Assume I am on a very big planet. I took a car ride and it's moving at 99.9% of the speed of light. While in the car all seems normal, so I even don't feel the speed. I decide to flip a coin, I take it out of my pocket and flip it in the air of the car, 90 degrees from the car movement and 90 degrees to the planet.

At the exact same time, a person at rest flips same coin, on the same planet.

Will both coins land at the same time? Does time dilation effects gravity?

  • 2
    $\begingroup$ at the same time, in which frame of reference? $\endgroup$ – John Dvorak Apr 21 '17 at 4:31
  • $\begingroup$ @JanDvorak What are my options here? $\endgroup$ – Ilya Gazman Apr 21 '17 at 4:33
  • $\begingroup$ @Ilya_Gazman There are infinitely many options. But main two options are your frame of reference and the frame of reference of the person at rest. $\endgroup$ – Danijel Apr 21 '17 at 5:22
  • 2
    $\begingroup$ You are missing the fact that if you are traveling at 99.9% of speed of light with respect to the planet and gravity holds you on the surface of the planet, that planet will be a blackhole :) $\endgroup$ – Ali Apr 21 '17 at 6:25
  • $\begingroup$ You might find the following SMBC comic interesting: smbc-comics.com/comic/moral-relativity . The same idea applies to coin toss; the person traveling fast will see the other person's coin toss take about 22 times less time than his own as can be calculate from the time dilation formula for special relativity. The person staying put will see the moving person's coin toss take about 22 times more time than his. $\endgroup$ – Echows Apr 21 '17 at 12:38

If the person at rest flips a coin a certain height in the air and they time it as taking 1 second to return to their hand then the person in the car will be able to perform the same experiment (assuming you can exactly reproduce the coin flip to the same height) and record the same result in their frame of reference (1 second).

The person at rest will observe the car travelling at 99.9% the speed of light to be effected by time dilation, e.g. the time in the car will appear to be travelling slower relative to their own frame of reference (that of rest). As such the person at rest will observe the flipped coin in the car to take longer than 1 second (as measured from the frame of reference at rest).

Note that I've avoided them flipping at the same time as they will be at different places when they start and end the flip and you can't tell if the events occur at the same time or not. Hopefully though the scenario above answers your question, the person at rest will observe the coin flip in the car to take longer to land than what they would expect if they had performed it in their frame of reference and they explain this by saying that time dilation effects the frame of reference moving close to the speed of light.

Further Notes:

  • I'm ignoring the big planet part and just assuming a uniform gravity throughout both frames of reference.

  • See XKCD's Relativistic Baseball for an idea of what happens if your car travels at relativistic speeds in an atmosphere.

  • Flipping a coin is quite random and hard to reproduce, you and the person at rest will need to practice this a LOT to get it exactly reproducible each time.

  • $\begingroup$ To sync me and my friend lets remove him. Let's say I been at rest and measured the time it takes to perform a flip of the coin, then I entered this car, waited for it to achieve the full speed of 0.9999C and perform the experiment again. Will I measure the same time? $\endgroup$ – Ilya Gazman Apr 21 '17 at 10:49
  • $\begingroup$ Yes. You will measure the same time (provided that the value of "g" is constant along your path) because the value of "g" is the same for all the Lorentz observers. $\endgroup$ – Tuhin Subhra Mukherjee Apr 21 '17 at 13:15
  • $\begingroup$ In theory yes, if everything is the same, e.g. angle and speed of release of the coin, gravity at start and end of coin toss, temperature, etc then it should give exactly the same result regardless of your inertial frame of reference. In practice every coin toss will differ and travelling at that speed gravity would have changed subtly from start and end of the coin toss. $\endgroup$ – Quantumplate Apr 21 '17 at 20:05

Not the answer you're looking for? Browse other questions tagged or ask your own question.