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Sam Woole

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Sam Woole

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reilly

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Regards,

Reilly Atkinson

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DrChinese

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Sam Woole said:Is cheating going on in Einstein’s theory of relativity?

Welcome to PhysicsForums!

Not sure exactly what you mean about cheating. Relativity (both SR and GR) has been very well tested through the years. John Baez has some good stuff at his physics site, including this:

What is the experimental basis of Special Relativity?

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Chronos

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Sam Woole

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Thanks to reilly, DrChinese, and Chronos for your replies to my question. I however think the problem was not so simple.

I wish all of you have access to the said textbook and then let us look into the thought experiment therein. The textbook authors claimed that the clocks will produce 2 different time intervals. But I believe otherwise and let me show you how.

If we simplify the graphics and ignore the light, the thought experiment can be shown as follows:

S__________R

P__________Q

where S, R, P, and Q are points on two reference frames. A clock will move from S to R. And at P and Q each there is another clock. When the clock at S departs, it must register a reading of time. Let me call it the departure

time "b". I believe the same departure time "b" will be recorded on another clock located at P. Don't you agree? (The textbook said, it was a light beam that departed. But it was all the same departure time, and that was why it was perfect all right not to mention the departure of light.)

Now the clock will record an arrival time when it arrives at R. Let us call it "e". And the other clock at Q certainly will record the same arrival time "e" according to the design of the thought experiment. To obtain the time interval t, it is to deduct the departure time from the arrival time:

t = e - b. This means that all 3 clocks would measure one identical time interval t, not two different time intervals claimed by the authors of the textbook.

Consequently the relativistic nature of time cannot be proven unless relativistic advocates resort to cheating.

I wish all of you have access to the said textbook and then let us look into the thought experiment therein. The textbook authors claimed that the clocks will produce 2 different time intervals. But I believe otherwise and let me show you how.

If we simplify the graphics and ignore the light, the thought experiment can be shown as follows:

S__________R

P__________Q

where S, R, P, and Q are points on two reference frames. A clock will move from S to R. And at P and Q each there is another clock. When the clock at S departs, it must register a reading of time. Let me call it the departure

time "b". I believe the same departure time "b" will be recorded on another clock located at P. Don't you agree? (The textbook said, it was a light beam that departed. But it was all the same departure time, and that was why it was perfect all right not to mention the departure of light.)

Now the clock will record an arrival time when it arrives at R. Let us call it "e". And the other clock at Q certainly will record the same arrival time "e" according to the design of the thought experiment. To obtain the time interval t, it is to deduct the departure time from the arrival time:

t = e - b. This means that all 3 clocks would measure one identical time interval t, not two different time intervals claimed by the authors of the textbook.

Consequently the relativistic nature of time cannot be proven unless relativistic advocates resort to cheating.

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Doc Al

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Why don't you scan it in?Sam Woole said:I wish all of you have access to the said textbook and then let us look into the thought experiment therein. The textbook authors claimed that the clocks will produce 2 different time intervals. But I believe otherwise and let me show you how.

To say "A clock moves from S to R" is very different from saying "A beam of light is emitted at S and arrives at R". Which is it? (I'll assume the latter.)If we simplify the graphics and ignore the light, the thought experiment can be shown as follows:

S__________R

P__________Q

where S, R, P, and Q are points on two reference frames. A clock will move from S to R. And at P and Q each there is another clock. When the clock at S departs, it must register a reading of time. Let me call it the departure

time "b". I believe the same departure time "b" will be recorded on another clock located at P. Don't you agree? (The textbook said, it was a light beam that departed. But it was all the same departure time, and that was why it was perfect all right not to mention the departure of light.)

You have two frames in relative motion, I presume. I assume S & R are in the "moving" frame. At the moment that the light is emitted at S, the point P passes by point S. Let's define the time read by the clocks at S and P to read the same

Not true at all! I assume the experiment is that the light arrives at point R just at the moment that point Q passes R. Each frame has its own clocks, presumably synchronized in their own frame: the moving frame has clocks at S and R; the other frame has clocks at P and Q. When the light arrives at R and Q, those clocks willNow the clock will record an arrival time when it arrives at R. Let us call it "e". And the other clock at Q certainly will record the same arrival time "e" according to the design of the thought experiment.

You seem to justTo obtain the time interval t, it is to deduct the departure time from the arrival time:

t = e - b. This means that all 3 clocks would measure one identical time interval t, not two different time intervals claimed by the authors of the textbook.

Where exactly is the cheating? You seem to have givenConsequently the relativistic nature of time cannot be proven unless relativistic advocates resort to cheating.

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JesseM

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What you are missing is the fact that in the rest frame of the clock which moved from S to R, the clocks that are resting at P and Q are out-of-sync, so the time interval between them is not the correct one in the first clock's rest frame (if you walked through a hallway and the clock on one end said 2:00 while the clock on the other end said 3:00, then you would not conclude it took you an hour to cross the hallway, you'd just conclude the second clock was ahead of the first one). This is known as "the relativity of simultaneity", and it's a pretty simple consequence of the idea that each frame should assume light moves at c. Just imagine a rocket going by you where a crewmember sets off a flash at the rocket's midpoint and synchronizes clocks on either end of the rocket by setting them to the same time at the moment the light from the flash hits them; since the rocket is moving forward in your frame, you will see that the clock at the front of the rocket is movingSam Woole said:Thanks to reilly, DrChinese, and Chronos for your replies to my question. I however think the problem was not so simple.

I wish all of you have access to the said textbook and then let us look into the thought experiment therein. The textbook authors claimed that the clocks will produce 2 different time intervals. But I believe otherwise and let me show you how.

If we simplify the graphics and ignore the light, the thought experiment can be shown as follows:

S__________R

P__________Q

where S, R, P, and Q are points on two reference frames. A clock will move from S to R. And at P and Q each there is another clock. When the clock at S departs, it must register a reading of time. Let me call it the departure

time "b". I believe the same departure time "b" will be recorded on another clock located at P. Don't you agree? (The textbook said, it was a light beam that departed. But it was all the same departure time, and that was why it was perfect all right not to mention the departure of light.)

Now the clock will record an arrival time when it arrives at R. Let us call it "e". And the other clock at Q certainly will record the same arrival time "e" according to the design of the thought experiment. To obtain the time interval t, it is to deduct the departure time from the arrival time:

t = e - b. This means that all 3 clocks would measure one identical time interval t, not two different time intervals claimed by the authors of the textbook.

Consequently the relativistic nature of time cannot be proven unless relativistic advocates resort to cheating.

You might want to take a look at a detailed example I worked out (with diagrams) on this thread, involving two rulers with clocks mounted at regular intervals sliding alongside each other, showing how in each ruler's rest frame it is the clocks on the other ruler which are running slow (and which are out-of-sync).

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Sam Woole

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In your reply to me you said: "When the light arrives at R and Q, those clocks will not read the same times."

S__________R

P__________Q

Why the clock at Q will NOT read the same arrival time as the clock at R did? You did not explain why. I believe otherwise, because this was a thought experiment. If all the clocks at S and P could be thought to read the same

departure time, the same "thought" logic must apply to the arrival point. Hence when the light beam arrived at R and Q, the clocks there would logically read the same arrival time.

Further, there could not be any synchronization problem either, simply because it was all a thought. In a thought an infinite number of clocks can be thought as synchronized perfectly. Imperfection can only happen in real experiments in the laboratory. If the clocks in different frames were synchronized in different ways, our thought can make compensations to them, adding or deducting a few nano-seconds to make difference disappear.

As a last argument of mine, even if the clocks at R and Q would not read the same arrival time as you alleged, the cheating was still here, simply because the textbook authors produced the different time intervals by means

of Newtonian physics. My logic was so glaringly clear that nobody could miss. Say I bought an orange and an apple from Mr. Newton's store, and I told my parents I bought them from Mr. Smith's store. There was no change

to the fact that there were an orange and an apple. But I cheated my parents about the source.

If my parents got sick after consuming the fruits, the blame would fall on Mr. Smith, not on Mr. Newton, as a result of my cheating.

As to JesseM's reply, I believe he had the same argument since he mentioned synchronization. My point was how to identify the cheat, not to tell what the light would do in different reference frames. As to your relativity of simultaneity, I shall respond separately after I have studied your details.

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JesseM

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But whose clocks do you add or deduct time from? Each observer has a set of clocks which are at rest with respect to himself and whichSam Woole said:Further, there could not be any synchronization problem either, simply because it was all a thought. In a thought an infinite number of clocks can be thought as synchronized perfectly. Imperfection can only happen in real experiments in the laboratory. If the clocks in different frames were synchronized in different ways, our thought can make compensations to them, adding or deducting a few nano-seconds to make difference disappear.

Did you look at the diagrams of the two ruler/clock systems I provided on the other thread I linked to? They make the symmetry of the situation pretty clear.

Where do textbook authors use Newtonian physics when justifying the ideas of relativity? I've never seen that before, can you quote the section you're talking about?Sam Woole said:As a last argument of mine, even if the clocks at R and Q would not read the same arrival time as you alleged, the cheating was still here, simply because the textbook authors produced the different time intervals by means

of Newtonian physics.

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Doc Al

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You do realize that this is aSam Woole said:Thank you, Al Doc. I am sorry that I did not give you enough information because I don't have the scanning device. But in my opening post I have made it clear between [clocks] and [equation]. We used clocks to measure time intervals, or we used equations to compute time intervals. These are two distinctly different operations, clocks and equations. If someone told me he would use clocks to do something and he switched to equations, of course he cheated me. This was exactly what had happened in the said textbook.

The reason why the clocks at S and P read the same at the instant that they pass each other is because we agreed to set them to read the same timeIn your reply to me you said: "When the light arrives at R and Q, those clocks will not read the same times."

S__________R

P__________Q

Why the clock at Q will NOT read the same arrival time as the clock at R did? You did not explain why. I believe otherwise, because this was a thought experiment. If all the clocks at S and P could be thought to read the same

departure time, the same "thought" logic must apply to the arrival point. Hence when the light beam arrived at R and Q, the clocks there would logically read the same arrival time.

You seem to have the wrong idea as to what a "thought experiment" means. It doesn't mean "just think whatever you want". It means that you set up a situation (one that may be impractical to do in real life) and apply real laws of physics to deduce what happens. You don't justFurther, there could not be any synchronization problem either, simply because it was all a thought. In a thought an infinite number of clocks can be thought as synchronized perfectly. Imperfection can only happen in real experiments in the laboratory. If the clocks in different frames were synchronized in different ways, our thought can make compensations to them, adding or deducting a few nano-seconds to make difference disappear.

A clear application of the assumptions of relativity (that the speed of light is invariant, for one) will show that the two frames

What "Newtonian Physics" was used? I suspect that what was used was something like: Distance = speed X time. That's just as true for Einstein as for Newton.As a last argument of mine, even if the clocks at R and Q would not read the same arrival time as you alleged, the cheating was still here, simply because the textbook authors produced the different time intervals by means

of Newtonian physics. My logic was so glaringly clear that nobody could miss. Say I bought an orange and an apple from Mr. Newton's store, and I told my parents I bought them from Mr. Smith's store. There was no change

to the fact that there were an orange and an apple. But I cheated my parents about the source.

Once again: Where's the cheating? To show "cheating" you should lay out, step by step, the actual argument made in the textbook (not just

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Sam Woole

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JesseM

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No one was asking you to transmit the full text (which would be illegal anyway), just to quote a paragraph or two where they mentioned Newtonian physics in connection to relativity, or perhaps scan a page if you don't feel like typing it.Sam Woole said:

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Chronos

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Chronos

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Footnote: I think you are profoundly wrong, Sam, but I like the way you think.

