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When to apply Lorentz transformations and laws of time dilations and length contractions: explanations
The magnetic field near a charging capacitorHelp Me Gain an Intuitive Understanding of Lorentz ContractionTime dilation and the speed of light in the twin paradoxLength Contraction ConfusionWhat would a nuclear explosion look like at relativistic speeds where time dilation and length contraction become significant?Homogeneity and isotropy and derivation of the Lorentz transformationsIs proper time and proper length relative or absolute in relation to two events?Clarification of Conclusion of Length Contraction (Without Lorentz Transformations)Relativity tangential light clockReconciling Length Contraction and Lorentz Transformation Interpretations of a Special Relativity ProblemLorentz Transformation Exercise confusion
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$begingroup$
In the laboratory we are observing the motion of a particle moving in the positive direction of the axis
$x$ of a frame system attached to it. At the initial moment, the particle is in the origin and in a time interval of $2,0$ ns covers a distance of $25$ cm. A spacecraft passes with velocity $v = 0,80$ c along the $x$ direction of the laboratory, in the positive direction, and from it is observed the motion of the same particle. Determine the average particle speeds in the two systems of reference. What interval of time and what distance would an observer placed on the spaceship?
Why I can not to use the formulas of time expansion and length contraction and I must applicate, necessary, Lorentz transformations?
homework-and-exercises special-relativity lorentz-symmetry
asked 8 hours ago
SebastianoSebastiano
3652 silver badges20 bronze badges
$endgroup$
add a comment |
$begingroup$
In the laboratory we are observing the motion of a particle moving in the positive direction of the axis
$x$ of a frame system attached to it. At the initial moment, the particle is in the origin and in a time interval of $2,0$ ns covers a distance of $25$ cm. A spacecraft passes with velocity $v = 0,80$ c along the $x$ direction of the laboratory, in the positive direction, and from it is observed the motion of the same particle. Determine the average particle speeds in the two systems of reference. What interval of time and what distance would an observer placed on the spaceship?
Why I can not to use the formulas of time expansion and length contraction and I must applicate, necessary, Lorentz transformations?
homework-and-exercises special-relativity lorentz-symmetry
asked 8 hours ago
SebastianoSebastiano
3652 silver badges20 bronze badges
$endgroup$
1
$begingroup$
I've added the homework-and-exercises tag. In the future, please use this tag on this type of question.
$endgroup$
– Ben Crowell
8 hours ago
$begingroup$
@BenCrowell Thank you very much. I sincerely do not attend this site also because of a bad appreciation of my questions. I put my soul to write them better. Can you see my previous question: physics.stackexchange.com/questions/487680/…. I use the translator. If there are unclear parts you are allowed to change what you want. Also for any errors.
$endgroup$
– Sebastiano
8 hours ago
$begingroup$
@G.Smith I am referring to the text of the exercise. I am not interested in what is going on. Rather because I have to use Lorentz's formulas instead of the classic ones of time dilation and length contraction. If my question is not clear you can edit it according to my comments.
$endgroup$
– Sebastiano
7 hours ago
$begingroup$
@G.Smith What should I write or change to make the question clearer? I use the translator.
$endgroup$
– Sebastiano
7 hours ago
$begingroup$
@G.Smith I have edited my question. I had only copied the text with a normal font the exercise and I inserted in bold my doubt, that is, from the exercise there is some term that tells me that I must necessarily apply the transformations of Lorentz?
$endgroup$
– Sebastiano
7 hours ago
add a comment |
$begingroup$
In the laboratory we are observing the motion of a particle moving in the positive direction of the axis
$x$ of a frame system attached to it. At the initial moment, the particle is in the origin and in a time interval of $2,0$ ns covers a distance of $25$ cm. A spacecraft passes with velocity $v = 0,80$ c along the $x$ direction of the laboratory, in the positive direction, and from it is observed the motion of the same particle. Determine the average particle speeds in the two systems of reference. What interval of time and what distance would an observer placed on the spaceship?
