# Relativity question(layperson level)

1. Sep 27, 2010

### troymclure

Ok i'm a physics layperson, who has a question about relativity which i was hopging someone could answer in a way that i could understand. Ie without using mathematics hopefully as i failed high school maths(well quit but heh). I did pass physics so i'm not completely unfamiliar with some of the basics but a maths free or maths lite answer to my query would be much appreciated.

Ok so question is.

I get that you can't accelerate faster then the speed of light from you inertial frame of reference. To do so requires an infiniate amount of energy to cause your infinite amount of mass to accelerate once you get too 99.999999% (etc) of light speed. I think i get inertial frames of reference as well.

However. i'm thinking that i don't cause i don't see how this leads to not being able to travel faster then light.

Ie (and i reliase i'm very very likely missing something very obvious here hence why i'm asking for some assistance) lets say you build a spaceship which is capable of travelling at .25% of light speed ie .25c Thats very quick, but well within the laws of physics. Now let say you leave earth accelerating to .25c you then dock at a space station which is orbiting a planet some distance away from earth. And the speed that that station moves at (including the speed of the star the planet and the stations various velocities, ie rotational) is .25c faster then the speed the earth moves at. IE you don't have to change speeds to dock at this station, (though you could say +- a few%). YOu are now at rest in your new inertial frame of reference? aren't you? You can leave the ship, walk around the space station, etc etc. Then you leave, and you accelerate again .25c upon leaving this station and then ytou arrive at another station and you do the same thing, rinse repeat say 5 times.

Aren't you then travelling at 1.25c? Now you've never accelerated to 1.25c you've only ever accelerated to .25c but relative to earth you are now travelling at 1.25c.

Aren't you?

So yeah basically if someone could in (hopefully) non mathematical terms explain why this doesn't work to me, twould be much appreciated. I think it's a problem with my understanding of inertial reference frames... but if it's all relative as soon as you stop accelerating aren't you then in a new reference frame? Ie once the airplane stops accelerating you can move freely around the cabin.

anyways any and all answers/replies much appreciated.

2. Sep 27, 2010

### Staff: Mentor

The problem is that speeds don't add like you think they do. For low speeds (low compared to the speed of light), speeds approximately add. For example if you threw a baseball at 50 mph on a train moving at 50 mph, then the speed of the ball with respect to the tracks would be 100 mph.

But for high speeds, one must 'add' them relativistically. If your rocket moved at 0.25c with respect to a space station, and the station moved at 0.25c with respect to earth, then it turns out that the speed of the rocket with respect to earth would be only 0.4c, not 0.5c. And the effect gets stronger the faster the speeds get. You can never 'add' two speeds less that c and get something greater than c.

3. Sep 27, 2010

### JDoolin

I think if you can understand visually, the concepts of scaling, rotation, and skew, you have a chance to understand relativity without using any math.

If you can realize that the Galilean Transformation (what you are used to thinking about velocity change) is a skew operation on a space-time diagram, but Special Relativity uses two scaling operations on a space-time diagram, which preserve the speed of light, then you have a good visual but nonmathematical basis for understanding Special Relativity.

4. Sep 27, 2010

### troymclure

Ahh awesome thanks muchly.

The way to add velocities in special relativity is the simple answer i was missing (now off to see if i can gronk how that works). Much appreciated.

and thanks about the how to visualise relativity without maths. Think visualisation makes the most sense as should be alot quicker then trying to learn a couple of years of maths syllabus. tq all.

5. Sep 27, 2010

### troymclure

OK actually another quick question if i may. Just in a bid to see if i do understand intertial frames. What makes an inertial frame?

Ie the above scenario is an inertial frame.. but wouldn't' simply turning off the engine of the space ship amount to the same thing? Ie as soon as it stops accelerating it's in a new inertial frame of reference. and wouldn't' that mean that even though you can't attain speeds faster then c you can at least attain speeds close to c without incurring high mass penalties so long as you only ever accelerate in fractions of c?

