Exploring the Speed of Light in Deep Space: A Scenario on a Space Station

In summary, the space station sends a beam of light in the same direction as the rocket ship is traveling. The light is measured to be going at .5 c.
  • #106
zoobyshoe said:
Somehow, I missed any statement of the results. What actual times do you get for the flashes from the train frame? Please show how you moved from - or + 346,410.15 to a value in seconds from t=0 for each flash. As I mourned to Doc Al, this number, 346,410.15 carries no meaning for me, + or -.

As I said (see previous post for derivation):

So at rest the length of the train is: 2 * 1.15 * L. Now why should this be so?

The train as measured in the embankment frame has length 2L. However, the train is in motion in the embankment frame, and thus appears contracted in that frame. So, since the value we started out with measures the contracted length of the (moving) train in the embankment frame, it makes sense that in the train frame, where the train is at rest, the length - no longer being contracted - is measured to be greater.​

So if we're setting L = 300,000km then the rest length of half the train is: x' = (1.15)(300,000km) = 345,000km.

That's one significance of the number.

So far no one's calculated a value for the time of the flashes in the train frame. It would be (for the flash at B):
[tex]
\beta\ =\ .5c/c\ =\ .5
[/tex]

[tex]
\gamma\ =\ \frac{1}{\sqrt{1 - (0.5)^2}}\ =\ \sqrt{4/3}\ =\ 1.15
[/tex]

[tex]
t'\ =\ \frac{-\gamma\beta L}{c}
\ =\ \frac{-1.15 \cdot 0.5 \cdot 300,000\ \textrm{km}}{300,000\ \textrm{km}/\textrm{s}}
\ =\ -0.575\ \textrm{s}
[/tex]​
So the event "flash at B" has coordinates
(x, t) = (L km, 0 s) = (300000 km, 0 s)
in the embankment frame, and coordinates
(x', t') = (345000 km, -0.575 s)
in the train frame.
 
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  • #107
Physicsfreek said:
Below is an excerpt (Chapter 12) from my upcoming book...

Please don't do that again. Your post is tangential to this topic. Please start a new topic if you are looking for feedback on your writing.
 
  • #108
zoobyshoe said:
By "at the same time" do you mean at a specific time coordinate such as T=0? Or something else?
That's exactly what I mean by "at the same time". If the train frame wishes to measure the length of something that's moving with the embankment frame, then measurements of the positions of the ends of that length must be made at the same time according to the train frame. Of course, since we already figured out length contraction, we know the answer will be [itex]L/\gamma[/itex] if L is the length as measured in the embankment frame.

This is what I have understood you to be saying about this since I raised my objection. However, I haven't been able to understand from you how the answer that is produced: -346, 410.16 and + 346,410.16, contains any useful information about the when and where according to the moving frame. If I observe that the A or B event took place in the embankment frame at t=0, x=+/-300,000 km. What does a value of -/+346,410.16 tell me about the moving frame?
It tells you where the event took place according to the train frame. The events are the flashing of the lights. We know where they took place according to the embankment: at x = +/- L = +/- 300,000 km. We use the LT to find out where they took place according to the train. To find out where, use the LT equation for x': note that x' depends on both x and t. It tells us that x' = +/- 346,410.16 km. That's where these flashes take place according to the train frame. That simply means that train observers measure the location of these flashes to occur at a distance of 346,410.16 km from the origin (M'). I believe you think this is somehow wrong because you are unconsciously thinking that the two frames are somehow "lined up" at T = 0. But, no, the only point they have in common is at 0,0: the point (in space-time) where M' passes M.

According to you, this number says something useful about the when and where of event A' or B' in the moving frame. If this number actually has any use, I can't see it. In fact, to me it looks like it is saying something both incorrect and useless about these events in the A' and B' frame. These numbers seem to me to say nothing at all about the when in the A' and B' frame, and to say something outright incorrect about the position A' and B' will have in the moving frame.
To find the when, use the LT for t': note that t' depends on both x and t.

