where is the distance?

[Sisyphus]

True, I didn't realize the Y axis was time. That doesn't make any sense.

[michel123456]

Is that the arrow I putted on the Y axis that makes the diagram flawed?

 

B' is in the past. I cannot change that.

The observation of B from A is confusing. in fact, a ray of light goes from B' to A. So the Arrow on the Y axis is wrong. O.K.? or is there anything else?

Edited January 6, 2010 by michel123456

[Sisyphus]

Ok. Since B hasn't moved relative to A between time 2 and time 0, there is nothing to calculate. The distance is observed as 6*10^8m, and it is 6*10^8m.

[michel123456]

I don't understand why you're calling it a "space part" and a "time part." That is just needlessly confusing. It's a calculation of spatial distance, full stop. Yes, there is a difference between the observed location and the present location. So what?

 

Well, maybe (surely) I am the one confusing things. If you read again the thread from the beginning, you may notice I am not talking about moving objects. The difference is not due to different locations. It is due to the time light takes to travel. When I take examples as stars or distant objects, I understand it is confusing. But I am only examining objects that don't move in regard to each other. Like a cup of tea standing at the edge of a table 600.000 km long. I have to completely rethink my last diagram I guess, and change the presentation. Working on it. Waiting for further comments.

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discarding the precedent diagram

[Sisyphus]

There isn't a difference, though. I made that comment when I thought both axes were spatial.

[michel123456]

Here a new one:

 

 

There is a very long table top, going from A to B, we want to measure. We are standing at point A. We putted a laser at point B, and we will measure the time elapsed for the beam until it reaches us in order to calculate the length of the table.

At time zero, we are at point A, the laser emits a light beam from B. The light beam travels through space & time and reaches us 2 seconds later. We are now at point A' (same location in space, 2 seconds later), the edge of the table is at point B' (same location than B, 2 seconds later). What have we measured?

I guess the A' B diagonal.

Something we could say as "6*10^8meters and 2 seconds"

[insane_alien]

you've measered both AB and BB' (or AA' or whatever)

 

you made two measurements, not one. therefore you have both the distance and the time it took to travel that distance.

[Sisyphus]

The diagonal is just an artifact of how you drew your graph, which is arbitrary. Just because you happen to draw it so that a second and a meter are represented by the same distance on a piece of paper (in orthogonal directions), doesn't mean they are directly comparable, let alone equal.

 

So yes, you are measuring two things. Time in seconds, and then distance in meters.

[jajrussel]

The diagram does not agree with your statement. The diagram says that no time elapsed on the first portion of the lasers journey. The diagram is not logical.

[michel123456]

So when I measure a distance, I measure 2 things?

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The diagram does not agree with your statement. The diagram says that no time elapsed on the first portion of the lasers journey. The diagram is not logical.

 

I don't think so.

 

I don't know how to explain my thoughts.

The laser beam is an example. If you replace the laser beam with a solid steel beam, you should obtain the same result, because inside the beam, the matter that makes the beam, cannot interact with itself faster than C. So if you had to measure the steel beam, you would measure the diagonal.

Edited January 6, 2010 by michel123456
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[insane_alien]

no, when you measure difference you measure one thing. but in the scenario you described you made two measurements.

[Sisyphus]

So when I measure a distance, I measure 2 things?

 

In this particular experiment, yes. Not as a rule. Here you're calculating distance based on known velocity and measured time.

[michel123456]

Is that too confusing?

 

In the diagram, the only measurement is time. Speed is known. Distance is calculated on the basis of speed by time.

The question is: what have we measured?

[Sisyphus]

Is that too confusing?

 

In the diagram, the only measurement is time. Speed is known. Distance is calculated on the basis of speed by time.

The question is: what have we measured?

 

Didn't you just say what you measured?

[michel123456]

Look: if you replace time and put distance instead, you obtain a simple spatial diagram, a regular euclidian triangle. If I asked you what have you measured, surely you would agree it is the hypothenuse. The trouble comes when time is inserted. And if I asked you: can we replace time with distance?, you surely would have answered yes. It is what we do on a Minkowski diagram.

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I can replace the 2 seconds with 6*10^8m.