6. Sep 27, 2010

### HallsofIvy

What do you mean by "accelerate in fractions of c"? First of all, c is a velocity, not an acceleration, so they have different units.

7. Sep 27, 2010

### troymclure

I mean accelerate the ship from say velocity of 0% of to 10% of c. Turning the engine off and thereby "locking in" a new inertial reference frame each time? (this is the bit i'd like some clarity on).

Ie does just turning the engine off and no longer accelerating cause a new a inertial reference frame? They are no longer accelerating, if someone were to say go for a spacewalk outside the ship, the ship would not move away from them and so it's a new inertial reference frame isn't it?

Which means that they could accelerate from 0% to 10% of c again. Then rinse/repeat as per previous example.

8. Sep 27, 2010

### JesseM

You don't have to stop to "lock in" a new inertial frame, in you can analyze any situation from any frame you wish at any time, and an accelerating ship will have an "instantaneous co-moving inertial frame" at any given instant (i.e. the frame where the ship has an instantaneous velocity of 0 at that instant). There is also a concept called "proper acceleration", which is the rate at which coordinate velocity is changing with coordinate time in the instantaneous co-moving inertial frame, and which also determines the G-force that will be felt inside the accelerating ship. Maintaining a constant proper acceleration isn't any harder in relativity than it would be in Newtonian physics (though in both cases it's difficult to accelerate at the same rate for a long period of time, since the longer you want to accelerate the more fuel you must carry onboard at the start, so the more massive your ship will be at the start and the more energy you will need to use up early on to accelerate the ship at that rate). However, in Newtonian physics the proper acceleration is the same as the coordinate acceleration (change in coordinate velocity per unit of coordinate time) in the frame where the ship started out at rest in before beginning the acceleration, but in relativity this is no longer true, a ship with constant proper acceleration will have a continually decreasing coordinate acceleration in the frame it started out at rest in, which explains why it can never reach the speed of light in that frame (or any other frame) even if it maintains the same proper acceleration forever.

9. Sep 27, 2010

### JDoolin

No. There's an old quote "Wherever you go, there you are." Roughly translated to relativity, that means, no matter how fast you accelerate, you're always stopped.

Also, changes in velocity are linear in rapidity. Input a change in rapidity between -infinity and infinity, and you get a change in velocity between -c and c.

10. Sep 27, 2010

### troymclure

Ah yep, that makes sense, was thinking it's an exponential increase in mass requirements as speeds approach C which it is, but even accelerating in .10% of c lots you'll still get the same exponential increase in mass requirements due to the way velocities are added in relativity. Which all makes sense which is great and much appreciated all. And good to know how inertial reference frames work in regards to accelerating objects. Still got too have a think about some of the concepts you've all posted(haven't quite gotten how it's possible adding two velocities which should be greater then c do not equal a velocity greater then c, well seen the maths, unsure how it's physically possible atm but i've read it's a counterintuitive concept so i'll likely learn how too wrap my head around it eventually), still as said thanks all i think i understand relativity a bit better now.

i think.

Last edited: Sep 27, 2010
11. Sep 27, 2010

### Passionflower

Of course very true.

An interesting aside is that acceleration in geometrized units becomes:

$${c^2 \over \alpha} \, \, [meters^{-1}]$$

$\alpha$ by itself is actually dimensionless and is a rate.

Last edited: Sep 27, 2010
12. Sep 27, 2010

### JesseM

Keep in mind that there's no objective notion of speed in relativity, so mass doesn't increase in any objective way either, you can only talk about mass increase relative to some choice of frame.
It may help to realize that the velocity addition formula is comparing speeds in two different frames--if you see a rocket moving at 0.6c in your own rest frame, and I see you moving at 0.8c in the same direction in my frame, then I will measure the speed of that rocket in my frame to be (0.6c + 0.8c)/(1 + 0.6*0.8) = 1.4c/1.48 = 0.946c. Each frame defines speed in terms of distance/time as measured by rulers and clocks at rest in that frame, but rulers at rest in one frame are length contracted in other frames, and a pair of synchronized clocks at rest in one frame are time dilated and out of sync. I gave a numerical example showing how two different frames would both measure the same light beam to have a speed of c using their own rulers and clocks in post #7 of this thread if you're interested.

