Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Imagine a space ship, loaded with all sorts of computer systems, traveling near the speed of light.

Electricity itself is very fast, and can reach speeds close the speed of light. (up to 99% according to wikipedia). So, what would happen to the electronic circuits in this spaceship?

Will the computers shut down, because electricity can't reach the components? Or are they just not related to each other and will the computers keep working perfectly?

share|improve this question

2 Answers 2

up vote 6 down vote accepted

You're dealing with an incomplete form of relativity.

In the frame of the spaceship, nobody will notice anything different, since all inertial frames are equivalent

In the "ground" frame, electricity would be moving at a different speed, by the relation $$\rm v_{e,ground}=\frac{v_{ship}+v_{electricity}}{1+\frac{v_{ship}v_{electricity}}{c^2}}$$

We cannot simply use relative velocities, we need to use the above equation. If you compare this with the time dilation of the system, the computers will all seem to be working the same, albeit slower, from your POV.

share|improve this answer
Thank you! I don't know anything of the maths behind it, I just know the basic concepts. This makes it hard for me to understand the kind of thing I was asking. So basically, if a calculation used to take 1 second, it might take 20 seconds now? –  Simon Verbeke Apr 11 '12 at 11:46
@Simon from the ship's point of view, it will take 1sec. From the POV of an external observer, it will take 30secs. –  Manishearth Apr 11 '12 at 12:11
Aha, thank you! –  Simon Verbeke Apr 11 '12 at 16:38

Your question is about what will be the speed of elektron? Is this sum of speeds of electron and the ship, or not? If it is a sum of these speeds, speed of elektron will be more than speed of light ? If not, then there will be a time which is slowing down, so,"can the electron reach to command in circuits? Is not it?

Firstly the speed of electron won't be more than speed of light acccording to passenger. It will be as usual for passenger. Acoording to observer, if possible, the speed of electron will decelarate by the slowing time and sum of the speeds of ship and electons will be again same as usual. So your computer will work properly.

According to my view, in reality, the time inside ship is under different physics rules. Because by giving more energy to accelarate the ship with it's increasing mass, (I'm thinking according to String theory), the strings are energized and extended, so they can compress the spatial dimensions (on the same route with accelaration). Also "the time" is a wave kind, sourced from universal expansion (it is again my idea, not a proved scientifical data). So time of the ship will be acting as waves. And you know the behaviour of waves while they are passing from different environment as usual. Their wave lenght and speeds are changing, after left this environment, they turn again their ordinary position. So, if you think "the time" as waves, which are also in the speed of light, limit of speed is the limit of these waves. And, if an object is accelarated in these waves, it will be limited at doppler barrier. Because all waves will be overlapping. (And their energy will transfer to mass)

share|improve this answer
Self-promotion should be transparently advertised. –  ACuriousMind Apr 22 at 15:12
The link is clearly advertizing a personal theory in violation of site policy on this topic, so I am going to remove it. (I should also mention that hiding a link to your own blog behind a URL obscurer is not doing you any favor in terms of the reception you'll receive here.) The rest of the answer is fairly incomplete and will likely continue to accumulate downvotes unless improved. –  dmckee Apr 22 at 23:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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