From the general direction and tone of your answers, it seems clear that you understand the difficulty I am having with these numbers. I am trying to phrase my difficulty more specifically in the hope you'll see how to solve it.
I can see that you are serious about learning this stuff, so I am happy to help. But please realize that jcsd and plover already know this stuff and they are trying to help. You are the one who needs to learn it. :smile: You will get there!
Specify which of the four you wish me to try for myself. I can use the one for x' and I can use the one for t' but there is no one of the four that gives an answer for both x' and t'. You are giving me one equation and claiming the answer is telling me both the x' and t' coordinates with one number.
Here's how it goes. An event is specified by a set of coordinates in the embankment frame: x, t. That tells where and when the event occurred in the embankment frame. Now use the LT to find where and when that same event occurred according to the train frame. You will need two equations: one that gives you x' and one that gives you t'. Here they are again:

[tex]x' = \gamma (x - vt)[/tex]
[tex]t' = \gamma (t - vx/c^2)[/tex]

jcsd gave:

[tex]\gamma=\frac {1}{\sqrt{1-\beta^2}}[/tex]

where:

[tex]\beta = \frac{u}{c}[/tex]
These are just the definitions of [itex]\gamma[/itex] and [itex]\beta[/itex].
This is exactly the same equation as:

[tex]t=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
No it's not. This is just a version of the the time dilation formula. It should be written as:
[tex]\Delta t=\frac{\Delta t'}{\sqrt{1-\frac{v^2}{c^2}}} = \gamma \Delta t'[/tex]
This is a specialized formula meant for a specific use: it is not a general Lorentz transformation. (For one thing, it only applies when [itex]\Delta x' = 0[/itex], which is the case for a moving clock.) This formula means: When a moving clock measures a time of [itex]\Delta t'[/itex], how much time will pass according to the observer's clocks [itex]\Delta t[/itex].
I need for you and jcsd to justify giving me the special application of the Lorentz Transformation for time dilation and telling me it has something useful to say about the x' coordinate of either A' or B'. Here's the primary problem I have with it: it doesn't address position, it only addresses time. The result, when it is applied to position, strikes me as downright useless.
Apply the real Lorentz transformation as given above. Don't try to take short cuts by applying other formulas that are derived from the LT, but only apply in specific situations. (Like the "time dilation" formula or the "length contraction" formula.)
 
  • #109
jcsd said:
What we've done is perform the Lorentz transformation completely, to do this you must consider all four dimensions (though saying that we've cut out two because they don't change as the direction of motion is orthogonal to them, for confirmation of this in the method examine the Lorentz transformation in full),
I follow this logic without problem. y'=y and z'=z so we don't need to treat these two coordinates. We are concerned only with the two that will change from one system to the other. These two are x' and t'.

Yet, at the moment we are concerned about, t=0 and t'=0' there is nothing in the train frame coordinate for time that is different from the embankment frame coordinate for time. x' is different. At this point x is the only coordinate that is different between the two frames due to relative motion at speed .5 c.

When you and Doc Al say you are considering time and position both at once, I grasp the concept, but since the time coordinate is 0 for both frames, the net result of considering both time and location for this moment t' = 0 should be that only position, the x coordinate, is left to treat in any way.

The position of A' from M' or B' from M' should turn out to be closer together by any means of calculation than it is in the M frame. Length contracts. Instead, to my disbelief and confusion, they seem to be farther apart.

Positions A and B, are indeed, located by means of length. To say they have a coordinate is meaningless unless we locate that coordinate with repect to some point =0. This you have done, by using L, length, as the distance from M--->A and M--->B. We have designated that length to be 300,000 km.

The Lorentz transformation you gave:

[tex]\gamma = \frac{1}{\sqrt{1-\beta^2}}[/tex]

gives us the figure 1.1547005 when we put .5 c in as the relative velocity of the train and the embankment.

1.1547005 times the L, 300,000, = 346,410.16

Since the coordinate system requires a length in order to define a position, it seems obvious to me that this number: 346,410.16 refers to a length. You may argue it also refers to a time because you have used the tensor Lorentz transformation, but since the time component at this point is 0, we are left only with length to transform.

Therefore, I am confused by the fact the length 300,000 km has become 346,410.16, i.e. a larger figure.




t doesn't equal gamma, gamma is just a unitless quantity that is useful in special relativity and it comes into the equation for length contraction too.
Bear the concept of context in mind. In the context of my post, the statement: t = [tex]\gamma[/tex] was completely accurate.
 