Why I can not to use the formulas of time expansion and length contraction and I must applicate, necessary, Lorentz transformations?
homework-and-exercises special-relativity lorentz-symmetry
asked 8 hours ago
SebastianoSebastiano
3652 silver badges20 bronze badges
$endgroup$
In the laboratory we are observing the motion of a particle moving in the positive direction of the axis
$x$ of a frame system attached to it. At the initial moment, the particle is in the origin and in a time interval of $2,0$ ns covers a distance of $25$ cm. A spacecraft passes with velocity $v = 0,80$ c along the $x$ direction of the laboratory, in the positive direction, and from it is observed the motion of the same particle. Determine the average particle speeds in the two systems of reference. What interval of time and what distance would an observer placed on the spaceship?
Why I can not to use the formulas of time expansion and length contraction and I must applicate, necessary, Lorentz transformations?
homework-and-exercises special-relativity lorentz-symmetry
homework-and-exercises special-relativity lorentz-symmetry
asked 8 hours ago
SebastianoSebastiano
3652 silver badges20 bronze badges
asked 8 hours ago
SebastianoSebastiano
3652 silver badges20 bronze badges
edited 7 hours ago
Sebastiano
asked 8 hours ago
SebastianoSebastiano
3652 silver badges20 bronze badges
asked 8 hours ago
SebastianoSebastiano
3652 silver badges20 bronze badges
asked 8 hours ago
SebastianoSebastiano
3652 silver badges20 bronze badges
3652 silver badges20 bronze badges
1
$begingroup$
I've added the homework-and-exercises tag. In the future, please use this tag on this type of question.
$endgroup$
– Ben Crowell
8 hours ago
$begingroup$
@BenCrowell Thank you very much. I sincerely do not attend this site also because of a bad appreciation of my questions. I put my soul to write them better. Can you see my previous question: physics.stackexchange.com/questions/487680/…. I use the translator. If there are unclear parts you are allowed to change what you want. Also for any errors.
$endgroup$
– Sebastiano
8 hours ago
$begingroup$
@G.Smith I am referring to the text of the exercise. I am not interested in what is going on. Rather because I have to use Lorentz's formulas instead of the classic ones of time dilation and length contraction. If my question is not clear you can edit it according to my comments.
$endgroup$
– Sebastiano
7 hours ago
$begingroup$
@G.Smith What should I write or change to make the question clearer? I use the translator.
$endgroup$
– Sebastiano
7 hours ago
$begingroup$
@G.Smith I have edited my question. I had only copied the text with a normal font the exercise and I inserted in bold my doubt, that is, from the exercise there is some term that tells me that I must necessarily apply the transformations of Lorentz?
$endgroup$
– Sebastiano
7 hours ago
add a comment |
1
$begingroup$
I've added the homework-and-exercises tag. In the future, please use this tag on this type of question.
$endgroup$
– Ben Crowell
8 hours ago
$begingroup$
@BenCrowell Thank you very much. I sincerely do not attend this site also because of a bad appreciation of my questions. I put my soul to write them better. Can you see my previous question: physics.stackexchange.com/questions/487680/…. I use the translator. If there are unclear parts you are allowed to change what you want. Also for any errors.
$endgroup$
– Sebastiano
8 hours ago
$begingroup$
@G.Smith I am referring to the text of the exercise. I am not interested in what is going on. Rather because I have to use Lorentz's formulas instead of the classic ones of time dilation and length contraction. If my question is not clear you can edit it according to my comments.
$endgroup$
– Sebastiano
7 hours ago
$begingroup$
@G.Smith What should I write or change to make the question clearer? I use the translator.
$endgroup$
– Sebastiano
7 hours ago
$begingroup$
@G.Smith I have edited my question. I had only copied the text with a normal font the exercise and I inserted in bold my doubt, that is, from the exercise there is some term that tells me that I must necessarily apply the transformations of Lorentz?
$endgroup$
– Sebastiano
7 hours ago
1
1
$begingroup$
I've added the homework-and-exercises tag. In the future, please use this tag on this type of question.
