- #106
Grimble
- 485
- 11
Ich said:It isn't traveling the same distance.
But it is, within the clock's frame of reference, and the speed of light has to be the same wherever it is observed from
Ich said:It isn't traveling the same distance.
Ich said:I still don't see it. The problem Einstein refers to arises if you say that w=c-v is the speed of light as measured by the carriage. It isn't.
w is the required velocity of light with respect to the carriage, and we have
w = c - v.
The velocity of propagation of a ray of light relative to the carriage thus comes out smaller than c.
Ich said:They didn't write what you read. There is no such thing as "observing the clock's frame of reference". You may observe the clock, recording time and date of such measurements as given by your frame of reference.
How are we to find the place and time of an event in relation to the train, when we know the place and time of the event with respect to the railway embankment? Is there a thinkable answer to this question of such a nature that the law of transmission of light in vacuo does not contradict the principle of relativity? In other words: Can we conceive of a relation between place and time of the individual events relative to both reference-bodies, such that every ray of light possesses the velocity of transmission c relative to the embankment and relative to the train? This question leads to a quite definite positive answer, and to a perfectly definite transformation law for the space-time magnitudes of an event when changing over from one body of reference to another.
Why don't you read it as saying:
"However to an observer who sees the clock pass at velocity v, the light takes more time to traverse the length of the clock when the pulse is traveling in the same direction as the clock, and it takes less time for the return trip." ?
Again: why don't you try to understand what they wrote instead of interpreting their every word? That has nothing to do with accusing you of being dim, but you definitely always skip the first step - reading and following their argument - and jump to the second: interpreting hat you have learned.
I think a reasonable first step would be if you draw a spacetime diagram of the described events; you can then easily read off distances, times, and velocities, and check with what the link is saying contrary to what you have assumed it was saying.
matheinste said:Hello Grimble.
As you say :-
Einstein wrote:
w is the required velocity of light with respect to the carriage, and we have
w = c - v.
The velocity of propagation of a ray of light relative to the carriage thus comes out smaller than c.
But if you follow the argument from beginning to end he explains that this relation cannot hold unless we abandon the principle of relativity, and as we know, he did not abandon it.
I have not had time to reread the translation of Einstein's work, but I recall that it is probably one of the best and simplest explantions of the theory. I think it may be unwise to put a different interpretation on what Einstein wrote, after all its his theory and subsequent events have so far proved him correct. He proposed the constancy of c in inertial frames and so how likely do you think it is that he would argue against it except to make a point of how the theory would not hold up if this proposition was untrue.
Matheinste.
The problem Einstein refers to arises if you say that w=c-v is the speed of light as measured by the carriage. It isn't.
Both sides of the moving rectangle are measured in coordinate light-seconds. When you want to compute a distance in any coordinate system, you only use that coordinate system's measures of distance. Again, if you want to know the distance between two points in a given frame, you take the coordinate distance between them on the x-axis which we can call dx, the coordinate distance on the y-axis dy, and the coordinate distance on the z-axis dz, then use the pythagorean formula [tex]\sqrt{dx^2 + dy^2 + dz^2}[/tex] to get the total distance between the points. The procedure is the same regardless of whether you are measuring the distance between objects (or parts of an object, like different corners of a square) that are at rest in this frame, or between objects that are in motion in this frame. Nowhere would the notion of "proper distance" enter into it at all.Grimble said:Yes, I can see what you are saying BUT you have neglected to say that the 8 light-seconds is measured in co-ordinate seconds, not proper seconds and how do you apply Euclidean/Newtonian physics to a rectangle whose sides are measured in different units?
JesseM said:I don't understand, why do you think the 1st postulate implies that the time should be constant? The 1st postulate implies that if you run the same experiment in different frames each frame will see the same result, so if you construct a light clock at rest in frame A and measures the time in frame A, it should give the same answer that you'd get if you constructed an identical light clock at rest in frame B and measured the time in frame B. But the 1st postulate does not say that if you construct a light clock at rest in frame A but in motion in frame B, and measure the time in frame B, that the result would be the same as if you construct a clock at rest in frame B and measure the time in frame B. In this case we are dealing with two clocks that have different velocities in frame B, but we are measuring both their ticking rates from the perspective of frame B--nothing about the 1st postulate suggests that their ticking rates should be identical.
Identical time in their own respective rest frames. They certainly do not keep identical time if you measure both from the perspective of a single inertial frame in which they have different speeds--that's exactly what I meant when I said above "In this case we are dealing with two clocks that have different velocities in frame B, but we are measuring both their ticking rates from the perspective of frame B--nothing about the 1st postulate suggests that their ticking rates should be identical." Do you think the 1st postulate implies their ticking rate should be identical even in this situation? If so, can you explain your reasoning?Grimble said:You are quite right when you say that the two clocks should keep identical time;
No, it definitely would not! Time dilation is what remains after you correct for transmission delays (i.e. correcting for the Doppler effect). For example, suppose the clock is moving at 0.6c, and in 2020 I see it next to a marker 10-light-years away from me (in my frame) with the clock showing a reading of 30 years, and then in 2036 I see it next to a marker 16-light-years-away from me with the clock showing a reading of 38 years. If I subtract off the light travel times (10 years for the first reading to travel 10 light-years from the clock to my eyes, and 16 years for the second reading to travel 16 light-years from the clock to my eyes), I will conclude that the clock "really" showed the first reading of 30 years in 2020-10=2010, and it "really" showed the second reading of 38 years in 2036-16=2020. So I will conclude that in the 10 years between 2010 and 2020, the clock itself only ticked forward by 8 years from 30 to 38, so it must have been slowed down by a factor of 0.8 in my frame. This is different from how much it appeared to be slowed down visually--visually it took 2036-2020=16 years to tick forward by 8 years, so it appeared to be running slow by a factor of 0.5, but this extra slowdown is just due to the Doppler effect (which is a consequence of the fact that light from different readings on the clock has different delays in reaching me since the clock's distance from me is changing).Grimble said:if the moving clock were to transmit a pulse of light each second, then allowing for the transmission time for the pulse of light, the stationary frame would still determine that the moving clock was 'ticking' at the same rate as its own clock.
