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mkline55
2015-Jun-18, 04:06 PM
1) According to length contraction theory, an observer watching a passing object will observe length contraction in the direction of motion. The amount of contraction can be calculated using the Lorentz transform. I understand that this effect does not apply to individual photons, but does it not apply to the distance between photons? For example, a transmitter sends one photon toward a receiver, then sends another one second later. Does the Lorentz transform not apply to the length of an imaginary line drawn between the photons? To the receiver they are still one light second apart, and are not received simultaneously. Why? Does the answer also apply to other relativistic particles?

2) Expansion theory says light gets red-shifted somewhat proportional to the distance traveled. Does this also mean that photons which originated one second apart on the same path will be farther than one light-second apart on reception? If so, then does this factor into the luminosity of distant galaxies we observe?

Shaula
2015-Jun-18, 04:24 PM
1) What frame is the receiver in? What frame is the transmitter in? What frame are you measuring the length between the photons in? You should see that in order to get your zero length you would have to be measuring this in the photon frame. It is not the fact that the frame is tied to a photon that is the issue, it is the fact that the frame is moving at c. I think you are conceptually mixing frames here.

mkline55
2015-Jun-18, 04:35 PM
1) What frame is the receiver in? What frame is the transmitter in? What frame are you measuring the length between the photons in? You should see that in order to get your zero length you would have to be measuring this in the photon frame. It is not the fact that the frame is tied to a photon that is the issue, it is the fact that the frame is moving at c. I think you are conceptually mixing frames here.

Receiver is a satellite. Sender is on Earth. Receive is me. Sender is my neighbor. I don't care as long as their distance is greater than 1 light second. Or use a billionth of a light second. Doesn't matter to the question. Relative to the speed of the photons, the relative motion between them is nearly zero. I am not measuring anything in the invalid photon frame. The question amounts to, since the photons are traveling at c relative to both the transmitter and the receiver, why is there a non-zero distance the between the photons for an observer to which they by definition must be moving at c? Lorentz applies? Yes? No? And, if that answer applies to photons because they move at c and because they have no valid frame for you to measure, then does the same rule apply to other relativistic particles?

Shaula
2015-Jun-18, 05:13 PM
The important bit is which frame are you measuring the distance between the photons? The distance between them only asymptotically approaches zero if the frame in which this is measured in is moving at c relative to the frame in which your initial distance was measured in. In your examples there is only a small relative speed difference between your transmitter and your receiver, so why would you expect a measured intraphoton distance of zero?

Jeff Root
2015-Jun-18, 05:17 PM
I'm not at all sure that this answers or even leads to an
answer to your question, but I'll point out that it is not
possible to detect a photon without interacting with it.
That basically means absorbing it. So if you want to
measure the distance between two photons, you have
to absorb both of them.

Do you have two photon detectors or just one? If you
only have one, the distance between the two detections,
in your frame of reference, is zero. Of course, some time
goes by between those two detections, also in your frame
of reference.

Hope that helps...

-- Jeff, in Minneapolis

Jeff Root
2015-Jun-18, 05:22 PM
Shaula,

mkline55 expects the distance between the photons to be
contracted to zero because they represent leading and trailing
points on an object moving past him at the speed of light.
A straightforward conundrum but I don't see the solution.

-- Jeff, in Minneapolis

Shaula
2015-Jun-18, 05:43 PM
Shaula,

mkline55 expects the distance between the photons to be contracted to zero because they represent leading and trailing points on an object moving past him at the speed of light. A straightforward conundrum but I don't see the solution.
I don't see the conundrum. Two photons separated by x metres is not an object. You are measuring the distance between them in your frame so why would you expect it to be zero? Put another way. I send a photon out, wait a year and send another one out. Would you expect the distance between them to be zero as measured in my frame at the time of the second photon being emitted?

mkline55
2015-Jun-18, 06:04 PM
I don't see the conundrum. Two photons separated by x metres is not an object. You are measuring the distance between them in your frame so why would you expect it to be zero? Put another way. I send a photon out, wait a year and send another one out. Would you expect the distance between them to be zero as measured in my frame at the time of the second photon being emitted?

It is not my expectation that the distance should be zero. I am asking why the Lorentz transform does not apply. Also, how are the two photons not an 'object' when a board or a box or any other 'object' is merely a bunch of very tiny particles separated by mostly empty space?

I'm working through Jeff's response:


Do you have two photon detectors or just one? If you
only have one, the distance between the two detections,
in your frame of reference, is zero. Of course, some time
goes by between those two detections, also in your frame
of reference.

So in the detector's frame, the distance between them actually is zero, they are just separated in time? That seems to put a different light on the definition of distance.

Jeff Root
2015-Jun-18, 07:29 PM
So in the detector's frame, the distance between them
actually is zero, they are just separated in time? That
seems to put a different light on the definition of
distance.
Yes, that's what I had in mind. I'm not sure it helps.

-- Jeff, in Minneapolis

Jeff Root
2015-Jun-18, 07:30 PM
I don't see the conundrum. Two photons separated
by x metres is not an object.
That doesn't matter. You should be able to imagine them
representing two points, fore-and-aft, on an object. Or on
two objects. Doesn't make any difference. The important
thing is that these two points are moving past you at light
speed. And I'm beginning to see the solution.

Whatever the distance is between the two points -- in their
frame of reference -- will appear contracted in your frame.
However, the problem is that light has no sensible frame.
The concept of a reference frame just can't be applied to
photons. So there is no distance to contract. There is no
length measurement to which the Lorentz formula can be
applied.

I think mkline55 had it backwards: He was thinking that
because the photons are moving past him at the speed of
light, the distance between them must contract to zero, if
the Lorentz formula applies. But what is that distance?
It would be the distance measured in the photon's frame,
and there is no such distance. It is, in effect, zero. But
the observer watching the photons pass by (so to speak),
can see a definite distance between them. How is that
possible? If you imagine a photon going past you, with a
metre stick right behind it, followed by a sceond photon,
the metre stick would measure the distance between the
photons in the frame of the photons and metre stick.
Since the photons must be moving at the speed of light
relative to the metre stick as well as relative to you, the
photons cannot hang around the ends of the metre stick
for long. But the metre stick is also moving at ludicrous
speed in the same direction as the photons, so it doesn't
hang around for long, either. So right at the instant that
the photons and metre stick pass by you, they are all in
the right arrangement: the two photons are a metre apart,
in the frame of the metre stick. But the metre stick should
be contracted to zero length. Which means the photons
should be on top of each other. Even if the metre stick
is moving past you at slightly less than the speed of light,
it should still be very short, and thus the photons should
almost be on top of each other.

I'm sure the solution is simple, but I still don't see it.

-- Jeff, in Minneapolis

Shaula
2015-Jun-18, 08:08 PM
It is not my expectation that the distance should be zero. I am asking why the Lorentz transform does not apply.
The Lorentz transform does apply. You are just not using it correctly. Think about what it actually means. The Lorentz transform converts between proper time/distance in one frame to co-ordinate time/distance in another. We have already established that the photon frame is not a valid one so which proper time and which co-ordinate time are you transforming between? It is the frame of the emitter and the receiver. And they are not moving at c relative to each other. So the Lorentz transform is applied to the distance between the photons in one frame to give you what the other frame will regard as the distance between them. It doesn't matter what speed the particles sent between them move at unless you are trying to make a measurement in the frame of the particles. In this thought experiment no one is.

mkline55
2015-Jun-18, 08:28 PM
The Lorentz transform does apply.

The observer simply measures the time between photons. It is one second. Using c, he then knows the photons were 1 light second apart in his frame of reference. Is that much correct?

You appear to balk at using photons. Second part of the original first question: What about other relativistic particles? Neutrinos? High-speed electrons or protons? Two of them traveling at 0.9999 c and one light second apart as measured by the same observer and of the transmitter. How far apart are they in their own now-valid reference frame?

Jeff Root
2015-Jun-18, 08:47 PM
Shaula,

You aren't understanding the problem. The relation
between emitter and receiver is completely irrelevant.
It is the relation between the fast-moving light and
the receiver that is being asked about.

-- Jeff, in Minneapolis

Reality Check
2015-Jun-18, 09:14 PM
In which case the answer is quite trivial. Particles (photons, neutrinos, electrons, protons, etc.) travelling at 1 second apart will be detected by any receiver at 1 second apart.

However mkline55 seems to be asking about a situation like
* Let Alice emit particles at one light second apart.
* Let Bob detect these particles and measure that they are one light second apart.
* Let Charles be an observer travelling with the particles.
What distance between particles does Charles measure?
The answer is one light second apart because the question does not involve any other observer.

Jeff Root
2015-Jun-19, 12:30 AM
Reality Check,

I'm sure this is quite simple, but it is not trivial. You did not
answer it correctly. Charles would not measure a distance of
one light-second between the photons. One way to answer
correctly would be to say that Charles cannot measure the
distance between the photons. Another way would be to say
that Charles cannot exist. He is an impossibility. Another
way would be to change the description of Charles slightly
so that he can exist and can measure the distance between
the photons. Instead of travelling with the photons, he is
travelling in the same direction as the photons at nearly the
speed of light relative to Alice and Bob. That is the same as
the metre stick I described in the last sentence of the long
paragraph in my post #10.

The question mkline55 is asking is straightforward and
quite simple. He is asking how to apply the Lorentz formula
to an object moving past the observer at the speed of light.
For any ordinary object, the speed would be less than that
of light. But in this question, the "object" is light itself.
So it isn't clear to me how to come up with a correct answer.

-- Jeff, in Minneapolis
.

WayneFrancis
2015-Jun-19, 01:26 AM
I think Shaula and Reality understand fine.

Given this thought experiment


Alice is 100 light seconds away from Bob
Charlie is travelling at .99c at Alice directly towards Bob
When Charlie passes Alice she emits a photon on a timer of every 50 seconds.
The first photon is immediately absorbed by Charlie
The second photon is absorbed by Charlie 99 seconds later according to Alice and Bob
Charlie detects the photon 13.96566862 seconds later. He measures the distance between Alice and Bob as 14.10673598 light seconds.
Charlie calculates that in his frame the photons are being emitted 7.05336799 seconds apart.
So what? As you point out there can never be an observer traveling at c so the distance can never be measured at 0.
The best you can do is continually speed up an observer and get the distance and time to asymptotically approach 0.
Once you talk about an observer traveling at c you are outside the domain of applicability of the model you are dealing with.
Extrapolating to that point does not give a meaningful result.

Shaula
2015-Jun-19, 04:45 AM
The observer simply measures the time between photons. It is one second. Using c, he then knows the photons were 1 light second apart in his frame of reference. Is that much correct?
Yup.


You appear to balk at using photons. Second part of the original first question: What about other relativistic particles? Neutrinos? High-speed electrons or protons? Two of them traveling at 0.9999 c and one light second apart as measured by the same observer and of the transmitter. How far apart are they in their own now-valid reference frame?
1/70 of a light second in the frame of the particles. 1 light second in the emitter frame. If the receiver frame is at rest with respect to the emitter frame then 1 second in that too. Note I didn't avoid using photons - I said that for the question you were asking it was not actually relevant that they were photons because what you needed to do was convert between the proper/co-ordinate distances in the emitter/receiver frames. Thus the speed of the particles was actually irrelevant.

Jeff Root
2015-Jun-19, 08:11 AM
Wayne,

The thing is, mkline55 didn't ask about an observer
travelling at .99 c, nor did he ask about an observer travelling
at c. He asked about light travelling at c relative to the
observer. Which is something we deal with constantly,
including right now as you read this. It is a question about
a completely real, everyday situation that nobody here has
answered yet.

-- Jeff, in Minneapolis

Shaula
2015-Jun-19, 08:26 AM
The thing is, mkline55 didn't ask about an observer travelling at .99 c, nor did he ask about an observer travelling at c. He asked about light travelling at c relative to the observer. Which is something we deal with constantly, including right now as you read this. It is a question about a completely real, everyday situation that nobody here has answered yet.
How about you pose it explicitly, laying out which measurements are made in which frames, which frames you are transforming between and what you think the issue is. Because I am pretty sure that you are just mixing frames here or treating the effect as some kind of absolute.

Jeff Root
2015-Jun-19, 08:50 AM
Things moving relative to you appear length-contracted
according to the Lorentz formula. So the distance between,
for example, the front and rear of a train is measured to be
less when the train is moving past the measurer at high
speed than when it is at rest relative to the measurer.
Replace the front and rear of the train with two photons,
moving at the speed of light past the measurer in the
same direction as the train was travelling. The distance
between the two photons should be length contracted
according to the Lorentz formula.

So mkline55's questions are:

Is the distance between the photons length contracted?
If so, by how much, and relative to what?
If not, why not?

-- Jeff, in Minneapolis
.

Shaula
2015-Jun-19, 12:55 PM
What is the length of the photon pair and in what frame is it measured?

That is the critical question I am trying to get you to answer. To avoid another go around: The distance between the two photons (equivalent to the proper length of the 'object') is measured in the emitter's frame. It is then remeasured in the receiver's frame. If those two frames are moving relative to each other then there will be a difference in the 'length' the receiver records. If there is no then there will not be. The Lorentz transforms relate proper to co-ordinate time/distance.

What I think is going on here is that you are mixing up the idea of a length contraction with the idea of an absolute shrinking. It doesn't matter how fast the particles are moving - in this example the 'proper length' that is being transformed is set in the emitter's frame of reference. The co-ordinate length is then being measured in the receiver's.

mkline55
2015-Jun-19, 02:18 PM
Things moving relative to you appear length-contracted
according to the Lorentz formula. So the distance between,
for example, the front and rear of a train is measured to be
less when the train is moving past the measurer at high
speed than when it is at rest relative to the measurer.
Replace the front and rear of the train with two photons,
moving at the speed of light past the measurer in the
same direction as the train was travelling. The distance
between the two photons should be length contracted
according to the Lorentz formula.

So mkline55's questions are:

Is the distance between the photons length contracted?
If so, by how much, and relative to what?
If not, why not?

-- Jeff, in Minneapolis
.

That is a correct interpretation of the question I asked.

I can accept that the photons have no valid reference frame of their own using existing physics models, because application of a Lorentz transform would put all photons on the same path with zero distance between. I also appreciate Shaula's patient calculation of 1/70 light second distance in the frame of the example particles.

What about the second question? It hasn't been addressed at all.


2) Expansion theory says light gets red-shifted somewhat proportional to the distance traveled. Does this also mean that photons which originated one second apart on the same path will be farther than one light-second apart on reception? If so, then does this factor into the luminosity of distant galaxies we observe?

Grey
2015-Jun-19, 03:00 PM
2) Expansion theory says light gets red-shifted somewhat proportional to the distance traveled. Does this also mean that photons which originated one second apart on the same path will be farther than one light-second apart on reception? If so, then does this factor into the luminosity of distant galaxies we observe?You're right, this one hasn't been addressed. The answer is yes, if photons are redshifted (whether from cosmological redshift, gravitational redshift, or Doppler redshift), they will also arrive farther apart in time, which will indeed result in a net reduction in luminosity.

Jeff Root
2015-Jun-19, 11:11 PM
What is the length of the photon pair and in what frame is it
measured?

That is the critical question I am trying to get you to answer.
To avoid another go around: The distance between the two
photons (equivalent to the proper length of the 'object') is
measured in the emitter's frame. It is then remeasured in the
receiver's frame. If those two frames are moving relative to
each other then there will be a difference in the 'length' the
receiver records. If there is no then there will not be. The
Lorentz transforms relate proper to co-ordinate time/distance.
...

It doesn't matter how fast the particles are moving - in this
example the 'proper length' that is being transformed is set
in the emitter's frame of reference. The co-ordinate length
is then being measured in the receiver's.
That is a sensible and useful thing to do, since it is actually
doable. But it doesn't answer the question. It answers a
different, more practical question.



What I think is going on here is that you are mixing up the
idea of a length contraction with the idea of an absolute
shrinking.
I don't think so. mkline55 is asking a naive question
about a situation that is right in front of him at this very
moment. He and I both know that the "practical" answer is
that the Lorentz equation doesn't work for anything moving
at the speed of light relative to the observer. But we aren't
looking for a practical answer. We want an answer which
explains the physics.

Saying that it doesn't matter how fast the particles are
moving shows that you are addressing a different question
than the one being asked. It is all about the speed of the
particles relative to the observer. Specifically, what is the
effect of the speed of the particles relative to the observer
on the observed distance between them? If there is no
effect, why is there no effect? If the particles were bullets
or atomic nuclei shot out of a particle accelerator, I think
you would say that it matters how fast the particles are
moving relative to the observer. But when the particles
are photons, you say it doesn't matter. Why?

-- Jeff, in Minneapolis

Jeff Root
2015-Jun-19, 11:22 PM
We want an answer which explains the physics.
I think the "physics" in this case is just kinematics, or
even just geometry.

-- Jeff, in Minneapolis

Shaula
2015-Jun-20, 05:20 AM
Saying that it doesn't matter how fast the particles are moving shows that you are addressing a different question than the one being asked. It is all about the speed of the
particles relative to the observer. Specifically, what is the effect of the speed of the particles relative to the observer on the observed distance between them? If there is no effect, why is there no effect? If the particles were bullets or atomic nuclei shot out of a particle accelerator, I think you would say that it matters how fast the particles are moving relative to the observer. But when the particles
are photons, you say it doesn't matter. Why?
I have answered the question posed. If you emitted two neutrons, two bullets, two nuclei, two pigs - whatever - one second apart in the emitter's frame then the change in the ratio of their measured separation would be a function of the relative velocity of the emitter and the receiver. Their speed only defines their separation in the emitter's frame and thus the proper length you are transforming. Their nature is irrelevant. The important thing that you are missing is that two particles emitted a second apart don't have an intrinsic proper length like a train does (because, as I said, they are not an 'object' with that intrinsic property). In the questions posed you have assumed they have - but this proper length is set by their speed in the emitter's frame and thus as you change the speed of the particles it changes. Thus the question as posed becomes one about transforming a proper length from one frame to a co-ordinate length in another and is thus only dependent on the relative velocity of the two frames.

Now once this proper length is set if you were to accelerate your objects to a new speed while maintaining their proper length - their measured separation in their own centre of mass rest frame then the observer would see that distance as smaller when it were measured. But this is the critical extra step - and one which you cannot do for photons because you cannot accelerate them.

When trying to formulate these questions it is essential that you keep it very, very tightly defined what lengths you are measuring, what you are transforming and what frames you are working in.

jartsa
2015-Jun-20, 09:02 AM
Maybe the photon formation is infinitely contracted: From infinite length to 1 light second.

Shaula
2015-Jun-20, 10:47 AM
Maybe the photon formation is infinitely contracted: From infinite length to 1 light second.
And this is based on what?

Grey
2015-Jun-20, 01:41 PM
I think WayneFrancis was on the right track when he introduces an observer travelling from Alice toward Bob. In the case mkline55 describes, Alice shoots a relativistic particle toward Bob. Let's change this so that it's not a photon, so we can talk meaningfully about it's rest frame and put our third observer in that frame; how about a baseball travelling at half the speed of light? Two seconds later, she launches another baseball toward Bob at the same speed. From Alice's reference frame, then, the two baseballs are one light-second apart. Bob is at rest relative to Alice, and they've synchronized their clocks (in their own common reference frame) so he'll measure the same distance between the baseballs as Alice, and he'll agree with her about when the baseballs were launched.

Now Charlie is travelling along with the baseballs, and has a measuring rod that is one light-second long in Alice's frame, so one end should be even with the first baseball, and the other end should be even with the second baseball. Since he's moving at high speed relative to Alice, she should see that ruler length contracted. So Charlie should measure the ruler as about 15% longer than one light-second, and since that's the rest frame of the ruler, that's its proper length, and also the proper length separating the baseballs: slightly longer than one light-second. So Alice does see the distance between the baseballs as length contracted from what would be measured in Charlie's frame, moving along with the baseballs. How is that possible, since they were fired exactly two seconds apart? Well, remember that from Charlie's perspective, Alice's clock is running slowly. So when she fired the second baseball after her clock registered two seconds, more than two seconds had passed for Charlie (and the baseball), by that same 15%.

Of course, Charlie (and the baseballs) will also measure the distance between Alice and Bob as shorter than Alice and Bob measure it, and they'll disagree about how long it takes from the baseballs to travel from Alice to Bob (or, from Charlie's perspective, how long it takes Bob to get there, since it's Alice and Bob that are moving). This gets reconciled by relativity of simultaneity: they'll all agree on what time Bob's clock reads when the baseballs and Bob meet, but Charlie will disagree with Alice and Bob's statement that Alice's clock and Bob's clock are synchronized.

Jeff Root
2015-Jun-20, 08:41 PM
The important thing that you are missing is that two
particles emitted a second apart don't have an intrinsic
proper length like a train does (because, as I said, they
are not an 'object' with that intrinsic property).
I do not see that the particles are any different from
particles that are part of an object. In every case we
are discussing, the particles are a fixed distance apart
in ANY given inertial frame. If the particles are bullets
or pigs, they remain a fixed distance apart, just like
the front and rear points on the train.

In what way are the particles different from two points
on a solid object, and how does that difference matter?

It is clear to me that with bullets or pigs, there is no
difference. But with photons ... that's the question.

-- Jeff, in Minneapolis

jartsa
2015-Jun-20, 10:03 PM
And this is based on what?

The photon formation should be infinitely contracted, because it moves at speed of light. The contracted photon formation should be one light seconds long.

Hornblower
2015-Jun-20, 11:38 PM
I do not see that the particles are any different from
particles that are part of an object. In every case we
are discussing, the particles are a fixed distance apart
in ANY given inertial frame. If the particles are bullets
or pigs, they remain a fixed distance apart, just like
the front and rear points on the train.

In what way are the particles different from two points
on a solid object, and how does that difference matter?

It is clear to me that with bullets or pigs, there is no
difference. But with photons ... that's the question.

-- Jeff, in MinneapolisLet me take a shot at this one. Let's emit the photons one microsecond apart, so they will be separated by 300 meters, a nice visualizable length like a typical train. Now let's have a hypothetical train pulled by the locomotive of our dreams going arbitrarily close to the speed of light. It will need to be zillions of meters long in its own frame to be length-contracted to 300 meters and spanning the positions of the photons as seen in our frame, but that is OK in a thought exercise provided v is the least bit less than c. If we try to apply the equation to a hypothetical massless train with v = c, it becomes indeterminate. We have gone outside the domain in which the equation has any meaning.

If someone asks why the observed separation of the photons does not go to zero, all I would say initially is, "Why should it?"

Jeff Root
2015-Jun-21, 01:39 AM
Let's emit the photons one microsecond apart, so they will
be separated by 300 meters, a nice visualizable length like
a typical train. Now let's have a hypothetical train pulled by
the locomotive of our dreams going arbitrarily close to the
speed of light. It will need to be zillions of meters long in its
own frame to be length-contracted to 300 meters and spanning
the positions of the photons as seen in our frame, but that is
OK in a thought exercise provided v is the least bit less than c.
If we try to apply the equation to a hypothetical massless train
with v = c, it becomes indeterminate. We have gone outside
the domain in which the equation has any meaning.

If someone asks why the observed separation of the photons
does not go to zero, all I would say initially is, "Why should it?"
The equation says it should, so why shouldn't it? If you start
with a train of any definite proper length, and have it go past
you at speed c, you should measure its length as zero. At any
lower speed, the length will be greater than zero.

-- Jeff, in Minneapolis

Hornblower
2015-Jun-21, 01:55 AM
The equation says it should, so why shouldn't it? If you start
with a train of any definite proper length, and have it go past
you at speed c, you should measure its length as zero. At any
lower speed, the length will be greater than zero.

-- Jeff, in Minneapolis

I stand by my opinion that you are attempting to apply the equation to something outside of its domain of applicability. As I see it the equation becomes indeterminate at v = c.

Shaula
2015-Jun-21, 04:13 AM
I do not see that the particles are any different from particles that are part of an object. In every case we are discussing, the particles are a fixed distance apart in ANY given inertial frame. If the particles are bullets or pigs, they remain a fixed distance apart, just like the front and rear points on the train.

In what way are the particles different from two points on a solid object, and how does that difference matter?

It is clear to me that with bullets or pigs, there is no difference. But with photons ... that's the question.
As I have said, you need to consider what you are measuring and what you are transforming. In the particle case you are not applying the transform to the proper length of a particle, you are applying it to the distance between them in the emitter's frame of reference which is set by their emission times in the emitter's frame of reference. In the case of the pig or the train what you are transforming is their proper length, as measured in their own CoM frame. So if you fired off two pigs a second apart at near light speed then the observed length (in the receiver's frame) of the gap between them would depend solely on the relative motions of the two frames, the length of the pigs (measured in the receiver's frame) would depend on the speed of the pigs.

I've repeated this many times but what you need to do with this problem is stop trying to think about it in concepts and start being precise about what you are measuring, what you are transforming.

Shaula
2015-Jun-21, 04:15 AM
The photon formation should be infinitely contracted, because it moves at speed of light. The contracted photon formation should be one light seconds long.
You are applying the Lorentz transform, and SR itself, outside its domain of applicability. So what theoretical model are you using to make these predictions?

tusenfem
2015-Jun-21, 12:51 PM
The photon formation should be infinitely contracted, because it moves at speed of light. The contracted photon formation should be one light seconds long.


jartsa, this seems to go dangerously towards ATM country, let's not go there in Q&A.

Ken G
2015-Jun-21, 07:32 PM
The answer's in post #29, you just have to take the mathematical limit as the speed goes to c. There's no particular difficulty applying the Lorentz transform to the speed of light, if you take a limit, rather than just plugging in v=c. You just say v=c*(1-x) and take x->0, but leave x in there-- just imagine it is getting smaller and smaller and watch what happens.

Now let's do that, with a chain of bullets moving along the same path in a line. We have to decide what we will keep fixed about that chain, and this is crucial to decide. From the above question, it sounds like we should keep the time between the bullets fixed, as reckoned from the sender/receiver frame. So let's call that t. Now crank up v=c*(1-x), and see what happens as x goes to zero. In the sender/receiver frame, the distance between bullets increases, it is L = (1-x)*ct. This just a coordinate distance, it is not the proper distance between the bullets in the frame moving with them. To get the proper distance, we Lorentz transform to the frame of the bullets, and get (1-x)*ct*g, where g is the Lorentz gamma, so g = 1/root(1-(1-x)2) which = 1/root(2x) in the limit as x gets small. Thus in that limit, the proper distance L' = ct/root(2x).

So we see something interesting that was in post 29, but I'm not sure was appreciated, because the debate has continued: the proper distance between the bullets, the distance in their own frame, gets larger with v, and without limit as v goes to c. I believe this is the source of the confusion above, that we often hear there is "length contraction" in the frame of the photons, such that the universe is contracted to a negligibly thin shell, which would make it seem like the bullets would have to seem to be right on top of each other in their own frame. But this is not the case, nor is it true that the universe is contracted to zero in the photon frame. What is contracted to zero in the photon frame is anything we regard as finite in our frame, but Minkowski space is infinite, so the photon, in the limit I describe, has plenty of room for finite proper distances in its own frame. Even more-- we see from this problem that the photon has room for infinite proper distances!

So what's happening is, it's all a question of what you hold fixed. If it is the time between bullets, in our frame, that we hold fixed, then the proper distance between bullets, in the frame where the bullets are a stationary chain, increases if we imagine larger bullet speeds. This trend continues without limit as v goes to c, so is not hard at all to extend to the limit of the photon frame-- the proper distance between bullets goes to infinity in that case.

Note the language we would use in our own frame to "explain" what's going on there. We are keeping the time between the bullets fixed as v increases, so our reckoning of the distance between them increases with v. Yet we also think the rulers of observers on the bullets are length contracted, so we can make sense of why those observers reckon the bullets as being even farther apart than we do: they are using shorter rulers. Nothing changes that as the speed goes to c, these trends just continue to increase, the latter one increasing without bound.

Also, there is no difficulty with imagining that rigid rods connect the bullets, all observers would agree the rods fit perfectly between bullets. The rods are also reckoned to "length contract." But that is what happens when going from the rods own frame, to our frame as sender/receiver, so to go the other way we have to "length expand." We don't normally talk about "length expansion" in special relativity because we normally regard the object's length in its own frame to be the reference for comparison, and then all other frames see it as contracted from that. But if we start in the sender/receiver frame and take that distance as the reference, then the length of the rods, and the distances between bullets, are both expanded in the bullet frame, relative to this reference value. That trend continues even in the limit as v goes to c.

In summary, if there were rigid rods in there, we'd reckon their length to be vt in our frame, and we could calculate that their length in their own frame would be longer than that by the Lorentz factor. For them to be moving close to c, the rods would need to be vastly long in their own frame. This is all because what we are keeping fixed is the time t between bullets, but we are used to keeping the rod length in its own frame fixed (say, a "ruler"). The Lorentz transform always applies, even in the limit as the speed goes to c, and we can easily see what happens in that limit: the distance between the photons goes infinite. This is simply a ramification of holding the time between photons fixed in the sender frame.

Jeff Root
2015-Jun-21, 10:06 PM
Ken's post pretty much supercedes this reply to Hornblower,
but I put too much time and effort into it not to post it...



I stand by my opinion that you are attempting to
apply the equation to something outside of its domain
of applicability.
Yes, but how do I know that?



As I see it the equation becomes indeterminate at v = c.
Yes, if we try to calculate the proper length of the train
from the observed length, we get no meaningful result.

But if we assume the proper length of the (whatever)
moving at c relative to us is 300 meters, then we
calculate a value of zero for the length, which may be
very annoying, but is meaningful. It doesn't force me
to believe that I've attempted to apply the equation to
something outside its domain of applicability.

Consider if the train is moving at .99999999 c instead
of at c. Then its speed is 299792455 m/s instead
of 299792458 m/s. That is a speed difference of just
3 m/s. I can easily jog that fast. I can go from standing
still to jogging at 3 m/s in about 2 seconds. I just did,
before I started typing this paragraph, to be sure. The
difference between 299792455 m/s and 299792458 m/s
is so small that I expect it would take very sensitive
equipment in carefully controlled conditions to detect
the difference. The relativistic Lorentz factor (gamma)
at 299792455 m/s is 7071. That means a train which
appears to be 300 meters long would have a proper
length of 2,121,300 meters, which is terribly long for a
train but way, way, way short of "zillions of meters long".
Can a speed difference of just 3 m/s -- so small that it
must be difficult to even detect -- make the difference
between a train which would easily fit within Minnesota
in the east-west direction and a train that couldn't fit
within the space between the most distant galaxies on
opposite sides of the sky, even if it was doubled back
on itself zillions of times? Can that tiny difference in
speed make the difference between a length I could
measure with just a little bit of effort, and a length that
cannot be defined or measured even in principle? How
close to c can the speed of the train get before it makes
the jump from having a measureable length to infinite
length? Eight nines doesn't do it, and that must be,
as I said, really hard to measure. Nine nines? Ten?
A million? A zillion? Three zillion? So what speed do
particles need to have to be beyond the domain of
applicability of the Lorentz equation? How can I
distinguish between a situation where I can apply the
Lorentz equation, and a situation where I can't? And
how can I know that distinction is appropriate?

-- Jeff, in Minneapolis

Hornblower
2015-Jun-22, 12:51 AM
Upon reading Ken G's post I retract my remarks about the equation becoming inapplicable. It was a memory lapse on my part about the use of limits.

But if we assume the proper length of the (whatever)
moving at c relative to us is 300 meters, then we
calculate a value of zero for the length, which may be
very annoying, but is meaningful. It doesn't force me
to believe that I've attempted to apply the equation to
something outside its domain of applicability.
Why should we assume a proper length of 300 meters? I see no relevance to the exercise, in which we observe a length of 300 meters as measured in the frame in which we, the emitter and the detectors are stationary or nearly so. We apply the equation to the observation to find out what the proper length would need to be, using the limit technique as Ken showed. For v = c that length becomes infinite, which I would take as mathematically valid but not physically meaningful.

Reality Check
2015-Jun-22, 01:19 AM
He is asking how to apply the Lorentz formula
to an object moving past the observer at the speed of light.

Then it is trivial - you cannot apply the Lorentz formula to an object moving past the observer at the speed of light.
As you pointed out for the Charles observer, it is a physical impossibility. The Lorentz formula is undefined for v = c because the Lorentz factor is 1/0.

Jeff Root
2015-Jun-22, 03:33 AM
But if we assume the proper length of the (whatever)
moving at c relative to us is 300 meters, then we
calculate a value of zero for the length, which may be
very annoying, but is meaningful. It doesn't force me
to believe that I've attempted to apply the equation to
something outside its domain of applicability.
Why should we assume a proper length of 300 meters?
Because I needed a proper length for the example, and
you provided a figure of 300 meters which you would
recognize when I re-purposed it.



I see no relevance to the exercise, in which we observe
a length of 300 meters as measured in the frame in which
we, the emitter and the detectors are stationary or nearly
so. We apply the equation to the observation to find out
what the proper length would need to be, using the limit
technique as Ken showed.
No, as I said very clearly, in the case I described, we are
applying the equation to the proper length in order to find
out how long the thing would appear to be as it passes by.
That is the more common use of the equation, by far.
It has two other advantages: 1) It more closely matches
the case of photons passing by (the original question of
the thread). Since we aren't able to measure the length
of photons, I had to choose an arbitrary length for the
example, so I used 300 meters. 2) I wasn't forced to try
to divide by zero.



For v = c that length becomes infinite, which I would take
as mathematically valid but not physically meaningful.
Hmmm. I think it must have some physical meaning, but
almost certainly not the obvious one that the length is in
any way actually infinite. Perhaps the physical meaning is
that photons do not have the property of length-- either
proper or relative.

-- Jeff, in Minneapolis

Jeff Root
2015-Jun-22, 04:02 AM
He is asking how to apply the Lorentz formula to
an object moving past the observer at the speed of light.
Then it is trivial - you cannot apply the Lorentz formula to
an object moving past the observer at the speed of light.
That is not trivial at all. Did you just make up that rule?
No? Then where did the rule come from? Why does the
rule exist? How do I know the rule is correct?



As you pointed out for the Charles observer, it is a physical
impossibility.
Charles is a physical impossibility. Light is a commonplace.
We want to know how to apply the Lorentz formula to light.
The "object" in the quote above is light, or the spatial
intervals between particles of light. If the Lorentz formula
cannot be applied to them, then we want to know why.

I would assume that particles of light have length. That
seems natural since light has the property of wavelength.
I would certainly expect that there is a definite spatial
interval between two particles of light travelling one after
the other in the same direction. I would expect to be able
to apply the Lorentz formula to that length or interval.
When I do, I have to begin by assuming some arbitrary
value for either the length or the interval. And then I get
the interesting but annoying result that nomatter what
value I assume, the result is zero. That tell me something.
What it tells me has been the subject of this thread. I'm
not at all sure it is ended yet, because Ken's post is pretty
hard to grok fully. But I think at least one conclusion is
pretty clear: An individual photon has no length.

-- Jeff, in Minneapolis

Shaula
2015-Jun-22, 04:17 AM
That is not trivial at all. Did you just make up that rule? No? Then where did the rule come from? Why does the rule exist? How do I know the rule is correct?
Postulate 2 of SR is that light always travels at c in any inertial reference frame. Therefore by taking the photon frame (in which photons can consider themselves stationary) you break postulate 2 and invalidate SR. If you allow other things to travel at c you find that c is no longer actually the speed of light but just a conversion factor between units of space and time. And anything else moving at c now comes under postulate 2 using this broader definition, since you cannot accelerate anything with a mass to c and anything without mass must move at c.

jartsa
2015-Jun-22, 06:22 AM
The photon constellation has some internal energy, so it has some rest mass, so it moves a little bit slower than c, so it's very contracted, but not infinitely contracted.

(The law that says that it's impossible to cool things to absolute zero tells us that it's impossible to create photon formation with no internal energy)

Ken G
2015-Jun-22, 08:50 AM
Perhaps the physical meaning is
that photons do not have the property of length-- either
proper or relative.One relevant physical meaning is that photons have spin 1, yet only two spin states, which appear in the form of left- and right-handed circular polarization. A nonrelativistic particle with spin 1 should have three spin states relative to the direction of motion in a frame in which the particle is moving, but not if it is moving at c-- then the length contraction in effect makes it impossible for the photon to spin perpendicular to the direction of motion, and you just get the two opposite spin states along the direction of motion.

Ken G
2015-Jun-22, 09:07 AM
The photon constellation has some internal energy, so it has some rest mass, so it moves a little bit slower than c, so it's very contracted, but not infinitely contracted.

(The law that says that it's impossible to cool things to absolute zero tells us that it's impossible to create photon formation with no internal energy)Photon "constellations" have no internal energy, in the way we currently model them as particles with zero rest mass that move at c. As usual, to see this, one must take a limit. Any collection of nonrelativistic particles would have a sum of rest masses, plus an internal kinetic energy, and we could see what that sum was by imagining a series of observers moving faster and faster. The energy of the collection of nonrelativistic particles relative to each observer in the series would reach a minimum and then start to rise again as the observer speed increased. That minimum would be the sum of the rest masses and the internal kinetic energy, and indeed it would be impossible in practice to get the internal kinetic energy to zero in a collection of particles. But there's no difficulty imagining a series of collections of nonrelativistic particles with less and less internal kinetic energy, and take the limit as it goes toward zero, just as we can set up a series of observers and take the limit as their speed approaches c. Limits like this present no difficulties, and are what we mean by zero internal kinetic energy and observers moving at c, even though neither of those states are physically realizable. Mathematically, the limits can be unbounded, but that's what we mean by the limits being infinite, or they can be zero.

The situation is quite different for a collection of photons. Here a series of observers at faster and faster speeds along the photon direction finds that the limit of the photon energies, relative to them, does not go to a minimum and then rise again-- it goes monotonically to zero. Thus the internal energy is zero, even though there is no frame in which the particles have zero energy-- it's a limit. It is true that you cannot have a collection of photons all at exactly the same energy and frequency, as this would violate the Heisenberg uncertainty principle (as well as be thermodynamically problematic, as you say), but you incorrectly assert this means the internal energy cannot be zero. We model photons as particles with zero rest mass, because we have never measured any rest mass for them, so as far as we know the limit of a series of observers with faster and faster speeds along the photon direction is that the photon collection will have zero energy in the limit as those observers go to c. This holds even when the photons are not all at the same energy and frequency.

Hornblower
2015-Jun-22, 12:59 PM
Why should we assume a proper length of 300 meters?

Because I needed a proper length for the example, and
you provided a figure of 300 meters which you would
recognize when I re-purposed it.

My first post:

Let me take a shot at this one. Let's emit the photons one microsecond apart, so they will be separated by 300 meters, a nice visualizable length like a typical train. Now let's have a hypothetical train pulled by the locomotive of our dreams going arbitrarily close to the speed of light. It will need to be zillions of meters long in its own frame to be length-contracted to 300 meters and spanning the positions of the photons as seen in our frame, but that is OK in a thought exercise provided v is the least bit less than c. If we try to apply the equation to a hypothetical massless train with v = c, it becomes indeterminate. We have gone outside the domain in which the equation has any meaning.

I thought it was clear from the context that I meant 300 meters as observed in our frame of reference. Since you apparently did not find it clear, I emphasize it here. I never assumed a train or anything else with a proper length of 300 meters. As before I retract the last sentence in response to Ken's post.

Reality Check
2015-Jun-22, 10:09 PM
That is not trivial at all. Did you just make up that rule? ....

The rule is that any number divided by zero is undefined, Jeff Root.
Putting v=c results in a Lorentz factor of 1/0. The Lorentz formula contains the Lorentz factor. It is undefined at v=c. Then it is trivial - you cannot apply the Lorentz formula to an object moving past the observer at the speed of light (v=c).
Not that you need this to measure the interval between 2 objects - just put up a screen and measure the time between the 2 objects impacting it.

Jeff Root, we agree that Charles is a physical impossibility - you do not have to repeat this also trivial fact :p!

"I would assume that particles of light have length" would be an partially incorrect assumption. Photons are treated as point particles (https://en.wikipedia.org/wiki/Point_particle#In_quantum_mechanics) since they have no known internal structure. As such their properties such as wavelength and spin are intrinsic properties. However their wave packet will have a nonzero volume which if we were doing a QM analysis could have an effect. But this thread is about SR.

Jeff Root
2015-Jun-22, 11:06 PM
I thought it was clear from the context that I meant
300 meters as observed in our frame of reference.
Since you apparently did not find it clear, I emphasize
it here. I never assumed a train or anything else with
a proper length of 300 meters.
It was clear. I understood that you meant the observed
length was 300 meters, not the proper length. As I said,
I "re-purposed" the figure you provided, since I needed a
proper length as a starting point, and 300 meters is a
nice value, as you said.

-- Jeff, in Minneapolis

Jeff Root
2015-Jun-22, 11:44 PM
The rule is that any number divided by zero is undefined
Yes, of course. But I didn't divide by zero. Because I
wasn't required to.



Putting v=c results in a Lorentz factor of 1/0. The Lorentz
formula contains the Lorentz factor. It is undefined at v=c.
Then it is trivial - you cannot apply the Lorentz formula
to an object moving past the observer at the speed of
light (v=c).
It isn't trivial at all. You are saying I can apply the Lorentz
formula to something moving past me at 299792455 m/s,
but not to something moving past me at 299792458 m/s,
just 3 m/s faster, a difference in speed that would be very
hard to measure. And to measure the length of something
moving past me that fast-- Wow! Super difficult!



we agree that Charles is a physical impossibility - you do
not have to repeat this also trivial fact :p!
I agree that that is trivial, but my point was that in
contrast to the impossibility of the imaginary observer
Charles, the light itself-- which the original question was
about-- is not impossible, and is in fact a commonplace.
So it is something that we should be able to describe
with the same formula that we describe slower things.
At least, that is what one would expect and assume
as a starting point of an analysis.



"I would assume that particles of light have length"
would be an partially incorrect assumption. Photons are
treated as point particles (https://en.wikipedia.org/wiki/Point_particle#In_quantum_mechanics)
since they have no known internal structure.
Yes, it is interesting that photons are treated as point
particles. I wonder if they actually are point particles.
I'm pretty sure that their length cannot be measured,
if they have length. But do they have length? That's
what I'd like to find out. I've been working on that
problem for about 30 years. Maybe someday I'll come
up with a way to answer that question.

Nobody who knows me ever accused me of being fast.
Photons and I are opposites in that regard. Opposites
attract.



As such their properties such as wavelength and spin
are intrinsic properties.
It would be reasonable for me to say that I have no
idea what "intrinsic properties" are in this context.
From the general meaning of the word "intrinsic", I
would say that the wavelength is exactly the opposite
of an intrinsic property: It is a property which is not
intrinsic to any photon, but only shows itself when an
observer observes a group of photons. I'd call it an
"extrinsic" property. If wavelength really is labeled
an "intrinsic" property, then I think it is very badly
labeled.

-- Jeff, in Minneapolis

John Mendenhall
2015-Jun-23, 02:38 AM
Yes, it is trivial, Jeff. Anything divided by zero is undefined. It doesn't matter how you arrived at zero. Zero is zero. Lorentz no longer applies

Shaula
2015-Jun-23, 02:44 AM
So it is something that we should be able to describe with the same formula that we describe slower things. At least, that is what one would expect and assume as a starting point of an analysis.
Only if you ignore where the formula came from and what is meant by a limit.

By your 'only 3m/s faster' analogy you would expect that you could use the same equations to describe a complete system in equilibrium at -273C as -276C.

Reality Check
2015-Jun-23, 02:50 AM
But I didn't divide by zero.

You did not divide by zero but you stated what you thought mkline55's question was:

He is asking how to apply the Lorentz formula to an object moving past the observer at the speed of light
and that statement requires dividing by zero.
BTW in hindsight the question is meaningless in SR because the Lorentz transformation ("formula") requires two observers and the question does not state what the other observer is - unless it is the object and we are back to Charles!

I suspect that a wavelength could be detected from a single photon at a time. Maybe direct a single photon at a time at an electron in a Penning trap and measure the electron change in energy as it absorbs a photon which would give you the photon wavelength. But it may not be practical.

mkline55
2015-Jun-23, 12:15 PM
You did not divide by zero but you stated what you thought mkline55's question was:

and that statement requires dividing by zero.
BTW in hindsight the question is meaningless in SR because the Lorentz transformation ("formula") requires two observers and the question does not state what the other observer is - unless it is the object and we are back to Charles!

I suspect that a wavelength could be detected from a single photon at a time. Maybe direct a single photon at a time at an electron in a Penning trap and measure the electron change in energy as it absorbs a photon which would give you the photon wavelength. But it may not be practical.

The question was 'why' not 'how'. And the 'object' in question is an imaginary line between two photons. Observations indicate photons maintain their spatial relationships even though you cannot divide by zero.

Shaula
2015-Jun-23, 02:05 PM
The question was 'why' not 'how'. And the 'object' in question is an imaginary line between two photons. Observations indicate photons maintain their spatial relationships even though you cannot divide by zero.
Yes, because as I have said that spatial relationship is set in the emitter's frame, not the photon frame.

mkline55
2015-Jun-23, 03:53 PM
Yes, because as I have said that spatial relationship is set in the emitter's frame, not the photon frame.

Agreed. The original issue is not so much that Lorentz does not apply, it is that some relationship does exist between two or more photons so that they can transfer the emitter's understanding of the time/distance separating them perfectly (when expansion, gravity and velocity are accounted for) to the receiver. In other words, the photons maintain space/time properties of separation even though the (inapplicable) math would have indicated they cannot.

Shaula
2015-Jun-23, 04:30 PM
Agreed. The original issue is not so much that Lorentz does not apply, it is that some relationship does exist between two or more photons so that they can transfer the emitter's understanding of the time/distance separating them perfectly (when expansion, gravity and velocity are accounted for) to the receiver. In other words, the photons maintain space/time properties of separation even though the (inapplicable) math would have indicated they cannot.
But the maths does not imply that. The transforms link co-ordinate length to proper length for inertial frames. The only reason people are getting zeroes and infinities is because they are applying a formula without its context (the model it is derived from), because its context would tell them that the formula does not tell them anything about the photon frame. There doesn't need to be any kind of relationship between the photons somehow preserving distance and times because the idea that you need that implicitly assumes that you are using the invalid transform to the photon frame. Don't do that transform and there is no need for the inter-photon relationship to even exist.

mkline55
2015-Jun-23, 06:17 PM
But the maths does not imply that. The transforms link co-ordinate length to proper length for inertial frames. The only reason people are getting zeroes and infinities is because they are applying a formula without its context (the model it is derived from), because its context would tell them that the formula does not tell them anything about the photon frame. There doesn't need to be any kind of relationship between the photons somehow preserving distance and times because the idea that you need that implicitly assumes that you are using the invalid transform to the photon frame. Don't do that transform and there is no need for the inter-photon relationship to even exist.

my bold
If there is no need for a relationship, why do they arrive in the same sequence at the receiver as they were sent from the transmitter? We receive light signals reliably now that originated billions of years before our solar system even existed. Yet, those signals still reflect their original sequence. How do you explain how photons maintain their relative positions in the billions of years when they went unobserved, when you cannot define their relationship to one another at any time between when they are created and when they are absorbed?

Shaula
2015-Jun-23, 06:31 PM
my bold
If there is no need for a relationship, why do they arrive in the same sequence at the receiver as they were sent from the transmitter? We receive light signals reliably now that originated billions of years before our solar system even existed. Yet, those signals still reflect their original sequence. How do you explain how photons maintain their relative positions in the billions of years when they went unobserved, when you cannot define their relationship to one another at any time between when they are created and when they are absorbed?
You can define their relationship to each other, perfectly simply. The only thing you cannot do is do that in the rest frame of the photons - because there is no such valid frame in SR. There is no need to transform into the photon frame because the photons were emitted in some emitter's frame and when you make a measurement it will not be in the photon frame. There is no need to perform an invalid transform of this type. You seem to be elevating the CoM frame of a particle to some kind of 'more true' frame than any other frame. It is not. The spatial and temporal relationship of the photons are well defined in all valid SR frames and you can use the relationship between these frames and the frame in which the photon was emitted to work out the spatial and temporal sequence for any arbitrary observer.

Ken G
2015-Jun-23, 08:36 PM
I think what is really being asked here is a kind of philosophical question, which is, if photons cannot have a rest frame, how can they exist? What does it mean to exist for something that has no rest frame? That is certainly a fair question, and its answer is going to go deeply into the various different choices we can have for a philosophy of science. For example, one answer is to take the viewpoint of the die-hard empiricist, that photons don't exist, they are conceptualizations that we form in our own minds, like all theoretical constructs that are not the things we actually measure, and no one said that a conceptualization needed to come with a rest frame. Another answer is to go to the opposite end of the spectrum of the die-hard rationalist, and say that theoretical constructs are all that exists, so photons don't need a rest frame to exist because they are not really "objects", because there is no such thing as an object, there is only the mathematics. The rationalist imagines that a "frame of reference" is a kind of illusion brought by the biological entities trying to do physics, the "real" physics is all in the invariants. Photons do have those-- their invariant rest mass is zero (in our current models), and their invariant proper time between sender and receiver is also zero. The mathematics is fine with this, it's only those darn "observers" that need to transform between their arbitrary coordinate systems that are producing all the problems in this thread.

Reality Check
2015-Jun-23, 09:15 PM
The question was 'why' not 'how'.
The answer to "why" is easy - you apply the Lorentz transformation to transform from one observer's coordinate system to another observer's coordinate system in SR, mkline55.
But your 'object" is not a pair of observers so the 'why' question is moot :eek:!

ETA: As to your "How do you explain how photons maintain their relative positions in the billions of years when they went unobserved" - that is basically Newton's First law (https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion)! They have the same velocity. They will not change positions relative to each other. Their separation will not change. This is even true for massive particles such as neutrinos.
In SR different observers with different velocities relative to each other will measure different separations between any series of objects.

mkline55
2015-Jun-24, 11:47 AM
The answer to "why" is easy - you apply the Lorentz transformation to transform from one observer's coordinate system to another observer's coordinate system in SR, mkline55.
But your 'object" is not a pair of observers so the 'why' question is moot :eek:!

ETA: As to your "How do you explain how photons maintain their relative positions in the billions of years when they went unobserved" - that is basically Newton's First law (https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion)! They have the same velocity. They will not change positions relative to each other. Their separation will not change. This is even true for massive particles such as neutrinos.
In SR different observers with different velocities relative to each other will measure different separations between any series of objects.

I am trying to get this as precise as possible. You say "Their separation will not change." As measured by whom? For any inertial frame, their separation depends on the inertial frame. For an accelerating frame, their separation changes. The only separation which cannot change is the one which requires using an invalid reference frame. Then there's that problem of expansion, which would require the photons to gradually increase their separation as well. Would it be more accurate to say their separation is inversely proportional to the change in energy as measured by the observer vs. the transmitter, or is that oversimplified as well?

Ken G
2015-Jun-24, 01:06 PM
Would it be more accurate to say their separation is inversely proportional to the change in energy as measured by the observer vs. the transmitter, or is that oversimplified as well?One key lesson of relativity is that "separation" as we normally use the term is not a statement of a physical truth, it is a way of talking about something. That means it requires various conventions of language, our coordinate choices, and these will need to be translated into other words for other observers, especially accelerating ones. So this is all handled in relativity by creating the concept of the "spacetime manifold," which unifies space and time into a realm of invariants that do not require translation between observers. It is certainly tempting to associate "physical truth" with these invariants, and "separation between particles", being a coordinate-dependent notion, is not one of them. Separation between events along a given spacetime path is one of them, but here "separation" does not mean in the sense of spatial distance, indeed it usually means the time elapsed on a clock following that path.

Reality Check
2015-Jun-24, 10:10 PM
I am trying to get this as precise as possible. You say "Their separation will not change." As measured by whom? ...

As measured by a single observer.
For any single inertial frame, their separation does not depend on the inertial frame.
For any single accelerating frame, their separation does not change.
You need at least 2 inertial frames or non-inertial frames for observers to measure different separations relative to each other (in a sense that this the Relativity part of Special and General Relativity).
If you have an actual SR example, mkline55, then you need to get this as precise as possible. For example:
* Alice is an observer ...
* Bob is an observer ...
* SR states that Alice will measure X and that Bob will measure Y (e.g. length contraction, i.e. contraction of the distance between a series of objects).

There's no problem of expansion, etc. because that is irrelevant to your question about SR. To get this as precise as possible, you are asking about SR, not GR.

mkline55
2015-Jun-25, 12:40 PM
For any single accelerating frame, their separation does not change.

Is that the same as Grey's response in #23?


The answer is yes, if photons are redshifted (whether from cosmological redshift, gravitational redshift, or Doppler redshift), they will also arrive farther apart in time, which will indeed result in a net reduction in luminosity.

Ken G
2015-Jun-25, 12:51 PM
For any single accelerating frame, their separation does not change.What do you mean by their separation?

mkline55
2015-Jun-25, 12:55 PM
I mean their distance in space or the the time that passes in the single observer's accelerating frame between the observation of the first and the observation of the second.

Grey
2015-Jun-25, 04:02 PM
Is that the same as Grey's response in #23?No, and it's not even right, I think, if I'm correctly understanding what Reality Check is saying. Reality Check may be trying to stress that, in a certain sense, you need multiple reference frames to talk meaningfully about what happens even in one reference frame. That's reasonable, since relativity (special or general) is really all about comparing different reference frames, and being able to figure out what you'd expect to measure in one reference frame based on what you measure in a different reference frame. There may be a certain sense in which, if you're strictly limited to only talking about a single reference frame, relativity doesn't even apply, because there's nothing to compare it to. But of course, the changes in actual, real measurements take place regardless of how we talk about them, or what reference frames we permit ourselves to talk about. And there will be measurable consequences of general relativity in an accelerating reference frame, whether or not we discuss any other reference frames.

So I'm reasonably certain that in an accelerating reference frame, the separation between two pulses of light travelling in the same direction as my acceleration will in fact change. From a special relativistic point of view, I would indeed need to take into account multiple reference frames to work this out, but with general relativity, just the one accelerating frame will do.

Let's set it up. I'm in a spaceship. Let's assume I have an infinitely long, infinitely rigid ruler attached to the nose, with light detectors stationed wherever I need them, so that I can get measurements as desired. I fire two pulses of light, one shortly after the other (I'm using short pulses, rather than just individual photons, so that I can have my detectors occasionally check on them without destroying the whole pulse). Then I start accelerating after them. I'll never catch them, of course, but they'll occasionally pass by the detectors on the ruler (the detectors could just be partially silvered mirrors, so that most of the pulse continues on, but some of it is reflected back to me; or they could be photoreceptors that absorb and analyze a small portion of each light pulse as it passes by and then transmit their results back to me), so I'll get information back about what the light pulses are doing. But from general relativity, if I'm feeling acceleration, but still considering myself in a stationary reference frame, then there must be a gravitational field pervading the universe. Since the two photons are in a gravitational field, they'll be redshifted based on how high they are in that gravitational field (their gravitational potential); since they're at different heights, they'll be redshifted by different amounts from each other (and also different from me, since I'm at still a different gravitational potential). And since they're redshifted, the distance between wave crests has changed, and likewise the distance between pulses has to change by the same amount.

This makes sense, because if we go back to looking at this from special relativity in multiple frames, two differently moving observers should disagree about the distance between the light pulses (as they will disagree about all other distances measured), and an accelerating observer is handled in special relativity by treating them as changing from one inertial frame to another (and then essentially taking the limit as the difference in velocity between each frame goes to zero).

This is obviously non-trivial, and it's easy to get it wrong even when you know what you're talking about. So somebody take a look at what I just said. Did I make any errors in my reasoning? ;)

jartsa
2015-Jun-25, 05:08 PM
Let's set it up. I'm in a spaceship. Let's assume I have an infinitely long, infinitely rigid ruler attached to the nose, with light detectors stationed wherever I need them, so that I can get measurements as desired. I fire two pulses of light, one shortly after the other (I'm using short pulses, rather than just individual photons, so that I can have my detectors occasionally check on them without destroying the whole pulse). Then I start accelerating after them. I'll never catch them, of course, but they'll occasionally pass by the detectors on the ruler (the detectors could just be partially silvered mirrors, so that most of the pulse continues on, but some of it is reflected back to me;



No. There's a redshift when the light climbs up, and a blueshift when the light falls down.




or they could be photoreceptors that absorb and analyze a small portion of each light pulse as it passes by and then transmit their results back to me), so I'll get information back about what the light pulses are doing.


Yes. Photoreseptors will report to you that the light is redshifting as it is climbing up.

Ken G
2015-Jun-25, 05:22 PM
This is obviously non-trivial, and it's easy to get it wrong even when you know what you're talking about. It sounded fine to me, though there are problems keeping long rulers rigid in an accelerating frame. For example, if you put the ruler out the back of your ship, and sent the light that way, your acceleration produces a "Rindler horizon" behind you, which forces you to imagine that the entire infinite ruler fits into a finite distance behind you. That certainly implies stresses no ruler can support! But it also implies that any naive meaning of the "separation" between photons cannot stay fixed, rather the Rindler horizon behind you will act kind of like the event horizon of a black hole, and the distance between the photons will be reckoned to pile up there. Rindler did a whole lot with frames in constant acceleration, sadly I've never spent the necessary time making sense of it all!

Grey
2015-Jun-25, 07:32 PM
It sounded fine to me, though there are problems keeping long rulers rigid in an accelerating frame.Yeah, I think that you'd have to imagine little rocket engines attached all along your ruler, all carefully coordinated in advance to keep the whole thing accelerating together (and it has to be coordinated in advance, because otherwise you run into issues of light travel signal delays in controlling those rockets). Mostly, I just wanted to establish a physical equivalent of the imaginary coordinate system that you always extend to infinity when you're talking about a reference frame, so that we could continue to measure the light pulses even after they're far away. It's often amusing just how tricky it can be to create that physically without running into problems.

Ken G
2015-Jun-25, 07:35 PM
Mostly, I just wanted to establish a physical equivalent of the imaginary coordinate system that you always extend to infinity when you're talking about a reference frame, so that we could continue to measure the light pulses even after they're far away. It's often amusing just how tricky it can be to create that physically without running into problems.Right, that''s also a problem encountered by the curious "Rindler horizon" that appears behind any accelerating object, which apparently does have physical consequences like bathing them in Unruh radiation (which is a whole lot like Hawking radiation), though I don't think that's ever been conclusively confirmed!

Reality Check
2015-Jun-25, 10:01 PM
What do you mean by their separation?
A source emits a series of individual particles with a velocity v at a time interval t. Their separation is then v*t. For example the 1 light-second for photons that has popped up in this thread.
A single observer can detect the particles and measure that separation. This will be v*t (unless we have the Doppler effect).

Reality Check
2015-Jun-25, 10:08 PM
I mean their distance in space or the the time that passes in the single observer's accelerating frame between the observation of the first and the observation of the second.
You mean you move the goal posts from SR to GR :p! The question is meaningless for a single observer in SR and GR.
ETA:
For SR you need to know the velocity between two inertial frames of reference in order to transform between the frames.
For GR you need to know the gravitational potential difference between two accelerating frames of reference in order to transform between the frames.

A source emits a series of individual particles with a velocity v at a time interval t. Their separation is then v*t according to the source. A single observer can detect the particles and measure that separation. We cannot say anything about what that measurement will be without knowing the velocity of the source relative to the observer or the gravitational field of the source relative to the observer. With v = 0 and the same gravitational potential the separation is v*t.
You have mentioned the expansion of the universe and there is similar meaningless without knowing the position of the source relative to the observer.

Reality Check
2015-Jun-25, 10:08 PM
Is that the same as Grey's response in #23?no

Shaula
2015-Jun-26, 12:37 AM
I am trying to get this as precise as possible. You say "Their separation will not change." As measured by whom? For any inertial frame, their separation depends on the inertial frame. For an accelerating frame, their separation changes. The only separation which cannot change is the one which requires using an invalid reference frame.
This is why you need to stop thinking in terms of separation and think in terms of relativistic invariants. Their spacetime interval will not change in any valid frame. This is what you should be considering rather than getting hung up on co-ordinate distances.

Jeff Root
2015-Jun-26, 02:41 AM
mkline55,

I'd be interested to see a summary from you of what aspects
of your question you feel have been adequately answered so
far, and which have not been. I'd like to know how your view
has changed, if it has changed.

-- Jeff, in Minneapolis

Ken G
2015-Jun-26, 02:52 AM
A source emits a series of individual particles with a velocity v at a time interval t. Their separation is then v*t.You have an accelerating observer-- what t are you talking about?

Reality Check
2015-Jun-26, 03:34 AM
I do not have an accelerating observer - mkline55 is the one who as an accelerating (as in GR) observer and I answered that:

A source emits a series of individual particles with a velocity v at a time interval t. Their separation is then v*t according to the source. A single observer can detect the particles and measure that separation. We cannot say anything about what that measurement will be without knowing the velocity of the source relative to the observer or the gravitational field of the source relative to the observer. With v = 0 and the same gravitational potential the separation is v*t.

AFAIK You can also treat an accelerating observer in SR but the response is the same - the question is meaningless because there is not enough information to answer it.

Ken G
2015-Jun-26, 03:44 AM
I do not have an accelerating observer - mkline55 is the one who as an accelerating (as in GR) observer and I answered that:
I was referring to the statement I quoted:

For any single accelerating frame, their separation does not change.
There you have an accelerating observer, but the separation between photons, if interpreted as a spatial distance between them, does indeed change. It changes both in time and with distance from the observer. As Grey was saying, intuitive concepts get very tricky enough in relativity, but they get downright hazardous when we have accelerating observers. It is a shame the full calculations are pretty hard to do, but without them, we risk being incorrect.

Anyway, I think the key question that was causing difficulty from the OP was "For example, a transmitter sends one photon toward a receiver, then sends another one second later. Does the Lorentz transform not apply to the length of an imaginary line drawn between the photons?"

The answer to that does not have anything to do with the fact that there is no rest frame for photons, the answer is the Lorentz transform does apply (at least in the limit), which means the distance the photons would reckon as being between them does get length contracted into whatever we, the sender and receiver, perceive as a finite distance. This means the distance between them in the photon frame is infinite, that's the part that was not being understood because of the misconception that the universe gets squeezed into a plane in the photon frame. Yes there is no actual photon frame, but these answers are easily given their correct meanings by imagining a chain of particles moving at a speed less than c, and watch what happens in the limit as that speed approaches c.

So the OP is actually raising an interesting point-- if you have a beam of photons at regular intervals, the photons would perceive the other photons as being infinitely far away. If you don't like talking about what photons perceive because of the singularities, just talk about regularly spaced bullets travelling near c, and note that the distance between the bullets that they perceive in their own frame gets arbitrarily large as they get arbitarily close to c. We don't usually think about this aspect of highly relativistic travel, since the objects the bullets would pass by would seem highly squashed into pancakes, so it seems everything is about squashing, not stretching. But the distances between the bullets would appear to stretch as we enter the bullet frame, and that stretching gets arbitrarily large if the speed is allowed to be arbitrarily close to c.

Reality Check
2015-Jun-28, 10:54 PM
So the OP is actually raising an interesting point-- if you have a beam of photons at regular intervals, the photons would perceive the other photons as being infinitely far away.
Any observer traveling at c has a problem with measuring anything because they will always measure the proper time between events as zero.
If we switch to the different example of bullets travelling at near c then we have a different situation.
* Alice fires a series of bullets with a certain separation. In Alice's frame, the separation is constant.
* Bob is an accelerating observer who is accelerating along the line of bullets with respect to Alice.
Bob can measure the distance between any two points. Bob will find that distance decreasing with increasing velocity (length contraction). AFAIK it does not matter what points Bob measures - the start and end of one bullet or the end of one bullet and the start of the next bullet. All length measurements will contract. So the length and separation of the bullets as measured by Bob will decrease as he accelerates.

Ken G
2015-Jun-29, 04:50 AM
Any observer traveling at c has a problem with measuring anything because they will always measure the proper time between events as zero.That's why you have to instead take a limit of frames with speeds approaching c. Then there is no difficulty.

If we switch to the different example of bullets travelling at near c then we have a different situation.
* Alice fires a series of bullets with a certain separation. In Alice's frame, the separation is constant.It is all a question of what you hold fixed as you consider higher and higher speeds. What is appropriate to this question is to hold the separation as constant, in Alice's frame, and regard Bob as being in the frame of the bullets, which we may first regard as a constant speed. Simply take Bob's speed, and the speed of the bullets, relative to Alice as being closer and closer to c. Not accelerating, just consider higher and higher speeds. Then ask what Bob sees as the separation between the bullets. It goes to infinity, without bound, as Bob's speed approaches c relative to Alice. That is all we can mean by the "photon frame", it presents no particular difficulties if we regard it as a limit.


* Bob is an accelerating observer who is accelerating along the line of bullets with respect to Alice.If Bob is accelerating, then by the time he reaches the bullet frame, the above answer holds as long as we talk about the bullets at Bob's location, so nothing in particular is gained by letting him accelerate. Generally speaking, it's best to just pick a frame and stick with it, acceleration confuses things.

If we instead take the bullets to be photons moving at c, then we can just consider faster and faster Bob's, approaching c. All the Bob's in this sequence will still regard the photons as getting farther and farther from each other, without bound as Bob approaches c. This is the surprising fact that the OP is asking about-- we normally think of Bob's fast speed as "length contracting" everything he perceives, but by making Bob's speed increase, he is actually reducing his speed relative to the bullets, so the length contraction is lessening as we consider faster and faster Bob's. By the time Bob reaches the putative "photon frame", via this limit, the length contraction has gone away completely, but that requires the photons be infinitely far from each other because they would need to be infinitely length contracted to Alice's perception of finite length. Again, if speaking in terms of these infinities, instead just treat Bob's speed as a variable and take the limit.


Bob can measure the distance between any two points. Bob will find that distance decreasing with increasing velocity (length contraction). No, the opposite is true. That's why the OP question is actually kind of interesting, it gives us a new kind of check on our relativistic intuition.

Reality Check
2015-Jun-29, 10:10 PM
That's why you have to instead take a limit of frames with speeds approaching c.

That is correct. But remember that the limit never gets to c and so you can never treat photons! There is no "photon frame" Photons always travel at c. You have stated that we are taking that limit as v tends to c. That limit never gets to c. You can never "get to infinity".

So what you want is:
* Alice fires a series of bullets at a speed v1
* Bob flies beside the bullets at a speed v1 relative to Alice and measures a separation.
Bob measures a smaller separation than Alice (length contraction).
Then
* Alice fires another series of bullets at a speed v2
* Bob flies beside the bullets at a speed v2 relative to Alice and measures a separation.
Bob measures a smaller separation than Alice (length contraction).
If v2 > v1 then that length contraction is greater. Keep on increasing v and then length contraction increases as v gets closer to (but can never reach) c.

Ken G
2015-Jun-30, 12:39 PM
That is correct. But remember that the limit never gets to c and so you can never treat photons!That is the nature of limits, you never arrive at them, but you can talk about them. So you can talk somewhat metaphorically about the "photon frame", just as you can talk metaphorically about an "inertial frame" (you can't arrive at inertial frames either). Metaphorical language based on some kind of mathematical approximation or limit is not unusual in physics! So I don't have any great problem talking about the "photon frame", though I certainly agree it involves singularities that cannot be taken literally, it is a language of limits. My real point here is not that there is actually a photon frame where the distance between photons is infinite, it is the somewhat surprising fact that if we perceive some chain of equally spaced bullets moving at very high speed, then in the frame of the bullets, those bullets are actually much farther apart than we perceive them in our frame, and that distance between them increases without limit if we fix the spacing of the bullets that we perceive, and imagine their speed is approaching c. This flies in the face of naive notions of "length contraction," because we are forgetting that it is we who are already perceiving that length contraction.


* Alice fires a series of bullets at a speed v1
* Bob flies beside the bullets at a speed v1 relative to Alice and measures a separation.
Bob measures a smaller separation than Alice (length contraction).That's what is incorrect, so that's what I view is the real point of the OP question-- not issues with what it means to talk about a photon frame.

Reality Check
2015-Jun-30, 09:27 PM
That is the nature of limits, you never arrive at them, but you can talk about them.
I hope we agree that we can never talk about v = c in the limit because that violates the definition of a limit.

An observer can never be at rest with respect to a photon even if they are massless and can travel at the speed of light. Thus you cannot have an observer with a photon. That makes a "photon frame" useless in physics.
I am driving my car at the speed of light and I turn on my headlights. What do I see? (http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/headlights.html)

Sometimes people persist: What would the world look like in the reference frame of a photon? What does a photon experience? Does space contract to two dimensions at the speed of light? Does time stop for a photon?. . . It is really not possible to make sense of such questions and any attempt to do so is bound to lead to paradoxes. There are no inertial reference frames in which the photon is at rest so it is hopeless to try to imagine what it would be like in one. Photons do not have experiences. There is no sense in saying that time stops when you go at the speed of light. This is not a failing of the theory of relativity. There are no inconsistencies revealed by these questions. They just don't make sense.

For example let such an observer exist and we see this is not a physically relevant frame because proper time is always zero for that frame (measured velocities are undefined!).

So if you want to talk metaphorically about the "photon frame", go ahead - just do not expect to do any physics with it :D!


That's what is incorrect,...
And your evidence that this is incorrect, Ken G?
I can see no difference between Bob measuring
* The distance from the start of a bullet and the end of that bullet.
* The distance from the end of a bullet and the start of the previous bullet (separation).
They are both length measurements and are both subject to length contraction (https://en.wikipedia.org/wiki/Length_contraction) even if the Wikipedia article only talks about a single object.

ETA: This Physics Forums thread has a post answering the question If two objects start moving near the speed of light at the same speed, is the length between the two objects contracted? (https://www.physicsforums.com/threads/length-contracting-between-objects.783381) (scroll to the bottom). Remember that the Lorentz transformation is a rotation of a world line closer to the light cone. Rotate the world lines for two objects at rest to be in a frame with v > 0 (Bob's frame) and you get a distance between then that is less.

Jeff Root
2015-Jun-30, 11:34 PM
I agree with Ken's analysis that the bullets are farther
apart in their own frame than an observer sees them to be;
that the effect increases without limit with increasing speed;
and that by extension, photons must be infinitely far apart
from each other "in their own frame". The only reason to say
that photons have no reference frame is that the math doesn't
work, but the math *does* work: It gives infinite distance
between photons, and zero distance between things those
photons are passing.

Ken and I had an argument about describing limits a couple
of weeks ago. He thought I didn't understand how it works.
I didn't follow up on that. I agreed with how he said limits
work, but I was just disagreeing with how he described how
they work. And if that sounds confused, I think that is why
Ken thought I didn't understand. I don't like my description
any more than I liked Ken's. But in this case I'm not seeing
anything wrong with how he described it. So I agree with
both his physics and his description of the physics.

-- Jeff, in Minneapolis

Ken G
2015-Jul-01, 01:15 AM
And your evidence that this is incorrect, Ken G?It's just the way relativity works. But you have to keep careful track of the reference frames you are using, and what is being held fixed as you consider different frames.


I can see no difference between Bob measuring
* The distance from the start of a bullet and the end of that bullet.
* The distance from the end of a bullet and the start of the previous bullet (separation).Yes, that's not the problem. It's fine to imagine that there are rigid rods connecting the bullets, that's just fine. My answer is the correct one whether we put rigid rods between the bullets, or not.


ETA: This Physics Forums thread has a post answering the question If two objects start moving near the speed of light at the same speed, is the length between the two objects contracted? (https://www.physicsforums.com/threads/length-contracting-between-objects.783381)That's the wrong question. That question refers to a situation where the distance between the objects is specified in their own frame, and asks if that distance is length contracted in a frame that sees them as moving. Of course the answer to that is "yes", indeed I already used that fact in the explanation I gave just above.

But this situation is quite different. Here the distance is set in the frame where the objects are moving, and asks what that distance is in the frame of the bullets! So that reverses the answer, the distance is then un-contracted, i.e., larger. The degree to which it is un-contracted is unbounded as the speed approaches c, because what is being held fixed is always the distance in the frame of the sender and receiver of the "bullets."

So what I'm saying is, the correct answer to the OP was not given, because the situation got muddied by the red herring of the impossibility of being in the photon frame. That frame is fine if regarded as a limit, the real problem was that the distance is increased in the object frame, not contracted. It is contracted going the other way.

Reality Check
2015-Jul-01, 02:35 AM
I agree with Ken's analysis that the bullets are farther
apart in their own frame than an observer sees them to be;
that the effect increases without limit with increasing speed;
and that by extension, photons must be infinitely far apart
from each other "in their own frame".

We agree with Ken's analysis that the bullets are farther apart in their own frame than an observer (say Bob) sees them to be.
But the effect "increases" toward a limit with increasing speed because this is Bob measuring length contraction. The limit of the separation is zero. By extension if Bob was traveling at the speed of light then the separation would be measured to be 0.
The math does not always work for photons. Say we have a hypothetical observer Charles sitting on a photon with a tiny clock. Charles will always measure the time according to his clock between events to be zero. Charles will know that any speed he calculates will be undefined. The math for speed does not work (produce a result).

Reality Check
2015-Jul-01, 03:00 AM
Yes, that's not the problem. ...derail into rigid rods snipped...

"Yes": So we agree that as Bob increases velocity relative to Alice, Bob measures the separation between bullets to decrease.

If two objects start moving near the speed of light at the same speed, is the length between the two objects contracted? (https://www.physicsforums.com/threads/length-contracting-between-objects.783381) answers the situation that that I proposed: "If Bob increases velocity then what happens to the separation Bob measures?" with what we both know (it decreases).


Here the distance is set in the frame where the objects are moving, and asks what that distance is in the frame of the bullets!
And as you noted easily answered:
* Two bullets exist travelling with a certain separation. Let Alice be an observer moving with the bullets. She measures a proper length of L0 between the bullets.
* Bob is travelling at a velocity v relative to the bullets. Bob measures the separation of the bullets. This is the observer length (the L in length contraction (https://en.wikipedia.org/wiki/Length_contraction)).
If Bob applies the Lorentz transformation to his measurement then what will he calculate? The Lorenz factor goes from 1 to infinity and multiplies the proper length. Bob thus calculates the separation to be L[SUB]0[/SUB and higher.

Ken G
2015-Jul-01, 03:28 AM
"Yes": So we agree that as Bob increases velocity relative to Alice, Bob measures the separation between bullets to decrease.
No, you are doing the wrong problem. Again: Alice sees the bullets as being in a relativistically moving equally spaced chain of given separation, because Alice has always been the observer in the "sender/receiver frame" from the OP question. Read your own post #14 and see what frame Alice is in, it made perfect sense to have her in that frame because that is what is given in the problem: that's where the OP specified the distance, and the OP is wondering what happens to that sender-frame specified distance in the frame of the objects themselves (be they photons, as in a limiting case, or bullets, as in a real case we could actually test and check that we are getting the limiting case correct). Bob is in a frame moving with the bullets, that was the whole point of introducing Bob (you originally called him Charles, but there was never a need to have a Charles and a Bob, so that just got reduced to Bob-- the usual Alice-and-Bob scenario, perhaps that's what created the confusion). The point is, we need someone in the frame of the bullets, so if we call him Bob, then Bob measures the separation between the bullets as larger than Alice. That's what the OP was asking! And the reason for this answer is that Alice is the one who regards the distance between the bullets (or the imaginary rigid rods between the bullets) as length contracted.

This is the whole point of the answer to the OP-- the OP is confused because of a misconception that as soon as we go into a frame moving with respect to Alice, everything has to get length contracted. That's what is not true-- if the moving frame is moving with the objects, as specified in the OP question, then we don't get length contraction, we remove length contraction. The whole issue with the impossibility of a photon frame is a complete red herring to this simple misconnect in intuition about length contraction.

The key is correctly keeping track of what frame the separation is given in, and what frame is moving with the bullets, so we can tell where it is appropriate to talk about length contraction. Don't change the problem into something it isn't, by suddenly claiming that Alice is the one in the bullet frame. The answer to the OP question is that if we replace the photons with very fast bullets, then in the frame of the bullets, they will see the bullets as farther apart than they are reckoned in the sender/receiver frame that is firing the bullets at fixed intervals. If we keep the separation between the bullets fixed in the sender frame, as specified in the OP, and imagine faster and faster versions of this same question, the distance between the bullets in the bullet frame will increase without bound as the bullets approach c in the sender frame. That's the answer.

What's important about it is it puts the lie to the usual picture you hear that photons exist in a kind of pancake-universe where there is no distance along their direction of motion, and no time either. Those are both wrong, as can be seen by looking at a limiting sequence as above. What gets turned into a pancake is anything that we regard as spherical in our reference frame, like galaxy clusters or some such thing. But anything moving with the photons, like rigid rods between them (again regarded in the limiting sequence), do not get length contracted in the photon frame. In fact, if the length of those rods is specified in our frame, they get infinitely long (in the limit) in the "photon frame." There is no problem with this language if understood as a limit. Thus, we should not say that photons live in a universe with no spatial extent along their direction of motion, we should say that they reckon no spatial extent to anything that we regard as stationary, or even anything we regard as not moving with the photons. Things we regard as moving with the photons, and having a finite spatial extent, are experienced as infinitely long to the photons.

We should also not say that photons experience no time because time is "frozen" in the photon frame, that's also the opposite of the correct limiting behavior. The limit of what happens to time is: nothing at all! Time is never affected when you are in the frame where time is being reckoned, that's the core principle of relativity. Nothing happens to this principle in the limit of a "photon frame", it's a perfectly well-behaved limit. So if there "really were" photons, and they "really existed" in an infinite Minkowski space, they "really would" experience time perfectly normally. However, anything that we regard as stationary, or even not moving with the photons, that we regard as happening in a finite time interval, the photons would regard as taking no time to happen. A photon would say that our clocks were out of synchronization by just the amount we regarded as that time interval, and our clocks never tick at all. That's the only degree to which "time stands still" for photons-- if we regard them as created and absorbed in a finite time, then the photons would regard their own life as being infinitely short. But if they are never absorbed in an infinite Minkowski space, they have no trouble living any timespan we can imagine. However, each photon would regard the next photon in the chain as never being emitted, because to a photon, all our clocks are stopped, and never tick at all. They see us as completely frozen in time, they do not see themselves as frozen in time (though they might well have an infinitely short life-- which is something kind of the opposite of being frozen in time, things happen incredibly fast!).

Of course this is all highly idealized-- we don't have an infinite Minkowski space, our model is that of a highly gravitational expanding universe, and it might not even be infinite itself, in which case our model in our frame would indeed be compressed into an infinitely thin pancake in the "photon frame." But not the distance between the photons-- a photon would see that distance as infinite, all our clocks as stopped, and things happening so fast for the photon that they are likely to be born and absorbed all in the span of the same instant. Things happening super fast is the opposite of time being frozen, photons would have a spectacularly frenetic life!

Reality Check
2015-Jul-01, 04:15 AM
No, you are doing the wrong problem.
Again you did not read the actual problem being addressed which is not the OP question
The problem I addressed and you agreed that separation contracts was:
* Alice fires a series of bullets
* Bob flies beside the bullets at a speed v relative to Alice and measures a separation.
Bob measures a smaller separation than Alice would measure (length contraction).

This is my post #14

However mkline55 seems to be asking about a situation like
* Let Alice emit particles at one light second apart.
* Let Bob detect these particles and measure that they are one light second apart.
* Let Charles be an observer travelling with the particles.
What distance between particles does Charles measure?
The answer is one light second apart because the question does not involve any other observer.

Or one interpretation of the OP in terms of bullets:


* Two bullets exist travelling with a certain separation. Let Alice be an observer moving with the bullets. She measures a proper length of L0 between the bullets.
* Bob is travelling at a velocity v relative to the bullets. Bob measures the separation of the bullets. This is the observer length (the L in length contraction (https://en.wikipedia.org/wiki/Length_contraction)).
If Bob applies the Lorentz transformation to his measurement then what will he calculate? The Lorenz factor goes from 1 to infinity and multiplies the proper length. Bob thus calculates the separation to be L[SUB]0[/SUB and higher.

Ken G
2015-Jul-01, 05:10 AM
Again you did not read the actual problem being addressed which is not the OP questionActually, I did read the problem, gave the right answer, and I think it is very much related to the OP question.

The problem I addressed and you agreed that separation contracts was:
* Alice fires a series of bullets
* Bob flies beside the bullets at a speed v relative to Alice and measures a separation.
Bob measures a smaller separation than Alice would measure (length contraction).That's the right problem, but the wrong answer. The answer I gave is the opposite: Bob measures a larger separation than Alice would measure, because Alice is getting the "length contracted" result. This is the key point-- one should not immediately imagine that motion of Bob relative to Alice will produce length contraction for Bob, length contraction is about the motion relative to the objects. Bob is in the object frame, hence Bob does not get any length contraction for those objects-- but Alice does. This is also the answer to the OP question, simply taken to the limit as v approaches c.


Or one interpretation of the OP in terms of bullets:Yes, replacing the photons with bullets is a step in the right direction, but you have to do that problem correctly, and then take the limit as v goes to c. That answers the OP-- the photons "see" an infinite separation between each other, and the reason is that a leading photon thinks a trailing photon is never fired at all, because it thinks all our clocks are stopped and never get to the next tick where the next photon could be fired.

