The length and width of the box are not different frames. It is possible to explain some scenarios using selected parts of relativity, for simplicity's sake. However if you want to be rigorous, the best way is to apply the Lorentz transform - this means doing some maths! This also means not discussing time dilation and length contraction individually any more, since both are quite selective aspects of the Lorentz transform.
I'm not sure what you're getting at here... Are you disputing that the distance travelled by the light beam is longer in our frame than in the box frame?
I don't have a problem with that at all .....nice diagram too. I think what I had in mind was some sort of photo sensitive crystal that measured 'c' over a constant distance with in the crystal. The crystal would become the ship or clocks absolute reference for time and compared to an atomic clock. ( sciuence fiction I guess) a solar power cell for instance could be reconfigured to give a velocity reading in dilated conditions. the assumption of 299792kps [et] flat distance would be a fixed figure not unlike the figure "pi"
I think my comment was ill-advised and unecessarilly complicates the issue at hand so please disregard it for the present.
Lorentz transform treatment of light clock: The x dimension is parallel to the length of the box. The y dimension is parallel to the width of the box. The z dimension is parallel to the height of the box. Distance units are metres. Time units are 1/299792000 seconds. Let us use the emission of the light as our origin event - In all frames, the light is emitted at: x=0, y=0, z=0, t=0 The event of interest is that of the light being received at the sensor. In the box frame: The light is detected at x=0 (the same point in the length dimension) y=1 (1 meter across in the width dimension) z=0 (the same point in the height dimension) t=1 (=1/299792000 seconds) Now, let us transform those coordinates to our Earth reference frame, which has a relative velocity of v. It will also be convenient to use the abbreviation B to refer to the boost factor = √(1 - v<sup>2</sup> / c<sup>2</sup>) We use the Lorentz transform to get our new coordinates: x' = (x - vt) / √(1 - v<sup>2</sup> / c<sup>2</sup>) y' = y z' = z t' = (t - vx/c<sup>2</sup>) / √(1 - v<sup>2</sup> / c<sup>2</sup>) x' = -v / B y' = 1 z' = 0 t' = 1 / B What does all this mean? In the box frame, the light was emitted at (0,0,0), and received 1 tick later at (0,1,0). In the Earth frame, the light was emitted at (0,0,0), and received 1/B ticks later at (-v/B,1,0). How far did the light travel in the box frame? That's easy - use pythagorus: s<sup>2</sup> = dx<sup>2</sup> + dy<sup>2</sup> + dz<sup>2</sup> = 0<sup>2</sup> + 1<sup>2</sup> + 0<sup>2</sup> = 1 s = 1 How far did the light travel in the Earth frame? s<sup>2</sup> = dx<sup>2</sup> + dy<sup>2</sup> + dz<sup>2</sup> = (v/B)<sup>2</sup> + 1<sup>2</sup> + 0<sup>2</sup> = 1 + (v/B)<sup>2</sup> s = √(1 + (v/B)<sup>2</sup>) In Earth's frame, the beam travels further than it does in the box frame
and at a guess, this discrepancy indicates the dilation on board our box doesn't it? Being mathamatically dysletic I wont attempt to follow your calculation...but the discrpency would be applied by earth to find the atomic tick rate of our box. is this correct?
and if we go back to the clock does the ratio stand as equal if assuming width as constant regardless of velocity......using you [y] dimension
That's right I forget the details of your setup... If we could somehow detect the tick rate in our frame of a clock attached to the box (after adjusting for signal delays), we would need to adjust that rate for time dilation to determine that clock's native tick rate.
Pete sorry for confusing, although it isn't surprising given the nature of our question. I woudl venture that many a physicist has found this quesstion perplexing and difficult to organise properly. a couple of thoughts occurred over night about our perpendicular beam. Firstly, some have considered tha the beam of light would be seen at a different angle from Earths perpective, [ep]. The light can not be seen to travel from any angle is in fact the case, as light can only be seen in reflection and not in transit. so to infer that a lightray is seen at an angle is I think an incorrect statement ( I don;t think that you are claiming this but I thought I'd note it for the record.) The important thing about all this is: The entire clock is one frame, but within this frame are three frames or aspects. whilst the box travels at velocity, the width frame is equivelant to the earth Frame. So the width distance is maintained as earth distance. Yet at the same time it's length is contracted. so we have Box frame = Earth distance , velocity distance, dilated time for all inner frames. So in your calculation that you have done you came to the conclusion that according to earth frame the light has travelled further than the box frame. This means that the transformation has some how ended up with an inverse result. I think because the dillated time of our box isbeing unsed to calculate distance rather than the distance being used to calculate time. This is an important point. the whole idea requires that time be calculated using distance not the other way round. The time it takes for our ray of light to travel the width of our box should be as per Earth metric but as we know it will be dilated. example in our box Sample [1] [ed] = 299792kms ( width ) Velocity = zero [dt] = 1 second [et] = 1 second Sample [2] [ed] = 299792kms ( width ) velocity = 0.2c [dt] = 1.02 seconds ( abstract ) [et] = 1 second [ fact ] Sample [3] [ed] = 299792kms (width) velocity = 0.5c [dt] = 1.2 seconds (abstract) [et] = 1 seconds [fact] and so on.... but all figures are achieved with in our boxs frames and not from external referrences even though the width is a [ed] so the transformations are relevant to the [ed] the [dt] and the figure we want is the [et] ( earth time ) but only acquired from dimensions with this our clock. the ray is fired and we get a figure of [dt] teh distance is [ed] which we know the [et] to be based on 299792kpsec. the figure [dt] is compared with our known figure [et] and a ratio is assessed. but all with in our box.