JesseM said:I'm not sure what you mean by "space and time not being absolute"--certainly it's true that the distance and time between a pair of events will vary depending on what frame you use, but there is nothing in SR that implies we can't use the Euclidean formula for the distance between events in a single frame, or that we can't use the ordinary kinematical formula that velocity = distance/time in each frame. The light clock thought-experiment makes use of these ordinary geometrical and kinematical rules which still apply in relativity, along with the uniquely relativistic notion that if the light is bouncing between mirrors at c in the light clock's own rest frame, it must also be bouncing between them at c in the frame where the light clock is in motion.
OK, but in that quote he was saying exactly the same thing I was saying when I said "certainly it's true that the distance and time between a pair of events will vary depending on what frame you use". Do you agree that there is no conflict between this statement and my other statement immediately after? Namely:Grimble said:I am referring to Einstein's statement in http://www.bartleby.com/173/11.html" where he says
THE RESULTS of the last three sections show that the apparent incompatibility of the law of propagation of light with the principle of relativity (Section VII) has been derived by means of a consideration which borrowed two unjustifiable hypotheses from classical mechanics; these are as follows:
1. The time-interval (time) between two events is independent of the condition of motion of the body of reference.
2. The space-interval (distance) between two points of a rigid body is independent of the condition of motion of the body of reference.
there is nothing in SR that implies we can't use the Euclidean formula for the distance between events in a single frame, or that we can't use the ordinary kinematical formula that velocity = distance/time in each frame. The light clock thought-experiment makes use of these ordinary geometrical and kinematical rules which still apply in relativity, along with the uniquely relativistic notion that if the light is bouncing between mirrors at c in the light clock's own rest frame, it must also be bouncing between them at c in the frame where the light clock is in motion.
JesseM said:Both sides of the moving rectangle are measured in coordinate light-seconds.
Yes, my reasoning is to follow what the 1st postulate says:JesseM said:Identical time in their own respective rest frames. They certainly do not keep identical time if you measure both from the perspective of a single inertial frame in which they have different speeds--that's exactly what I meant when I said above "In this case we are dealing with two clocks that have different velocities in frame B, but we are measuring both their ticking rates from the perspective of frame B--nothing about the 1st postulate suggests that their ticking rates should be identical." Do you think the 1st postulate implies their ticking rate should be identical even in this situation? If so, can you explain your reasoning?
or as he says in http://www.bartleby.com/173/5.html"The Principle of Relativity – The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems in uniform translatory motion relative to each other.
If K is a Galileian co-ordinate system, then every other co-ordinate system K' is a Galileian one, when, in relation to K, it is in a condition of uniform motion of translation. Relative to K' the mechanical laws of Galilei-Newton hold good exactly as they do with respect to K. 2
We advance a step farther in our generalisation when we express the tenet thus: If, relative to K, K' is a uniformly moving co-ordinate system devoid of rotation, then natural phenomena run their course with respect to K' according to exactly the same general laws as with respect to K. This statement is called the principle of relativity (in the restricted sense).
JesseM said:OK, but in that quote he was saying exactly the same thing I was saying when I said "certainly it's true that the distance and time between a pair of events will vary depending on what frame you use". Do you agree that there is no conflict between this statement and my other statement immediately after? Namely:there is nothing in SR that implies we can't use the Euclidean formula for the distance between events in a single frame, or that we can't use the ordinary kinematical formula that velocity = distance/time in each frame. The light clock thought-experiment makes use of these ordinary geometrical and kinematical rules which still apply in relativity, along with the uniquely relativistic notion that if the light is bouncing between mirrors at c in the light clock's own rest frame, it must also be bouncing between them at c in the frame where the light clock is in motion.
but it applies them across frames.The light clock thought-experiment makes use of these ordinary geometrical and kinematical rules which still apply in relativity
JesseM said:No, it definitely would not! Time dilation is what remains after you correct for transmission delays (i.e. correcting for the Doppler effect). For example, suppose the clock is moving at 0.6c, and in 2020 I see it next to a marker 10-light-years away from me (in my frame) with the clock showing a reading of 30 years, and then in 2036 I see it next to a marker 16-light-years-away from me with the clock showing a reading of 38 years. If I subtract off the light travel times (10 years for the first reading to travel 10 light-years from the clock to my eyes, and 16 years for the second reading to travel 16 light-years from the clock to my eyes), I will conclude that the clock "really" showed the first reading of 30 years in 2020-10=2010, and it "really" showed the second reading of 38 years in 2036-16=2020. So I will conclude that in the 10 years between 2010 and 2020, the clock itself only ticked forward by 8 years from 30 to 38, so it must have been slowed down by a factor of 0.8 in my frame. This is different from how much it appeared to be slowed down visually--visually it took 2036-2020=16 years to tick forward by 8 years, so it appeared to be running slow by a factor of 0.5, but this extra slowdown is just due to the Doppler effect (which is a consequence of the fact that light from different readings on the clock has different delays in reaching me since the clock's distance from me is changing).
matheinste said:With regards to the first point the wording seems over complicated and still confuses me. Clocks just show time
Perhaps I can give examples, in my view, of faulty and correct reasoning with regard to the often used example of the muon's lifetime as an aid to illustrating time dialtion. These two methods lead to exactly the opposite outcome.
Let the lab frame be regarded as the stationary frame and the muon's frame the moving frame with repect to it. We can use the values of 2 microseconds 60 microseconds as being the figures used for the decay times of the muon measured by clocks in the muon and lab frame respectivley. Both explanations are non rigorous.
WRONG reasoning:- The muon's lifetime of 2 micoseconds its own frame is extended to 60 microseconds in the lab frame. This is an example of time dilation this shows that the number of seconds which the muon lives is dilated, made bigger, to 60 microseconds.
Now bear in mind the definition which says that a moving clock viewed from a stationary frame runs slow, and reason as follows.
CORRECT reasoning:- The muon has a lifetime as measured in its own frame, the time measured by a clock carried with it, its proper time, of 2 microseconds. This is an invariant and is the same for everyone, it cannot be changed. In the lab frame this is measured as 60 microseconds. This is an example of time dilation and shows that 2 microseconds in the muon's frame takes 60 microseconds to pass in the lab frame. So the time it takes 2 microseconds to pass in the moving frame is dilated to 60 microseconds as viewed from the stationary frame. That is, the preiod is extended.
Remember that although the lab frame measures 60 microseconds, the lab observers still agree that the muon's clock reads a proper, invariant time of 2 microseconds.