Jeff Root
2015-Jul-01, 06:10 AM
I originally didn't get that the photons would "see" infinite
distance between themselves, but once you showed how
that results by analyzing it as the limiting case, it became
obvious. I kind of disagree with how you described the
photon's view of space and time in post #91, but totally
agree with the physics and the method of analysis.

-- Jeff, in Minneapolis

Grey
2015-Jul-01, 06:19 PM
Sorry, Reality Check, but Ken G is indeed describing the situation correctly. I have one tiny quibble about the introduction of a third observer, which I think was actually important, rather than being an unnecessary complication. Taking a look back at the OP (I've added the emphasis):


1) According to length contraction theory, an observer watching a passing object will observe length contraction in the direction of motion. The amount of contraction can be calculated using the Lorentz transform. I understand that this effect does not apply to individual photons, but does it not apply to the distance between photons? For example, a transmitter sends one photon toward a receiver, then sends another one second later. Does the Lorentz transform not apply to the length of an imaginary line drawn between the photons? To the receiver they are still one light second apart, and are not received simultaneously. Why? Does the answer also apply to other relativistic particles?It seems pretty clear that the two observers mkline55 had in mind initially were the transmitter and the receiver; mkline55 clarifies in later posts that these two are stationary with respect to each other. So if these are our original Alice and Bob, it seems like mkline55 was imagining that, because the photons are moving, Bob should somehow measure the distance between them differently than Alice does. So the first issue was to try to clarify that Alice and Bob will measure all distances the same, because they are stationary with respect to each other. The fact that some of the endpoints of those distances may be moving doesn't affect that in any way, and Alice and Bob would not observe any length contraction at all when comparing their measurements. I think it makes sense to leave these as our Alice and Bob to make that side of things clear, since I think that was mkline55's original misunderstanding.

The third observer (Charles) was then introduced moving along with the objects Alice is firing (and those were changed from photons to some physical object so that Charles could actually be moving along with them), so that we could talk about an observer in this scenario who would measure different distances, and discuss how those distances would be different from what Alice and Bob would both measure.

And it's true that Charles will measure a longer distance, and that the reason is because the question is phrased in the reverse of most relativistic story problems. Usually, we're given some object (a spaceship, say), and told how long that object is in its own rest frame, and then we try to work out how long it will appear to be in some frame moving relative to that. In that case, it always appears shorter. But in this scenario, we have the length specified in a frame moving relative to the object (the "object" in this case is the two bullets, possibly with a rigid rod suspended between them, or possibly just an imaginary ruler between them; the result doesn't depend on whether there's a physical rod there or not), and we have to figure out how long the object will be in its own rest frame. That's just the reverse of the typical calculation, and so we find that Charles will measure the distance to be longer than Alice does.

Reality Check, if you're still having a hard time seeing it, think about a parallel situation. Alice takes a spaceship that has some specific length (but I'm not telling what). She starts accelerating it to relativistic speed, and as she does so, she notices that the length she measures for the ship becomes shorter as it moves faster. Eventually, when it's travelling at 90% of the speed of light, she notices that she measures it as being exactly 100 meters long, and decides that's good enough. How long was Alice's original measurement of the ship before it started moving? And how long will it appear to Charles, who's been on the spaceship the whole time? Will he measure any difference in length once the ship is coasting at full speed, as compared with the length before it started accelerating?

Ken G
2015-Jul-01, 07:29 PM
It seems pretty clear that the two observers mkline55 had in mind initially were the transmitter and the receiver; mkline55 clarifies in later posts that these two are stationary with respect to each other. So if these are our original Alice and Bob, it seems like mkline55 was imagining that, because the photons are moving, Bob should somehow measure the distance between them differently than Alice does.Yes, I agree the question does reveal a confusion about that. That confusion then got superceded when it was not clear that the receiver was in the same frame as the sender, so it made sense to talk about Bob as the receiver, but when it became clear that Bob and Alice were in the same frame, it seemed we only needed Bob for the frame of the moving particles. But you're right-- since there was a confusion about the receiver at first, I should have just left it Alice and Bob in the same frame, seeing all the same things, and Charles in the "photon frame", seeing the infinite distance between photons. That should answer everything in the OP, and is a bit counterintuitive at first as well!

Interestingly, it seems the answer to the OP is simply that the Lorentz transform is only used to change reference frames, so since the receiver is in the same frame as the sender, the Lorentz transform simply doesn't come up at all, and none of its attributes are relevant. But since the confusion in the question involved transforming (unnecessarily) to the photon frame, and that by itself stimulated some wrong answers that had to be navigated, it ended up being a thread with an interesting lesson all the same!

Reality Check
2015-Jul-01, 09:48 PM
That's the right problem, but the wrong answer. ...

Then your answer is wrong, Ken G. The key point is the information in the Alice & Bob example (http://cosmoquest.org/forum/showthread.php?p=2297836#post2297836) and SR.
SR states that all observers moving with a speed that is not zero relative to another observer measure lengths that are contracted. No observer ever measures that an object's length expands. No observer moving with the bullets means that we cannot say anything about the length contraction relative to the bullets. The Wikipedia the Alice & Bob example (https://en.wikipedia.org/wiki/Length_contraction]length contraction[/URL] article has an implicit observer moving with the object to measure the proper length. Even if we ignore that implicit observer we still have no knowledge of what the proper length is between the bullets because there is no one measuring it in [URL="http://cosmoquest.org/forum/showthread.php?p=2297836#post2297836).
We do not know Alice's speed relative to the bullets so we can never say what Alice's measurement contraction is relative to the bullets.
We do not know Bob's speed relative to the bullets so we can never say what the Bob's measurement contraction is relative to the bullets.

Bob is an observer moving with a speed that is not zero relative to Alice. Thus Bob measures lengths that are contracted relative to Alice. If Alice measures a separation of x then Bob measures a separation y < x. That is all was can say with the information in the Alice & Bob example (http://cosmoquest.org/forum/showthread.php?p=2297836#post2297836).
If we take the limit as of Bob's speed relative to Alice going toward c then Bob's s measurements of the separation relative to Alice are more and more contracted, i.e. tend to zero. Photons never "see" an infinite separation between each other because they "see" no separation between each other. But the question is meaningless since proper length is undefined - read the definition of length contraction and There are no inertial reference frames in which the photon is at rest so it is hopeless to try to imagine what it would be like in one (http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/headlights.html). Remember that it is a postulate of SR that the speed of light in vacuum is c in every inertial reference frame.

If we know Alice's speed relative to the bullets then we say something about the different scales of contraction relative to the bullets, e.g.
Alice has a speed vA relative to the bullets.
Bob has a speed vB relative to the bullets (e.g. add Bob's speed relative to Alice to her speed relative to the bullets).
If vA = vB than Alice and Bob measure the same length contraction.
If vA < vB than Alice measures less contraction relative to the bullets than Bob.
If vA > vB than Alice measures more contraction relative to the bullets than Bob.

Grey
2015-Jul-02, 12:04 AM
If we know Alice's speed relative to the bullets then we say something about the different scales of contraction relative to the bullets, e.g.
Alice has a speed vA relative to the bullets.
Bob has a speed vB relative to the bullets (e.g. add Bob's speed relative to Alice to her speed relative to the bullets).
If vA = vB than Alice and Bob measure the same length contraction.
If vA < vB than Alice measures less contraction relative to the bullets than Bob.
If vA > vB than Alice measures more contraction relative to the bullets than Bob.You're almost there. Since Alice is firing the bullets, we can assume that vA is large (obviously, unless these are relativistic bullets, we won't notice any of this, so vA is close to the speed of light). And since we've said that Bob is travelling along with the bullets, vB must be zero. So, as you've said, with vA > vB (much larger, actually), Alice measures more contraction between the bullets than Bob does. Therefore, Alice will measure a shorter distance than Bob, and conversely, Bob will measure a larger distance than Alice.

Grey
2015-Jul-02, 12:11 AM
Interestingly, it seems the answer to the OP is simply that the Lorentz transform is only used to change reference frames, so since the receiver is in the same frame as the sender, the Lorentz transform simply doesn't come up at all, and none of its attributes are relevant. But since the confusion in the question involved transforming (unnecessarily) to the photon frame, and that by itself stimulated some wrong answers that had to be navigated, it ended up being a thread with an interesting lesson all the same!I agree!

Reality Check
2015-Jul-02, 09:34 PM
...Therefore, Alice will measure a shorter distance than Bob, and conversely, Bob will measure a larger distance than Alice
I am "totally there" :) because in my example Alice is not assumed to have fired the bullets and Bob is not assumed to be traveling with the bullets.

You are right, Grey: Add the assumptions that Alice fires the bullets and that Bob is travelling with the bullets and that is the vA > vB case in my post (http://cosmoquest.org/forum/showthread.php?157384-Photon-races&p=2297947#post2297947) with vB = 0 and vA > 0 (relativistic if we want to look at the limit of vA tending to c).

Grey
2015-Jul-02, 09:57 PM
I am "totally there" :) because in my example Alice is not assumed to have fired the bullets and Bob is not assumed to be traveling with the bullets.Well, I'm a little perplexed, because looking back at the thread, you're actually the one who introduced Alice and Bob by name (http://cosmoquest.org/forum/showthread.php?157384-Photon-races&p=2295522#post2295522), and you explicitly stated that Alice was the one firing particles, whatever they may be. Since then, I thought we'd all been assuming that Alice is the one firing particles, and I'm not sure when you decided that she wasn't. There has been some confusion about Bob and Charles. In your post linked above, you actually had Bob as the one receiving the particles, and therefore seemingly stationary with respect to Alice, and introduced a third observer, Charles, who was moving with the particles. Now it seems like you've gone to just Alice and Bob moving with respect to each other (that confusion of which observers were which was exactly why I suggested here (http://cosmoquest.org/forum/showthread.php?157384-Photon-races&p=2297922#post2297922) that we should leave the observers as they were originally named, rather than shifting around the labels; that only adds to the confusion). So maybe this has just been a misunderstanding about which observer was which.


You are right, Grey: Add the assumptions that Alice fires the bullets and that Bob is travelling with the bullets and that is the vA > vB case in my post (http://cosmoquest.org/forum/showthread.php?157384-Photon-races&p=2297947#post2297947) with vB = 0 and vA > 0 (relativistic if we want to look at the limit of vA tending to c).Cool, so we're agreed that if Alice fires some relativistic bullets, and a second observer (maybe his name is Bob, maybe his names is Charles ;)) happens to passing by at exactly the same speed and direction as those bullets, then Alice will measure the distance between those bullets to be shorter than Bob (or Charles), and conversely, Bob (or Charles!) will measure the distance to be greater than Alice does?

Reality Check
2015-Jul-02, 10:37 PM
Well, I'm a little perplexed,..

That was another example. In the current example
* We do not know Alice's speed relative to the bullets so we can never say what Alice's measurement contraction is relative to the bullets (http://cosmoquest.org/forum/showthread.php?157384-Photon-races&p=2297947#post2297947).
* Bob measures lengths (need not be not hit by the bullets!), e.g. maybe by bouncing a laser off the bullets.
* Bob can have any speed and thus need not be stationary with respect to Alice or the bullets.
* There is no Charles.

And yes - we agree on the basics of SR, Grey :D!
The length contraction that observers measure gets greater with their speed relative to the object being measured. So an Alice with v > 0 relative to the bullets will measure greater contraction than a Bob with v = 0 relative to the bullets and vice versa.

Grey
2015-Jul-02, 10:56 PM
That was another example. In the current example
* We do not know Alice's speed relative to the bullets so we can never say what Alice's measurement contraction is relative to the bullets (http://cosmoquest.org/forum/showthread.php?157384-Photon-races&p=2297947#post2297947). I can't speak for everyone else in the thread, but I'm pretty sure that nobody but you was thinking that way. For me, and I think everyone else, we all assumed that Alice was the one firing the particles, and that therefore they were moving at high speed relative to her, and that a second observer (with some confusion about the name) was moving relative to the bullets. And in fact, even in your post just prior (http://cosmoquest.org/forum/showthread.php?157384-Photon-races&p=2297838#post2297838) to the one you linked here, you said


Again you did not read the actual problem being addressed which is not the OP question
The problem I addressed and you agreed that separation contracts was:
* Alice fires a series of bullets
* Bob flies beside the bullets at a speed v relative to Alice and measures a separation.
Bob measures a smaller separation than Alice would measure (length contraction).So clearly here, you were thinking that Alice was the one firing the bullets, that Bob was flying along beside the bullets, and you seemed convinced that Bob would measure a smaller separation than Alice. And, as we've now agreed, that's not true.


And yes - we agree on the basics of SR, Grey :D!
The length contraction that observers measure gets greater with their speed relative to the object being measured. So an Alice with v > 0 relative to the bullets will measure greater contraction than a Bob with v = 0 relative to the bullets and vice versa.Well, I'm happy to hear you say that, but it sounded quite a bit like you were saying just the opposite yesterday, which is why I was trying to make sure I was quite clear on what you're saying now. ;)

Reality Check
2015-Jul-03, 03:34 AM
I can't speak for everyone else in the thread, ...

I can only speak for what I wrote in my post containing the example which I thought clearly did not have Alice firing the bullets or Bob moving with the bullets. Maybe I should have swapped to Charles and Diana to make the point clear :D. The generalization at the end though allows for both assumptions.

Ken G
2015-Jul-03, 03:45 AM
Right, but the problem was the incorrect answer right from the start, when you introduced Alice in post #14. If we go back to the conventions of that post, which are good conventions, then we have:

However mkline55 seems to be asking about a situation like
* Let Alice emit particles at one light second apart.
* Let Bob detect these particles and measure that they are one light second apart.
* Let Charles be an observer travelling with the particles.
What distance between particles does Charles measure?
The answer is one light second apart because the question does not involve any other observer.
So that's what's wrong, Charles does not measure the particles as being one light second apart, he measures their separation to be much larger than that, and indeed infinitely large in the limit as the particles are getting close to being photons. This became the issue of interest in the thread, because the answer to the OP question is much simpler-- there's no Lorentz contraction when you consider only Alice and Bob, and they are in the same frame. But there is when you consider Charles, only it does not make Charles think the distances are length contracted (or the same, as it sounds like you were implying in post #14), it makes Alice think the lengths are length contracted, and since Alice's lengths are the ones being stipulated by the sender/receiver setup, it means Charles is getting a kind of length expansion.

That's what's interesting about this problem, because usually we just have one object and we think of it in its own frame, so length expansion isn't a useful notion compared to length contraction. But when it is the space between a chain of objects, it is quite normal to talk about that space in a frame where the objects are moving, and that's when the concept of "length expansion" does indeed become relevant, though it's more of a removal of length contraction. We rarely see language like that, but it would look like saying "since Charles does not see the spaces between the particles as being length contracted, he gets a larger separation than Alice does." Contrast this with the kind of common expectation we often see that "if Charles is moving relative to Alice, everything he sees will be length contracted relative to Alice." This thread shows us that this statement is false, and the place where it completely breaks down is in regard to objects moving with Charles-- like the photons of the OP.

Swift
2015-Jul-03, 03:50 AM
This is obviously an extended discussion, not a simple Q&A, so moved from Q&A to Astronomy

Ken G
2015-Jul-03, 12:46 PM
Another interesting point is that if we introduce a fourth observer, Daphne, who has arbitrary motion relative to Alice so is not necessarily in the frame of the bullets, then Daphne might see a larger separation between them, or a smaller one. But Daphne will always reckon more time as elapsing between the firing of the bullets than Alice does, given that Alice is firing them. Not the time interval between the passing of the bullets at Daphne's location, that could again be shorter or longer than Alice gets, but the time interval between the firing of the bullets, that's what Daphne will always reckon as longer. So we can say that Daphne's time will always be a dilated version of Alice's time for "things happening to Alice," like firing the bullets, but we cannot say that Daphne's lengths will always be a contracted version of Alice's lengths, that's only guaranteed for things that are stationary for Alice-- only then would we have the analog of a "length that is happening to Alice." So what I mean is, if one wants to interpret length contraction and time dilation as "real physical effects on the system", one must then also regard the "reality" of the distance between the bullets as being a happening that is defined in the bullet frame, whereas the "reality" of the time between firing the bullets as being a happening that is defined in Alice's frame.

Reality Check
2015-Jul-05, 11:54 PM
So that's what's wrong.
Alice fired the particles. Charles will have a larger speed than Alice as he is travelling with the particles. Charles will measure a length contraction relative to Alice (and Alice will measure a length contraction relative to Charles). Thus Charles will measure a smaller separation relative to Alice that will go to zero as his speed relative to Alice tends to c.

