OT : Relativity question.

What's your traction coefficient?

It depends how carefully you designed your speedometer.

It should take into account the apparent Lorentz contraction of the space you are passing through,

the time dilation that you experience at 0.87c - both 50% at that velocity, as you have obviously worked out.

Your tyres (and wheels) would fly apart at that velocity, so you aren't timing wheel revolutions.

Doppler shift would work - the 21 cm hydrogen line might be worth using, both looking ahead and behind - but there's more going on than just Doppler shift.

No.

No measurement of any object's velocity can be >c. The relative speed of tw o objects in a single frame of reference can approach 2c, e.g. galaxies at the edge of the observable Universe that appear 180 degrees apart in our sk y.

That apparently magic 0.87c number is where gamma (1/sqrt(1-v**2/c**2) equa ls 2, so that an object's kinetic energy mc**2*(gamma -1) equals its rest m ass energy mc**2. But that's not important at all--the KE of a 1 TeV proton is over 1000 times its rest mass energy. Nothing special happens when gamm a=2.

Of course a baseball moving at 0.87c has a kinetic energy of 3.4 megatons, so probably you'd run out of juice before you got going that fast. ;)

Cheers

Phil Hobbs

Even if I went out earlier and placed posts every mile which my speedo then counted as I flew past?

Cheers

It doesn't know about that, it's just timing 'wheel rotations'.

Yes, it was just an illustration. Timing marker posts previously placed every mile along the route would do. If I've travelled what I knew to be 5 light years and I've aged 4 years, then my own 'biological speedo' would infer 1.25c. Would my mechanical one show 1.25c too?

Cheers

If you are just counting them then yes you will determine a speed >c.

But the distance between them will appear foreshortened so the

For those desiring an explanation... :-)

Tim

Indeed. There is a rather beautiful derivation of the special relativity equations that works by considering nothing more complicated than the mutual events of two metre rules approaching each other at a significant speed and passing close enough together.

In the first metre A-B rules frame of reference it is 1m long and observes the other fast approaching one C-D to be Lorentz contracted.

The mutual events are therefore as follows:

A--------B C-----D B&C same place

A--------B C-----D D&B same place

A--------B C-----D A&C same place

A--------B A&D same place C-----D

That gives you the 4 mutual events in AB's reference frame.

But since the laws of physics are the same in all inertial frames the observer on C-D sees his rod as 1m long in his rest frame and A-B's as Lorentz contracted. It is left to the reader to finish this off.

Combining these mutual event observations with a symmetry argument allows the SR equations to be derived from scratch with nothing more than (admittedly very tedious) high school algebra.

I would love to know which elementary relativity text this elegant proof was in - it was not in a textbook that I ever actually owned.

If you don't care about general relativity, then try "Spacetime Physics" by Taylor and Wheeler.

(used copies at abebooks for less money)

George H.

I've already set up mileposts...

If my ship was 1 mile long when I built it, with milepost detectors fore and aft, will both detectors fire at the same time as measured by me in the middle of the ship no matter what my speed is?

Cheers

No. A spacetime event is only uniquely defined if you are at it. Any other position and by moving you can alter the timing of events within certain limits.

You have to put your observers at each end of the spaceship (or 1m rod) in the case of the example I gave in an adjacent response. You are on the right track but you have to understand that to observe something you actually have to be there. You can't watch from a distance because the light travel time will affect what you perceive to be happening.

So what /will/ happen as I speed up (I'm in the middle of a one mile long ship listening to its fore and aft milepost detectors), will the front detector start to appear to fire before the rear or vice-versa?

Cheers

My fave is the clock made from a light pulse bouncing back and forth from m irrors at the ends of a moving metre stick. With the stick oriented perpend icular to its motion, in the lab frame the light has to traverse the hypote nuse and therefore the clock runs slow. Pythagoras gives you the equation f or time dilation.

With the stick oriented parallel to its motion, making the clock run at tha t same slow rate gives you Lorentz contraction.

Cheers

Phil Hobbs

But then they aren't a mile apart in your reference frame, so it isn't a va lid measurement. Velocity is displacement / time in a single reference fram e. Your proposed scheme measures lab frame distance / spaceship time, which isn't a useful number and certainly isn't the ship's velocity.

Cheers

Phil Hobbs

Relativity says (among other things), that the equations work just as well from any frame of reference (with complications if that frame is accelerated), and that nothing can travel faster than c.

So if you're sitting still and the road you're on is traveling at 0.87c underneath you, what will your speedometer show?

Well, it's surely a useful number in that it allows me to calculate how much time I have remaining before I arrive, just like my car speedo?

Cheers

I suppose, if it works both ways, if the road was marked with a line every mile when it was stationary, and I timed these when it was moving, assuming them to still be 1 mile apart, I would measure the road's speed as c.

If, as been suggested, the marks are no longer 1 mile apart and I modify the speedo reading accordingly then it's 0.87c.

So the faster you go between two points, the shorter the distance?

Cheers

As has been suggested, hell. Try as the theory shows, and as innumerable experiments have proven.

It's complercated, and sometimes extremely counter-intuitive. Start with the Wikipedia page on relativity, and go from there.

Or this video:

Nice video. To be honest I've never gotten the twin paradox. (But that's a whole 'nother discussion.)

George H.