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I'm trying to understand Special Theory of Relativity through reading Feynman's lectures. In chapter 15 Feynman gives example of clock: rod of 1m length with mirrors at the ends. Light goes from first mirror to second, reflects to first mirror. I understand for this example why observer on the earth and observer in a moving spaceship with this type of clock would observe different time. Feynman also says for every clock in a moving spaceship obesrver on the spaceship would observe the same time, or in other way he would be able to understand that he is moving and find his velocity. So, I hope that it will help me to understand.

For example, I'm moving on platform with speed v in X direction, and I throwing a ball directly in Y direction(up) with speed u. Observer standing at the groung will see ball with speed (v;u). For me it would be just (0;u). Second example. I throwing a ball directly in X direction(moving on same platform with speed v) with speed u. Observer and the ground will notice speed of v + u for ball. I'll notice u.

The thing I dont understand here is following: if I replace ball with light source then, according to what I read in Feynman's lectures it wont recieve any impact from moving platforme, it will not recieve for c + v. Just c. For every example. Why?

Second question, returning to clocks... Let's say we got clock that works in this way: some machine that throwing a ball up, and when ball falls down and hit his initial state, this is how we define a 1 second. Obviously this clock need a force of gravity to work, so let's say that this machine also creates this force in some amazing way(in other words ball always affected by some force mass_of_ball * acceleration).

And we took this clock with that lihgt clock on spaceship. We know how(according to Feynman) light clocks get working slowly while moving, comparing to rest state. I cant figure out how ball-clock time will fit that time with light-clock. I cant see why ball clock get slowly while moving. Ball just recieves speed in other direction, that dont affect his falling.

Sorry for mess in thoughts and language.

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First of all, for your first question - why the speed of light always 'c', no matter where we are (either on the moving body or on any rest 'ground'). I should say this is the result of experiment, not imagination or assumption of Einstein, Feynman, or anybody else. The famous experiment is called Michelson-Morley experiment, you can refer to the Wikipage for detailed introduction of the experiment: Micheson-Morley Experiment. The basic idea of this experiment is: assuming the speed of light 'c' is specifically relative to something (historically, the 'something' here was called aether, from what I remember), then we know our earth is rotating around the self-axis, thus the speed of light, of course, is NOT 'c', if our previous assumption ('c' is specifically relative to aether) stands. Then using the facility designed by Michelson and Morley, we should observe some corresponding effect, which is the result of the changing of speed of light relative to our earth. I won't bother with what the effect will be like (detailed information can be found in the link at the bottom), but I can tell the result: no expected effect was observed if our previous assumption stands! That's it, it seems that we have to accept the experimental FACT that speed of light is always 'c' - relative to the observer on spacecraft, or relative to our earth, or relative to anything moving on the earth, or, or, relative to ANYTHING in the universe. Actually, it was just the result of the famous Michelson-Morley experiment (and some other following experiments - see the link below) that directly leads to the invention of the special theory of relativity. Moreover, the invariance of speed of light is one of the two basic principles of special theory of relativity.

Then for the second question, I will try to explain it, but maybe not direct answer to your question (sorry about that, I really cannot think of direct way of explanation to help you). When we are talking about TIME, we actually mean the interval (along time axis) between two EVENTS. So thinking about your problem, where we have the ball rising up and falling down. So the two events, in this case, are 'ball leaves the surface - starting to rise up' and 'ball comes back to the surface - finishing falling down'. Then two observers - one moves together with the machine you mentioned and the other one stands on the rest 'ground' - observe exactly the same two events, are the time (interval) measured by these two guys going to be different? The answer is: Yes. Why? Why? Why? Well, again, I should say this is what we have to accept as the FACT. However, it's worth a bit explanation, and now we need to go back to the experimental FACT - the speed of light does not change from frame to frame (ignoring the influence of medium, e.g. from air to water, and all the other effects). It is based on this FACT that people (Lorentz should be one of them, I think) obtained the invariant quantity called space-time interval for ANY frames. The definition is:

