- #1

- 87

- 1

Thanks,

Ian.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Ian
- Start date

- #1

- 87

- 1

Thanks,

Ian.

- #2

- 248

- 0

1) The kinetic energy of what?

2) Why did you post this in the**general math** forum?

2) Why did you post this in the

- #3

Njorl

Science Advisor

- 267

- 13

E

p=mv

substitute and simplify to E=mc

then use the binomial theorem. The first term of the binomial expansion will be the rest energy, the next term will be the commonly used (1/2)mv

Njorl

- #4

- 248

- 0

- #5

Njorl

Science Advisor

- 267

- 13

Originally posted by suyver

I agree. I don't think there is a standard terminology for the next few terms of the expansion. I suppose they could be called the first through fifth order corrections to the classical formula for kinetic energy.

Njorl

- #6

ahrkron

Staff Emeritus

Gold Member

- 756

- 1

Originally posted by Ian

Can someone please tell me what the first six terms are in the classical expression for kinetic energy.

Thanks,

Ian.

They do not correspond to any named quantity in classical physics.

This needs not be surprising. You can write any function as a linear combination of other functions in many ways (much in the way you can use different coordinate systems for describing vectors).

It turns out that, when you expand the K.E. in powers of v/c, you gain some understanding of why things look like classical mechanics when v/c<<1, but the (infinitely many) higher-order terms do not have a wide enough use to get them a name of their own; actually, had relativity not prompted this peculiar expansion of KE in terms of powers of (v/c), those terms would probably never had shown up elsewhere.

- #7

Integral

Staff Emeritus

Science Advisor

Gold Member

- 7,212

- 56

I guess for the most part I would simply call them, ignored!

Share: