Light clock moving to demonstrate time dilation

In summary: But the two observers always agree on which direction the ray went! So the argument that the photon takes a longer path back to the mirror because it crosses more world-lines doesn't seem to be very strong.
  • #1
RK1992
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Hello all, having decided that I wish to apply to Oxford to study Physics (as well as Imperial and, I am still considering Natural Sciences at Cambridge), I have been informed that extra reading and independent study would be advised, so I'm delving into the world of relativity (I'm a first year AS student so we've done nothing on it yet).

I've stumbled upon an explanation of why time dilation happens but the explanation seems odd to me. We've been told that time dilation happens in GCSEs etc but never had any sort of explanation as to why. This analogy (I don't think it's a proof but I haven't seen it referred to as a proof or an analogy so I'm assuming it's an analogy) doesn't appear to prove anything to me...

Say you have a stationary light clock with the photon bouncing (I'm presuming you all know the experiment I;m talking about - http://galileoandeinstein.physics.virginia.edu/lectures/srelwhat.html has it around half way down if you don't).

The photon will bounce off one mirror at the angle it hits it - so straight back off the mirror. If you then move the mirror fast enough, to say the right, then the photon will have bounced off one mirror but the other mirror will now be to the right of the photon's path before the photon has arrived and the photon will not reflect back to the other mirror, it will just continue off into space because there is no mirror to reflect, so how can the photon take a "longer path" back to the mirror.

Is this an example of the analogy breaking down or am I missing some knowledge of how photons move? Does the photon reflect off a surface at an angle if the surface is moving? Or is it just a flaw in the analogy?

I accept that time dilation happens - experimental data proves it as such, but is there an analogy which would actually explain this phenomenon without any flaws? Is this actually a flaw or is it my lack of knowledge?

Thanks in advance.
 
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  • #2
RK1992 said:
Is this actually a flaw or is it my lack of knowledge?

In my opinion, yours is a god question and it's not a flaw of the theory, but it's a flaw of some explanations. In any frame the photon hits the target (the local top mirror), because it is aimed at it. But what “aiming at” means is usually missed out in the explanations, I do not know why. I’ll reproduce what I said in another thread regarding this and maybe someone more knowledgeable can comment:

It has to do with the fact that light is created in an instrument that reproduces, at a smaller scale, the trajectory that it should follow afterwards, in the outer world (…). Photons are created initially in random directions but only those that follow the line between the two extremes of the instrument succeed in coming out through the hole at the exit. Hence they maintain outside the trajectory that they followed inside, the successful trajectory, one enabling them to hit the target again (in this case, the top mirror). In particular, lasers, instead of relying on luck, are good at producing many photons with the right trajectory: they generate very little diverging beams that follow a thin straight line because their mechanism favours that photons bounce between a bottom and top mirror and create by stimulated emission new photons that acquire the same direction.

But this explains why the photon hits the target, i.e. why it takes the direction of its source. It does not explain why it does not take the speed of the source. The classical explanation was an analogy with sound: light travels through a medium (aether) just like sound through air. SR rejects the aether and prefers not to complicate its life tampering with root-causes that can't be proved: it chooses instead a geometric analogy = the universe is a 4-dimension continuum (spacetime) where time can be treated as a relative value just like space.
 
  • #3
RK1992 said:
Is this an example of the analogy breaking down or am I missing some knowledge of how photons move? Does the photon reflect off a surface at an angle if the surface is moving? Or is it just a flaw in the analogy?

Observers in different frames of reference always agree on whether world-lines intersect. Therefore you're guaranteed that if the observer at rest with respect to the clock sees the light ray's world-line as intersecting the world-line of the mirror, you're guaranteed that the same will be true in other frames.

The agreement on intersection of world-lines is very fundamental, and basically trumps any other argument. However, I think you can also see this explicitly just by considering the ordinary laws of ray optics. The ray that was emitted in a certain direction according to one observer is emitted in another direction according to the other observer; that's why it reaches the mirror the first time. (This is called aberration of light.) The zigzag path has equal angles of incidence and reflection.
 
  • #4
I'm really not too sure about these explanations... from what you're saying:

Either the photon reflects off one mirror at an angle which is not equal to the angle of incidence thus defying the laws of optics and it has effectively predicted the future by reflecting to where the other mirror face will be.

Or the photon has bounced off the mirror face with an angle of reflection equal to the angle of incidence but has then changed its path as the mirror on the other side has moved because of the fact that it's predetermined that the photon must bounced between the mirrors because their "world-lines" have crossed or "because it was aimed at it" - if I kick a ball at a door and the door opens, the ball doesn't curve and hit the door "because I aimed it there".

Or the photon takes a curved path. I'm assuming this is correct because of space-time curvature which I've heard of (although I've not yet read anything which I believe would allow me to understand the idea). But even then, how and why does it take a curved path?

World-lines seem to me like a "weasel word" here... because a photon was aimed at a mirror once, it must bounce off and head towards the other infinitely regardless of the mirror's location? I can accept this if the clock is stationary and the angle of incidence/reflection remains at 90o. This, however sounds wrong to me... I'm struggling to believe that if I was in a spaceship traveling near to the speed of light next to another a reflective panel on the side side and I aimed a torch at the panel on the other spaceship then I would see a bright patch on the reflective panel.

How does the photon change its path after it's started moving? I thought photons traveled in straight lines.
 
  • #5
RK1992 said:
Either the photon reflects off one mirror at an angle which is not equal to the angle of incidence thus defying the laws of optics
No, it doesn't do that.

RK1992 said:
Or the photon has bounced off the mirror face with an angle of reflection equal to the angle of incidence but has then changed its path as the mirror on the other side has moved because of the fact that it's predetermined that the photon must bounced between the mirrors because their "world-lines" have crossed or "because it was aimed at it" - if I kick a ball at a door and the door opens, the ball doesn't curve and hit the door "because I aimed it there".
No, it doesn't change its path. The aberration effect is simply a difference between the direction of propagation of the light ray as seen by observers in two different frames. If you're riding a bike in the rain, the vertically falling raindrops appear to you to be falling at some non-vertical angle.

RK1992 said:
Or the photon takes a curved path.
No, it doesn't do this.
 
  • #6
So what does the photon do? How can the photon take a zigzag path if it doesn't do any of those?
 
  • #7
RK1992 said:
So what does the photon do? How can the photon take a zigzag path if it doesn't do any of those?

It gets reflected by the mirrors. By "changing its path," did you mean to include reflection? Yes, it does change its path for that reason.
 
  • #8
I meant literally turning a corner when I said changing its path.

But why would the photon reflect off the mirror at an angle not equal to the angle at which it struck the mirror? I can appreciate it doing so when the mirror is at a constant velocity but when the photon is between the mirrors and the mirror then moves how can the photon all of a sudden be heading on a different path?
 
  • #9
RK1992 said:
I meant literally turning a corner when I said changing its path.
I don't follow what you mean by this. Do you mean in flight, or on reflection?

RK1992 said:
But why would the photon reflect off the mirror at an angle not equal to the angle at which it struck the mirror?
It doesn't.
 
  • #10
RK1992 said:
but when the photon is between the mirrors and the mirror then moves how can the photon all of a sudden be heading on a different path?

Are you imagining the mirrors as beginning at rest and then starting to move? That isn't what's going on. The mirrors are not accelerating. We're just viewing the non-accelerating mirrors in two different frames of reference.
 
  • #11
bcrowell said:
I don't follow what you mean by this. Do you mean in flight, or on reflection?
I meant in flight, but that's wrong so scrap that.

bcrowell said:
It doesn't.
But it would have to to strike the other mirror.

bcrowell said:
Are you imagining the mirrors as beginning at rest and then starting to move? That isn't what's going on. The mirrors are not accelerating. We're just viewing the non-accelerating mirrors in two different frames of reference.
That's what I described in the first post...
But even if they aren't accelerating and are traveling at constant velocity then to get to a constant velocity from rest and for the
photon to continue bouncing then there would need to be at least one case of the laws of optics being broken, right?

http://www.spikedhumor.com/articles/83031/Time_Travel_Einstein_s_Big_Idea_Very_Thought_Provoking_.html [Broken]

This video is what made me think the photons must reflect at impossible angles... does that mean this video when he moves the clock from side to side is wrong?
 
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  • #12
RK1992 said:
But even if they aren't accelerating and are traveling at constant velocity then to get to a constant velocity from rest and for the
photon to continue bouncing then there would need to be at least one case of the laws of optics being broken, right?

No. In frame A, the mirrors are at rest and have *always been* at rest. In frame B, the mirrors are moving and have *always been* moving.
 
  • #13
This is a poor example to muse upon, since reflection of light does not work in the way you think it does.
 
  • #14
Ah, I watched the video whose link you posted and understand now your concern.

In my opinion, you should not pay any attention to that video. The very fact that it talks about time travel discredits it. (Long to explain...)

The light clock or photon clock thought experiment doesn't go like that. It goes like bcrowell has explained. Imagine two observers armed with their respective light clocks, one on a train the other on the platform, moving relative to each other. When they are lined up, each of them fires his respective photon from his respective bottom mirror to his respective top mirror. The platform photon hits the platform top mirror. The train photon hits the train top mirror. But not the other way round.

What is shown in the video cannot happen. If the train were at rest with the platform and the train photon were fired and only then, while the photon travels upwards, the train accelerated and the train top mirror moved away, the photon would miss the top mirror and hit somewhere behind it (precisely where it was aiming at when it was fired!). If it then reflected, it would do so respecting the law of reflection, with the same angle of reflection as the angle of incidence.
 
  • #15
bcrowell said:
No. In frame A, the mirrors are at rest and have *always been* at rest. In frame B, the mirrors are moving and have *always been* moving.
Okay that's sorted - in the original post I mentioned them starting at rest as in the video then accelerating away. I should've been clearer.

TcheQ said:
This is a poor example to muse upon, since reflection of light does not work in the way you think it does.
Are you saying that the laws about incidence and reflection are different to how I know them or that I'm missing fine details?

Saw said:
Ah, I watched the video whose link you posted and understand now your concern.

In my opinion, you should not pay any attention to that video. The very fact that it talks about time travel discredits it. (Long to explain...)

The light clock or photon clock thought experiment doesn't go like that. It goes like bcrowell has explained. Imagine two observers armed with their respective light clocks, one on a train the other on the platform, moving relative to each other. When they are lined up, each of them fires his respective photon from his respective bottom mirror to his respective top mirror. The platform photon hits the platform top mirror. The train photon hits the train top mirror. But not the other way round.

What is shown in the video cannot happen. If the train were at rest with the platform and the train photon were fired and only then, while the photon travels upwards, the train accelerated and the train top mirror moved away, the photon would miss the top mirror and hit somewhere behind it (precisely where it was aiming at when it was fired!). If it then reflected, it would do so respecting the law of reflection, with the same angle of reflection as the angle of incidence.

Okay, that helps a lot - your last paragraph is exactly the same scenario I tried to explain in the first post but in much clearer terms.

Thanks for the help everyone :D
 
  • #16
RK1992 said:
Are you saying that the laws about incidence and reflection are different to how I know them or that I'm missing fine details?

It is a bit of both, but let's keep it simple (no need to go into laborious mathematical explanations). When you look at reflection on an atomic level you see that an image or photon undergoes a number of processes before it becomes what we observe as "reflection". Pretty complicated stuff that's not usually touched on mathematically till 3rd year physics (since understanding the process is a Solid State Physics or Quantum Mechanics question). I thought the complexity of the example case detracted from what you were trying to ask.

One thing you will learn via physics is the most valuable tool of an ability to decide what and what are not reliable sources of information (as others have stated, those of us who (think we) understand relativity would try to find an example from a video that states scientific facts, and not propagating myths like mass traveling faster than the speed of light)
 
  • #17
TcheQ said:
It is a bit of both, but let's keep it simple (no need to go into laborious mathematical explanations). When you look at reflection on an atomic level you see that an image or photon undergoes a number of processes before it becomes what we observe as "reflection". Pretty complicated stuff that's not usually touched on mathematically till 3rd year physics (since understanding the process is a Solid State Physics or Quantum Mechanics question). I thought the complexity of the example case detracted from what you were trying to ask.

One thing you will learn via physics is the most valuable tool of an ability to decide what and what are not reliable sources of information (as others have stated, those of us who (think we) understand relativity would try to find an example from a video that states scientific facts, and not propagating myths like mass traveling faster than the speed of light)

Okay, I see your point now. And yes, I see what you mean about physics in general - it's a subject which teaches lessons in many ways whilst answering questions - that's probably why it fascinates me so much. I love trying to find out why things happen in the real world My 6th form college has a list of books for extended reading - I've just ordered 6 not so easy pieces by Feynman because it appears to deal with relativity and special relativity. Is it likely to be over my head considering I've done no work on relativity in school or college thus far?
 
  • #18
RK1992 said:
Okay, I see your point now. And yes, I see what you mean about physics in general - it's a subject which teaches lessons in many ways whilst answering questions - that's probably why it fascinates me so much. I love trying to find out why things happen in the real world My 6th form college has a list of books for extended reading - I've just ordered 6 not so easy pieces by Feynman because it appears to deal with relativity and special relativity. Is it likely to be over my head considering I've done no work on relativity in school or college thus far?

I don't think that relativity requires anything more of a person than any other discipline - people who keep their eyes and ears open have no problem understanding evolution, a lunar eclipse, the subject matter just happens to be less practical and more theoretical. THis was an issue before high speed computers could be employed to run simulations.

To understand the math of a four dimenstional tensor that governs relativity might require another 5 or 6 years of mathematics, but that is unimportant to an initial understanding of the concepts.

Many people (myself included) have a problem wrapping their head around what relativity is about, and what it's implications are for how we observe the universe, and more importantly, how it might be useful to us (such as GPS satellite data).

I continually review my knowledge (like any good scientist) in this and other areas (since it is easy to forget), and it always helps to have the same information come from different sources. In the end it all comes together.

Some videos of lectures on some basic concepts of relativity - they are first year courses, and there are related videos by the same lecturers you can view at your own discretion


and example of a lecturer structure on the same content, with heavy mathematics
http://www.youtube.com/view_play_list?p=CCD6C043FEC59772&search_query=stanford+relativity
General

Special
 
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  • #19
TcheQ said:
I don't think that relativity requires anything more of a person than any other discipline - people who keep their eyes and ears open have no problem understanding evolution, a lunar eclipse, the subject matter just happens to be less practical and more theoretical. THis was an issue before high speed computers could be employed to run simulations.

To understand the math of a four dimenstional tensor that governs relativity might require another 5 or 6 years of mathematics, but that is unimportant to an initial understanding of the concepts.

Many people (myself included) have a problem wrapping their head around what relativity is about, and what it's implications are for how we observe the universe, and more importantly, how it might be useful to us (such as GPS satellite data).

I continually review my knowledge (like any good scientist) in this and other areas (since it is easy to forget), and it always helps to have the same information come from different sources. In the end it all comes together.

Some videos of lectures on some basic concepts of relativity - they are first year courses, and there are related videos by the same lecturers you can view at your own discretion


and example of a lecturer structure on the same content, with heavy mathematics
http://www.youtube.com/view_play_list?p=CCD6C043FEC59772&search_query=stanford+relativity
General

Special


Wow, thanks for those links. I've watched the first and it was fine and the second wasn't too maths-y until my internet broke.

And the task set at the end of the first is to show that the speed of light is constant regardless of one's motion using the Lorentz Transformation but I'm not seeing how you do that using the lorentz transformation? I thought the whole point was the lorentz transformation was derived assuming that c is constant in all frames?
 
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  • #20
RK1992 said:
And the task set at the end of the first is to show that the speed of light is constant regardless of one's motion using the Lorentz Transformation but I'm not seeing how you do that using the lorentz transformation? I thought the whole point was the lorentz transformation was derived assuming that c is constant in all frames?

There are lots of different ways of deriving the Lorentz transformation. There are various sets of postulates you can take, and various arguments you can base on a given set of postulates. Here is an example that doesn't assume constant c: http://www.lightandmatter.com/html_books/genrel/ch02/ch02.html [Broken] From the point of view of this derivation, constancy of c is something that is derived from the Lorentz transformation, and in fact once you've discovered this universal velocity c, it takes a little more work to convince yourself that light must also travel at that velocity.

The modern way of thinking about this is that the c in relativity is not the speed of light, it's the maximum speed of cause and effect. Light just happens to travel at that velocity. (And this is only assuming that light has zero rest mass, which experiments could prove tomorrow to be false tomorrow. See, e.g., R.S. Lakes, "Experimental limits on the photon mass and cosmic magnetic vector potential", Physical Review Letters , 1998, 80, 1826-1829, http://silver.neep.wisc.edu/~lakes/mu.html )
 
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  • #21
RK1992 said:
Wow, thanks for those links. I've watched the first and it was fine and the second wasn't too maths-y until my internet broke.

And the task set at the end of the first is to show that the speed of light is constant regardless of one's motion using the Lorentz Transformation but I'm not seeing how you do that using the lorentz transformation? I thought the whole point was the lorentz transformation was derived assuming that c is constant in all frames?

I couldn´t watch the videos, but I can tell you my view: All SR formulas are derived (on paper) assuming that the speed of light is the same in all frames (=c) and that time and lengths are instead frame-dependent. The (mathematical) reasoning is: you freeze c and you allow t and x to change. That's not a demonstration. It's a postulate.

At the beginning, that caused me some reluctance, because many papers say that they are going to "demonstrate" (on paper!) one thing or the other. And later, somewhere else, yes, they tell you that the other thing is "proved" by applying the very same formulas. Thus the logic looks circular.

But if you think of it, that is not really a flaw, if put into the right context. Working on paper, one *always* proceeds that way. You first "assume or postulate" that something is going to happen. In this case: all observers will measure the same speed for light. Once you have that mental picture or film of a hypothetical reality, you express it with formulas or drawings (geometry) and extract consequences that you might have never thought of. With the outcome (in mathematical or geometric language), you make predictions: if my postulate is true, then it will happen such and such. And finally it is experiment what actually "proves" whether your assumption was correct or not, what refutes or validates the theory.

For your tranquility, classical mechanics did not proceed in a different manner. For example, it was assumed that time (instead of light speed) was going to be measured as the same value by all observers (just like simultaneity or the length of objects). That is also a pure postulate, there was no proof for that. Consequently, it was thought, since the distance traversed by an object is going to be forcefully different in two frames moving relative to each other, its speed (different distance/same time) had to be different, too, no matter whether we talked about a ball or a light pulse. Experiments confirmed that approach for normal objects moving at Earth-like speeds.

However, experiments under more demanding conditions (relativistic speeds) have so far proved that SR's assumption (the invariant thing is c, not time or lengths) is preferable.

A different thing is whether the assumption is more or less random or is based on some deeper thought about the root causes of the phenomenon. Personally I think that sooner or later one should look for those root causes, but there are diverging views on that...
 
  • #22
RK1992 said:
Hello all, having decided that I wish to apply to Oxford to study Physics (as well as Imperial and, I am still considering Natural Sciences at Cambridge), I have been informed that extra reading and independent study would be advised, so I'm delving into the world of relativity (I'm a first year AS student so we've done nothing on it yet).

I've stumbled upon an explanation of why time dilation happens but the explanation seems odd to me. We've been told that time dilation happens in GCSEs etc but never had any sort of explanation as to why. This analogy (I don't think it's a proof but I haven't seen it referred to as a proof or an analogy so I'm assuming it's an analogy) doesn't appear to prove anything to me...

Say you have a stationary light clock with the photon bouncing (I'm presuming you all know the experiment I;m talking about - http://galileoandeinstein.physics.virginia.edu/lectures/srelwhat.html has it around half way down if you don't).

The photon will bounce off one mirror at the angle it hits it - so straight back off the mirror. If you then move the mirror fast enough, to say the right, then the photon will have bounced off one mirror but the other mirror will now be to the right of the photon's path before the photon has arrived and the photon will not reflect back to the other mirror, it will just continue off into space because there is no mirror to reflect, so how can the photon take a "longer path" back to the mirror.

Is this an example of the analogy breaking down or am I missing some knowledge of how photons move? Does the photon reflect off a surface at an angle if the surface is moving? Or is it just a flaw in the analogy?

I accept that time dilation happens - experimental data proves it as such, but is there an analogy which would actually explain this phenomenon without any flaws? Is this actually a flaw or is it my lack of knowledge?

Thanks in advance.

Let's keep it really simple. The emission is a multiple photon event. Each observer sees a different photon, the one that has the correct angle to intercept the mirror.
 
  • #23
phyti said:
Let's keep it really simple. The emission is a multiple photon event. Each observer sees a different photon, the one that has the correct angle to intercept the mirror.

Would that not evade the whole point the thought experiment is trying to make?
 
  • #24
I think it's a flawed thought experiment.

While moving at a constant velocity, the photon would need to be directed at an angle to reach the other mirror, bounce off at angle, and so on..
This means the photon is traveling the same distance in space, and taking the same amount of time, from either frame of reference.

Therefore I see no need for time dilation, or anything else, to explain what's happening here..
 
  • #25
Would it not be better to think of it, not as a photon, but as a ray of light which radiates in all directions? Maybe to account for the angles.
 
  • #26
Saw said:
All SR formulas are derived (on paper) assuming that the speed of light is the same in all frames (=c) and that time and lengths are instead frame-dependent. The (mathematical) reasoning is: you freeze c and you allow t and x to change. That's not a demonstration. It's a postulate.

That's incorrect. See the link in #20. Other treatments that don't take constant c as a postulate:

Morin, Introduction to Classical Mechanics, Cambridge, 1st ed., 2008

Rindler, Essential Relativity: Special, General, and Cosmological, 1979, p. 51
 
  • #27
bcrowell said:
That's incorrect. See the link in #20. Other treatments that don't take constant c as a postulate:

Morin, Introduction to Classical Mechanics, Cambridge, 1st ed., 2008

Rindler, Essential Relativity: Special, General, and Cosmological, 1979, p. 51

Did you read what I said in my post? I like "Light and matter" site and will read the link carefully. But, in logical terms, the task seems impossible: deriving a formula where c is a constant without assuming that c is constant...? How do you do it: you start by writing c and c' and in some step you forget yourself and start writing c everywhere?
 
  • #28
sneh said:
I think it's a flawed thought experiment.