Matheinste.
You bolded the second half of my sentence but then ignored the first half, taking the meaning out of context. I first said "In this case we are dealing with two clocks that have different velocities in frame B, but we are measuring both their ticking rates from the perspective of frame B"--so when I then said "nothing about the 1st postulate suggests that their ticking rates should be identical", I was clearly talking about their ticking rates in frame B, not their ticking rates in their own respective rest frames. Hopefully you'd agree that nothing about the first postulate suggests that their ticking rates should be identical in frame B, given that one is at rest in frame B and the other is not?Grimble said:And if, natural phenomena run their course with respect to K' according to exactly the same general laws as with respect to K, then time and distance must be identical. (A muon's half life cannot be different and if time is identical distance has to be also cf. the speed of light)JesseM said:Identical time in their own respective rest frames. They certainly do not keep identical time if you measure both from the perspective of a single inertial frame in which they have different speeds--that's exactly what I meant when I said above "In this case we are dealing with two clocks that have different velocities in frame B, but we are measuring both their ticking rates from the perspective of frame B--nothing about the 1st postulate suggests that their ticking rates should be identical." Do you think the 1st postulate implies their ticking rate should be identical even in this situation? If so, can you explain your reasoning?
That is; they will keep the same 'Proper Time' within their own frames of reference.
So in two independent Inertial Frames of Reference, identical clocks will keep identical time.
But yes, if viewed by an independent observer they will shew different times."
Grimble
How do you think the light clock thought-experiment applies these rules "across frames"? The light clock thought-experiment derives the slowed down rate of ticking of the moving light clock using only a single frame, namely the frame in which the light clock is moving--the derivation only uses velocities and distances and times which are measured in the coordinates of that frame.Grimble said:No, because but it applies them across frames.JesseM said:The light clock thought-experiment makes use of these ordinary geometrical and kinematical rules which still apply in relativity
I'm not sure what you mean "the total duration in absolute terms". Certainly if you're talking about "proper time", meaning the time as measured by a clock moving along with the object (in this case the muon), then it is true that there is no disagreement between frames about the proper time between two events on the object's worldline (like the muon being created and then decaying, which you can average for many muons to derive the half-life). But what does this point about proper times have to do with the light clock derivation of the time dilation equation, an equation which deals with coordinate time in the frame where the light clock is moving, not proper time?Grimble said:As I said earlier, identical clocks in Inertial frames of reference will keep identical time.
It is only when one is observed from the other that time dilation occurrs.
Time dilation is the phenomenon where the time observed from one frame is different from that observed from the other.
So the observer from the other frame will see the time transformed in unit size and number of units but the total duration in absolute terms has to be the same - the half-life of the muon cannot change, only how it is measured can, as demonstrated by the afore mentioned experiment where the half-life was extended to 65secs, 65 transformed seconds that are [itex]\frac{1}{29.4}[/itex] of the laboratory seconds.
From this, and some of your comments before, it is obvious that you have no idea what a reference frame is good for, or even how "speed" is defined. You should get familiar with this basic stuff in relativity before you move on.Ich said:It isn't traveling the same distance.
But it is, within the clock's frame of reference, and the speed of light has to be the same wherever it is observed from
Ich said:Sorry, I missed a few posts of yours.
From this, and some of your comments before, it is obvious that you have no idea what a reference frame is good for, or even how "speed" is defined. You should get familiar with this basic stuff in relativity before you move on.
So I invite you to draw that spacetime diagram of the quoted situation, and post it here along with the derivation of light travel time and speed of light. You'll encounter some points where you don't know how to proceed; it would be most fruitful if we could help you exactly with these points.
JesseM said:How do you think the light clock thought-experiment applies these rules "across frames"? The light clock thought-experiment derives the slowed down rate of ticking of the moving light clock using only a single frame, namely the frame in which the light clock is moving--the derivation only uses velocities and distances and times which are measured in the coordinates of that frame.
Do you understand that just because I am observing a clock which is moving relative to myself, does not mean that I need to use any frames other than my own rest frame to analyze its behavior? That talking about the properties of an object which is moving relative to me (like the time on a moving clock) does not in any way imply I am comparing multiple frames, I can analyze these properties just fine using nothing but my own rest frame? A frame is just a coordinate system after all, I can perfectly well keep track of the way the position coordinate of the moving object changes with coordinate time using just the coordinates of my rest frame.
JesseM said:You bolded the second half of my sentence but then ignored the first half, taking the meaning out of context. I first said "In this case we are dealing with two clocks that have different velocities in frame B, but we are measuring both their ticking rates from the perspective of frame B"--so when I then said "nothing about the 1st postulate suggests that their ticking rates should be identical", I was clearly talking about their ticking rates in frame B, not their ticking rates in their own respective rest frames. Hopefully you'd agree that nothing about the first postulate suggests that their ticking rates should be identical in frame B, given that one is at rest in frame B and the other is not?
Didn't you talk abouthttp://www.pa.msu.edu/courses/2000spring/PHY232/lectures/relativity/contraction.html" ?I have been picturing the stationary clock as being placed on Einstein's embankment and the moving clock riding on his train; am I misreading this situation?
Yes, distance/time if both are measured in the same reference frame. See the difference to your statements?As for speed, I take that as non-directional (for speed with direction is velocity?) and it is distance/time.
Einstein wasn't analyzing time dilation in the train/embankment thought-experiment, he was analyzing the relativity of simultaneity, and since the relativity of simultaneity is all about how simultaneity differs between two frames of course he needed to look at the thought-experiment from the perspective of both frames. But that doesn't mean every analysis of a moving object requires multiple frames.Grimble said:I'm sorry if I am getting confused here, but as I have just said in my last post I was visualising this as Einstein's embankment and moving train where he thought it necessary to use separate co-ordinate frames. Was he over complicating it when he could have worked it all in relation to the embankment, is that what you are saying? Or am I becoming confused again?
