What is physically different about a moving vs still object in space?
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?
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?
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
New contributor
add a comment |
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?
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?
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
New contributor
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
2 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
2 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
1 hour 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
1 hour ago
add a comment |
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?
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?
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
New contributor
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?
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?
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
special-relativity inertial-frames observers
New contributor
New contributor
edited 1 hour ago
Qmechanic♦
101k121831149
101k121831149
New contributor
asked 2 hours ago
Jack Nicholson
1235
1235
New contributor
New contributor
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
2 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
2 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
1 hour 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
1 hour ago
add a comment |
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
2 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
2 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
1 hour 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
1 hour 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
2 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
2 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
2 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
2 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
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.
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
1 hour ago
add a comment |
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.
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
2 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
2 hours ago
@nielsnielsen I understand you completely. Is it just that we don't have the tools or scientific knowledge to be able to work it out? I cannot, in my head, grasp the idea that a moving object has zero differences to a stationary one. Or is that because everything is relative? I.e. the stationary one is the same as the moving one? --still doesn't sound right to me-- (@inertialobserver, upvoted so far, thank you - still trying to grasp it for the moment)
– Jack Nicholson
1 hour 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
1 hour 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
45 mins ago
|
show 3 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Jack Nicholson is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f451267%2fwhat-is-physically-different-about-a-moving-vs-still-object-in-space%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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
1 hour ago
add a comment |
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.
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
1 hour ago
add a comment |
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.
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.
answered 1 hour 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
1 hour ago
add a comment |
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
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
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
1 hour ago
add a comment |
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.
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
2 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
2 hours ago
@nielsnielsen I understand you completely. Is it just that we don't have the tools or scientific knowledge to be able to work it out? I cannot, in my head, grasp the idea that a moving object has zero differences to a stationary one. Or is that because everything is relative? I.e. the stationary one is the same as the moving one? --still doesn't sound right to me-- (@inertialobserver, upvoted so far, thank you - still trying to grasp it for the moment)
– Jack Nicholson
1 hour 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
1 hour 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
45 mins ago
|
show 3 more comments
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.
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
2 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
2 hours ago
@nielsnielsen I understand you completely. Is it just that we don't have the tools or scientific knowledge to be able to work it out? I cannot, in my head, grasp the idea that a moving object has zero differences to a stationary one. Or is that because everything is relative? I.e. the stationary one is the same as the moving one? --still doesn't sound right to me-- (@inertialobserver, upvoted so far, thank you - still trying to grasp it for the moment)
– Jack Nicholson
1 hour 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
1 hour 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
45 mins ago
|
show 3 more comments
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.
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.
edited 1 hour ago
answered 2 hours ago
InertialObserver
1,406517
1,406517
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
2 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
2 hours ago
@nielsnielsen I understand you completely. Is it just that we don't have the tools or scientific knowledge to be able to work it out? I cannot, in my head, grasp the idea that a moving object has zero differences to a stationary one. Or is that because everything is relative? I.e. the stationary one is the same as the moving one? --still doesn't sound right to me-- (@inertialobserver, upvoted so far, thank you - still trying to grasp it for the moment)
– Jack Nicholson
1 hour 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
1 hour 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
45 mins ago
|
show 3 more comments
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
2 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
2 hours ago
@nielsnielsen I understand you completely. Is it just that we don't have the tools or scientific knowledge to be able to work it out? I cannot, in my head, grasp the idea that a moving object has zero differences to a stationary one. Or is that because everything is relative? I.e. the stationary one is the same as the moving one? --still doesn't sound right to me-- (@inertialobserver, upvoted so far, thank you - still trying to grasp it for the moment)
– Jack Nicholson
1 hour 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
1 hour 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
45 mins 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
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
2 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
2 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
2 hours ago
@nielsnielsen I understand you completely. Is it just that we don't have the tools or scientific knowledge to be able to work it out? I cannot, in my head, grasp the idea that a moving object has zero differences to a stationary one. Or is that because everything is relative? I.e. the stationary one is the same as the moving one? --still doesn't sound right to me-- (@inertialobserver, upvoted so far, thank you - still trying to grasp it for the moment)
– Jack Nicholson
1 hour ago
@nielsnielsen I understand you completely. Is it just that we don't have the tools or scientific knowledge to be able to work it out? I cannot, in my head, grasp the idea that a moving object has zero differences to a stationary one. Or is that because everything is relative? I.e. the stationary one is the same as the moving one? --still doesn't sound right to me-- (@inertialobserver, upvoted so far, thank you - still trying to grasp it for the moment)
– Jack Nicholson
1 hour 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
1 hour 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
1 hour 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
45 mins 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
45 mins ago
|
show 3 more comments
Jack Nicholson is a new contributor. Be nice, and check out our Code of Conduct.
Jack Nicholson is a new contributor. Be nice, and check out our Code of Conduct.
Jack Nicholson is a new contributor. Be nice, and check out our Code of Conduct.
Jack Nicholson is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f451267%2fwhat-is-physically-different-about-a-moving-vs-still-object-in-space%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
2 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
2 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
1 hour 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
1 hour ago