$endgroup$
– Ben Crowell
8 hours ago
$begingroup$
I've added the homework-and-exercises tag. In the future, please use this tag on this type of question.
$endgroup$
– Ben Crowell
8 hours ago
$begingroup$
@BenCrowell Thank you very much. I sincerely do not attend this site also because of a bad appreciation of my questions. I put my soul to write them better. Can you see my previous question: physics.stackexchange.com/questions/487680/…. I use the translator. If there are unclear parts you are allowed to change what you want. Also for any errors.
$endgroup$
– Sebastiano
8 hours ago
$begingroup$
@BenCrowell Thank you very much. I sincerely do not attend this site also because of a bad appreciation of my questions. I put my soul to write them better. Can you see my previous question: physics.stackexchange.com/questions/487680/…. I use the translator. If there are unclear parts you are allowed to change what you want. Also for any errors.
$endgroup$
– Sebastiano
8 hours ago
$begingroup$
@G.Smith I am referring to the text of the exercise. I am not interested in what is going on. Rather because I have to use Lorentz's formulas instead of the classic ones of time dilation and length contraction. If my question is not clear you can edit it according to my comments.
$endgroup$
– Sebastiano
7 hours ago
$begingroup$
@G.Smith I am referring to the text of the exercise. I am not interested in what is going on. Rather because I have to use Lorentz's formulas instead of the classic ones of time dilation and length contraction. If my question is not clear you can edit it according to my comments.
$endgroup$
– Sebastiano
7 hours ago
$begingroup$
@G.Smith What should I write or change to make the question clearer? I use the translator.
$endgroup$
– Sebastiano
7 hours ago
$begingroup$
@G.Smith What should I write or change to make the question clearer? I use the translator.
$endgroup$
– Sebastiano
7 hours ago
$begingroup$
@G.Smith I have edited my question. I had only copied the text with a normal font the exercise and I inserted in bold my doubt, that is, from the exercise there is some term that tells me that I must necessarily apply the transformations of Lorentz?
$endgroup$
– Sebastiano
7 hours ago
$begingroup$
@G.Smith I have edited my question. I had only copied the text with a normal font the exercise and I inserted in bold my doubt, that is, from the exercise there is some term that tells me that I must necessarily apply the transformations of Lorentz?
$endgroup$
– Sebastiano
7 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let distance be $D=25cm$, you are asking why we can't just use $D'=D/gamma$ (length contraction) and $t'=gamma t$ for $t=2ns$ (time dilation) so that we calculate average speed in spaceship's frame as $u'=D'/t'=u/gamma^2$ where u indicate velocity of particle in the lab frame? If it's your question then it's an easy one, intuitively speaking it's because spaceship moves in $x$ direction as well, same as particle. If spaceship were to move in y axis, you could use time dilation to conclude that speed is $u'_x=u/gamma$ but now that spaceship moves in x direction, you should consider this fact that spaceship will ascribe his speed to particle too!
answered 6 hours ago
ParadoxyParadoxy
2502 silver badges10 bronze badges
$endgroup$
$begingroup$
Surely there is my upvote. Could you please edit your question a little clearer? If a bar is stopped in an inertial $Sigma'$ system with $v$ speed in the direction of $x$ motion, an observer placed in $Sigma'$ sees the bar moving. Here I apply the length contraction formulas with the Lorentz factor $gamma$. Why do you refer to the $y$ direction?
$endgroup$
– Sebastiano
6 hours ago
$begingroup$
@Sebastiano forget y direction. that was just another scenario. if we drop your typos in the comment, maybe you are asking if we were to analyse scenario in particle's frame, we would only need to use length contraction, why it cant be done in the case of spaceship too? As i said it's because we should consider the relative velocity between particle and spaceship, in the case of laboratory and particle we need not to do that because velocity of lab's frame is zero
$endgroup$
– Paradoxy
6 hours ago
add a comment |
$begingroup$
Why I can not to use the formulas of time expansion and length contraction and I must applicate, necessary, Lorentz transformations?