JesseM said:If you think any frames other than the observer's rest frame are used in analyzing the light clock, can you point out the specific step in the analysis where you think this happens? For example, do we need any frames other than the observer's frame to figure out the distance the mirrors have traveled horizontally in a given time if we know their velocity v? Do we need any frames other than the observer's frame to use the pythagorean theorem to figure out the diagonal distance the light must travel if we know the horizontal distance traveled by the mirrors (just v*t, where t is the time between the light hitting the top and bottom mirror and v is the horizontal velocity of the mirrors) and the vertical distance h between them? Do we need any frames other than the observer's frame to figure out the time T that would be required in order to ensure that the diagonal distance D = [tex]\sqrt{v^2 T^2 + h^2}[/tex] will satisfy D/T = c? (making use of the second postulate which says light must move at c in every frame, including the observer's frame, along with the ordinary kinematical rule that speed = distance/time)
From all this, you can conclude that the time T between ticks of the light clock in the observer's frame must be equal to [tex]T = \frac{h}{\sqrt{c^2 - v^2}}[/tex]. Only here do you have to consider another frame if you want to derive the time dilation formula from this--you have to figure out what the time t between ticks would be in the light clock's own rest frame, and obviously if h is the vertical distance between mirrors this would be t = h/c (here you do need to make an argument to show the vertical height h will be the same in both frames, that there will be no length contraction perpendicular to the direction of motion). Then if you divide T/t you get the gamma factor [tex]\frac{1}{\sqrt{1 - v^2/c^2}}[/tex]. But this is just simple division, when deriving the time between ticks in each frame you can work exclusively with the coordinates of that frame and not worry about other frames.
Ich said:Didn't you talk abouthttp://www.pa.msu.edu/courses/2000spring/PHY232/lectures/relativity/contraction.html" ?
The velocity W of the man relative to the embankment is here replaced by the velocity of light relative to the embankment. w is the required velocity of light with respect to the carriage, and we have
w = c - v.
The velocity of propagation of a ray of light relative to the carriage thus comes out smaller than c.
But this result comes into conflict with the principle of relativity set forth in Section V. For, like every other general law of nature, the law of the transmission of light in vacuo must, according to the principle of relativity, be the same for the railway carriage as reference-body as when the rails are the body of reference. But, from our above consideration, this would appear to be impossible. If every ray of light is propagated relative to the embankment with the velocity c, then for this reason it would appear that another law of propagation of light must necessarily hold with respect to the carriage—a result contradictory to the principle of relativity.
JesseM said:Again, do you understand the difference between talking about an object moving relative to you and talking about a frame moving relative to you? If I am at rest on the embankment, I can perfectly well analyze the behavior of a moving train, or a moving light clock aboard the train, using only the embankment frame, without ever making reference to the train's own rest frame--do you disagree with that?
If not, note that this is in fact the sort of thing I did with the moving light clock, figuring out the time between ticks using only the observer's frame, not the light clock's own rest frame.
Galileo or Newton would not have assumed that the light must be traveling at c in both their own frame and the clock's rest frame--in fact they would have assumed that if it is traveling at c in one of those frames, it must be traveling at a different speed in the other frame, in such a way that both frames end up agreeing on the time between ticks.Grimble said:as indeed Galileo and Newton would have done! But what has it to do with SR?
OK, so we agree that the time between ticks for the moving light clock can be derived using only the observer's frame. Then as I pointed out in that earlier post, if you want to derive the time dilation equation which compares time in the clock's rest frame to time in the observer's frame, that's when you do have to bring in some assumptions about multiple frames:Grimble said:Let me address the previous post you quote.
We have the horizontal distance = vT,
The vertical distance h (which, if the clock is ticking seconds in this frame = 1 x c)
and the diagonal distance D, where [tex]D = \sqrt{{v^2}{T^2} + {h^2}}[/tex]
And you quite rightly say that only one frame is needed to reach this point. In fact it could be done using either frame.
Consider, if you will, that if we were to use the light clock's own frame of reference, then we should still say that the oberver, traveling at v relative to the clock would still see the diagonal [tex]D = \sqrt{{v^2}{T^2} + {h^2}}[/tex].
I have no problem with this at all, nor with the conclusion that:
T is the time for light to travel from the mirror to the observer while
t is the time for the light to travel back to the source
and that if the clock is ticking seconds in its own frame of reference then it will take γ seconds, measured in that same frame, to reach the observer.
The assumption that the vertical height between mirrors is the same in both frames does involve thinking about multiple frames in SR, as does the assumption that if the height is h in the clock's rest frame, the time between ticks must be t = h/c in that frame.JesseM said:From all this, you can conclude that the time T between ticks of the light clock in the observer's frame must be equal to [tex]T = \frac{h}{\sqrt{c^2 - v^2}}[/tex]. Only here do you have to consider another frame if you want to derive the time dilation formula from this--you have to figure out what the time t between ticks would be in the light clock's own rest frame, and obviously if h is the vertical distance between mirrors this would be t = h/c (here you do need to make an argument to show the vertical height h will be the same in both frames, that there will be no length contraction perpendicular to the direction of motion). Then if you divide T/t you get the gamma factor [tex]\frac{1}{\sqrt{1 - v^2/c^2}}[/tex]. But this is just simple division, when deriving the time between ticks in each frame you can work exclusively with the coordinates of that frame and not worry about other frames.
You have it backwards here. The time in the observer's frame will be the greater time, not the lesser time (and the gamma factor γ is always greater than 1). So, if the time in the clock's frame is 1 second, the time in the observer's frame will be gamma seconds (and if the time in the clock's frame is T seconds, the time in the observer's frame is T*gamma seconds)Grimble said:What we have to determine here, is how to resolve the difference in what the observer sees, between the time in his own frame of 1 second and that he observes in the clock's frame of γ seconds.
That's not good terminology, since both frames are "inertial" ones in the terminology of relativity. Also it's not as if the time in the observer's frame is intrinsically the one that's been "transformed", you can equally well start out with the time in the observer's frame and then use the Lorentz transformation to derive the time in the clock's frame. Better terminology would just be to give names to the two frames, like "clock's frame" and "observer's frame", or just use different notation to refer to them like unprimed t vs. primed t'.Grimble said:We transform the time observed in the clock's frame using the Lorentz Transformation, and thus we find that T, in transformed units( in order to avoid any dispute over exactly what we are referring to I shall refer to them as transformed units and inertial units) is equal to 1 second inertial time.