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jtbell

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You may simply be one of those people who likes to learn things "by debate", so to speak, using an argumentative skeptical style. In many fields this is common, and you'll even find this sort of thing in physics, especially in true "cutting edge" areas.

The problem is that on the Net there are many anti-relativity cranks and crackpots who are convinced that they are right and mainstream scientists are all wrong, and who argue their views very aggressively and persistently. Anyone who comes onto an Internet discussion group and tries to discuss relativity in a similar style is likely to be treated like one of those people, regardless of their actual motives.

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Sam Woole

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Why? Because the spatial separation of the events can affect the time intervals measured by the observers."

The textbook then gives a thought experiment involving an observer Sally on a train moving at v relative to another observer Sam on the ground, represented in two figures. I do not know how to draw these figures. Let me describe the first figure and see whether you can visualize.

Sally will send a light pulse from her clock toward the ceiling of the cabin where she is in. To her, the light travels along the height (D) of her cabin, and comes down along the same height (D) and arrives at her clock. The textbook says: "Sally measures a certain time interval t between the two events, related to the distance 2D from source to mirror by t = 2D/c."

The quote immediately above has two parts. The first part means Sally would obtain information from her clock, reading a departure time and an arrival time; the second part means she needs not read, because the time interval is related to the distance traveled by light.

Don't you think these two actions, saying one thing and doing a different thing, constituted a lie?

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Sam Woole said:Sally will send a light pulse from her clock toward the ceiling of the cabin where she is in. To her, the light travels along the height (D) of her cabin, and comes down along the same height (D) and arrives at her clock. The textbook says: "Sally measures a certain time interval t between the two events, related to the distance 2D from source to mirror by t = 2D/c."

The quote immediately above has two parts. The first part means Sally would obtain information from her clock, reading a departure time and an arrival time; the second part means she needs not read, because the time interval is related to the distance traveled by light.

Don't you think these two actions, saying one thing and doing a different thing, constituted a lie?

Your passage described a light-clock.

The duration between emission and reception, t, represents "1 tick" of that clock... a perfectly good unit of time. One could regard the first emission as the time reading of "0 ticks".

Upon the first reception, if the source reflects back or re-emits a light signal, waiting for a second reception yields a second tick.

The duration of a tick is proportional to the separation, D, between the "mirror" on the ceiling and the source. That's what t=2D/c means... a relationship relating t and D.. not a definition of t. (If you want a clock with a finer resolution, reduce the separation D.)

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JesseM

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Why is it a lie? It just says that the time interval she actuallySam Woole said:

Why? Because the spatial separation of the events can affect the time intervals measured by the observers."

The textbook then gives a thought experiment involving an observer Sally on a train moving at v relative to another observer Sam on the ground, represented in two figures. I do not know how to draw these figures. Let me describe the first figure and see whether you can visualize.

Sally will send a light pulse from her clock toward the ceiling of the cabin where she is in. To her, the light travels along the height (D) of her cabin, and comes down along the same height (D) and arrives at her clock. The textbook says: "Sally measures a certain time interval t between the two events, related to the distance 2D from source to mirror by t = 2D/c."

The quote immediately above has two parts. The first part means Sally would obtain information from her clock, reading a departure time and an arrival time; the second part means she needs not read, because the time interval is related to the distance traveled by light.

Don't you think these two actions, saying one thing and doing a different thing, constituted a lie?

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Sam Woole said:Sally will send a light pulse from her clock toward the ceiling of the cabin where she is in. To her, the light travels along the height (D) of her cabin, and comes down along the same height (D) and arrives at her clock. The textbook says: "Sally measures a certain time interval t between the two events, related to the distance 2D from source to mirror by t = 2D/c."

The quote immediately above has two parts. The first part means Sally would obtain information from her clock, reading a departure time and an arrival time; the second part means she needs not read, because the time interval is related to the distance traveled by light.

Don't you think these two actions, saying one thing and doing a different thing, constituted a lie?

I'm sorry, but I don't see anything wrong with this quote. All it says is that the time that Sally reads on the Clock and 2D/c will be the same amount of time. It's like saying that if I start a stopwatch as someone starts to drive away from me at 60mph, then that person turns around after 10 miles and comes back, my stop watch will read 20 min upon his return. If he turns around after 20 miles, my stopwatch will read 40 min, and if he turns around after 5 miles it will read 10 min. There is a direct relation to the time I read on my stopwatch and the distance he drives before turning around.

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Sam Woole

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It seemed that JesseM and Janus and most people believed that Sally's clock would produce a time interval equal to the one produced by Newton's equation.

First of all my allegation had nothing to do with the equality between time intervals. My allegation was, if the textbook authors eventually used Newton's equation to compute the time interval, they should not in the beginning have said they would use clocks to do the job. They need not set up any clocks at all. Too much ado about nothing. They should not use the word "measure". They should just say they will compute. As a result the first paragraph I quoted in the former post of mine should read like this:

"If observers who move relative to each other COMPUTE the time interval between two events, they generally will find different results. Why? Because the spatial separation of the events can affect the time intervals COMPUTED by the observers."

The change of one action to another (from measure to compute) has illogical consequences. When computation has produced a shorter time interval, the explanation we can give is that we have put in (into Newton's equation) a shorter distance. I believe we cannot say, the reason for the equation to have produced a shorter time interval is because the clock ticked too slow. Can we accept such a logic?

But in my view this was exactly what was happening in teaching the relativity theory. When a shorter time interval was obtained from Newton's equation, we were told the clock was malfunctioning, or time dilated.

Such a practice (a shorter distance producing a shorter time interval) means time dilation has been used to prove the existence of time dilation. Is this logic sound?

Secondly, if the textbook authors believed that the clocks would produce a time interval equal to the one computed from Newton's equation, they were obliged to demonstrate HOW. Or at least they should have said a few words

about it.

Let us represent your belief like this:

(e - b) = t = 2D/c...wherein e is the arrival time, b the departure time, shown on clocks; t is the resulted time interval, which equals the computed time interval 2D/c.

As is customary that two reference frames are moving at v relative to each other. As a result when the light pulse departs and arrives at the clocks, the two frames have created a distance. Let us call this distance g. This

means, to create the distance g the frames have taken the same time interval t as the light pulse has done. Don't you agree? If you do, the situation can be represented with this: t = g/v. This equation must be true especially if we

refer to Janus example of stopwatches and cars.

Since you believed that the measured time interval (e - b) is equal to the computed time interval 2D/c, it means

(e - b) = t = 2D/c = g/v. This equation means that

both frames have obtained the one and only time interval t, whether the observers used the measuring technique or the computation one.

But the textbook told us that there would be two "different results." So, there must be something wrong? Don't you agree?

First of all my allegation had nothing to do with the equality between time intervals. My allegation was, if the textbook authors eventually used Newton's equation to compute the time interval, they should not in the beginning have said they would use clocks to do the job. They need not set up any clocks at all. Too much ado about nothing. They should not use the word "measure". They should just say they will compute. As a result the first paragraph I quoted in the former post of mine should read like this:

"If observers who move relative to each other COMPUTE the time interval between two events, they generally will find different results. Why? Because the spatial separation of the events can affect the time intervals COMPUTED by the observers."

The change of one action to another (from measure to compute) has illogical consequences. When computation has produced a shorter time interval, the explanation we can give is that we have put in (into Newton's equation) a shorter distance. I believe we cannot say, the reason for the equation to have produced a shorter time interval is because the clock ticked too slow. Can we accept such a logic?

But in my view this was exactly what was happening in teaching the relativity theory. When a shorter time interval was obtained from Newton's equation, we were told the clock was malfunctioning, or time dilated.

Such a practice (a shorter distance producing a shorter time interval) means time dilation has been used to prove the existence of time dilation. Is this logic sound?

Secondly, if the textbook authors believed that the clocks would produce a time interval equal to the one computed from Newton's equation, they were obliged to demonstrate HOW. Or at least they should have said a few words

about it.

Let us represent your belief like this:

(e - b) = t = 2D/c...wherein e is the arrival time, b the departure time, shown on clocks; t is the resulted time interval, which equals the computed time interval 2D/c.

As is customary that two reference frames are moving at v relative to each other. As a result when the light pulse departs and arrives at the clocks, the two frames have created a distance. Let us call this distance g. This

means, to create the distance g the frames have taken the same time interval t as the light pulse has done. Don't you agree? If you do, the situation can be represented with this: t = g/v. This equation must be true especially if we

refer to Janus example of stopwatches and cars.

Since you believed that the measured time interval (e - b) is equal to the computed time interval 2D/c, it means

(e - b) = t = 2D/c = g/v. This equation means that

both frames have obtained the one and only time interval t, whether the observers used the measuring technique or the computation one.

But the textbook told us that there would be two "different results." So, there must be something wrong? Don't you agree?

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JesseM

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What are you talking about when you say "Newton's equation"? Do you just mean speed = distance/time? This equation is not specifically Newtonian, it's just as true in relativity as in Newtonian physics.Sam Woole said:It seemed that JesseM and Janus and most people believed that Sally's clock would produce a time interval equal to the one produced by Newton's equation.

Even if you feel the clocks and rulers areSam Woole said:First of all my allegation had nothing to do with the equality between time intervals. My allegation was, if the textbook authors eventually used Newton's equation to compute the time interval, they should not in the beginning have said they would use clocks to do the job. They need not set up any clocks at all. Too much ado about nothing. They should not use the word "measure". They should just say they will compute.

But what they mean here by "compute" is "subtract the measured time of the first event according to that observer's clocks from the measured time of the second event", so clocks are still involved. If Sally is on the train and Sam is on the ground, and Sam uses his own clocks to mark the time that the light beam leaves the floor and the time it returns to the floor (he will need two separate synchronized clocks for this since the train is moving while the clocks need to be at rest relative to him, and each clock needs to be next to the event it's measuring so that you don't have to worry about delays due to the finite speed of light), then if he subtracts the measured time on his own clocks that the light beam left the floor from the measured time on his own clocks that it returned to the floor, he'll get a larger time interval between these two events than Sally got using her onboard clock (which means from his point of view, Sally's clock must have been ticking slow). But he also measures the light beam to have taken a longer path between the floor and the ceiling, since the train was moving horizontally and thus the light beam was moving diagonally in his frame rather than purely vertically; if he divides the distance he measured by the time he measured, he will also find that the light beam traveled at c in his frame.Sam Woole said:As a result the first paragraph I quoted in the former post of mine should read like this:

"If observers who move relative to each other COMPUTE the time interval between two events, they generally will find different results. Why? Because the spatial separation of the events can affect the time intervals COMPUTED by the observers."

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Sam Woole

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JesseM said:What are you talking about when you say "Newton's equation"? Do you just mean speed = distance/time? This equation is not specifically Newtonian, it's just as true in relativity as in Newtonian physics.

Yes, I did mean that. I used to called t=d/v Newton's equation. I now realized I should not. I also apologize for the incorrect use of the word "compute". Anyway, reading clocks is an act very different from using the equation.

Let us make our discussion short. JesseM's language implied that he has access to the said textbook. If so, I believe he would be able to judge these 3 cases. (1) The two reference frames have created a distance; (2) A time interval will be obtained when this distance is put into the equation (t = d/v). (3) And the time interval so obtained is the time interval needed by the frames for creating that distance, and is equal to the one taken by light for its round trip. Do you think these 3 cases justified?

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JesseM

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I didn't mean to imply that, I just was inferring the textbook authors' meaning from the quotes you provided.Sam Woole said:Let us make our discussion short. JesseM's language implied that he has access to the said textbook.

What do you mean they "created a distance"? Each reference frame measures the distance between events using their own set of rulers, I don't know if that has any relation to what you're talking about here.Sam Woole said:If so, I believe he would be able to judge these 3 cases. (1) The two reference frames have created a distance;

If you know the object's velocity in your frame, I guess you could do this, but usually it's the other way around--you have clocks at two locations along the object's path, and you note the times on each clock at the moment the object passes them and subtract one from the other to get a time interval, then you divide by the distance between the two clocks.Sam Woole said:(2) A time interval will be obtained when this distance is put into the equation (t = d/v).

I still don't understand what it means for a time interval "create" a distance, a distance is just something you measure on a ruler. And the time interval for an object to cross a given distance will only be equal to the one for light to cross that distance if the object is traveling at light speed, if it's going slower then the time interval will be larger. But I'm probably misunderstanding what you're trying to say--maybe you could give me a numerical example of what you're talking about?Sam Woole said:(3) And the time interval so obtained is the time interval needed by the frames for creating that distance, and is equal to the one taken by light for its round trip. Do you think these 3 cases justified?

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vanesch

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Sam Woole said:Yes, I did mean that. I used to called t=d/v Newton's equation. I now realized I should not. I also apologize for the incorrect use of the word "compute". Anyway, reading clocks is an act very different from using the equation.

Sam, if I may jump in, as others have already argued, you may have a difficulty understanding what is mean to be a "thought experiment". In fact, a thought experiment is not an experiment at all, but a way of reasoning, which COULD eventually correspond to an experimental situation. It is, in a certain way, "armchair philosophy" But what's the purpose ? The purpose is to analyse the consequences of a certain set of (imagined or real) laws of nature. Now, in special relativity, we do the following:

1) we still assume that, for an observer, space is 3-dim Euclidean space (meaning, you can use all high school geometry), we call it E^3_observer

2) we still assume that LOCALLY an observer has a clock that gives him "time". Whether this is a real clock, or a mathematical variable you call t_observer, doesn't matter of course.

With points 1 and 2, we can now define of course things like "a trajectory" (for an observer), which is a mapping from t_observer into E^3_observer, and from such a trajectory, we can define a velocity which is de derivative of this trajectory to t_observer. All this makes sense WITH RESPECT TO AN OBSERVER.

Up to here we are on par with Newtonian physics, except that:

3-Newton) Newton assumes that for all observers, the t_observer are equal

3-Relativity) Relativity assumes that c is a constant for all inertial observers.

From the 3-Relativity point, you can now DEDUCE that the time seen on the watch of Sally, when her light pulse bounced up and down, MUST BE equal to 2 D/c

Indeed, the light pulse is going to cover a distance 2 D (this we can deduce, from Sally's point of view, because of 1) and because of 3-Relativity, we know that the speed of light is c, so we have to have that c = 2 D / time-seen-by-Sally

- #25

Sam Woole

- 55

- 0

JesseM said:I didn't mean to imply that, I just was inferring the textbook authors' meaning from the quotes you provided. What do you mean they "created a distance"? Each reference frame measures the distance between events using their own set of rulers, I don't know if that has any relation to what you're talking about here. If you know the object's velocity in your frame, I guess you could do this, but usually it's the other way around--you have clocks at two locations along the object's path, and you note the times on each clock at the moment the object passes them and subtract one from the other to get a time interval, then you divide by the distance between the two clocks. I still don't understand what it means for a time interval "create" a distance, a distance is just something you measure on a ruler. And the time interval for an object to cross a given distance will only be equal to the one for light to cross that distance if the object is traveling at light speed, if it's going slower then the time interval will be larger. But I'm probably misunderstanding what you're trying to say--maybe you could give me a numerical example of what you're talking about?

I am sorry it was all my fault due to lack of information, specifically lack of graphics about the thought experiment. Let us drop this very textbook and go to Einstein's writing available online.

I believe people cannot succeed in their attempt to prove that time is relative unless they cheat. This belief of mine included Einstein himself as the cheater. To find out whether there is cheating in the theory of relativity,

and whether Einstein cheated, we shall refer to Chapter 9 of Einstein's own book: Relativity, The Special and General Theory, available at http://www.ivorix.com/en/einstein/ [Broken].

With easy access to Einstein's own words, we are in a better position to exchange opinions than when I could not provide you enough information from the textbook Fundamentals of Physics named in my first posting.

In the said chapter, Einstein designed a thought experiment to prove that simultaneity is relative. He described it in these words:

"Up to now our considerations have been referred to a particular body of reference, which we have styled a railway embankment. We suppose a very long train traveling along the rails with the constant velocity v and in

the direction indicated in Fig 1. People traveling in this train will with advantage use the train as a rigid reference-body (co-ordinate system); they regard all events in reference to the train. " And the Figure looks like this:

Fig. 1.

v→

A ____________________m ____________________B

A ____________________M____________________ B

Einstein continued:

"Then every event which takes place along the line also takes place at a particular point of the train. Also the definition of simultaneity can be given relative to the train in exactly the same way as with respect to the embankment. As a natural consequence, however, the following question arises :

"Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? We shall show directly that the answer must be in the negative.

"When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length AB of the embankment. But the events A and B also correspond to positions A and B on the train. Let m be the mid-point of the distance AB on the traveling train.

"Just when the flashes (as judged from the embankment) of lightning occur, this point m naturally coincides with the point M but it moves towards the right in the diagram with the velocity v of the train. "

Here I drew the following figure to represent Einstein's description, Figure 2:

A__________________m__________________B

A____________________M___________________B

In Einstein's book, there was no figure like this showing a total of 4 events A, A, B, B. But his description of the graphics implied explicitly that there were 2 events on the train and 2 events on the embankment. Yet the above graphics may or may not agree with Einstein's later ideas. We shall encounter this doubtful situation in the last paragraph of my post.

