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Just use the relativistic formula for addition of velocities. If you don't know it, look at a book.

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Ok i looked at the book and got an answer of 0.9944c(approx).Just use the relativistic formula for addition of velocities. If you don't know it, look at a book.

How can i justify this with the relative displacement after one second?

Doc Al

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What do you mean? According to A, C moves 0.9944c*1s closer in one second.How can i justify this with the relative displacement after one second?

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Sorry i din't mean it.Initally let all of them have same x co-ordinates =0; According to A,what will be the positions of B & C in after 1 second in A's frame?What do you mean? According to A, C moves 0.9944c*1s closer in one second.

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HallsofIvy

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Doc Al

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According to A, B has a speed of -.9c. So B will be at x = -.9c*1sec (~ -2.7 10^8 m) after one second. Similarly, C will be at position x = -.9944c*1sec (~ -3 10^8 m).Initally let all the points have same x co-ordinates =0; According to A,what will be the positions of B & C in after 1 second in A's frame?

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Is there any length contraction to be considered or not?According to A, B has a speed of -.9c. So B will be at x = -.9c*1sec (~ -2.7 10^8 m) after one second. Similarly, C will be at position x = -.9944c*1sec (~ -3 10^8 m).

Doc Al

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Each observer will view the "metersticks" used by the others to be contracted. But you don't need to explicitly use that fact to figure out the coordinates of each at a given time, since you are not changing reference frames.Is there any length contraction to be considered or not?

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Ok what about if B and C are long rods with length 'c' in their frames?Each observer will view the "metersticks" used by the others to be contracted. But you don't need to explicitly use that fact to figure out the coordinates of each at a given time, since you are not changing reference frames.

Doc Al

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If B and C carry long rods (held parallel to the direction of motion) that have length [itex]L_0[/itex] in their own frames, then the length of those rodsOk what about if B and C are long rods with length 'c' in their frames?

[tex]L = L_0/\gamma[/tex]

Where:

[tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]

And v is the relative velocity of B or C with respect to A (discussed earlier).

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This shortening will be from one end or it would be symmetrical from centre of Rod?If B and C carry long rods (held parallel to the direction of motion) that have length [itex]L_0[/itex] in their own frames, then the length of those rodsas measured by Awill be:

[tex]L = L_0/\gamma[/tex]

Where:

[tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]

And v is the relative velocity of B or C with respect to A (discussed earlier).

Doc Al

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I don't understand what you mean. All segments of the rod will be uniformly "shortened".This shortening will be from one end or it would be symmetrical from centre of Rod?

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I think he's asking if the rod will also shrink in dimensions not parallel to the direction of travel.I don't understand what you mean. All segments of the rod will be uniformly "shortened".

Regards,

Bill

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Ok.i will come to the point.Consider A,B and C are at rest to each other.B & C are each carrying a rod of length 'L'.This rods are made of 'n' number of molecules in x direction.Now let us say the spacing from centre to centre of molecules are 's'.Let us mark it on C's Rod.I don't understand what you mean. All segments of the rod will be uniformly "shortened".

Now B,with the rod goes for a trip in x direction.He came back to cross his initial position with a relative velocity of 0.9c(in x-direction).The rod now is contracted as per A.Does the spacing between the molecules of the rod in B's hand change,with respect to the spacing marked on C's rod(in A's frame).?

If so,can we say the space itself is contracted ?

Doc Al

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I don't quite understand what you're asking, especially in that last sentence. (Not sure what you mean by "spacing marked on C's rod in A's frame".)Ok.i will come to the point.Consider A,B and C are at rest to each other.B & C are each carrying a rod of length 'L'.This rods are made of 'n' number of molecules in x direction.Now let us say the spacing from centre to centre of molecules are 's'.Let us mark it on C's Rod.

Now B,with the rod goes for a trip in x direction.He came back to cross his initial position with a relative velocity of 0.9c(in x-direction).The rod now is contracted as per A.Does the spacing between the molecules of the rod in B's hand change,with respect to the spacing marked on C's rod(in A's frame).?

In any case, I'll rephrase what I already said. You can imagine each rod as having markings every meter (or other interval of your choice). According to frame A, the markings on B and C are closer together by the factor given in post #11. (Similar statements can be made about measurements according to any of the three frames.)

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Actualy i mean the centre to centre distance of molecules(or atoms of the rod)I don't quite understand what you're asking, especially in that last sentence. (Not sure what you mean by "spacing marked on C's rod in A's frame".)

In any case, I'll rephrase what I already said. You can imagine each rod as having markings every meter (or other interval of your choice). According to frame A, the markings on B and C are closer together by the factor given in post #11. (Similar statements can be made about measurements according to any of the three frames.)

I will try to explain my question with another example.A,B,C and D are four persons.Initialy they are all at rest with respect to each other.

B,Cand D are in a line in x direction.

C is in the middle of B and D

C is holding a long bar whose end are just touching B and D.

Now this B,C and D started a Journey in x direction with same velocity(they are at rest with respect to each other).

They all came back in the x direction to cross A with 0.9c velocity with respect to the stationary observer A.

Do A see the ends of the rod still touching B and D or not?