What? I thought you were using "inertial" time to refer to time in the clock's frame, but the time in the clock's frame cannot be both T seconds and gamma seconds. If the time between ticks of the clock in the clock's frame is T seconds, then the time between ticks of that same clock in the observer's frame (what you were calling 'transformed' time) would be gamma*T seconds.Grimble said:But if T inertial seconds = γ seconds in (inertial) time, then
Don't understand these either. And why are you using three times--1 second, T seconds, and gamma seconds? If we have only two frames to consider there should be only two times involved. If the time in the clock's frame is 1 second than the time in the observer's frame will be gamma seconds, while if we say more generally that the time in the clock's frame is T seconds (i.e. not assuming the distance between mirrors is 1 light-second), then the time in the observer's frame is gamma*T seconds.Grimble said:T in transformed seconds = γ seconds in transformed time
yet
T in transformed seconds = 1 second in inertial time
therefore
γ seconds in transformed time = 1 second in inertial time
If t is supposed to be the time in the clock's own rest frame and t' is the time in the observer's frame, then this formula is wrong, you should be multiplying by gamma rather than dividing by it: [tex]t' = t * \gamma[/tex].Grimble said:or in the more usual terms used [tex]{t^'} = \frac{t}{\gamma}[/tex]
This is the correct formula for length contraction if L is the object's length in its own rest frame and L' is the length in the observer's frame, but you can see that it's not exactly analogous to the correct formula for time dilation I wrote above, the formula for time dilation involves multiplying by gamma while the formula for length contraction involves dividing by gamma.Grimble said:which is what we should expect as it reflects the analagous formula for length contraction [tex]{L^'} = \frac{L}{\gamma}[/tex]
Nope, the fact that the speed of light is c in both frames cannot be derived from the length contraction and time dilation formulas alone, you also have to take into account the relativity of simultaneity. See my post here for a numerical example of how to take into account all three factors to show that two frames will both measure a light beam to move at c.Grimble said:from which we can see that if [tex]c = \frac{L}{t}[/tex] then [tex]c = \frac{L^'}{t^'}[/tex] which of course it has to do.
Again I'm confused--wasn't inertial time supposed to be time in the clock's own rest frame? Why would you use gamma to find that time? If the mirrors are 1 light-second apart, the time in the clock's rest frame will be 1 second, naturally. I suppose you're free to imagine that the mirrors are gamma light-seconds apart in the clock's rest frame, but there's no reason why relativity demands this, the mirrors can be set to any distance apart you wish in the clock's rest frame.Grimble said:If we say that v = 0.866c then [itex] \gamma = 2 [/itex] and if we apply this to our scenario above
we find that the time for light to traverse the diagonal path = [itex]\gamma[/itex] = 2 seconds inertial time
Yes, if the vertical distance between the mirrors happens to be 2 light seconds, and the mirrors are moving at 0.866c in the observer's frame, then the observer will see the diagonal distance as 4 light seconds and the time between ticks as 4 seconds. But again there's no reason why the vertical distance had to be 2 light seconds, you could have equally well said it was some other arbitrary distance like 3.5 light seconds, in which case the time between ticks in the observer's frame would be 2*3.5 = 7 seconds.Grimble said:= [itex]2\gamma[/itex] (or [itex]{\gamma^2}[/itex]) [itex] = 4 [/itex]seconds transformed time
and the corresponding diagonal distance will be 4 transformed light seconds
The observer isn't "observing the clocks reference frame", he's just observing the clock, and measuring it from the perspective of his own inertial frame. He doesn't have to know anything about the Lorentz transformation whatsoever, he can just measure the time between ticks directly using synchronized clocks at rest in his own frame (or else he can figure out what the time between ticks must be given the vertical distance between the mirrors, the speed at which the clock is traveling which allows him to calculate the diagonal distance using the Pythagorean theorem, and the assumption based on the second postulate that the light must move at c in his frame). If he uses a network of synchronized clocks at rest in his frame, then he just has to note the time T1 on a clock of his that was right next to the bottom mirror when the light was emitted ('right next to' so that he is assigning times using only local measurements and he doesn't have to worry about delays between when an event happens and when a signal from the event reaches one of his clocks), and then note the time T2 on a clock of his that was right next to the top mirror when the light reached it, and the time between ticks in the observer's frame will just be T2 - T1. If the two mirrors happen to be 2 light-seconds apart vertically, and the light clock is moving at 0.866c, then T2 - T1 will equal 4 seconds as you say, but you can see that there was no need for the observer to make use of the Lorentz transformation to find this time (although of course it will agree with the time predicted by the Lorentz transformation if you start with the time in the clock's rest frame and then transform into the observer's frame).Grimble said:so the observer observing the clocks reference frame will see the light take 4 transformed seconds to cover the 4 transformed light seconds; that is the equivalent of 2 seconds to cover 2 lightseconds inertial time. (all at the speed of c)
Grimble said:[...] an identical clock in the oberver's frame of reference [...]
Rasalhague said:Are you clear on the fact that if there are two clocks, each moving with some contant velocity relative to the other, both clocks are in all frames of reference, in the sense that both can be described using the language of special relativity with respect to any inertial frame of reference? I think what you have in mind here is "an identical clock which is at rest in the observer's rest frame" (i.e. an identical clock at rest with respect to this observer, not moving with respect to the observer, at a constant distance from the observer--whether located in the same place as the observer or in any other place, so long as there are no tidal effects of gravity present there).
JesseM said:Again, do you understand the difference between talking about an object moving relative to you and talking about a frame moving relative to you? If I am at rest on the embankment, I can perfectly well analyze the behavior of a moving train, or a moving light clock aboard the train, using only the embankment frame, without ever making reference to the train's own rest frame--do you disagree with that?
If not, note that this is in fact the sort of thing I did with the moving light clock, figuring out the time between ticks using only the observer's frame, not the light clock's own rest frame.
You have it backwards here. The time in the observer's frame will be the greater time, not the lesser time (and the gamma factor γ is always greater than 1). So, if the time in the clock's frame is 1 second, the time in the observer's frame will be gamma seconds (and if the time in the clock's frame is T seconds, the time in the observer's frame is T*gamma seconds)
If t is supposed to be the time in the clock's own rest frame and t' is the time in the observer's frame, then this formula is wrong, you should be multiplying by gamma rather than dividing by it: {itex] {t^'} = t*\gamma[/itex].
