar to the Work-Energy theorem)

$$\vec f \cdot \vec v = \frac{d (K.E.)}{dt}$$

and integrate that.

It's not a simple calculation and there may be tricks that I left out of my outline. It's possible, but not particularly easy.

Thanks for the information. From what you have said, I think I am over-complicating the problem and that they are just looking for something as simple as us comparing the two formulas to see that they are not equal. However, I do like your suggestion here and am going to try it out later just for practice to see what I can get.

 

[collinsmark]

However, I do like your suggestion here and am going to try it out later just for practice to see what I can get.

Great! Be prepared for some trickiness though.

One of the things you'll start with evaluating what $\dot \gamma$ is.

$$\dot \gamma = = \frac{d}{dt} \left\{ \gamma \right\} = \frac{d}{dt} \left\{ \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \right\}$$

and note that

$$\frac{d}{dt} \left\{ \vec v \right\} = \vec a$$

where $\vec a$ is the 3-acceleration.

In the process you'll come across things like:

$$\frac{d}{dt} \left\{ \frac{v^2}{c^2} \right\} = \frac{2 \left( \vec v \cdot \vec a \right)}{c^2}$$

where you start with all scalars but the derivatives involve vectors and dot products of vectors.

All of those instances of the acceleration 3-vectors, $\vec a$ fall out through the magic of algebra, before the end of the derivation. As a matter of fact, it's very rewarding when they do, and it's a very rewarding derivation overall. But be prepared that in the interim, things can get, shall we say, "tricky."