The time dilation and length contraction formulas are derived from the Lorentz transform using some specific assumptions. For time dilation the assumption is that the clock is at rest in one of the frames. For length contraction the assumptions are that the rod is at rest in one of the frames and that the length is constant.
In this problem you cannot use the length contraction or time dilation formulas because the assumptions are not met. The particle is not at rest in either the lab or rocket frames.
As an aside, I recommend that new students of relativity not use the time dilation and length contraction formulas at all. They are too easy to misapply as in this problem. Just use the Lorentz transform, it will automatically simplify when appropriate, and you will avoid misapplying them when they are not appropriate.
answered 5 hours ago
DaleDale
8,0011 gold badge9 silver badges35 bronze badges
$endgroup$
add a comment |
$begingroup$
Well, firstly, think of how the length contraction formula came. We used $∆t'=0$ in the inverse Lorentz transformation, right?
Why did we did that? Because, to measure a length, you have to measure the both end at the same time. But, here the space time coordinate measured at the same time. First coordinate is measured at $t=0$, and second one is at $t=2$ ns. So length contraction is not valid here.
I myself faced the exact same problem at the first days of my learning relativity. I recommend you to go through the derivative of length contraction again.
edited 4 hours ago
Sebastiano
3652 silver badges20 bronze badges
answered 6 hours ago
Sk ShafayatSk Shafayat
492 bronze badges
$endgroup$
$begingroup$
I have not understood: I recommend you to go through the derivative of length contraction again. +1
$endgroup$
– Sebastiano
6 hours ago
$begingroup$
@Sebastiano i think he meant "derivation of length contraction" , yeah i also recommend that.
$endgroup$
– Paradoxy
6 hours ago
$begingroup$
I have vote this....and others :-)
$endgroup$
– Sebastiano
6 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let distance be $D=25cm$, you are asking why we can't just use $D'=D/gamma$ (length contraction) and $t'=gamma t$ for $t=2ns$ (time dilation) so that we calculate average speed in spaceship's frame as $u'=D'/t'=u/gamma^2$ where u indicate velocity of particle in the lab frame? If it's your question then it's an easy one, intuitively speaking it's because spaceship moves in $x$ direction as well, same as particle. If spaceship were to move in y axis, you could use time dilation to conclude that speed is $u'_x=u/gamma$ but now that spaceship moves in x direction, you should consider this fact that spaceship will ascribe his speed to particle too!
answered 6 hours ago
ParadoxyParadoxy
2502 silver badges10 bronze badges
$endgroup$
$begingroup$
Surely there is my upvote. Could you please edit your question a little clearer? If a bar is stopped in an inertial $Sigma'$ system with $v$ speed in the direction of $x$ motion, an observer placed in $Sigma'$ sees the bar moving. Here I apply the length contraction formulas with the Lorentz factor $gamma$. Why do you refer to the $y$ direction?
$endgroup$
– Sebastiano
6 hours ago
$begingroup$
@Sebastiano forget y direction. that was just another scenario. if we drop your typos in the comment, maybe you are asking if we were to analyse scenario in particle's frame, we would only need to use length contraction, why it cant be done in the case of spaceship too? As i said it's because we should consider the relative velocity between particle and spaceship, in the case of laboratory and particle we need not to do that because velocity of lab's frame is zero
$endgroup$
– Paradoxy
6 hours ago
add a comment |
$begingroup$
Let distance be $D=25cm$, you are asking why we can't just use $D'=D/gamma$ (length contraction) and $t'=gamma t$ for $t=2ns$ (time dilation) so that we calculate average speed in spaceship's frame as $u'=D'/t'=u/gamma^2$ where u indicate velocity of particle in the lab frame? If it's your question then it's an easy one, intuitively speaking it's because spaceship moves in $x$ direction as well, same as particle. If spaceship were to move in y axis, you could use time dilation to conclude that speed is $u'_x=u/gamma$ but now that spaceship moves in x direction, you should consider this fact that spaceship will ascribe his speed to particle too!
answered 6 hours ago
ParadoxyParadoxy
2502 silver badges10 bronze badges
$endgroup$
$begingroup$
Surely there is my upvote. Could you please edit your question a little clearer? If a bar is stopped in an inertial $Sigma'$ system with $v$ speed in the direction of $x$ motion, an observer placed in $Sigma'$ sees the bar moving. Here I apply the length contraction formulas with the Lorentz factor $gamma$. Why do you refer to the $y$ direction?
