# Frame of Reference

1. Sep 23, 2004

### DrWarezz

Hi all,
I'm pretty young, and thus don't understand any of the relativity theory, well; it's more of a case of "I don't understand WHY..", if you know what I mean \

Basically, going to the rule that no mass can equal/breach the speed of light, well, I'm confused here. Not about the most obvious of; "WHY?" (I'll tackle that another time I think :uhh: ), but, what do you measure an objects velocity relative to?

For example:

Planet A (Moving < way) Planet B ('still') Mass C (Moving > way)

Let's say that 'Mass C' 'took off' from planet B. And is travelling at over half the speed of light (possible). And Planet A, is moving in the opposite direction to the Mass C at over half the speed of light (possible). Well, a problem occurs here; Mass C's velocity is over the speed of light, relative to Planet A.

I understand this is a common question, but I hope that someone can explain to me, ie; what you measure the masses velocity relative to? And clear up any other 'mess'?

[r.D]

2. Sep 23, 2004

### Gonzolo

Even though A and C each move over half the speed of light in opposite directions relative to B, the speed of C relative to A is still less than the speed of light. Funky huh? Counter-intuitive, totally unobvious, yet Einstein made this totally coherent mathematically. Let us know if you want the equations.

3. Sep 23, 2004

### DrWarezz

Thanks - [r.D]

4. Sep 23, 2004

### Staff: Mentor

5. Sep 23, 2004

### Gonzolo

Let v and w be the speeds of A and C relative to B, then the speed of C relative to A (or vice-versa) is u :

$$u = \frac{v + w}{1 + \frac{vw}{c^2}}$$

Clearly, with speeds we are used to (much less than c), the denominator is nearly 1, so that what we have learned prior to relativity is still the best way to calculate in most practical cases. (don't confuse c = speed of light and C = your flying object)

Last edited by a moderator: Sep 23, 2004
6. Sep 24, 2004

### DrWarezz

Thanks, I'm still not clear though
I don't quite understand why the speed of C relative to A is still less than c

I've found a website using Doc Als' link, which explains einsteins velocity addition, but I haven't read it yet. I will later -- I'm guessing that that will explain it a bit more clearly?

Thanks alot,
[r.D]

7. Sep 24, 2004

### Gonzolo

Yup. If you insert v = 0.5c (A going half the speed of light relative to B) and w = 0.5c (C going half the speed of light relative to B), you get u = 0.8c (C going 80% the speed of light relative to A, not 100% as a simple addition would predict). Any other v and w you insert that is less than c will give u < c.

8. Sep 25, 2004

### DrWarezz

Thanks alot, Gonzolo!
Is the reason for why no object relative to any other object can breach the speed of light related to Time Dilation? If so, could you elaborate a little on how?

Thanks alot, have a good weekend,
[r.D]

9. Sep 25, 2004

### DrWarezz

Oh yes, another thing:
You say that the simple addition works with small numbers/velocities. What range of numbers would you say are 'small'? ie; What number onwards should you use this theory, etc, on?
ta - [r.D]

10. Sep 25, 2004

### Gonzolo

This is all intimately related. So is "length contraction" - a 10-ft dragster car going near c might only be 8 ft long for the spectator.

All these effects are consequences of the postulate ( = an educated guess = hyposthesis, = the basis of Special Relativity) that the speed of light c is always the same for anyone in any situation. This means that for the spectator, the speed of the light emitted from the dragster's headlights is c, not "dragster speed + c". And for the driver, it is also c.

Now suppose there is a light on the left of the dragster that sends a pulse towards a mirror on the right and comes back. For the driver, it takes a time t = carwidth/c to go back and forth. But for the spectator, the lightpulse will have moved obliquely so that it will have travelled a greater distance. Since c = d/t is a constant, the time it has taken is also greater. So for the spectator, intervals of time (the time it takes for light to go from the left to the right and back) in the car are greater than they are for the driver. Each reflexion can be considered to be the tick of a clock. This is time dilatation. I believe it has been verified experimentally with very sensitive (atomic) clocks on board aircrafts (not cars) flying for an extended period of time and comparing these to identical atomic clocks on the ground.

The speed equation is another consequence of the postulate. I won't demonstrate it now, but as you can see, putting the dragster speed w in the equation and the speed of light v = c, it will give :

$$u = \frac{c + w}{1 + \frac{cw}{c^2}} = (\frac{c + w}{1 + \frac{w}{c}})\frac{c}{c} = \frac{c + w}{c + w}c = c$$

... still c. Adding a velocity to c, still gives c, just like I said earlier.

For your second question, note the number $$\frac{vw}{c^2}$$ in the speed equation. When v and w are each 0.1 c = 30 000 000 m/s, the whole denominator becomes 1.01 (= 1 would mean regular speed additions, from the numerator). So below about 10 000 000 m/s, it's no use taking about special relativity. Experiments can be done with aircrafts only because atomic clocks are extremely sensitive.

