What is physically different about a moving vs still object in space?












4














If I have two asteroids. One dead still in space and one whizzing by at 10,000mph. What is the difference between the two, physically?



If I freeze time and look at the two of them - what differences would they have? How could one tell that one was moving really quickly, and the other not?



enter image description here



Is there some sort of Quantum difference with the particles in front or behind the asteroid?



Does it have a different gravitational force on spacetime surrounding it?



enter image description here



No images online seem to suggest this as Earth moves quickly through space but has a simple, aligned bump beneath.



I'm just a programmer who is interested in Physics & has been watching too many Leonard Susskind lectures recently.










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  • 1




    Why do you expect there to be a difference? Physics is the same in all frames moving at constant speed with respect to each other, so you're free to look at the moving object in its rest frame.
    – jacob1729
    4 hours ago










  • @jacob1729 So there would be no way to tell that one was moving and the other wasn't if you froze time and tried to measure them both? There are zero physical differences? (I'm not saying the Asteroid itself will look or be different, but either, the particles surrounding them / spacetime / vacuum energy?)
    – Jack Nicholson
    4 hours ago












  • There is no absolute frame of reference which determines that some objects are still and some are moving. You can only talk about things moving relative to some other thing (like in S. McGrew's answer). This isn't just a feature of Einstein's relativity, it goes back to Galileo.
    – PM 2Ring
    3 hours ago








  • 1




    I think the comments so far have missed the point of the question. There is no physical reason why two things should be the same or look the same when both are observed in the same reference frame.
    – Andrew Steane
    2 hours ago
















4














If I have two asteroids. One dead still in space and one whizzing by at 10,000mph. What is the difference between the two, physically?



If I freeze time and look at the two of them - what differences would they have? How could one tell that one was moving really quickly, and the other not?



enter image description here



Is there some sort of Quantum difference with the particles in front or behind the asteroid?



Does it have a different gravitational force on spacetime surrounding it?



enter image description here



No images online seem to suggest this as Earth moves quickly through space but has a simple, aligned bump beneath.



I'm just a programmer who is interested in Physics & has been watching too many Leonard Susskind lectures recently.










share|cite|improve this question









New contributor




Jack Nicholson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    Why do you expect there to be a difference? Physics is the same in all frames moving at constant speed with respect to each other, so you're free to look at the moving object in its rest frame.
    – jacob1729
    4 hours ago










  • @jacob1729 So there would be no way to tell that one was moving and the other wasn't if you froze time and tried to measure them both? There are zero physical differences? (I'm not saying the Asteroid itself will look or be different, but either, the particles surrounding them / spacetime / vacuum energy?)
    – Jack Nicholson
    4 hours ago












  • There is no absolute frame of reference which determines that some objects are still and some are moving. You can only talk about things moving relative to some other thing (like in S. McGrew's answer). This isn't just a feature of Einstein's relativity, it goes back to Galileo.
    – PM 2Ring
    3 hours ago








  • 1




    I think the comments so far have missed the point of the question. There is no physical reason why two things should be the same or look the same when both are observed in the same reference frame.
    – Andrew Steane
    2 hours ago














4












4








4


2





If I have two asteroids. One dead still in space and one whizzing by at 10,000mph. What is the difference between the two, physically?



If I freeze time and look at the two of them - what differences would they have? How could one tell that one was moving really quickly, and the other not?



enter image description here



Is there some sort of Quantum difference with the particles in front or behind the asteroid?



Does it have a different gravitational force on spacetime surrounding it?



enter image description here



No images online seem to suggest this as Earth moves quickly through space but has a simple, aligned bump beneath.



I'm just a programmer who is interested in Physics & has been watching too many Leonard Susskind lectures recently.










share|cite|improve this question









New contributor




Jack Nicholson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











If I have two asteroids. One dead still in space and one whizzing by at 10,000mph. What is the difference between the two, physically?



If I freeze time and look at the two of them - what differences would they have? How could one tell that one was moving really quickly, and the other not?



enter image description here



Is there some sort of Quantum difference with the particles in front or behind the asteroid?



Does it have a different gravitational force on spacetime surrounding it?



enter image description here



No images online seem to suggest this as Earth moves quickly through space but has a simple, aligned bump beneath.



I'm just a programmer who is interested in Physics & has been watching too many Leonard Susskind lectures recently.







special-relativity inertial-frames observers






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Jack Nicholson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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Check out our Code of Conduct.









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edited 2 hours ago









Qmechanic

101k121831150




101k121831150






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asked 4 hours ago









Jack Nicholson

1235




1235




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Jack Nicholson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Jack Nicholson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.








  • 1




    Why do you expect there to be a difference? Physics is the same in all frames moving at constant speed with respect to each other, so you're free to look at the moving object in its rest frame.
    – jacob1729
    4 hours ago










  • @jacob1729 So there would be no way to tell that one was moving and the other wasn't if you froze time and tried to measure them both? There are zero physical differences? (I'm not saying the Asteroid itself will look or be different, but either, the particles surrounding them / spacetime / vacuum energy?)
    – Jack Nicholson
    4 hours ago












  • There is no absolute frame of reference which determines that some objects are still and some are moving. You can only talk about things moving relative to some other thing (like in S. McGrew's answer). This isn't just a feature of Einstein's relativity, it goes back to Galileo.
    – PM 2Ring
    3 hours ago








  • 1




    I think the comments so far have missed the point of the question. There is no physical reason why two things should be the same or look the same when both are observed in the same reference frame.
    – Andrew Steane
    2 hours ago














  • 1




    Why do you expect there to be a difference? Physics is the same in all frames moving at constant speed with respect to each other, so you're free to look at the moving object in its rest frame.
    – jacob1729
    4 hours ago










  • @jacob1729 So there would be no way to tell that one was moving and the other wasn't if you froze time and tried to measure them both? There are zero physical differences? (I'm not saying the Asteroid itself will look or be different, but either, the particles surrounding them / spacetime / vacuum energy?)
    – Jack Nicholson
    4 hours ago












  • There is no absolute frame of reference which determines that some objects are still and some are moving. You can only talk about things moving relative to some other thing (like in S. McGrew's answer). This isn't just a feature of Einstein's relativity, it goes back to Galileo.
    – PM 2Ring
    3 hours ago








  • 1




    I think the comments so far have missed the point of the question. There is no physical reason why two things should be the same or look the same when both are observed in the same reference frame.
    – Andrew Steane
    2 hours ago








1




1




Why do you expect there to be a difference? Physics is the same in all frames moving at constant speed with respect to each other, so you're free to look at the moving object in its rest frame.
– jacob1729
4 hours ago




Why do you expect there to be a difference? Physics is the same in all frames moving at constant speed with respect to each other, so you're free to look at the moving object in its rest frame.
– jacob1729
4 hours ago












@jacob1729 So there would be no way to tell that one was moving and the other wasn't if you froze time and tried to measure them both? There are zero physical differences? (I'm not saying the Asteroid itself will look or be different, but either, the particles surrounding them / spacetime / vacuum energy?)
– Jack Nicholson
4 hours ago






@jacob1729 So there would be no way to tell that one was moving and the other wasn't if you froze time and tried to measure them both? There are zero physical differences? (I'm not saying the Asteroid itself will look or be different, but either, the particles surrounding them / spacetime / vacuum energy?)
– Jack Nicholson
4 hours ago














There is no absolute frame of reference which determines that some objects are still and some are moving. You can only talk about things moving relative to some other thing (like in S. McGrew's answer). This isn't just a feature of Einstein's relativity, it goes back to Galileo.
– PM 2Ring
3 hours ago






There is no absolute frame of reference which determines that some objects are still and some are moving. You can only talk about things moving relative to some other thing (like in S. McGrew's answer). This isn't just a feature of Einstein's relativity, it goes back to Galileo.
– PM 2Ring
3 hours ago






1




1




I think the comments so far have missed the point of the question. There is no physical reason why two things should be the same or look the same when both are observed in the same reference frame.
– Andrew Steane
2 hours ago




I think the comments so far have missed the point of the question. There is no physical reason why two things should be the same or look the same when both are observed in the same reference frame.
– Andrew Steane
2 hours ago










2 Answers
2






active

oldest

votes


















4














I'll choose to interpret the question as specifying that one of the objects is stationary with respect to an observer, and the other object is moving with respect to the observer. Then the question goes on to ask if the observer can discern any difference between the two in an instantaneous "snapshot" of the two objects.



It's really a good question, and the answer is yes, there is a discernable difference.



Suppose that the objects are charged particles with no intrinsic magnetic moments. The observer will see no magnetic field around the "stationary" particle, but will see a magnetic field around the "moving" particle.



Even in the case of uncharged particles, there is a difference. An observer who sees a mass moving relative to himself sees an additional field besides the gravitational field of the mass. The additional field is a consequence of general relativity and is analogous to the magnetic field that's observed around a moving charged particle.






share|cite|improve this answer





















  • It's important to note though that there does exist a frame (CM frame) in which both planets will have the identical snapshot of their E&B fields, and that even though you will be able to say that the planet is moving from the combination of $E$ fields under LTs that they both will have the same equations of motion that govern a nearby particles behavior.
    – InertialObserver
    3 hours ago












  • I wonder how meaningful it is to talk about fields in a "snapshot of reality", as the value of a field can only be observed by its effect on particles, in particular, their acceleration. On the other hand, the concept of snapshot is not well-defined here, and everything is a field in the end anyway.
    – M. Winter
    1 hour ago



















1














I will only cover the case of special relativistic effects



In the following discussion I take the frame of either one of the asteroids. It doesn't matter which one we take.



In order to be able to tell anything from a snapshot due to relativistic effects we need to have a relativistic $gamma$ factor.



However, since $v = 10,000$ mph this corresponds to a



$$ beta = frac{v}{c} = 1.5 times 10^{-5}. $$



This will correspond to a lorentz factor of



$$gamma = 1.00000000011 $$



which means that the prospect of distinguishing the moving (in our chosen frame) asteroid due to any of our relativistic machinery (i.e. length contraction, doppler shift, time dilation etc.) is a lost cause.



Other possibilities: I don't immediately see any other realistic possibilities for two bodies that have every property in common except for their speed (in your frame of reference). Indeed if you took the CM frame (which is inertial) they would be moving precisely at the same speed! Perhaps this is a more elegant argument that the one I just gave above, but it at least reinforces the conclusion.



Adding to the previous paragraph: Of course if you knew that the proper radius $R_0$ of the planets beforehand you could in principle be able to tell that they were both moving in the $CM$ frame. However, again, the $gamma$ factor would be too small to make this a reasonable measurement.






share|cite|improve this answer























  • If it wasn't an image and you could go over to the asteroids and make measurements, would you still not be able to tell? If I resume time, one object will continue moving while the other will stay still - how does it have that memory? Surely there is something physical surrounding it that keeps its motion that could be measured?
    – Jack Nicholson
    3 hours ago








  • 1




    Jack, it's actually worse than you think. if you are out in space and observing those two asteroids, there is no way physically possible for you to determine which of the two is standing still and which is moving. If you hitched a ride on first one then the other, and performed any physics experiment you can think of while riding on each, the results would show no differences at all. Out in space, experiencing uniform, unaccelerated motion and being at rest are indistinguishable. This is the crux of the law of special relativity.
    – niels nielsen
    3 hours ago






  • 1




    @JackNicholson It’s because you have no right to call yourself stationary and the other thing moving. For all you know if you’re on the “stationary” asteroid, you’re moving towards the “moving asteroid” and it’s at rest. This is what’s meant by the “pop science” everything is relative ordeal.
    – InertialObserver
    3 hours ago












  • The reference to the Lorentz factor and length contraction threatens to be badly misleading. If an asteroid in motion (relative to me) stops dead all at once (relative to me), its length as measured in my frame does not change, so there's no effect for me to observe. (Its length in its own initial frame changes, because in that frame the "front" and "back" ends of the asteroid stopped moving at different times --- but the question is about what the "stationary" observer can discern.)
    – WillO
    2 hours ago










  • The length contraction will be apparent if I take a picture. If it stops obviously things get more complicated. But the length contraction will be visible by taking a picture if it's moving relativistically.
    – InertialObserver
    2 hours ago











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2 Answers
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active

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active

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active

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4














I'll choose to interpret the question as specifying that one of the objects is stationary with respect to an observer, and the other object is moving with respect to the observer. Then the question goes on to ask if the observer can discern any difference between the two in an instantaneous "snapshot" of the two objects.



It's really a good question, and the answer is yes, there is a discernable difference.



Suppose that the objects are charged particles with no intrinsic magnetic moments. The observer will see no magnetic field around the "stationary" particle, but will see a magnetic field around the "moving" particle.



Even in the case of uncharged particles, there is a difference. An observer who sees a mass moving relative to himself sees an additional field besides the gravitational field of the mass. The additional field is a consequence of general relativity and is analogous to the magnetic field that's observed around a moving charged particle.






share|cite|improve this answer





















  • It's important to note though that there does exist a frame (CM frame) in which both planets will have the identical snapshot of their E&B fields, and that even though you will be able to say that the planet is moving from the combination of $E$ fields under LTs that they both will have the same equations of motion that govern a nearby particles behavior.
    – InertialObserver
    3 hours ago












  • I wonder how meaningful it is to talk about fields in a "snapshot of reality", as the value of a field can only be observed by its effect on particles, in particular, their acceleration. On the other hand, the concept of snapshot is not well-defined here, and everything is a field in the end anyway.
    – M. Winter
    1 hour ago
















4














I'll choose to interpret the question as specifying that one of the objects is stationary with respect to an observer, and the other object is moving with respect to the observer. Then the question goes on to ask if the observer can discern any difference between the two in an instantaneous "snapshot" of the two objects.



It's really a good question, and the answer is yes, there is a discernable difference.



Suppose that the objects are charged particles with no intrinsic magnetic moments. The observer will see no magnetic field around the "stationary" particle, but will see a magnetic field around the "moving" particle.



Even in the case of uncharged particles, there is a difference. An observer who sees a mass moving relative to himself sees an additional field besides the gravitational field of the mass. The additional field is a consequence of general relativity and is analogous to the magnetic field that's observed around a moving charged particle.






share|cite|improve this answer





















  • It's important to note though that there does exist a frame (CM frame) in which both planets will have the identical snapshot of their E&B fields, and that even though you will be able to say that the planet is moving from the combination of $E$ fields under LTs that they both will have the same equations of motion that govern a nearby particles behavior.
    – InertialObserver
    3 hours ago












  • I wonder how meaningful it is to talk about fields in a "snapshot of reality", as the value of a field can only be observed by its effect on particles, in particular, their acceleration. On the other hand, the concept of snapshot is not well-defined here, and everything is a field in the end anyway.
    – M. Winter
    1 hour ago














4












4








4






I'll choose to interpret the question as specifying that one of the objects is stationary with respect to an observer, and the other object is moving with respect to the observer. Then the question goes on to ask if the observer can discern any difference between the two in an instantaneous "snapshot" of the two objects.



It's really a good question, and the answer is yes, there is a discernable difference.



Suppose that the objects are charged particles with no intrinsic magnetic moments. The observer will see no magnetic field around the "stationary" particle, but will see a magnetic field around the "moving" particle.



Even in the case of uncharged particles, there is a difference. An observer who sees a mass moving relative to himself sees an additional field besides the gravitational field of the mass. The additional field is a consequence of general relativity and is analogous to the magnetic field that's observed around a moving charged particle.






share|cite|improve this answer












I'll choose to interpret the question as specifying that one of the objects is stationary with respect to an observer, and the other object is moving with respect to the observer. Then the question goes on to ask if the observer can discern any difference between the two in an instantaneous "snapshot" of the two objects.



It's really a good question, and the answer is yes, there is a discernable difference.



Suppose that the objects are charged particles with no intrinsic magnetic moments. The observer will see no magnetic field around the "stationary" particle, but will see a magnetic field around the "moving" particle.



Even in the case of uncharged particles, there is a difference. An observer who sees a mass moving relative to himself sees an additional field besides the gravitational field of the mass. The additional field is a consequence of general relativity and is analogous to the magnetic field that's observed around a moving charged particle.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









S. McGrew

6,3872925




6,3872925












  • It's important to note though that there does exist a frame (CM frame) in which both planets will have the identical snapshot of their E&B fields, and that even though you will be able to say that the planet is moving from the combination of $E$ fields under LTs that they both will have the same equations of motion that govern a nearby particles behavior.
    – InertialObserver
    3 hours ago












  • I wonder how meaningful it is to talk about fields in a "snapshot of reality", as the value of a field can only be observed by its effect on particles, in particular, their acceleration. On the other hand, the concept of snapshot is not well-defined here, and everything is a field in the end anyway.
    – M. Winter
    1 hour ago


















  • It's important to note though that there does exist a frame (CM frame) in which both planets will have the identical snapshot of their E&B fields, and that even though you will be able to say that the planet is moving from the combination of $E$ fields under LTs that they both will have the same equations of motion that govern a nearby particles behavior.
    – InertialObserver
    3 hours ago












  • I wonder how meaningful it is to talk about fields in a "snapshot of reality", as the value of a field can only be observed by its effect on particles, in particular, their acceleration. On the other hand, the concept of snapshot is not well-defined here, and everything is a field in the end anyway.
    – M. Winter
    1 hour ago
















It's important to note though that there does exist a frame (CM frame) in which both planets will have the identical snapshot of their E&B fields, and that even though you will be able to say that the planet is moving from the combination of $E$ fields under LTs that they both will have the same equations of motion that govern a nearby particles behavior.
– InertialObserver
3 hours ago






It's important to note though that there does exist a frame (CM frame) in which both planets will have the identical snapshot of their E&B fields, and that even though you will be able to say that the planet is moving from the combination of $E$ fields under LTs that they both will have the same equations of motion that govern a nearby particles behavior.
– InertialObserver
3 hours ago














I wonder how meaningful it is to talk about fields in a "snapshot of reality", as the value of a field can only be observed by its effect on particles, in particular, their acceleration. On the other hand, the concept of snapshot is not well-defined here, and everything is a field in the end anyway.
– M. Winter
1 hour ago




I wonder how meaningful it is to talk about fields in a "snapshot of reality", as the value of a field can only be observed by its effect on particles, in particular, their acceleration. On the other hand, the concept of snapshot is not well-defined here, and everything is a field in the end anyway.
– M. Winter
1 hour ago











1














I will only cover the case of special relativistic effects



In the following discussion I take the frame of either one of the asteroids. It doesn't matter which one we take.



In order to be able to tell anything from a snapshot due to relativistic effects we need to have a relativistic $gamma$ factor.



However, since $v = 10,000$ mph this corresponds to a



$$ beta = frac{v}{c} = 1.5 times 10^{-5}. $$



This will correspond to a lorentz factor of



$$gamma = 1.00000000011 $$



which means that the prospect of distinguishing the moving (in our chosen frame) asteroid due to any of our relativistic machinery (i.e. length contraction, doppler shift, time dilation etc.) is a lost cause.



Other possibilities: I don't immediately see any other realistic possibilities for two bodies that have every property in common except for their speed (in your frame of reference). Indeed if you took the CM frame (which is inertial) they would be moving precisely at the same speed! Perhaps this is a more elegant argument that the one I just gave above, but it at least reinforces the conclusion.



Adding to the previous paragraph: Of course if you knew that the proper radius $R_0$ of the planets beforehand you could in principle be able to tell that they were both moving in the $CM$ frame. However, again, the $gamma$ factor would be too small to make this a reasonable measurement.






share|cite|improve this answer























  • If it wasn't an image and you could go over to the asteroids and make measurements, would you still not be able to tell? If I resume time, one object will continue moving while the other will stay still - how does it have that memory? Surely there is something physical surrounding it that keeps its motion that could be measured?
    – Jack Nicholson
    3 hours ago








  • 1




    Jack, it's actually worse than you think. if you are out in space and observing those two asteroids, there is no way physically possible for you to determine which of the two is standing still and which is moving. If you hitched a ride on first one then the other, and performed any physics experiment you can think of while riding on each, the results would show no differences at all. Out in space, experiencing uniform, unaccelerated motion and being at rest are indistinguishable. This is the crux of the law of special relativity.
    – niels nielsen
    3 hours ago






  • 1




    @JackNicholson It’s because you have no right to call yourself stationary and the other thing moving. For all you know if you’re on the “stationary” asteroid, you’re moving towards the “moving asteroid” and it’s at rest. This is what’s meant by the “pop science” everything is relative ordeal.
    – InertialObserver
    3 hours ago












  • The reference to the Lorentz factor and length contraction threatens to be badly misleading. If an asteroid in motion (relative to me) stops dead all at once (relative to me), its length as measured in my frame does not change, so there's no effect for me to observe. (Its length in its own initial frame changes, because in that frame the "front" and "back" ends of the asteroid stopped moving at different times --- but the question is about what the "stationary" observer can discern.)
    – WillO
    2 hours ago










  • The length contraction will be apparent if I take a picture. If it stops obviously things get more complicated. But the length contraction will be visible by taking a picture if it's moving relativistically.
    – InertialObserver
    2 hours ago
















1














I will only cover the case of special relativistic effects



In the following discussion I take the frame of either one of the asteroids. It doesn't matter which one we take.



In order to be able to tell anything from a snapshot due to relativistic effects we need to have a relativistic $gamma$ factor.



However, since $v = 10,000$ mph this corresponds to a



$$ beta = frac{v}{c} = 1.5 times 10^{-5}. $$



This will correspond to a lorentz factor of



$$gamma = 1.00000000011 $$



which means that the prospect of distinguishing the moving (in our chosen frame) asteroid due to any of our relativistic machinery (i.e. length contraction, doppler shift, time dilation etc.) is a lost cause.



Other possibilities: I don't immediately see any other realistic possibilities for two bodies that have every property in common except for their speed (in your frame of reference). Indeed if you took the CM frame (which is inertial) they would be moving precisely at the same speed! Perhaps this is a more elegant argument that the one I just gave above, but it at least reinforces the conclusion.



Adding to the previous paragraph: Of course if you knew that the proper radius $R_0$ of the planets beforehand you could in principle be able to tell that they were both moving in the $CM$ frame. However, again, the $gamma$ factor would be too small to make this a reasonable measurement.






share|cite|improve this answer























  • If it wasn't an image and you could go over to the asteroids and make measurements, would you still not be able to tell? If I resume time, one object will continue moving while the other will stay still - how does it have that memory? Surely there is something physical surrounding it that keeps its motion that could be measured?
    – Jack Nicholson
    3 hours ago








  • 1




    Jack, it's actually worse than you think. if you are out in space and observing those two asteroids, there is no way physically possible for you to determine which of the two is standing still and which is moving. If you hitched a ride on first one then the other, and performed any physics experiment you can think of while riding on each, the results would show no differences at all. Out in space, experiencing uniform, unaccelerated motion and being at rest are indistinguishable. This is the crux of the law of special relativity.
    – niels nielsen
    3 hours ago






  • 1




    @JackNicholson It’s because you have no right to call yourself stationary and the other thing moving. For all you know if you’re on the “stationary” asteroid, you’re moving towards the “moving asteroid” and it’s at rest. This is what’s meant by the “pop science” everything is relative ordeal.
    – InertialObserver
    3 hours ago












  • The reference to the Lorentz factor and length contraction threatens to be badly misleading. If an asteroid in motion (relative to me) stops dead all at once (relative to me), its length as measured in my frame does not change, so there's no effect for me to observe. (Its length in its own initial frame changes, because in that frame the "front" and "back" ends of the asteroid stopped moving at different times --- but the question is about what the "stationary" observer can discern.)
    – WillO
    2 hours ago










  • The length contraction will be apparent if I take a picture. If it stops obviously things get more complicated. But the length contraction will be visible by taking a picture if it's moving relativistically.
    – InertialObserver
    2 hours ago














1












1








1






I will only cover the case of special relativistic effects



In the following discussion I take the frame of either one of the asteroids. It doesn't matter which one we take.



In order to be able to tell anything from a snapshot due to relativistic effects we need to have a relativistic $gamma$ factor.



However, since $v = 10,000$ mph this corresponds to a



$$ beta = frac{v}{c} = 1.5 times 10^{-5}. $$



This will correspond to a lorentz factor of



$$gamma = 1.00000000011 $$



which means that the prospect of distinguishing the moving (in our chosen frame) asteroid due to any of our relativistic machinery (i.e. length contraction, doppler shift, time dilation etc.) is a lost cause.