$endgroup$
– Sebastiano
6 hours ago
$begingroup$
@Sebastiano forget y direction. that was just another scenario. if we drop your typos in the comment, maybe you are asking if we were to analyse scenario in particle's frame, we would only need to use length contraction, why it cant be done in the case of spaceship too? As i said it's because we should consider the relative velocity between particle and spaceship, in the case of laboratory and particle we need not to do that because velocity of lab's frame is zero
$endgroup$
– Paradoxy
6 hours ago
add a comment |
$begingroup$
Let distance be $D=25cm$, you are asking why we can't just use $D'=D/gamma$ (length contraction) and $t'=gamma t$ for $t=2ns$ (time dilation) so that we calculate average speed in spaceship's frame as $u'=D'/t'=u/gamma^2$ where u indicate velocity of particle in the lab frame? If it's your question then it's an easy one, intuitively speaking it's because spaceship moves in $x$ direction as well, same as particle. If spaceship were to move in y axis, you could use time dilation to conclude that speed is $u'_x=u/gamma$ but now that spaceship moves in x direction, you should consider this fact that spaceship will ascribe his speed to particle too!
answered 6 hours ago
ParadoxyParadoxy
2502 silver badges10 bronze badges
$endgroup$
Let distance be $D=25cm$, you are asking why we can't just use $D'=D/gamma$ (length contraction) and $t'=gamma t$ for $t=2ns$ (time dilation) so that we calculate average speed in spaceship's frame as $u'=D'/t'=u/gamma^2$ where u indicate velocity of particle in the lab frame? If it's your question then it's an easy one, intuitively speaking it's because spaceship moves in $x$ direction as well, same as particle. If spaceship were to move in y axis, you could use time dilation to conclude that speed is $u'_x=u/gamma$ but now that spaceship moves in x direction, you should consider this fact that spaceship will ascribe his speed to particle too!
answered 6 hours ago
ParadoxyParadoxy
2502 silver badges10 bronze badges
answered 6 hours ago
ParadoxyParadoxy
2502 silver badges10 bronze badges
answered 6 hours ago
ParadoxyParadoxy
2502 silver badges10 bronze badges
answered 6 hours ago
ParadoxyParadoxy
2502 silver badges10 bronze badges
2502 silver badges10 bronze badges
$begingroup$
Surely there is my upvote. Could you please edit your question a little clearer? If a bar is stopped in an inertial $Sigma'$ system with $v$ speed in the direction of $x$ motion, an observer placed in $Sigma'$ sees the bar moving. Here I apply the length contraction formulas with the Lorentz factor $gamma$. Why do you refer to the $y$ direction?
$endgroup$
– Sebastiano
6 hours ago
$begingroup$
@Sebastiano forget y direction. that was just another scenario. if we drop your typos in the comment, maybe you are asking if we were to analyse scenario in particle's frame, we would only need to use length contraction, why it cant be done in the case of spaceship too? As i said it's because we should consider the relative velocity between particle and spaceship, in the case of laboratory and particle we need not to do that because velocity of lab's frame is zero
$endgroup$
– Paradoxy
6 hours ago
add a comment |
$begingroup$
Surely there is my upvote. Could you please edit your question a little clearer? If a bar is stopped in an inertial $Sigma'$ system with $v$ speed in the direction of $x$ motion, an observer placed in $Sigma'$ sees the bar moving. Here I apply the length contraction formulas with the Lorentz factor $gamma$. Why do you refer to the $y$ direction?