any time measurement in our box will be dilated but if teh distance is known to be an earth distance then th edilated time will show itself to be dilated and yet appear to the box to be normal. so if the captain of our clock saw a time recorded of 1.2 seconds for every 299792kms and knows that his width is uncontracted and the same as earth he then knows that his earth time is 1 second thus his [et clock] will read a tick rate faster than his [dt clock].
I created this little story to show the thinking again. Light Clock - Absolute time- Relativity We have a Space ship, travelling at a velocity to be determined but certainly relativistic. It is in deep space some where in a galaxy next to ours. The ship and everything on board was built on Earth. The captain of this ship is sitting at his flight console and amongst all his gadgets he has two clocks. One clock is atomically determined and the other is light velocity determined. At the moment his atomic clock is ticking over at what appears to be the same rate it always has, including when it was tested and calibrated on earth. The light Velocity clock how ever is ticking 20% faster. This indicates the dilation effects on board using the Earth ( in another galaxy as our reference frame) and from this difference the ships velocity can be calculated. ( relative to Earth ) In the middle of this ship is a chamber that is 0.299792 kms wide, the width of our ship and perpendicular to the ships velocity. At random within a range of a 10 second time periods 5 pulses of light are fired of from a laser and the time it takes for our light to reach the opposite side of our chamber (0.299792kms away ) is recorded. On Earth this time was 0.000001 seconds but here at this velocity it is calculated at 0.0000012 seconds by our atomic clock indicating that time dilation is currently 20% of Earth time rate. The captain sees the difference and calculates a tick rate ratio of 1.2 :1. The captain knows that he is aging slower than those persons left behind on Earth by 20%. for every 365 earth days he is aging 292 days, after 30 Earth years he has aged only 24years if he maintains this velocity. The captain decides to contact a fellow captains ship nearby that is travelling at 0.7c and asks him to describe his clocks. The fellow Captain replies that the atomic clock hasn’t changed since leaving Earth but his light clock is ticking at a rate of 1.3 times his atomic clock a ration of tick rates being 1.3:1. The captain of our first ship does a quick calculation and noted that his fellow captains ship is travelling at a velocity that is 10% faster than he is. And that for every 365 days on Earth his fellow captain is aging only 255.5 days. The light clock in the ship is constructed in a way that allows a distance measured by Earth as 0.299792 kms and because contraction affects only effect the length of the ship ( Ships bow pointed in the direction of velocity) this Earth measure (width) is not subject to contraction, and is true to it’s original dimension. Any light beam fired across it’s length must travel this “fixed” distance. However time rate is relative to velocity and thus dilation occurs in the measurement of light velocity over this fixed distance. The beam of light length and velocity is constant. ‘c’ is ‘c’. The dilation means that the time used to measure the velocity is dilated but the distance isn’t contracted. Thus light is measured as in this example as1.2 dilated seconds for every 299792 kms. or .0000012 seconds for 0.299792kms. So every time the pulse of light hits the sensor the time is extrapolated back to 1 second per 299792kms. If our atomic clock measures 1.2 seconds for something that should take only 1 second it is clear that the ship and it’s atomic clock are dilated by a factor of 20%. The light clock is absolute in it’s reference regardless of velocity. Normally when doing this exercise length and time are proportionally effected by velocity, so light would always be measured by our atomic clock as 1 second per 299792kms. but because the distance factor has not changed the atomic clock is dilated but distance isn’t changed thus light speed is measured to be slower relative to the ship but the same relative to Earth measure. An Absolute time reference for our ship is created and established using Earth distances on board as the preferred frame. The width of our chamber is the same as it was when it was built on Earth regardless of the ships velocity. Now is this possible?
The beam will still travel different distances in the two frames. This isn't to do with special relativity, this is much easier than that. Imagine you're in the back seat of a bus throwing a ball across the to me. The bus is travelling at 20m/s relative to the road. There is 2m between you and me. It takes 1 second for the ball to get from you to me. In the reference frame of the bus, the ball goes how far? 2 meters, right? In the reference frame of the road, there is still only 2m between you and me, but how far does the ball go now? It must be more than 2 meters, because the bus and everything in it has moved forward 20m while the ball is in transit.
I see your point, but is this relevant to our clock and it's reference to distance of width is all that matters. if the width is fixed and moving at the velocity of the rest of the box and light is invariant then as far as the box is concerned the light has travelled a relatively fixed distance, and it is the box that uses this distance to find it's dilation.... now this is a main sticky point yes? as far as earth is concened the distance is more but as far as the box is concerned the distance is the same regardless of velocity.....can we agree that as far as the boxes frame is concerned that distance has been maintained? yes or no?
