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Who Should Measure Speed


Maine farmer

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According to Special Relativity, the speed measured by the observer on the spaceship and the stationary observer agree.

 

What they disagree about is the distance between points in the direction of travel of the spaceship. For example, if the spaceship is traveling at 0.99 c from Earth to Alpha Centuri 4.37 ly away as measured from Earth, the spaceship observer would observe the distance to be [math]\sqrt{1 – 0.99^2} \cdot 4.37 \dot=[/math] 0.616 ly. Thus, while the Earth observer’s clock says the trip took 4.15 years, the spaceship observer’s says it 0.62.

 

It could be useful for the spaceship observer to convert their proper speed (which can’t exceed c) into an unlimited “pseudo-speed” by dividing by the Lorentz term, eg: vproper 0.99 c : vpseudo 7.02 c.

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So essentially, the speed of light, although a finite number, could be considered to be the same as infinite velocity, essentially being at every point at the same time?

Yes, I think that’s a good way to say it, though I’d be a careful with the “every point at the same time” part. Length contraction occurs only in the direction of travel, so an observer moving the speed of light would be at every point directly in front and behind his direction of travel, not at points in other directions.
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  • 2 weeks later...

If being at every point in the direction of travel at the same time, every point could be considered to be but one point, and observed time would be zero, then the speed of light could "feel" the same as zero velocity?

You were careful in the 1st post to say the spaceship’s speed is “near light speed” rather than “at light speed”. Don’t stop now!

 

As long as the speed is even slightly less than c, the “feel” of the length contraction doesn’t lead to any weirdness where the length of the universe in some direction is zero. The universe can be made to appear bizarrely squeezed, but never squeezed down to just 2 spatial dimensions.

 

Also, remember that the heart and soul of Special Relativity is that the speed of light in vacuum is constant. So, even on a spacecraft traveling nearly the speed of light, a measurement of the speed of light in vacuum in any direction on the ship finds it to be the usual 299792458 m/s.

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According to Special Relativity, the speed measured by the observer on the spaceship and the stationary observer agree.

 

What they disagree about is the distance between points in the direction of travel of the spaceship. For example, if the spaceship is traveling at 0.99 c from Earth to Alpha Centuri 4.37 ly away as measured from Earth, the spaceship observer would observe the distance to be [math]\sqrt{1 – 0.99^2} \cdot 4.37 \dot=[/math] 0.616 ly. Thus, while the Earth observer’s clock says the trip took 4.15 years, the spaceship observer’s says it 0.62.

 

It could be useful for the spaceship observer to convert their proper speed (which can’t exceed c) into an unlimited “pseudo-speed” by dividing by the Lorentz term, eg: vproper 0.99 c : vpseudo 7.02 c.

 

There could be other ways to address the problem without psuedo speeds CraigD.

 

If points A and B were both emitting visual timing signals (like the old 'on the third stroke the time will be x, bing, bing, bing') 5 years prior to the trip then the spaceship wouldn't need to have a clock to calculate its location as its location would be a given from the timing signals it would receive from both points A and B during a return journey. A psuedo spaceship time based on A and B's timing signals will preserve the speed and physical location without distortion. The timers at both points could be synchronized prior so there would always be at least one non length contracted timing signal so the ship can get an accurate distance and time wrt a point on a line between A and B. Wherever the spaceship is it will be in receipt of an incoming signal from the timer in its direct line of travel and this signal will have traveled to the ship at c from its emission point. The calculated time is then the total synchronization time (i.e. one cycle from A to B at c) minus the undistorted incoming time signal. 

 

I have seen SR based verfication of solutions (O. Gron) to relativistic rolling wheel problems where a length contraction of the position of a point on a wheel was used to determine the emission point of light that then travels in a straight line at c to a stationary observer on the road. This solution could also be reversed by moving the road at a relativistic speed and sending light from the observer at the various angles at the start time so that they reached their respective emission points on the relativistic rolling wheel as the road (and observer) moved past. 

 

If we used robots on the spaceships I bet their warranties will be based on our time not spaceship time, LOL.