While moving at a constant velocity, the photon would need to be directed at an angle to reach the other mirror, bounce off at angle, and so on..
This means the photon is traveling the same distance in space, and taking the same amount of time, from either frame of reference.

Therefore I see no need for time dilation, or anything else, to explain what's happening here..

The thought experiment assumes there is a single photon emitted, say from a train and observed in the train and ground frame, or if you prefer two photons, i.e. train photon emitted from the train and observed in both train and ground frame and ground photon emitted from the ground and observed in both ground and train frame.

Let us call the frame from which each photon is emitted the "local" frame and the other the "foreign frame".

What is clear and unavoidable is that the local frame observes that its local photon traverses from bottom to top mirror a vertical trajectory, while the foreign frame observes a diagonal pat. Obviously, the first is shorter.

What is not necessarily true, unless one provides more information about the underlying physical proceses, is that the foreign photon is faster or slower than the local photon.

The thought experiment simply "assumes" that the foreign photon is slower and by the way assumes it bilaterally. Thus you get some equation, which in itself does not prove anything. First, it only accounts for time dilation. In other thought experiments you have to add to that length contraction, relativity of simultaneity... And anyhow, what proves the validity of the whole system is real-life experiment, not the thought experiments themselves.

danielatha4 said:
Would it not be better to think of it, not as a photon, but as a ray of light which radiates in all directions? Maybe to account for the angles.

You mean a wave, not a ray, don´t you? Well, that way you ensure that there is light hitting its target anyhow. But you don't need to do that. A laser produces a beam that is very little divergent and in spite of that it hits the target. How does it manage to if its target is flying away? I tried to explain in post #2: basically, the laser beam only exists if it has the right direction.
 
  • #29
Saw said:
Did you read what I said in my post?

Yes, I did read your post.

Saw said:
I like "Light and matter" site and will read the link carefully. But, in logical terms, the task seems impossible: deriving a formula where c is a constant without assuming that c is constant...? How do you do it: you start by writing c and c' and in some step you forget yourself and start writing c everywhere?

Maybe we could resume the discussion after you take a look at the derivation.
 
  • #30
bcrowell said:
Maybe we could resume the discussion after you take a look at the derivation.

Bcrowell, the text is very dense, full of technicalities and most interesting. But with regard to this particular point..., I don't think it manages to derive the LTs without postulating that the speed of light is the same for both frames.

The author says “For convenience, let's adopt time and space units in which c=1”. Assuming the formulas are derived and right and useful, it is actually very convenient to equate c with 1 (the so called geometric or natural units). But if one uses that trick in a “derivation”, one should warn what it means, since it’s a major step. What it means is that, in frame A, a rod is said to be 1 light-second long if light takes 1 second to traverse its length (or, rather, 2 s to complete the round-trip). And if that same rod is now moving relative to A, that is to say, it’s at rest with frame B, is it still 1 light-second long? It wouldn’t if light didn’t take 1 s to traverse its length. But we assume it does, so c = 1 also in frame B. Conclusion: the very fact of using that convention (c=1) equates to assuming, from the outset of the derivation, that both frames measure the same speed for light.
 
  • #31
Saw said:
Bcrowell, the text is very dense, full of technicalities and most interesting. But with regard to this particular point..., I don't think it manages to derive the LTs without postulating that the speed of light is the same for both frames.

The author says “For convenience, let's adopt time and space units in which c=1”. Assuming the formulas are derived and right and useful, it is actually very convenient to equate c with 1 (the so called geometric or natural units). But if one uses that trick in a “derivation”, one should warn what it means, since it’s a major step. What it means is that, in frame A, a rod is said to be 1 light-second long if light takes 1 second to traverse its length (or, rather, 2 s to complete the round-trip). And if that same rod is now moving relative to A, that is to say, it’s at rest with frame B, is it still 1 light-second long? It wouldn’t if light didn’t take 1 s to traverse its length. But we assume it does, so c = 1 also in frame B. Conclusion: the very fact of using that convention (c=1) equates to assuming, from the outset of the derivation, that both frames measure the same speed for light.

39:00 onwards, he deals with how you arrive at the conclusion c=c', using c=1 for simplicity
 
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  • #32
TcheQ said:
39:00 onwards, he deals with how you arrive at the conclusion c=c', using c=1 for simplicity


TcheQ, thanks for pointing at the relevant part of the video.

I think the lecturer himself confirms the point, although what he says does help to explain it better.

First, he considers that a light beam has, in an unprimed frame, a speed = c, whatever it is. Without special choice of units, that means that:

x (distance traversed by the beam in a given time = L = the length of a rod where that distance is marked) = ct (speed of light x that time lapse).

