If two objects approach at 75 percent the speed of light

http://einstein.stanford.edu/content/relativity/q1021.html

by mavdi • 11 years ago

70 47 11 years ago

einstein.stanford.edu

alberte• 11 years ago

The question should be "If two objects approach at 75 percent the speed of light relative to me then what is their speed relative to each other"

dalke• 11 years ago

Or to be even more exact, "... relative to me and I am between them", since speed is scalar; the ships could be meeting each other at a 30 degree angle.

hexasquid• 11 years ago

Get out of the way?

EricSu• 11 years ago

I agree...I know this is supposed to be about their relativity to each other but the way my mind interpreted the wording was "Why are neither of the two objects traveling at 1.5 times the speed of light?" and I immediately thought "Well because it's relative and you can't say that you driving 60mph and another car going 60 mph in the opposite direction means you're driving at 120mph because you're still driving 60mph

edit: Because the question doesn't specify the speed relative to what exactly, I automatically thought about the speeds of the objects relative to themselves rather than to each other or a 3rd party.

tcgv• 11 years ago

If my car is travelling at 60 mph and there is a car in the opposite direction also travelling at 60 mph (according to a inertial frame of reference) then classical mechanics allows me to say that from my point of view the other car is indeed traveling at 120 mph. But special relativity only allows me to say that the other car is travelling at 119.9999999999990394 mph from my point of view. The higher the speed, the greater the difference in the results from "classical mechanics" and "special relativity" will be.

dm2• 11 years ago

Here are several related images and graphs:

https://www.google.com/search?q=relative+velocity+time+dilat...

https://www.google.com/search?q=velocity+special+relativity&...

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel.ht...

s_kilk• 11 years ago

You will be removed from the universe, for violating it's laws.

coldpie• 11 years ago

Hello, friend! "It's" is a contraction, usually for "it is" or "it has." "Its" is the possessive form of "it," just like "hers" and "his." "Its" is used to show that something is owned by "it." In this case, the laws belong to the universe, so you want the possessive form: "for violating its laws."

Thanks for reading!

moron4hire• 11 years ago

Sometimes my phone autocorrects it incorrectly on me, and I haven't noticed before I tap the "reply" button.

astrobe_• 11 years ago

So if those objects are on a collision course, they will crash on each other at 0.96c instead of 1.5c?

marcosdumay2• 11 years ago

Yes, but not problem, they'll release all the energy they've used to accelerate so the collision will be exactly as amazing.

learnstats22• 11 years ago

So, if two objects approach each other at 60mph, they are actually travelling at 119.9994mph relative to each other?

donkeyd• 11 years ago

I'd like to say 'No' from intuition, but it is probably flawed.

marcosdumay2• 11 years ago

Are you sure you didn't mean a few (about 10) nines somewhere?

learnstats22• 11 years ago

ah, in fact I did! I forgot to square c :)

mavdi• 11 years ago

No, much closer to 120.

120 / (1 + (60 / 670616629) * (60 / 670616629)) ~= 119.9999999999990394

snowwrestler• 11 years ago

The difference between this speed and 120 mph is less than half a picometer per a second, which illustrates why we don't typically notice relativistic effects at highway speeds. :-)

amirmc• 11 years ago

"Some problems that we all have in understanding how nature works, have more to do with our conviction that our intuitions are correct, despite the fact that our intuitions are often flawed in areas of experience outside of our normal environment."

This really stood out to me. It's applicable to more than just the current example.

zak_mc_kracken• 11 years ago

True. What makes this model compelling is that our flawed intuitions are very good at describing the world we live in, which is macroscopic and non relativistic.

Letting go of these intuitions is a requirement to become a competent physicist.

Retra• 11 years ago

You replace them with new intuitions about what kinds of transformations are allowed. (Like unitarity.)

Steuard5• 11 years ago

This explanation is correct, but I find it pretty empty of intuition: "Here's a magic formula! Look, it adds up funny." I've taught relativity quite a bit recently, and I much prefer to build intuition using "space-time diagrams" (or "Minkowski diagrams") showing coordinate systems for two different observers.

Tom Moore's Six Ideas That Shaped Physics textbook (Unit R) does a very nice job of this. But here's a really bargain-basement handout that I made many years ago to explain the idea before I started using Moore's text for this. (I'd probably do it a little differently now.) http://www.slimy.com/~steuard/teaching/classes/spacetime.pdf

My discussion of velocity addition there (at the bottom of page three out of four in that PDF) is much too terse for this context, I'm afraid. But if you draw my example there carefully and measure in the tilted coordinates as described earlier, you'll start to see why the leftmost observer still measures the rightmost observer moving less than the speed of light.

bkcooper• 11 years ago

Another good text for this approach (and it's also relatively cheap) is Mermin's It's About Time.