Edited by LaurieAG
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There could be other ways to address the problem without psuedo speeds CraigD.

The original question was “should the speed be measured by stationary observation, or by observation on the spacecraft?”. Since the answer is “they agree”, we don’t need any other notions to answer it.

 

The problem with this answer it that, for spacecraft with speeds a sizable fraction of the speed of light, some intuitive assumptions about speed gleaned from our lifetimes dealing with speeds only a minute fraction of the speed of light begin to falter. In particular, our intuitive notion that the time it takes to get from point A to point B is [math]t = \frac{d_{(A, B )}}{v}[/math], where [math]d_{(A, B )}[/math] is the distance between point A and B as measured by an observer at rest relative to both points and [math]v[/math] is the mover’s speed, falters.

 

Introducing [math]v_{pseudo} = \frac{v}{\sqrt{1-v^2}}[/math] (where speed is in units of c) shows how easy it is to invent a new metric that restores the troubled assumption. Just like the speedometer on an ordinary car, passengers could divide the distance between their origin and destination by the reading from their spaceship’s pseudo-speedometer, and know how long the trip was going to be.

 

A consequence of this simple equation is that, unlike [math]v[/math] which must be between 0 and 1 c, there’s no upper limit on [math]v_{pseudo}[/math]. Our hypothetical spaceship passengers might be quite comfortable saying they were “cruising at 3.1 times the speed of light”, referring to this pseudo-speed rather than an equivalent true speed of about 0.9517 c.

 

If points A and B were both emitting visual timing signals (like the old 'on the third stroke the time will be x, bing, bing, bing') 5 years prior to the trip then the spaceship wouldn't need to have a clock to calculate its location as its location would be a given from the timing signals it would receive from both points A and B during a return journey.

This could work. It’s essentially a “2 sat GPS”. Its 1-dimensional position formula is much simpler than our real-world, 4+ sat GPS’s 3-D position formula. It’s just

[math]d = \frac{d_{(A, B )} -t_A +t_B}{2}[/math]

where [math]t_A[/math] and [math]t_B[/math] are the time values of the messages received by the ship from the senders at point A and B, in units of the time it takes light to travel from A to B.

 

As you note, Laurie, the receiver doesn’t even need its own clock, just to compare the times it gets from the clocks at A and B.

 

A psuedo spaceship time based on A and B's timing signals will preserve the speed and physical location without distortion. The timers at both points could be synchronized prior so there would always be at least one non length contracted timing signal so the ship can get an accurate distance and time wrt a point on a line between A and B.

I’m not sure what you mean by “length contracted timing signal”. Assuming they’re like the “the time is now” audio signals from old-fashioned services like at US phone number 202-762-1401, or the high-precision binary digital ones from GPS satelites, the timing signal is simply conveying a number (and, for our 2 sat solution, to work, something that lets us tell if it came from A or B . It’s not a length, so can’t experience length contraction.

 

If we used robots on the spaceships I bet their warranties will be based on our time not spaceship time, LOL.

LOL, ‘cause the fastest spaceships flown by us to date, the 2 Helios probes, launched in 1975, which reach 0.000234 c, are less than 30 seconds younger than they’d be if they never left Earth. So Earth time and ship time are practically identical.

 

If we ever manage to make much, much faster spaceships, this won’t be the case. Let’s consider a science fictional example.

 

Suppose, for some reason, we want to shoot robotic spaceships between Earth and Barnard's Star, 5.963 ly, using a supper-efficient space rail gun/catcher that quickly give them a speed of v=0.999 c, vpseudo=22.34 c. Let’s say our robots come with a 5 year bumper-to-bumper warrantee. If we use shipboard/robot time, the one-way trip measures less than 98 days, and our robot is under warrantee for 9 round trips. But if we’re required to use Earth/manufacturer’s time, it’s out-of-warrantee before the end of the 1st one-way trip.

 

Of course, if we had the technology to shoot robots around at 0.999 c, we’d likely be able to make ‘em good enough that warrantees were a relic of ancient history, but at such speeds, time dilation is dramatic.

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