# Twin Paradox speed of light

Twin Paradox Can anyone clarify and or correct the following for me? A space ship is flying at speed v equal to 0.8 times the speed of light. Within the ship are three stations, a transmitter at station A, station B directly across the ship from station A and station C directly forward of station A. The distance AB is noted as d and the distance AC equals AB. A photon is transmitted from A towards B at the speed of light c. At the same instant another photon is transmitted from A towards C. The transit time t for the photons is measured inside the space craft as d divided by c in both cases. A stationary observer notes the transit of the photon from A to B and measuring the transit calculates the distance to be Tab multiplied by the speed of light c and also calculates the distance travelled by the space craft as v * Tab and concludes that : d = Tab * (c^2-v^2)^0.5 And as d = t*c t = Tab * (1- (v/c)^2)^0.5 For v/c = 0.8, (1- (v/c)^2)^0.5 = 0.6 t = 0.6 * Tab

This is the Lorentz equation for time dilation on which the twin paradox is based. Now consider what the observer sees of the photon travelling from A to C. Firstly Lorentz would contend that there is a shortening of lengths in the direction of travel and as a consequence the observer sees the distance A to C as d multiplied by (1- (v/c)^2)^0.5 hence the observed distance of travel is v*Tab + d*(1- (v/c)^2)^0.5 and: Tac = (v*Tab + d*(1- (v/c)^2)^0.5)/c Tac = 1.93333*t If this sum is done for a photon travelling from C to A the numbers are even harder to understand Tca = (-v*Tab + d*(1- (v/c)^2)^0.5)/c Tac = -0.73333*t As the time of arrival within the space ship is the same in all three cases then should not the observed time of arrival also be the same for all three cases not three different times and particularly the later one cannot be negative. Please will someone resolve my paradox.

I do not find easy to understand your calculations, but can give you an explanation which is not based on specific distances. It is easy to see why the observer inside the ship will perceive the events as simultaneous and the one outside the ship will not. First, notice that for every observer the speed of light is the same, c. So the observer on the ships perceives that the two lengths, being equal, and also the speeds being equal, will result in the two photons reaching B and C at the same time. The observer outside the ship will also perceive the two speed being c, but the distances the photons need to cover to reach B and C will be different. This is because station B is approaching the point where station A emitted its photon, and station B is receding (moving forward buy getting farther away). So this observer will perceive the event as non-simultaneous, the 1st photon reaching station A before second reaches station B.

Things are much simpler here if one thinks in terms of spacetime events and their coordinates in the relatively moving reference frames.

As best as I can tell, there are three events of interest:

• Event A: two photons are emitted from station A
• Event B: one of the photons is received at station B
• Event C: the other photon is received at station C

Stations A, B, & C are at rest with respect to each other and, in the inertial frame of reference in which the stations are at rest, which we will call the primed frame, the spatial separation of events AC equals the spatial separation of events AB which, in this frame, has the value $d$.

In another inertial reference frame, which we will call the unprimed frame, the stations are observed to have speed $v = 0.8c$ along the line through stations A & C. The line through stations A & B is perpendicular to the direction of motion.

In the rest frame of the stations, the events B & C are simultaneous since

$$\Delta t'_{ab} = \frac{d}{c} = \Delta t'_{ac}$$

So, in this rest frame, we can assign coordinates to the three events as follows:

$$A: t' = 0, x' = 0, y' = 0$$

$$B: t' = \frac{d}{c}, x' = 0, y' = d$$

$$C: t' = \frac{d}{c}, x' = d, y' = 0$$

Now, via the Lorentz transformation, we find the coordinates in the unprimed frame (assuming standard configuration) using the following equations:

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

$$x = \frac{x' + vt'}{\sqrt{1 - \frac{v^2}{c^2}}}$$

$$y = y'$$

Thus, in the unprimed frame, the coordinates of the events are

$$A: t = 0, x = 0, y = 0$$

$$B: t = \frac{5}{3}\frac{d}{c}, x = \frac{4}{3}d, y = d$$

$$C: t = 3\frac{d}{c}, x = 3d, y =0$$

So, as expected, in the unprimed framed, the events B & C are not simultaneous. This is the well known relativity of simultaneity.

Also, note that, in either coordinates, the intervals $\Delta s_{AB}$ and $\Delta s_{AC}$ are light-like as required where

$$\Delta s = c^2\Delta t^2 - \Delta x^2 - \Delta y^2 = c^2\Delta t'^2 - \Delta x'^2 - \Delta y'^2$$

Your calculation refers to time intervals between two events inside the ship, as seen from a different reference frame (Earth?), and in Special Relativity this is equivalent to what would be seen from the ship if in the other reference frame the same experiment was being performed.

However this is not the correct way to solving the paradox, this paradox arises in the frame of Special Relativity and cannot be solved in it. You need General Relativity, and the reason is that for any movement where you return to the place of origin, some turning forces need to take place*. That means you need to change your direction and that implies that the system is no longer an inertial one. I insist on this: you need to verify the applicability of the postulates: Special Relativity applies to inertial reference frames (Wikipedia)**.

In the frame of General Relativity, gravitational forces (and other forces due to the Equivalence Principle) will affect the time passing by slowing it, and this is an absolute phenomenon. That means that living in the surface of Earth is already slowing our time passing with respect to what would be living in space.

So, the paradox is solved if you define how the travel in the ship will be, lets say you travel in a perfect circular trajectory. If is large enough that the centripetal force is smaller than Earth's on-surface force, the traveller in the ship arrives older. If the centripetal force is larger (shorter trajectory) the the ship traveller will be younger, and of course you could also tune it to be the same. For more complex trajectories you would have to integrate the effect of the steering forces perceived in the ship during the travel.

*This is true at least in a 3D Euclidean space, since deformed spaces might allow you to return to origin without changing direction.

**An innertial reference frame is reference frame which is static or moving with a constant velocity. This implies that velocity cannot change direction either (see Wikipedia).

• You do not need general relativity to explain the twin paradox. In some cases GR offers a handy way, but it is not needed. You do need to add up the proper time along the paths of the participants---which you can do with only special relativity and is always sufficient to explain the situation. Jul 17, 2014 at 14:21
• As mentionned above, Special Relativity is only applicable to inertial systems (please see Wikipedia), and this is not the case if you return to the starting point. It is a common misconception when the theory is not well reviewed. In fact trying to use Special Relativity to understand this situation of the twins is precisely what raises the paradox, and again the mistake is forgetting the the postulates: only for innertial systems. Jul 17, 2014 at 15:15
• Nope. Special relativity is all about physics in flat Minkowski space. Things like time dilation are only good as formulae for inertial frame, but the concept of proper time $\tau = \int \sqrt{\mathrm{d}x^\mu\mathrm{d}x_\mu}$ is all that is needed to resolve the "paradoxa", and it is perfectly well defined in SR. Jul 17, 2014 at 15:29
• For serious considerations see Twins paradox where references are given. Jul 17, 2014 at 15:31
• Yes. Do look at the wikipedia article "The general relativity solution for a static homogeneous gravitational field and the special relativity solution for finite acceleration produce identical results." and "the issue is how long the world-lines are, not how bent". Jul 17, 2014 at 19:27