Steuard5• 11 years ago

If you've got way too much time to spare, I've got a set of topic review videos that I made for various classes on this stuff. The quality's not great, but seeing some worked examples could be interesting. https://www.youtube.com/playlist?list=PLQcPPNE9JfpiF4GpUSOGm...

Retric• 11 years ago

IMO, a much more natural approach is to look at things from the perspective of light emitted. Specifically black body radiation. Red/Blue shifting correlates with changing temperatures. And temperature works as a natural proxy for time.

The other thing that confuses people is presenting things from an 'independent' perspective in the diagram.

figure8• 11 years ago

If I'm not mistaken, the derivation of the black-body radiation energy distribution does not use special relativity at all, only statistical mechanics and quantum mechanics. How can it be used to explain special relativity?

madaxe_again2• 11 years ago

A thought experiment I've always enjoyed is the idea of two infinitely long rods, each moving at near as damnit c, perpendicular to one another, with a point of contact.

At what speed does the point of contact move, relative to the reference frame of either bar, or to a static external observer?

Steuard5• 11 years ago

So, do I understand you correctly: relative to the "static observer", the rods are both moving "sideways", as if the x-axis of a graph were moving toward +y and the y-axis were moving toward +x?

In that case, the "static observer" would see the point of contact moving at very nearly sqrt(2) times the speed of light (along the x-y diagonal). But that's okay, because no physical object (or bit of information) is being carried along at that intersection point. (If you put some little spindle at the intersection to be carried along, you'd suddenly have Problems.)

On the other hand, from the reference frame of one moving bar, the intersection point appears to be moving almost infinitely fast! If the "static observer" measured speeds u and v for the two rods (as fractions of light speed, so u and v are both close to 1), then an observer moving with one rod will see the intersection point move at speed v/sqrt(1-u^2). If we write u = 1 - ε for some very small ε, then the intersection point's speed in this frame is roughly v/sqrt(2ε). (The intuitive reason for this result is interesting: an observer on one moving rod will consider the other rod to be almost parallel, at a very small angle (rather than perpendicular), and still rushing toward her at nearly the speed of light.)

lamontcg• 11 years ago

I like the problem of "you are travelling at 0.75c and you turn on your headlights, how fast do you see the photons travelling and how fast does a stationary observer see the photons travelling?" better.

Then the answer is just the speed of light in both cases, and that's really the crux of the problem of special relativity and that the speed of light is constant in all references frames. Then you work backwards to see that the observer that is moving must have time dilation effects.

I also liked the geometrical approach to Minkowski space-time a lot better than just chewing through formulas -- like Taylor and Wheeler's Space-Time Physics book. Drawing light cones and various grids and curves of constant space or time measurements relative to different observers made it all a lot more concrete to me.

amelius• 11 years ago

Here is a phenomenon that is truly fascinating: https://www.youtube.com/watch?v=wteiuxyqtoM

hliyan• 11 years ago

I find this explanation pretty intuitive:

1. I left Earth at 0.75c and haven't accelerated or decelerated since

2. I receive a message from Earth that an object is heading my way at 0.75c (far enough away for them to message me at 1c before it hits me)

3. I look where they point and see the object approaching at 0.96c

4. I'm not confused because I fully realize that relative to Earth, my brain and my instruments are running slower due to time dilation. Speed is a time-related measure (d/t) -- you can't use relative distance and absolute time to calculate (d/t). You use relative values for both. So the speed I get when I divide by my t is different to what Earth gets when it divides by it's t.

The problem arises when people accidentally assume an absolute vantage point to observe the two objects. A scientist on Earth will see the gap between the 2 objects close at 1.5c, but he'll know better than to make the usual assumption we make on Earth -- that everyone else will see the gap close at the same rate (because time will be slower for some observers).

yAak• 11 years ago

Yay, your explanation makes perfect sense to me! This part: "...relative to Earth, my brain and my instruments are running slower due to time dilation" helped a lot with clarity and now it's intuitive for me as well.

The linked page just managed to confuse me, so thanks for posting this!

skierscott• 11 years ago

Nothing can exceed the speed of light. This equation represents that.

I also seem to remember that causality is not preserved. This means that if A happens before B in one frame, it's possibly that B happens before A in another frame.