  • #110
Just a quick comment, as I know all will be clear when you read my previous post! :rolleyes:
zoobyshoe said:
Yet, at the moment we are concerned about, t=0 and t'=0' there is nothing in the train frame coordinate for time that is different from the embankment frame coordinate for time. x' is different. At this point x is the only coordinate that is different between the two frames due to relative motion at speed .5 c.

When you and Doc Al say you are considering time and position both at once, I grasp the concept, but since the time coordinate is 0 for both frames, the net result of considering both time and location for this moment t' = 0 should be that only position, the x coordinate, is left to treat in any way.
No no no! You assume that the times are the same in both frames. Not so! They only have one point where they agree: that point where/when M' passes M. That point is 0,0 in both frames. But imagine a clock at A and a clock at B. While the embankment frame will say that the flashes occurred at t = 0 according to their clocks, the train frame sees those clocks as being out of synch! That's a key result of the "Einstein train gedanken", quantitatively explained by the LT. In the train frame, those flashes do not occur at the same time.

The position of A' from M' or B' from M' should turn out to be closer together by any means of calculation than it is in the M frame. Length contracts. Instead, to my disbelief and confusion, they seem to be farther apart.

Positions A and B, are indeed, located by means of length. To say they have a coordinate is meaningless unless we locate that coordinate with repect to some point =0. This you have done, by using L, length, as the distance from M--->A and M--->B. We have designated that length to be 300,000 km.
Again, you are talking about the length between A and M, but we are talking about the distance between events that happen at different times! Not the same thing at all!
 
  • #111
The spped of light is independant of motion and the same for all observers.
i.e no matter how fast you travel it will always catch up with you at the same rate.
for example the speed of light is 670 million mph(not S.I units i know) if you try to outrun the light at 600million mph it will still gain on you at a rate of 670million mph

I hope this helps
 
  • #112
Gentlemen,

I do not believe, given the amount of math a person seems to need to know to understand this situation, that I'm going to be able to grasp it.

It would be nice if I understood Relativity as you do, but since I have no actual need of it in my life, this is turning out to be more trouble than it is worth simply to satisfy a curiosity.

I think that if four intelligent, educated people working patiently over several days, can't get me over my confusion, then it is probably a sure diagnosis of a defective Relativity lobe in my brain. I'm disapointed to discover I suffer from this malady, but there is nothing to be done.

So, I am retiring from this thread. I would like to thank you all for your efforts.

Zooby
 
  • #113
Doc Al makes a vital point t= 0 and t'=0 are not equivalent only (x,t) and (x',t') are equivalent that is because we are dealing with two different co-ordinate systems.

imagine the transformation of a co-ordiante system on a piece of graphing paper:

x' = x + 5

y' = y + 1

The Lorentz tranfomration is simlair to this, by changing refernce frames we are just choosing a new co-ordinarte system in spacetime. I choose the above example for simplicity but in actual fact the Lorentz transformation is more like rotating the axes about a point and you won't go far wrong as visualizng the Lorentz transformation as a rotation about a point in 4-dimensional spacetime.
 
  • #114
book recommendations

zoobyshoe said:
I do not believe, given the amount of math a person seems to need to know to understand this situation, that I'm going to be able to grasp it.
To derive and understand the LT and the famous SR effects (time dilation, length contraction, and simultaneity) all you need is a bit of algebra. (Forget about 4-vectors and tensors... for now. :smile:)
I think that if four intelligent, educated people working patiently over several days, can't get me over my confusion, then it is probably a sure diagnosis of a defective Relativity lobe in my brain. I'm disapointed to discover I suffer from this malady, but there is nothing to be done.
I think part* of the problem is that you are jumping into the middle of a problem before learning the basics. What I would recommend--the next time the relativity bug bites you--is to pick up a book geared to the beginner. One that takes you step by step, anticipates student misconceptions and thinking traps, and is focused on teaching the basics, not covering everything there is to know about SR. Here are a few that I like:
Space and Time in Special Relativity by N. David Mermin. This is probably the easiest place to start.
Space Time Physics by E. Taylor and J. Wheeler. A classic.
A Traveler's Guide to Space Time by Thomas Moore. I only recently discovered this one (even though it's 10 years old), but I like it.​
Check these out in your library and see if one suits your learning style. (Every one will require work.) Based on the questions you'ved asked, I recommend Mermin.

