# Special Relativity: why is length contraction the inverse of time dilation factor?

## Homework Statement

Hello
I actually just need help with an explanation rather than equations (I hope this is OK).

In my essay I need to explain why the length contraction in Special Relativity is the inverse of the time dilation factor. My explanation is below, but I'm not convinced that I have totally understood it. Could someone please check that I have the correct understanding?

## The Attempt at a Solution

''Imagine the pilot on the spaceship flying past the stationary observer on the space station. After a short distance there a two satellites, a large distance apart hovering in space. The stationary observer on the space station watches the rocket fly past the first satellite, and then travel towards the second satellite at velocity v, passing it at time t. The pilot however, sitting still in her rocket passes the first satellite and then watches the second satellite move towards her at velocity v, passing it at time t’.

From the observer on the space station’s point of view, the distance between the two satellites is L = vt, but the distance from the pilot’s point of view is L’ = vt’.

The time from the pilot’s reference frame is less than the time from the observer on the space station’s reference frame, but as they both agree on the length of time it takes to get from one place to another, the distance travelled must also be different to compensate. The distance between the two satellites appears shorter to the pilot.
L=vt = L'=vt'
L'/L = vt'/vt

These ratios are equivalent but L and L’ are different, and t and t’ are different, and so to make the equations equivalent one must be the inverse of the other.''

Related Introductory Physics Homework Help News on Phys.org

Well think about it carefully, I'm not so sure you're going on the right lines here;

If you take the moving reference frame; what does this 'see' happening to the distance between it and the 2nd space station when at the first - moving at velocity v? Write this down numerically.

However obviously, the observer does not see this happening, they can't, therefore what do they 'see' happening to the time taken? Write this down numerically.

Once you've done this, think carefully, this is what they've both seen, but the same event has been described, therefore both must be correct.

Therefore - why must length contraction be the inverse in order for both of the equations to be simultaneous?

Hi
Thanks for replying. I'm not actually a physics student - I'm doing a 'cross curricular' assignment, and I've learned this topic by reading books (i.e. I don't have lessons on it) so I haven't got a solid grasp on it and probably have some vital knowledge missing.

I've tried to answer your questions, but I'm not sure if I've just repeated my original question in different words.

The moving reference frame sees the 2nd station approaching him and so the distance is getting shorter. The 2nd station is approaching him at velocity v. If he is moving at velocity 0.90c then it would be approaching him at 0.90c and it would take say 3 minutes to reach him. I'm not sure how to write this down numerically but I'll have a go:
L' = 3/0.9c

The motionless observer sees the observer 'in motion' moving towards the second station at velocity 0.90c. This motionless observer would see the the rocket arrive at the second station also after 3 minutes on his own watch, but if he looked at the clock on the rocket it would appear (to the motionless observer) that it took say 2 minutes.

L = 2/0.9c

For these equations to be simultaneous L' must be smaller than L so that it compensates for the difference in time. To compensate exactly it must be the same size (but the inverse). So its just algebra - if one side of the equation is multiplied by gamma, then the other side of the equation must be divided by gamma.