$endgroup$
– Sebastiano
6 hours ago
$begingroup$
@Sebastiano forget y direction. that was just another scenario. if we drop your typos in the comment, maybe you are asking if we were to analyse scenario in particle's frame, we would only need to use length contraction, why it cant be done in the case of spaceship too? As i said it's because we should consider the relative velocity between particle and spaceship, in the case of laboratory and particle we need not to do that because velocity of lab's frame is zero
$endgroup$
– Paradoxy
6 hours ago
$begingroup$
Surely there is my upvote. Could you please edit your question a little clearer? If a bar is stopped in an inertial $Sigma'$ system with $v$ speed in the direction of $x$ motion, an observer placed in $Sigma'$ sees the bar moving. Here I apply the length contraction formulas with the Lorentz factor $gamma$. Why do you refer to the $y$ direction?
$endgroup$
– Sebastiano
6 hours ago
$begingroup$
Surely there is my upvote. Could you please edit your question a little clearer? If a bar is stopped in an inertial $Sigma'$ system with $v$ speed in the direction of $x$ motion, an observer placed in $Sigma'$ sees the bar moving. Here I apply the length contraction formulas with the Lorentz factor $gamma$. Why do you refer to the $y$ direction?
$endgroup$
– Sebastiano
6 hours ago
$begingroup$
@Sebastiano forget y direction. that was just another scenario. if we drop your typos in the comment, maybe you are asking if we were to analyse scenario in particle's frame, we would only need to use length contraction, why it cant be done in the case of spaceship too? As i said it's because we should consider the relative velocity between particle and spaceship, in the case of laboratory and particle we need not to do that because velocity of lab's frame is zero
$endgroup$
– Paradoxy
6 hours ago
$begingroup$
@Sebastiano forget y direction. that was just another scenario. if we drop your typos in the comment, maybe you are asking if we were to analyse scenario in particle's frame, we would only need to use length contraction, why it cant be done in the case of spaceship too? As i said it's because we should consider the relative velocity between particle and spaceship, in the case of laboratory and particle we need not to do that because velocity of lab's frame is zero
$endgroup$
– Paradoxy
6 hours ago
add a comment |
$begingroup$
Why I can not to use the formulas of time expansion and length contraction and I must applicate, necessary, Lorentz transformations?
The time dilation and length contraction formulas are derived from the Lorentz transform using some specific assumptions. For time dilation the assumption is that the clock is at rest in one of the frames. For length contraction the assumptions are that the rod is at rest in one of the frames and that the length is constant.
In this problem you cannot use the length contraction or time dilation formulas because the assumptions are not met. The particle is not at rest in either the lab or rocket frames.
As an aside, I recommend that new students of relativity not use the time dilation and length contraction formulas at all. They are too easy to misapply as in this problem. Just use the Lorentz transform, it will automatically simplify when appropriate, and you will avoid misapplying them when they are not appropriate.
answered 5 hours ago
DaleDale
8,0011 gold badge9 silver badges35 bronze badges
$endgroup$
add a comment |
$begingroup$
Why I can not to use the formulas of time expansion and length contraction and I must applicate, necessary, Lorentz transformations?
The time dilation and length contraction formulas are derived from the Lorentz transform using some specific assumptions. For time dilation the assumption is that the clock is at rest in one of the frames. For length contraction the assumptions are that the rod is at rest in one of the frames and that the length is constant.
In this problem you cannot use the length contraction or time dilation formulas because the assumptions are not met. The particle is not at rest in either the lab or rocket frames.
As an aside, I recommend that new students of relativity not use the time dilation and length contraction formulas at all. They are too easy to misapply as in this problem. Just use the Lorentz transform, it will automatically simplify when appropriate, and you will avoid misapplying them when they are not appropriate.
answered 5 hours ago
DaleDale
8,0011 gold badge9 silver badges35 bronze badges
$endgroup$
add a comment |
$begingroup$
Why I can not to use the formulas of time expansion and length contraction and I must applicate, necessary, Lorentz transformations?
The time dilation and length contraction formulas are derived from the Lorentz transform using some specific assumptions. For time dilation the assumption is that the clock is at rest in one of the frames. For length contraction the assumptions are that the rod is at rest in one of the frames and that the length is constant.
In this problem you cannot use the length contraction or time dilation formulas because the assumptions are not met. The particle is not at rest in either the lab or rocket frames.
