# Hi, can't find the way to debunk this

1. May 5, 2007

### polystethylene

Hello all, new to the forum, so I hope this question isn't out of place or I'm contravening some rules, apologies if i am.

Anyway, I got linked to this...

http://www.webspawner.com/users/relativity/ [Broken]

Now, I had a SR course in the 1st year of uni, and I'm revising for a cosmology exam this year, (my third), so I'm comfortable with SR and its workings.

Problem is, I can't find the reasoning to debunk this rubbish, I know the solution is probably simple =/ The best I can think of is this:

In 1-D: one observer travelling at 0.6c towards a beam of light, observed from our 'rest' frame. Gallilean relativity says in the frame of the observer @ 0.6c, the light beam should be observed to be 1.6c, but is actually c as we all know. Therefore, the apparent length contraction and time dilation as seen by us at rest act to 'reduce' this to c. This is what is described in the link above. I suspect it could be this reasoning that causes the issue?

Anyway, if the observer @ 0.6c is moving away from the beam, galillean rel. says observed speed by moving guy is (1-0.6)c = 0.4c. The argument in the link says 'how can time dilation and length contraction act to increase this to c?

Well I'm thinking, taken from the rest frame of the observer @ 0.6c, we appear to move at 0.6c towards the light beam, i.e. the situation that allows length contraction and time dilation to correct for c.

I feel like in engaging in thinking about this (and it's seriously been bugging me), I've overlooked something fundamental that makes even the very proposal of this problem ridiculous. I just can't spot it.

Sorry that was a long and rambling first post, but I'd love to put my mind to rest on this one.

Thanks

Last edited by a moderator: May 2, 2017
2. May 5, 2007

### Staff: Mentor

I have to go off to commencement exercises shortly so I don't have time yet to write something out fully, but the resolution to "paradoxes" such as this almost invariably lies in the relativity of simultaneity.

Time dilation and length contraction do not give a complete account of situations like this. You have to include relativity of simultaneity along with them. Or you can use the Lorentz transformation, which incorporates all three effects and can be used to derive them.

3. May 5, 2007

### polystethylene

Thanks for the quick response. I see how it's essentially the same problem as the light bulb in the middle of a train carriage. Seeing as how that's always used a preliminary introduction to the relativity of simultaneity I figured that the lorentz transforms should take care of this basic issue, I just can't see how. I think I might go scribble some stuff down and see if I can intuit the answers from the LT's. Meant to be revising laser physics today and I'm wasting time held up on this... It's bloody annoying lol.

edit: what's the LT to use for this, should I use the velocity transforms as they incorporate both the length and time changes?

4. May 5, 2007

### Staff: Mentor

If you consider the fundamental principle from which length contraction, time dilation, and the relativity of simulaneity are derived--that the speed of light is the same for all inertial observers--cased closed. You move towards a light source: you measure the speed of the light to be c; you move away from the light source: you measure the speed of the light be c.

Viewed in terms of adding velocities: If the light source moves at some speed v with respect to you, and the speed of light with respect to that moving frame is c (which it must be), then the speed of the light with respect to you can be found by adding the speeds. But you must do it relativistically:

$$V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}$$

Plug in the numbers and you'll always get c as the speed of light with respect to you.

5. May 5, 2007

### Staff: Mentor

The Lorentz transformation is the pair of equations that connects the spacetime coordinates $(x,t)$ of an event in one inertial reference frame with its coordinates $(x^{\prime},t^{\prime})$ in another reference frame.

$$x^{\prime} = \gamma (x - vt)$$

$$t^{\prime} = \gamma \left( t - \frac{vx}{c^2} \right)$$

where

$$\gamma = \frac {1} {\sqrt{1-v^2/c^2}}$$

(One can derive the equations for length contraction, time dilation and relativity of simultaneity from these.)

Let one event be the emission of the light at a suitable $(x_1,t_1)$ in the "rest" frame, and let the other event be the arrival or absorption of the light at $(x_2,t_2)$ in that frame. Pick suitable values for $x_1$, $x_2$, and $t_1$, and calculate $t_2$ by assuming that the light travels at speed c in the "rest" frame.