Last edited by a moderator: Sep 26, 2004
11. Sep 26, 2004

### DrWarezz

Wicked. Thanks alot for all the help, Gonzolo.
I think I get it :D

ta,
[r.D]

12. Sep 26, 2004

### Gonzolo

That should be t = 2*carwidth/c. :grumpy: But the reasoning remains the same.

13. Sep 27, 2004

### aekanshchumber

This what theory of relativity all about. the calculation you did was in accordance to he newtonian mechenics. according to thery of relativity , on order to keep c the maximum speed the time start dilating also change occurs in distance too to keep 'c' the maximum speed.

14. Sep 27, 2004

### DrWarezz

Thanks.

Is there a reason why c = 299 792 458 ms, that we know?
Like; why not: 299 792 459 ms? (assuming it's actually not). Is there some sort of formula which explains this velocity?

Thank you.
[r.D]

15. Sep 27, 2004

### Gonzolo

Not that I know of. I explain it to myself by saying that we wrongly chose the lengths of a meter and a second. You can do all of physics with c = 1 "some other speed unit" (1 light-year per year or whatever). The physics is independant of the units.

You could say that $$c = 1/ \sqrt{\epsilon \mu}$$, but this only displaces the question to why do $$\epsilon$$ and $$\mu$$ have their own values? Physicists are usually better off using c to explain the rest of physics, rather than the other way around.

Last edited by a moderator: Sep 27, 2004
16. Sep 28, 2004

### DrWarezz

Thanks :) lol

Oh, also: This is kind of embarrassing to ask, but; What is 'mass' exactly? What differs it from matter, and what happens when an objects mass increases? \

Thanks alot,
[r.D]

17. Sep 28, 2004

### Gonzolo

Trying to answer "what is mass" can open up quite a pandora's box. Here are a few related notions, depending on which theory you use :

= related to inertia (classical, Newton etc.)
= quantity of matter (classical, Newton etc.)
= depends on the number of atoms, nucleons (this displaces the question to what is the mass of a nucleon?)
= caused by Higgs boson (particle physicist's domain)
= causes space-time curvature (general relativist's domain)

Someone else may want to answer how space-time curvature and Higg's boson are compatible, this fringes on the limits on what is observable and thus on the limits of what is solidly explainable. Whether there is an "exact" answer, I do not think so.

In special relativity, there are distincly two types of mass : rest mass and relativistic mass. Rest mass is exactly "whatever mass is that we are used to." - an electron has a rest mass of about 10E-31 kg, classically as in relativity. The relation between rest mass and relativistic mass is :

$$m = rest mass/\sqrt{1-(\frac{v}{c})^2}$$

So if you accelerate the electron to 0.9 c, its mass is about double or so. 0.99 c : triple etc. (not exactly, but you get the idea). A rough way of putting it is that instead of gaining speed, a particle gains in relativistic mass when it nears c (except within its own frame of course, where it stays = rest mass)

A photon can also be said to have relativistic mass (since $$E = mc^2$$, $$m = E/c^2 = photon energy/c^2$$). But a photon has no rest mass. Trying to find the rest mass of a photon with the first equation gives c/c in the denominator, but dividing by zero is undefined, this implies that the rest mass of something that goes c is undefined. The definition $$m = E/c^2$$ remains valid though for a photon's relativistic mass.

Last edited by a moderator: Sep 28, 2004
18. Sep 28, 2004

### Gecko

sorry, but the only way i could get .8c in your equation was to take out the $$c^2$$ from the bottom fraction. if you where to put in .5c in both v and w like you gave the example of, you would get $$u = \frac{1}{1.25}$$ which = .8 i tried using the $$c^2$$ in the bottom and couldnt get .8. i also tried using 2\2 on the bottom equation to get a similar fraction without using big numbers. like i multiplied .25 (.5 x .5) by 2 and get .5. then i squared 2 and get 4. .5\4 is the same as 1\8 which is .125 and 1\1.125 doesnt = .8. if you could explain how you get it, it'd help out alot. thanks

Last edited: Sep 28, 2004
19. Sep 28, 2004

### Gecko

never mind. i figured it out. i messed up when i tried to use 2\2. instead, i should have just used one which would have gotten me .25/1. then you would add the 1 and get 1/1.25. sorry, forget the question.

20. Sep 28, 2004

### Gonzolo

Just for fun (I'm getting use to this Latex!) :

$$u = \frac{v + w}{1 + \frac{vw}{c^2}} = (\frac{0.5c + 0.5c}{1 + \frac{0.5c * 0.5c}{c^2}}) = (\frac{c}{1 + \frac{0.25c^2}{c^2}}) = (\frac{c}{1 + 0.25}}) = \frac{1}{1.25}c = 0.8c$$

Last edited by a moderator: Sep 28, 2004