But no it doesn't, just because someone at sometime in the distant past decided to use the labels length-contraction and time-dilation implying one increases while the other decreases people have been making the mistake of thinking the formulae are opposites, while they are in fact analagous.This is the correct formula for length contraction if L is the object's length in its own rest frame and L' is the length in the observer's frame, but you can see that it's not exactly analogous to the correct formula for time dilation I wrote above, the formula for time dilation involves multiplying by gamma while the formula for length contraction involves dividing by gamma.
Nope, the fact that the speed of light is c in both frames cannot be derived from the length contraction and time dilation formulas alone, you also have to take into account the relativity of simultaneity. See my post here for a numerical example of how to take into account all three factors to show that two frames will both measure a light beam to move at c.
Grimble said:Inertial units (time and space) are those measured locally (by a local observer within that same frame of reference) in any inertial frame of reference. That they are the same in all inertial frames of reference (Galilean frames) is a demand of the first postulate.
Grimble said:Transformed units (time and space) are those same inertial units, transformed using the Lorentz equations, which is how they will appear when observed from another inertial frame of reference and are a function of their relative velocity.
Grimble said:(I am not saying that one is the clock and the other is the observer, I am saying that those terms are dependent on where they are measured from.
Inertial units (time and space) are those measured locally (by a local observer within that same frame of reference) in any inertial frame of reference. That they are the same in all inertial frames of reference (Galilean frames) is a demand of the first postulate.
Rasalhague said:What do you mean by "local"? The "same frame of reference" as what? What you you mean by "in the same frame of reference [...] in any inertial frame of reference"?
I don't get it. When you say "within the same frame of reference", "same" as what, exactly? And if they are measured relative to a particular frame, then why do you say "they are the same in all inertial frames"?Grimble said:Let me start by defining the terms I use.
Firstly I do not use the terms primed and unprimed as I have seen these used both ways round and swapped so many times that their use, for me at least, has been compromised;
Inertial units (time and space) are those measured locally (by a local observer within that same frame of reference) in any inertial frame of reference. That they are the same in all inertial frames of reference (Galilean frames) is a demand of the first postulate.
The first postulate obviously doesn't demand that the distance and time between a given pair of events be the same when different inertial observers measure it using their own ruler/clock systems. If that's not what you meant, then what are you saying would be the same in all inertial frames of reference?Grimble said:That they are the same in all inertial frames of reference (Galilean frames) is a demand of the first postulate.
Again, can you explain how this terminology applies to my above example? It's true, for example, that if you know that the A frame assigned the red firecracker explosion coordinates (x=346.2 meters, t=1 microsecond), then if you just plug these coordinates into the Lorentz transformation (with gamma = 2 and v = 0.866c), you can deduce that the B frame will assign the red firecracker explosion coordinates (x'=173.1 meters, t'=0 microseconds). Does this make the latter set of coordinates "transformed units", even though they are just what B found using his own inertial ruler/clock system? And note that of course you can also go in reverse--if at first you only know that B assigned the red firecracker explosion coordinates (x'=173.1 meters, t'=0 microseconds), then you can apply the Lorentz transformation to that to deduce that A assigned this same explosion the coordinates (x=346.2 meters, t=1 microsecond). So can every measurement be seen as both inertial units and transformed units, depending on what data you start with and then apply the Lorentz transformation to? If not, then again, please explain the difference between "inertial units" and "transformed units" in terms of the example I have given with the firecrackers and the two inertial ruler/clock systems.Grimble said:Transformed units (time and space) are those same inertial units, transformed using the Lorentz equations, which is how they will appear when observed from another inertial frame of reference and are a function of their relative velocity.
What does that have to do with the first postulate? The first postulate in no way demands that the time between the events (light hitting bottom mirror) and (light hitting top mirror) be the same in both frames (just like it didn't demand that the time between the blue and red firecracker explosions in my example above should be the same in both frames), if it did then relativity would violate the first postulate. The first postulate just demands that the laws of physics obey the same equations in both frames.Grimble said:and Galileo and Newton would have agreed that the increased distance traveled by the light would have meant that the speed of light would have increased but the time would have remained constant (relativity principle = 1st postulate)
Here you are speculating about what Newton and Galileo would have said about physics that didn't come along until well after they were dead. When I talked about what Newton and Galileo would have said, I didn't mean to talk about what they might have said if they had lived to see new ideas long after their time, I just meant to talk about what is true in classical pre-relativistic physics. In classical physics the 2nd postulate is just false, you can't have any object that has the same speed in all inertial frames.Grimble said:and they would have agreed that the time would have had to increase if the speed were to remain constant (light speed principle = 2nd postulate)
If you are somehow under the impression that the first postulate says different frames should agree on the time and distance between events, and that Einstein says we should comply with that, you are badly misunderstanding the meaning of the first postulate, which again is just about the general equations for the laws of physics, not about the time and distances between a specific pair of events. In fact even in classical physics the distance between a pair of events can be different in two different inertial frames, although in classical physics (unlike in relativity) the time between a pair of events is the same in all inertial frames.Grimble said:But Einstein would have said No! We must comply with Both Postulates.
Yes, although of course in the rest frame of the clock, v=0 so the path from one mirror to another is purely vertical rather than diagonal in this frame.Grimble said:We have the horizontal distance = vT,
The vertical distance h (which, if the clock is ticking seconds in this frame = 1 x c)
and the diagonal distance D, where [tex]D = \sqrt{{v^2}{T^2} + {h^2}}[/tex]
And you quite rightly say that only one frame is needed to reach this point. In fact it could be done using either frame.
You seem to be confused about what physicists mean when they talk about "using" a given frame of reference--it means that you analyze things using only the distance and time coordinates of that frame (along with coordinate-invariant things like proper times and statements about pairs of events that locally coincide), and don't refer to the coordinates of any other frame. So, it's an incorrect usage of the lingo to say that you can "use the light clock's own frame of reference" to deduce what coordinate distance the light traveled in the observer's frame (though you can use the light clock's frame to figure out what markings and clock readings on the observer's ruler/clock system would line up with the events of the light hitting the bottom and top mirrors).Grimble said:Consider, if you will, that if we were to use the light clock's own frame of reference, then we should still say that the observer, traveling at v relative to the clock would still see the diagonal [tex]D = \sqrt{{v^2}{T^2} + {h^2}}[/tex].