This is quite basic SR: all length measurements contract toward 0 for an observer with a speed v relative to another observer as that speed v tends to c.

So there is nothing much interesting except an assertion that measured lengths such as separations go to infinity as v goes to c. There is no "length expansion" in SR. You seem to think that SR somehow knows that an observer is measuring a length that is a separation of 2 objects rather than a length of an object :D. If you have a source backing this up then I would like to see it.

Ken G
2015-Jul-06, 01:40 AM
Alice fired the particles. Charles will have a larger speed than Alice as he is travelling with the particles.I know, that's why I had to correct the situation as it was being described above.


Charles will measure a length contraction relative to Alice (and Alice will measure a length contraction relative to Charles).No, that's what's wrong. Length contraction is not a global attribute of a reference frame, it is an attribute of the motion of objects within a reference frame. One should not say that "Charles will measure a length contraction relative to Alice", but one can say "Charles will measure a length contraction relative to Alice of objects moving with Alice." For other objects, we will need more information to say anything about the length contraction that Charles may or may not observe "relative to Alice."
There is no "length expansion" in SR. I explained exactly what I meant by that, you can just look at my posts above. I stand by the accuracy of all of it.

Reality Check
2015-Jul-06, 03:24 AM
No, that's what's wrong. Length contraction is not a global attribute of a reference frame, it is an attribute of the motion of objects within a reference frame.

That is doubly wrong, Ken G.
1. I never say that length contraction is "a global attribute of a reference frame" because I know basic SR where
2. Length contraction (https://en.wikipedia.org/wiki/Length_contraction) is an attribute of the measurements of length between reference frames. The Wikipedia article has the implicit frame of the object and the explicit frame of the observer.

Charles will measure a length contraction relative to Alice (and Alice will measure a length contraction relative to Charles):
* They are moving at a speed relative of v relative to each other.
* Alice has a speed v relative to Charles.
* Charles has the same speed v relative to Alice.
* Relative to each other they will measure the same length contraction because the contraction is symmetrical: https://en.wikipedia.org/wiki/Length_contraction#Symmetry

Relative to the particles they will measure different length contractions (not so far imaginary expansions) because Charles has v (relative to the particles) =0 and Alice has v (relative to the particles) > 0.

Repeating an unsupported assertion about "length expansion" in SR does not support that assertion :eek:!
6 July 2015 Ken G: Please scientific sources stating that in SR that two observers will measure "length expansion" that will tend to infinity as their relative speed v tends to c.

Ken G
2015-Jul-06, 04:30 AM
I'm not sure why you'd want to argue this point. You have indeed made several flatly false claims in this thread, and then you come back and deny it. But did you not say:

However mkline55 seems to be asking about a situation like
* Let Alice emit particles at one light second apart.
* Let Bob detect these particles and measure that they are one light second apart.
* Let Charles be an observer travelling with the particles.
What distance between particles does Charles measure?
The answer is one light second apart because the question does not involve any other observer.
That's wrong, Charles certainly does not measure the particles to be one light second apart, he measures a much larger separation, indeed there is no bound to the separation he would measure as his speed, and that of the particles he is moving with, approaches c, relative to Alice-- as in the construction of the scenario.

Did you not also say:

* Alice fires a series of bullets with a certain separation. In Alice's frame, the separation is constant.
* Bob is an accelerating observer who is accelerating along the line of bullets with respect to Alice.
Bob can measure the distance between any two points. Bob will find that distance decreasing with increasing velocity (length contraction).
Well that just isn't right, I felt it necessary to correct it because otherwise it would be assumed that the forum was agreeing with these statements. And you still don't seem to understand the point here, because even now you say:

Charles will measure a length contraction relative to Alice (and Alice will measure a length contraction relative to Charles):
* They are moving at a speed relative of v relative to each other.
* Alice has a speed v relative to Charles.
* Charles has the same speed v relative to Alice.
* Relative to each other they will measure the same length contraction because the contraction is symmetrical: https://en.wikipedia.org/wiki/Length...ction#Symmetry
What you are not seeing is that this symmetry only applies if Charles is talking about the length of objects moving with Alice, and Alice is talking about the length of objects moving with Charles. It does not apply if Charles and Alice are talking about the same objects, but that is the situation we have in this thread. The situation is certainly not symmetric if Alice is shooting out particles, and Charles is moving with them-- in that situation, Alice will interpret a length-contracted distance between the objects, and Charles will not. Or, if one inteprets Alice as being authoritative about the distance between the particles, on the grounds that she is the one firing them, then we must assert that Charles detects the length between them as either un-contracted, or expanded, depending on how authoritative we consider Alice's stance to be. But the one thing that we certainly cannot say is that both Alice and Charles will interpret the length between the particles as being contracted in a symmetric way. That would be just plain wrong, and I certainly do not need a source on relativity to show that.

mkline55
2015-Jul-06, 04:05 PM
Since I started this, I should give some feedback. Thanks to all who provided input to the discussion. I was actually satisfied with the answers to the first question at post #22.

I was unable to rejoin the discussion for the past week and was surprised to see it still active. After quickly reading through most of the chain, I kept missing something. Did anyone mention that the non-photon relativistic particles that were separated by 1 light second in the originator's reference frames would see that originator as very, very flat? I'm not sure how the particles would measure the distance between themselves in their own frame - meaning by what ruler, but the rest of the universe through which they are passing (except for other particles moving in the same direction with similar velocity) should appear to be very much length contracted. So, while the originator might say a distant galaxy is a billion light years away, in the frame of the highly-relativistic particles they are just a few nanometers away according to their ruler.

Ken G
2015-Jul-06, 05:12 PM
Since I started this, I should give some feedback. Thanks to all who provided input to the discussion. I was actually satisfied with the answers to the first question at post #22. Then you should have kept reading, because that answer was wrong! It is actually just the opposite of that, if the vt in the frame of Alice, shooting the particles, is 1 light second, then in Shaula's example, the distance between the particles in their own frame will be 70 light seconds, not 1/70 of a light second. So you actually asked a pretty good question, it can fool people because we are not used to framing the question of length contraction that way. Showing that has been the entire purpose of my posts to this thread, to demonstrate why that answer is wrong-- it comes down to a mistaken idea that simply going into a moving frame "length contracts everything." That's just wrong, length contraction is not a relationship between any two reference frames, it is a relationship between a reference frame and a specific object (or the specific frame moving with that object). Thus length contraction is a one-way relationship, and does not exhibit the symmetries of the Lorentz transformation. Of course it comes from the Lorentz transformation, but it is not all the Lorentz transformation, it is the Lorentz transformation in a specific application that does not apply in every situation. In particular, it does not apply for the person moving with the particles, there is only length contraction for Alice not Charles.

I'm not sure how the particles would measure the distance between themselves in their own frame - meaning by what ruler, but the rest of the universe through which they are passing (except for other particles moving in the same direction with similar velocity) should appear to be very much length contracted. Yes, Alice would appear very length contracted (to 1/70 of her normal width), but the distance between the particles would appear expanded relative to how Alice sees them-- Alice would see 1 light second, the particles themselves would reckon their separation as 70 light seconds. So notice we encounter both the number 1/70 of a distance, and 70 times a distance, in this same problem, the way you set it up. That's the part that was not correctly answered originally, and is what makes it a better question than was appreciated.


So, while the originator might say a distant galaxy is a billion light years away, in the frame of the highly-relativistic particles they are just a few nanometers away according to their ruler.Yes.

mkline55
2015-Jul-06, 05:59 PM
With the speed of 0.9999 c as measured by Alice, and 1 second apart as measured in her reference frame, and traveling in the same direction, how much time in the frame of the particles does it take a photon emitted by the first particle to reach the second particle?

Ken G
2015-Jul-06, 07:24 PM
Using the Einstein simultaneity convention, that time will be the separation over c. The Lorentz factor for .9999 c is 70.7. The separation can be found by removing the length contraction reckoned by Alice from her measurement of 1 light second, yielding 70.7 light seconds as the separation in the frame of the particles. Hence, the time you asked is 70.7 seconds. That's why I said that in the limit as the speed goes to c and the particles become photons, the separation they perceive (if photons could perceive) between the particles goes infinite.

Jeff Root
2015-Jul-06, 07:51 PM
Reality Check,

I think that if you carefully go through Ken's arguement from
the beginning, and compare it to your own, you will finally see
how you screwed up. An observer moving with the high-speed
particles sees them as farther apart than does an observer who
is not moving with them. That's the expansion Ken refers to.
He made it clear from the start, and it looks to me to be a
completely appropriate way to describe what happens.

-- Jeff, in Minneapolis

mkline55
2015-Jul-06, 08:16 PM
It seems odd that it's called 'contraction' when you show it as 'expansion'. Opposites sides of the same coin, I expect. One might expect the particles to measure their own separation as 1/70 of what Alice measures it to be, but it appears that the reverse is true. I expect then that the particles also measure two objects which are stationary relative to Alice to be 70 times farther apart in Alice's frame of reference than in the particles' frame of reference. Hence, the appearance from the particle frame of reference that Alice is very flat, or length contracted.

Reality Check
2015-Jul-06, 09:24 PM
It seems odd that it's called 'contraction' when you show it as 'expansion'.
The real oddity is that Ken G and Jeff Root seem to share the same unsupported point of view
* the length contraction only applies to physical objects, not the separation between objects.
* There is an unnamed law of physics that makes the separation between objects as measured by an observer traveling at a speed v relative to the objects go to infinity as v goes to c.
6 July 2015 Ken G: Please scientific sources stating that in SR that two observers will measure "length expansion" that will tend to infinity as their relative speed v tends to c. (http://cosmoquest.org/forum/showthread.php?157384-Photon-races&p=2298589#post2298589)

I have the support of basic SR - length contraction (https://en.wikipedia.org/wiki/Length_contraction) apples to all measurements taken by an observer in one inertial frame of reference moving relative to an object or series of objects. An observer could even measure a part of a bullet and get length contraction!

Reality Check
2015-Jul-06, 09:38 PM
... It does not apply if Charles and Alice are talking about the same objects, but that is the situation we have in this thread.

The situation is certainly symmetric if Alice is shooting out particles, and Charles is moving with them. They have a velocity v relative to each other and that is all that appears in Length contraction (https://en.wikipedia.org/wiki/Length_contraction).
Alice will measure (not "interpret") a length contracted by gamma relative to Charles.
Charles will measure a length contracted by gamma relative to Alice.
A fundamental point in SR is that there are no "authoritative" observers eek:!

Also: Charles will measure the proper length between the bullets "relative to himself". That length will be longer than Alice's measurements relative to Charles.

P.S. 6 July 2015 Ken G: Please scientific sources stating that in SR that two observers will measure "length expansion" that will tend to infinity as their relative speed v tends to c. (http://cosmoquest.org/forum/showthread.php?157384-Photon-races&p=2298589#post2298589)

Jeff Root
2015-Jul-06, 09:55 PM
Reality Check,

I repeat: You need to re-read Ken's argument.

Both of your asterisked statements are completely wrong.
The statement in your request to Ken is equally wrong.

Aside from a minor typo, your final statements are correct.
Those statements do not conflict with what Ken and I say.
The entire problem is that you do not understand what Ken
and I are saying. We have both said it very clearly. You just
need to read it more carefully. Take 20 minutes to read it
instead of 20 seconds.

-- Jeff, in Minneapolis

Reality Check
2015-Jul-06, 10:00 PM
An observer moving with the high-speed
particles sees them as farther apart than does an observer who
is not moving with them.
Jeff Root,
I know that basic SR states that an observer moving with the high-speed particles measures them as having a certain separation. As you know that is not expansion - that is the proper length of the separation between the particles (the length measured by an observer moving with the particles). It is a constant. It does not vary. It does not "expand".

An observer moving with a different speed from that observer will measure a contracted length. Where Ken G goes wrong is claiming that this contracted length goes to infinity as v goes to c. The Lorentz factor increases as v goes to c. The measured length between particles goes to zero. That is not expansion, that is contraction :eek:!
But he can easily show that I am wrong by answering 6 July 2015 Ken G: Please scientific sources stating that in SR that two observers will measure "length expansion" that will tend to infinity as their relative speed v tends to c. (http://cosmoquest.org/forum/showthread.php?157384-Photon-races&p=2298589#post2298589)
or even just doing an explicit example, e.g.
Observer 1: Moving with the bullets and measures a "proper separation" S1.
Observer 2: Moving with a speed v2 relative to the bullets and measures a separation S2.
Observer 3: Moving with a speed v3 that is > v[SUB]2[/SUB relative to the bullets and measures a separation S3.
My claim is that length contraction means that S3 < S2 < S1. Easily verified
* Let S1 = 1 meter.
* Let v2 = 0.6c. This has a Lorentz factor of 1.25. Thus S2 = 1/1.25 = 0.8 meters.
* Let v3 = 0.866c. This has a Lorentz factor of 2.000. Thus S3 = 1/2 = 0.5 meters.
Do you agree Jeff Root? If not then where is the error?

Reality Check
2015-Jul-06, 10:05 PM
Reality Check,

I repeat: You need to re-read Ken's argument.

Jeff Root,

I have re-read Ken G's unsupported assertion many times and it violates SR every time that I read it. Lengths contract. Length contraction increases with increasing v. Every observer not moving with the bullets measures the separation to contract. That separation contraction increases with increasing v. So unless Ken G means contraction by "expansion" he is wrong!

ETA: Let us say we ignore that length contraction applies to all measured lengths. Let us say that only the physical length of the bullets contact. Then we could claim that the separation increases with increasing v. But it does not increase to infinity! It increases to a limit of the separation of the bullets + the proper length of the bullets (a finite amount).

Ken G
2015-Jul-06, 10:38 PM
It seems odd that it's called 'contraction' when you show it as 'expansion'. The question is all about using the terminology correctly. Length contraction is a term applied to how we regard the length of an object when we go from the frame of the object to any other frame. Using that phrase in any other situation would be an incorrect use of the term. In particular, it would certainly be incorrect usage when going from Alice's frame to Charles frame, since then we are going from not the frame of the object to the frame of the object! In such a situation, a far more correct use of language would be length un-contraction, or if you prefer, expansion. However, note that the "expansion" is simply undoing the length contraction perceived by Alice, as I said.


Opposites sides of the same coin, I expect.Indeed, but the correct answer about the perceived lengths is still the correct answer, whatever words we choose to describe it.

One might expect the particles to measure their own separation as 1/70 of what Alice measures it to be, but it appears that the reverse is true.Precisely, that's why your question is an interesting one.

I expect then that the particles also measure two objects which are stationary relative to Alice to be 70 times farther apart in Alice's frame of reference than in the particles' frame of reference.Yes.

Hence, the appearance from the particle frame of reference that Alice is very flat, or length contracted.Exactly. There is some value in regarding Alice as being authoritative about her "own width", in which case we can say the particles see her as "length contracted", but again this is only in the special case of transforming out of the frame in which Alice is stationary. Using the concept in any other situation can lead to the errors we saw above.

Ken G
2015-Jul-06, 10:52 PM
I have re-read Ken G's unsupported assertion many times and it violates SR every time that I read it. Lengths contract.I really don't understand why you think "lengths contract" is a useful claim, when missing all the necessary qualifiers to make it actually correct. I provided those qualifiers above, and showed how incorrect answers given above were led astray by the idea that length contraction just automatically occurs any time you go to a frame moving relative to the original one. Indeed, I don't need to show that, because you have by now said it yourself:

Every observer not moving with the bullets measures the separation to contract. That's what I said so many posts ago, to support the fact that the observers moving with the bullets will perceive the largest separation between them, in stark contrast with your claims that I quoted above. What's more, if the observer (Charles) moving with the bullets sees the largest separation, than that separation is expanded relative to the given length of 1 light second, which was the constraint imposed by Alice when she decided how long to wait between firing the bullets.


Observer 1: Moving with the bullets and measures a "proper separation" S1.
Observer 2: Moving with a speed v2 relative to the bullets and measures a separation S2.
Observer 3: Moving with a speed v3 that is > v[SUB]2[/SUB relative to the bullets and measures a separation S3.
My claim is that length contraction means that S3 < S2 < S1. Easily verified
* Let S1 = 1 meter.That's all obvious, yet it seems like you conveniently forgot everything that we have been talking about. The problem is, S1 is not the appropriate reference length that is being controlled, here 1 light second (read the problem above). Again: I said that if we set the distances to be 1 light second in Alice's frame, as has always been the case in this problem, then the distances in Charles' frame will be larger than 1 light second. Hence, S1 is larger than 1 light second. Indeed, in the limit as the bullets become photons, the length increases without bound, just as I said. You are just not keeping track of the given meaning of S1, and hence missing what is interesting in this question-- and how badly the concept of "length contraction" can mislead us if we are too cavalier about how we apply the notion-- like claiming "lengths contract" without a clear understanding of the necessary qualifiers.