[Sisyphus]

No you wouldn't be measuring the hypotenuse. To make an analogy where you replace the time in that graph with another axis of distance:

 

You want to know the length of AB. You have a physical wedge of a known angle*, angle A-B-A1. You lay it down across AB, pointy side at A. You measure the vertical height of the wedge at B, i.e. length BB1. What you have then is angle-angle-side, and from that you can calculate length AB.

 

However, thank you for finally stating what you're trying to get at, Minkowski diagrams. That's not going to make any sense without general relativity, and there's not going to be any point if you're only dealing with one reference frame.

 

*Or to more closely parallel velocity as distance/time, "height gained per unit width"

[jajrussel]

The diagram is not logical. Your statement was that it took 2 seconds for the light to get there, never mind the physics. The diagram does not show those two seconds. You can not make a logical statement that agrees with the diagram. Use light, or use a stick. Any measure is an agreement of two measures. You can not state that one is correct if the other does not agree. Your statement also alludes that I can physically measure the distance to something that is not there. I haven't figured out how to do this using light, or a stick. How do you bounce light off of something that is not there? How do I measure it with a stick?

[michel123456]

No you wouldn't be measuring the hypotenuse. To make an analogy where you replace the time in that graph with another axis of distance:

 

You want to know the length of AB. You have a physical wedge of a known angle*, angle A-B-A1. You lay it down across AB, pointy side at A. You measure the vertical height of the wedge at B, i.e. length BB1. What you have then is angle-angle-side, and from that you can calculate length AB.

 

However, thank you for finally stating what you're trying to get at, Minkowski diagrams. That's not going to make any sense without general relativity, and there's not going to be any point if you're only dealing with one reference frame.

 

*Or to more closely parallel velocity as distance/time, "height gained per unit width"

 

Sorry i can't get it. If I replace 2 seconds with 2*10^8m, the "physical wedge of a known angle" is 45 degrees, as in a Minkowski diagram. And surely I want to get the AB distance, and I know how to get it geometrically. My question is why am I so sure that the measurement I made IS the AB distance, and not the hypotenuse. I repeat myself, if we were not talking about time, SOL, and all that stuff, it would be a very simple triangle as we learn at school.

 

So, step by step.

At the beginning, we are at point A.

At the end of the measurement, we are at point A' (as if we moved a distance of 600.000 km). What are we measuring? My answer is: the hypotenuse.

Is that so wrong?

[Sisyphus]

Again: The measurement you make is the "distance" (time elapsed in the original, or vertical height in the analogy) AA1. You then calculate AB.

[jajrussel]

No, you are not measuring a hypotenuse, You have formed a equilateral triangle. You have measured the side of a equilateral triangle. Also, which the diagram does not show.

[michel123456]

Again: The measurement you make is the "distance" (time elapsed in the original, or vertical height in the analogy) AA1.

 

Agree.

 

You then calculate AB.

 

THAT is the question. If i don't know what I measured, I will make a mistake.

..........

I know I look stubborn.

And i know the possible influence of my question (and your answer) on other fields of physics. I know also that there is 99,99% of chance being wrong, but I'd like to put it clearly into my head. Till now, I still think we are measuring the diagonal.

[jajrussel]

As it stands the only thing you have measured is how far A has moved.

[michel123456]

No, you are not measuring a hypotenuse, You have formed a equilateral triangle. You have measured the side of a equilateral triangle. Also, which the diagram does not show.

 

YES. Good point. I can understand that. i suppose you meant isosceles, not equilateral.

 

But even in this case, light traveled the diagonal. No?

[jajrussel]

I am certain that I know what I meant. I do not agree that light followed the diagonal. The diagonal you show does not agree with your statement. The only indication I see of movement is an implication. Place B where it belongs on the diagram, and see the truth, or leave it where it is and ignore the truth.

[michel123456]

I am certain that I know what I meant. I do not agree that light followed the diagonal. The diagonal you show does not agree with your statement. The only indication I see of movement is an implication. Place B where it belongs on the diagram, and see the truth, or leave it where it is and ignore the truth.

 

I am not playing a game. I am not a priest either. Show me where is B, and I'll stop. Apologizing if you wish.