Huh? What do you mean when you say "it will take [itex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/itex] seconds, measured in that same frame, to reach the observer"? First of all, the time dilation formula [tex]t' = t*\sqrt{1 - v^2/c^2}[/tex] is normally understood purely in terms of relating the time between ticks in the clock's rest frame to the time between ticks in the frame of an observer moving relative to the clock, the idea that it should have something to do with the time for the light of a tick to reach an observer moving relative to the clock as measured in the clock's own rest frame appears to be an idea unique to you. Second, it's not even clear what you mean when you talk about the time for it to reach an observer--won't this be totally dependent on how far the observer is from the clock? If the observer is right next to the source at the bottom mirror at the moment the light has traveled back to the source from the top mirror, then if it takes 1 second for the light to travel from the source to the top mirror and back to the source, that must mean it also takes 1 second for the light to travel from the source to the top mirror and down to the observer in this frame (of course if we knew the time on the observer's clock when the light left the source according to this frame's definition of simultaneity, we could use this frame to calculate the time on the observer's clock when the light returns to the source, and it might be different than 1 second--is this the sort of thing you were getting at?)Grimble said:I have no problem with this at all, nor with the conclusion that:
T is the time for light to travel from the mirror to the observer while
t is the time for the light to travel back to the source
and that if the clock is ticking seconds in its own frame of reference then it will take [itex]γ = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/itex] seconds, measured in that same frame, to reach the observer. . . . . . .(1)
Here you seem to be saying that the γ seconds is supposed to be the time in the observer's own frame, but before you said that "if the clock is ticking seconds in its own frame of reference then it will take [itex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/itex] seconds, measured in that same frame, to reach the observer". Is γ seconds supposed to be the time between two events in the clock's own frame, or the time between two events in the observer's frame?Grimble said:But the first postulate also requires that an identical clock, stationary in the observer's frame of reference would also be ticking seconds, identical seconds, as they are both inertial frames of reference, or Galilean frames as Einstein termed them. . . . . . .(2)
What we have to determine here, is how to resolve the difference in what the observer sees, between the 1 second that his own identical clock takes to tick and the γ seconds that he observes the moving clock take for each tick.
Please consider that your understanding of SR and what Einstein meant may just be badly confused. Einstein never introduced any distinction between "inertial time" and "transformed time", in SR we just talk about times in different inertial frames, and the Lorentz transformation just distance and time intervals (or distance and time coordinates of individual events) of one inertial frame to the intervals between the same events in a different inertial frame. Also, in SR there is no notion that the first postulate demands that the time and distance intervals between a given pair of events be the same in different inertial frames.Grimble said:Thankfully Einstein gave us the way to resolve the difference in what the observer sees,:
We transform the time observed in the clock's frame using the Lorentz Transformation, and thus we find that T, in transformed units is equal to 1 second inertial time. . . . . . (3)
The notion of one frame "viewing" another also is not part of SR. Each frame is used to analyze things in terms of the coordinates of that frame alone, and then the Lorentz transformation relates an "analysis-wholly-in-frame-A" to an "analysis-wholly-in-frame-B". For example, in my example above with two firecracker explosions, the wholly-in-frame-A analysis gives the distance between them as 346.2 meters and the time between them as 1 microsecond, while the wholly-in-frame B analysis gives the distance between them as 173.1 meters and the time between them as 0 microseconds. If we start out knowing the wholly-in-frame-A intervals, we can plug them into the Lorentz transformation to deduce the wholly-in-frame-B intervals like so:Grimble said:Yes, indeed, one could take the time from any inertial frame (they are all, by definition equal, after all) and transform it (transformed time being that in one frame viewed from another).
Again I see no reason why the gamma formula would have anything to do with the "distance to the observer" in the clock's rest frame, you're either confused about what these equations mean in ordinary SR or you're trying to introduce your own novel ideas which are not part of mainstream SR. And the observer's distance from the mirror (which one? Top or bottom?) is constantly changing in this frame, so what moment do you want to talk about the distance from observer to mirror, anyway?Grimble said:But, in the initial scenario and using only one frame, Ti seconds = γti seconds in time( where γ or [itex]\frac{1}{1-\frac{v^2}{c^2}}[/itex] is the ratio of the distance between the mirrors and that from the mirror to the observer, the diagonal distance)
Again, nothing in mainstream SR corresponds to your distinction between "inertial" time and "transformed" time as far as I can tell, are you trying to introduce new ideas here or are you under the impression that what you are saying is a part of regular SR? Either way you haven't clearly explained what "inertial time" and "transformed time" are supposed to mean, please show how these terms would apply to a specific numerical example like my example with the two firecracker explosions.Grimble said:then
[itex]{T_t} = \gamma {t_t}[/itex] seconds in transformed time ...
If different people write it differently it's just because their definition of the primed and unprimed frame is different. If unprimed is the time between a pair of events (like the ticks of a clock) in the frame where those events happen at the same position in space (as would be true in the clock's rest frame if the events represent ticks of a clock), and primed is the time between the same pair of events in a frame moving at speed v relative to the first frame, then the correct formula is 1) above--in the case of clock ticks, the time is greater in the frame where the clock is moving than in the clock's rest frame. On the other hand, if you use primed to be the time in the frame where the events happen at the same position (like the clock's rest frame), and unprimed to represent the time in the frame moving at v relative to the first, then 2) would be the correct formula.Grimble said:I have seen the time dilation formula written (and used) in both forms:
1) [tex]{t^'} = t * \gamma[/tex]
2) [tex]{t^'} = \frac{t}{\gamma}[/tex]
1) is the more commonly used but my research indicates that it is 2) that is the correct one for the following reasons
Grimble said:
- It is the one Einstein formulated;
- It is the one derived from the Lorentz equations;
How does it reverse the terms? If follows the more common convention where unprimed is the rest frame of the clock, and therefore derives your formula 1), T' = gamma*T.Grimble said:[*]A http://www.answers.com/topic/special-relativity#Time_dilation_and_length_contraction" reverses the terms and thereby reverses the derived formula;
He did not show that x = ct and x'=ct' are general relations which hold for events of arbitrary coordinates, this was just supposed to be the equation of motion for a light beam which was released from the x=0 at t=0 (which also is assumed to coincide with x'=0 and t'=0 in the primed frame in the Lorentz transformation). It's easy to show using the Lorentz transformation that if you pick an event on the worldline of this light beam which occurs at coordinates x=cT and t=T in the unprimed frame (which satisfies x = ct), then in the primed frame this same event has coordinates:Grimble said:[*]Einstein himself, proved
"[URL="[PLAIN]http://www.bartleby.com/173/11.html"[/URL] that [itex]{x} = c{t}[/itex] and that [itex]{x^'} = c{t^'}[/itex] so the formulae for Length contraction and time dilation have to be analagous or these two equations cannot both be correct;
Time dilation and length contraction are conceptually different things when illustrated on a Minkowski diagram--time dilation deals with the time between a single pair of events in two different frames, while length contraction does not deal with the distance between a single events in two different frames, rather it deals with the distance between two parallel worldlines at a single moment in two different frames (with 'at a single moment' depending on each frame's definition of simultaneity). It would be possible to come up with a spatial analogue for time dilation which deals with the distance between a single pair of events, in which case the equation would look just like the time dilation equation, and likewise to come up with a temporal analogue for length contraction which deals with the time between two parallel spacelike surfaces at a single position in two different frames, in which case the equation would look just like the length contraction equation. If you're interested you can take a look at the diagram I drew which neopolitan posted in post #5 of this thread, where we were discussing the issue of whether it's meaningful to talk about "time contraction" or "length dilation" (I don't really recommend reading the whole thread though, it went on a lot of tangents).Grimble said:[*]Minkowski space time shews quite conclusively exactly how length contraction and time dilation are, in fact, the same process and could, incidentally, have been termed 'length dilation' and 'time contraction' and all those terms would have been correct! (I will shew this later).