Good luck to you. We'll be here when you're ready.


* another part of the problem is that each of us "helpers" is approaching the problem slightly differently--that can only add to the confusion. Some of us are so used to doing these kinds of problems, that it is hard to remember how a complete SR novice feels. There are "advanced" (not really, but I'm sure it looks that way) methods for calculating SR effects--using 4-vector algebra and tensors--that make these problems trivial. But you don't need that to do these problems or to learn the basics.
 
  • #115
Yep, I should of said that 4-vectors and tensors aren't necessary in anyway to understanding SR, they're just useful tools (infact SR is a good way of introducing tensors).

Zoobyshoe SR is not beyond you, infact it's a very logical and easy to follow theory provided you follow it correctly.
 
  • #116
Doc Al said:
* another part of the problem is that each of us "helpers" is approaching the problem slightly differently--that can only add to the confusion. Some of us are so used to doing these kinds of problems, that it is hard to remember how a complete SR novice feels. There are "advanced" (not really, but I'm sure it looks that way) methods for calculating SR effects--using 4-vector algebra and tensors--that make these problems trivial. But you don't need that to do these problems or to learn the basics.
Another challenge, not necessarily in this thread, is the somewhat disjointed nature of the 'conversation', as well as an unavoidable reliance on a somewhat limiting medium (PF is great for in many ways, but real-time discussion of good diagrams, for example, isn't one of them). There's also - necessarily - a limitation in being able to match the 'teaching' to each learner's own best learning styles.
 
  • #117
I was trying to work out a way to set up a list of agreed upon points and assumptions for the thread so people wouldn't have to construct a different context for each poster, but I never finished writing up the suggestions. Plus, of course, there's no real support for that kind of thing in the forum software.

I agree with jcsd, there's no reason this should be beyond you. The scenario that's been worked out here may have become too tangled to be useful at the moment, but writers have come up with many different examples to illustrate the foundations of SR. Working through the logic of one of these other examples might be all that's necessary to bring SR into focus for you.

If you are a visual learner, the Einstein essay, good as it is, is probably not the best place to start as it is so sparsely illustrated. More recent treatments tend to have a lot of diagrams to show the different steps of the logic, e.g. the Wheeler book mentioned above does this. I'm not familiar with the Mermin book, but Mermin is a terrific writer - so even without Doc Al's recommendation I would expect a treatment of these issues by him to be worth a look.
 
  • #118
Hmmm, I see I'm not needed anymore, I got to page three, and i saw what i think then was the aspect that zoobyshoe seemed to miss.

I came up with the following, before it got to all that mathematics ( i couldn't follow as clearly as i'd have liked)

A & B are the spaceships. to maintain compatability, just drop A and only look at references to B

Lightning flashes at points A & B when Rocket is at point M as seen from Observer at point X ( at rest with A & B, located Parallel to M, at some distance to A-M-B)

the time for light from A & B to reach point M is 1 second.

Rocket is moving away from point A, towards point B at .5c (stated because it's the most important fact.

in 1 second, light from B has passed the rocket and reaches point M
in 1 second, Light from A has reached point M, the rocket is another .5 Second away in terms of distance to travel at light speed, so the Observer on the rocket has seen the flash from B, but has yet to see the flash from A.

from this I conclude the observer sees the flash from B at .66 Seconds from M and Flash from A at 1.66 seconds from M

while the Observer at X see's both flashes after one second (slightly longer than one second, as X is part of a Rightangle triangle AMX)
 
  • #119
Forgot to add, an observer at point X would see the Flashes at the same instant s/he saw the rocket halfway between M & B. I got the impression Zoo was thinking they all saw the flashes when the rocket was at point M.

the lightning hit when both saw the rocket at point M, not both saw the light when the rocket was at point M.
 
  • #120
Ouch. I tried to follow the needlepoint in this thread. My head hurts. Simplification. When a photon is emitted, everything in its path 'freezes'. No matter what speed the receiver is traveling, the photon will reach it in exactly d/c seconds [according to the receiver's clock]. Call it time dilation, length contraction, or whatever. It will get there right on 'time'.
 
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