Einstein continued:

"If an observer sitting in the position m in the train did not possesses this velocity, then he would remain permanently at M, and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated. Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A."

Here I believe a cheat has been committed: "he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A." To appreciate my charge, let me repeat Figure 2:

A__________________m__________________B

A____________________M___________________B

When Einstein said m is "hastening towards the beam of light coming from B", which B was he talking about?

If he was talking about the B on the train (m's reference frame), I think he lied against his light-speed principle. If he was talking about the B on the embankment (M's reference frame), I think he lied again, against his own regulation: " People traveling in this train will with advantage use the train as a rigid reference-body (co-ordinate system); they regard all events in reference to the train. "

I believe my thinking above is correct meaning Einstein did cheat.

Einstein continued his cheating by saying:

"Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A."

I believe otherwise. I believe observers taking the railway train as their reference-body cannot come to the conclusion as Einstein did, but rather they will conclude that lightning flashes A and B emitted simultaneously.

My belief can be justified like this:

Since Am = mB, so that Am/c = mB/c = t.

But Einstein continued and said:

"We thus arrive at the important result:

"Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity)."

To me this statement of Einstein's was a big lie. First of all the first word "events" was false or highly contradictory, because, according to Einstein's description of his thought experiment, there were a total of 4 events, two on the train frame and another two on the embankment frame. I believe we as students of his theory are entitled to be told which events he was talking about (again?). If he was talking about only two events, which two?

When you try to identify them, I believe you shall find that the entire sentence representing Einstein's "important result" does not make sense at all. If you do not find so, I am glad to learn from you and retract all my charges.

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- #26

Sam Woole

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vanesch said:Sam, if I may jump in, as others have already argued, you may have a difficulty understanding what is mean to be a "thought experiment". In fact, a thought experiment is not an experiment at all, but a way of reasoning, which COULD eventually correspond to an experimental situation. It is, in a certain way, "armchair philosophy" But what's the purpose ? The purpose is to analyse the consequences of a certain set of (imagined or real) laws of nature.

vanesch, welcome. You got it right regarding my difficulty in understanding the relativity theory. I am just a leisure reader, not a student of physics. If I am wrong, I am readily happy to change.

Certainly you are perfect right in saying that the thought experiment is a kind of reasoning. I think I do understand this point. But the point I was trying to make was, can we lie in the reasoning process. If we lied, of course truth was sacrificed in the reasoning process.

Now, in special relativity, we do the following:

1) we still assume that, for an observer, space is 3-dim Euclidean space (meaning, you can use all high school geometry), we call it E^3_observer

2) we still assume that LOCALLY an observer has a clock that gives him "time". Whether this is a real clock, or a mathematical variable you call t_observer, doesn't matter of course.

From the 3-Relativity point, you can now DEDUCE that the time seen on the watch of Sally, when her light pulse bounced up and down, MUST BE equal to 2 D/c

Indeed, the light pulse is going to cover a distance 2 D (this we can deduce, from Sally's point of view, because of 1) and because of 3-Relativity, we know that the speed of light is c, so we have to have that c = 2 D / time-seen-by-Sally

I think your deduction process is a lie, even though you obtained the same time interval as 2D/c. If you told your students you would use clocks to obtain the time interval but you did not do so, you lied. Whether you have obtained the true time interval or not, you lied in the reasoning or deduction process.

Secondly, I did challenge the truth of the time interval t = 2D/c. Due to my inability to quote enough information from the textbook, I was unable to demonstrate fully to my own satisfaction so as to support my challenge. I propose we move to Einstein's book which will let me quote enough information to facilitate our discussion. Please go to my new reply below.

- #27

JesseM

Science Advisor

- 8,518

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I think the idea is that the train and the embankment are arbitrarily close in the vertical dimension--If you like, you could imagine that they are both 1-dimensional and that there is no distance at all between them. So at the moment the train and the embankment were lined up, there would be no separation between the two "As" in your picture or between the two "Bs", and there are only two events, the first lightning strike (which occurs at the same time and place as the two lined-up As) and the second one (which occurs at the same time and place as the two lined-up Bs).Sam Woole said:I am sorry it was all my fault due to lack of information, specifically lack of graphics about the thought experiment. Let us drop this very textbook and go to Einstein's writing available online.

I believe people cannot succeed in their attempt to prove that time is relative unless they cheat. This belief of mine included Einstein himself as the cheater. To find out whether there is cheating in the theory of relativity,

and whether Einstein cheated, we shall refer to Chapter 9 of Einstein's own book: Relativity, The Special and General Theory, available at http://www.ivorix.com/en/einstein/ [Broken].

With easy access to Einstein's own words, we are in a better position to exchange opinions than when I could not provide you enough information from the textbook Fundamentals of Physics named in my first posting.

In the said chapter, Einstein designed a thought experiment to prove that simultaneity is relative. He described it in these words:

"Up to now our considerations have been referred to a particular body of reference, which we have styled a railway embankment. We suppose a very long train traveling along the rails with the constant velocity v and in

the direction indicated in Fig 1. People traveling in this train will with advantage use the train as a rigid reference-body (co-ordinate system); they regard all events in reference to the train. " And the Figure looks like this:

Fig. 1.

v?

A ____________________m ____________________B

A ____________________M____________________ B

Einstein continued:

"Then every event which takes place along the line also takes place at a particular point of the train. Also the definition of simultaneity can be given relative to the train in exactly the same way as with respect to the embankment. As a natural consequence, however, the following question arises :

"Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? We shall show directly that the answer must be in the negative.