If we then choose to measure distance in light seconds, the expression becomes, by definition (a light-second = distance traversed by light in a second):

x = t

So far, so good. The choice of units is innocuous. No major step. Just a convenient approach.

Second, he wonders about the coordinates of that very same light beam in a primed frame, moving relative to the other. Without choice of units and without any special assumption, the equation will be:

x' = c't'

Now it’s time to make your choice of units. Can you simply say, like we did before…

x’ = t’?

Well, classical mechanics would say you can’t. Its reasoning would be:

(a) If the rod whose length we are using as reference is the same one as before, then we have x’ = x.
(b) In the primed frame the beam does not travel at c, but at (c-v) in the go trip (since the target is escaping), (c+v) in the return trip (since it is heading towards its target) and at the average between the two in the round trip.
(c) With (b)’s assumption, no matter which reference you take (light speed at the go trip, at the return trip or at the average of the round trip), if you make the calculation, you’ll conclude that the time that the beam takes to traverse the length of the rod (x=x’=L=L’) is never x light-second.

Hence with the classical assumptions you *cannot* say that x’=t’!

Of course you can change the assumptions. You can postulate that c = c’ and so x’=t’, even if as a consequence of that x’ may be different from x and t’ different from t. If you take that step (a major step, by the way), then all the rest follows. But without that postulate, you go nowhere. The lecturer somehow recognizes it when he states that, in a later part of the derivation, x^2-t^2=0 does not necessarily imply x’^2-t’^2=0, but he could have made it clearer.

To sum up: c = c’ is not a conclusion you arrive at but a postulate you derive consequences from.

Unlike what oversimplified explanations suggest, you cannot change a physical theory (you cannot shift from classical to special relativity) just with algebra or geometry, unless the original theory contained some algebraic or geometric mistake. And I wouldn’t advise anyone to tell Newton that in his face. If you wish to change, you can (and I think you must) but you need a new physical assumption for that purpose, not just math or drawings.
 
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  • #33
Saw said:
Bcrowell, the text is very dense, full of technicalities and most interesting. But with regard to this particular point..., I don't think it manages to derive the LTs without postulating that the speed of light is the same for both frames.

The author says “For convenience, let's adopt time and space units in which c=1”. Assuming the formulas are derived and right and useful, it is actually very convenient to equate c with 1 (the so called geometric or natural units). But if one uses that trick in a “derivation”, one should warn what it means, since it’s a major step. What it means is that, in frame A, a rod is said to be 1 light-second long if light takes 1 second to traverse its length (or, rather, 2 s to complete the round-trip). And if that same rod is now moving relative to A, that is to say, it’s at rest with frame B, is it still 1 light-second long? It wouldn’t if light didn’t take 1 s to traverse its length. But we assume it does, so c = 1 also in frame B. Conclusion: the very fact of using that convention (c=1) equates to assuming, from the outset of the derivation, that both frames measure the same speed for light.

No, that's incorrect. The existence of a constant c has already been established at that point. Setting c=1 after that is simply a choice of units.

If you still have any doubts about the possibility of deriving the Lorentz transformation without assuming a constant speed of light, please take a look at the two references I gave in #26. Amazon will probably let you see the relevant parts with their "look inside" feature. Wolfgang Rindler is an extremely well known relativist. I really don't think he's deluding himself when he says you can derive the Lorentz transformations without assuming constant c.

Critical thinking is great, but when I cite multiple published references to demonstrate a particular point, the burden then falls on you to explain why so many published sources are saying the same thing. I think at a minimum you either need to (a) read all the references and explain why *all* of them are wrong, or (b) demonstrate that there is some published controversy on this point among researchers in the field.
 
  • #34
TcheQ said:
39:00 onwards, he deals with how you arrive at the conclusion c=c', using c=1 for simplicity


I watched this lecture and all he does is show that the Lorentz transformations have the property that the speed of light is constant in all inertial frames while the Newtonian (Galliean) transformations do not have that property and on that basis rejects the Newtonian transformation. At time 43:00 he states that he is looking for a transformation that preserves the constancy of the speed of light and then picks the Lorentz transformation (without any derviation or postulates shown) because it has the desired property. It is not surprising that using a transformation that was formulated with the initial assumption that the speed of light is constant in all frames should predict that the speed of light is constant in all frames.
bcrowell said:
From the point of view of this derivation, constancy of c is something that is derived from the Lorentz transformation, and in fact once you've discovered this universal velocity c, it takes a little more work to convince yourself that light must also travel at that velocity.
As above, if you start with a transformation that was formulated with the initial assumption that the speed of light is constant in all frames then it is not surprising that the transformation should predict that the speed of light is constant in all frames. Circular reasoning.
Saw said:
Did you read what I said in my post? I like "Light and matter" site and will read the link carefully. But, in logical terms, the task seems impossible: deriving a formula where c is a constant without assuming that c is constant...? How do you do it: you start by writing c and c' and in some step you forget yourself and start writing c everywhere?
Essentially I am agreeing with Saw that most derivations that do not have the constanty of the speed of light explicity stated as an assumption or postulate, have it impicitly assumed at the outset or have a conditional that it is a required outcome. However, it might not be "impossible" to have such a derivation but it would need a large and unreasonable set of alternative postulates.