13. Sep 28, 2010

### troymclure

Ah cheers JesseM that does help the gronking a bit. It really is all relative. (in regards to the mass thing) and i'll have a further perusal of your post again though a cursory look does seem to click over a few more gears so much appreciated one and all.

14. Dec 21, 2010

### Rap

Remember velocity is relative. There is no such thing as just velocity, only velocity with respect to some frame. If you are in frame A and accelerate to 0.10C then that velocity is with respect to frame A. If you call this new frame B, and then accelerate to 0.10C, that will be the velocity with respect to B. But the new velocity with respect to A will not be 0.10C+0.10C=0.20C, because the velocities do not add. The new velocity with respect to A will be less than 0.20C. And if you do the exact math, you will see that no number of such steps will ever get you going faster than C.

15. Mar 2, 2011

### troymclure

Then how can I ever move at all? Because why is my starting frame A "special".

Eg. Lets say i build a rocket ship on earth and travel to alpha centauri going at .9c. Thing is, I use only specific atoms to build the rocket ship these atoms all come from big bang. And as the big bang has a different reference frame to that of the earths... what happens then?

"but in relativity this is no longer true, a ship with constant proper acceleration will have a continually decreasing coordinate acceleration in the frame it started out at rest in"

Ie this quote. the frame it started out at rest in. That's not the earth is it? If you don't need to be at rest to "lock-in" an intertial reference frame then the inertial reference frame of all matter should be the big bang?

16. Mar 2, 2011

### JesseM

Sorry if my phrasing was confusing, but when I said "started out at rest in" I didn't mean for it to refer to any objective starting point in the history of ship or its particles. You can pick any point on the ship's worldline (whether the moment when it first sets out from Earth or the moment it has already been accelerating away from Earth for 10 years according to its own clock) and there will be a frame where it's instantaneously at rest at that point, then you can arbitrarily define that time as the "starting time" t=0 in that frame (the time an observer at rest in that frame first turns on their stopwatch, say) and then use the relativistic rocket equations http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken] to find the ship's velocity at later times in that frame. Of course you could also just derive modified equations which don't assume the accelerating ship had a velocity of 0 at t=0, but this assumption does make the math a bit simpler.

Last edited by a moderator: May 5, 2017
17. Mar 2, 2011

### troymclure

Cheers JesseM.

Also is the big bang the intitial reference frame for most of the matter in the universe? And does that actually affect anything?

18. Mar 3, 2011

### HallsofIvy

The "big bang" did NOT occur at any specific point in space. Space itself was created by the big bang. Every point in space can be taken as "where the big bang happened".

19. Mar 3, 2011

### lugita15

Let's consider the early universe at time t. Then at time t+dt the universe will have a new points in space. But doesn't there exist a natural isomorphism between some subset of the universe at t+dt and the full universe at time t? Generalizing, if s<t, doesn't there exist a natural isomorphism between some subset of the universe at time t and the full universe at time s? Call that subset U_s(t). Then if we are living at time T now, couldn't we say that the "location of the big bang" is the limit of U_s(T) as s goes to 0? Even if such a limit is not well-defined, we can still make s sufficiently small that the volume of U_s(T) is negligible.

Or am I mistaken? Could the universe have been infinite right after the big bang?

20. Mar 3, 2011

### Staff: Mentor

Think about this a bit. Consider the expanding circle x²+y²=t². At time t how many points are on the circle? At time t+dt how many points are on the circle? Which points at t+dt are "new points"? What is the natural isomorphism between the subset of the universe at t+dt and the full universe at t?