# A dark part of special relativity(at least for me)

1. Jun 7, 2009

### ShayanJ

Hi
For some days I was looking for the reason that relativity said the speed of light is the maximum speed.At last after a lot of thinking I have noticed the equations below:

limv$$\rightarrow$$c$$\frac{M_{0}}{\sqrt{1-\frac{v2}{c2}}}$$=$$\infty$$
lim m$$\rightarrow$$$$\infty$$ $$\frac{F}{m}$$=0

So if sth moved with the speed of light,It couldn't change its speed.It means that not only the speed of light is the maximum speed but also When sth reaches this speed if will move with the speed of light for ever.because acceleration that means the change in velocity is 0 so that thing can't increase nor decrease its speed.
Is my reasoning right and if yes,have you noticed about it?

2. Jun 7, 2009

### ShayanJ

And don't you have sth easier and faster than this latex?
It took me ten minutes to write this post!

3. Jun 7, 2009

### sylas

Yes, well done. It a feature of relativity that anything which moves less than light speed will always move less than light speed; and anything which moves at light speed can never move at any other speed. Another point is... particles will non-zero rest mass move less than light speed. Particles with zero rest mass move at light speed.

LaTeX is designed to allow writing production quality typesetting of mathematics. It takes a bit of getting used to, but once you know it, it's the best there is for this task. You should put the whole formulae in LaTeX; not just bits of it.

If you don't know LaTeX, you can try just writing ascii. But in the meantime, just to help, here are your formulae rewritten a bit. (Note that a superscript in LaTeX is written x^2.) Click on the formula to see the LaTeX code.

\begin{align*} & \lim_{v \to c} \frac{M_0}{\sqrt{1-\frac{v^2}{c^2}}} = \infty \\ & \lim_{m \to \infty} \frac{F}{m} = 0 \end{align*}​

Last edited: Jun 7, 2009
4. Jun 7, 2009

### HallsofIvy

Well, that happens when you don't know Latex! Practice, practice, practice!

$$\lim_{v\to c}\frac{M_{0}}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$\lim_{m\to\infty} \frac{F}{m}= 0$$
That took me only a few seconds.

Your only error is in thinking that it is possible to an object with non-zero mass to accelerate to c! Since nothing with mass can ever get to c so "what happens at c" is not relevant. It is true that light itself, which has mass 0, can neither accelerate nor decelerate and always moves at "c".

Darn! Sylas got in while I was typing!

5. Jun 7, 2009

### ShayanJ

does it have a mathematical reason that nothing with non zero rest mass can accelerate to c?

6. Jun 7, 2009

### sylas

Yes. You can't give something infinite energy.

7. Jun 7, 2009

### ShayanJ

And I have a problem with photons too.I read that they're rest mass is 0.So we have:

$$\lim_{m \rightarrow 0} a \ = \ \lim_{m \rightarrow 0} \frac{F}{m} \ = \ \infty$$

but Light has a constant speed.

8. Jun 7, 2009

### sylas

The equation for energy of a particle is
$$E^2 = (pc)^2 + (mc^2)^2$$​

The E is total energy, and p is momentum, and m is rest mass. A photon has no rest mass, so it reduces to
$$E = pc$$​

The fundamental quantities are energy and momentum. You are much better to think in terms of these than in terms of mass.

Cheers -- sylas

9. Jun 7, 2009

### ShayanJ

I still don't understand.because for getting to c you don't need an impossible acceleration.the same goes for force.so how does the energy gets infinite?

10. Jun 7, 2009

### sylas

The momentum of a particle with non-zero rest mass m and velocity v is
$$\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$$​

The total energy is
$$\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$$​

As you show in your first post, this diverges to infinite values as v approaches c.

Note that F = ma is only true at sub-relativistic velocity. But $F = \frac{dp}{dt}$ is always true. As you apply arbitrarily large forces, you can get momentum as large as you like... but never infinite.