As an aside, I recommend that new students of relativity not use the time dilation and length contraction formulas at all. They are too easy to misapply as in this problem. Just use the Lorentz transform, it will automatically simplify when appropriate, and you will avoid misapplying them when they are not appropriate.
answered 5 hours ago
DaleDale
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Why I can not to use the formulas of time expansion and length contraction and I must applicate, necessary, Lorentz transformations?
The time dilation and length contraction formulas are derived from the Lorentz transform using some specific assumptions. For time dilation the assumption is that the clock is at rest in one of the frames. For length contraction the assumptions are that the rod is at rest in one of the frames and that the length is constant.
In this problem you cannot use the length contraction or time dilation formulas because the assumptions are not met. The particle is not at rest in either the lab or rocket frames.
As an aside, I recommend that new students of relativity not use the time dilation and length contraction formulas at all. They are too easy to misapply as in this problem. Just use the Lorentz transform, it will automatically simplify when appropriate, and you will avoid misapplying them when they are not appropriate.
answered 5 hours ago
DaleDale
8,0011 gold badge9 silver badges35 bronze badges
answered 5 hours ago
DaleDale
8,0011 gold badge9 silver badges35 bronze badges
answered 5 hours ago
DaleDale
8,0011 gold badge9 silver badges35 bronze badges
answered 5 hours ago
DaleDale
8,0011 gold badge9 silver badges35 bronze badges
8,0011 gold badge9 silver badges35 bronze badges
add a comment |
add a comment |
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Well, firstly, think of how the length contraction formula came. We used $∆t'=0$ in the inverse Lorentz transformation, right?
Why did we did that? Because, to measure a length, you have to measure the both end at the same time. But, here the space time coordinate measured at the same time. First coordinate is measured at $t=0$, and second one is at $t=2$ ns. So length contraction is not valid here.
I myself faced the exact same problem at the first days of my learning relativity. I recommend you to go through the derivative of length contraction again.
edited 4 hours ago
Sebastiano
3652 silver badges20 bronze badges
answered 6 hours ago
Sk ShafayatSk Shafayat
492 bronze badges
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I have not understood: I recommend you to go through the derivative of length contraction again. +1
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– Sebastiano
6 hours ago
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@Sebastiano i think he meant "derivation of length contraction" , yeah i also recommend that.
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– Paradoxy
6 hours ago
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I have vote this....and others :-)
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– Sebastiano
6 hours ago
add a comment |
$begingroup$
Well, firstly, think of how the length contraction formula came. We used $∆t'=0$ in the inverse Lorentz transformation, right?
Why did we did that? Because, to measure a length, you have to measure the both end at the same time. But, here the space time coordinate measured at the same time. First coordinate is measured at $t=0$, and second one is at $t=2$ ns. So length contraction is not valid here.
I myself faced the exact same problem at the first days of my learning relativity. I recommend you to go through the derivative of length contraction again.
edited 4 hours ago
Sebastiano
3652 silver badges20 bronze badges
answered 6 hours ago
Sk ShafayatSk Shafayat
492 bronze badges
$endgroup$
$begingroup$
I have not understood: I recommend you to go through the derivative of length contraction again. +1
$endgroup$
– Sebastiano
6 hours ago
$begingroup$
@Sebastiano i think he meant "derivation of length contraction" , yeah i also recommend that.
$endgroup$
– Paradoxy
6 hours ago
$begingroup$
I have vote this....and others :-)
$endgroup$
– Sebastiano
6 hours ago
add a comment |
$begingroup$
Well, firstly, think of how the length contraction formula came. We used $∆t'=0$ in the inverse Lorentz transformation, right?
Why did we did that? Because, to measure a length, you have to measure the both end at the same time. But, here the space time coordinate measured at the same time. First coordinate is measured at $t=0$, and second one is at $t=2$ ns. So length contraction is not valid here.
