# A newbie question on time dilation formula

1. Apr 22, 2004

### Xiox

Hi, I'm new to this forum, and I don't really consider myself as "worfy" to post here... and nor do I try to claim that Einstein was wrong (I would certainly slap myself if I did)

Anyway, I have currently finished to read Relativity by Einstein (I heard that is not he best book on that subject, though), and I'm wondering about the time dilation formula. When K' moves with a constant velocity v relatively to K, and when light is travelling along the y'-axis from origo, how does it reach a particular point L on it. I mean doesn't light move perpendiculary to y-axis and not along the x'-axis (so that when I stand on the origo of K' I don't see light reach L) when an observer who stands on the origo of K'? I mean doesn't light travell perpendiculary to the x-axis but not along the y'-axis. I'm sorry but I have wicked imagination.

I am thinking from K, don't I?

Last edited: Apr 22, 2004
2. Apr 22, 2004

### Severian596

whoa...I've examined plenty of spacetime problems, but this one is the most confusing. What is your question? Please clearly state that, and please clearly explain which frame of reference you're asking the question from, then explain the situation. I (and perhaps some others here) will do my best to respond.

3. Apr 22, 2004

### Xiox

I am sorry to confuse you...

(Note that I already presume that the statement bellow is a fallacious.)

I was thinking (on one of those nights) about the Lorentz transformations (url: http://cmtw.harvard.edu/Courses/Phys16/l1_latex/l1_latex.html). Then, we have two reference frames, K, the one that is unprimed, and K' the primed one that is moving with a constant velocity v relatively to K. At the time t = 0 (I tell this rather for myself than for you) all of the axis of K and K' (the two "Cartesian co-ordinate systems") interfere with each other. We have a flash-bulb at the origo of K', so it is pointed upwords at a particular point L' on the y'-axis where a mirror is situated, which points at the origo of K'.

When K' moves relatively to K, the flush-bulb is emitting light, so it then travells a distance d' (2L) with the velocity of light c; the time being t' = 2L/c (again for myself...). This is seen from an observer situated at the origo of K'.

The question is strangely enough not about that is seen from K on K', but that is seen from the observer from K', from K' (and more wicked does it get). So, perceptibly, I've already answered to my own question (...), but still I don't quite dig it. If I stand on the origo of K' and see the light, doesn't the light that is travelling from the origo move along the y'-axis? (Note this is really not a question, and you people are not my only victims.) I mean, since the light is travelling parallell (1) to the y'-axis, shouldn't it not travell along the x'-axis - wrong, yes? So, finally, my question is: Is it not so that the light is merely another frame of reference, like the same action from K set is? The first thing I thought about was that, accoring the principle of GR, a train can both move in a tunnel relatively to the tunnel as well as the tunnel can move reciprocally from the point of view of the train. But, (besides that I can't quite believe that I'm still writting this) cannot the light be resembled with the referece of K to K'?

(1): I wrote "parallell" but the question - my primary goal on my clasess is to delude teaches - is palpably: Does the light travell along the y'-axis? (Doesn't it, then, start from the origo of K' and "bend" towards the negative x'-area and postive y'-area? I am thinking that K' is moving and that the light is not, not with the positive co-ordinates of x', that is) As may be stated the answer is obviously not, but why? (And it is here supposed that some of you folks are writting the famous "that's-how-simple-tis"-answers..)

Once again I'm sorry, I haven't read so much more than "Relativity" and wouldn't mind to see you recommending some other books, perhaps about GR?

Answer: well... I just got it, since the light is emitted while K' is moving, it is also keeping its intertia along with K'... ("constant velocities", wasn't it? How cruell I am, making you read all this!)

Last edited by a moderator: Apr 20, 2017
4. Apr 22, 2004

### Severian596

Holy crap, Xiox! Okay a pointer for you: for my sake, try to complete one thought each sentence instead of including parenthesis for every tangent thought that crosses your mind while you type. But if I understood you, you're asking the following:

With respect to an observer at the origin of K', why doesn't light that's directed parallel to the y' axis "drag" toward the negative x' direction between the light's point of origin and some destination point P on y'?

Common sense says 'no' because let's says K' is a car and someone at the origin tosses a ball straight up. There's no apparent "drag" unless the ball encounters air resistance. However this is using Newtonian velocity addition, which doesn't apply to light (as its speed should be the same no matter what speed the source travels at). Is this what you're wondering about?

Last edited: Apr 22, 2004
5. Apr 22, 2004

### Severian596

You don't have to apologize. If I didn't want to read it I wouldn't!! :)

I recommend NOT reading anything about GR until you master SR. GR is a superset of special relativity. It doesn't make sense trying to understand the superset before you understand the subset.

I'm happy to recommend this excellent publication by David W. Hogg: http://physics.nyu.edu/hogg/sr/sr.pdf

6. Apr 22, 2004

### Xiox

I am happy as long as I can elaborate my thoughts and understand myself.

It seems that my apprently copiusly exhuberated phlegmatic brain is unble to consentrate at this point. But, alas, it seems that you have slightly missunderstood me. Nope, I didn't wonder about the constacy of velocity of the light, bur rather the exact position of of the reflection. But you did understand the question correctly! Although I am still a little confused...

So, the light-beam is travelling along y' while K' is moving... and the velocity of light is independent of the state of the motion of the condition of the body of reference... and now it hits me that the light is not "bended" with reference to K', if the light is a straight line and y' = 0 and L' = 0, L' being an arbitrary point on y'-axis, form another line, then firing of the antecedent one... with the condition that if the two lines' points coincide.

Still... () if the poing L' on y' in K' is moving faster than light as light is simultaniusly emitted from the origo, then it is possible for the reflection not to occur (hmm?), but I just was merely ruminating that since v < c, then the reflection wouldn't conceivably seem to be quite "exact" without the condition v = c, which is of course impossible.

Thanx! As you may have noticed questioning things is quite a way for me to learn, and unfortunately I can scarcely ask such question to my friends in the school.

Last edited: Apr 22, 2004
7. Apr 22, 2004

### Severian596

What time is it wherever you are? Are you up late on a LOT of coffee?

You said some things that didn't make sense to me, like y' = 0. But I think I'm just tuning into your ramblings. If you're answering your own questions through thought experiment and you're happy with what you learn, it's cool with me. I'll stay tuned if you have a question.