Doc Al

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It all depends on how B, C, and D accelerate to reach their final speeds. If they do it in such a way as to maintain a constant proper length between themselves, then their distances will be uniformly Lorentz contracted as seen by A--and everyone will agree that the ends of the rod still touch B and D. (Note that this is equivalent to having A accelerate to speed 0.9c, while the others remain at rest in their original inertial frame.)Actualy i mean the centre to centre distance of molecules(or atoms of the rod)

I will try to explain my question with another example.A,B,C and D are four persons.Initialy they are all at rest with respect to each other.

B,Cand D are in a line in x direction.

C is in the middle of B and D

C is holding a long bar whose end are just touching B and D.

Now this B,C and D started a Journey in x direction with same velocity(they are at rest with respect to each other).

They all came back in the x direction to cross A with 0.9c velocity with respect to the stationary observer A.

Do A see the ends of the rod still touching B and D or not?

Whether the ends of the rod still touch B and D (or not) will be a fact that all observers will agree upon.

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In fact the space itself is contracted isn't it?.It all depends on how B, C, and D accelerate to reach their final speeds. If they do it in such a way as to maintain a constant proper length between themselves, then their distances will be uniformly Lorentz contracted as seen by A--and everyone will agree that the ends of the rod still touch B and D. (Note that this is equivalent to having A accelerate to speed 0.9c, while the others remain at rest in their original inertial frame.)

Whether the ends of the rod still touch B and D (or not) will be a fact that all observers will agree upon.

If only the rod was contracted,the ends would not have touch B and D.

The contraction of space leads to an ambiguity.Because the formula for velocity(derived from time dilation) doesn't take into consideration of this reduced distance.

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Doc Al

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Yes, all distances are contracted.In fact the space itself is contracted isn't it?.

If only the rod was contracted,the ends would not have touch B and D.

What ambiguity in what formula? Velocity = Distance/Time works just fine as long as you stick to a single frame.The contraction of space leads to an ambiguity.Because the formula for velocity(derived from time dilation) doesn't take into consideration of this reduced distance.

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Ok let us stick to one stationary frame from where the observation are made.Yes, all distances are contracted.

What ambiguity in what formula? Velocity = Distance/Time works just fine as long as you stick to a single frame.

Consider another object moving with a velocity of 0.9c (all movements are in x direction)w.r.t the frame.

For this statement to be true, the object,after 1 seconds should be at a distance of 0.9 x (3E+8) =( 2.7E+8)meter.

Since all distances are contracted,the equvalent distance of (2.7E+8) as observed from stationary frame ,after contraction will be(1.305E+8)meter -(using length contraction formula).

This means he is observing the other object moving (1.305E+8)meter in x direction in one second in his clock.

Now if we calculate the velocity with this observed distance,the velocity will be

Please note that,the distance and time here are as observed from the stationary frame.

This is a loop.no escape out of it.

Doc Al

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That's correct. The object will have moved that distance (d = vt)Ok let us stick to one stationary frame from where the observation are made.

Consider another object moving with a velocity of 0.9c (all movements are in x direction)w.r.t the frame.

For this statement to be true, the object,after 1 seconds should be at a distance of 0.9 x (3E+8) =( 2.7E+8)meter.

He's not moving with respect to himself, so why would he use the length contraction formula?? The purpose of the length contraction formula (and other consequences of the Lorentz transformations) is to be able to relate measurements made in one frame to corresponding measurements made in another. In this case, all measurements were made in the same frame, so no such transformations are needed.Since all distances are contracted,the equvalent distance of (2.7E+8) as observed from stationary frame ,after contraction will be(1.305E+8)meter -(using length contraction formula).

Incorrect.This means he is observing the other object moving (1.305E+8)meter in x direction in one second in his clock.

Now if we calculate the velocity with this observed distance,the velocity will be0.435c.

Please note that,the distance and time here are as observed from the stationary frame.

You are trapped in a loop of your own making, I'm afraid.This is a loop.no escape out of it.

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Length contraction, according to Hendrik Lorentz, is the physical phenomenon of a decrease in length detectedHe's not moving with respect to himself, so why would he use the length contraction formula?? The purpose of the length contraction formula (and other consequences of the Lorentz transformations) is to be able to relate measurements made in one frame to corresponding measurements made in another. In this case, all measurements were made in the same frame, so no such transformations are needed.

The above is from Wikipedia.

I am confused.Realy confused.

Please explain me why the observer in rest frame should not use the length contraction formula for observing other Object.Moreover you have no disagrement that all the distance there are also contracted.

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Doc Al

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There's nothing wrong with that quote, though I would strongly recommend getting a real textbook if you want to learn relativity. (Don't try to learn physics via Wiki!)The above is from Wikipedia.

I am confused.Realy confused

Think of it this way. I'm in my stationary frame which extends forever in all directions. I've already got distances mapped out and marked all over the place. In particular, I've got a marker at the x = 2.7E+8 meter position which can detect when something passes by it.

This fast moving object moves past the origin at t = 0 (according to my clock, of course) and 1 second later (again, according to me) it passes the above-mentioned marker. So I deduce that it must have been moving at 0.9c. That's what it

Of course, another frame--such as that of the moving object--will have a different set of measurements.

Doc Al

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If the stationary observer were to measurePlease explain me why the observer in rest frame should not use the length contraction formula for observing other Object.

As measured from a moving frame, not a stationary one.Moreover you have no disagrement that all the distance there are also contracted.

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