For example let us take this set of postulates or initial assumptions:

1) Objects length contract with relative motion, in accord with the Lorentz transformations.
2) Clocks with motion relative to the observer are measured to run slower than clocks at rest with the observer, in accord with the Lorentz transformations.
3) Simultaneity is relative, in accord with the Einstein simultaneity equations.
4) Velocities add according to the Relativistic velocity addition equations.

Starting with all those assumptions we could probably claim that the constanty of the speed of light in all inertial reference frames is an inevitable logical conclusion if we accept that the above 4 postulates are self evident and reasonable assumptions supported by the available experimental evidence. However at the time Lorentz, Einstein, Fitzgerald and others were formulating the Lorentz transformations, none of the above 4 postulates were self evident and reasonable initial assumptions. In fact they were conclusions that were hard to swallow at the time because there was no direct experimetal evidence they were true and it was certainly way outside of everyday experience. Now if someone had doubts about the validity of the constancy of the speed of light, it would be difficult to argue they must accept it as inevitable conclusion, because everyone accepts that length contraction is self evident, obvious and inalienable fact.

On the other hand if someone had doubts about time dilation or length contraction and it was pointed out that if they accept that the speed of light is measured to be the same and constant in all inertial reference frames (and the laws of physics are the same in all IRFs) they would probably have to concede they are inevitable logical outcomes. After all there was already experimetal evidence that the speed of light is constant and independent of the motion of the source and Maxwell's equations also strongly suggested it.

If you claim that the Lorentz transformation can be derived without any assumtion of the constancy of the speed of light you should make it clear what initial assumptions you are making. However, it would be unreasonable and circular to derive the Lorentz tranformations from the postulate "the Lorentz transformations are a correct description of time, space and motion in nature".

Basically I think you need to show that the Lorentz transformations can be derived without any assumption of the constancy of the speed of light. One way of doing this is it to pick random transformations, until you find one that has the property that the speed of light is constant for inertial observers. Once you have done that it would be unreasonable to claim that the speed of light MUST be constant because your randomly picked transformation predicts it. For example I could pick random transformations until I find one that satifies the condition that the speed of light is observer dependent and then claim that "proves" the speed of light is not constant, which is of course utter rubbish, because I have effectively made it an initial condition that the speed of light is not constant.
 
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  • #35
bcrowell said:
No, that's incorrect. The existence of a constant c has already been established at that point. Setting c=1 after that is simply a choice of units.

I think we may have a little misunderstanding here. What I am saying, really, is not that the choice of units itself is the postulate. The postulate is assuming the speed of light is the same in all frames, with whatever units you play. If you choose units so that c=1 in one frame, then the postulate is assuming that c=1 in all frames. You may want to look at post #32 where I try to express the idea more at length than what you just quoted. Do you agree to that?

bcrowell said:
If you still have any doubts about the possibility of deriving the Lorentz transformation without assuming a constant speed of light, please take a look at the two references I gave in #26. Amazon will probably let you see the relevant parts with their "look inside" feature. Wolfgang Rindler is an extremely well known relativist. I really don't think he's deluding himself when he says you can derive the Lorentz transformations without assuming constant c.

Where does he say so? In page 45 he states that:

Newton’s axiom t=t’ would lead to (…) the GT. Instead, we now appeal to Einstein’s Law of propagation of light. According to it, x=ct and x’=ct’ are valid simultaneously, being descriptions of the same light signal in S and S

And then he goes on to derive the LTs.

That’s just what I was saying: to derive the LTs you have to leave aside the classical assumption (t=t’) and rely on a different assumption (c=c’, at the expense of admitting that t≠t’ and x≠x’).

I am not aware that there is any “published controversy” on this issue. In fact, I don’t think it’s controversial. From what I have read in the forum, it is usually commented that the habitual thought experiments are not proofs of SR, but ways to quantitatively derive its equations, assuming of course that you buy its postulates (which are true or not as ruled by experiment).

Edit: I agree with kev's post.
 
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