Other possibilities: I don't immediately see any other realistic possibilities for two bodies that have every property in common except for their speed (in your frame of reference). Indeed if you took the CM frame (which is inertial) they would be moving precisely at the same speed! Perhaps this is a more elegant argument that the one I just gave above, but it at least reinforces the conclusion.



Adding to the previous paragraph: Of course if you knew that the proper radius $R_0$ of the planets beforehand you could in principle be able to tell that they were both moving in the $CM$ frame. However, again, the $gamma$ factor would be too small to make this a reasonable measurement.






share|cite|improve this answer














I will only cover the case of special relativistic effects



In the following discussion I take the frame of either one of the asteroids. It doesn't matter which one we take.



In order to be able to tell anything from a snapshot due to relativistic effects we need to have a relativistic $gamma$ factor.



However, since $v = 10,000$ mph this corresponds to a



$$ beta = frac{v}{c} = 1.5 times 10^{-5}. $$



This will correspond to a lorentz factor of



$$gamma = 1.00000000011 $$



which means that the prospect of distinguishing the moving (in our chosen frame) asteroid due to any of our relativistic machinery (i.e. length contraction, doppler shift, time dilation etc.) is a lost cause.



Other possibilities: I don't immediately see any other realistic possibilities for two bodies that have every property in common except for their speed (in your frame of reference). Indeed if you took the CM frame (which is inertial) they would be moving precisely at the same speed! Perhaps this is a more elegant argument that the one I just gave above, but it at least reinforces the conclusion.



Adding to the previous paragraph: Of course if you knew that the proper radius $R_0$ of the planets beforehand you could in principle be able to tell that they were both moving in the $CM$ frame. However, again, the $gamma$ factor would be too small to make this a reasonable measurement.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 3 hours ago

























answered 4 hours ago









InertialObserver

1,411517




1,411517












  • If it wasn't an image and you could go over to the asteroids and make measurements, would you still not be able to tell? If I resume time, one object will continue moving while the other will stay still - how does it have that memory? Surely there is something physical surrounding it that keeps its motion that could be measured?
    – Jack Nicholson
    3 hours ago








  • 1




    Jack, it's actually worse than you think. if you are out in space and observing those two asteroids, there is no way physically possible for you to determine which of the two is standing still and which is moving. If you hitched a ride on first one then the other, and performed any physics experiment you can think of while riding on each, the results would show no differences at all. Out in space, experiencing uniform, unaccelerated motion and being at rest are indistinguishable. This is the crux of the law of special relativity.
    – niels nielsen
    3 hours ago






  • 1




    @JackNicholson It’s because you have no right to call yourself stationary and the other thing moving. For all you know if you’re on the “stationary” asteroid, you’re moving towards the “moving asteroid” and it’s at rest. This is what’s meant by the “pop science” everything is relative ordeal.
    – InertialObserver
    3 hours ago












  • The reference to the Lorentz factor and length contraction threatens to be badly misleading. If an asteroid in motion (relative to me) stops dead all at once (relative to me), its length as measured in my frame does not change, so there's no effect for me to observe. (Its length in its own initial frame changes, because in that frame the "front" and "back" ends of the asteroid stopped moving at different times --- but the question is about what the "stationary" observer can discern.)
    – WillO
    2 hours ago










  • The length contraction will be apparent if I take a picture. If it stops obviously things get more complicated. But the length contraction will be visible by taking a picture if it's moving relativistically.
    – InertialObserver
    2 hours ago


















  • If it wasn't an image and you could go over to the asteroids and make measurements, would you still not be able to tell? If I resume time, one object will continue moving while the other will stay still - how does it have that memory? Surely there is something physical surrounding it that keeps its motion that could be measured?
    – Jack Nicholson
    3 hours ago








  • 1




    Jack, it's actually worse than you think. if you are out in space and observing those two asteroids, there is no way physically possible for you to determine which of the two is standing still and which is moving. If you hitched a ride on first one then the other, and performed any physics experiment you can think of while riding on each, the results would show no differences at all. Out in space, experiencing uniform, unaccelerated motion and being at rest are indistinguishable. This is the crux of the law of special relativity.
    – niels nielsen
    3 hours ago






