# Mass growing with speed

1. May 30, 2017

### Karol

1. The problem statement, all variables and given/known data
Fast electrons in synchrotron approach the speed of light:

I don't get to the "one part in 8,000,000"

2. Relevant equations
Momentum: $~P=mv=\frac{m_0v}{\sqrt{1-v^2/c^2}}$

3. The attempt at a solution
$$\frac{v^2}{c^2}=1-\frac{1}{4E6}~\rightarrow~\frac{v}{c}=\frac{1}{1.000,000,5}$$
$$\rightarrow~v=\frac{c}{1.000,000,5}$$
$$c-v=c\left( 1-\frac{1}{1.000,000,5} \right) \neq c\cdot \frac{1}{8,000,000}$$

2. May 30, 2017

### scottdave

1 part in 8 million is 1/8000000. But since it differs from 1 by this much, you have 7,999,999 / 8,000,000

3. May 30, 2017

### haruspex

How do you get that? What is the Taylor expansion of (1-¼10-6)?

4. May 31, 2017

### Karol

I executed $~1-\frac{1}{4E6}$
What do you mean by (1-¼1016)? maybe you made a mistake, do you mean the Taylor expansion for $~\sqrt{1-\frac{1}{4E6}}~$?
Taylor expansion:
$$f(x+h)=f(x)+hf'(x)+\frac{h^2}{2!}f''(x)+...$$
It has a function f(x) in it and i don't, i just have numbers, so how will i take derivative?

5. May 31, 2017

### haruspex

Yes, sorry, the - in the exponent was a typo. Should have been (1-¼10-6)½. And I meant binomial expansion. Other than that....

6. May 31, 2017

### scottdave

I am not sure why you need Taylor Series to approximate it. You could do this: let r = (v/c). Then we have (1 - r^2) = (1 + r)(1 - r).
Since v differs from c by 1 part in 8 million, then (1 + r) is approx 2 {since r is approx 1}, and (1 - r) is 1 divided by 8 million. So you have: (1 - r^2) is approx 2/(8 million) = 1/(4 million).
The square root of that is 1/2000. You could run this in reverse, if you know the 2000 and want to determine the 1/(8 million).

Last edited: May 31, 2017
7. May 31, 2017

### hilbert2

I've seen the Taylor polynomial approximation used in quantum mechanics if there's a need to include the relativistic "mass increase" in the form of a lowest order perturbation in the Schrödinger equation, but I'm not sure why that is done here.

8. May 31, 2017

### PeroK

This is all based on, for small $\epsilon$:

$\frac{1}{1 \pm \epsilon} = 1 \mp \epsilon$

And

$(1 + \epsilon)^2 = 1 + 2\epsilon$

In particular, if $\frac{v}{c} = 1- \epsilon$, then $\frac{v^2}{c^2} = 1- 2\epsilon$ and $1 - \frac{v^2}{c^2} = 2\epsilon$

9. May 31, 2017

### Ray Vickson

The positive solution of $1/\sqrt{1-\beta^2} = 2000$ is $\beta = \sqrt{3,999,999}/2000$, so $1 - \beta = 0.1250 \times 10^{-6} = 1/8,000,000.$

Actually, that last should be $1/7,999,999.500000031250$, approximately

Last edited: May 31, 2017
10. May 31, 2017

### Karol

But in binomial expansion the exponents are positive integers
Why?
$$c-v=\frac{1}{8E6},~1-r=1-\frac{v}{c}=\frac{c-v}{c}=\frac{1/8E6}{c}=\frac{1}{c\cdot 8E6}$$
But you don't use $\frac{1}{1 \pm \epsilon} = 1 \mp \epsilon$, you logically and correctly say $\frac{v}{c} = 1- \epsilon$
$$\frac{v}{c} = 1- \epsilon~\rightarrow~\frac{v^2}{c^2}=(1-\epsilon)^2=1-2\epsilon+\epsilon^2 \approx 1-2\epsilon$$
$$1-\frac{v^2}{c^2}=\frac{1}{4E6}=2\epsilon~\rightarrow~\epsilon=\frac{1}{8E6}$$
$$\frac{v}{c}=1-\frac{1}{8E6},~~c-v=\frac{c}{8E8}$$

11. May 31, 2017

### PeroK

That looks right to me.

12. May 31, 2017

### scottdave

You should notice something that when you see: c - v = 1/(8 E6). On the left side, you have speed (Length/Time), and the right side is dimensionless. So something didn't get put in, correctly. Now with this: c-v = c/(8 E6), both sides have the same dimensions.

13. May 31, 2017

### Karol

c-v = c/(8 E6) has correct dimensions and, as PeroK said, is probably the correct answer but how do you explain "v differs from c by one part in 8,000,000"?
I expected to see c-v=1/8,000,000

14. May 31, 2017

### PeroK

Well, you expected wrong. That equation is dimensionally nonsense.

15. May 31, 2017

### Karol

How did you get that?
$$\frac{1}{\sqrt{1-\beta^2}}=2000~\rightarrow~\sqrt{1-\beta^2}=0.0005~\rightarrow~\beta=0.999,999,875$$

16. May 31, 2017

### haruspex

Which differs from 1 by one part in 8 million.

17. May 31, 2017

### haruspex

Please see my correction in post #5.

18. May 31, 2017

### Ray Vickson

I am lazy in my old age, so I just let Maple solve the problem for me, and then evaluated the solution using 20-digit numerical computations.

Of course, the solution of $1/\sqrt{1-\beta^2} = \gamma$ is $\beta = \sqrt{1-\gamma^{-2}},$ which expands out as
$$\beta = 1 -\frac{1}{2 \gamma^2} -\frac{1}{8 \gamma^4} - \cdots,$$
by applying the series expansion for $\sqrt{1-x}$ for small $x = 1/\gamma^2.$ If you keep only the $1/\gamma^2$ term you get exactly 1 part in 8,000,000 because we are using $\gamma = 2000.$

19. May 31, 2017

### scottdave

What if it said something like v differed from c by 1%, would it be clear that v = 0.99c ?
Instead of 1% they could say v differs from c by 1 part in 100. It means the same thing.

So instead of 1 part in 8million difference, they could say differ by 0.0000125%

20. Jun 1, 2017

### Karol

Thank you very much scottdave, PeroK, Ray, Hllbert2 and Haruspex