Jeff Root
2015-Jul-06, 11:02 PM
I know that basic SR states that an observer moving with the
high-speed particles measures them as having a certain separation.
As you know that is not expansion - that is the proper length of
the separation between the particles (the length measured by an
observer moving with the particles). It is a constant. It does not
vary. It does not "expand".
We do not disagree on those points.



An observer moving with a different speed from that observer will
measure a contracted length.
And we do not disagree on that point.



Where Ken G goes wrong is claiming that this contracted
length goes to infinity as v goes to c.
What Ken said is correct. In my own words:

When the separation as seen by an observer not moving with
the particles is held constant, and the speed v of the particles
relative to that observer goes to c, the separation as seen by
an observer moving with the particles goes to infinity.



The Lorentz factor increases as v goes to c. The measured length
between particles goes to zero. That is not expansion, that is
contraction :eek:!
And that is what is measured by an observer not moving with
the particles, which are not maintained at a constant separation
relative to that observer.

You are railing against something neither Ken nor I ever said.



But he can easily show that I am wrong by answering ...

Do you agree Jeff Root? If not then where is the error?
The error is in your understanding of what Ken and I said.

-- Jeff, in Minneapolis

Reality Check
2015-Jul-06, 11:19 PM
I'm afraid you are only digging yourself deeper. ...

Ken G: Do the following. Read the definition of length contraction. Read that it includes speed not velocity. Read that it applies to any observes moving with a speed relative to other observers - not only the ones moving faster :eek:. Charles happened to be moving slower than Alice and so will measure a length that is contracted relative to Alice's measurement.

There is one not measured length that will increase with increasing speed. If Alice calculated the proper length (https://en.wikipedia.org/wiki/Proper_length) between the particles then she will calculate that the proper length increases as her speed increases. On the other hand, if Alice calculated the proper distance (https://en.wikipedia.org/wiki/Proper_length) between the particles then she would get a constant value.

Ken G
2015-Jul-06, 11:22 PM
I don't need to read anything, other than the problem being solved. Yup, it's just what I thought, see your own post #14 please. I streamlined my point in post #124 above, to get to just the key issues. Read it again, it spells out the whole situation, and how conveniently you have changed the meaning of S1 to try and change history in this thread.
Read that it applies to any observes moving with a speed relative to other observers - not only the ones moving faster . Charles happened to be moving slower than Alice and so will measure a length that is contracted relative to Alice's measurement.You have just given an argument that correctly concludes that S1 is the largest separation, and a simple rereading of the thread shows that it is the one Charles gets. So you should see the ramifications of that, and how it flatly contradicts what you just said. I suppose we can always make our own answer correct if we simply change the meaning of every word we used. But the words are really clear: Alice fires the bullets a time t apart, moving at speed v, producing a separation in her frame of vt. No reinterpretation of history is going to change the simple fact that the constraint is being applied in Alice's frame, and interpreted in Charles' frame. Yes you can get any answer you like if you feel free to change all that.

Reality Check
2015-Jul-06, 11:42 PM
In my own words:
When the separation as seen by an observer not moving with
the particles is held constant, and the speed v of the particles
relative to that observer goes to c, the separation as seen by
an observer moving with the particles goes to infinity.

Your own words are wrong, Jeff Root.
For this case when the separation as measured by an observer not moving with the particles is held to be constant then the separation that they calculate for the particles rest frame between the particles goes to infinity as their speed relative to the particles goes to c. But that is not the "separation as seen by an observer moving with the particles". The separation as measured by any observer moving with the particles is constant for that observer (proper length (https://en.wikipedia.org/wiki/Proper_length)).

ETA: That word constant first appears in this thread in your post.

Reality Check
2015-Jul-06, 11:44 PM
I don't need to read anything, other than the problem being solved.....
S1 is largest, S2 is smaller, S3 is smaller still. And SN goes to zero as v goes to c. So what?
That is basic SR - length contraction increases with speed.

Jeff Root
2015-Jul-07, 12:33 AM
In my own words:

When the separation as seen by an observer not moving with
the particles is held constant, and the speed v of the particles
relative to that observer goes to c, the separation as seen by
an observer moving with the particles goes to infinity.
Your own words are wrong, Jeff Root.
For this case when the separation as measured by an observer
not moving with the particles is held to be constant then the
separation that they calculate for the particles rest frame
between the particles goes to infinity as their speed relative
to the particles goes to c.
Yes, that is what Ken and I have been saying!



But that is not the "separation as seen by an observer moving
with the particles".
Yes it is.



The separation as measured by any observer moving with the
particles is constant for that observer
Now I think you need to read more carefully what *you*
have written.

The separation as measured by any observer moving with the
particles goes to infinity as the speed of the particles relative
to some other observer goes to c, for a given separation as
measured by that other observer.



ETA: That word constant first appears in this thread in your post.
It first appears in your post #82:



* Alice fires a series of bullets with a certain separation.
In Alice's frame, the separation is constant.
That information is correct, appropriate, and relevant.
It is exactly what I was referring to as "constant".

-- Jeff, in Minneapolis

Reality Check
2015-Jul-07, 02:02 AM
Yes, that is what Ken and I have been saying!

The "is held constant" looks new to me. Can you cite where you and Ken G state that "the separation as seen by an observer not moving with the particles is held constant" before your post (http://cosmoquest.org/forum/showthread.php?157384-Photon-races&p=2298732#post2298732)?

Repeating an error does not change the physics that the separation as measured by any observer moving with the particles is constant for that observer. It does not matter how many times that Charles measures the separation between the particle, Charles will always measure the same value :eek:! But let us check that
Jeff Root:
Let Charles be an observer sitting on a particle in a series of particles all travelling at the same speed of v = 0 relative to him.
At t = t0, Charles measures the separation from his particle to the next particle, e.g. bounces a photon off it.
At t = t1, Charles measures the separation from his particle to the next particle, e.g. bounces a photon off it.
Are these two measurements the same for Charles or different?
(ignore that for a real measurement the photon will have given the next particle a tiny boost in the first measurement)

The word "constant" appears in that post as does the word "a" :D!
But: Where in this post does Alice change her speed, Jeff Root? Is Alice an "observer moving with the particles"?

Any observer traveling at c has a problem with measuring anything because they will always measure the proper time between events as zero.
If we switch to the different example of bullets travelling at near c then we have a different situation.
* Alice fires a series of bullets with a certain separation. In Alice's frame, the separation is constant.
* Bob is an accelerating observer who is accelerating along the line of bullets with respect to Alice.
Bob can measure the distance between any two points. Bob will find that distance decreasing with increasing velocity (length contraction). AFAIK it does not matter what points Bob measures - the start and end of one bullet or the end of one bullet and the start of the next bullet. All length measurements will contract. So the length and separation of the bullets as measured by Bob will decrease as he accelerates.
It is Bob who is the accelerating observer. The length contraction will increase as you and I know.

Ken G
2015-Jul-07, 02:33 AM
It's hopeless Jeff, the goal posts are just going to keep moving. In any event, mkline55's last post shows he now understands the situation, so there's not much more to say, unless there are other specific questions. In summary, here are the answers to the questions that have come up:
1) There is no need for any Lorentz transformation, or any length contraction, if we are simply talking about the distance between the bullets in the frame of the sender and receiver, given that they are in the same frame.
2) If the universe is expanding, that will increase the received distances between bullets, by whatever factor the universe expands during the propagation of the bullets, assuming both sender and receiver are in the comoving frame.
3) If there is an observer moving with the bullets, that observer will reckon the distance between the bullets as larger than what Alice reckons as she fires them.
4) In the limit as the bullets are photons, the distance between them will go infinite, given that Alice is firing them at a fixed time interval t. Of course the photon frame is officially an impossible frame, but we can still consider the limit.
5) If an observer starts in Alice's frame, and gradually accelerates up to the speed of the bullets, all the while they will reckon the distance between the bullets at the observer's location as increasing.
6) Above all, this means that "length contraction" is not something that categorically happens to all things when you go from a stationary frame to a moving frame, it is something that happens to objects that go from being regarded as stationary to being regarded as moving.

Jeff Root
2015-Jul-07, 05:24 AM
The "is held constant" looks new to me. Can you cite where you
and Ken G state that "the separation as seen by an observer not
moving with the particles is held constant" before your post (http://cosmoquest.org/forum/showthread.php?157384-Photon-races&p=2298732#post2298732)?
Starting with the original post of the thread:



For example, a transmitter sends one photon toward a
receiver, then sends another one second later. Does the
Lorentz transform not apply to the length of an imaginary
line drawn between the photons?
This specifies a constant distance of one light-second.

Ken gave a detailed answer in his first post, post #38.
I wanted to quote relevant parts, but there is too much.
He talks explicitly about "what we will keep fixed":



The answer's in post #29, you just have to take the
mathematical limit as the speed goes to c. ...
In post #83, Ken used the word "constant":



It is all a question of what you hold fixed as you consider
higher and higher speeds. What is appropriate to this
question is to hold the separation as constant ...
My bold.

You have been missing the point, I think because you are
so sure you know what the point is that you aren't taking
in what you are reading.



Repeating an error does not change the physics that the
separation as measured by any observer moving with the
particles is constant for that observer. It does not
matter how many times that Charles measures the
separation between the particle, Charles will always
measure the same value :eek:!
We are describing an experimental setup and what the
results of running it would be. You are imagining a similar
but different setup, and criticizing what we say the results
would be because they are different from what you think
they would be. Over and over and over and over and over.

Read what Ken wrote, and pay attention while you read it.
Don't assume you understand what you are reading before
you read it, because it is completely clear that you do not.



But let us check that Jeff Root:
Let Charles be an observer sitting on a particle in a series of
particles all travelling at the same speed of v = 0 relative to him.
At t = t0, Charles measures the separation from his particle to
the next particle, e.g. bounces a photon off it.
At t = t1, Charles measures the separation from his particle to
the next particle, e.g. bounces a photon off it.
Are these two measurements the same for Charles or different?
(ignore that for a real measurement the photon will have given
the next particle a tiny boost in the first measurement)
They are the same.



The word "constant" appears in that post as does the word "a" :D!
By "that post", I presume you mean your post #82.

As I said, the word "constant" in your post #82 referred to
exactly what I referred to as "constant" in post #125, making
it relevant and appropriate, unlike the word "a". Why you
think such silliness is called for, I can only speculate.



But: Where in this post does Alice change her speed, Jeff Root?
Alice does not "change her speed".

The speed of the particles being emitted by Alice, relative to
Alice, has been explicitly specified as varying in many posts.
Whenever you read "as the speed goes to c" (Ken in post #38)
or "as they get arbitarily close to c" (Ken in post #81) or
"closer and closer to c" (Ken, post #83), or "the speed v of the
particles relative to that observer goes to c" (Jeff, post #125),
what is meant is that the speed of the particles emitted by
Alice, relative to Alice, is considered as greater and greater
in successive cases, approaching c as the limit. We could
equivalently say that Alice's speed changes, but we have not
done that, in part because there was no reason to.



Is Alice an "observer moving with the particles"?
No. As defined by you in post #14, Alice emits the particles.



It is Bob who is the accelerating observer.
You defined Bob as stationary relative to Alice in post #14,
so Alice and Bob are equivalent and redundant observers,
neither of whom is accelerating.



The length contraction will increase as you and I know.
What you "know" seems to be unique to you.

-- Jeff, in Minneapolis
.

Hornblower
2015-Jul-07, 02:52 PM
Let's go back to post 29.


I think WayneFrancis was on the right track when he introduces an observer travelling from Alice toward Bob. In the case mkline55 describes, Alice shoots a relativistic particle toward Bob. Let's change this so that it's not a photon, so we can talk meaningfully about it's rest frame and put our third observer in that frame; how about a baseball travelling at half the speed of light? Two seconds later, she launches another baseball toward Bob at the same speed. From Alice's reference frame, then, the two baseballs are one light-second apart. Bob is at rest relative to Alice, and they've synchronized their clocks (in their own common reference frame) so he'll measure the same distance between the baseballs as Alice, and he'll agree with her about when the baseballs were launched.

Now Charlie is travelling along with the baseballs, and has a measuring rod that is one light-second long in Alice's frame, so one end should be even with the first baseball, and the other end should be even with the second baseball. Since he's moving at high speed relative to Alice, she should see that ruler length contracted. So Charlie should measure the ruler as about 15% longer than one light-second, and since that's the rest frame of the ruler, that's its proper length, and also the proper length separating the baseballs: slightly longer than one light-second. So Alice does see the distance between the baseballs as length contracted from what would be measured in Charlie's frame, moving along with the baseballs. How is that possible, since they were fired exactly two seconds apart? Well, remember that from Charlie's perspective, Alice's clock is running slowly. So when she fired the second baseball after her clock registered two seconds, more than two seconds had passed for Charlie (and the baseball), by that same 15%.

Of course, Charlie (and the baseballs) will also measure the distance between Alice and Bob as shorter than Alice and Bob measure it, and they'll disagree about how long it takes from the baseballs to travel from Alice to Bob (or, from Charlie's perspective, how long it takes Bob to get there, since it's Alice and Bob that are moving). This gets reconciled by relativity of simultaneity: they'll all agree on what time Bob's clock reads when the baseballs and Bob meet, but Charlie will disagree with Alice and Bob's statement that Alice's clock and Bob's clock are synchronized.
As I see it Grey succeeded in cutting through the fog of confusion about frames of reference and gave a correct answer. Later Ken G correctly showed how to use limits as v approaches c to handle cases in which simply plugging c into an equation makes it indeterminate. I am baffled by the persistence of the confusion among some of the posters here. It is as if they are seeing what they expect to see rather than what is actually printed in the correct answers.

Grey
2015-Jul-07, 08:39 PM
As I see it Grey succeeded in cutting through the fog of confusion about frames of reference and gave a correct answer. Later Ken G correctly showed how to use limits as v approaches c to handle cases in which simply plugging c into an equation makes it indeterminate. I am baffled by the persistence of the confusion among some of the posters here. It is as if they are seeing what they expect to see rather than what is actually printed in the correct answers.What is frustrating to me is that we seem to be moving backward. We had agreement, and then we seemed to lose it again.

Reality Check, back in this post (http://cosmoquest.org/forum/showthread.php?157384-Photon-races&p=2298086#post2298086) you seemed to agree that if Alice would measure a shorter distance between the bullets than a second observer (maybe named Bob or Charles) who is at rest with respect to the bullets.


Cool, so we're agreed that if Alice fires some relativistic bullets, and a second observer (maybe his name is Bob, maybe his names is Charles ;)) happens to passing by at exactly the same speed and direction as those bullets, then Alice will measure the distance between those bullets to be shorter than Bob (or Charles), and conversely, Bob (or Charles!) will measure the distance to be greater than Alice does?

And yes - we agree on the basics of SR, Grey :D!
The length contraction that observers measure gets greater with their speed relative to the object being measured. So an Alice with v > 0 relative to the bullets will measure greater contraction than a Bob with v = 0 relative to the bullets and vice versa.

That would correspond to the third case in your post here (http://cosmoquest.org/forum/showthread.php?157384-Photon-races&p=2297947#post2297947).


If we know Alice's speed relative to the bullets then we say something about the different scales of contraction relative to the bullets, e.g.
Alice has a speed vA relative to the bullets.
Bob has a speed vB relative to the bullets (e.g. add Bob's speed relative to Alice to her speed relative to the bullets).
If vA = vB than Alice and Bob measure the same length contraction.
If vA < vB than Alice measures less contraction relative to the bullets than Bob.
If vA > vB than Alice measures more contraction relative to the bullets than Bob.I asked a couple times to be sure, and you assured me that was definitely the case, and laughed at me for asking repeatedly. But let me ask once more. ;) Here's the situation again. Alice shoots some relativistic bullets. Charles is flying by at just the right velocity so that the bullets are stationary with respect to him. Do you still agree that Alice (who now has some substantial velocity with respect to the bullets) will measure a shorter distance between them than Charles (who is in the rest frame of the bullets)?