Use the Lorentz transformation to calculate the coordinates of the emission and absorption points in the "moving" frame, and calculate the speed of the light in that frame as

$$c^{\prime} = \frac {x_2^{\prime} - x_1^{\prime}} {t_2^{\prime} - t_1^{\prime}}$$

Last edited: May 5, 2007
6. May 5, 2007

### polystethylene

Thanks for all the responses, I clearly need to resit that SR course as I'm getting lost here. I'm still baffled but I think I need to go away and work this through.
I still can't get that link's reasoning out of my head though. Why isn't a time contraction and a length dilation necessary to redress the balance? Part of me thinks his point makes sense, the other part of me thinks why the hell should it matter what direction the person travels, SR is isotropic right? I think I actually have a headache.

Thankyou again tho

7. May 5, 2007

### robphy

It may help to draw a spacetime diagram of the situation.

8. May 5, 2007

### Staff: Mentor

You posted while I was in the middle of revising mine. LaTeX for equations doesn't show in preview mode, so I had to debug the equations by re-editing the post and re-loading the page several times. I think my instructions are pretty explicit now.

And now I really do have to go... before the commencement lineup marches without me!

9. May 5, 2007

### polystethylene

Thank you, I'll carry out those instructions and see if I can resolve it, cheers!

10. May 5, 2007

### polystethylene

What would any of you say directly to this greg alexander guy to explain away his little theory, in a qualitative sense? Just out of interest?

11. May 5, 2007

### Staff: Mentor

I would say that time dilation and length contraction alone are not sufficient to explain why you measure the speed of light to be c in all cases. As jtbell stated, the missing piece in all such "paradoxes" is the relativity of simultaneity. Mr Alexander is not really giving an argument, he's just waving his hands. I wonder how he would explain how two observers in relative motion both observe each other's clocks to go slow?

To get a handle on this problem you need to picture it carefully. Imagine you and your co-moving observers are all spread out along the x-axis. Then imagine two space ships moving along the x-axis with speed v with respect to your frame. In the moving frame of the ships, they measure the distance between the two ships; you, of course, measure a shortened distance. They shoot a beam of light from one ship to the other and measure how long it takes; realize that you measure a different time for the light to go from ship to ship. Realize further that they consider their clocks to be synchronized, but you disagree: you measure their clocks to be out of synch--the clock in the lead ship lags behind the other. (Note that when the light shines in one direction--from ship A to ship B--it moves towards the lagging clock; but when shined in the other direction it moves away from the lagging clock. That makes a difference!)

The ship observers, using their measurements of distance and time, figure the speed of the light beam to be c. To figure what you would measure for distance and time between the two events--(1) Ship A sends the light pulse, (2) Ship B receives the light pulse--you can't just take the ship values and slap on time dilation and length contraction. For one thing, the ships move while the light travels; for another, their time measurements--according to you--were taken with unsynchronized clocks operating slowly. But after all is said and done, using your measurements of distance and time you will also measure the speed of the light pulse to be c.

If you analyze the problem using the LT (or using the addition of velocities formula derived from the LT) as jtbell described in detail, all three "effects"--time dilation, length contraction, and the relativity of simultaneity--are included automatically. (But it's a fun exercise to solve directly just using the properties of moving clocks and metersticks.)

12. May 5, 2007

### polystethylene

Thank you, I think that really elucidated the importance of the relativity of simultaneity. I can't understand why the two clocks for the ships would be out of sync though, if their relative velocity is 0 ?

Also, can I just say that I'm very impressed with this forum, I've received lots of help already and I only posted this morning. It seems to be a great resource, full of friendly helpful people. No doubt I'll be coming back here a few times over the next few weeks as I've got 7 more exams :S

13. May 5, 2007

### Chris Hillman

So you want to "argue" with a crank?

I assume that's a rhetorical question, and that you aren't planning to actually try to contact this fellow. But if I'm wrong about that, someone should point out that by definition, you can't explain anything to a crank, so unless you are a masochist you'll only be wasting your time and energy:
http://en.wikipedia.org/w/index.php?title=Crank_(person)&oldid=54987381

14. May 5, 2007

### JesseM

They are synchronized in the ship's own frame, but not in the frame where the ship is moving. This is because every observer synchronizes clocks in his own frame using the "Einstein synchronization convention", which is based on the assumption that light travels at the same speed (c) in all directions in that frame. So, using this assumption, the observer on the ship could synchronize two clocks at the front and back of the ship by setting off a flash at the midpoint between them, and then setting them to read the same time (say, 12 noon) at the moment the light from the flash reaches them. But if the second observer who sees the ship moving assumes that both pulses of light move the same speed in his frame, he must necessarily conclude that back clock received light from the flash before the front clock, because the back clock was in motion towards the position where the flash was set off, while the front clock was in motion away from it.

