Period of pendulum in a relativistic train moving on a flat earth?

Well according to that old Alexander problem, the gravitational acceleration aboard the train is:

g/(1-v²/c²) = (25/16) g

Generally the period is T = 2π√(l/g)

On the earth, T_earth = 2 sec

On train, T_train = 1.6 sec

I’ll give it a go …

Assuming we are talking of the equivalent simple (theoretical) pendulum of 1m with a time for one complete cycle (return to its starting point and direction of swing) governed by T = 2*pi*sqrt(L/g), then T = 2.00649 seconds (using excel).

http://en.wikipedia.org/wiki/Pendulum

http://en.wikipedia.org/wiki/Standard_gravity

When considering clocks, most people take the time for 1 swing as the period, rather than the two swings needed to complete a cycle as most would understand it. I assume the reason for this is that mechanical clocks normally give impulse to the pendulum in both directions (ticking twice in the process). So it is convenient mechanically, if you want a clock to record seconds, to make the pendulum of a length equivalent to the theoretical length of 994mm. If the wheel that gives the impulse (‘scape wheel) has 30 teeth (each tooth acting twice), then it will turn once a minute and its arbour can be extended through the dial to carry a seconds hand.

http://en.wikipedia.org/wiki/Seconds_pendulum

Obviously, if we use this definition of the period of swing, then T would be half the above value; T = 1.003205 seconds. For a seconds pendulum T =1

Now for the difficult bit! I know quite a lot about clocks, rather less about relativity!

Assuming a seconds pendulum, T = 1; my understanding is that an observer on the train, travelling with the pendulum ,would observe the same time; T still equals 1.

However, an observer on the earth would record the pendulum swinging at a slower rate. How much slower? Luckily, whilst looking for a way to answer, I came across this!

http://www.1728.com/reltivty.htm

By inputting 0.6 and selecting ‘c’=1, I got a Relativistic Change Factor of 1.25. So, if I’ve got this right, the observer on the earth would record a time of 1.25 seconds for a swing of the train-board pendulum.

To sum up:

For 1 swing, seconds pendulum, observed on the train T = 1, observed from the earth T = 1.25

For 1 swing, 1m pendulum, observed on the train T = 1.003205, observed from the earth T = 1.254006

If you understand the period as the time to complete a full cycle (a return to the starting point and direction), you would need to double these figures.

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Now, your questions are usually beyond my capabilities. I wonder what it is that I haven't thought of ...

In case it's significant, I have assumed the pendulum is swinging at right angles to the direction of motion of the train, as seen by the train-board observer. :-)

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Edit: Thinking this through, the above would apply as the pendulum passes the earth-based observer. At other times the Doppler Effect would also come into play.

Edit 2: Having read the other answers and links, I suspect there IS something else. Gravity aboard the train increases? That's a new one to me! So, would the increase in g oppose the decrease in T on the train? Would the train-board pendulum keep the same time as an earth-based one, as viewed by the earth-based observer? Would an observer on the train see the train-board clock running faster?

Would I be better sticking with global warming?