"When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length AB of the embankment. But the events A and B also correspond to positions A and B on the train. Let m be the mid-point of the distance AB on the traveling train.

"Just when the flashes (as judged from the embankment) of lightning occur, this point m naturally coincides with the point M but it moves towards the right in the diagram with the velocity v of the train. "

Here I drew the following figure to represent Einstein's description, Figure 2:

A__________________m__________________B

A____________________M___________________B

In Einstein's book, there was no figure like this showing a total of 4 events A, A, B, B. But his description of the graphics implied explicitly that there were 2 events on the train and 2 events on the embankment.

He's discussing things from the frame of the embankment (in the train's own frame it is at rest so it isn't 'hastening toward' anything), so it'd be the bottom B. Note that as time passes the top m will be moving to the right relative to the bottom M.Sam Woole said:Einstein continued:

"If an observer sitting in the position m in the train did not possesses this velocity, then he would remain permanently at M, and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated. Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A."

Here I believe a cheat has been committed: "he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A." To appreciate my charge, let me repeat Figure 2:

A__________________m__________________B

A____________________M___________________B

When Einstein said m is "hastening towards the beam of light coming from B", which B was he talking about?

Huh? That quote only refers to how "people traveling in this train" will measure things, but clearly M is on the embankment, not on the train, so in M's frame it will appear that m is moving towards the location of the lightning strike on the right and away from the location of the lightning strike on the left.Sam Woole said:If he was talking about the B on the train (m's reference frame), I think he lied against his light-speed principle. If he was talking about the B on the embankment (M's reference frame), I think he lied again, against his own regulation: " People traveling in this train will with advantage use the train as a rigid reference-body (co-ordinate system); they regard all events in reference to the train. "

Impossible, since the light from lightning strike B will hit m before the light from lightning strike A does (as you can see by considering things from M's frame). So if m assumes that both signals traveled at the same speed, and if he also measures both to have happened at the same distance from him, he can only explain this by saying the strike at B happened before the strike at A.Sam Woole said:Einstein continued his cheating by saying:

"Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A."

I believe otherwise. I believe observers taking the railway train as their reference-body cannot come to the conclusion as Einstein did, but rather they will conclude that lightning flashes A and B emitted simultaneously.

I assume "Am" represents the distance from A to m, and "mB" represents the distance from m to B? It is indeed true that m will judge these distances to be equal, and will therefore judge both signals to have taken the sameSam Woole said:My belief can be justified like this:

Since Am = mB, so that Am/c = mB/c = t.

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- #28

vanesch

Staff Emeritus

Science Advisor

Gold Member

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Sam Woole said:I think your deduction process is a lie, even though you obtained the same time interval as 2D/c. If you told your students you would use clocks to obtain the time interval but you did not do so, you lied.

A 'clock' in a thought experiment is a mathematical variable, of course.

If I say: let us imagine I throw a rock down from a 10m high tower, and I calculate what time this rock would take to fall from 10m (in Newtonian physics), are you going to accuse me of lying because I didn't use a rock, but white chalk on the blackboard ?

- #29

Severian596

- 286

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I believe otherwise. I believe observers taking the railway train as their reference-body cannot come to the conclusion as Einstein did, but rather they will conclude that lightning flashes A and B emitted simultaneously.

My belief can be justified like this:

Since Am = mB, so that Am/c = mB/c = t.

Here you justify your belief by claiming that the individual on the train will measure his (or her) distance from event A and event B as the same distance (given by your Am = mB). This assumption is incorrect, and in fact Einstein's illustration hinges on the fact that Am

To me this statement of Einstein's was a big lie. First of all the first word "events" was false or highly contradictory, because, according to Einstein's description of his thought experiment, there were a total of 4 events, two on the train frame and another two on the embankment frame.

There were only TWO events total. Two lightning flashes observed by two observers does not create any new events. Do two lightning flashes observed by 4 observers result in 8 events? No. The events are what they are. If two observers hear a stone hit the pavement, it doesn't mean two stones hit the pavement. So the observer on the train and the observer on the embankment are in fact observing the same events.

Okay, here it goes, an analogy using balls. You and I are at the start of the 100m dash. You can run faster than I can. The starter fires his pistol and we're off and running! A person awaits us at the finish line, and he's got a baseball in his hand. At some point during the race he throws the ball towards us (opposite of our direction of motion). Who will the ball reach first? It will reach you first, because you're faster than me and you're closer to the finish line.

With that in mind, how do you measure WHEN an event takes place? The closest you can come is to measure the time that an event took place by observing when the light from that event reaches you. Prior to the moment when a beam of light that occurred AT THE LOCATION OF THAT EVENT, and at the SAME MOMENT AS THAT EVENT reaches you, you had no knowledge of the event. Your friend with his cell phone standing at the event couldn't call you on his cell phone, because radio waves travel at c, so his call would reach you after the event's flash of light. The fastest that information can travel is c.

How do you synchronize two clocks that are at rest with respect to each other? You put an LED light midway between the two clocks, and tell each clock to turn on when they "see" the LED turn on. The light beam from the LED travels one half of the distance between the clocks and hits each clock "simultaneously", and so the clocks are synchronized.

But what about us in our 100m dash? Let's dumb-down our ability to measure the time of event (the thrown baseball). Let's assume that we can't SEE the person throw the ball, let's say we can only feel the wind from the ball fly by our face. You end up measuring the time that the event occurred (the ball flew by your face) when you were at the 60 meter mark, I measure the time when I was at the 40 meter mark. It took you t seconds to reach the 60 meter mark and that's when you measure the event. It took me t seconds to reach some mark, and y more seconds to reach the 40 meter mark when I felt the ball go by. Here are the times when we measured the event of the thrown baseball:

You measured event time = t

I measured event time = t + y

The event was not simultaneous for us. According to you, the ball was thrown sooner (after less time). The lightning bolt analogy is the EXACT same idea, just with two balls traveling at the speed of light, one from the finish line and one from the starting line. Two events! Two observers traveling at different speeds, different measured times for the events.