$s^2 = -c^2(\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$

To better express my idea (and also help your understanding, I hope), I will give an example as following:

enter image description here

In the figure, there are two observers in one-dimension frames - one is standing on the ground, and the other is moving relative to the ground with the speed of v. Let's call the axis x-axis, and the frame for observer-1 and observer-2 is named frame-1 (for which we use the notion $x$ and $t$) and frame-2 (for which we use the notion of $x^{'}$ and $t^{'}$), respectively. Moreover, both observer-1 and observer-2 stays at their own origins. Now, let's imagine two bulb emitting light (as shown in the figure) one after another. Right at the beginning, bulb-1 emits light at the origin of frame-1, and just at that moment, observer-2 is also at the origin of frame-1 (which means the origins of frame-1 and frame-2 coincides with each other). Then we record the position and time for the event of bulb-1 emitting light, in two frames. For frame-1, it is: $x_0 = 0,\ t_0 = 0$, and for frame-2, it is: $x^{'} = 0,\ t^{'} = 0$. Then after some moments, observer-2 (remember? observer-2 is moving with speed of v, in frame-1, or as seen from observer-1) arrives at the position where bulb-2 is placed in frame-1, and JUST AT THAT EXACT MOMENT, bulb-2 emits light. Then we write down the position and time for the event of bulb-2 emitting light. For frame-1, it is: $x_1 = x,\ t_1 = t$, and for frame-2, it is: $x_1^{'} = 0,\ t_1^{'} = t^{'}$. You guess what? The two events happens at the same place as seen by observer-2 (or we say, in frame-2)!

So now, we have two events, which are analogue to the two events in your question as I mentioned above. What we suspect is: is $t$ (the time interval for these two events observed in frame-1, or by observer-1) and 't^{'}' (the time interval for these two events observed in frame-2, or by observer-2), the same? Or different? Let's have a look. As is already given above, we have the invariant quantity s from frame to frame, thus we have:

$-c^2t^2 + x^2 = -c^2t^{'2} + 0$

What else do we have? Remember observer-2 arrives at bulb-2 at the exact moment when bulb-2 emits light? So definitely we should have:

$vt = x$

By replacing x into the previous equation, we have:

$-c^2t^2 + v^2t^2 = -c^2t^{'2}$

Rearranging the above equation, it is easy to get:

$t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}t^{'} = \gamma t^{'}$

So, so, so, finally, without assuming anything except accepting the FACT that speed of light doesn't change from frame to frame (based on which we then have the invariant quantity space-time interval), we obtain the result that for the two events that we cannot visually 'feel' any difference when changing the observing frame, the time interval observed in two different frames (one is moving relative to another), is indeed different! - They are linked up by the so called $\gamma$ factor, as you can see from the above formula.

What else can we say? Well, I should say, the difference of time interval observed in different frames (moving relative to each other) between two events, is some kind of PROPERTY of our space, and time. I am afraid, we have to accept it, maybe, no other choice.

OK, that's my personal explanation for your question, which, I hope, could help you. And sorry for any ambiguity of language or typos.

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  • $\begingroup$ Thank you! It gives insight about such thing as time and other things to think about. Kind of sad though that some things we should take on faith. For now. $\endgroup$ – Il'ya Zhenin Jan 19 '15 at 18:40
  • $\begingroup$ Maybe there is some deep reason for all such kind of peculiar behaviour of light. However for me, this is what I can see, for now. I will really appreciate if someone could give an idea about what's hidden behind such kind of behaviour of light.:) $\endgroup$ – Ai-Peng1990 Jan 19 '15 at 18:59
  • $\begingroup$ If you dont mind, I'll ask something else that appeard in my head a day ago :) Speed of light is constant in every inertial system in every experiment because when we trying to measure speed of light with our tool they(tools) get slowed and kind of shrinked(not sure about that) and this portion of slowing is just enough to "maintain" speed of light constant. And second thing. In a formula of Lorenz transformation for time is $c$. It isn't actually speed of light, it is a maximal speed "allowed" in our Universe, it just happend to be equal speed of light? $\endgroup$ – Il'ya Zhenin Jan 23 '15 at 13:19
  • $\begingroup$ Sorry for the late reply. First of all, I don't think the experiment result of speed of light does not change is something to do with some kind of shrinking. If we have to say something is strange, then I should say the only thing that is strange is light itself. Just a few days ago, BBC reported that Scottish physicists successfully obtained speed of light less than c, in vacuum. So, light is indeed strange, isn't it? $\endgroup$ – Ai-Peng1990 Feb 2 '15 at 21:51
  • $\begingroup$ For the second question, from what I learned about relativity theory, I should say the the 'c' in Lorentz transformation is the speed of light, but not the imagined some kind of maximum 'allowed' in the universe. Why? Let's imagine a simple example: I sit in a stadium watching a football match, you are somewhere else flying. Both of us have the right to say I (or you) am not moving, you (or I) are moving. Now if from my point of view you are flying with speed of c, what will happen? If I see someone scores at some moment, what does that mean to you? Nothing happens! $\endgroup$ – Ai-Peng1990 Feb 2 '15 at 22:12

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