I myself faced the exact same problem at the first days of my learning relativity. I recommend you to go through the derivative of length contraction again.
edited 4 hours ago
Sebastiano
3652 silver badges20 bronze badges
answered 6 hours ago
Sk ShafayatSk Shafayat
492 bronze badges
$endgroup$
Well, firstly, think of how the length contraction formula came. We used $∆t'=0$ in the inverse Lorentz transformation, right?
Why did we did that? Because, to measure a length, you have to measure the both end at the same time. But, here the space time coordinate measured at the same time. First coordinate is measured at $t=0$, and second one is at $t=2$ ns. So length contraction is not valid here.
I myself faced the exact same problem at the first days of my learning relativity. I recommend you to go through the derivative of length contraction again.
edited 4 hours ago
Sebastiano
3652 silver badges20 bronze badges
answered 6 hours ago
Sk ShafayatSk Shafayat
492 bronze badges
edited 4 hours ago
Sebastiano
3652 silver badges20 bronze badges
edited 4 hours ago
Sebastiano
3652 silver badges20 bronze badges
edited 4 hours ago
Sebastiano
3652 silver badges20 bronze badges
3652 silver badges20 bronze badges
answered 6 hours ago
Sk ShafayatSk Shafayat
492 bronze badges
answered 6 hours ago
Sk ShafayatSk Shafayat
492 bronze badges
answered 6 hours ago
Sk ShafayatSk Shafayat
492 bronze badges
492 bronze badges
$begingroup$
I have not understood: I recommend you to go through the derivative of length contraction again. +1
$endgroup$
– Sebastiano
6 hours ago
$begingroup$
@Sebastiano i think he meant "derivation of length contraction" , yeah i also recommend that.
$endgroup$
– Paradoxy
6 hours ago
$begingroup$
I have vote this....and others :-)
$endgroup$
– Sebastiano
6 hours ago
add a comment |
$begingroup$
I have not understood: I recommend you to go through the derivative of length contraction again. +1
$endgroup$
– Sebastiano
6 hours ago
$begingroup$
@Sebastiano i think he meant "derivation of length contraction" , yeah i also recommend that.
$endgroup$
– Paradoxy
6 hours ago
$begingroup$
I have vote this....and others :-)
$endgroup$
– Sebastiano
6 hours ago
$begingroup$
I have not understood: I recommend you to go through the derivative of length contraction again. +1
$endgroup$
– Sebastiano
6 hours ago
$begingroup$
I have not understood: I recommend you to go through the derivative of length contraction again. +1
$endgroup$
– Sebastiano
6 hours ago
$begingroup$
@Sebastiano i think he meant "derivation of length contraction" , yeah i also recommend that.
$endgroup$
– Paradoxy
6 hours ago
$begingroup$
@Sebastiano i think he meant "derivation of length contraction" , yeah i also recommend that.
$endgroup$
– Paradoxy
6 hours ago
$begingroup$
I have vote this....and others :-)
$endgroup$
– Sebastiano
6 hours ago
$begingroup$
I have vote this....and others :-)
$endgroup$
– Sebastiano
6 hours ago
add a comment |
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I've added the homework-and-exercises tag. In the future, please use this tag on this type of question.
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– Ben Crowell
8 hours ago
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@BenCrowell Thank you very much. I sincerely do not attend this site also because of a bad appreciation of my questions. I put my soul to write them better. Can you see my previous question: physics.stackexchange.com/questions/487680/…. I use the translator. If there are unclear parts you are allowed to change what you want. Also for any errors.
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– Sebastiano
8 hours ago
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@G.Smith I am referring to the text of the exercise. I am not interested in what is going on. Rather because I have to use Lorentz's formulas instead of the classic ones of time dilation and length contraction. If my question is not clear you can edit it according to my comments.
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– Sebastiano
7 hours ago
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@G.Smith What should I write or change to make the question clearer? I use the translator.
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– Sebastiano
7 hours ago
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@G.Smith I have edited my question. I had only copied the text with a normal font the exercise and I inserted in bold my doubt, that is, from the exercise there is some term that tells me that I must necessarily apply the transformations of Lorentz?
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– Sebastiano
7 hours ago