- #30

JesseM

Science Advisor

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Actually, in this thought-experiment Am does equal mB. Suppose the observer on the ruler is carrying a ruler, and he's sitting on the "0" marking of the ruler. If the observer on the embankment sees the lightning flash at B as occurring next to marking "x" on the train-observer's ruler, won't he see lightning flash A occurring next to marking "-x" on the train-observer's ruler? After all, the train-observer's ruler shrinks symmetrically, and you don't have to worry about how his clocks are synchronized when checking which markings the flashes occur next to.Severian596 said:Here you justify your belief by claiming that the individual on the train will measure his (or her) distance from event A and event B as the same distance (given by your Am = mB). This assumption is incorrect, and in fact Einstein's illustration hinges on the fact that Amdoes NOT equalmB as far as the individual on the train is concerned. Keep reading and I'll explain why with (yet another) analogy.

You can check this using the Lorentz transform. Say that in the embankment-observer's frame, flash A happens at x = -10 light seconds, t = 0 seconds, while flash B happens at x = 10 light-seconds, t = 0. If the train is moving at 0.6c, then you can use the Lorentz transform to see the coordinates of these events in the train's frame:

x' = 1.25(x - 0.6c*t)

t' = 1.25(t - 0.6c*x/c^2)

Plugging in the coordinates of flash A, you get x' = -12.5 light seconds, t' = 7.5 seconds; plugging in the coordinates of flash B, you get x' = 12.5 light seconds, t' = -7.5 seconds. So, both flashes do occur 12.5 light-seconds away from the center of the coordinate system where the train-observer is sitting, although they occur at different times, 15 seconds apart. Since they both occur 12.5 light-seconds away, the light from both flashes will take 12.5 seconds to reach the train-observer in his own frame, so Sam Woole's equation is correct if t is understood to be a time-interval rather than an actual time-coordinate; in the train-observer's frame the flashes reach him at different times because they happened at different times, not because the time-interval between the flash happening and the light reaching him is different in the two cases.

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- #31

Severian596

- 286

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Thanks very much for the correction.

- #32

Sam Woole

- 55

- 0

vanesch said:A 'clock' in a thought experiment is a mathematical variable, of course.

If I say: let us imagine I throw a rock down from a 10m high tower, and I calculate what time this rock would take to fall from 10m (in Newtonian physics), are you going to accuse me of lying because I didn't use a rock, but white chalk on the blackboard ?

Of course I could not accuse you so. If you have said you will use a clock to measure the drop (a departure time) and fall (arrival time) of the rock, and you did not use the clock, I and anybody else could accuse you of lying. Furthermore, if you tell your teacher you will use clock A to measure, and then you used clock B, and in your report to the teacher you wrote down clock A, I believe you lied, even though the two clocks were identical.

- #33

JesseM

Science Advisor

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I looked it over, another possible confusion is that in your baseball analogy you seem to be mixing up the time that an event happens in a given observer's coordinate system with the time he receives aSeverian596 said:

Thanks very much for the correction.

- #34

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- 1,923

in post #17 (just after your transcription of the text you first referenced),

I responded with

Do you agree that this is a clock? and that the clock just measured a time of "1 tick"? Note: I am making no explicit references to the separation d.

- #35

Sam Woole

- 55

- 0

Let me apologize again for the Figure 2 I posted. It does not look like I wanted it, and as a result my language might be misunderstood. Let me try again and see whether it would keep the form I wanted.

Figure 2:

... A __________m__________B

A __________M__________B.

which shows the train has moved a distance to the right; it no longer coincides with the embankment.

Figure 2:

... A __________m__________B

A __________M__________B.

which shows the train has moved a distance to the right; it no longer coincides with the embankment.

Severian596 said:Hi Sam, I have followed your discussion thus far and maybe I can help. Maybe not! But maybe...I'll quote your reply to Einstein.

I thank you for coming to help a novice.

Here you justify your belief by claiming that the individual on the train will measure his (or her) distance from event A and event B as the same distance (given by your Am = mB). This assumption is incorrect, and in fact Einstein's illustration hinges on the fact that Amdoes NOT equalmB as far as the individual on the train is concerned. Keep reading and I'll explain why with (yet another) analogy.

I think you were wrong. First, JesseM has shown by higher math that they do equal to each other. Second, Einstein's words were: "Let m be the mid-point of the distance A B on the traveling train." Please note, Einstein needed the midpoint, Am = mb, so as to deduce whether the departures of light were simultaneous. If Am does not equal mB, how can he deduce?

There were only TWO events total. Two lightning flashes observed by two observers does not create any new events. Do two lightning flashes observed by 4 observers result in 8 events? No. The events are what they are. If two observers hear a stone hit the pavement, it doesn't mean two stones hit the pavement. So the observer on the train and the observer on the embankment are in fact observing the same events.

Here my Figure 2 will play a part. The foregoing words of yours changed my figure 2 into a new one as follows:

Figure 2:

... __________m__________

A __________M__________B.

where observer m on the train frame would use the two events A and B in the embankment frame to deduce whether the events A and B are simultaneous. I do not believe any deduction from such a situation would make sense.

Furthermore, Einstein expressly said: "People traveling in this train will with advantage use the train as a rigid reference-body (co-ordinate system); they regard all events in reference to the train. Then every event which takes place along the line also takes place at a particular point of the train." Observers cannot jump fences.

I have read another demonstration online where the author used meteors instead of lightnings used by Einstein. He said two meteors would strike simultaneously on both ends of the train and the embankment resulting in two damages on both ends of the train, for the observers on the train to see; and two damages on the embankment for the observers on the ground to see. This demonstration means, other people were understanding Einstein's thought experiment in the same way like me, 4 events, not 2.

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