Do you agree that if [tex]\Delta t[/tex] represents the time between ticks of a clock in the clock's rest frame, and [tex]\Delta t'[/tex] represents the time between ticks in a frame moving at speed v relative to the clock, then the correct formula is [tex]\Delta t' = \gamma * \Delta t[/tex]? And do you agree that if L represents the distance between either end of an object in the object's rest frame, and L' represents the distance between either end of that object in a frame moving at speed v relative to the object (with the positions of each end measured simultaneously in whatever frame is measuring the distance), then the correct formula is [tex]L' = \frac{L}{\gamma}[/tex]?Grimble said:But no it doesn't, just because someone at sometime in the distant past decided to use the labels length-contraction and time-dilation implying one increases while the other decreases people have been making the mistake of thinking the formulae are opposites, while they are in fact analagous.
Nope, you cannot show that the speed of light is the same in two frames using only length contraction and time dilation, you have to take into account the relativity of simultaneity too.Grimble said:from which we can see that if [tex]c = \frac{L}{t}[/tex] then [tex]c = \frac{L^'}{t^'}[/tex] which of course it has to do.
JesseM said:Nope, the fact that the speed of light is c in both frames cannot be derived from the length contraction and time dilation formulas alone, you also have to take into account the relativity of simultaneity. See my post here for a numerical example of how to take into account all three factors to show that two frames will both measure a light beam to move at c.
Obviously you can derive the fact that a light beam moves at c in all frames using the Lorentz transformation, since the Lorentz transformation was itself derived using the second postulate and since time dilation, length contraction and the relativity of simultaneity can all be derived from it too. But we were talking about your claim (which you repeat above) that somehow one could combine the length contraction equation and the time dilation equation alone to get the invariance of c, without making use of the full Lorentz transformation equations, and also without making use of the relativity of simultaneity. It's this claim which doesn't make any sense (on the other hand, if you use the time dilation equation along with what I called the 'spatial analogue of time dilation', which unlike length contraction deals with the distance between a single pair of events, then you can combine these two equations to get the conclusion that distance/time for two events on the worldline of a lightbeam must always equal c).Grimble said:Yet Einstein did it directly from the Lorentz transformation formula, very simply. It isn't complicated. See paragraph 6
Grimble said:I think that what I wrote was quite clear - (by a local observer within THAT same frame of reference)
Grimble said:I was defining what I meant by the terms I used.
By inertial units I mean units within an inertial frame of reference, as measured and referred to within that same frame of reference, in order to avoid anyone claiming that they could be anything else.
By transformed units I mean units that have been transformed by Lorentz transformations.
Grimble said:and Galileo and Newton would have agreed that the increased distance traveled by the light would have meant that the speed of light would have increased but the time would have remained constant (relativity principle = 1st postulate) and they would have agreed that the time would have had to increase if the speed were to remain constant (light speed principle = 2nd postulate)
But Einstein would have said No! We must comply with Both Postulates.
Grimble said:And you quite rightly say that only one frame is needed to reach this point. In fact it could be done using either frame.
Grimble said:1) is the more commonly used but my research indicates that it is 2) that is the correct one for the following reasons:
JesseM said:I don't get it. When you say "within the same frame of reference", "same" as what, exactly? And if they are measured relative to a particular frame, then why do you say "they are the same in all inertial frames"?.
JesseM said:Huh? What do you mean when you say "it will take [itex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/itex] seconds, measured in that same frame, to reach the observer"? First of all, the time dilation formula [tex]t' = t*\sqrt{1 - v^2/c^2}[/tex] is normally understood purely in terms of relating the time between ticks in the clock's rest frame to the time between ticks in the frame of an observer moving relative to the clock, the idea that it should have something to do with the time for the light of a tick to reach an observer moving relative to the clock as measured in the clock's own rest frame appears to be an idea unique to you.
I don't see how it changes the meaning. I repeat the question, with that word changed:Grimble said:Before I say anything else will somebody, anybody, please read what I have written?
I DID NOT write "within the same frame of reference" I wrote "within that same frame of reference" - bold used for emphasis!
Which if read as written changes the meaning somwhat.
JesseM said:I don't get it. When you say "within that same frame of reference", "same" as what, exactly? And if they are measured relative to a particular frame, then why do you say "they are the same in all inertial frames"?Grimble said:Inertial units (time and space) are those measured locally (by a local observer within that same frame of reference) in any inertial frame of reference. That they are the same in all inertial frames of reference (Galilean frames) is a demand of the first postulate.
No problem, and sorry for misquoting (and yeah, feel free to just call me Jesse), but as I said I don't understand why "the same" vs. "that same" makes a difference in how I should interpret your statement. Again, "same" as what?Grimble said:PS. I apologise Jesse (if I may call you that?) but yours is the second of two replies that have made the same misquote of what I had written...