# A noob question about relativity.

1. Mar 1, 2008

### sei6i

Hey everyone. This is my first post, and I apologize if my question is too stupid for this forum.

I bought Steven Hawking's book a few years ago but was too lazy to read it up till now.
A few pages later I read that speed of light is constant in in every frame of reference( sorry for the non-physics English). I decided to stop reading until I understand that fully and I imagined a few examples. I want to share them with you and ask you if I'm getting this or if I'm completely off. Hope I'm not too boring.

Say I and my friend are on spaceships next to each other. 6 lightseconds away (same distance from both spaceships) someone decides to send a Ben Affleck picture via wireless. Now a picture wont upset my friend, so he doesn't do anything,
but It will upset me so I decide to run away from the event in the opposite direction.
If light moves with a constant speed for any observer does this mean
that no matter how fast I move Ben Affleck will always get to me in 6 seconds.

If this is true then suppose an event happens in all directions, like a star exploding. Does this mean that the time when the waves from that star wipe out everyone depends on, if you look at it this way, the distance they were away from the star at the moment of explosion,
no matter how fast they are moving. Does this apply for change in trajectory or acceleration? Sounds pretty counter-intuitive so I'm probably getting everything totally wrong.

Maybe my example is flawed. So please tell me if I got the absolute basics at least.
Train coming towards me flashes its light in my direction. For me the speed of light is C, not C+(train speed), right?

2. Mar 1, 2008

### Staff: Mentor

I think you need to keep reading. You don't have the entire story yet. If you are moving relative to your friend, the distance is not a constant 6 light-seconds, so the time it takes light to traverse the distance will be greater than 6 seconds. That does not mean you would measure the speed of light to be variable. The distance and time are variable in a ratio that keeps the speed of light the same.

3. Mar 1, 2008

### CompuChip

If you are running away from the event, it will still take the light from that event 6 seconds to get to where you were before you started running. Once it got there, you will have moved from that position, so it will still have to travel a bit farther to reach you.

I don't know if this will help, but I have the feeling I can best explain it mathematically (since I don't know how well you are at mathematics, I'll try to keep the formulas to a minimum). Your friend is 6 light seconds (ls) away from the event, so the light -- travelling at 1 lightsecond per second (ls / s) will reach him in a time
$$t = \frac{6 \ ls}{1 \ ls/s} = 6 \ s$$
This is just the basic physics formula: time equals distance divided by velocity.
Now suppose you are running away at a constant speed v and the light takes a time t to reach you. Let me apply that same formula again, but this time, the distance the light has to cross is not (6 ls), but (6 ls + v t) because in the time t the light was travelling, you have moved (v t) away from it. The velocity of the light, on the other hand, is not, as you may expect, (1 ls/s + v) but it is still (1 ls/s) (that's the principle of relativity at work for you). So the formula for the time now gives
$$t = \frac{6 \ ls + v t }{1 \ ls/s}$$
If you solve this formula for t, you will get
$$t = \frac{6 ls}{1 ls/s - v}$$
and since the denominator is now smaller than 1 ls/s (for example, if you are running away at one quarter the speed of light: v = 1/4 ls/s, it will be 3/4 ls/s) the time will be larger than just 6 seconds (if v = 1/4 ls/s, it will be 8 seconds); also: if your velocity approaches the speed of light, it will take the light longer and longer to overtake you (but as long as you run slower than the light bundle, it will reach you). If you could run at exactly the speed of light (which is impossible, by the way) the light would always stay 6 lightseconds behind you).

4. Mar 1, 2008

### _Mayday_

What book is this? I've just read a very simple book aimed at newcomers to physics. It's called, Blackholes, wormholes and time machines, it was a nice simple introduction to this part of physics. I am 17, would this book the OP has suggested be easy enough for me to comprehend, or would there be better, maybe easier books to understand that I should read before?

_Mayday_

5. Mar 1, 2008

### malawi_glenn

Hawking -
A Brief History of Time

6. Mar 1, 2008

### sei6i

I'm majoring ship navigation in college so we are studying a ridiculous amount of mathematics and mechanics, but ships are too slow to care about light speed so noone bothers to teach relativity.

But from what you explained the picture is no different from a Newtonian mechanic's equasions.
In other words, this is how my stationary friend believes i saw the signal: 8 seconds, which would also be true for me, but with different parameters.

What I first understood was that from my reference frame light would travel with C and reach me in 6 seconds since it was 6l/s away when the event started.

From what you wrote I make out that the result would still be like Newtonian physics only multiplied on a constant that would make light speed C and thus slowing time down so the event for didn't just seem like occurring 8 seconds before, it actually did.
(the constant being the time slowage and space shrinkage that is true for this experiment only)

Am I on the right track here?

Last edited: Mar 1, 2008
7. Mar 1, 2008

### _Mayday_

Thanks Malawi, is it quite difficult to read? Does it start from the basics and build up? I am interested int he topic, but if the book is far too difficult I do not want to scare myself. :tongue:

Goodluck with your exams next week! If I don't speak to you before or see you in chat, hope it all goes well! =]

8. Mar 1, 2008

### sei6i

The book has only one formula and is targeted for the average Joe in physics.

I don't know if you got the relativity concept, and sorry for posting a useless link if you did, but this youtube vid was pretty helpful to me.

This guy helped me get a better idea on how a fast moving object sees the outer world:

Last edited by a moderator: Sep 25, 2014
9. Mar 1, 2008

### tiny-tim

Hey sei6i! Welcome to PF!

Yes, that's the right track. With problems like this, work out time and distance for the stationary observer first (which is easiest), and then apply the Lorentz transformations.

(btw, cool guys call them "time dilation", "length contraction", and the "gamma factor" or just "$$\gamma$$" )

10. Mar 2, 2008

### CompuChip

There are some difference, but at least they will look quite similar. Actually, this is precisely what one would expect, considering the fact that Newtonian mechanics is just a special case of special relativity. More rigorously, if you take the mathematical limit of a special relativistic formula for velocities being much smaller than the speed of light (e.g. v/C -> 0) you will recover the Newtonian formula's.

On your clock, you would measure 8 seconds. On his clock, he would measure 8 seconds. The "problem" is what he would measure on your clock

Actually, it's not really a constant but a function of velocity. If the velocity is very small (as in, approximately up to 0.5 C) its value is almost equal to 1 and nothing changes.