I also seem to remember that if it's possible to break the speed of light, it'd be possible for B to fire before A (if A triggers B, all in one frame).

Steuard5• 11 years ago

As you say, event order is not preserved: two events that appear simultaneous to one observer can easily appear in different orders to other observers.

But causality itself is safe in relativity: if a signal from event A could possibly reach event B (given the speed of light limit), then every reference frame will agree that A comes before B. The only events whose relative order is indeterminate are those separated by enough distance (relative to elapsed time that one couldn't influence the other: events who are not in each other's "light cones".

spacehome• 11 years ago

Causality is always preserved. If A happens in the past light cone of B in one frame, then it happens in the past light cone of B in all frames.

If, on the other hand, neither A nor B are in each other's past light cones, then there are reference frames where A or B happen 'first', but 'first' is in quotes, because this difference in time is really an artifact of a coordinate choice, and not physically meaningful.

The best way to think about time relationships is that events (points in spacetime) have a partial order given by the relationship 'is in the past lightcone of'. Some pairs of points have this relationship one way or the other, and it's transitive. However some pairs of points don't have this relationship, and there's no meaningful way to compare them. Attempting to do so is inappropriately forcing a classical viewpoint onto a data model that doesn't support it.

tlogan• 11 years ago

The explanation really does not help people who are not familiar with this topic.

I find that the following Susskind's lectures explain this subject really well: https://www.youtube.com/playlist?list=PLDDFE71BA2DE55505

Basically, the key of understanding of special relativity is to start with: Maxwell's equations are correct. But these equations also need to be correct in every reference frame - even if speed of light is constant. How to fix that?

octatoan• 11 years ago

That's a nice way to put it!

peter303• 11 years ago

Unsatisfactory answer, though mathematically correct. The fuller explanation considers what constitutes "measurement" in a physics with a speed limit (lightspeed). We measure "intervals", i.e. lengths between to points of space or time. We'll see the intervals stetched/contracted by the Lorentz factor in another frame moving at a high constant velocity.

z3t4• 11 years ago

If there actually where something moving faster then light, and you measured the speed, it would still appear to be moving at the speed of light!

Say you had two space ships with FTL capability that would race from our solar system to the finish line located near Alpha Centauri. How would you know witch one came first?

sqeaky• 11 years ago

Depends on what you mean by FTL. Most Scifi canons take FTL to mean "ignores relativity", like Star Trek and Star Wars.

In this context someone using a relativistic viewpoint would simply see the ship disappear, then appear somewhere else before they left (using a huge telescope or whatever and calculating the time difference).

The results of return trip are totally based on the interactions of FTL with relativity as we know it. It could be totally ridiculous.

Imagine the a ship here leaves for Alpha Centauri (AC) at hyperspeed or warp 9. For this discussion lets assume that the travel takes 10 minutes. But for light it would take 5 years (rounding).

This functionally would be travel 5 years into the past. So earthbound observatories would see the ship right now, because it left the Falcon/Enterprise 5 years ago from a relativistic stance. Then the ship could return and see its own light given off at alpha centauri?

What does this look like from the viewpoint of observatories on Alpha Centauri(AC)?

Thinking about it, it would have to look the same. By definition FTL is faster than light, but light is attached to time. So the trip from Sol to AC must either go back 5 years or otherwise reconcile our 5 year clock skew.

I don't see how any reconciliation mechanic makes any sense.

But yeah you are right that a ship that simply accelerates so much that its riders think and measure their own speed as faster than light are observed by outsiders as hugging light speed.

strags• 11 years ago

I found this extraordinarily helpful:

http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf

The Lorentz transform allows you to calculate exactly how time/distance differ between two reference frames.

The above link derives the transform from first principles - assuming only spatial symmetry and the invariance of the speed of light in different reference frames. There are other derivations, but I found this one the most straightforward.

ck2• 11 years ago

What really blows my mind about it all is we didn't even have the math to figure that out until 100 years ago (Einstein 1905)

btilly• 11 years ago

Useless trivia.

Did you know that fact that speeds do not simply add together at relativistic velocities was directly measured in 1851? The result of that measurement was the belief that there was some sort of "partial frame dragging" of the ether. The exact details of which got more and more confusing as experimenters looked farther into it.

See http://en.wikipedia.org/wiki/Fizeau_experiment for details.

return0• 11 years ago

Don't forget this year is the centennial of the General theory of Relativity

If two objects approach at 75 percent the speed of light

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