11. Jun 7, 2009

### HallsofIvy

In saying
$$\frac{F}{0}= \infty$$
you are assuming that F is not 0. What force are you assuming is acting on light?

12. Jun 7, 2009

### HallsofIvy

Are you talking about light? No one here has said that light has infinite energy.

If you are talking about a particle with non-zero mass, then you can't get to c! That's what we have been saying. So the energy of a non-zero mass particle does not "get infinite".

13. Jun 7, 2009

### ShayanJ

I may seem stupid but as I know $$p=\gamma m_{0} v$$ and $$m_{0}$$ is the rest mass.if it is,so pc must become 0,too.Hey its really confusing!

14. Jun 7, 2009

### sylas

You can't use that formula for a massless particle, because γ is undefined. (It diverges to infinite.) So you are multiplying a zero by an undefined infinite value.

For a massless particle, the velocity is c, and the momentum is just E/c, where E is its energy.

15. Jun 7, 2009

### ShayanJ

Is it the planck momentum?

16. Jun 7, 2009

### DrGreg

The equation

$$F = ma$$​

isn't valid in relativity. The correct version is

$$F = \frac{dp}{dt} = \frac{d}{dt}\left( \frac{mv}{\sqrt{1 - v^2/c^2}} \right)$$​

where m is non-zero rest mass.

Unless I've made a mistake, this works out to be

$$F = \frac {ma} {(1 - v^2/c^2)^{3/2}}$$​

So, if m and a are constant, the force really does diverge to infinity. (In practice, it's more likely F and m would be constant, so a would converge to zero.)

Last edited: Jun 7, 2009
17. Jun 7, 2009

### sylas

I'm not familiar with that term... but the energy of a photon is hf and its momentum is hf/c, where f is the frequency.

18. Jun 7, 2009

### ShayanJ

19. Jun 7, 2009

### sylas

Ah. Thank you. That is a unit of momentum in the Planck system of units. It's like an alternative to metric for ubernerds.

I'm not presuming any particular system of units; just giving the general laws which apply, regardless of what units you like.

20. Jun 7, 2009

### Cyosis

Slight correction:
$$F=\frac{d}{dt}(\gamma m v)=\gamma m a +m v \frac{d \gamma}{dt}=\left(\frac{v^2}{c^2}\frac{1}{(1-\frac{v^2}{c^2})^{3/2}}+\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\right)m a$$

Either way the conclusion remains the same.

21. Jun 7, 2009

### DrGreg

Actually

$$\frac{v^2/c^2}{1-v^2/c^2} + 1 = \frac{1}{1-v^2/c^2}$$​

so we are in agreement!

(l left out the intermediate steps in my calculation and just gave the final answer.)

22. Jun 7, 2009

### sylas

Keep going.
\begin{align*} \frac{v^2}{c^2}\frac{1}{(1-\frac{v^2}{c^2})^{3/2}}+\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} &= \frac{\frac{v^2}{c^2}}{(1-\frac{v^2}{c^2})^{3/2}}+\frac{1-\frac{v^2}{c^2}}{(1-\frac{v^2}{c^2})^{3/2}} \\ &= \frac{1}{(1-\frac{v^2}{c^2})^{3/2}} \end{align*}​

PS. By the way, I believe the quantity
$$\frac{a}{(1-\frac{v^2}{c^2})^{3/2}} = \gamma^3 a$$​
is called proper acceleration, and is an invariant... corresponding the acceleration experienced by the moving body in its accelerated reference frame.

23. Jun 7, 2009

### Cyosis

Ugh, it looks like I am particular sharp tonight!

I understood that, but I thought you had forgotten to apply the product rule, silly me. Anyway it's good we agree in the end!

24. Jun 7, 2009

### ShayanJ

You know,I just read it in wikipedia and I haven't noticed that its a unit.

25. Jun 7, 2009

### DrGreg

(As a minor aside, I actually worked it out by doing the substitution $v = c \tanh \phi$, which might ring a bell with you from another thread!)