  • 1




    @JackNicholson It’s because you have no right to call yourself stationary and the other thing moving. For all you know if you’re on the “stationary” asteroid, you’re moving towards the “moving asteroid” and it’s at rest. This is what’s meant by the “pop science” everything is relative ordeal.
    – InertialObserver
    3 hours ago












  • The reference to the Lorentz factor and length contraction threatens to be badly misleading. If an asteroid in motion (relative to me) stops dead all at once (relative to me), its length as measured in my frame does not change, so there's no effect for me to observe. (Its length in its own initial frame changes, because in that frame the "front" and "back" ends of the asteroid stopped moving at different times --- but the question is about what the "stationary" observer can discern.)
    – WillO
    2 hours ago










  • The length contraction will be apparent if I take a picture. If it stops obviously things get more complicated. But the length contraction will be visible by taking a picture if it's moving relativistically.
    – InertialObserver
    2 hours ago
















If it wasn't an image and you could go over to the asteroids and make measurements, would you still not be able to tell? If I resume time, one object will continue moving while the other will stay still - how does it have that memory? Surely there is something physical surrounding it that keeps its motion that could be measured?
– Jack Nicholson
3 hours ago






If it wasn't an image and you could go over to the asteroids and make measurements, would you still not be able to tell? If I resume time, one object will continue moving while the other will stay still - how does it have that memory? Surely there is something physical surrounding it that keeps its motion that could be measured?
– Jack Nicholson
3 hours ago






1




1




Jack, it's actually worse than you think. if you are out in space and observing those two asteroids, there is no way physically possible for you to determine which of the two is standing still and which is moving. If you hitched a ride on first one then the other, and performed any physics experiment you can think of while riding on each, the results would show no differences at all. Out in space, experiencing uniform, unaccelerated motion and being at rest are indistinguishable. This is the crux of the law of special relativity.
– niels nielsen
3 hours ago




Jack, it's actually worse than you think. if you are out in space and observing those two asteroids, there is no way physically possible for you to determine which of the two is standing still and which is moving. If you hitched a ride on first one then the other, and performed any physics experiment you can think of while riding on each, the results would show no differences at all. Out in space, experiencing uniform, unaccelerated motion and being at rest are indistinguishable. This is the crux of the law of special relativity.
– niels nielsen
3 hours ago




1




1




@JackNicholson It’s because you have no right to call yourself stationary and the other thing moving. For all you know if you’re on the “stationary” asteroid, you’re moving towards the “moving asteroid” and it’s at rest. This is what’s meant by the “pop science” everything is relative ordeal.
– InertialObserver
3 hours ago






@JackNicholson It’s because you have no right to call yourself stationary and the other thing moving. For all you know if you’re on the “stationary” asteroid, you’re moving towards the “moving asteroid” and it’s at rest. This is what’s meant by the “pop science” everything is relative ordeal.
– InertialObserver
3 hours ago














The reference to the Lorentz factor and length contraction threatens to be badly misleading. If an asteroid in motion (relative to me) stops dead all at once (relative to me), its length as measured in my frame does not change, so there's no effect for me to observe. (Its length in its own initial frame changes, because in that frame the "front" and "back" ends of the asteroid stopped moving at different times --- but the question is about what the "stationary" observer can discern.)
– WillO
2 hours ago




The reference to the Lorentz factor and length contraction threatens to be badly misleading. If an asteroid in motion (relative to me) stops dead all at once (relative to me), its length as measured in my frame does not change, so there's no effect for me to observe. (Its length in its own initial frame changes, because in that frame the "front" and "back" ends of the asteroid stopped moving at different times --- but the question is about what the "stationary" observer can discern.)
– WillO
2 hours ago












The length contraction will be apparent if I take a picture. If it stops obviously things get more complicated. But the length contraction will be visible by taking a picture if it's moving relativistically.
– InertialObserver
2 hours ago




The length contraction will be apparent if I take a picture. If it stops obviously things get more complicated. But the length contraction will be visible by taking a picture if it's moving relativistically.
– InertialObserver
2 hours ago










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