15. May 5, 2007

### matheinste

Hi Polystethylene

This is an extract fron your quoted source

"""If an observer with velocity v heads towards a beam of light one would have expected that the measurable velocity of the light beam would have been c + v. However according to the Special Theory of Relativity because time slows down and length decreases with velocity, the measured velocity of the beam would still be c. In other words a change in space and time for the observer slowed the new velocity of c + v back down to c again"""

The measured velocity of the beam is c. Time dilation and length contraction have nothing to do with it. They are, to use a phrase from later in the quoted article, Red Herrings.

I hope this helps.

Matheinste.

16. May 5, 2007

### JesseM

I disagree, time dilation and length contraction (and the relativity of simultaneity) are necessary to explain why the measured velocity is c if you calculate things from the perspective of a frame where the person doing the measuring is moving at velocity v.

Here's an example I came up with on another thread:

Last edited: May 5, 2007
17. May 5, 2007

### matheinste

Hi JesseM.

I hope I haven't jumped in out of my depth but I thought the velocity of a light beam measured by an observer relative to himself was always c and everything else follows as a consequence of that.

If the calculation is done by an observer who measures the velocity of an object ( or another observer ) as 0.6c relative to him in one direction and a light beam moving, of course at c, in the opposite direction, then this first observer will assign a relative velocity between the light and the object of 1.6c. So in this case we still do not have to use time dilation or length comtraction as the velocities are added classically.

Matheinste.

18. May 5, 2007

### JesseM

Yes, you're right that time dilation, length contraction, and the relativity of simultaneity are all consequences of the two fundamental postulates of relativity, namely that the laws of physics work the same way in all inertial frames, and the speed of light is c in all inertial frames. Still, conceptually it can be helpful to reverse things and show how, if each observer measures the speed of light using rulers and synchronized clocks at rest in their own frame, then time dilation, length contraction and the relativity of simultaneity all work together to ensure that each observer measures the speed as c.
Yes, but suppose the observer moving at 0.6c (from the perspective of the first observer) has a ruler at rest with respect to him, and two clocks at rest on either end of the ruler which have been synchronized in their own rest frame. If we note the time t1 on the left clock as the light beam passes it, and then note the time t2 on the right clock as the light beam passes it, and the ruler's length in its own rest frame is d, then it must be true that d/(t2 - t1) = c. And the events of the light beams passing each clock are physical ones, so every frame should predict the same thing about the time on each clock, even in frames where the clocks are ticking at a slowed-down rate and are not in sync with each other. So this is basically what I was doing in my example above, showing that even if we do our calculations in a frame where the ruler and clocks are moving, we still find that d/(t2 - t1) = c.

19. May 5, 2007

### nrqed

Hi. By now others may have explained everything to you but I wanted to point out a key point. The main problem, as I see it, is that it is NOT correct to simply use the length contraction and time dilation formula to find the speed in different frames by doing a simple ratio. It's more subtle than that.

The point is that the length contraction formula $L' = L/\gamma$ applies to length measurements, that is measurements taken at the same time in frame S'. This formula cannot be applied to calculate the speed of an object in a frame because the time of the "departure" of the object and the time of its "arrival" are not equal. So one must use the full Lorentz transformation, not just length contraction and time dilation.

To make the point clear, you can think of a purely classical example. Let's say someone throws a baseball from the back of a train toward the front. To calculate teh speed of the baseball in the frame S of someone on the ground, you can't simply take the length of the train and divide that by the time it took for the ball to get from the back to the front! Why? because the ball did not simply travel the lenght of the train, as seen from the guy on the ground!
If the ball is thrown in the same direction as the motion of the train, the distance travelled will be larger than the length of the train, and smaller if the ball was thrown opposite to the motion of the train.

So, in SR, even if you have the length of the train as measured in S and you have the time for the ball to get from the back to the front (still as measured in S), you still can't get the speed of the ball by dividing the two.

Hope this helps

Patrick

Last edited by a moderator: May 2, 2017
20. May 5, 2007

### Staff: Mentor

JesseM already answered this, but I'll repeat for emphasis. The two ship clocks are synchronized in their own frame; when observed from another frame, they will be seen as unsynchronized. That's the meaning of the "relativity of simultaneity".